philpot mom 2nd ch07-11 ism

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7.1 For the cantilever beam and loading shown,

(a) Derive equations for the shear force V and the

bending moment M for any location in the beam.

(Place the origin at point A.)

(b) Plot the shear-force and bending-moment

diagrams for the beam using the derived functions. Fig. P7.1

Solution

Beam equilibrium:

0

0

0

2

0

0

02

2

y y

y

A A

A

F A w L

A w L

LM M w L

w LM

Section a-a:

0

0 0

0

0 ( )

0yF w L w x V

V w L w x w L x

2

0

0 0

2 2

0 0 2 20

00 ( )2

02 2

2 2

a a

w L xM w Lx w x M

w L w xM

wL x wL Lxw x

(b) Shear-force and bending-moment diagrams




7.2 For the simply supported beam shown,





diagrams for the beam using the derived functions.

Fig. P7.2

Solution

Beam equilibrium:

0

( ) 0

and

y y y

A y

y y

F A C P

M Pa C a b

Pa PbC A

a b a b

Section a-a:

For the interval 0 ≤ x < a:

0

0

y y

a a y

PbV

a b

PbM x

a b

PbF A V V

a b

PbM A x M x M

a b

Section b-b:

For the interval a ≤ x < b:

0y y

PbF A P V P V

a b

PaV

a b

(

( )

) ( ) 0b b y

PbM A x P x a M x P x a M

a

Pb

b

M x P x aa b










Solution

Beam equilibrium:

0

02 2

2 2

y a b y

y a b

C a b C

C a b

F w a w b C

C w a w b

a bM w a b w b M

a bM w a b w b

Section a-a:


2

2

0

02

y a

a a

a

aa

V w xF w x V

xM w x M

w xM

Section b-b:


2

2 2

0

02 2

y a b

b b a

b

b

a

b

aV w a w x a

F w a w x a V

a

w x aaM w a x

x aM w a x w x a M




7.4 For the simply supported beam subjected to the

loading shown,






Fig. P7.4

Solution

Beam equilibrium:

2

2

( ) 02 2

( 2 )

2( )

( ) 02 2

(2 )

2( )

C a b y

a by

A a b y

a by

a bM w a b w b A a b

w a a b w bA

a b

a bM w a w b a C a b

w a w b a bC

a b

Section a-a:


2

2 22

0

02

2

( 2 )

2( )

( 2 )

2 2( )

y y a a y

a a y a

a ba

ya ba a

F A w x V V w x A

xM A x w x M

w xM

w a a b w bw x

a b

w x w a a b w bA

a bx x

Section b-b:


2( 2 )( )

2( ) 2( )

( ) 0

( )

a ba

y a b

a b

b

y

y

F A w a w x a V

V A w a w x

w a a b w bw a w x a

a b a b

a

22 2

( 2 )

2

02 2

2 22( ) 2(2 )

b b y a b

ay a ba b

b

a x aM A x w a x w x a M

x aaM A x w

x aaw a a b w bx w a x w

a b aa x w

b










Solution

Beam equilibrium:

0 0

2

0 0

02 2

02 3 6

y y y

B B B

w L w LF B B

w L L w LM M M

Section a-a:

0

2

0

3

0

0 02

02 3

2

6

y

a a

w xw x xF V

L

w x x xM M

L

VL

w xM

L





7.6 For the simply supported beam shown,






(c) Determine the location and the magnitude of the

maximum bending moment.

Fig. P7.6

Solution

Beam equilibrium:

0 0

0 0

02 3 6

20

2 3 3

B y y

A y y

w L L w LM A L A

w L L w LM B L B

Section a-a:

2

0 0 0

3

0 0

2

0 0

3

0 0

0

6 2

6

02 6 2

02

6

3 6 6

y y

a a y

w x x w L w xF A V V

L L

w x x x w Lx w xM A x M M

L

w L w xV

L

w x w LxM

L

L





(c) Location and magnitude of maximum bending moment:

The maximum bending moment is located at the location where V = 0. Therefore, the maximum

bending moment occurs at:

2

0 0

2 2

0 0

06

0.577350

2

2 6 3 3

w L w xV

L

w x w L L

LL

Lx Ans.

Substitute this value of x into the bending moment equation to determine the moment magnitude:

3

0 0

3

0 2

0

0

max

6 6

(0.577350 )0.06415

(0.577

60

350 )

6

w x w LxM

L

w L w L LM

Lw L Ans.





loading shown,






(c) Report the maximum bending moment and its

location.

Fig. P7.7

Solution

Beam equilibrium:

(50 kN)(3 m) (75 kN)(6 m) (10 m) 0

60 kN

50 kN 75 kN 0

65 kN

A y

y

y y y

y

M D

D

F A D

A

Section a-a:

For the interval 0 ≤ x < 3 m:

65 kN 0

(6

65 kN

(65 5 kNk ) N) 0

y y

a a y

V

M

F A V V

M A x M x M x

Section b-b:

For the interval 3 m ≤ x < 6 m:

50 kN 65 kN 50 k

15 k

0

N

Ny yF A V

V

V

(50 kN)( 3 m)

(65 kN) (50 kN)( 3 m

(15 kN) 150 kN m

) 0

-

b b yM A x x

M x

M

x x M

Section c-c:


50 kN 75 kN

65 kN 50 kN 75 kN 0

60 kN

y yF A V

V

V

(50 kN)( 3 m) (75 kN)( 6 m)

(65 kN) (50 kN)( 3 m) (75 kN)(

(60

6

kN) 600 kN-m

m) 0

c c yM A x x x M

x x x M

M x





(c) Maximum bending moment

and its location

Mmax = 240 kN-m @ x = 6 m





loading shown,






(c) Report the maximum positive bending moment,

the maximum negative bending moment, and their

respective locations.

Fig. P7.8

Solution

Beam equilibrium:

(20 kN)(2 m) (60 kN)(6 m) (8 m) 0

40 kN

20 kN 60 kN 0

40 kN

B y

y

y y y

y

M D

D

F B D

B

Section a-a:


2020 kN 0

(20 k

kN

N (20 kN)) 0

y

a a

F

MM M

VV

xx

Section b-b:


20 kN

(20 kN) 80 kN-m

20 kN 20 kN 40 kN 0

(20 kN) ( 2 m)

(20 kN) (40 kN)( 2 m) 0

y y

b b y

F B V V

M x B x M

x M

x

x

V

M




Section c-c:


20 kN 60 kN

20 kN 40 kN 60 k

40 kN

N 0

y yF B

V

V

V

(20 kN) ( 2 m) (60 kN)( 8 m)

(20 kN) (40 kN)( 2 m) (60 kN)( 8 m)

(40 kN) 400 kN-

0

m

c c yM x B x x

M

M

x x M

x

x



and its location

Mmax-positive = 80 kN-m @ x = 8 m

Mmax-negative = –40 kN-m @ x = 2 m





loading shown,









Fig. P7.9

Solution

Beam equilibrium:

(7 kips/ft)(30 ft)(15 ft) (21 ft) 0

150 kips

(7 kips/ft)(30 ft) 0

60 kips

C y

y

y y y

y

M B

B

F B C

C

Section a-a:

For the interval 0 ≤ x < 9 ft:

2

(7 kips/ft) 0

(7 kips/ft)( )

(7 kips/ft)

(7 kips/f )

20

2

t

y

a a

F x V

x

V x

xM x MM

Section b-b:

For the interval 9 ft ≤ x < 30 ft:

(7 kips/ft) (7 kips/ft) 150 kips 0

(7 kips/ft) 150 kips

y yF x B V x V

V x

2

(7 kips/ft)( ) ( 9 ft)2

(7 kips/ft)( ) (150 kips)( 9 ft) 02

(7 kips/ft)(150 kips) 1,350 kips

2

b b y

xM x B

M x

x M

xx x M

x






and its location

Mmax-positive = 257.14 kip-ft @ x = 21.43 ft

Mmax-negative = –283.50 kip-ft @ x = 9 ft










Solution

Beam equilibrium:

(4 kips/ft)(8 ft) 0

32 kips

(4 kips/ft)(8 ft)(12 ft) 0

384 kip-ft

y y

y

C C

C

F C

C

M M

M

Section a-a:


2

(4 kips/ft)

4 kips

(4 kips/f

/f

t) 0

(4 kips/ftt

2)( ) 0

2

y

a a

F x V

xM x M

V x

M x

Section b-b:


(4 kips/ft)(8 ft) 0

(4 kips/ft)(8 ft)( 4 ft

32 kips

(32 kips) 128 ki f

0

p t

)

-

y

b b

F V

M x M

V

M x





loading shown,






(c) Report the maximum bending moment and its

location.

Fig. P7.11

Solution

Beam equilibrium:

(42 kips)(10 ft) (6 kips/ft)(20 ft)(20 ft)

(30 ft) 0

94 kips

A

y

y

M

C

C

42 kips (6 kips/ft)(20 ft) 0

68 kips

y y y

y

F A C

A

Section a-a:


68 kips

(

68 kips 0

(68 kips 6) 0 8 kips)

y y

a a y

F A V V

M A x M Mx

V

M x

Section b-b:


42 kips (6 kips/ft)( 10 ft)

68 kips 42 kips (6 kips/ft)( 10

(6 kips

ft)

/ft) 86

0

kips

y yF A x V

x V

V x

2

2

10 ft(42 kips)( 10 ft) (6 kips/ft)( 10 ft)

2

6 kips/ft(68 kips) (42 kips)( 10 ft) ( 10 f

3 86 120 kip

t2

-

)

ft

0

b b y

xM A x x x M

x x x

M

M

x x






and its location

Mmax = 736.33 kip-ft @ x = 14.33 ft





loading shown,









Fig. P7.12

Solution

Beam equilibrium:

180 kN-m (9 m) (36 kN)(12 m) 0

28 kN

36 kN 0

8 kN

A y

y

y y y

y

M C

C

F A C

A

Section a-a:


8 kN 0

(

8 kN

(8 kN)8 kN) 0

y y

a a y

V

M

F A V V

M xA x M x M

Section b-b:


8 kN

(8 kN) 180 kN-m

8 kN 0

180 kN-m

(8 kN) 180 kN-m 0

y y

x y

F A V V

M A x M

M

x

V

x

M

Section c-c:


8 kN 28 kN 0

36 kN

y y yF A C V V

V

( 9 m) 180 kN-m

(8 kN) (28 kN)( 9

(36 kN) 432

m) 180 kN

kN

-m 0

-m

c c y yM A x C

M

x M

x

x

x M






and its location

Mmax-positive = 32 kN-m @ x = 4 m

Mmax-negative = –148 kN-m @ x = 4 m










Fig. P7.13

Solution

Beam equilibrium:

Section a-a: For the interval 0 ≤ x < 8 ft:

Section b-b: For the interval 8 ft ≤ x < 14 ft:

5 kips/ft 6 ft 0

30 kips

120 kip-ft 5 kips/ft 6 ft 3 ft 0

210 kip-ft

y y

y

C C

C

F C

C

M M

M

-

0

120 kip-

0 kips

120 kip- tft 0 f

y

a a

VF V

MM M

-

2

5 kips/ft 8 ft 0

8 ft120 kip-ft 5 kips/

5 40

ft 8 ft =02

kips

2.5 40 280 kip-f t

y

b b

F x V

xM x M

V x

M x x




(b) Shear-force and bending-moment diagrams:










Fig. P7.14

Solution

Beam equilibrium:



16 kips/ft 9 ft 17 kips 0

2

44 kips

16 kips/ft 9 ft 6 ft

2

17 kips 14 ft 0

400 kip-ft

y y

y

A A

A

F A

A

M M

M

2

1 6 1 644 kips 0

2 9 2 9

1 6

2 9 3

1 6 400 kip-ft 44 kips 0

2

44 ki s

9

p

3

3

y y

a a A y

x xF A x V x V

x xM M A x x M

x xx x

xV

M

3

44 400 kip-ft9

x

M x

16 kips/ft 9 ft

2

1 44 kips 6 kips/ft 9 ft

17 ki

02

ps

y yF A V

V

V




-

16 kips/ft 9 ft 6 ft

2

1 400 kip-ft 6 kips/ft 9 ft 6 ft 044

1

kips2

7 238 kip-f t

b b A yM M A x x M

M x

x x M






loading shown,









Fig. P7.15

Solution

Beam equilibrium:



2

61.03 kips 07 kips/ft 7 kips/ft

250 kip-ft7 kips/ft2

250 kip-ft 061.03 kips 7

7 61.03 kips

3.5 61.

kips/ft2

3 0

y y

b b y

F A x V x V

xM A x x M

xx x M

V x

M x 250 kip-ftx

250 kip-ft 7 kips/ft 25 ft 12.5 ft

17 ft 0

113.97 kips

7 kips/ft 25 ft

113.97 kips 7 kips/ft 25 ft 0

61.03 kips

A

y

y

y y y

y

y

M

C

C

F A C

A

A

2

61.03 kips 07 kips/ft 7 kips/ft

7 kips/ft2

061.03 kips 7

7 61.03 kips

3.5 61.03

kips/ft2

kip- t f

y y

a a y

F A x V x V

xM A x x

V x

M

M

xx

x

M

x

x




Section c-c: For the interval 17 ft ≤ x < 25 ft:

7 kips/ft

61.03 kips 113.97 kips 07 kips/ft

( 17 ft) 7 kips/ft2

250 kip-ft

(17 ft)175 kips 113.97 kips 7 kips/ft2

7 175 kips

y y y

c c y y

F A C x V

x V

xM A x C x x

M

xx

V

x

x

2

250 kip-ft 0

3.5 175 2,187.5 kip -ftM x x

M


(c) Maximum bending moment and its

location

Mmax-positive = 266.04 kip-ft

@ x = 8.72 ft

Mmax-nagative = –224.0 kip-ft

@ x = 17 ft




7.16 Use the graphical method to construct the

shear-force and bending-moment diagrams for the

beam shown. Label all significant points on each

diagram and identify the maximum moments (both

positive and negative) along with their respective

locations. Clearly differentiate straight-line and

curved portions of the diagrams.

Fig. P7.16

Solution

Beam equilibrium:

Shear-force and bending-moment diagrams:

28 kips 4 ft 42 kips 8 ft 14 ft 0

32.00 kips

34 kips 56 kips

32.00 kips 28 kips 42 kips 0

38.00 kips

A y

y

y y y

y

y

M D

D

F A D

A

A











Fig. P7.17

Solution

Beam equilibrium:


35 kN 4 m 45 kN 8 m

15 kN 14 m 10 m 0

71 kN

35 kN 45 kN 15 kN

71 kN 35 kN 45 kN 15 kN 0

24 kN

A

y

y

y y y

y

y

M

D

D

F A D

A

A











Fig. P7.18

Solution

Beam equilibrium:


15 kips 25 kips 0

10 kips

15 kips 9 ft 25 kips 3 ft 0

60 kip-ft

y y

y

C C

C

F C

C

M M

M











Fig. P7.19

Solution

Beam equilibrium:


12 ft 6 ft 18 ft 010 kips/ft

40 kips

10 kips/ft 12 ft

40 kips 10 kips/ft 12 ft 0

80 kips

A y

y

y y y

y

y

M C

C

F A C

A

A











Fig. P7.20

Solution

Beam equilibrium:


4.5 kips/ft 12 ft 9 ft 12 ft 0

40.50 kips

4.5 kips/ft 12 ft

40.50 kips 4.5 kips/ft 12 ft 0

13.50 kips

A y

y

y y y

y

y

M C

C

F A C

A

A











Fig. P7.21

Solution

Beam equilibrium:


40 kN/m 3 m 50 kN 0

70 kN

40 kN/m 3 m 1.5 m

(50 kN)(3 m) 60 kN-m 0

90 kN-m

y y

y

A A

A

F A

A

M M

M











Fig. P7.22

Solution

Beam equilibrium:


28 kips 9 kips/ft 5 ft 0

17 kips

28 kips 8 ft 9 kips/ft 5 ft 2.5 ft 0

111.5 kip-ft

y y

y

C C

C

F C

C

M M

M











Fig. P7.23

Solution

Beam equilibrium:


10 kips/ft 9 ft 4.5 ft 6 ft 20 ft 0

47.25 kips

10 kips/ft 9 ft 0

42.75 kips

A y

y

y y y

y

M D

D

F A D

A











Fig. P7.24

Solution

Beam equilibrium:


4.5 kips/ft 7 ft 7 ft

38 kips 14.5 ft 18 ft 0

18.36 kip

4.5 kips/ft 7 ft 38 kips

18.36 kips 4.5 kips/ft 7 ft 38 kips 0

11.86 kips

A

y

y

y y y

y

y

M

E

E

F A E

A

A











Fig. P7.25

Solution

Beam equilibrium:


60 kN 2.5 m 45 kN/m 5 m 7.5 m

9 m 0

156.67 kN

60 kN 45 kN/m 4 m

156.67 kN 60 kN 45 kN/m 4 m 0

83.33 kN

A

y

y

y y y

y

y

M

D

D

F A D

A

A











Fig. P7.26

Solution

Beam equilibrium:


10 kips 0

10 kips

10 kips 10 ft 60 kip-ft 0

40 kip-ft

y y

y

C C

C

F C

C

M M

M











Fig. P7.27

Solution

Beam equilibrium:


2 kN 11 kN 0

13 kN

50 kN-m 2 kN 3.5 m

11 kN 6 m 0

23 kN-m

y y

y

A A

A

F A

A

M M

M











Fig. P7.28

Solution

Beam equilibrium:


66 kN-m 96 kN-m 12 m 0

13.50 kN

13.50 kN 0

13.50 kN

A y

y

y y y y

y

M D

D

F A D A

A










curved portions of the diagrams. Fig. P7.29

Solution

Beam equilibrium:


80 kN-m 25 kN/m 6 m 3 m

50 kN-m 6 m 0

70 kN

25 kN/m 6 m

70 kN 25 kN/m 6 m 0

80 kN

A

y

y

y y y

y

y

M

B

B

F A B

A

A











Fig. P7.30

Solution

Beam equilibrium:


25 kN-m 15 kN 8 m

7 kN/m 3 m 13.5 m 12 m 0

31.54 kN

15 kN 7 kN/m 3 m

31.54 kN 15 kN 7 kN/m 3 m 0

4.46 kN

A

y

y

y y y

y

y

M

D

D

F A D

A

A




7.31 Draw the shear-force and bending-moment

diagram for the beam shown in Fig. P7.31. Assume

the upward reaction provided by the ground to be

uniformly distributed. Label all significant points

on each diagram. Determine the maximum value of

(a) the internal shear force and (b) the internal

bending moment. Fig. P7.31

Solution

Beam equilibrium:

2 kips/ft 8 ft 25 kips 25 kips 16 ft 0

4.125 kips/ft

yF w

w

Shear-force and bending-moment diagrams

(a) Maximum value of internal shear force:

V = 16.50 kips @ x = 4 ft Ans.

V = −16.50 kips @ x = 12 ft Ans.

(b) Maximum value of internal bending

moment:

M = 33 kip-ft @ x = 4 ft, 12 ft Ans.




7.32 Draw the shear-force and bending-moment

diagram for the beam shown in Fig. P7.32. Assume

the upward reaction provided by the ground to be

uniformly distributed. Label all significant points

on each diagram. Determine the maximum value of

(a) the internal shear force and (b) the internal

bending moment. Fig. P7.32

Solution

Beam equilibrium:

40 kN/m 1 m 50 kN

40 kN/m 1 m 4 m 0

32.5 kN/m

yF

w

w


(a) Maximum value of internal shear force:

V = ±25 kN @ x = 2 m Ans.

(b) Maximum value of internal bending moment:

M = −4.62 kN-m @ x = 1.23 m

M = −4.62 kN-m @ x = 2.77 m

Mmax = 5.00 kN-m Ans.







diagram and identify the maximum moments along

with their respective locations. Additionally:

(a) Determine V and M in the beam at a point

located 0.75 m to the right of B.

(b) Determine V and M in the beam at a point

located 1.25 m to the left of C.

Fig. P7.33

Solution

Beam equilibrium:


Shear force V and bending moment M

at specific locations:

(a) At x = 3.75 m,

V = 177.5 kN Ans.

M = 1,167 kN-m Ans.

(b) At x = 13.75 m,

V = −323 kN Ans.

M = 442 kN-m Ans.

125 kN 3 m 50 kN/m 12 m 9 m

15 m 0

385.00 kN

125 kN 50 kN/m 12 m

385.00 kN 125 kN 50 kN/m 12 m 0

340.00 kN

A

y

y

y y y

y

y

M

C

C

F A C

A

A








with their respective locations. Additionally:

(a) Determine V and M in the beam at a point

located 0.75 m to the right of B.



Fig. P7.34

Solution

Beam equilibrium:




(a) At x = 3.75 m,

V = 85.5 kN Ans.

M = 30.4 kN-m Ans.

(b) At x = 7.75 m,

V = −74.5 kN Ans.

M = 52.4 kN-m Ans.

15 kN 3 m 40 kN/m 6 m 3 m

18 kN 10 m 6 m 0

142.50 kN

15 kN 40 kN/m 6 m 18 kN

142.5 kN 15 kN 40 kN/m 6 m 18 kN 0

130.50 kN

B

y

y

y y y

y

y

M

C

C

F B C

B

B




7.35 Use the graphical method to construct the shear-

force and bending-moment diagrams for the beam

shown. Label all significant points on each diagram

and identify the maximum moments along with their

respective locations. Additionally:

(a) Determine V and M in the beam at a point located

0.75 m to the right of B.



Fig. P7.35

Solution

Beam equilibrium:




(a) At x = 3.75 m,

V = 91.3 kN Ans.

M = 199.2 kN-m Ans.

(b) At x = 6.75 m,

V = −103.8 kN Ans.

M = 180.5 kN-m Ans.

25 kN/m 3 m 1.5 m

65 kN/m 5 m 2.5 m 5 m 0

185.00 kN

25 kN/m 3 m 65 kN/m 5 m

185.00 kN 25 kN/m 3 m

65 kN/m 5 m 0

65.00 kN

B

y

y

y y y

y

y

M

C

C

F B C

B

B













Fig. P7.36

Solution

Beam equilibrium:




(a) At x = 3.25 m,

V = −21.3 kN Ans.

M = 16.09 kN-m Ans.

(b) At x = 4.75 m,

V = 31.25 kN Ans.

M = 23.6 kN-m Ans.

75 kN 60 kN 35 kN/m 6 m 0

75.00 kN

75 kN 6 m 60 kN 3.5 m 120 kN-m

35 kN/m 6 m 3 m 0

90.00 kN-m

y y

y

C

C

C

F C

C

M

M

M












located 1.25 m to the left of D.

Fig. P7.37

Solution

Beam equilibrium:




(a) At x = 4.5 m,

V = 93.7 kN Ans.

M = 23.9 kN-m Ans.

(b) At x = 10.75 m,

V = −125.1 kN Ans.

M = 75.7 kN-m Ans.

52 kN 3 m 35 kN/m 9 m 4.5 m

150 kN-m 36 kN 12 m 9 m 0

204.83 kN

52 kN 35 kN/m 9 m 36 kN

204.83 kN 52 kN 35 kN/m 9 m 36 kN 0

198.17 kN

B

y

y

y y y

y

y

M

D

D

F B D

B

B













Fig. P7.38

Solution

Beam equilibrium:




(a) At x = 5.0 m,

V = 54.0 kN Ans.

M = −44.0 kN-m Ans.

(b) At x = 12.25 m,

V = −47.3 kN Ans.

M = −49.5 kN-m Ans.

25 kN/m 3.5 m 1.75 m

25 kN/m 12.5 m 6.25 m 80 kN 5.5 m

20 kN 12.5 m 10 m 0

161.00 kN

25 kN/m 16 m 80 kN 20 kN

161.00 kN 25 kN/m 16 m

80 kN 20 kN 0

B

y

y

y y y

y

M

D

D

F B D

B

179.00 kNyB













Fig. P7.39

Solution

Beam equilibrium:




(a) At x = 3.50 m,

V = 285 kN Ans.

M = 63.8 kN-m Ans.

(b) At x = 7.75 m,

V = −190.0 kN Ans.

M = 331 kN-m Ans.

160 kN 2 m 50 kN/m 2 m 1 m

50 kN/m 2 m 1 m

120 kN/m 5 m 4.5 m 7 m 0

340 kN

160 kN 50 kN/m 4 m

120 kN/m 5 m

340 kN 160 kN 50 kN/m 4 m

B

y

y

y y y

y

M

D

D

F B D

B

120 kN/m 5 m 0

620 kNyB








with their respective locations. Clearly differentiate

straight-line and curved portions of the diagrams.

Fig. P7.40

Solution

Beam equilibrium:


225 kN-m 120 kN/m 4 m 2 m

60 kN/m 2.5 m 8.75 m 7.5 m 0

273 kN

120 kN/m 4 m 60 kN/m 2.5 m

273 kN 120 kN/m 4 m

60 kN/m 2.5 m 0

357 kN

A

y

y

y y y

y

y

M

C

C

F A C

A

A










Fig. P7.41

Solution

Beam equilibrium:



5 kips/ft 5 ft 2.5 ft 25 kips 10 ft

15 ft 0

17.67 kips

5 kips/ft 8 ft 25 kips

17.67 kips 5 kips/ft 8 ft 25 kips 0

B

y

y

y y y

y

M

E

E

F B E

B

47.33 kipsyB










Fig. P7.42

Solution

Beam equilibrium:



17 kips 12 ft 9 ft 0

62.56 kips

8 kips/ft 9 ft 17 kips

62.56 kips 8 kips/ft 9 ft 17 kips 0

26.44 kips

B

y

y

y y y

y

y

M

C

C

F B C

B

B










Fig. P7.43

Solution

Beam equilibrium:


6 kips/ft 30 ft 15 ft 60 kips 10 ft

60 kips 20 ft 3 kips/ft 10 ft 35 ft

90 kip-ft 30 ft 0

62.00 kips

6 kips/ft 30 ft 10 ft3 kips/ft

60 kips 60 k

A

y

y

y y y

M

D

D

F A D

ips

62.00 kips 6 kips/ft 30 ft 10 ft3 kips/ft

60 kips 60 kips 0

28.00 kips

y

y

A

A








respective locations. Clearly differentiate straight-line

and curved portions of the diagrams.

Fig. P7.44

Solution

Beam equilibrium:


5 kips 5 ft 2 kips/ft 20 ft 10 ft 25 kip-ft

15 kips 8 ft 10 kips 23 ft 20 ft 0

23 kips

5 kips 2 kips/ft 20 ft 15 kips 10 kips

23 kips 5 kips 2 kips/ft 20 ft

B

y

y

y y y

y

M

D

D

F B D

B

15 kips 10 kips 0

17 kipsyB










Fig. P7.45

Solution

Beam equilibrium:


50 kN/m 2 m 1 m 20 kN 2 m

25 kN/m 3 m 3.5 m 50 kN 5 m 0

47.50 kN-m

50 kN/m 2 m 20 kN

25 kN/m 3 m 50 kN 0

55 kN

A A

A

y y

y

M M

M

F A

A










Fig. P7.46

Solution

Beam equilibrium:

20 kips 15 ft 6 kips/ft 8 ft 11 ft

12 kips/ft 7 ft 3.5 ft

70 kips 7 ft 0

32.00 kip-ft

20 kips 6 kips/ft 8 ft 70 kips

12 kips/ft 7 ft 0

C

C

C

y

y

M

M

M

F

C

C 42.00 kipsy











Fig. P7.47

Solution

Beam equilibrium:

4,000 lb-ft 800 lb/ft 4 ft 2 ft

9,000 lb-ft 600 lb/ft 10 ft 10 ft

3,600 lb 10 ft 15 ft 0

5,640 lb

B

y

y

M

E

E

800 lb/ft 4 ft 600 lb/ft 10 ft 3,600 lb

5,640 lb 800 lb/ft 4 ft 600 lb/ft 10 ft 3,600 lb 0

7,160 lb

y y y

y

y

F B E

B

B











Fig. P7.48

Solution

Beam equilibrium:

400 kN-m 500 kN-m 600 kN-m

60 kN 2 m 1 m 120 kN/m 4 m 4 m

60 kN/m 2 m 7 m 150 kN 6 m 8 m 0

530 kN

B

y

y

M

E

E

60 kN 2 m 120 kN/m 4 m

60 kN 2 m 150 kN

530 kN 60 kN 2 m 120 kN/m 4 m

60 kN 2 m 150 kN

340 kN

y y y

y

y

F B E

B

B











Fig. P7.49

Solution

Beam equilibrium:

Consider free-body diagram of DE:

55 kN/m 3 m 1.5 m 3 m 0

82.5 kN

55 kN/m 3 m

82.5 kN 55 kN/m 3 m 0

82.5 kN

D y

y

y y y

y

y

M E

E

F D E

D

D

Consider free-body diagram of ABCD:

60 kN-m 75 kN/m 5 m 2.5 m

100 kN 2.5 m 5 m 3.5 m

60 kN-m 75 kN/m 5 m 2.5 m

100 kN 2.5 m 82.5 kN 5 m 3.5 m 0

440 kN

75 kN/m 5 m 100 kN

A

y y

y

y

y y y y

M

D C

C

C

F A C D

A 440 kN 75 kN/m 5 m 100 kN 82.5 kN 0

117.5 kN

y

yA










Fig. P7.50

Solution

Beam equilibrium:

Consider free-body diagram of ABC:

500 lb/ft 10 ft 5 ft 15 ft 0

1,666.67 lb

500 lb/ft 10 ft

1,666.67 lb 500 lb/ft 10 ft 0

3,333.33 lb

A y

y

y y y

y

y

M C

C

F A C

A

A

Consider free-body diagram of CDE:

1,200 lb

1,666.67 lb 1,200 lb 0

2,866.67 lb

10 ft 1, 200 lb 8 ft

1,666.67 lb 10 ft 1,200 lb 8 ft 0

26,266.67 lb-ft

y y y

y

y

E y E

E

E

F C E

E

E

M C M

M

M










Fig. P7.51

Solution

Beam equilibrium:

1

2

1

2

1

2

70 kN/m 7 m (3.5 m)

70 kN/m 3 m 8 m

55 kN 10 m 7 m 0

443.57 kN

70 kN/m 7 m

70 kN/m 3 m 55 kN

443.57 kN 70 kN/m 7 m

70 kN/m 3 m 55 kN 0

206.43 kN

A

y

y

y y y

y

y

M

B

B

F A B

A

A










straight-line and curved portions of the diagrams. Fig. P7.52

Solution

Beam equilibrium:

1

2

1

2

1

2

8 ft6 kips/ft 8 ft

3

4 kips/ft 15 ft (14.5 ft) 17 ft 0

47.41 kips

6 kips/ft 8 ft

4 kips/ft 15 ft

47.41 kips 6 kips/ft 8 ft

4 kips/ft 15 ft 0

36.59 k

B

y

y

y y y

y

y

M

D

D

F B D

B

B ips











Fig. P7.53

Solution

Beam equilibrium:

1

2

1

2

(9 kips)(4 ft)

4 kips/ft 9 ft (13 ft) 0

270.00 kip-ft

(9 kips) 4 kips/ft 9 ft 0

27.00 kips

A

A

A

y y

y

M

M

M

F A

A











Fig. P7.54

Solution

Beam equilibrium:

1

2

1

2

(25 kN)(6 m)

30 kN/m 3 m 3 m 0

285.00 kN-m

25 kN 30 kN/m 3 m 0

70 kN

D

D

D

y y

y

M

M

M

F D

D










straight-line and curved portions of the diagrams. Fig. P7.55

Solution

Beam equilibrium:

12

12

12

(6 kips/ft)(22 ft)(11 ft)

2(8 ft)(9 kips/ft)(8 ft) 22 ft (22 ft) 0

3

110.73 kips

(6 kips/ft)(22 ft) (9 kips/ft)(8 ft)

(110.73 kips) (6 kips/ft)(22 ft) (9 k

A

y

y

y y y

y

M

B

B

F A B

A ips/ft)(8 ft) 0

57.27 kipsyA





7.56 For the beam and loading shown,

(a) Use discontinuity functions to write the

expression for w(x). Include the beam reactions in

this expression.

(b) Integrate w(x) to twice to determine V(x) and

M(x).

(c) Use V(x) and M(x) to plot the shear-force and

bending-moment diagrams.

Fig. P7.56

Solution

Beam equilibrium:

(180 lb)(2 ft) (450 lb)(6 ft) (9 ft) 0

340 lb

180 lb 450 lb 0

290 lb

A y

y

y y y

y

M D

D

F A D

A

Load function w(x):

1 1 1 1

( ) 290 lb 0 ft 180 lb 2 ft 450 lb 6 ft 340 lb 9 ftw x x x x x Ans.

Shear-force function V(x) and bending-moment function M(x):

0 0 0 0

( ) 290 lb 0 ft 180 lb 2 ft 450 lb 6 ft 340 lb 9 ftV x x x x x Ans.

1 1 1 1

( ) 290 lb 0 ft 180 lb 2 ft 450 lb 6 ft 340 lb 9 ftM x x x x x Ans.








this expression.


M(x).



Fig. P7.57

Solution

Beam equilibrium:

(10 kN)(2.5 m) (35 kN)(3 m) (5 m) 0

16 kN

10 kN 35 kN 0

29 kN

B y

y

y y y

y

M D

D

F B D

B

Load function w(x):

1 1 1 1

( ) 10 kN 0 m 29 kN 2.5 m 35 kN 5.5 m 16 kN 7.5 mw x x x x x Ans.


0 0 0 0

( ) 10 kN 0 m 29 kN 2.5 m 35 kN 5.5 m 16 kN 7.5 mV x x x x x Ans.

1 1 1 1

( ) 10 kN 0 m 29 kN 2.5 m 35 kN 5.5 m 16 kN 7.5 mM x x x x x Ans.








this expression.


M(x).



Fig. P7.58

Solution

Beam equilibrium:

(30 kN)(3 m) (20 kN)(7 m)

(15 kN)(15 m) (10 m) 0

45.5 kN

30 kN 20 kN 15 kN 0

19.5 kN

B

y

y

y y y

y

M

D

D

F A D

A

Load function w(x):

1 1 1

1 1

( ) 19.5 kN 0 m 30 kN 3 m 20 kN 7 m

45.5 kN 10 m 15 kN 15 m

w x x x x

x x Ans.


0 0 0

0 0

( ) 19.5 kN 0 m 30 kN 3 m 20 kN 7 m

45.5 kN 10 m 15 kN 15 m

V x x x x

x x Ans.

1 1 1

1 1

( ) 19.5 kN 0 m 30 kN 3 m 20 kN 7 m

45.5 kN 10 m 15 kN 15 m

M x x x x

x x Ans.








this expression.


M(x).



Fig. P7.59

Solution

Beam equilibrium:

5 kN 0

5 kN

(5 kN)(6 m) 20 kN-m 0

10 kN-m

y y

y

C C

C

F C

C

M M

M

Load function w(x):

1 2 1 2

( ) 5 kN 0 m 20 kN-m 3 m 5 kN 6 m 10 kN-m 6 mw x x x x x Ans.


0 1 0 1

( ) 5 kN 0 m 20 kN-m 3 m 5 kN 6 m 10 kN-m 6 mV x x x x x Ans.

1 0 1 0

( ) 5 kN 0 m 20 kN-m 3 m 5 kN 6 m 10 kN-m 6 mM x x x x x Ans.








this expression.


M(x).



Fig. P7.60

Solution

Beam equilibrium:

(35 kN/m)(2 m) 0

70 kN

(35 kN/m)(2 m)(4 m) 0

280 kN-m

y y

y

A A

A

F A

A

M M

M

Load function w(x):

1 2 0 0

( ) 70 kN 0 m 280 kN-m 0 m 35 kN/m 3 m 35 kN/m 5 mw x x x x x Ans.


0 1 1 1

( ) 70 kN 0 m 280 kN-m 0 m 35 kN/m 3 m 35 kN/m 5 mV x x x x x Ans.

1 0 2 235 kN/m 35 kN/m

( ) 70 kN 0 m 280 kN-m 0 m 3 m 5 m2 2

M x x x x x Ans.








this expression.


M(x).



Fig. P7.61

Solution

Beam equilibrium:

(25 kN)(4 m)(2 m) (32 kN)(6 m) (8 m) 0

49 kN

(25 kN)(4 m) 32 kN 0

83 kN

A y

y

y y y

y

M D

D

F A D

A

Load function w(x):

1 0 0

1 1

( ) 83 kN 0 m 25 kN/m 0 m 25 kN/m 4 m

32 kN 6 m 49 kN 8 m

w x x x x

x x Ans.


0 1 1

0 0

( ) 83 kN 0 m 25 kN/m 0 m 25 kN/m 4 m

32 kN 6 m 49 kN 8 m

V x x x x

x x Ans.

1 2 2

1 1

25 kN/m 25 kN/m( ) 83 kN 0 m 0 m 4 m

2 2

32 kN 6 m 49 kN 8 m

M x x x x

x x Ans.








this expression.


M(x).



Fig. P7.62

Solution

Beam equilibrium:

(3,000 lb)(5 ft) 8,000 lb-ft

(800 lb)(7 ft)(12.5 ft) (20 ft) 0

3,850 lb

3,000 lb (800 lb)(7 ft) 0

4,750 lb

A

y

y

y y y

y

M

E

E

F A E

A

Load function w(x):

1 1 2

0 0 1

( ) 4,750 lb 0 ft 3,000 lb 5 ft 8,000 lb-ft 5 ft

800 lb/ft 9 ft 800 lb/ft 16 ft 3,850 lb 20 ft

w x x x x

x x x Ans.


0 0 1

1 1 0

( ) 4,750 lb 0 ft 3,000 lb 5 ft 8,000 lb-ft 5 ft

800 lb/ft 9 ft 800 lb/ft 16 ft 3,850 lb 20 ft

V x x x x

x x x Ans.

1 1 0

2 2 1

( ) 4,750 lb 0 ft 3,000 lb 5 ft 8,000 lb-ft 5 ft

800 lb/ft 800 lb/ft9 ft 16 ft 3,850 lb 20 ft

2 2

M x x x x

x x x Ans.








this expression.


M(x).



Fig. P7.63

Solution

Beam equilibrium:

(800 lb/ft)(12 ft) (800 lb)(6 ft) 0

14,400 lb

(800 lb/ft)(12 ft)(6 ft)

(800 lb)(6 ft)(21 ft) 0

158,400 lb-ft

y y

y

A

A

A

F A

A

M

M

M

Load function w(x):

1 2 0

0 0 0

( ) 14,400 lb 0 ft 158,400 lb-ft 0 ft 800 lb-ft 0 ft

800 lb/ft 12 ft 800 lb/ft 18 ft 800 lb/ft 24 ft

w x x x x

x x x Ans.


0 1 1

1 1 1

( ) 14,400 lb 0 ft 158,400 lb-ft 0 ft 800 lb-ft 0 ft

800 lb/ft 12 ft 800 lb/ft 18 ft 800 lb/ft 24 ft

V x x x x

x x x Ans.

1 0 2

2 2 2

800 lb-ft( ) 14,400 lb 0 ft 158,400 lb-ft 0 ft 0 ft

2

800 lb/ft 800 lb/ft 800 lb/ft12 ft 18 ft 24 ft

2 2 2

M x x x x

x x x Ans.








this expression.


M(x).



Fig. P7.64

Solution

Beam equilibrium:

12 kN-m (18 kN/m)(2 m)(2 m) (5 m) 0

12 kN

(18 kN/m)(2 m) 0

24 kN

A y

y

y y y

y

M D

D

F A D

A

Load function w(x):

1 2 0

0 1

( ) 24 kN 0 m 12 kN-m 0 m 18 kN/m 1 m

18 kN/m 3 m 12 kN 5 m

w x x x x

x x Ans.


0 1 1

1 0

( ) 24 kN 0 m 12 kN-m 0 m 18 kN/m 1 m

18 kN/m 3 m 12 kN 5 m

V x x x x

x x Ans.

1 0 2

2 1

18 kN/m( ) 24 kN 0 m 12 kN-m 0 m 1 m

2

18 kN/m3 m 12 kN 5 m

2

M x x x x

x x Ans.








this expression.


M(x).



Fig. P7.65

Solution

Beam equilibrium:

12

12

12


2(8 ft)(9 kips/ft)(8 ft) 22 ft (22 ft) 0

3

110.73 kips

(6 kips/ft)(22 ft) (9 kips/ft)(8 ft)

(110.73 kips) (6 kips/ft)(22 ft) (9 k

A

y

y

y y y

y

M

B

B

F A B

A ips/ft)(8 ft) 0

57.27 kipsyA

Load function w(x):

1 0 1 0

1 1 0

( ) 57.27 kips 0 ft 6 kips/ft 0 ft 110.73 kips 22 ft 6 kips/ft 22 ft

9 kips/ft 9 kips/ft22 ft 30 ft 9 kips/ft 30 ft

8 ft 8 ft

w x x x x x

x x x Ans.


0 1 0 1

2 2 1

( ) 57.27 kips 0 ft 6 kips/ft 0 ft 110.73 kips 22 ft 6 kips/ft 22 ft


2(8 ft) 2(8 ft)

V x x x x x

x x x Ans.

1 2 1 2

3 3 2

6 kips/ft 6 kips/ft( ) 57.27 kips 0 ft 0 ft 110.73 kips 22 ft 22 ft

2 2

9 kips/ft 9 kips/ft 9 kips/ft22 ft 30 ft 30 ft

6(8 ft) 6(8 ft) 2

M x x x x x

x x x Ans.







this expression.


M(x).



Fig. P7.66

Solution

Beam equilibrium: 1

2

1

2

(20 kN/m)(3 m) (30 kN/m)(3 m) 0

105 kN

(20 kN/m)(3 m)(2.5 m)

(30 kN/m)(3 m)(2 m) 0

240 kN-m

y y

y

C

C

C

F C

C

M

M

M

Load function w(x):

0 1 0 1

0 1 2

30 kN/m 30 kN/m( ) 20 kN/m 0 m 0 m 20 kN/m 3 m 3 m

3 m 3 m

30 kN/m 3 m 105 kN 4 m 240 kN-m 4 m

w x x x x x

x x x Ans.


1 2 1 2

1 0 1


2(3 m) 2(3 m)

30 kN/m 3 m 105 kN 4 m 240 kN-m 4 m

V x x x x x

x x x Ans.

2 3 2 3

2 1 0

20 kN/m 30 kN/m 20 kN/m 30 kN/m( ) 0 m 0 m 3 m 3 m

2 6(3 m) 2 6(3 m)

30 kN/m3 m 105 kN 4 m 240 kN-m 4 m

2

M x x x x x

x x x Ans.








this expression.


M(x).

(c) Determine the maximum bending moment in

the beam between the two simple supports.

Fig. P7.67

Solution

Beam equilibrium: 1

2

1

2

9 kN-m (18 kN/m)(3 m)(1 m) (3 m) 0

6 kN

(18 kN/m)(3 m) 0

21 kN

B y

y

y y y

y

M C

C

F B C

B

Load function w(x):

2 1 0

1 1 1

( ) 9 kN-m 0 m 21 kN 1 m 18 kN/m 1 m

18 kN/m 18 kN/m1 m 4 m 6 kN 4 m

3 m 3 m

w x x x x

x x x Ans.


1 0 1

2 2 0

( ) 9 kN-m 0 m 21 kN 1 m 18 kN/m 1 m

18 kN/m 18 kN/m1 m 4 m 6 kN 4 m

2(3 m) 2(3 m)

V x x x x

x x x Ans.

0 1 2

3 3 1

18 kN/m( ) 9 kN-m 0 m 21 kN 1 m 1 m

2

18 kN/m 18 kN/m1 m 4 m 6 kN 4 m

6(3 m) 6(3 m)

M x x x x

x x x Ans.




Maximum bending moment:

Mmax = 5.66 kN-m at x = 2.59 m Ans.








this expression.


M(x).



Fig. P7.68

Solution

Beam equilibrium: 1

2

1

2

(5 kips/ft)(9 ft)(6 ft) (14 ft) 0

9.64 kips

(5 kips/ft)(9 ft) 0

12.86 kips

A y

y

y y y

y

M C

C

F A C

A

Load function w(x):

1 1 1

0 1

5 kips/ft 5 kips/ft( ) 12.86 kips 0 ft 0 ft 9 ft

9 ft 9 ft

5 kips/ft 9 ft 9.64 kips 14 ft

w x x x x

x x Ans.


0 2 2

1 0


2(9 ft) 2(9 ft)

5 kips/ft 9 ft 9.64 kips 14 ft

V x x x x

x x Ans.

1 3 3

2 1


6(9 ft) 6(9 ft)

5 kips/ft9 ft 9.64 kips 14 ft

2

M x x x x

x x Ans.





Mmax = 58.3 kip-ft at x = 6.80 ft Ans.








this expression.


M(x).



Fig. P7.69

Solution

Beam equilibrium:

1

2

1

2


21 ft(9 kips/ft)(21 ft) 6 ft (16 ft) 0

3

82.41 kips

(5 kips/ft)(6 ft) (9 kips/ft)(21 ft) 0

42.09 kips

A

y

y

y y y

y

M

C

C

F A C

A

Load function w(x):

1 0 0 0

1 1 1

( ) 42.09 kips 0 ft 5 kips/ft 0 ft 5 kips/ft 6 ft 9 kips/ft 6 ft

9 kips/ft 9 kips/ft6 ft 82.41 kips 16 ft 27 ft

21 ft 21 ft

w x x x x x

x x x Ans.


0 1 1 1

2 0 2

( ) 42.09 kips 0 ft 5 kips/ft 0 ft 5 kips/ft 6 ft 9 kips/ft 6 ft


2(21 ft) 2(21 ft)

V x x x x x

x x x Ans.

1 2 2 2

3 1 3

5 kips/ft 5 kips/ft 9 kips/ft( ) 42.09 kips 0 ft 0 ft 6 ft 6 ft

2 2 2


6(21 ft) 6(21 ft)

M x x x x x

x x x Ans.





Mmax = 170.9 kip-ft at x = 7.39 ft Ans.








this expression.


M(x).



Fig. P7.70

Solution

Beam equilibrium:

1

2

1

2

(25 kN/m)(4.0 m)(4.5 m)

2(4.0 m)(45 kN/m)(4.0 m) 2.5 m (8 m) 0

3

114.38 kN

(25 kN/m)(4.0 m) (45 kN/m)(4.0 m) 0

75.63 kN

A

y

y

y y y

y

M

D

D

F A D

A

Load function w(x):

1 0 0 1

1 0 1

45 kN/m( ) 75.63 kN 0 m 25 kN/m 2.5 m 25 kN/m 6.5 m 2.5 m

4.0 m

45 kN/m6.5 m 45 kN/m 6.5 m 114.38 kN 8 m

4.0 m

w x x x x x

x x x Ans.


0 1 1 2

2 1 0

45 kN/m( ) 75.63 kN 0 m 25 kN/m 2.5 m 25 kN/m 6.5 m 2.5 m

2(4.0 m)

45 kN/m6.5 m 45 kN/m 6.5 m 114.38 kN 8 m

2(4.0 m)

V x x x x x

x x x Ans.

1 2 2 3

3 2 1

25 kN/m 25 kN/m 45 kN/m( ) 75.63 kN 0 m 2.5 m 6.5 m 2.5 m

2 2 6(4.0 m)

45 kN/m 45 kN/m6.5 m 6.5 m 114.38 kN 8 m

6(4.0 m) 2

M x x x x x

x x x Ans.





Mmax = 275 kN-m at x = 4.57 m Ans.








this expression.


M(x).



Fig. P7.71

Solution

Beam equilibrium:

1

2

1

2

2(7.0 m)(30 kN/m)(7.0 m)(3.5 m) (40 kN/m)(7.0 m)

3

(50 kN/m)(2.0 m)(1.0 m) (5.5 m) 0

234.24 kN

(30 kN/m)(7.0 m) (40 kN/m)(7.0 m)

(50 kN/m)(2.0 m) 0

215.

C

y

y

y y y

y

M

B

B

F B C

C 76 kN

Load function w(x):

0 0 1

1 0 1

1 0 0

40 kN/m( ) 30 kN/m 0 m 40 kN/m 0 m 0 m

7.0 m

40 kN/m234.24 kN 1.5 m 30 kN/m 7 m 7 m

7.0 m

215.76 kN 7 m 50 kN/m 7.0 m 50 kN/m 9.0 m

w x x x x

x x x

x x x Ans.


1 1 2

0 1 2

0 1 1

40 kN/m( ) 30 kN/m 0 m 40 kN/m 0 m 0 m

2(7.0 m)

40 kN/m234.24 kN 1.5 m 30 kN/m 7 m 7 m

2(7.0 m)

215.76 kN 7 m 50 kN/m 7.0 m 50 kN/m 9.0 m

V x x x x

x x x

x x x Ans.

2 2 3

1 2 3

1 2 2

30 kN/m 40 kN/m 40 kN/m( ) 0 m 0 m 0 m

2 2 6(7.0 m)

30 kN/m 40 kN/m234.24 kN 1.5 m 7 m 7 m

2 6(7.0 m)

50 kN/m 50 kN/m215.76 kN 7 m 7.0 m 9.0 m

2 2

M x x x x

x x x

x x x Ans.





Mmax = 86.6 kN-m at x = 4.00 m Ans.








this expression.


M(x).



Fig. P7.72

Solution

Beam equilibrium:

1

2

1

2

2(4.5 m)(60 kN)(1.5 m) (90 kN/m)(4.5 m)

3

(6.5 m) 0

79.62 kN

60 kN (90 kN/m)(4.5 m) 0

182.88 kN

B

y

y

y y y

y

M

D

D

F B D

B

Load function w(x):

1 1 1

1 0 1

90 kN/m( ) 60 kN 0 m 182.88 kN 1.5 m 1.5 m

4.5 m

90 kN/m6 m 90 kN/m 6 m 79.62 kN 8 m

4.5 m

w x x x x

x x x Ans.


0 0 2

2 1 0

90 kN/m( ) 60 kN 0 m 182.88 kN 1.5 m 1.5 m

2(4.5 m)

90 kN/m6 m 90 kN/m 6 m 79.62 kN 8 m

2(4.5 m)

V x x x x

x x x Ans.

1 1 3

3 2 1

90 kN/m( ) 60 kN 0 m 182.88 kN 1.5 m 1.5 m

6(4.5 m)

90 kN/m 90 kN/m6 m 6 m 79.62 kN 8 m

6(4.5 m) 2

M x x x x

x x x Ans.





Mmax = 197.2 kN-m at x = 5.01 m Ans.





8.1 During fabrication of a laminated timber arch, one of the 10 in. wide by 1 in. thick Douglas fir [E =

1,900 ksi] planks is bent to a radius of curvature of 40 ft. Determine the maximum bending stress

developed in the plank.

Solution

From Eq. (8.3):

1,900 ksi

( 0.5 in.) 1.979 ksi(40 ft)(12 in./f

1.979 kt

si)

x

Ey

Ans.




8.2 A high-strength steel [E = 200 GPa] tube having an outside diameter of 80 mm and a wall thickness

of 3 mm is bent into a circular curve having a 52-m radius of curvature. Determine the maximum

bending stress developed in the tube.

Solution

From Eq. (8.3):

200,000 MPa

( 80 mm / 2) 153.846 MPa(52 m)(1,000 mm/

153.8 MPm)

ax

Ey

Ans.




8.3 A high-strength steel [E = 200 GPa] band saw blade wraps around a pulley that has a diameter of

450 mm. Determine the maximum bending stress developed in the blade. The blade is 12-mm wide and

1-mm thick.

Solution

The radius of curvature of the band saw blade is:

450 mm 1 mm

225.5 mm2 2

From Eq. (8.3):

200,000 MPa

( 0.5 mm) 443.459 MPa225.5 mm

443 MPax

Ey

Ans.




8.4 The boards for a concrete form are to be bent into a circular shape having an inside radius of 10 m.

What maximum thickness can be used for the boards if the normal stress is not to exceed 7 MPa?

Assume that the modulus of elasticity for the wood is 12 GPa.

Solution

The radius of curvature of the concrete form is dependent on the board thickness:

10,000 mm2

t

From Eq. (8.3):

12,000 MPa

7 MPa2

10,000 mm2

x

E ty

t

Solve for t:

12,000 MPa 7 MPa 10,000 mm2 2

6,000 70,000 3.5

5,996.5

11.67 mm

70,000

t t

t t

t

t

Ans.




8.5 A beam having a tee-shaped cross section is subjected to equal 12 kN-m bending moments, as

shown in Fig. P8.5a. The cross-sectional dimensions of the beam are shown in Fig. P8.5b. Determine:

(a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus

about the z axis.

(b) the bending stress at point H. State whether the normal stress at H is tension or compression.

(c) the maximum bending stress produced in the cross section. State whether the stress is tension or

compression.

Fig. P8.5a Fig. P8.5b

Solution

(a) Centroid location in y direction: (reference axis at bottom of tee shape)

Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

top flange 2,500.0 162.5 406,250.0

stem 3,750.0 75.0 281,250.0

6,250.0 mm2

687,500.0 mm

3

3

2

687,500.0 mm

6,250.0 mm110.0 mm

i i

i

y Ay

A

(measured upward from bottom edge of stem) Ans.

Moment of inertia about the z axis:

Shape IC d = yi – y d²A IC + d²A

(mm4) (mm) (mm

4) (mm

4)

top flange 130,208.33 52.50 6,890,625.00 7,020,833.33

stem 7,031,250.00 −35.00 4,593,750.00 11,625,000.00

Moment of inertia about the z axis (mm4) = 18,645,833.33

418,656,000 mmzI Ans.




Section moduli:

4

3

top

top

4

3

bot

bot

3

18,645,833.33 mm286,858.974 mm

(175 mm 110 mm)

18,645,833.33 mm169,507.576 mm

110 mm

169,500 mm

z

z

IS

c

IS

c

S

Ans.

(b) Bending stress at point H: (y = 175 mm − 25 mm − 110 mm = 40 mm)

4

(12 kN-m)(40 mm)(1,000 N/kN)(1,000 mm/m)

18,654,833.33 mm

25.743 MPa 25.7 MPa (C)

x

z

M y

I

Ans.

(c) Maximum bending stress:

The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the

cross section is at y = +65 mm, and the bottom of the cross section is at y = −110 mm. The larger

bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom

of the cross section.

4

(12 kN-m)( 110 mm)(1,000 N/kN)(1,000 mm/m)

18,654,833.33 mm

70.793 MPa 70.8 MPa (T)

x

z

M y

I

Ans.




8.6 A beam is subjected to equal 6.5 kip-ft bending moments, as shown in Fig. P8.6a. The cross-

sectional dimensions of the beam are shown in Fig. P8.6b. Determine:


about the z axis.

(b) the bending stress at point H, which is located 2 in. below the z centroidal axis. State whether the

normal stress at H is tension or compression.


compression.


Solution

(a) Centroid location in y direction: (reference axis at bottom of shape)

Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

left side 8.0 4.0 32.0

top flange 4.0 7.5 30.0

right side 8.0 4.0 32.0

20.0 in.2

94.0 in.

3

3

2

94.0 i4.70

n.

20.0 i i

nn

..

i i

i

y Ay

A

(measured upward from bottom edge of section) Ans.



(in.4) (in.) (in.

4) (in.

4)

left side 42.667 −0.700 3.920 46.587

top flange 0.333 2.800 31.360 31.693

right side 42.667 −0.700 3.920 46.587

Moment of inertia about the z axis (in.4) = 124.867

4124.9 in.zI Ans.




Section moduli:

43

top

top

43

bot

bot

3

124.867 in.37.8384 in.

(8 in. 4.7 in.)

124.867 in.26.5674 in.

4.7 in

26.6 in.

.

z

z

IS

c

IS

c

S

Ans.

(b) Bending stress at point H: (y = −2 in.)

4

( 6.5 kip-ft)( 2 in.)(12 in./ft)

124.867 in.

1,249 p 1,249 psi ( )i Cs

x

z

M y

I

Ans.



cross section is at y = +3.30 in., and the bottom of the cross section is at y = −4.7 in. The larger bending

stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom of the

cross section.

4

( 6.5 kip-ft)( 4.7 in.)(12 in./ft)

124.867 in

2,940 psi (

.

2,935. i C)9 ps

x

z

M y

I

Ans.




8.7 A beam is subjected to equal 470 N-m bending moments, as shown in Fig. P8.7a. The cross-



about the z axis.



compression.


Solution

(a) Centroid location in y direction: (reference axis at bottom of U shape)

Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

left side 400.0 25.0 10,000.0

bottom flange 272.0 4.0 1,088.0

right side 400.0 25.0 10,000.0

1,072.0 mm2

21,088.0 mm

3

3

2

21,088.0 mm

1,072.0 mm19.67 mm

i i

i

y Ay

A

(measured upward from bottom edge of section) Ans.



(mm4) (mm) (mm

4) (mm

4)

left side 83,333.33 5.33 11,356.56 94,689.89

bottom flange 1,450.67 −15.67 66,803.30 68,253.96

right side 83,333.33 5.33 11,356.56 94,689.89

Moment of inertia about the z axis (mm4) = 257,633.75

4257,600 mmzI Ans.




Section moduli:

4

3

top

top

4

3

bot

bot

3

257,633.75 mm8,494.814 mm

(50 mm 19.672 mm)

257,633.75 mm13,096.708 mm

19.672

8,495 mm

mm

z

z

IS

c

IS

c

S

Ans.

(b) Bending stress at point H: (y = 8 mm − 19.672 mm = −11.672 mm)

4

(470 N-m)( 11.672 mm)(1,000 mm/m)

257,633.75 m

21.

m

21 3 MPa.293 (T) MPa

x

z

M y

I

Ans.



cross section is at y = +30.328 mm, and the bottom of the cross section is at y = −19.672 mm. The larger

bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the top of

the cross section.

4

(470 N-m)(30.328 mm)(1,000 mm/m)

257,633.7

55.3 MPa (

5 mm

55.328 C) MPa

x

z

M y

I

Ans.




8.8 A beam is subjected to equal 17.5 kip-ft bending moments, as shown in Fig. P8.8a. The cross-



about the z axis.


(c) the bending stress at point K. State whether the normal stress at K is tension or compression.

(d) the maximum bending stress produced in the cross section. State whether the stress is tension or

compression.


Solution


Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

top flange 12.0000 13.0000 156.0000

web 20.0000 7.0000 140.0000

bottom flange 20.0000 1.0000 20.0000

52.0000 in.2

316.0000 in.

3

3

2

316.0 in.6.077 in.

52.06.08 in.

in.

i i

i

y Ay

A

(measured upward from bottom edge of bottom

flange) Ans.



(in.4) (in.) (in.

4) (in.

4)

top flange 4.000 6.923 575.148 579.148

web 166.667 0.923 17.041 183.708

bottom flange 6.667 -5.077 515.503 522.170

Moment of inertia about the z axis (in.4) = 1,285.026

Ans.




Section Moduli

43

4

3

6.0769 in.

14 in. 6.0769 in. 7.9231 in.

1,285.026 in.211.460 in.

6.0769 in.

1,285.026 in.162.188 in.

7.9231 in.

The controlling section modulus is the smaller of the

bot

top

zbot

bot

ztop

top

c

c

IS

c

IS

c

3

two values; theref

162.2

ore,

in.S Ans.

Bending stress at point H:

From the flexure formula:

4

( 17.5 kip-ft)(7.9231 in. 2 in.)(12 in./ft)967.9544 psi

1,285.0256 i968 psi (T)

n.x

z

M y

I

Ans.

Bending stress at point K:

From the flexure formula:

4

( 17.5 kip-ft)( 6.0769 in. 2 in.)(12 in./ft)666.2543 psi

1,285.026 in666 psi

.(C)x

z

M y

I

Ans.

Maximum bending stress

Since ctop > cbot, the maximum bending stress occurs at the top of the flanged shape. From the flexure

formula:

4

( 17.5 kip-ft)(7.9231 in.)(12 in./ft)1,294.8 psi

1,285.021,295 psi

6 n.(T)

ix

z

M y

I

Ans.

Also, note that the same maximum bending stress magnitude can be calculated with the section

modulus:

3

(17.5 kip-ft)(12 in./ft)1,294.8 psi

162.1,295 ps

1877 in.i

x

M

S Ans.

The sense of the stress (either tension or compression) would be determined by inspection.




8.9 The cross-sectional dimensions of a beam are

shown in Fig. P8.9.

(a) If the bending stress at point K is 43 MPa (C),

determine the internal bending moment Mz acting

about the z centroidal axis of the beam.

(b) Determine the bending stress at point H. State

whether the normal stress at H is tension or

compression.

Fig. P8.9

Solution

Centroid location in y direction: (reference axis at bottom of double-tee shape)

Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

top flange 375.0 47.5 17,812.5

left stem 225.0 22.5 5,062.5

right stem 225.0 22.5 5,062.5

825.0 mm2

27,937.5 mm

3

3

2

27,937.5 mm33.864 mm 33.9 mm

825.0 mm

i i

i

y Ay

A

(measured upward from bottom of section)



(mm4) (mm) (mm

4) (mm

4)

top flange 781.250 13.636 69,731.405 70,512.655

left stem 37,968.750 −11.364 29,054.752 67,023.502

right stem 37,968.750 −11.364 29,054.752 67,023.502


(a) Determine bending moment:

At point K, y = 50 mm − 5 mm − 33.864 mm = 11.136 mm. The bending stress at K is x = −43 MPa;

therefore, the bending moment magnitude can be determined from the flexure formula:

2 4( 43 N/mm )(204,559.659 mm )

11.136 mm

789,850.765 N- 790 N-mmm

x

z

x z

M y

I

IM

y

Ans.

(b) Bending stress at point H:

At point H, y = −33.864 mm. The bending stress is computed with the flexure formula:

4

(789,850.765 N-mm)( 33.864 mm)130.755 MPa

2130.8 MPa (

04,559.659 mmT)x

z

M y

I

Ans.





shown in Fig. P8.10.

(a) If the bending stress at point K is 2,600 psi (T),

determine the internal bending moment Mz acting

about the z centroidal axis of the beam.

(b) Determine the bending stress at point H. State


compression.

Fig. P8.10

Solution

Centroid location in y direction: (reference axis at bottom of inverted-tee shape)

Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

bottom flange 0.56250 0.12500 0.07031

stem 0.56250 1.37500 0.77344

1.12500 in.2

0.84375 in.

3

3

2

0.84375 in.0.750 in.

1.1250 in.

i i

i

y Ay

A

(measured upward from bottom edge of section)



(in.4) (in.) (in.

4) (in.

4)

bottom flange 0.00293 −0.62500 0.21973 0.22266

stem 0.23730 0.62500 0.21973 0.45703


(a) Determine bending moment:

At point K, y = 2.50 in. − 0.75 in. = 1.750 in. The bending stress at K is x = +2,600 psi; therefore, the

bending moment magnitude can be determined from the flexure formula:

4(2,600 psi)(0.67967 in. )

1.750 in.

1,009.820 lb-in. 1,010 lb-in. 84.2 lb-ft

x

z

x z

M y

I

IM

y

Ans.

(b) Bending stress at point H:

At point H, y = −0.75 in. The bending stress is computed with the flexure formula:

4

( 1,009.820 lb-in.)( 0.75 in.)1,114.286 psi

01,114 psi

.67969 i(C)

n.x

z

M y

I

Ans.




8.11 The cross-sectional dimensions of a box-

shaped beam are shown in Fig. P8.11. If the

maximum allowable bending stress is b =

15,000 psi, determine the maximum internal

bending moment Mz magnitude that can be

applied to the beam.

Fig. P8.11

Solution

Moment of inertia about z axis:

3 3

4(3 in.)(2 in.) (2.5 in.)(1 in.)1.791667 in.

12 12zI

Maximum internal bending moment Mz:

4(15,000 psi)(1.791667 in. )

26,875 lb-in.1 in.

2,240 lb-ft

zx

z

x z

M c

I

IM

c

Ans.





shown in Fig. P8.12. The internal bending moment

about the z centroidal axis is Mz = +2.70 kip-ft.

Determine:

(a) the maximum tension bending stress in the

beam.

(b) the maximum compression bending stress in the

beam.

Fig. P8.12

Solution

Centroid location in y direction: (reference axis at bottom of shape)

Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

left stem 2.000 2.000 4.000

top flange 2.500 3.750 9.375

right stem 2.000 2.000 4.000

6.500 in.2

17.375 in.

3

3

2

17.375 in.2.673 in.

6.500 in.

i i

i

y Ay

A

(measured upward from bottom edge of section)



(in.4) (in.) (in.

4) (in.

4)

left stem 2.66667 −0.67308 0.90607 3.57273

top flange 0.05208 1.07692 2.89941 2.95149

right stem 2.66667 −0.67308 0.90607 3.57273


(a) Determine maximum tension bending stress:

For a positive bending moment, tension bending stresses will be created below the neutral axis.

Therefore, the maximum tension bending stress will occur at point K (i.e., y = −2.673 in.):

4

(2.70 kip-ft)( 2.673 in.)(12 in./ft)8.578 ksi

10.09696 in.8.58 ksi (T)x

z

M y

I

Ans.

(b) Determine maximum compression bending stress:

For a positive bending moment, compression bending stresses will be created above the neutral axis.

Therefore, the maximum compression bending stress will occur at point H (i.e., y = 4 in. − 2.673 in. =

1.327 in.):

4

(2.70 kip-ft)(1.327 in.)(12 in./ft)4.258 ksi

10.09694.26 ksi

6 in. (C)x

z

M y

I Ans.






(a) If the bending stress at point K is 35.0 MPa (T),

determine the bending stress at point H. State


compression.

(b) If the allowable bending stress is b = 165 MPa,

determine the magnitude of the maximum bending

moment Mz that can be supported by the beam.

Fig. P8.13

Solution



(mm4) (mm) (mm

4) (mm

4)

top flange 540,000.000 160.000 184,320,000.000 184,860,000.000

web 32,518,666.667 0.000 0.000 32,518,666.667

bottom flange 540,000.000 −160.000 184,320,000.000 184,860,000.000


(a) At point K, y = −90 mm, and at point H, y = −175 mm. The bending stress at K is x = +35 MPa, and

the bending stress is distributed linearly over the depth of the cross section. Therefore, the bending

stress at H can be found from the ratio:

175 mm

(35.0 MPa) 68.056 MPa90

6

8.1 MPa (T)mm

H K

H K

HH K

K

y y

y

y

Ans.

(b) Maximum internal bending moment Mz:

2 4(165 N/mm )(402,238,667 mm )

379,253,600 N-mm175 m

3 Nm

79 k -m

zx

z

x z

z

M c

I

IM

c

Ans.






(a) If the bending stress at point K is 9.0 MPa (T),

determine the bending stress at point H. State


compression.

(b) If the allowable bending stress is b = 165 MPa,

determine the magnitude of the maximum bending

moment Mz that can be supported by the beam.

Fig. P8.14

Solution



(mm4) (mm) (mm

4) (mm

4)

left flange 9,720,000 0 0 9,720,000

web 31,680 0 0 31,680

right flange 9,720,000 0 0 9,720,000

Moment of inertia about the z axis (mm4) = 19,471,680

(a) At point K, y = −60 mm, and at point H, y = +90 mm. The bending stress at K is x = +9.0 MPa, and

the bending stress is distributed linearly over the depth of the cross section. Therefore, the bending

stress at H can be found from the ratio:

90 mm

(9.0 MP 13.50 a) 13.50 MPa60 m

MPa (Cm

)

H K

H K

HH K

K

y y

y

y

Ans.

(b) Maximum bending moment Mz:

2 4(165 N/mm )(19,471,680 mm )

35,698,080 N-mm90 mm

35.7 kN-m

zx

z

x z

M c

I

IM

c

Ans.






about the z centroidal axis is Mz = −1.55 kip-ft.

Determine:

(a) the maximum tension bending stress in the beam.


beam.

Fig. P8.15

Solution

Centroid location in y direction:

Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

top flange 8.0 4.5 36.0

left web 3.0 2.5 7.5

left bottom flange 3.0 0.5 1.5

right web 3.0 2.5 7.5

right bottom flange 3.0 0.5 1.5

20.0 in.2

54.0 in.

3

3

2

54.0 in.2.70 in.

20.0 in.

i i

i

y Ay

A

(measured upward from bottom edge of bottom flange)



(in.4) (in.) (in.

4) (in.

4)

top flange 0.6667 1.8000 25.9200 26.5867

left web 2.2500 −0.2000 0.1200 2.3700

left bottom flange 0.2500 −2.2000 14.5200 14.7700

right web 2.2500 −0.2000 0.1200 2.3700

right bottom flange 0.2500 −2.2000 14.5200 14.7700


(a) Maximum tension bending stress:

For a negative bending moment, the maximum tension bending stress will occur at the top surface of the

cross section. From the flexure formula, the bending stress at the top surface is:

4

( 1.55 kip-ft)(5.0 in. 2.70 in.)(12 in./ft)0.7028 ksi

60.8667 i703 p

nsi T)

. (x

z

M y

I

Ans.

(b) Maximum compression bending stress:

The maximum compression bending stress will occur at the bottom surface of the cross section. From

the flexure formula, the bending stress at the bottom surface is:

4

( 1.55 kip-ft)( 2.70 in.)(12 in./ft)0.8251 ksi

60.8667 in.825 psi (C)x

z

M y

I

Ans.






about the z centroidal axis is Mz = +270 lb-ft.

Determine:

(a) the maximum tension bending stress in the beam.


beam.

Solution Fig. P8.16


Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

bottom flange 0.40625 0.06250 0.02539

left web 0.28125 1.25000 0.35156

left top flange 0.09375 2.43750 0.22852

right web 0.28125 1.25000 0.35156

right top flange 0.09375 2.43750 0.22852

1.15625 in.2

1.18555 in.

3

3

2

1.18555 in.1.0253 in.

1.15625 in.

i i

i

y Ay

A

(measured upward from bottom edge of bottom flange)



(in.4) (in.) (in.

4) (in.

4)

bottom flange 0.000529 −0.962838 0.376617 0.377146

left web 0.118652 0.224662 0.014196 0.132848

left top flange 0.000122 1.412162 0.186956 0.187079

right web 0.118652 0.224662 0.014196 0.132848

right top flange 0.000122 1.412162 0.186956 0.187079


(a) Maximum tension bending stress:

For a positive bending moment of Mz = +270 lb-ft, the maximum tension bending stress will occur at the

bottom surface of the cross section (i.e., y = −1.0253 in.). From the flexure formula, the bending stress

at the bottom of the cross section is:

4

(270 lb-ft)( 1.0253 in.)(12 in./ft)3,26 3,270 psi6.446 psi

1.016999(

i)

n. Tx

z

M y

I

Ans.


The maximum compression bending stress will occur at the top surface of the cross section (i.e., y = 2.50

in. − 1.0253 in. = 1.4747 in.). From the flexure formula, the bending stress at the top of the cross

section is:

4

(270 lb-ft)(1.4747 in.)(12 in./ft)4,69 4,700 psi8.164 psi

1.016999

in.(C)x

z

M y

I Ans.




8.17 Two vertical forces are applied to a simply supported beam (Fig. P8.17a) having the cross section

shown in Fig. P8.17b. Determine the maximum tension and compression bending stresses produced in

segment BC of the beam.


Solution


Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

top flange 3,000.0 167.5 502,500.0

stem 1,440.0 80.0 115,200.0

4,440 mm2

617,700 mm

3

3

2

617,700 mm139.1216 mm

4,440 mm

i i

i

y Ay

A

(measured upward from bottom edge of stem)



(mm4) (mm) (mm

4) (mm

4)

top flange 56,250.00 28.38 2,415,997.08 2,472,247.08

stem 3,072,000.00 −59.12 5,033,327.25 8,105,327.25






The maximum moment occurs between B and C. The moment magnitude is 12 kN-m.

Maximum tension bending stress:

For a positive bending moment, the maximum tension bending stress will occur at the bottom surface of

this cross section. From the flexure formula, the bending stress at the bottom of the tee stem is:

6 4

(12 kN-m)( 139.1216 mm)(1,000 N/kN)(1,000 mm/m)

10.5776 10 m157.8 MPa )

m(Tx

z

M y

I

Ans.

Maximum compression bending stress:

The maximum compression bending stress will occur at the top of the flange:

6 4

(12 kN-m)(175 mm 139.1216 mm)(1,000 N/kN)(1,000 mm/m)

10.5776 10 mm

40.7 MPa 40.7 MPa (C)

x

z

M y

I

Ans.




8.18 Two vertical forces are applied to a simply supported beam (Fig. P8.18a) having the cross section

shown in Fig. P8.18b. Determine the maximum tension and compression bending stresses produced in

segment BC of the beam.


Solution


Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

left stem 0.7500 1.5000 1.1250

bottom flange 0.5000 0.1250 0.0625

right stem 0.7500 1.5000 1.1250

2.000 in.2

2.3125 in.

3

3

2

2.3125 in.1.1563 in.

2.000 in.

i i

i

y Ay

A

(measured upward from bottom edge of stem)



(in.4) (in.) (in.

4) (in.

4)

left stem 0.56250 0.34375 0.08862 0.65112

bottom flange 0.00260 −1.03125 0.53174 0.53434

right stem 0.56250 0.34375 0.08862 0.65112






The maximum moment occurs between B and C. The moment magnitude is 600 lb-ft.


For a positive bending moment, the maximum tension bending stress will occur at the bottom surface of

this cross section at y = −1.1563 in. From the flexure formula, the bending stress at the bottom of the U

shape is:

4

(600 lb-ft)( 1.1563 in.)(12 in./ft)4,533 4,530 psi (.053 psi

1.83659 in.T)x

z

M y

I

Ans.

Maximum compression bending stress:

The maximum compression bending stress will occur at the top of the U shape, where y = 1.8438 in.:

4

(600 lb-ft)(1.8438 in.)(12 in./ft)7,228 7,230 psi .265 psi

1(

.83659 in.C)x

z

M y

I Ans.




8.19 A WT230 × 26 standard steel shape is used to support the loads shown on the beam in Fig. P8.19a.

The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch of

the cross section (Fig. P8.19b). Consider the entire 4-m length of the beam and determine:

(a) the maximum tension bending stress at any location along the beam, and

(b) the maximum compression bending stress at any location along the beam.


Solution

Section properties

From Appendix B: 6 416.7 10 mmzI


Maximum bending moments

positive M = 13.61 kN-m

negative M = −20.00 kN-m

Bending stresses at max positive moment

2

6 4

2

6 4

(13.61 kN-m)(60.7 mm)(1,000)

16.7 10 mm

49.5 MPa (C)

(13.61 kN-m)( 164.3 mm)(1,000)

16.7 10 mm

133.9 MPa (T)

x

x

Bending stresses at max negative moment

2

6 4

2

6 4

( 20 kN-m)(60.7 mm)(1,000)

16.7 10 mm

72.7 MPa (T)

( 20 kN-m)( 164.3 mm)(1,000)

16.7 10 mm

196.8 MPa (C)

x

x

(a) Maximum tension bending stress

(b) Maximum compression bending stress

133.9 MPa (T)

196.8 MPa (C)

Ans.

Ans.




8.20 A WT305 × 41 standard steel shape is used to support the loads shown on the beam in Fig. P8.20a.

The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch of

the cross section (Fig. P8.19b). Consider the entire 10-m length of the beam and determine:




Solution

Section properties

From Appendix B: 6 448.7 10 mmzI






2

6 4

2

6 4

(45.84 kN-m)(88.9 mm)(1,000)

48.7 10 mm

83.7 MPa (C)

(45.84 kN-m)( 211.1 mm)(1,000)

48.7 10 mm

198.7 MPa (T)

x

x


2

6 4

2

6 4

( 24 kN-m)(88.9 mm)(1,000)

48.7 10 mm

43.8 MPa (T)

( 24 kN-m)( 211.1 mm)(1,000)

48.7 10 mm

104.0 MPa (C)

x

x



198.7 MPa (T)

104.0 MPa (C)

Ans.

Ans.




8.21 A steel tee shape is used to support the loads shown on the beam in Fig. P8.21a. The dimensions of

the shape are shown in Fig. P8.21b. Consider the entire 24-ft length of the beam and determine:




Solution


Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

top flange 24.0000 19.2500 462.0000

stem 13.8750 9.2500 128.3438

37.875 in.2

590.3438 in.

3

3

2

590.3438 in.15.5866 in. (from bottom of shape to centroid)

37.8750 in.

4.4134 in. (from top of shape to centroid)

i i

i

y Ay

A



(in.4) (in.) (in.

4) (in.

4)

top flange 4.5000 3.6634 322.0861 326.5861

stem 395.7266 −6.3366 557.1219 952.8484







positive M = 100.75 kip-ft

negative M = −68.00 kip-ft


4

4

(100.75 kip-ft)(4.4134 in.)(12 in./ft)

1, 279.4345 in.

4.17 ksi (C)

(100.75 kip-ft)( 15.5866 in.)(12 in./ft)

1, 279.4345 in.

14.73 ksi (T)

x

x


4

4

( 68 kip-ft)(4.4134 in.)(12 in./ft)

1,279.4345 in.

2.81 ksi (T)

( 68 kip-ft)( 15.5866 in.)(12 in./ft)

1,279.4345 in.

9.94 ksi (C)

x

x



14.73 ksi (T)

9.94 ksi (C)

Ans.

Ans.




8.22 A flanged wooden shape is used to support the loads shown on the beam in Fig. P8.22a. The

dimensions of the shape are shown in Fig. P8.22b. Consider the entire 18-ft length of the beam and

determine:




Solution


Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

top flange 20.0 11.0 220.0

web 16.0 6.0 96.0

bottom flange 12.0 1.0 12.0

48.0 in.2

328.0 in.

3

3

2


48.0 in.


i i

i

y Ay

A



(in.4) (in.) (in.

4) (in.

4)

top flange 6.667 4.167 347.222 353.889

web 85.333 –0.833 11.111 96.444

bottom flange 4.000 –5.833 408.333 412.333







positive M = 10,580 lb-ft

negative M = −8,400 lb-ft


4

4

(10,580 lb-ft)(5.1667 in.)(12 in./ft)

862.667 in.

760.4 psi (C)

(10,580 lb-ft)( 6.8333 in.)(12 in./ft)

862.667 in.

1,005.6 psi (T)

x

x


4

4

( 8,400 lb-ft)(5.1667 in.)(12 in./ft)

862.667 in.

603.7 psi (T)

( 8,400 lb-ft)( 6.8333 in.)(12 in./ft)

862.667 in.

798.5 psi (C)

x

x


(b) Maximum compression bending stres

1,006 psi (T)

799 psi (C)s

Ans.

Ans.




8.23 A channel shape is used to support the loads shown on the beam in Fig. P8.23a. The dimensions of

the shape are shown in Fig. P8.23b. Consider the entire 12-ft length of the beam and determine:




Solution


Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

left stem 3.000 3.000 9.000

top flange 5.500 5.750 31.625

right stem 3.000 3.000 9.000

11.500 in.2

49.625 in.

3

3

2


11.500 in.


i i

i

y Ay

A



(in.4) (in.) (in.

4) (in.

4)

left stem 9.0000 −1.3152 5.1894 14.1894

top flange 0.1146 1.4348 11.3223 11.4369

right stem 9.0000 −1.3152 5.1894 14.1894







positive M = 8,850 lb-ft

negative M = −9,839 lb-ft


4

4

(8,850 lb-ft)(1.6848 in.)(12 in./ft)

39.8157 in.

4,494 psi (C) 4.49 ksi (C)

(8,850 lb-ft)( 4.3152 in.)(12 in./ft)

39.8157 in.

11,510 psi (T) 11.51 ksi (T)

x

x


4

4

( 9,839 lb-ft)(1.6848 in.)(12 in./ft)

39.8157 in.

4,996 psi (T) 5.00 ksi (T)

( 9,839 lb-ft)( 4.3152 in.)(12 in./ft)

39.8157 in.

12,796 psi (C) 12.80 ksi (C)

x

x



11.51 ksi (T)

12.80 ksi (C)

Ans.

Ans.




8.24 A W360 × 72 standard steel shape is used to support the loads shown on the beam in Fig. P8.24a.

The shape is oriented so that bending occurs about the weak axis as shown in Fig. P8.24b. Consider the

entire 6-m length of the beam and determine:




Solution

Section properties

From Appendix B: 6 421.4 10 mm 204 mmz fI b





Since the shape is symmetric about the z axis,

the largest bending stresses will occur at the

location of the largest moment magnitude –

either positive or negative. In this case, the

largest bending stresses will occur where the

moment magnitude is 31.50 kN-m.

Bending stresses at maximum moment

2

6 4

(31.50 kN-m)( 204 mm/2)(1,000)

21.4 10 mm

150.1 MPa (T) and (C)

x



150.1 MPa (T)

150.1 MPa (C)

Ans.

Ans.




8.25 A 1.00-in.-diameter solid steel

shaft supports loads PA = 180 lb and PC

= 240 lb as shown in Fig. P8.25.

Assume L1 = 5 in., L2 = 16 in., and L3 =

8 in. The bearing at B can be idealized

as a roller support and the bearing at D

can be idealized as a pin support.

Determine the magnitude and location

of the maximum bending stress in the

shaft.

Fig. P8.25

Solution

Section properties

4 4 4(1.00 in.) 0.049087 in.64 64

I D



positive M = 980 lb-in.

negative M = −900 lb-in.

Since the circular cross section is symmetric

about the z axis, the largest bending stresses

will occur at the location of the largest moment

magnitude – either positive or negative. In this

case, the largest bending stresses will occur at

C, where the moment magnitude is 980 lb-in.


4

(980 lb-in.)( 1.00 in./2)

0.049087

9

in.

,980 psi

x

Ans.




8.26 A 30-mm-diameter solid steel

shaft supports loads PA = 1,400 N and

PC = 2,600 N as shown in Fig. P8.26.

Assume L1 = 100 mm, L2 = 200 mm,

and L3 = 150 mm. The bearing at B can

be idealized as a roller support and the

bearing at D can be idealized as a pin

support. Determine the magnitude and

location of the maximum bending stress

in the shaft.

Fig. P8.26

Solution

Section properties

4 4 4(30 mm) 39,760.8 mm64 64

I D



positive M = 162,857 N-mm

negative M = −140,000 N-mm





case, the largest bending stresses will occur

where the moment magnitude is 162,857 N-

mm.


4

(162,857 N-mm)( 30 mm/2)

39,760.8 mm

61.4 MPa

x

Ans.




8.27 A 20-mm-diameter solid steel shaft

supports loads PA = 500 N, PC = 1,750

N, and PE = 500 N as shown in Fig.

P8.27. Assume L1 = 90 mm, L2 = 260

mm, L3 = 140 mm, and L4 = 160 mm.

The bearing at B can be idealized as a

roller support and the bearing at D can

be idealized as a pin support. Determine

the magnitude and location of the

maximum bending stress in the shaft. Fig. P8.27

Solution

Section properties

4 4 4(20 mm) 7,853.9816 mm64 64

zI D



positive M = 91,500 N-mm

negative M = −80,000 N-mm






C, where the moment magnitude is 91,500 N-

mm.


4

(91,500 N-mm)( 20 mm/2)

7,853.9816

1

mm

16.5 MPa

x

Ans.




8.28 A 1.75-in.-diameter solid steel shaft

supports loads PA = 250 lb, PC = 600 lb,

and PE = 250 lb as shown in Fig. P8.28.

Assume L1 = 9 in., L2 = 24 in., L3 = 12

in., and L4 = 15 in. The bearing at B can

be idealized as a roller support and the



location of the maximum bending stress

in the shaft. Fig. P8.28

Solution

Section properties

4 4 4(1.75 in.) 0.460386 in.64 64

I D



positive M = 1,550 lb-in.

negative M = −3,750 lb-in.






support D, where the moment magnitude is

3,750 lb-in.


4

( 3,750 lb-in.)( 1.75 in./2)

0.460386 in.

7,130 psi

x

Ans.




8.29 A HSS12 × 8 × 1/2 standard steel shape

is used to support the loads shown on the

beam in Fig. P8.29. The shape is oriented so

that bending occurs about the strong axis.

Determine the magnitude and location of the

maximum bending stress in the beam.

Fig. P8.29

Solution

Section properties

From Appendix B: 4333 in. 12 in.zI d



positive M = 124.59 kip-ft

negative M = −72.00 kip-ft





largest bending stresses will occur at C, where

the moment magnitude is 124.59 kip-ft.

Bending stresses at max moment magnitude

4

(124.59 kip-ft)( 12 in./2)(12 in./ft)

333 in.26.9 ksix Ans.




8.30 A W410 × 60 standard steel shape

is used to support the loads shown on

the beam in Fig. P8.30. The shape is

oriented so that bending occurs about

the strong axis. Determine the

magnitude and location of the

maximum bending stress in the beam. Fig. P8.30

Solution

Section properties

From Appendix B: 6 4216 10 mm 406 mmzI d



positive M = 50 kN-m

negative M = −70 kN-m





largest bending stresses will occur between B

and C, where the moment magnitude is 70 kN-

m.

Bending stresses at max moment magnitude

2

6 4

(70 kN-m)( 406 mm/2)(1,000)

216 10 m65.8 MPa

mx Ans.




8.31 A solid steel shaft supports loads

PA = 200 lb and PD = 300 lb as shown

in Fig. P8.31. Assume L1 = 6 in.,

L2 = 20 in., and L3 = 10 in. The bearing

at B can be idealized as a roller support

and the bearing at C can be idealized

as a pin support. If the allowable

bending stress is 8 ksi, determine the

minimum diameter that can be used for

the shaft.

Fig. P8.31

Solution


Maximum bending moment magnitude

M = 3,000 lb-in.

Minimum required section modulus

33,000 lb-in.0.375 in.

8,000 psi

x

x

M

S

MS

Section modulus for solid circular section

3

32

dS

Minimum shaft diameter

330.375 in.

1.

32

563 in.

d

d

Ans.




8.32 A solid steel shaft supports loads

PA = 500 N and PD = 400 N as shown

in Fig. P8.32. Assume L1 = 200 mm,

L2 = 660 mm, and L3 = 340 mm. The

bearing at B can be idealized as a roller

support and the bearing at C can be

idealized as a pin support. If the

allowable bending stress is 25 MPa,

determine the minimum diameter that

can be used for the shaft.

Fig. P8.32

Solution



M = 136,000 N-mm


3

2

136,000 N-mm5,440 mm

25 N/mm

x

x

M

S

MS

Section modulus for solid circular section

3

32

dS

Minimum shaft diameter

335,440 m

38. m

m3

1 m

2

d

d

Ans.




8.33 A simply supported wood beam (Fig. P8.33a) with a span of L = 20 ft supports a uniformly

distributed load of w = 800 lb/ft. The allowable bending stress of the wood is 1,400 psi. If the aspect

ratio of the solid rectangular wood beam is specified as h/b = 1.5 (Fig. P8.33b), determine the minimum

width b that can be used for the beam.


Solution


Also, see Example 7-3 for shear-force and bending-moment diagram development.

Maximum bending moment

2 2

max

(800 lb/ft)(20 ft)

8 8

40,000 lb-ft 480,000 lb-in.

wLM


3480,000 lb-in.342.8571 in.

1, 400 psi

x

x

M

S

MS

Section modulus for solid rectangular section

3 2/12

/ 2 6

I bh bhS

c h

The aspect ratio of the solid rectangular wood beam is specified as h/b = 1.5; therefore, the section

modulus can be expressed as:

2 2 3

3(1.5 ) 2.250.3750

6 6 6

bh b b bS b

Minimum allowable beam width

3 30.3750 342.8571

9.71 in

.

in.b

b

Ans.

The corresponding beam height h is

/ 1.5 1.5 1.5(9.71 in.) 14.57 in.h b h b




8.34 A simply supported wood beam (Fig. P8.34a) with a span of L = 14 ft supports a uniformly

distributed load of w. The beam width is b = 6 in. and the beam height is h = 10 in. (Fig. P8.34b). The

allowable bending stress of the wood is 900 psi. Determine the magnitude of the maximum load w that

may be carried by the beam.


Solution

Moment of inertia for rectangular cross section about horizontal centroidal axis

3 3

4(6 in.)(10 in.)500 in.

12 12

bhI

Maximum allowable moment

4(900 psi)(500 in. )

90,000 lb-in. 7,500 lb-ft5 in.

xx

IMcM

I c


Also, see Example 7-3 for shear-force and bending-moment diagram development.

Determine distributed load intensity

Equate the moment expression from the bending-

moment diagram to the maximum allowable moment

that can be applied to the rectangular cross section:

2

max 7,500 lb-ft8

wLM

Solve for the maximum distributed load w that can be

applied to the 14-ft simple span:

2 2

2

(14 ft)7,500 lb-ft

8 8

8(7,500 lb-ft)

(14 ft)306 lb/ft

wL w

w

Ans.




8.35 A cantilever timber beam (Fig. P8.35a) with a span of L = 2.5 m supports a uniformly distributed

load of w = 4 kN/m. The allowable bending stress of the wood is 9 MPa. If the aspect ratio of the solid

rectangular timber is specified as h/b = 0.5 (Fig. P8.35b), determine the minimum width b that can be

used for the beam.


Solution

Maximum moment magnitude:

The maximum bending moment magnitude in the cantilever beam occurs at support A:

2 2

6

max

(4 kN/m)(2.5 m)12.5 kN-m 12.5 10 N-mm

2 2

wLM


6

6 3

2

12.5 10 N-mm1.3889 10 mm

9 N/mm

x

x

M

S

MS


3 2/12

/ 2 6

I bh bhS

c h

The aspect ratio of the solid rectangular wood beam is specified as h/b = 0.5; therefore, the section

modulus can be expressed as:

2 2 3

3(0.5 ) 0.250.0416667

6 6 6

bh b b bS b

Minimum allowable beam width

3 6 30.0416667 1.3889 10 mm

321.83 mm 322 mm

b

b

Ans.

The corresponding beam height h is

/ 0.5

0.5 0.5(321.83 mm) 161 mm

h b

h b




8.36 A cantilever timber beam (Fig. P8.36a) with a span of L = 3 m supports a uniformly distributed

load of w. The beam width is b = 300 mm and the beam height is h = 200 mm (Fig. P8.36b). The

allowable bending stress of the wood is 6 MPa. Determine the magnitude of the maximum load w that

may be carried by the beam.


Solution


3 2 2

6 3/12 (300 mm)(200 mm)2 10 mm

/ 2 6 6

I bh bhS

c h

Maximum allowable bending moment:

2 6 3 6

allow (6 N/mm )(2 10 mm ) 12 10 N-mmx x

MM S

S

Maximum bending moment in cantilever span:

The maximum bending moment magnitude in the cantilever beam occurs at support A:

2

max2

wLM

Maximum distributed load:

2

allow

6

allow

allow 2 2

2

2 2(12 10 N-mm)2.67 N/mm

[(3 m)(1,000 mm2.67 kN/

/ )m

m ]

wLM

Mw

L

Ans.




8.37 The beam shown in Fig. P8.37 will be

constructed from a standard steel W-shape

using an allowable bending stress of 24 ksi.

(a) Develop a list of five acceptable shapes

that could be used for this beam. On this list,

include the most economical W10, W12,

W14, W16, and W18 shapes.

(b) Select the most economical W shape for

this beam. Fig. P8.37

Solution



M = 90 kip-ft


3(90 kip-ft)(12 in./ft)45 in.

24 ksi

x

x

M

S

MS

(a) Acceptable steel W-shapes

3

3

3

3

3

W10 45, 49.1 in.

W12 40, 51.5 in.

W14 34, 48.6 in.

W16 31, 47.2 in.

W18 35, 57.6 in.

S

S

S

S

S

(b) Most economical W-shape

W16 31 Ans.





constructed from a standard steel W-shape using

an allowable bending stress of 165 MPa.

(a) Develop a list of four acceptable shapes that

could be used for this beam. Include the most

economical W360, W410, W460, and W530

shapes on the list of possibilities.

(b) Select the most economical W shape for this

beam.

Fig. P8.38

Solution



M = 206.630 kN-m


2

3 3

2

(206.63 kN-m)(1,000)1,252 10 mm

165 N/mm

x

x

M

S

MS


3 3

3 3

3 3

3 3

W360 79, 1,270 10 mm

W410 75, 1,330 10 mm

W460 74, 1,460 10 mm

W530 66, 1,340 10 mm

S

S

S

S


W530 66 Ans.





constructed from a standard steel W-shape using

an allowable bending stress of 165 MPa.

(a) Develop a list of four acceptable shapes that

could be used for this beam. Include the most

economical W360, W410, W460, and W530


(b) Select the most economical W shape for this

beam. Fig. P8.39

Solution



M = 238.57 kN-m


2

3 3

2

(238.57 kN-m)(1,000)1,446 10 mm

165 N/mm

x

x

M

S

MS


3 3

3 3

3 3

3 3

W360 101, 1,690 10 mm

W410 85, 1,510 10 mm

W460 74, 1,460 10 mm

W530 74, 1,550 10 mm

S

S

S

S


or W460 74 W530 74 Ans.




8.40 The beam shown in Fig. P8.40 will be constructed from a standard

steel W-shape using an allowable bending stress of 165 MPa.

(a) Develop a list of four acceptable shapes that could be used for this

beam. Include the most economical W310, W360, W410, and W460


(b) Select the most economical W shape for this beam.

Fig. P8.40

Solution

Maximum moment magnitude:

The maximum bending moment magnitude occurs at the base of the cantilever beam:

max

6

1 1(15 kN)(3.0 m) (40 kN/m)(3.0 m) (3.0 m)

2 3

105.0 kN-m 105.0 10 N-mm

M


2

3 3

2

(105.0 kN-m)(1,000)636 10 mm

165 N/mm

x

x

M

S

MS


3 3

3 3

3 3

3 3

W310 60, 844 10 mm

W360 44, 688 10 mm

W410 46.1, 773 10 mm

W460 52, 944 10 mm

S

S

S

S


W360 44 Ans.





constructed from a standard steel HSS-shape

using an allowable bending stress of 30 ksi.

(a) Develop a list of three acceptable shapes that

could be used for this beam. On this list, include

the most economical HSS8, HSS10, and HSS12

shapes.

(b) Select the most economical HSS-shape for this

beam.

Fig. P8.41

Solution



M = 45.56 kip-ft


3(45.56 kip-ft)(12 in./ft)18.22 in.

30 ksi

x

x

M

S

MS

(a) Acceptable steel HSS shapes

3

3

3

3

HSS8 none are acceptable

HSS10 4 3 / 8, 20.8 in.

HSS10 6 3 / 8, 27.4 in.

HSS12 6 3 / 8, 35.9 in.

HSS12 8 3 / 8, 43.7 in.

S

S

S

S

(b) Most economical HSS shape

HSS10 4 3 / 8 Ans.




8.42 A composite beam is fabricated by bolting two 3 in. wide × 12 in. deep timber planks to the sides

of a 0.50 in. × 12 in. steel plate (Fig. P8.42b). The moduli of elasticity of the timber and the steel are

1,800 ksi and 30,000 ksi, respectively. The simply supported beam spans a distance of 20 ft and carries

two concentrated loads P, which are applied at the quarter points of the span (Fig. P8.42a).

(a) Determine the maximum bending stresses produced in the timber planks and the steel plate if P = 3

kips.

(b) Assume that the allowable bending stresses of the timber and the steel are 1,200 psi and 24,000 psi,

respectively. Determine the largest acceptable magnitude for concentrated loads P. (You may neglect

the weight of the beam in your calculations.)

Fig. P8.42a

Fig. P8.42b

Solution

Let the timber be denoted as material (1) and the steel plate as material (2). The modular ratio is:

2

1

30,000 ksi16.6667

1,800 ksi

En

E

Transform the steel plate (2) into an equivalent amount of wood (1) by multiplying its width by the

modular ratio: b2, trans = 16.6667(0.50 in.) = 8.3333 in. Thus, for calculation purposes, the 12 in. × 0.50

in. steel plate is replaced by a wood board that is 12 in. deep and 8.3333-in. thick.

Moment of inertia about the horizontal centroidal axis


(in.4) (in.) (in.

4) (in.

4)

timber (1) 864 0 0 864

transformed steel plate (2) 1,200 0 0 1,200

Moment of inertia about the z axis = 2,064 in.4

Maximum bending moment in beam for P = 3 kips The maximum bending moment in the simply supported beam with two 3-kip concentrated loads is:

max (3 kips)(5 ft) 15 kip-ft 180 kip-in.M

Bending stress in timber (1) From the flexure formula, the maximum bending stress in timber (1) is:

1 4

(180 kip-in.)( 6 in.)0.5233 ksi

2,523

064 in.psi

My

I Ans.




Bending stress in steel plate (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the

maximum bending stress in steel plate (2) is:

2 4

(180 kip-in.)( 6 in.)(16.6667) 8.7209 ksi

2,064 8,720

n p

.s

ii

Myn

I Ans.

Determine maximum P

If the allowable bending stress in the timber is 1,200 psi, then the maximum bending moment that may

be supported by the beam is:

4

11 max

(1.200 ksi)(2,064 in. )412.80 kip-in.

6 in.

IMyM

I y

If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may be

supported by the beam is:

4

22 max

(24.00 ksi)(2,064 in. )495.36 kip-in.

(16.667)(6 in.)

IMyn M

I ny

Note: The negative signs were omitted in the previous two equations because only the moment

magnitude is of interest here.

From these two results, the maximum moment that the beam can support is 412.80 kip-in. The

maximum concentrated load magnitude P that can be supported is found from:

max

max

(5 ft)

412.80 kip-in.

5 ft (5 ft)(12 in./ft)6.88 kips

M P

MP Ans.




8.43 The cross section of a composite beam that

consists of 4-mm-thick fiberglass faces bonded to a 20-

mm-thick particleboard core is shown in Fig. P8.43.

The beam is subjected to a bending moment of 55 N-m

acting about the z axis. The elastic moduli for the

fiberglass and the particleboard are 30 GPa and 10 GPa,

respectively. Determine:

(a) the maximum bending stresses in the fiberglass

faces and the particleboard core.

(b) the stress in the fiberglass at the joint where the two

materials are bonded together.

Fig. P8.43

Solution

Let the particleboard be denoted as material (1) and the fiberglass as material (2). The modular ratio is:

2

1

30 GPa3

10 GPa

En

E

Transform the fiberglass faces into an equivalent amount of particleboard by multiplying their width by

the modular ratio: b2, trans = 3(50 mm) = 150 mm. Thus, for calculation purposes, the 50 mm × 4 mm

fiberglass faces are replaced by particleboard faces that are 150-mm wide and 4-mm thick.



(mm4) (mm) (mm

4) (mm

4)

transformed fiberglass top face 800.00 12.00 86,400.00 87,200.00

particleboard core 33,333.33 0 0 33,333.33

transformed fiberglass bot face 800.00 12.00 86,400.00 87,200.00

Moment of inertia about the z axis = 207,733.33 mm4

Bending stress in particleboard core (1) From the flexure formula, the maximum bending stress in the particleboard core is:

1 4

(55 N-m)( 10 mm)(1,000 mm/m)

207,733.33 2.65 MPa

mm

My

I Ans.

Bending stress in fiberglass faces (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the

maximum bending stress in the fiberglass faces (2) is:

2 4

(55 N-m)( 14 mm)(1,000 mm/m)(3)

207,733.33 mm11.12 MPa

Myn

I Ans.

Bending stress in fiberglass (2) at interface At the interface between the particleboard and the fiberglass, y = ±10 mm:

2 4

(55 N-m)( 10 mm)(1,000 mm/m)(3)

207,733.33 m7.94

m MPa

Myn

I Ans.




8.44 A composite beam is made of two brass [E =

100 GPa] plates bonded to an aluminum [E = 75

GPa] bar, as shown in Fig. P8.44. The beam is

subjected to a bending moment of 1,750 N-m acting

about the z axis. Determine:

(a) the maximum bending stresses in the brass

plates and the aluminum bar.

(b) the stress in the brass at the joints where the two

materials are bonded together.

Fig. P8.44

Solution

Let the aluminum be denoted as material (1) and the brass as material (2). The modular ratio is:

2

1

100 GPa1.3333

75 GPa

En

E

Transform the brass plates into an equivalent amount of aluminum by multiplying their width by the

modular ratio: b2, trans = 1.3333(50 mm) = 66.6666 mm. Thus, for calculation purposes, the 50 mm × 10

mm brass plates are replaced by aluminum plates that are 66.6666-mm wide and 10-mm thick.



(mm4) (mm) (mm

4) (mm

4)

transformed top brass plate 5,555.55 20 266,666.40 272,221.95

aluminum bar 112,500.00 0 0 112,500.00

transformed bot brass plate 5,555.55 –20 266,666.40 272,221.95

Moment of inertia about the z axis = 656,943.90 mm4

Bending stress in aluminum bar (1) From the flexure formula, the maximum bending stress in the aluminum bar is:

1 4

(1,750 N-m)( 15 mm)(1,000 mm/m)

656,943.90 mm40.0 MPa

My

I Ans.

Maximum bending stress in brass plates (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the

maximum bending stress in the brass plates (2) is:

2 4

(1,750 N-m)( 25 mm)(1,000 mm/m)(1.3333)

656,943.90 m88.8 MPa

m

Myn

I Ans.

Bending stress in brass plates (2) at interface At the interface between the brass plates and the aluminum bar, y = ±15 mm:

2 4

(1,750 N-m)( 15 mm)(1,000 mm/m)(1.3333)

656,943.90 m53.3 MPa

m

Myn

I Ans.




8.45 An aluminum [E = 10,000 ksi] bar is bonded to a steel [E = 30,000 ksi] bar to form a composite

beam (Fig. P8.45b). The composite beam is subjected to a bending moment of M = +300 lb-ft about the

z axis (Fig. P8.45a). Determine:

(a) the maximum bending stresses in the aluminum and steel bars.

(b) the stress in the two materials at the joint where they are bonded together.


Solution

Denote the aluminum as material (1) and denote the steel as material (2). The modular ratio is:

2

1

30,000 ksi3

10,000 ksi

En

E

Transform the steel bar (2) into an equivalent amount of aluminum (1) by multiplying its width by the

modular ratio: b2, trans = 3(2.00 in.) = 6.00 in. Thus, for calculation purposes, the 2.00 in. × 0.75 in. steel

bar is replaced by an aluminum bar that is 6.00-in. wide and 0.75-in. thick.

Centroid location of the transformed section in the vertical direction

Shape Width b Height h Area Ai

yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

aluminum bar (1) 2.00 0.50 1.00 0.25 0.2500

transformed steel bar (2) 6.00 0.75 4.50 0.875 3.9375

5.50

4.1875

3

2

4.1875 in.

5.500.7614 in.

in.

i i

i

y Ay

A (measured upward from bottom edge of section)



(in.4) (in.) (in.

4) (in.

4)

aluminum bar (1) 0.02083 –0.5114 0.2615 0.2823

transformed steel bar (2) 0.2109 0.1136 0.05811 0.2690

Moment of inertia about the z axis = 0.5514 in.4




(a) Maximum bending stress in aluminum bar (1) From the flexure formula, the maximum bending stress in aluminum bar (1) is:

1 4

(300 lb-ft)( 0.7614 in.)(12 in./ft)

0.5514 in4,970 psi (T)

.

My

I Ans.

(a) Maximum bending stress in steel bar (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the

maximum bending stress in steel bar (2) is:

2 4

(300 lb-ft)(1.250 in. 0.7614 in.)(12 in./ft)(3)

0.5514 i9,570 psi (C)

n.

My

I Ans.

(b) Bending stress in aluminum bar (1) at interface

1 4

(300 lb-ft)(0.50 in. 0.7614 in.)(12 in./ft)

0.5514 in.1,706 psi (T)

My

I Ans.

(b) Bending stress in steel bar (2) at interface

2 4

(300 lb-ft)(0.50 in. 0.7614 in.)(12 in./ft)(3)

0.5514 in.5,120 psi (T)

My

I Ans.




8.46 An aluminum [E = 10,000 ksi] bar is bonded to a steel [E = 30,000 ksi] bar to form a composite

beam (Fig. P8.46b). The allowable bending stresses for the aluminum and steel bars are 20 ksi and 30

ksi, respectively. Determine the maximum bending moment M that can be applied to the beam.


Solution

Denote the aluminum as material (1) and denote the steel as material (2). The modular ratio is:

2

1

30,000 ksi3

10,000 ksi

En

E

Transform the steel bar (2) into an equivalent amount of aluminum (1) by multiplying its width by the

modular ratio: b2, trans = 3(2.00 in.) = 6.00 in. Thus, for calculation purposes, the 2.00 in. × 0.75 in. steel

bar is replaced by an aluminum bar that is 6.00-in. wide and 0.75-in. thick.



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

aluminum bar (1) 2.00 0.50 1.00 0.25 0.2500

transformed steel bar (2) 6.00 0.75 4.50 0.875 3.9375

5.50

4.1875

3

2

4.1875 in.

5.500.7614 in.

in.

i i

i

y Ay




(in.4) (in.) (in.

4) (in.

4)

aluminum bar (1) 0.02083 –0.5114 0.2615 0.2823

transformed steel bar (2) 0.2109 0.1136 0.05811 0.2690





(a) Maximum bending moment magnitude based on allowable aluminum stress Based on an allowable bending stress of 20 ksi for the aluminum, the maximum bending moment

magnitude that be applied to the cross section is:

4

11

(20 ksi)(0.5514 in. )14.484 kip-in.

0.7614 in.

My IM

I y (a)

Maximum bending moment magnitude based on allowable steel stress Based on an allowable bending stress of 30 ksi for the steel, the maximum bending moment magnitude

that be applied to the cross section is:

4

22

(30 ksi)(0.5514 in. )11.285 kip-in.

(3)(1.25 in. 0.7614 in.)

My In M

I n y (b)


From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the

cross section is

max 11.285 kip 940- in l ft. b-M Ans.




8.47 Two steel [E = 30,000 ksi] plates are securely

attached to a Southern pine [E = 1,800 ksi] timber

to form a composite beam (Fig. P8.47). The

allowable bending stress for the steel plates is

24,000 psi and the allowable bending stress for the

Southern pine is 1,200 psi. Determine the maximum

bending moment that can be applied about the

horizontal axis of the beam.

Fig. P8.47

Solution

Denote the timber as material (1) and denote the steel as material (2). The modular ratio is:

2

1

30,000 ksi16.6667

1,800 ksi

En

E

Transform the steel plates into an equivalent amount of timber by multiplying their width by the

modular ratio: b2, trans = 16.6667(8 in.) = 133.3333 in. Thus, for calculation purposes, the 8 in. × 0.25 in.

steel plates can be replaced by wood plates that are 133.3333-in. wide and 0.25-in. thick.



(in.4) (in.) (in.

4) (in.

4)

transformed steel plate at top 0.1736 8.125 2,200.52 2,200.694

timber (1) 3,413.3333 0 0 3,413.333

transformed steel plate at bottom 0.1736 –8.125 2,200.52 2,200.694

Moment of inertia about the z axis = 7,814.72 in.4

(a) Maximum bending moment magnitude based on allowable Southern pine stress Based on an allowable bending stress of 1,200 psi for the Southern pine timber, the maximum bending

moment magnitude that be applied to the cross section is:

4

11

(1.200 ksi)(7,814.72 in. )1,172.208 kip-in.

8 in.

IMyM

I y (a)

Maximum bending moment magnitude based on allowable steel stress Based on an allowable bending stress of 24,000 psi for the steel plates, the maximum bending moment

magnitude that be applied to the cross section is:

4

22

(24 ksi)(7,814.72 in. )1,364.021 kip-in.

(16.6667)( 8.25 in.)

IMyn M

I n y (b)


From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the

cross section is

max 1,172.208 kip 97.7 -in ki t. p-fM Ans.




8.48 A simply supported composite beam 5 m long carries a uniformly distributed load w (Fig. P8.48a).

The beam is constructed of a Southern pine [E = 12 GPa] timber, 200 mm wide by 360 mm deep, that is

reinforced on its lower surface by a steel [E = 200 GPa] plate that is 150 mm wide by 12 mm thick (Fig.

P8.48b).

(a) Determine the maximum bending stresses produced in the timber and the steel if w = 12 kN/m.

(b) Assume that the allowable bending stresses of the timber and the steel are 9 MPa and 165 MPa,

respectively. Determine the largest acceptable magnitude for distributed load w. (You may neglect the

weight of the beam in your calculations.)


Solution

Let the timber be denoted as material (1) and the steel plate as material (2). The modular ratio is:

2

1

200 GPa16.6667

12 GPa

En

E

Transform the steel plate (2) into an equivalent amount of wood (1) by multiplying its width by the

modular ratio: b2, trans = 16.6667(150 mm) = 2,500 mm. Thus, for calculation purposes, the 150 mm ×

12 mm steel plate is replaced by a wood board that is 2,500-mm wide and 12-mm thick.



yi

(from bottom) yi Ai

(mm) (mm) (mm2) (mm) (mm

3)

timber (1) 200 360 72,000 192 13,824,000

transformed steel plate (2) 2,500 12 30,000 6 180,000

102,000

14,004,000

3

2

14,004,000 mm

102,000 mm137.294 mm

i i

i

y Ay




(mm4) (mm) (mm

4) (mm

4)

timber (1) 777,600,000 54.71 215,476,817 993,076,817

transformed steel plate (2) 360,000 –131.29 517,144,360 517,504,360

Moment of inertia about the z axis =

1,510,581,176 mm4

= 1.5106 ×109 mm

4




Bending moment in beam for w = 12 kN/m The bending moment in the simply supported beam with a uniformly distributed load of 12 kN/m is:

2 2

6

max

(12 kN/m)(5 m)37.5 kN-m 37.5 10 N-mm

8 8

wLM

Bending stress in timber (1) From the flexure formula, the maximum bending stress in timber (1) is:

6

1 9 4

(37.5 10 N-mm)(372 mm 137.294 mm)

1.5106 10 mm5.83 MPa (C)

My

I Ans.

Bending stress in steel plate (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the

maximum bending stress in steel plate (2) is:

6

2 9 4

(37.5 10 N-mm)( 137.294 mm)(16.6667)

1.5106 10 m56.8 MPa (

mT)

My

I Ans.

Determine maximum w

If the allowable bending stress in the timber is 9 MPa, then the maximum bending moment that may be


2 9 4

611 max

(9 N/mm )(1.5106 10 mm )57.925 10 N-mm

(372 mm 137.294 mm)

IMyM

I y

If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may be


2 9 4

622 max

(165 N/mm )(1.5106 10 mm )108.926 10 N-mm

(16.6667)(137.294 mm)

IMyn M

I ny



From these two results, the maximum moment that the beam can support is 57.925×106 N-mm. The

maximum distributed load magnitude w that can be supported is found from:

2

max

6

max

2 2

8

8 8(57.925 10 N-mm)(1 m/1000 mm)18,536 N/m

(5 m18.54 kN/m

)

wLM

Mw

L Ans.




8.49 A glue-laminated timber beam is reinforced by carbon fiber reinforced plastic (CFRP) material

bonded to its bottom surface. The cross section of the composite beam is shown in Fig. P8.49b. The

elastic modulus of the wood is E = 12 GPa and the elastic modulus of the CFRP is 112 GPa. The simply

supported beam spans 6 m and carries a concentrated load P at midspan (Fig. P8.49a).

(a) Determine the maximum bending stresses produced in the timber and the CFRP if P = 4 kN.

(b) Assume that the allowable bending stresses of the timber and the CFRP are 9 MPa and 1,500 MPa,

respectively. Determine the largest acceptable magnitude for concentrated load P. (You may neglect the

weight of the beam in your calculations.)

Fig. P8.49a

Fig. P8.49b

Solution

Denoted the timber as material (1) and denote the CFRP as material (2). The modular ratio is:

2

1

112 GPa9.3333

12 GPa

En

E

Transform the CFRP into an equivalent amount of wood by multiplying its width by the modular ratio:

b2, trans = 9.3333(40 mm) = 373.33 mm. Thus, for calculation purposes, the 40 mm × 3 mm CFRP is

replaced by a wood board that is 373.33-mm wide and 3-mm thick.



yi

(from bottom) yi Ai


3)

timber (1) 90 250 22,500 128 2,880,000

transformed CFRP (2) 373.33 3 1,120 1.5 1,680

23,620

2,881,680

3

2

2,881,680 mm

23,620 mm122.00 mm

i i

i

y Ay




(mm4) (mm) (mm

4) (mm

4)

timber (1) 117,187,500 6.00 810,000 117,997,500

transformed CFRP (2) 840 –120.50 16,262,680 16,263,520

Moment of inertia about the z axis =

134,261,020 mm4

= 134.261 ×106 mm

4




Maximum bending moment in beam for P = 4 kN The maximum bending moment in the simply supported beam with a concentrated load of 4 kN at

midspan is:

6

max

(4 kN)(6 m)6 kN-m 6 10 N-mm

4 4

PLM

(a) Bending stress in timber (1) From the flexure formula, the maximum bending stress in timber (1) is:

6

1 6 4

(6 10 N-mm)(253 mm 122.00 mm)

134.261 15.85 MPa (C)

0 mm

My

I Ans.

(a) Bending stress in CFRP (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the

maximum bending stress in the CFRP is:

6

2 6 4

(6 10 N-mm)( 122.00 mm)(9.3333)

134.261 10 m50.9 MPa (

mT)

My

I Ans.

(b) Determine maximum P

If the allowable bending stress in the timber is 9 MPa, then the maximum bending moment that may be


2 6 4

611 max

(9 N/mm )(134.261 10 mm )9.224 10 N-mm

(253 mm 122.00 mm)

IMyM

I y

If the allowable bending stress in the CFRP is 1,500 MPa, then the maximum bending moment that may


2 6 4

622 max

(1,500 N/mm )(134.261 10 mm )176.867 10 N-mm

(9.3333)(122.00 mm)

IMyn M

I ny



From these two results, the maximum moment that the beam can support is 9.224×106 N-mm. The

maximum concentrated load magnitude P that can be supported is found from:

max

6

max

4

4 4(9.224 10 N-mm)(1 m/1000 mm)6,149 N

(6 m)6.15 kN

PLM

MP

L Ans.




8.50 Two steel plates, each 4 in. wide and 0.25 in.

thick, reinforce a wood beam that is 3 in. wide and

8 in. deep. The steel plates are attached to the

vertical sides of the wood beam in a position such

that the composite shape is symmetric about the z

axis, as shown in the sketch of the beam cross

section (Fig. P8.50). Determine the maximum

bending stresses produced in both the wood and the

steel if a bending moment of Mz = +50 kip-in is

applied about the z axis. Assume Ewood = 2,000 ksi

and Esteel = 30,000 ksi.

Fig. P8.50

Solution

Let the wood be denoted as material (1) and the steel plates as material (2). The modular ratio is:

2

1

30,000 ksi15

2,000 ksi

En

E

Transform the steel plates (2) into an equivalent amount of wood (1) by multiplying the plate

thicknesses by the modular ratio: b2, trans = 15(0.25 in.) = 3.75 in. (each). Thus, for calculation purposes,

each 4 in. × 0.25 in. steel plate is replaced by a wood board that is 4-in. tall and 3.75-in. wide.

Centroid location: Since the transformed section is doubly symmetric, the centroid location is found

from symmetry.

Moment of inertia about the z centroidal axis


(in.4) (in.) (in.

4) (in.

4)

wood beam (1) 128 0 0 128

two transformed steel plates (2) 40 0 0 40

Moment of inertia about the z axis = 168 in.4

Bending stress in wood beam (1) From the flexure formula, the maximum bending stress in wood beam (1) is:

1 4

(50 kip-in.)(4 in.)

11.190 ksi 1,190 ps

68 in.iz

z

M c

I Ans.

Bending stress in steel plates (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the

maximum bending stress in the steel plates (2) is:

2 4

(50 kip-in.)(2 in.)(15)

168 in.8.93 ksi 8,930 psiz

z

M cn

I Ans.




8.51 A glue-laminated timber beam is reinforced by carbon fiber reinforced plastic (CFRP) material

bonded to its bottom surface. The cross section of the composite beam is shown in Fig. P8.51b. The

elastic modulus of the wood is 1,700 ksi and the elastic modulus of the CFRP is 23,800 ksi. The simply

supported beam spans 24 ft and carries two concentrated loads P, which act at the quarter-points of the

span (Fig. P8.51a). The allowable bending stresses of the timber and the CFRP are 2,400 psi and

175,000 psi, respectively. Determine the largest acceptable magnitude for the concentrated loads P.

(You may neglect the weight of the beam in your calculations.)

Fig. P8.51a

Fig. P8.51b

Solution

Denoted the timber as material (1) and denote the CFRP as material (2). The modular ratio is:

2

1

23,800 ksi14

1,700 ksi

En

E

Transform the CFRP into an equivalent amount of wood by multiplying its width by the modular ratio:

b2, trans = 14(3 in.) = 42 in. Thus, for calculation purposes, the 3 in. × 0.125 in. CFRP is replaced by a

wood board that is 42-in. wide and 0.125-in. thick.



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

timber (1) 5.5 12 66 6.125 404.25

transformed CFRP (2) 42.0 0.125 5.25 0.0625 0.3281

71.25

404.5781

3

2

404.5781 in.

71.25 in.5.6783 in.

i i

i

y Ay




(in.4) (in.) (in.

4) (in.

4)

timber (1) 792 0.4467 13.1703 805.170

transformed CFRP (2) 0.00684 –5.6158 165.5697 165.577





Determine maximum P

If the allowable bending stress in the timber is 2,400 psi, then the maximum bending moment that may


4

11 max

(2.40 ksi)(970.747 in. )361.393 kip-in.

(12.125 in. 5.6783 in.)

IMyM

I y

If the allowable bending stress in the CFRP is 175,000 psi, then the maximum bending moment that may


4

22 max

(175 ksi)(970.747 in. )2,137 kip-in.

(14)(5.6783 in.)

IMyn M

I ny



From these two results, the maximum moment that the beam can support is 351.393 kip-in. = 30.116

kip-ft. The maximum concentrated load magnitude P that can be supported is found from:

max

max

(6 ft)

30.116 kip-ft

6 ft 6 ft5.02 kips

M P

MP Ans.




8.52 A steel pipe assembly supports a

concentrated load of P = 22 kN as shown in

Fig. P8.52. The outside diameter of the pipe is

142 mm and the wall thickness is 6.5 mm.

Determine the normal stresses produced at

points H and K.

Fig. P8.52

Solution

Section properties

2 2 2 2 2

4 4 4 4 4

2 142 mm 2(6.5 mm) 129 mm

(142 mm) (129 mm) 2,766.958 mm4 4

(142 mm) (129 mm) 6,364,867 mm64 64

z

d D t

A D d

I D d

Internal forces and moments

22 kN 22,000 N

(22,000 N)(370 mm) 8,140,000 N-mmz

F

M

Stresses

axial 2

bending 4

22,000 N7.951 MPa (C)

2,766.958 mm

(8,140,000 N-mm)(142 mm/2)90.802 MPa

6,364,867 mm

z

z

F

A

M c

I

Normal stress at H

By inspection, the bending stress at H will be compression; therefore, the normal stress at H is:

7.951 MPa 90.802 MPa 98.753 98.8 MPa (C)MPaH Ans.

Normal stress at K

By inspection, the bending stress at K will be tension; therefore, the normal stress at K is:

7.951 MPa 90.802 MPa 82.851 82.9 MPa (T)MPaK Ans.




8.53 The screw of a clamp exerts a compressive

force of 350 lb on the wood blocks. Determine the

normal stresses produced at points H and K. The

clamp cross-sectional dimensions at the section of

interest are 1.25 in. by 0.375 in. thick.

Fig. P8.53

Solution

Section properties

2

34

(0.375 in.)(1.250 in.) 0.468750 in.

(0.375 in.)(1.250 in.)0.061035 in.

12z

A

I


350 lb

(350 lb)(3.75 in. 1.25 in./2) 1,531.25 lb-in.z

F

M

Stresses

axial 2

bending 4

350 lb746.667 psi (T)

0.468750 in.

(1,531.25 lb-in.)(1.250 in./2)15,680.0 psi

0.061035 in.

z

z

F

A

M c

I

Normal stress at H

By inspection, the bending stress at H will be tension; therefore, the normal stress at H is:

746.667 psi 15,680 psi 16,426.667 psi 16,430 psi (T)H Ans.

Normal stress at K

By inspection, the bending stress at K will be compression; therefore, the normal stress at K is:

746.667 psi 15,680 psi 14,933.333 ps 14,930 i psi (C)K Ans.




8.54 Determine the normal stresses produced at points H and K of the pier support shown in Fig. P8.54a.

Fig. P8.54a Fig. P8.54b Cross section a–a

Solution

Section properties

2

39 4

(250 mm)(500 mm) 125,000 mm

(250 mm)(500 mm)2.60417 10 mm

12z

A

I


250 kN 400 kN 650 kN

(250 kN)(3.25 m) (400 kN)(2.25 m) 87.50 kN-mz

F

M

Stresses

axial 2

2

bending 9 4

650,000 N5.20 MPa (C)

125,000 mm

(87.5 kN-m)(500 mm/2)(1,000)8.40 MPa

2.60417 10 mm

z

z

F

A

M c

I

Normal stress at H

By inspection, the bending stress at H will be tension; therefore, the normal stress at H is:

5.20 MPa 8.40 MPa 3.20 MP 3.20 MPa (T)aH Ans.

Normal stress at K

By inspection, the bending stress at K will be compression; therefore, the normal stress at K is:

5.20 MPa 8.40 MPa 13.60 MPa 13.60 MPa (C)K Ans.




8.55 A tubular steel column CD supports

horizontal cantilever arm ABC, as shown in Fig.

P8.55. Column CD has an outside diameter of

10.75 in. and a wall thickness of 0.365 in.

Determine the maximum compression stress at

the base of column CD.

Fig. P8.55

Solution

Section properties

2 2 2 2 2

4 4 4 4 4

2 10.750 in. 2(0.365 in.) 10.020 in.

(10.750 in.) (10.020 in.) 11.908 in.4 4

(10.750 in.) (10.020 in.) 160.734 in.64 64

z

d D t

A D d

I D d


700 lb 900 lb 1,600 lb

(700 lb)(13 ft) (900 lb)(23 ft) 29,800 lb-ft 357,600 lb-in.

F

M

Stresses

axial 2

bending 4

1,600 lb134.36 psi (C)

11.908 in.

(357,600 lb-in.)(10.75 in./2)11,958.27 psi

160.734 in.

F

A

M c

I

Maximum compression stress at base of column

compression 134.36 psi 11,958.27 psi 12,092.63 12.09 ksi ( psi C) Ans.




8.56 Determine the normal stresses acting at points H and K for the structure shown in Fig. P8.56a. The

cross-sectional dimensions of the vertical member are shown in Fig. P8.56b.

Fig. P8.56a

Fig. P8.56b Cross section

Solution

Section properties

2

34

(4 in.)(8 in.) 32 in.

(4 in.)(8 in.)170.6667 in.

12z

A

I


1,200 lb 2,800 lb 4,000 lb

(1,200 lb)(12 in. 8 in./2) 19,200 lb-in.z

F

M

Stresses

axial 2

bending 4

4,000 lb125 psi (C)

32 in.

(19,200 lb-in.)(8 in./2)450 psi

170.6667 in.

z

z

F

A

M c

I

Normal stress at H


125 psi 450 psi 575 p 575 psi (C)siH Ans.

Normal stress at K


125 psi 450 psi 325 ps 325 psi (T)iK Ans.




8.57 A W18 × 35 standard steel shape is

subjected to a tension force P that is applied

15 in. above the bottom surface of the wide-

flange shape as shown in Fig. P8.57. If the

tension normal stress of the upper surface of

the W-shape must be limited to 18 ksi,

determine the allowable force P that may be

applied to the member.

Fig. P8.57

Solution

Section properties (from Appendix B)

2

4

Depth 17.7 in.

10.3 in.

510 in.z

d

A

I

Stresses

axial 2

2

bending 4 4 4

10.3 in.

(15 in. 17.7 in./2)(17.7 in./2) (6.15 in.)(8.85 in.) (54.4275 in. )

510 in. 510 in. 510 in.

z

z

F P

A

M c P P P

I

Normal stress on the upper surface of the W-shape

The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses.

Since these stresses are expressed in terms of the unknown force P, the tension normal stress is given

by:

2

upper surface 2 4

2 2

2

(54.4275 in. )

10.3 in. 510 in.

(0.097087 in. 0.106721 in. )

(0.203808 in. )

P P

P

P

The normal stress on the upper surface of the W-shape must be limited to 18 ksi; therefore,

2

2

(0.203808 in. ) 18 ksi

18 ksi

0.203808 in88.3 kip

.s

P

P

Ans.




8.58 A WT305 × 41 standard steel shape is

subjected to a tension force P that is applied 250

mm above the bottom surface of the tee shape, as

shown in Fig. P8.58. If the tension normal stress

of the upper surface of the WT-shape must be

limited to 150 MPa, determine the allowable

force P that may be applied to the member.

Fig. P8.58

Solution

Section properties (from Appendix B)

2

6 4

Depth 300 mm

Centroid 88.9 mm (from flange to centroid)

5,230 mm

48.7 10 mmz

d

y

A

I

Stresses

4 2

axial 2(1.9120 10 mm )

5,230 mm

F PP

A

bending 6 4

6 4

4 2

(250 mm 88.9 mm)(300 mm 88.9 mm)

48.7 10 mm

(161.1 mm)(211.1 mm)

48.7 10 mm

(6.9832 10 mm )

z

z

M c P

I

P

P

Normal stress on the upper surface of the WT-shape

The tension normal stress on the upper surface is equal to the sum of the axial and bending stresses.

Since these stresses are expressed in terms of the unknown force P, the tension normal stress is given

by:

4 2 4 2

upper surface

4 2

(1.9120 10 mm ) (6.9832 10 mm )

(8.8953 10 mm )

P P

P

The normal stress on the upper surface of the WT-shape must be limited to 150 MPa; therefore,

4 2

2

4 2

(8.8953 10 mm ) 150 MPa

150 N/mm168,629 N

8.8953 10 mm168.6 kN

P

P

Ans.




8.59 A pin support consists of a vertical

plate 60 mm wide by 10 mm thick. The pin

carries a load of 1,200 N. Determine the

normal stresses acting at points H and K for

the structure shown in Fig. P8.59.

Fig. P8.59

Solution

Section properties

2

34

(60 mm)(10 mm) 600 mm

(60 mm)(10 mm)5,000 mm

12

A

I


1,200 N

(1,200 N)(30 mm 10 mm/2) 42,000 N-mm

F

M

Stresses

axial 2

bending 4

1,200 N2.00 MPa (T)

600 mm

(42,000 N-mm)(10 mm/2)42.00 MPa

5,000 mm

F

A

M c

I

Normal stress at H


40.0 MPa (C2.00 MPa 42.00 MPa 40.00 ) MPaH Ans.

Normal stress at K


2.00 MPa 42.00 MPa 44.00 MPa 44.0 MPa (T)K Ans.




8.60 The tee shape shown in Fig. P8.60b is used as a short post to support a load of P = 4,600 lb. The

load P is applied at a distance of 5 in. from the surface of the flange, as shown in Fig. P8.60a. Determine

the normal stresses at points H and K, which are located on section a–a.

Fig. P8.60a

Fig. P8.60b Cross-sectional dimensions

Solution

Centroid location in x direction:

Shape width b height h Area Ai

xi

(from left) xi Ai

(in.) (in.) (in.2) (in.) (in.

3)

flange 12 2 24 1 24

stem 2 10 20 7 140

44 in.2

164 in.

3

3

2

164 in.3.7273 in. (from left side to centroid)

44 in.

8.2727 in. (from right side to centroid)

i i

i

x Ax

A


Shape IC d = xi – x d²A IC + d²A

(in.4) (in.) (in.

4) (in.

4)

flange 8 −2.7273 178.5160 186.5160

stem 166.6667 3.2727 214.2113 380.8790



4,600 lb

(4,600 lb)(5 in. 3.7273 in.) 40,145.455 lb-in.z

F

M




Stresses

axial 2

,bending 4

,bending 4

4,600 lb104.545 psi

44 in.

(40,145.455 lb-in.)( 3.7273 in.)263.720 psi

567.3940 in.

(40,145.455 lb-in.)(8.2727 in.)585.329 psi

567.3940 in.

zH

z

zK

z

F

A

M x

I

M x

I

Normal stress at H

104.545 psi 263.720 psi 368.265 p 368 psi (C)siH Ans.

Normal stress at K

104.545 psi 585.329 psi 480.784 p 481 psi (si T)K Ans.




8.61 The tee shape shown in Fig. P8.61b is used as a short post to support a load of P. The load P is

applied at a distance of 5 in. from the surface of the flange, as shown in Fig. P8.61a. The tension and

compression normal stresses in the post must be limited to 1,000 psi and 800 psi, respectively.

Determine the maximum magnitude of load P that satisfies both the tension and compression stress

limits.

Fig. P8.61a

Fig. P8.61b Cross-sectional dimensions

Solution



xi

(from left) xi Ai

(in.) (in.) (in.2) (in.) (in.

3)

flange 12 2 24 1 24

stem 2 10 20 7 140

44 in.2

164 in.

3

3

2

164 in.3.7273 in. (from left side to centroid)

44 in.

8.2727 in. (from right side to centroid)

i i

i

x Ax

A



(in.4) (in.) (in.

4) (in.

4)

flange 8 −2.7273 178.5160 186.5160

stem 166.6667 3.2727 214.2113 380.8790



(5 in. 3.7273 in.) (8.7273 in.)z

F P

M P P




Stresses

2

axial 2

2

,bending 4

2

,bending 4

(0.022727 in. )44 in.

(8.7273 in.) ( 3.7273 in.)(0.057331 in. )

567.3940 in.

(8.7273 in.) (8.2727 in.)(0.127246 in. )

567.3940 in.

zH

z

zK

z

F PP

A

M x PP

I

M x PP

I

Compression stress limit (at H)

2 2 2

2

(0.022727 in. ) (0.057331 in. ) (0.080058 in. )

(0.080058 in. ) 800 psi

9,992.76 lb

H P P P

P

P

Tension stress limit (at K)

2 2 2

2

(0.022727 in. ) (0.127246 in. ) (0.104519 in. )

(0.104519 in. ) 1,000 psi

9,567.64 lb

K P P P

P

P

Maximum magnitude of load P

max 9,570 lbP Ans.




8.62 The tee shape shown in Fig. P8.62b is used as a post that supports a load of P = 25 kN. Note that

the load P is applied 400 mm from the flange of the tee shape, as shown in Fig. P8.62a. Determine the

normal stresses at points H and K.

Fig. P8.62a Fig. P8.62b Cross-sectional dimensions

Solution



xi

(from left) xi Ai


3)

stem 20 130 2,600 65 169,000

flange 120 20 2,400 140 336,000

5,000

505,000

3

2

505,000 mm101.0 mm (from left side to centroid)

5,000 mm

49.0 mm (from right side to centroid)

i i

i

x Ax

A



(mm4) (mm) (mm

4) (mm

4)

stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67

flange 80,000.00 39.0 3,650,400.00 3,730,400.00



25 kN 25,000 N

(25,000 N)(400 mm 49.0 mm) 11,225,000 N-mmz

F

M




Stresses

axial 2

,bending 4

,bending 4

25,000 N5 MPa

5,000 mm

( 11,225,000 N-mm)( 101.0 mm)105.35 MPa

10,761,666.67 mm

( 11,225,000 N-mm)(49.0 mm)51.11 MPa

10,761,666.67 mm

zH

z

zK

z

F

A

M x

I

M x

I

Normal stress at H

5 MPa 105.35 M 100.4 P MPa (Ta )H Ans.

Normal stress at K

5 MPa 51.11 M 56.1 MPa (CPa )K Ans.




8.63 The tee shape shown in Fig. P8.63b is used as a post that supports a load of P, which is applied 400

mm from the flange of the tee shape, as shown in Fig. P8.63a. The tension and compression normal

stresses in the post must be limited to 165 MPa and 80 MPa, respectively. Determine the maximum

magnitude of load P that satisfies both the tension and compression stress limits.


Solution



xi

(from left) xi Ai


3)

stem 20 130 2,600 65 169,000

flange 120 20 2,400 140 336,000

5,000

505,000

3

2


5,000 mm


i i

i

x Ax

A



(mm4) (mm) (mm

4) (mm

4)

stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67

flange 80,000.00 39.0 3,650,400.00 3,730,400.00



(400 mm 49.0 mm) (449.0 mm)z

F P

M P P




Stresses

4 2

axial 2

3 2

,bending 4

3 2

,bending 4

(2 10 mm )5,000 mm

( 449 mm) ( 101.0 mm)(4.21394 10 mm )

10,761,666.67 mm

( 449 mm) (49.0 mm)(2.04439 10 mm )

10,761,666.67 mm

zH

z

zK

z

F PP

A

M x PP

I

M x PP

I

Tension stress limit (at H)

4 2 3 2

3 2

3 2 2

(2 10 mm ) (4.21394 10 mm )

(4.01394 10 mm )

(4.01394 10 mm ) 165 N/mm

41,106.7 N

H P P

P

P

P

Compression stress limit (at K)

4 2 3 2 3 2

3 2 2

(2 10 mm ) (2.04439 10 mm ) (2.24439 10 mm )

(2.24439 10 mm ) 80 N/mm

35,644.43 N

K P P P

P

P

Maximum magnitude of load P

max 35.6 kNP Ans.




8.64 The tee shape shown in Fig. P8.64b is used as a post that supports a load of P = 25 kN, which is

applied 400 mm from the flange of the tee shape, as shown in Fig. P8.64a. Determine the magnitudes

and locations of the maximum tension and compression normal stresses within the vertical portion BC of

the post.


Solution



xi

(from left) xi Ai


3)

stem 20 130 2,600 65 169,000

flange 120 20 2,400 140 336,000

5,000

505,000

3

2


5,000 mm


i i

i

x Ax

A



(mm4) (mm) (mm

4) (mm

4)

stem 3,661,666.67 −36.0 3,369,600.00 7,031,266.67

flange 80,000.00 39.0 3,650,400.00 3,730,400.00



(25 kN)cos35 20.4788 kN 20,478.8 N (vertical component)

(25 kN)sin 35 14.3394 kN 14,339.4 N (horizontal component)

at B (20,478.8 N)(400 mm 49.0 mm) 9,194,981.2 N-mm

at C (20,478.8 N)(400 m

z

z

F

V

M

M

m 49.0 mm) (14,339.4 N)(1,200 mm) 8,012,298.8 N-mm




Normal stress at H at location B

axial 2

,bending 4

20,478.8 N4.0958 MPa

5,000 mm

( 9,194,981.2 N-mm)( 101.0 mm)86.2964 MPa

10,761,666.67 mm

4.0958 MPa 86.2964 MPa 82.2 MPa

zH

z

H

F

A

M x

I

Normal stress at H at location C

,bending 4

(8,012,298.8 N-mm)( 101.0 mm)75.1967 MPa

10,761,666.67 mm

4.0958 MPa 75.1967 MPa 79.3 MPa

zH

z

H

M x

I

Normal stress at K at location B

,bending 4

( 9,194,981.2 N-mm)(49.0 mm)41.8666 MPa

10,761,666.67 mm

4.0958 MPa 41.8666 MPa 46.0 MPa

zK

z

K

M x

I

Normal stress at K at location C

,bending 4

(8,012,298.8 N-mm)(49.0 mm)36.4816 MPa

10,761,666.67 mm

4.0958 MPa 36.4816 MPa 32.4 MPa

zK

z

K

M x

I

Maximum tension stress

max tension 82.2 MP at locaa (T) tion B Ans.

Maximum compression stress

max compression 79.3 at loMPa cat (C) ion C Ans.




8.65 A beam with a box cross section is subjected to

a resultant moment magnitude of 2,100 N-m acting

at the angle shown in Fig. P8.65. Determine:

(a) the maximum tension and the maximum

compression bending stresses in the beam.

(b) the orientation of the neutral axis relative to the

+z axis. Show its location on a sketch of the cross

section.

Fig. P8.65

Solution

Section properties

3 34

3 34

(90 mm)(55 mm) (80 mm)(45 mm)640,312.5 mm

12 12

(55 mm)(90 mm) (45 mm)(80 mm)1, 421, 250.0 mm

12 12

y

z

I

I

Moment components

(2,100 N-m)sin30 1,050 N-m

(2,100 N-m)cos30 1,818.65 N-m

y

z

M

M

(a) Maximum bending stresses

For a shape having at least one axis of symmetry, Eq. (8-24) can be used to determine bending stresses.

Compute normal stress at y = 45 mm, z = 27.5 mm:

4 4

(1,050 N-m)(27.5 mm)(1,000 mm/m) ( 1,818.65 N-m)(45 mm)(1,000 mm/m)

640,312.5 mm 1,421,250.0 mm

45.0952 MPa 57.5827 MPa

102.6 102.7 MPa (T)779 MPa

y zx

y z

M z M y

I I

Ans.

Compute normal stress at y = −45 mm, z = −27.5 mm:

4 4

(1,050 N-m)( 27.5 mm)(1,000 mm/m) ( 1,818.65 N-m)( 45 mm)(1,000 mm/m)

640,312.5 mm 1,421,250.0 mm

45.0952 MPa 57.5827 MPa

102 102.7 MPa (C.6779 MPa )

y zx

y z

M z M y

I I

Ans.

(b) Orientation of neutral axis

For a shape having at least one axis of symmetry, Eq. (8-25) can be used to determine the orientation of

the neutral axis:




4

4

(1,050 N-m)(1,421,250.0 mm )tan 1.2815

( 1,818.65 N-m)(640

52.03

,312.5 mm )

(i.e., 52.03 CCW from axis)

y z

z y

M I

M I

z Ans.




8.66 The moment acting on the cross section of the

T-beam has a magnitude of 22 kip-ft and is oriented

as shown in Fig. P8.66. Determine:

(a) the bending stress at point H.

(b) the bending stress at point K.

(c) the orientation of the neutral axis relative to the


section.

Fig. P8.66

Solution

Section properties



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

top flange 7.00 1.25 8.7500 8.375 73.28125

stem 0.75 7.75 5.8125 3.875 22.52344

14.5625

95.80469

3

2


14.5625 in.


i i

i

y Ay

A



(in.4) (in.) (in.

4) (in.

4)

top flange 1.1393 1.7961 28.2273 29.3666

stem 29.0928 −2.7039 42.4956 71.5884


Moment of inertia about the y axis:

3 3

4(1.25 in.)(7.00 in.) (7.75 in.)(0.75 in.)36.0016 in.

12 12yI

Moment components

(22 kip-ft)cos55 12.6187 kip-ft 151.4242 kip-in.

(22 kip-ft)sin55 18.0213 kip-ft 216.2561 kip-in.

y

z

M

M




(a) Bending stress at H


To compute the normal stress at H, use the (y, z) coordinates y = 2.4211 in. and z = −3.5 in.:

4 4

( 151.4242 kip-in.)( 3.50 in.) ( 216.2561 kip-in.)(2.4211 in.)

36.0016 in. 100.9550 in.

14.7211 ksi 5.1862 ksi

19.9 19.91 ksi (T)074 ksi

y zx

y z

M z M y

I I

Ans.

(b) Bending stress at K

To compute the normal stress at K, use the (y, z) coordinates y = −6.5789 in. and z = 0.375 in.:

4 4

( 151.4242 kip-in.)(0.375 in.) ( 216.2561 kip-in.)( 6.5789 in.)

36.0016 in. 1

15.67 ksi (C)

00.9550 in.

1.5773 ksi 14.0927 ksi

15.6700 ksi

y zx

y z

M z M y

I I

Ans.

(c) Orientation of neutral axis


the neutral axis:

4

4

( 151.4242 kip-in.)(100.9550 in. )tan 1.9635

( 216.2561 kip-i

63.0

n.)(36.0016 in. )

(i.e., 63.01 CW from ax1 is)

y z

z y

M I

M I

z Ans.




8.67 A beam with a box cross section is subjected to

a resultant moment magnitude of 75 kip-in. acting

at the angle shown in Fig. P8.67. Determine:



(c) the maximum tension and the maximum

compression bending stresses in the beam.

(d) the orientation of the neutral axis relative to the


section.

Fig. P8.67

Solution

Section properties

3 34

3 34

(4 in.)(6 in.) (3.25 in.)(5.25 in.)32.8096 in.

12 12

(6 in.)(4 in.) (5.25 in.)(3.25 in.)16.9814 in.

12 12

y

z

I

I

Moment components

(75 kip-in.)cos 20 70.4769 kip-in.

(75 kip-in.)sin 20 25.6515 kip-in.

y

z

M

M



To compute the normal stress at H, use the (y, z) coordinates y = −2.0 in. and z = −3.0 in.:

4 4

(70.4769 kip-in.)( 3.0 in.) (25.6515 kip-in.)( 2.0 in.)

32.8096 in. 1

3.42 ksi (C)

6.9814 in.

6.4442 ksi 3.0211 ksi

3.4231 ksi

y zx

y z

M z M y

I I

Ans.


To compute the normal stress at K, use the (y, z) coordinates y = 2.0 in. and z = 3.0 in.:

4 4

(70.4769 kip-in.)(3.0 in.) (25.6515 kip-in.)(2.0 in.)

32.8096 in. 16.9814 in.

6.4442 ksi 3.0211 ksi

3.4231 k 3.42 s ksi (T)i

y zx

y z

M z M y

I I

Ans.




(c) Maximum bending stresses

The maximum tension normal stress occurs at the (y, z) coordinates y = −2.0 in. and z = 3.0 in.:

4 4

(70.4769 kip-in.)(3.0 in.) (25.6515 kip-in.)( 2.0 in.)

32.8096 in. 16.9814 in.

6.4442 ksi 3.0211 ksi

9.46 9.47 ksi (T53 ksi )

y zx

y z

M z M y

I I

Ans.

The maximum compression normal stress occurs at the (y, z) coordinates y = 2.0 in. and z = −3.0 in.:

4 4

(70.4769 kip-in.)( 3.0 in.) (25.6515 kip-in.)(2.0 in.)

32.8096 in. 16.9814 in.

6.4442 ksi 3.0211 ksi

9.4653 k 9.47 ksi (i C)s

y zx

y z

M z M y

I I

Ans.

(d) Orientation of neutral axis


the neutral axis:

4

4

(70.4769 kip-in.)(16.9814 in. )tan 1.4220

(25.6515 kip-in.)

54.88

(32.8096 in. )

(i.e., 54.88 CW from axis)

y z

z y

M I

M I

z Ans.





wide-flange beam has a magnitude of M = 12 kN-m and

is oriented as shown in Fig. P8.68. Determine:



(c) the orientation of the neutral axis relative to the +z

axis. Show its location on a sketch of the cross section.

Fig. P8.68

Solution

Section properties



(mm4) (mm) (mm

4) (mm

4)

top flange 59,062.5 97.5 29,944,687.5 30,003,750

web 4,860,000 0 0 4,860,000

bottom flange 59,062.5 −97.5 29,944,687.5 30,003,750



3 3

4(15 mm)(210 mm) (180 mm)(10 mm)2 23,167,500 mm

12 12yI

Moment components

6

6

(12 kN-m)sin 35 6.8829 kN-m 6.8829 10 N-mm

(12 kN-m)cos35 9.8298 kN-m 9.8298 10 N-mm

y

z

M

M



To compute the normal stress at H, use the (y, z) coordinates y = 105 mm and z = −105 mm:

6 6

4 4

(6.8829 10 N-mm)( 105 mm) (9.8298 10 N-mm)(105 mm)

23,167,500 mm 64,

47.1 MPa (C)

867,500 mm

31.1948 MPa 15.9114 MPa

47.1062 MPa

y zx

y z

M z M y

I I

Ans.





To compute the normal stress at K, use the (y, z) coordinates y = −105 mm and z = 105 mm:

6 6

4 4

(6.8829 10 N-mm)(105 mm) (9.8298 10 N-mm)( 105 mm)

23,167,500 mm 64,867,500 mm

31.1948 MPa 15.9114 MPa

47. 47.1 1062 M MPa (P T)a

y zx

y z

M z M y

I I

Ans.

(b) Orientation of neutral axis


the neutral axis:

4

4

(6.8829 kN-m)(64,867,500 mm )tan 1.9605

(9.8298 kN-m)(23,

62.

167,500 mm )

(i.e., 62.98 CW from axis8 )9

y z

z y

M I

M I

z Ans.




8.69 For the cross section shown in Fig. P8.69,

determine the maximum magnitude of the bending

moment M so that the bending stress in the wide-

flange shape does not exceed 165 MPa.

Fig. P8.69

Solution

Section properties



(mm4) (mm) (mm

4) (mm

4)

top flange 59,062.5 97.5 29,944,687.5 30,003,750

web 4,860,000 0 0 4,860,000

bottom flange 59,062.5 −97.5 29,944,687.5 30,003,750



3 3

4(15 mm)(210 mm) (180 mm)(10 mm)2 23,167,500 mm

12 12yI

Moment components

sin 35 cos35y zM M M M


The maximum tension bending stress should occur at point K, which has the (y, z) coordinates y = −105

mm and z = 105 mm:

4 4

sin35 (105 mm) cos35 ( 105 mm)165 MPa

23,167,500 mm 64,867,500 mm

y zx

y z

M z M y M M

I I

6 3 6 3 2

6

2.59957 10 mm 1.32595 10 mm 165 N/mm

42.0 kN-42.0327 10 N-mm m

M

M Ans.




8.70 The unequal-leg angle is subjected to a bending

moment of Mz = 20 kip-in. that acts at the orientation

shown in Fig. P8.70. Determine:




compression bending stresses in the cross section.

(d) the orientation of the neutral axis relative to the +z

axis. Show its location on a sketch of the cross section.

Fig. P8.70

Solution

Section properties



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

upright leg 0.375 4.000 1.5000 2.00 3.00

bottom leg 2.625 0.375 0.9844 0.1875 0.18457

2.4844

3.18457

3

2

3.18457 in.1.2818 in.(from bottom of shape to centroid)

2.4844 in.


i i

i

y Ay

A

Centroid location in z direction:

Shape Area Ai

zi

(from right edge) zi Ai

(in.2) (in.) (in.

3)

upright leg 1.5000 0.1875 0.2813

bottom leg 0.9844 1.6875 1.6612

2.4844

1.94243

3

2

1.94243 in.0.7818 in. (from right edge of shape to centroid)

2.4844 in.

2.2182 in. (from left edge of shape to centroid)

i i

i

z Az

A



(in.4) (in.) (in.

4) (in.

4)

upright leg 2.000 0.7182 0.77372 2.7737

bottom leg 0.011536 −1.0943 1.17881 1.1903






Shape IC d = zi – z d²A IC + d²A

(in.4) (in.) (in.

4) (in.

4)

upright leg 0.017578 −0.5943 0.52979 0.5474

bottom leg 0.565247 0.9057 0.80750 1.3727

Moment of inertia about the y axis (in.4) = 1.9201

Product of inertia about the centroidal axes:

Shape Iy’z’ yc zc Area Ai yc zc Ai Iyz

(in.4) (in.) (in.) (in.

2) (in.

4) (in.

4)

upright leg 0 0.7182 −0.5943 1.5000 −0.6402 −0.6402

bottom leg 0 −1.0943 0.9057 0.9844 −0.9757 −0.9757

Product of inertia (in.4) = −1.6159


Since the angle shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the

bending stresses. Equation (8.22) will be used here. Note that the bending moment component about

the y axis is zero (i.e., My = 0); therefore, the first term in Eq. (8.22) is eliminated. To compute the

normal stress at H, use (y, z) coordinates of y = 2.7182 in. and z = −0.4068 in.:

2

4 4

4 4 4 2

5

8

(1.9201 in. )(2.7182 in.) ( 1.6159 in. )( 0.4068 in.)(20 kip-in.)

(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )

4.5619 in.(20 kip-in.)

5.0001 in.

18.2469 ks

y yz

x z

y z yz

I y I zM

I I I

18.25 ksii (C) Ans.


To compute the normal stress at K, use (y, z) coordinates of y = −0.9068 in. and z = 2.2182 in.:

2

4 4

4 4 4 2

5

8

(1.9201 in. )( 0.9068 in.) ( 1.6159 in. )(2.2182 in.)(20 kip-in.)

(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )

1.8432 in.(20 kip-in.)

5.0001 in.

7.3728 ksi

y yz

x z

y z yz

I y I zM

I I I

7.37 ksi (C) Ans.





Since the angle shape has no axis of symmetry, Eq. (8.23) must be used to determine the orientation of

the neutral axis:

4

4

(20 kip-in.)( 1.6159 in. )tan 0.8416

(20 kip-in.)(1.9201 in. )

(i.e., 40.08 CCW from axi0 s8 )40.

y z z yz

z y y yz

M I M I

M I M I

z Ans.

(c) Maximum bending stresses

Sketch the orientation of the neutral axis. By inspection, the points on the angle cross section that are

farthest from the neutral axis are point H and the corner of the angle. The bending stress at H has

already been computed. To compute the normal stress at the corner of the angle, use (y, z) coordinates

of y = −1.2818 in. and z = −0.7818 in.

2

4 4

4 4 4 2

5

8

(1.9201 in. )( 1.2818 in.) ( 1.6159 in. )( 0.7818 in.)(20 kip-in.)

(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )

3.7245 in.(20 kip-in.)

5.0001 in.

14.8977 ksi

y yz

x z

y z yz

I y I zM

I I I

14.90 ksi (T)

Therefore, the maximum compression bending stress is:

18.25 ksi (C)x Ans.

and the maximum tension bending stress is:

14.90 ksi (T)x Ans.






moment M so that the bending stress in the unequal-

leg angle shape does not exceed 24 ksi.

Fig. P8.71

Solution

Section properties



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

upright leg 0.375 4.000 1.5000 2.00 3.00

bottom leg 2.625 0.375 0.9844 0.1875 0.18457

2.4844

3.18457

3

2

3.18457 in.1.2818 in.(from bottom of shape to centroid)

2.4844 in.


i i

i

y Ay

A


Shape Area Ai

zi


(in.2) (in.) (in.

3)

upright leg 1.5000 0.1875 0.2813

bottom leg 0.9844 1.6875 1.6612

2.4844

1.94243

3

2

1.94243 in.0.7818 in. (from right edge of shape to centroid)

2.4844 in.

2.2182 in. (from left edge of shape to centroid)

i i

i

z Az

A



(in.4) (in.) (in.

4) (in.

4)

upright leg 2.000 0.7182 0.77372 2.7737

bottom leg 0.011536 −1.0943 1.17881 1.1903







(in.4) (in.) (in.

4) (in.

4)

upright leg 0.017578 −0.5943 0.52979 0.5474

bottom leg 0.565247 0.9057 0.80750 1.3727




(in.4) (in.) (in.) (in.

2) (in.

4) (in.

4)

upright leg 0 0.7182 −0.5943 1.5000 −0.6402 −0.6402

bottom leg 0 −1.0943 0.9057 0.9844 −0.9757 −0.9757

Product of inertia (in.4) = −1.6159

Orientation of neutral axis

Since the angle shape has no axis of symmetry, it is helpful to determine the orientation of the neutral

axis from Eq. (8.23) before beginning the stress calculations:

4

4

(20 kip-in.)( 1.6159 in. )tan 0.8416

(20 kip-in.)(1.9201 in. )

(i.e., 40.08 CCW from axi0 s8 )40.

y z z yz

z y y yz

M I M I

M I M I

z

Allowable moments based on maximum tension and compression bending stresses


farthest from the neutral axis are point H and the corner of the angle. To compute the normal stress at

H, use (y, z) coordinates of y = 2.7182 in. and z = −0.4068 in.:

4 4

2 4 4 4 2

53

8

(1.9201 in. )(2.7182 in.) ( 1.6159 in. )( 0.4068 in.)

(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )

4.5619 in.( 0.9124 in. )

5.0001 in.

y yz

x z z

y z yz

z z

I y I zM M

I I I

M M

Therefore, based on an allowable bending stress of 24 ksi at H, the maximum magnitude of Mz is:




3(0.9124 in. ) 24 ksi

26.3054 kip-in.

z

z

M

M (a)

To compute the normal stress at the corner of the angle, use (y, z) coordinates of y = −1.2818 in. and z =

−0.7818 in.

4 4

2 4 4 4 2

53

8

(1.9201 in. )( 1.2818 in.) ( 1.6159 in. )( 0.7818 in.)

(1.9201 in. )(3.9640 in. ) ( 1.6159 in. )

3.7245 in.(0.7449 in. )

5.0001 in.

y yz

x z z

y z yz

z z

I y I zM M

I I I

M M

Therefore, based on the bending stress at the corner of the angle, the maximum magnitude of Mz is:

3(0.7449 in. ) 24 ksi

32.2197 kip-in.

z

z

M

M (b)

Maximum bending moment Mz

Compare the results in Eqs. (a) and (b) to find that the maximum bending moment that can be applied to

the angle shape is:

26.3 kip-in.zM Ans.




8.72 The moment acting on the cross section of

the unequal-leg angle has a magnitude of M = 20

kip-in. and is oriented as shown in Fig. P8.72.

Determine:





(d) the orientation of the neutral axis relative to

the +z axis. Show its location on a sketch of the

cross section.

Fig. P8.72

Solution


Shape IC d = yi – y Area Ai d²A IC + d²A

(mm4) (mm) (mm

2) (mm

4) (mm

4)

top flange 130,208.3 112.5 2,500 31,640,625.0 31,770,883.3

web 10,666,666.7 0 3,200 0 10,666,666.7

bottom flange 130,208.3 −112.5 2,500 31,640,625.0 31,770,883.3



Shape IC d = zi – z Area Ai d²A IC + d²A

(mm4) (mm) (mm

2) (mm

4) (mm

4)

top flange 2,083,333.3 −42.0 2,500 4,410,000 6,493,333.3

web 68,266.7 0 3,200 0 68,266.7

bottom flange 2,083,333.3 42.0 2,500 4,410,000 6,493,333.3

Moment of inertia about the y axis (mm4) = 13,054,933.3


Shape yc zc Area Ai yc zc Ai Iyz

(mm) (mm) (mm2) (mm

4) (mm

4)

top flange 112.5 −42.0 2,500 −11,812,500 −11,812,500

web 0 0 3,200 0 0

bottom flange −112.5 42.0 2,500 −11,812,500 −11,812,500

Product of inertia (mm4) = −23,625,000

Moment components

6

6

(40 kN-m)sin15 10.3528 kN-m 10.3528 10 N-mm

(40 kN-m)cos15 38.6370 kN-m 38.6370 10 N-mm

y

z

M

M





Since the zee shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the

bending stresses. Equation (8.21) will be used here.

2 2

6 4 6 4

4 4 4 2

6

( 38.6370 10 N-mm)(13,054,933.3 mm ) ( 10.3528 10 N-mm)( 23,625,000 mm )

(13,054,933.3 mm )(74,208,333.3 mm ) ( 23,625,000 mm )

( 10.3528 10 N-

z y y yz y z z yz

x

y z yz y z yz

M I M I y M I M I z

I I I I I I

y

4 6 4

4 4 4 2

3 3

mm)(74,208,333.3 mm ) ( 38.6370 10 N-mm)( 23,625,000 mm )

(13,054,933.3 mm )(74,208,333.3 mm ) ( 23,625,000 mm )

(0.63271 N/mm ) (0.35197 N/mm )

z

y z

To compute the normal stress at H, use (y, z) coordinates of y = 125 mm and z = −92 mm:

3 3(0.63271 N/mm )(125 mm) (0.35197 N/mm )( 92 mm)

46.7073 MPa 46.7 MPa (T)

x

Ans.


To compute the normal stress at K, use (y, z) coordinates of y = −125 mm and z = 92 mm:

3 3(0.63271 N/mm )( 125 mm) (0.35197 N/mm )(92 mm)

46.7073 MPa 46.7 MPa (C)

x

Ans.


Since the zee shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis

from Eq. (8.23) to help identify points of maximum stress.

4 4

4 4

tan

( 10.3528 kN-m)(74,208,333.3 mm ) ( 38.6370 kN-m)( 23,625,000 mm )

( 38.6370 kN-m)(13,054,933.3 mm ) ( 10.3528 kN-m)( 23,625,000 mm )

0.55629

(i.e., 29.09 CCW f29.09 rom

y z z yz

z y y yz

M I M I

M I M I

axis)z




(c) Maximum tension and compression bending stresses

Sketch the orientation of the neutral axis. By inspection, the points on the zee cross section that are

farthest from the neutral axis are on the top and bottom surfaces at the outside corners of the web. To

compute bending stresses at the upper point, use (y, z) coordinates of y = 125 mm and z = 8 mm:

3 3(0.63271 N/mm )(125 mm) (0.35197 N/mm )(8 mm)

81.9045 MPa Maximum tensi81.9 MP on benda (T) ing stress

x

Ans.

To compute bending stresses at the lower point, use (y, z) coordinates of y = −125 mm and z = −8 mm:

3 3(0.63271 N/mm )( 125 mm) (0.35197 N/mm )( 8 mm)

81.9045 MPa Maximum compre81.9 ssionMPa (C bending stre s) s

x

Ans.





unequal-leg angle has a magnitude of 14 kN-m and

is oriented as shown in Fig. P8.73. Determine:





(d) the orientation of the neutral axis relative to the


section.

Fig. P8.73

Solution

Section properties



yi

(from bottom) yi Ai


3)

horizontal leg 150 19 2,850 190.50 542,925.0

vertical leg 19 181 3,439 90.50 311,229.5

6,289

854,154.5

3

2

854,154.5 mm135.82 mm (from bottom of shape to centroid)

6,289 mm

64.18 mm (from top of shape to centroid)

i i

i

y Ay

A


Shape Area Ai

zi


(mm2) (mm) (mm

3)

horizontal leg 2,850 75.0 213,750.0

vertical leg 3,439 9.5 32,670.5

6,289

246,420.5

3

2

246,420.5 mm39.18 mm (from right edge of shape to centroid)

6,289 mm

110.82 mm (from left edge of shape to centroid)

i i

i

z Az

A



(mm4) (mm) (mm

4) (mm

4)

horizontal leg 85,737.50 54.68 8,522,088.15 8,607,825.65

vertical leg 9,388,756.58 −45.32 7,062,503.99 16,451,260.58







(mm4) (mm) (mm

4) (mm

4)

horizontal leg 5,343,750.00 35.82 3,656,188.87 8,999,938.87

vertical leg 103,456.58 −29.68 3,029,990.78 3,133,447.36

Moment of inertia about the y axis (mm4) = 12,133,386.23



(mm4) (mm) (mm) (mm

2) (mm

4) (mm

4)

horizontal leg 0 54.68 35.82 2,850 5,582,117.16 5,582,117.16

vertical leg 0 −45.32 −29.68 3,439 4,625,790.65 4,625,790.65

Product of inertia (mm4) = 10,207,907.81

Since the angle shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the


2 2

6 4

4 4 4 2

6 4

4

(14 10 N-mm)(12,133,386.23 mm )

(12,133,386.23 mm )(25,059,086.23 mm ) (10,207,907.81 mm )

(14 10 N-mm)(10,207,907.81 mm )

(12,133,386.23 mm )(25

z y y yz y z z yz

x

y z yz y z yz

M I M I y M I M I z

I I I I I I

y

4 4 2

3 3

,059,086.23 mm ) (10,207,907.81 mm )

( 0.84997 N/mm ) (0.71509 N/mm )

z

y z

To compute the normal stress at H, use (y, z) coordinates of y = 45.18 mm and z = 110.82 mm:

3 3( 0.84997 N/mm )(45.18 mm) (0.71509 N/mm )(110.82 mm)

40.8444 MPa 40.8 MPa (T)

x

Ans.


To compute the normal stress at K, use (y, z) coordinates of y = 64.18 mm and z = −39.18 mm:

3 3( 0.84997 N/mm )(64.18 mm) (0.71509 N/mm )( 39.18 mm)

82.5685 MP 82.6 MPa (a C)

x

Ans.



axis from Eq. (8.23) to help identify points of maximum stress.

4

4

tan

(14 kN-m)(10,207,907.81 mm )

(14 kN-m)(12,133,386.23 mm )

0.84

40.0

131

(i.e., 40.07 CW from axis)7

y z z yz

z y y yz

M I M I

M I M I

z






farthest from the neutral axis are on the top corner (at K) and on the inside corner of the vertical leg.

To compute bending stresses at the lower point, use (y, z) coordinates of y = −135.82 mm and z = −20.18

mm:

3 3( 0.84997 N/mm )( 135.82 mm) (0.71509 N/mm )( 20.18 mm)

101.0 MPa (T101.0129 MPa Maximum tension bendi) ng stress

x

Ans.

The maximum compression bending stress is

3 3( 0.84997 N/mm )(64.18 mm) (0.71509 N/mm )( 39.18 mm)

82.5685 MPa Maximum compre82.6 ssionMPa (C) bending stress

x

Ans.




8.74 The moment acting on the cross section of

the zee shape has a magnitude of M = 4.75 kip-ft

and is oriented as shown in Fig. P8.74. Determine:





(d) the orientation of the neutral axis relative to

the +z axis. Show its location on a sketch of the

cross section.

Fig. P8.74

Solution



(in.4) (in.) (in.

2) (in.

4) (in.

4)

top flange 0.0260 2.75 1.25 9.4531 9.4792

web 3.6458 0 1.75 0 3.6458

bottom flange 0.0260 −2.75 1.25 9.4531 9.4792




(in.4) (in.) (in.

2) (in.

4) (in.

4)

top flange 0.6510 1.075 1.25 1.4445 2.0956

web 68,266.7 0 1.75 0 0.0179

bottom flange 0.6510 −1.075 1.25 1.4445 2.0956




(in.) (in.) (in.2) (in.

4) (in.

4)

top flange 2.75 1.075 1.25 3.6953 3.6953

web 0 0 1.75 0 0

bottom flange −2.75 −1.075 1.25 3.6953 3.6953

Product of inertia (in.4) = 7.3906







2 2

4 4

4 4 4 2 4 4

( 4.75 kip-ft)(12 in./ft)(4.2091 in. ) ( 4.75 kip-ft)(12 in./ft)(7.3906 in. )

(4.2091 in. )(22.6042 in. ) (7.3906 in. ) (4.2091 in. )(22.6042 in. )

z y y yz y z z yz

x

y z yz y z yz

M I M I y M I M I z

I I I I I I

y4 2

3 3

(7.3906 in. )

(5.92065 kips/in. ) (10.39584 kips/in. )

z

y z

To compute the normal stress at H, use (y, z) coordinates of y = 3 in. and z = 2.325 in.:

3 3(5.92065 kips/in. )(3 in.) (10.39584 kips/in. )(2.325 in.)

6.4084 ksi 6.41 ksi (C)

x

Ans.


To compute the normal stress at K, use (y, z) coordinates of y = −2.50 in. and z = −2.325 in.:

3 3(5.92065 kips/in. )( 2.50 in.) (10.39584 kips/in. )( 2.325 in.)

9.3687 k 9.37 ksi (T)si

x

Ans.


Since the zee shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis

from Eq. (8.23) to help identify points of maximum stress.

4

4

tan

( 4.75 kip-ft)(7.3906 in. )

( 4.75 kip-ft)(4.2091 in. )

1.7

60.34

559

(i.e., 60.34 CW from axis)

y z z yz

z y y yz

M I M I

M I M I

z







compute bending stresses at the upper point, use (y, z) coordinates of y = 3 in. and z = −0.175 in.:

3 3(5.92065 kips/in. )(3 in.) (10.39584 kips/in. )( 0.175 in

19.58 ksi (T

.)

19.5812 ksi Maximum tension bending) stress

x

Ans.

To compute bending stresses at the lower point, use (y, z) coordinates of y = −3 in. and z = 0.175 in.:

3 3(5.92065 kips/in. )( 3 in.) (10.39584 kips/in. )(0.175 in.)

19.5812 ksi Maximum compre19.58 ssion bending strksi e(C) ss

x

Ans.






moment M so that the bending stress in the zee

shape does not exceed 24 ksi.

Fig. P8.75

Solution



(in.4) (in.) (in.

2) (in.

4) (in.

4)

top flange 0.0260 2.75 1.25 9.4531 9.4792

web 3.6458 0 1.75 0 3.6458

bottom flange 0.0260 −2.75 1.25 9.4531 9.4792




(in.4) (in.) (in.

2) (in.

4) (in.

4)

top flange 0.6510 1.075 1.25 1.4445 2.0956

web 68,266.7 0 1.75 0 0.0179

bottom flange 0.6510 −1.075 1.25 1.4445 2.0956




(in.) (in.) (in.2) (in.

4) (in.

4)

top flange 2.75 1.075 1.25 3.6953 3.6953

web 0 0 1.75 0 0

bottom flange −2.75 −1.075 1.25 3.6953 3.6953

Product of inertia (in.4) = 7.3906

Bending stresses in the section


bending stresses. Equation (8.21) will be used here. For this problem, My = 0 and from the sketch, Mz is

observed to be negative. The bending stress in the zee cross section is described by:




2 2

4 4

4 4 4 2 4 4 4 2

4

(4.2091 in. ) (7.3906 in. )

(4.2091 in. )(22.6042 in. ) (7.3906 in. ) (4.2091 in. )(22.6042 in. ) (7.3906 in. )

(0.103871 in. ) (

z y y yz y z z yz

x

y z yz y z yz

z z

z

M I M I y M I M I z

I I I I I I

M My z

M y 4

4 4

0.182383 in. )

(0.103871 in. ) (0.182383 in. )

z

z

M z

M y z

Orientation of neutral axis


axis from Eq. (8.23) before beginning the stress calculations:

4

4

(7.3906 in. )tan 1.7559

(4.2091 i

60.

n. )

(i.e., 60.34 CW from axi3 s)4

y z z yz z

z y y yz z

M I M I M

M I M I M

z

Allowable moments based on maximum tension and compression bending stresses



compute bending stresses at the upper point, coordinates of y = 3 in. and z = −0.175 in. are used. Set the

bending stress at this point to the 24-ksi allowable bending stress and solve for the moment magnitude:

4 4

3

(0.103871 in. )(3 in.) (0.182383 in. )( 0.175 in.) 24 ksi

24 ksi69.86287 kip-in.

05.82 kip-ft

.343530 in.

x z

z

M

M Ans.




8.76 A stainless-steel spring (shown in Fig.

P8.76) has a thickness of ¾ in. and a change in

depth at section B from D = 1.50 in. to d = 1.25

in. The radius of the fillet between the two

sections is r = 0.125 in. If the bending moment

applied to the spring is M = 2,000 lb-in.,

determine the maximum normal stress in the

spring.

Fig. P8.76

Solution

From Figure 8.18

0.125 in. 1.50 in.

0.10 1.20 1.691.25 in. 1.25 in.

r DK

d d

Moment of inertia at minimum depth section:

3

4(0.75 in.)(1.25 in.)0.122070 in.

12I

Nominal bending stress at minimum depth section:

nom 4

(2,000 lb-in.)(1.25 in./2)10.2400 ksi

0.122070 in.

My

I

Maximum bending stress:

max nom 1.69(10.2400 ksi) 17.3056 ksi 17.31 ksiK Ans.




8.77 An alloy-steel spring (shown in Fig. P8.77)

has a thickness of 25 mm and a change in depth at

section B from D = 75 mm to d = 50 mm. If the

radius of the fillet between the two sections is r =

8 mm, determine the maximum moment that the

spring can resist if the maximum bending stress in

the spring must not exceed 120 MPa.

Fig. P8.77

Solution

From Figure 8.18

8 mm 75 mm

0.16 1.50 1.5750 mm 50 mm

r DK

d d

Determine maximum nominal bending stress:

maxnom

120 MPa76.4331 MPa

1.57K


3

4(25 mm)(50 mm)260,416.67 mm

12I


2 4

nommax

(76.4331 N/mm )(260,416.67 mm )796,178.3 N-mm

50 mm/2796 N-m

IM

y Ans.




8.78 The notched bar shown in Fig. P8.78 is

subjected to a bending moment of M = 300 N-m.

The major bar width is D = 75 mm, the minor bar

width at the notches is d = 50 mm, and the radius

of each notch is r = 10 mm. If the maximum

bending stress in the bar must not exceed 90 MPa,

determine the minimum required bar thickness b.

Fig. P8.78

Solution

From Figure 8.17

10 mm 75 mm

0.20 1.50 1.7650 mm 50 mm

r DK

d d

Determine maximum nominal bending stress:

maxnom

90 MPa51.1364 MPa

1.76K

Minimum bar thickness b:

nom 3 2

2 2 2

nom

( /2) 6

/12

6 6(300 N-m)(1,000 mm/m)

(51.1364 N/mm1

)(50 mm)4.08 mm

M y M d M

I bd bd

Mb

d Ans.




8.79 The machine part shown in Fig. P8.79 is made of

cold-rolled 18-8 stainless steel (see Appendix D for

properties). The major bar width is D = 1.50 in., the

minor bar width at the notches is d = 1.00 in., the radius

of each notch is r = 0.125 in., and the bar thickness is b

= 0.25 in. Determine the maximum safe moment M

that may be applied to the bar if a factor of safety of 2.5

with respect to failure by yield is specified.

Fig. P8.79

Solution

From Figure 8.17

0.125 in. 1.50 in.

0.125 1.50 2.051.00 in. 1.00 in.

r DK

d d


3

4(0.25 in.)(1.00 in.)0.020833 in.

12I


From the specified factor of safety and the yield stress of the material, the allowable bending stress is:

allow

165 ksi66 ksi

FS 2.5

Y

Thus, the maximum allowable bending moment can be determined from:

allow

4

allowmax

(66 ksi)(0.020833 in. )1.3415 kip-in.

(2.05)(1.00 in./11

2)1.8 lb-ft

MyK

I

IM

K y Ans.




8.80 The shaft shown in Fig. P8.80 is supported

at each end by self-aligning bearings. The major

shaft diameter is D = 2.00 in., the minor shaft

diameter is d = 1.50 in., and the radius of the

fillet between the major and minor diameter

sections is r = 0.125 in. The shaft length is L =

24 in. and the fillets are located at x = 8 in. and x

= 16 in. Determine the maximum load P that

may be applied to the shaft if the maximum

normal stress must be limited to 24,000 psi.

Fig. P8.80

Solution

From Figure 8.20

0.125 in. 2.00 in.

0.083 1.33 1.781.50 in. 1.50 in.

r DK

d d

Moment of inertia at minimum diameter section:

4 4(1.50 in.) 0.248505 in.64

I


allow

4

allowmax

(24,000 psi)(0.248505 in. )4,467.50 lb-in.

(1.78)(1.50 in./2)

MyK

I

IM

K y

Bending moment at x = 8 in.:

(8 in.) (4 in.)2 2

P PM x P

Maximum load P:

(4 in.) 4,467.50 lb-in.

1,116.88 lb 1,117 lb

P

P Ans.

Check stress at midspan:

midspan

4 4

midspan 4

(1,116.88 lb)(24 in.)6,701.28 lb-in.

4 4

(2.00 in.) 0.785398 in.64

(6,701.28 lb-in.)(2.00 in./2)8,532 psi 24,000 psi OK

0.785398 in.

PLM

I

M y

I




8.81 The C86100 bronze (see Appendix D for

properties) shaft shown in Fig. P8.81 is supported

at each end by self-aligning bearings. The major

shaft diameter is D = 40 mm, the minor shaft

diameter is d = 25 mm, and the radius of the fillet

between the major and minor diameter sections is r

= 5 mm. The shaft length is L = 500 mm and the

fillets are located at x = 150 mm and x = 350 mm.

Determine the maximum load P that may be

applied to the shaft if a factor of safety of 3.0 with

respect to failure by yield is specified.

Fig. P8.81

Solution

From Figure 8.20

5 mm 40 mm

0.20 1.60 1.4825 mm 25 mm

r DK

d d


4 4(25 mm) 19,174.76 mm64

I


yield

allow

331 MPa110.33 MPa

FS 3.0

allow

2 4

allowmax

(110.33 N/mm )(19,174.76 mm )114,357.58 N-mm

(1.48)(25 mm/2)

MyK

I

IM

K y

Bending moment at x = 150 mm:

(150 mm) (75 mm)2 2

P PM x P

Maximum load P:

(75 mm) 114,357.58 N-mm

1,524.77 N 1,525 N

P

P Ans.

Check stress at midspan:

midspan

4 4

midspan 4

(1,524.77 N)(500 mm)190,596.25 N-mm

4 4

(40 mm) 125,663.71 mm64

(190,596.25 N-mm)(40 mm/2)30.33 MPa 110.33 MPa OK

125,663.71 mm

PLM

I

M y

I




8.82 The machine shaft shown in Fig. P8.82 is

made of 1020 cold-rolled steel (see Appendix D

for properties). The major shaft diameter is D =

1.000 in., the minor shaft diameter is d = 0.625 in.,

and the radius of the fillet between the major and

minor diameter sections is r = 0.0625 in. The fillet

is located at x = 4 in. from C. If a load of P = 125

lb is applied at C, determine the factor of safety

with respect to failure by yield in the fillet at B.

Fig. P8.82

Solution

For 1020 cold-rolled steel:

62,000 psiY

From Figure 8.20

0.0625 in. 1.000 in.

0.10 1.6 1.740.625 in. 0.625 in.

r DK

d d


4 4(0.625 in.) 0.0074901 in.64

I

Bending moment at x = 4 in.:

(125 lb)(4 in.) 500 lb-in.M Px

Maximum bending stress:

max 4

(500 lb-in.)(0.625 in./2)(1.74) 36,297.7 psi

0.0074901 in.

MyK

I

Factor of safety:

max

62,000 psiFS

36,297.7 psi1.708Y Ans.




8.83 The machine shaft shown in Fig. P8.83 is

made of 1020 cold-rolled steel (see Appendix D

for properties). The major shaft diameter is D = 30

mm, the minor shaft diameter is d = 20 mm, and

the radius of the fillet between the major and minor

diameter sections is r = 3 mm. The fillet is located

at x = 90 mm from C. Determine the maximum

load P that can be applied to the shaft at C if a

factor of safety of 1.5 with respect to failure by

yield is specified for the fillet at B.

Fig. P8.83

Solution

From Figure 8.20

3 mm 30 mm

0.10 1.5 1.5820 mm 20 mm

r DK

d d


4 4(20 mm) 7,853.98 mm64

I


allow

427 MPa284.6667 MPa

FS 1.5

Y

allow

2 4

allowmax

(284.6667 N/mm )(7,853.98 mm )141,504.2261 N-mm

(1.58)(20 mm/2)

MyK

I

IM

K y

Bending moment at x = 90 mm:

(90 mm)M Px P

Maximum load P:

(90 mm) 141,504.2261 N-mm

1,572.3 N 1,572 N

P

P Ans.




8.84 The grooved shaft shown in Fig. P8.84 is made

of C86100 bronze (see Appendix D for properties).

The major shaft diameter is D = 50 mm, the minor

shaft diameter at the groove is d = 34 mm, and the

radius of the groove is r = 4 mm. Determine the

maximum allowable moment M that may be applied

to the shaft if a factor of safety of 1.5 with respect to

failure by yield is specified.

Fig. P8.84

Solution

From Figure 8.19

4 mm 50 mm

0.20 1.471 1.9634 mm 34 mm

r DK

d d


4 4(34 mm) 65,597.24 mm64

I


allow

331 MPa220.6667 MPa

FS 1.5

Y

allow

2 4

allowmax

(220.6667 N/mm )(65,597.24 mm )

(1.96)(34 mm/2)

434,427. 435 N-m N mm 4 -

MyK

I

IM

K y

Ans.




9.1 For the following problems, a beam segment subjected to internal bending moments at sections A

and B is shown along with a sketch of the cross-sectional dimensions. For each problem:

(a) Sketch a side view of the beam segment and plot the distribution of bending stresses acting at

sections A and B. Indicate the magnitude of key bending stresses on the sketch.

(b) Determine the resultant forces acting in the x direction on the specified area at sections A and B and

show these resultant forces on the sketch.

(c) Is the specified area in equilibrium with respect to forces acting in the x direction? If not, determine

the horizontal force required to satisfy equilibrium for the specified area and show the location and

direction of this force on the sketch.

Consider area (1) of the 20-in.-long beam segment, which is subjected to internal bending moments of

MA = 24 kip-ft and MB = 28 kip-ft.

Fig. P9.1a Beam segment Fig. P9.1b Cross-sectional dimensions

Solution



(in.4) (in.) (in.

4) (in.

4)

left web 864.000 0.000 0.000 864.000

top flange 12.505 10.250 1,287.016 1,299.521

bottom flange 12.505 –10.250 1,287.016 1,299.521

right web 864.000 0.000 0.000 864.000





(a) Bending stress distribution

(b) Resultant forces acting on area (1)

On section A, the resultant force on area (1) in the x direction is

1

(798.699 psi 565.745 psi)(3.5 in.)(3.5 in.) 8,357. 8.36 kips (C)227 lb2

AF Ans.

and on section B, the horizontal resultant force on area (1) is

1

(931.816 psi 660.036 psi)(3.5 in.)(3.5 in.) 9,750. 9.75 kips (C)098 lb2

BF Ans.

(c) Equilibrium of area (1)

8,357.227 lb 9,750.098 lb 1,392.87

1.393 kip

1 b

s

l 0x

H

F

F Ans.

The horizontal shear force is directed from section A toward section B at the interface between area (1)

and the web elements.














MA = 700 lb-ft and MB = 400 lb-ft.


Solution

Centroid location in y direction: (reference axis at bottom of tee shape)


yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

top flange 4.5 1.0 4.50 6.50 29.25

stem 1.0 6.0 6.00 3.00 18.00

10.50 47.25

3

2

47.25 in.4.50 in.

10.50 in.

i i

i

y Ay

A (measured upward from bottom edge of shape)



(in.4) (in.) (in.

4) (in.

4)

top flange 0.375 2.000 18.000 18.375

stem 18.000 –1.500 13.500 31.500








1

(421.053 psi 252.632 psi)(4.5 in.)(1 in.) 1,515.792 lb 1,516 lb (2

C)AF Ans.


1

(240.602 psi 144.361 psi)(4.5 in.)(1 in.) 866.167 lb 866 lb (2

C)BF Ans.


1,515.792 lb 866.167 lb 649.625 lb 0

650 lb

x

H

F

F Ans.

The horizontal shear force is directed from section B toward section A at the interface between area (1)

and the stem.













Consider area (1) of the 500-mm-long beam segment, which is subjected to internal bending moments of

MA = −5.8 kN-m and MB = −3.2 kN-m.


Solution



yi

(from bottom) yi Ai


3)

top flange 160 30 4,800 285 1,368,000

left stem 20 270 5,400 135 729,000

right stem 20 270 5,400 135 729,000

15,600 2,826,000

3

2

2,826,000 mm181.154 mm

15,600 mm

i i

i

y Ay




(mm4) (mm) (mm

4) (mm

4)

top flange 360,000 103.846 51,763,160.3 52,123,160.3

left stem 32,805,000 –46.154 11,503,035.3 44,308,035.3

right stem 32,805,000 –46.154 11,503,035.3 44,308,035.3








1

(4.898 MPa 3.661 MPa)(160 mm)(30 mm) 20,542 20.5 kN (T)N2

AF Ans.


1

(2.702 MPa 2.020 MPa)(160 mm)(30 mm) 11,3 11.33 kN (T)34 N2

BF Ans.


20,542 N 11,334 N 9,209 N

9.21 kN

0x

H

F

F Ans.


and the stems.














MA = −3,300 lb-ft and MB = −4,700 lb-ft.


Solution



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

left flange (1) 1.50 3.50 5.25 10.25 53.8125

right flange (2) 1.50 3.50 5.25 10.25 53.8125

central stem 1.50 12.00 18.00 6.00 108.0000

28.50 215.6250

3

2

215.625 in.7.5658 in.

28.50 in.

i i

i

y Ay




(in.4) (in.) (in.

4) (in.

4)

left flange (1) 5.3594 2.6842 37.8262 43.1856

right flange (2) 5.3594 2.6842 37.8262 43.1856

central stem 216.0000 –1.5658 44.1305 260.1305








1

(506.765 psi 106.767 psi)(1.5 in.)(3.5 in.) 1,610.522 lb 1,611 lb 2

(T)AF Ans.


1

(721.756 psi 152.061 psi)(1.5 in.)(3.5 in.) 2,293.773 lb 2,290 lb 2

(T)BF Ans.


1,610.522 lb 2,293.773 lb 683.252 lb 0

683 lb

x

H

F

F Ans.


and the stem.














MA = −42 kip-in. and MB = −36 kip-in.


Solution



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

top flange (1) 6 2 12 11 132

bottom flange (2) 10 2 20 1 20

web 2 8 16 6 96

48 248

3

2

248 in.5.1667 in.

48 in.

i i

i

y Ay




(in.4) (in.) (in.

4) (in.

4)

top flange (1) 4.0000 5.8333 408.3333 412.3333

bottom flange (2) 6.6667 –4.1667 347.2222 353.8889

web 85.3333 0.8333 11.1111 96.4444








1

(332.690 psi 235.317 psi)(6 in.)(2 in.) 3,408.0 3.41 kips (43 ) lb T2

AF Ans.


1

(251.546 psi 154.173 psi)(6 in.)(2 in.) 2,921.1 2.92 kips (80 ) lb T2

BF Ans.


3,408.043 lb 2,921.180 lb 486.863 lb

0.487 ki s

0

p

x

H

F

F Ans.


and the web.













Consider area (2) of the beam segment shown in Problem 9.5.


Solution



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

top flange (1) 6 2 12 11 132

bottom flange (2) 10 2 20 1 20

web 2 8 16 6 96

48 248

3

2

248 in.5.1667 in.

48 in.

i i

i

y Ay




(in.4) (in.) (in.

4) (in.

4)

top flange (1) 4.0000 5.8333 408.3333 412.3333

bottom flange (2) 6.6667 –4.1667 347.2222 353.8889

web 85.3333 0.8333 11.1111 96.4444








1

(251.546 psi 154.173 psi)(10 in.)(2 in.) 4,057.1 4.06 kips (95 ) lb C2

AF Ans.


1

(215.611 psi 132.149 psi)(10 in.)(2 in.) 3,477.5 3.48 kips (95 ) lb C2

BF Ans.


4,057.195 lb 3,477.595 lb 579.599 l

0.580 p

b

0

ki s

x

H

F

F Ans.


and the web.













Consider area (1) of the 300-mm-long beam segment, which is subjected to internal bending moments of

MA = 7.5 kN-m and MB = 8.0 kN-m.


Solution



yi

(from bottom) yi Ai


3)

left stiff (1) 40 90 3,600 275 990,000

flange (2) 150 40 6,000 300 1,800,000

right stiff (3) 40 90 3,600 275 990,000

stem 40 280 11,200 140 1,568,000

24,400 5,348,000

3

2

5,348,000 mm219.180 mm

24,400 mm

i i

i

y Ay







(mm4) (mm) (mm

4) (mm

4)

left stiff (1) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87

flange (2) 800,000.00 80.8197 39,190,916.42 39,990,916.42

right stiff (3) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87

stem 73,173,333.33 –79.1803 70,218,672.40 143,392,005.73





1

(3.589 MPa 0.385 MPa)(40 mm)(90 mm) 7,153.755 N 7.15 kN (C2

)AF Ans.


1

(3.828 MPa 0.411 MPa)(40 mm)(90 mm) 7,630.672 N 7.63 kN (C2

)BF Ans.


7,153.755 N 7,630.672 N 476.917 N

0.477 kN

0x

H

F

F Ans.


and area (2).













Combine areas (1), (2), and (3) of the beam segment shown in Problem 9.7.


Solution



yi

(from bottom) yi Ai


3)

left stiff (1) 40 90 3,600 275 990,000

flange (2) 150 40 6,000 300 1,800,000

right stiff (3) 40 90 3,600 275 990,000

stem 40 280 11,200 140 1,568,000

24,400 5,348,000

3

2

5,348,000 mm219.180 mm

24,400 mm

i i

i

y Ay




(mm4) (mm) (mm

4) (mm

4)

left stiff (1) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87

flange (2) 800,000.00 80.8197 39,190,916.42 39,990,916.42

right stiff (3) 2,430,000.00 55.8197 11,217,008.87 13,647,008.87

stem 73,173,333.33 -79.1803 70,218,672.40 143,392,005.73








1

(3.589 MPa 0.385 MPa)(40 mm)(90 mm) 7,153.755 N2

AF


1

(3.828 MPa 0.411 MPa)(40 mm)(90 mm) 7,630.672 N2

BF

Resultant forces acting on area (3)

The forces acting on area (3) are identical to those acting on area (1).

Resultant forces acting on area (2)


1

(3.589 MPa 2.165 MPa)(150 mm)(40 mm) 17,262.855 N2

AF


1

(3.828 MPa 2.309 MPa)(150 mm)(40 mm) 18,413.697 N2

BF

Resultant forces acting on combined areas (1), (2), and (3)

On section A, the resultant force on combined areas (1), (2), and (3) is

2(7,153.755 N) 17,262.855 N 31,570.36 31.6 kN (C3 N )AF Ans.


2(7,630.672 N) 18,413.697 N 33,675.05 33.7 kN (C4 N )BF Ans.

(c) Equilibrium of combined areas (1), (2), and (3)

31,570.363 N 33,675.054 N 2,104.691 N 0

2.10 kN

x

H

F

F Ans.


and the stem of the tee.




9.9 A 1.6-m long cantilever beam supports a concentrated load of 7.2 kN, as shown below. The beam is

made of a rectangular timber having a width of 120 mm and a depth of 280 mm. Calculate the maximum

horizontal shear stresses at points located 35 mm, 70 mm, 105 mm, and 140 mm below the top surface

of the beam. From these results, plot a graph showing the distribution of shear stresses from top to

bottom of the beam.

Fig. P9.9a Cantilever beam Fig. P9.9b Cross-sectional dimensions

Solution

Shear force in cantilever beam:

V = 7.2 kN = 7,200 N

Shear stress formula:

V Q

I t

Section properties:

3

6 4(120 mm)(280 mm)219.52 10 mm

12I

t = 120 mm

Distance below top

surface of beam y Q

35 mm 105 mm 514,500 mm3 140.6 kPa

70 mm 70 mm 882,000 mm3 241 kPa

105 mm 35 mm 1,102,500 mm3 301 kPa

140 mm 0 mm 1,176,000 mm3 321 kPa




9.10 A 14-ft long simply supported timber beam carries a 6-kip concentrated load at midspan, as shown

in Fig. P9.10a. The cross-sectional dimensions of the timber are shown in Fig. P9.10b.

(a) At section a–a, determine the magnitude of the shear stress in the beam at point H.

(b) At section a–a, determine the magnitude of the shear stress in the beam at point K.

(c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the

14-ft span length.

(d) Determine the maximum tension bending stress that occurs in the beam at any location within the

14-ft span length.

Fig. P9.10a Simply supported timber beam Fig. P9.10b Cross-sectional

dimensions

Solution

Section properties:

3

4(6 in.)(15 in.)1,687.5 in. 6 in.

12I t

(a) Shear stress at H:

3

3

4

(6 in.)(3 in.)(6 in.) 108 in.

(3,000 lb)(108 in. )

(1,687.500 in. )(632.0 ps

.i

in )

Q

V Q

I t

Ans.

(b) Shear stress at K:

3

3

4

(6 in.)(1 in.)(7 in.) 42 in.

(3,000 lb)(42 in. )

(1,687.500 in. )(6 in.)12.44 psi

Q

V Q

I t

Ans.




(c) Maximum shear stress at any location:

3

max

3

4

(6 in.)(7.5 in.)(3.75 in.) 168.75 in.

(3,000 lb)(168.75 in. )

(1,687.500 in. )(6 in.)50.0 psi

Q

V Q

I t

Ans.

(d) Maximum bending stress at any location:

max

4

21 kip-ft 21,000 lb-ft

(21,000 lb-ft)(7.5 in.)(12 in./ft)

1,687.51,120 psi (T) and (C

00 in.)x

M

M c

I

Ans.




9.11 A 5-m long simply supported timber beam carries a uniformly distributed load of 12 kN/m,

as shown in Fig. P9.11a. The cross-sectional dimensions of the beam are shown in Fig. P9.11b.



(c) Determine the maximum horizontal shear stress that occurs in the beam at any location

within the 5-m span length.

(d) Determine the maximum compression bending stress that occurs in the beam at any location

within the 5-m span length.

Fig. P9.11a Simply supported timber beam Fig. P9.11b Cross-sectional dimensions

Solution

Section properties:

3

6 4(100 mm)(300 mm)225 10 mm 100 mm

12I t

(a) Shear stress magnitude at H:

3

3

6 4

(100 mm)(90 mm)(105 mm)

945,000 mm

(18,000 N)(945,000 mm )

(225 10 mm )(100 m

756 kP

m)

a

Q

V Q

I t

Ans.

(b) Shear stress magnitude at K:

3

3

6 4

(100 mm)(40 mm)(130 mm)

520,000 mm

(18,000 N)(520,000 mm )

(225 10 mm )(100 m

416 kP

m)

a

Q

V Q

I t

Ans.





3

max

3

6 4

(100 mm)(150 mm)(75 mm) 1,125,000 mm

(30,000 N)(1,125,000 mm )

(225 10 mm )(100 mm)1,500 kPa

Q

V Q

I t

Ans.

(d) Maximum compression bending stress at any location:

max

6 4

37.5 kN-m

(37.5 kN-m)(150 mm)(1,000 N/kN)(1,000 mm/m)

225 10 mm

25.0 MPa 25.0 MPa (C)

x

M

M y

I

Ans.




9.12 A 5-m long simply supported timber beam carries two concentrated loads, as shown in Fig. P9.12a.

The cross-sectional dimensions of the beam are shown in Fig. P9.12b.



(c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the 5-

m span length.

(d) Determine the maximum compression bending stress that occurs in the beam at any location within

the 5-m span length.


dimensions

Solution

Section properties:

3

6 4(150 mm)(450 mm)1,139.1 10 mm 150 mm

12I t

(a) Shear stress magnitude at H:

3

3

6 4

(150 mm)(150 mm)(150 mm)

3,375,000 mm

(39,200 N)(3,375,000 mm )

(1,139.1 10 mm )(150 m

7

m)

74 kPa

Q

V Q

I t

Ans.

(b) Shear stress magnitude at K:

3

3

6 4

(150 mm)(100 mm)(175 mm)

2,625,000 mm

(39,200 N)(2,625,000 mm )

(1,139.1 10 mm )(150 m

6

m)

02 kPa

Q

V Q

I t

Ans.





3

max

3

6 4

(150 mm)(225 mm)(112.5 mm) 3,796,875 mm

(39,200 N)(3,796,875 mm )

(1,139.1 10 mm )(150 m871 kP

m)a

Q

V Q

I t

Ans.

(d) Maximum bending stress at any location:

max

6 4

39.2 kN-m

(39.2 kN-m)(225 mm)(1,000 N/kN)(1,000 mm/m)

1,139.1 10 mm

7.74296 MPa 7,740 kPa (C)

x

M

M y

I

Ans.




9.13 A laminated wood beam consists of eight 2 in. × 6-in. planks glued together to form a section 6 in.

wide by 16 in. deep, as shown in Fig. P9.13a. If the allowable strength of the glue in shear is 160 psi,

determine:

(a) the maximum uniformly distributed load w that can be applied over the full length of the beam if the

beam is simply supported and has a span of 20 ft.

(b) the shear stress in the glue joint at H, which is located 4 in. above the bottom of the beam and 3 ft

from the left support. Assume the beam is subjected to the load w determined in part (a).

(c) the maximum tension bending stress in the beam when the load of part (a) is applied.


dimensions

Solution

Section properties:

3

4(6 in.)(16 in.)2,048 in. 6 in.

12I t

(a) Maximum Q:

3(6 in.)(8 in.)(4 in.) 192 in.Q

Maximum shear force V:

4

3

160 psi

(160 psi)(2,048 in. )(6 in.)10,240 lb

192 in.

V Q

I t

V

Maximum distributed load w:

max

max

10,240 lb2

2(10,240 lb)

20 ft1,024 lb/ft

wLV

w

Ans.




(b) Shear force at x = 3 ft:

10,240 lb (1,024 lb/ft)(3 ft) 7,168 lbV

3

3

4

(6 in.)(4 in.)(6 in.) 144 in.

(7,168 lb)(144 in. )

(2,048 in. )(6 in.)84.0 psi

Q

V Q

I t

Ans.

(c) Maximum tension bending stress at any location:

2 2

max

4

(1,024 lb/ft)(20 ft)51,200 lb-ft

8 8

(51,200 lb-ft)( 8 in.)(12 in./ft)

2,048 in.2,400 psi (T)x

wLM

M y

I

Ans.




9.14 A 5-ft long simply supported wood beam carries a concentrated load P at midspan, as shown in Fig.

P9.14a. The cross-sectional dimensions of the beam are shown in Fig. P9.14b. If the allowable shear

strength of the wood is 80 psi, determine the maximum load P that may be applied at midspan. Neglect

the effects of the beam’s self weight.


Solution

Section properties:

3

4(6 in.)(10 in.)500 in. 6 in.

12I t

Maximum Q:

3(6 in.)(5 in.)(2.5 in.) 75 in.Q


4

3

80 psi

(80 psi)(500 in. )(6 in.)3,200 lb

75 in.

V Q

I t

V

Maximum concentrated load P:

max

max

3,200 lb2

2(3,200 lb) 6,400 lb

PV

P

Ans.




9.15 A wood beam supports the loads shown in Fig. P9.15a. The cross-sectional dimensions of the beam

are shown in Fig. P9.15b. Determine the magnitude and location of:

(a) the maximum horizontal shear stress in the beam.

(b) the maximum tension bending stress in the beam.


Solution

Section properties:

3 3

4(75 mm)(240 mm) (20 mm)(100 mm)2 89,733,333 mm

12 12I

(a) Maximum shear force:

Vmax = 9.54 kN = 9,540 N @ support A

Check shear stress at neutral axis:

3

(75 mm)(120 mm)(60 mm)

2(20 mm)(50 mm)(25 mm) 590,000 mm

Q

3

4

(9,540 N)(590,000 mm )545 kPa

(89,733,333 mm )(115 mm)

VQ

I t

Check shear stress at top edge of cover plates:

3(75 mm)(70 mm)(85 mm) 446,250 mmQ

3

4

(9,540 N)(446,250 mm )633 kPa

(89,733,333 mm )(75 mm)

VQ

I t

Maximum shear stress in beam:

,max 633 kPaH Ans.

(b) Maximum bending moment:

Mmax = 6.49 kN-m (between support A and point B)


4

(6.49 kN-m)( 120 mm)(1,000 N/kN)(1,000 mm/m)

89,733,333 mm

8.67905 MPa 8,680 kPa (T)

x

M y

I

Ans.





supports loads PA = 1.5 kN and PC = 3.0 kN,

as shown in Fig. P9.16. Assume L1 = 150

mm, L2 = 300 mm, and L3 = 225 mm. The

bearing at B can be idealized as a roller

support and the bearing at D can be idealized

as a pin support. Determine the magnitude

and location of:

(a) the maximum horizontal shear stress in

the shaft.

(b) the maximum tension bending stress in

the shaft.

Fig. P9.16

Solution

Section properties:

4 4

4

3 3

3

(50 mm)64 64

306,796.158 mm

(50 mm)

12 12

10,416.667 mm

I D

DQ

Maximum shear force magnitude:

Vmax = 1.71 kN (between B and C)

Maximum bending moment magnitude:

Mmax = 289.29 kN-mm (at C)

(a) Maximum horizontal shear stress:

3

4

(1,710 N)(10,416.667 mm )

(306,796.158 mm )(50 mm)

(at neutral axis 1.161 MPa between and )

V Q

I t

B C

Ans.

(b) Maximum tension bending stress:

4

(289.29 kN-mm)( 50 mm/2)(1,000 N/kN)

306,796.15

23.6 MPa (T)

8 mm

23.574 MPa (on bottom of shaft at )

x

M y

I

C

Ans.





supports loads PA = 400 lb and PC = 900 lb,

as shown in Fig. P9.17. Assume L1 = 6 in.,

L2 = 12 in., and L3 = 8 in. The bearing at B

can be idealized as a roller support and the



location of:

(a) the maximum horizontal shear stress in

the shaft.

(b) the maximum tension bending stress in

the shaft.

Fig. P9.17

Solution

Section properties:

4 4

4

3 3

3

(1.25 in.)64 64

0.119842 in.

(1.25 in.)

12 12

0.162760 in.

I D

DQ


Vmax = 480 lb (between B and C)


Mmax = 3,360 lb-in. (at C)


3

4

(480 lb)(0.162760 in. )

(0.119842 in. )(1.25 in.)

521.519 psi (at neutral axis b522 etween and )psi

V Q

I t

B C

Ans.


4

(3,360 lb-in.)( 1.25 in./2)

0.119842 in.

17, 17,520 psi (T523.022 psi (on bottom of) shaft at )

x

M y

I

C

Ans.




9.18 A 1.00-in.-diameter solid steel shaft supports loads PA = 200

lb and PD = 240 lb, as shown in Fig. P9.18. Assume L1 = 2 in.,

L2 = 5 in., and L3 = 4 in. The bearing at B can be idealized as a

pin support and the bearing at C can be idealized as a roller

support. Determine the magnitude and location of:

(a) the maximum horizontal shear stress in the shaft.

(b) the maximum tension bending stress in the shaft.

Fig. P9.18

Solution

Section properties:

4 4

4

3 3

3

(1.00 in.)64 64

0.049087 in.

(1.00 in.)

12 12

0.083333 in.

I D

DQ


Vmax = 272 lb (between B and C)


Mmax = 960 lb-in. (at C)


3

4

(272 lb)(0.083333 in. )

(0.049087 in. )(1.00 in.)

461.762 psi (at neutral axis b462 etween and )psi

V Q

I t

B C

Ans.


4

(960 lb-in.)( 1.00 in./2)

0.049087 in.

9,7 9,780 psi (T)78.480 psi (on bottom of shaft at )

x

M y

I

C

Ans.




9.19 A 20-mm-diameter solid steel shaft supports loads PA = 900

N and PD = 1,200 N, as shown in Fig. P9.19. Assume L1 = 50

mm, L2 = 120 mm, and L3 = 90 mm. The bearing at B can be

idealized as a pin support and the bearing at C can be idealized as

a roller support. Determine the magnitude and location of:

(a) the maximum horizontal shear stress in the shaft.

(b) the maximum compression bending stress in the shaft.

Fig. P9.19

Solution

Section properties:

4 4

4

3 3

3

(20 mm)64 64

7,853.982 mm

(20 mm)

12 12

666.667 mm

I D

DQ


Vmax = 1,275 N (between B and C)


Mmax = 108,000 N-mm (at C)


3

4

(1,275 N)(666.667 mm )

(7,853.982 mm )(20 mm)

5.411 MPa (at neutral axi5.41 M s between and )Pa

V Q

I t

B C

Ans.


4

(108,000 N-mm)(20 mm/2)

7,853.982 mm

1 137.5 MPa (C)37.510 MPa (on top of shaft at )

x

M y

I

C

Ans.





supports loads PA = 600 lb, PC = 1,600 lb, and

PE = 400 lb, as shown in Fig. P9.20. Assume

L1 = 6 in., L2 = 15 in., L3 = 8 in., and L4 = 10

in. The bearing at B can be idealized as a roller

support and the bearing at D can be idealized

as a pin support. Determine the magnitude and

location of:

(a) the maximum horizontal shear stress in the

shaft.

(b) the maximum tension bending stress in the

shaft.

Fig. P9.20

Solution

Section properties:

4 4

4

3 3

3

(1.25 in.)64 64

0.119842 in.

(1.25 in.)

12 12

0.162760 in.

I D

DQ


Vmax = 1,060.9 lb (between C and D)


Mmax = 4,487 lb-in. (at C)


3

4

(1,060.9 lb)(0.162760 in. )

(0.119842 in. )(1.25 in.)

1,152.632 psi (at neutral axis1,153 p between ansi d )

V Q

I t

C D

Ans.


4

(4,487 lb-in.)( 1.25 in./2)

0.119842 in.

23, 23,400 psi (T400.309 psi (on bottom of) shaft at )

x

M y

I

C

Ans.





supports loads PA = 1,000 N, PC = 3,200 N,

and PE = 800 N, as shown in Fig. P9.21.

Assume L1 = 80 mm, L2 = 200 mm, L3 = 100

mm, and L4 = 125 mm. The bearing at B can be

idealized as a roller support and the bearing at

D can be idealized as a pin support. Determine

the magnitude and location of:

(a) the maximum horizontal shear stress in the

shaft.

(b) the maximum tension bending stress in the

shaft.

Fig. P9.21

Solution

Section properties:

4 4

4

3 3

3

(25 mm)64 64

19,174.760 mm

(25 mm)

12 12

1,302.083 mm

I D

DQ


Vmax = 2,200 N (between C and D)


Mmax = 120,000 N-mm (at C)


3

4

(2,200 N)(1,302.083 mm )

(19,174.760 mm )(25 mm)

(at neutral axis b5.98 M etween and )Pa

V Q

I t

C D

Ans.


4

(120,000 N-mm)( 25 mm/2)

19,174.760 mm

7 78.2 MPa 8.228 MPa (on bottom of shaft at )( T)

x

M y

I

C

Ans.




9.22 A 3-in. standard steel pipe (D = 3.500 in.; d = 3.068 in.) supports a concentrated load of P = 900 lb,

as shown in Fig. P9.22a. The span length of the cantilever beam is L = 3 ft. Determine the magnitude of:

(a) the maximum horizontal shear stress in the pipe.

(b) the maximum tension bending stress in the pipe.

Fig. P9.22a Cantilever beam Fig. P9.22b Pipe cross section

Solution

Section properties:

4 4 4 4 4

3 3 3 3 3

[ ] [(3.500 in.) (3.068 in.) ] 3.017157 in.64 64

1 1[ ] [(3.500 in.) (3.068 in.) ] 1.166422 in.

12 12

I D d

Q D d


Vmax = 900 lb


Mmax = (900 lb)(3 ft)(12 in./ft) = 32,400 lb-in.


3

4

(900 lb)(1.166422 in. )805.410 psi

(3.017157 in. )(3.500 805 psi

in. 3.068 in.)

VQ

I t

Ans.


4

( 32,400 lb-in.)(3.500 in./2)18,792.529 psi

3.018,790 psi (T

1715 in)

7 .x

M y

I

Ans.




9.23 A steel pipe (D = 170 mm; d = 150 mm) supports a concentrated load of P as shown in Fig. P9.23a.

The span length of the cantilever beam is L = 1.2 m.

(a) Compute the value of Q for the pipe.

(b) If the allowable shear stress for the pipe shape is 75 MPa, determine the maximum load P than can

be applied to the cantilever beam.

Fig. P9.23a Cantilever beam Fig. P9.23b Pipe cross section

Solution

(a) Section properties:

4 4 4 4 4

3 3 3 3 3 3

[ ] [(170 mm) (150 mm) ] 16,147,786.239 mm64 64

1 1[ ] [(170 mm) (150 mm) ] 128,166.667 mm

12128,170

12 mm

I D d

Q D d

Ans.

(b) Maximum load P:

2 4

max 3

75 MPa

(75 N/mm )(16,147,786.239 mm )(170 mm 150 mm)188,986 N

128,166.667 mm

V Q

I t

V

For the cantilever beam shown here, V = P; therefore,

max max 189.0 kNP V Ans.




9.24 A concentrated load P is applied to the upper end of a 1-m long

pipe, as shown in Fig. P9.24. The outside diameter of the pipe is D =

114 mm and the inside diameter is d = 102 mm.

(a) Compute the value of Q for the pipe.

(b) If the allowable shear stress for the pipe shape is 75 MPa, determine

the maximum load P than can be applied to the cantilever beam.

Fig. P9.24

Solution

(a) Section properties:

4 4 4 4 4

3 3 3 3 3

[ ] [(114 mm) (102 mm) ] 2,977,287 mm64 64

1 1[ ] [(114 mm) 35,028 (102 m mm) ]

12 1m

2

I D d

Q D d

Ans.

(b) Maximum load P:

2 4

max 3

75 MPa

(75 N/mm )(2,977,287 mm )(114 mm 102 mm)76,498 N

35,028 mm

V Q

I t

V


max max 76.5 kNP V Ans.




9.25 A concentrated load of P = 6 kips is applied to the upper end of a

4-ft long pipe, as shown in Fig. P9.25. The pipe is an 8 in. standard

steel pipe, which has an outside diameter of D = 8.625 in. and an

inside diameter of d = 7.981 in. Determine the magnitude of:

(a) the maximum vertical shear stress in the pipe.

(b) the maximum tension bending stress in the pipe.

Fig. P9.25

Solution

Section properties:

4 4 4 4 4

3 3 3 3 3

[ ] [(8.625 in.) (7.981 in.) ] 72.489241 in.64 64

1 1[ ] [(8.625 in.) (7.981 in.) ] 11.104874 in.

12 12

I D d

Q D d


Vmax = 6 kips = 6,000 lb


Mmax = (6,000 lb)(4 ft)(12 in./ft) = 288,000 lb-in.

(a) Maximum vertical shear stress:

3

4

(6,000 lb)(11.104874 in. )1,427.268 psi

(72.489241 in. )(8.625 in. 7.9811,427 p

in.)si

VQ

I t

Ans.


4

(288,000 lb-in.)(8.625 in./2)17,133.578 psi

72.4817,130 psi

924T

.(

1 in)x

M c

I Ans.




9.26 The cantilever beam shown in Fig. P9.26a is subjected to a concentrated load of P = 38 kips. The

cross-sectional dimensions of the wide-flange shape are shown in Fig. P9.26b. Determine:

(a) the shear stress at point H, which is located 4 in. below the centroid of the wide-flange shape.

(b) the maximum horizontal shear stress in the wide-flange shape.


Solution


Shape Width b Height h IC d = yi – y d²A IC + d²A

(in.) (in.) (in.4) (in.) (in.

4) (in.

4)

flange 6.75 0.455 0.0530 6.7725 140.8683 140.9213

web 0.285 13.090 53.2700 0.0000 0.0000 53.2700

flange 6.75 0.455 0.0530 –6.7725 140.8683 140.9213



3

3

4

0.455 in.(6.75 in.)(0.455 in.) 7 in.

2

7 in. 0.455 in. 4 in.(0.285 in.)(7 in. 0.455 in. 4 in.) 4 in. 24.6243 in.

2

(38 kips)(24.6243 in. )9.7974 ksi

(335.1125 in. )(0.285 in.)9.80 ksi

H

H

Q

Ans.

(b) Maximum horizontal shear stress:

max

3

3

max 4

0.455 in.(6.75 in.)(0.455 in.) 7 in.

2

7 in. 0.455 in.(0.285 in.)(7 in. 0.455 in.) 26.9043 in.

2

(38 kips)(26.9043 in. )10.7046 ksi

(335.1125 in. )(0.2810.70 k

5 is

)i

n.

Q

Ans.




9.27 The cantilever beam shown in Fig. P9.27a is subjected to a concentrated load of P. The cross-

sectional dimensions of the wide-flange shape are shown in Fig. P9.27b.

(a) Compute the value of Q that is associated with point K, which is located 2 in. above the centroid of

the wide-flange shape.

(b) If the allowable shear stress for the wide-flange shape is 14 ksi, determine the maximum

concentrated load P than can be applied to the cantilever beam.


Solution



(in.) (in.) (in.4) (in.) (in.

4) (in.

4)

flange 6.75 0.455 0.0530 6.7725 140.8683 140.9213

web 0.285 13.090 53.2700 0.0000 0.0000 53.2700

flange 6.75 0.455 0.0530 –6.7725 140.8683 140.9213


(a) Q associated with point K:

3

0.455 in.(6.75 in.)(0.455 in.) 7 in.

2

7 in. 0.455 in. 2 in.(0.285 in.)(7 in. 0.455 26.33in. 2 in.) 2 i 43 inn

2..

KQ

Ans.

(b) Maximum load P:

max

3

0.455 in.(6.75 in.)(0.455 in.) 7 in.

2

7 in. 0.455 in.(0.285 in.)(7 in. 0.455 in.) 26.9043 in.

2

Q

maxmax

4

max 3

14 ksi

(14 ksi)(335.1125 in. )(0.285 in.)49.6983 kips

26.9043 in.

VQ

It

V


max max 49.7 kipsP V Ans.




9.28 The cantilever beam shown in Fig. P9.28a is subjected to a concentrated load of P. The cross-

sectional dimensions of the rectangular tube shape are shown in Fig. P9.28b.

(a) Compute the value of Q that is associated with point H, which is located 90 mm above the centroid

of the rectangular tube shape.

(b) If the allowable shear stress for the rectangular tube shape is 125 MPa, determine the maximum

concentrated load P than can be applied to the cantilever beam.


Solution



(mm4) (mm) (mm

4) (mm

4)

outer rectangle 195,312,500 0.000 0.000 195,312,500

inner rectangle −143,077,428 0.000 0.000 −143,077,428


(a) Q associated with point H:

3

250 mm 8 mm(150 mm)(8 mm)

2 2

250 mm8 mm 90 mm

250 mm 22(8 mm) 8 mm 90 mm 90 mm

189,912 mm

2 2

HQ

Ans.

(b) Maximum load P:

max

3

250 mm 8 mm(150 mm)(8 mm)

2 2

250 mm8 mm

250 mm 22(8 mm) 8 mm 254,712 mm2 2

Q

max

max

2 4

max 3

125 MPa

(125 N/mm )(52,235,072 mm )(2 8 mm)410,150 N 410.15 kN

254,712 mm

VQ

It

V


max max 410 kNP V Ans.




9.29 The cantilever beam shown in Fig. P9.29a is subjected to a concentrated load of P = 175 kN. The

cross-sectional dimensions of the rectangular tube shape are shown in Fig. P9.29b. Determine:

(a) the shear stress at point K, which is located 50 mm below the centroid of the rectangular tube shape.

(b) the maximum horizontal shear stress in the rectangular tube shape.


Solution



(mm4) (mm) (mm

4) (mm

4)

outer rectangle 195,312,500 0.000 0.000 195,312,500

inner rectangle −143,077,428 0.000 0.000 −143,077,428


(a) Shear stress at K:

3

250 mm 8 mm(150 mm)(8 mm)

2 2

250 mm8 mm 50 mm

250 mm 22(8 mm) 8 mm 50 mm 50 mm2 2

234,712 mm

KQ

3

4

(175,000 N)(234,712 mm )49.1463 MPa

(52,235,072 mm )(2 849.1

mmPa

)MK

Ans.


max

3

250 mm 8 mm(150 mm)(8 mm)

2 2

250 mm8 mm

250 mm 22(8 mm) 8 mm 254,712 mm2 2

Q

3

max 4

(175,000 N)(254,712 mm )53.3341 MPa

(52,235,072 mm )(2 8 53.3 MP

)a

mm

Ans.




9.30 The internal shear force V at a certain section of an

aluminum beam is 8 kN. If the beam has a cross section

shown in Fig. P9.30, determine:

(a) the shear stress at point H, which is located 30 mm

above the bottom surface of the tee shape.

(b) the maximum horizontal shear stress in the tee shape.

Fig. P9.30

Solution


Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

top flange 375.0 72.5 27,187.5

stem 350.0 35.0 12,250.0

725.0 mm2

39,437.5 mm

3

3

2

39,437.5 mm54.397 mm (from bottom of shape to centroid)

725.0 mm


i i

i

y Ay

A



(mm4) (mm) (mm

4) (mm

4)

top flange 781.250 18.103 122,900.565 123,681.815

stem 142,916.667 −19.397 131,679.177 274,595.843



3

3

4

(5 mm)(30 mm)(39.397 mm) 5,909.550 mm

(8,000 N)(5,909.550 mm )

(398,277.658 mm )(5 mm23.7 MP

)a

H

H

Q

Ans.


At neutral axis:

3

max

3

max 4

(5 mm)(54.397 mm)(27.199 mm) 7,397.720 mm

(8,000 N)(7,397.720 mm )

(398,277.658 mm )(5 mm)29.7 MPa

Q

Ans.




9.31 The internal shear force V at a certain section of a

steel beam is 80 kN. If the beam has a cross section


(a) the shear stress at point H, which is located 30 mm

below the centroid of the wide-flange shape.

(b) the maximum horizontal shear stress in the wide-

flange shape.

Fig. P9.31

Solution



(mm4) (mm) (mm

4) (mm

4)

top flange 59,062.5 97.5 29,944,687.5 30,003,750.0

web 4,860,000.0 0.0 0.0 4,860,000.0

bottom flange 59,062.5 −97.5 29,944,687.5 30,003,750.0



3

3

4

(210 mm)(15 mm)(97.5 mm) (10 mm)(60 mm)(60 mm) 343,125 mm

(80,000 N)(343,125 mm )

(64,867,500 mm )(10 mm)42.3 MPa

H

H

Q

Ans.


At neutral axis:

3

max

3

max 4

(210 mm)(15 mm)(97.5 mm) (10 mm)(90 mm)(45 mm) 347,625 mm

(80,000 N)(347,625 mm )

(64,867,500 mm )(10 mm4

)2.9 MPa

Q

Ans.




9.32 The internal shear force V at a certain section

of a steel beam is 110 kips. If the beam has a cross

section shown in Fig. P9.32, determine:

(a) the value of Q associated with point H, which is

located 2 in. below the top surface of the flanged

shape.

(b) the maximum horizontal shear stress in the

flanged shape.

Fig. P9.32

Solution


Shape Area Ai

yi

(from bottom) yi Ai

(in.2) (in.) (in.

3)

top flange 5.0 11.5 57.5

web 10.0 6.0 60.0

bottom flange 8.0 0.5 4.0

23.0 in.2

121.5 in.

3

3

2


23.0 in.


i i

i

y Ay

A



(in.4) (in.) (in.

4) (in.

4)

top flange 0.4167 6.2174 193.2798 193.6964

web 83.3333 0.7174 5.1465 88.4798

bottom flange 0.6667 −4.7826 182.9868 183.6534


(a) Q at point H:

3(5 in.)(1 in.)(6.2174 in.) (1 in.)(1 in.)(5.2174 36.304in.) 4 in.Q Ans.


At neutral axis:

3

max

3

max 4

(5 in.)(1 in.)(6.2174 in.) (1 in.)(5.7174 in.)(2.8587 in.) 47.4313 in.

(110 kips)(47.4313 in. )

(465.8297 in. )(1 in.)11.20 ksi

Q

Ans.




9.33 The internal shear force V at a certain section of a

steel beam is 75 kips. If the beam has a cross section


(a) the shear stress at point H, which is located 2 in.

above the bottom surface of the flanged shape.

(b) the shear stress at point K, which is located 4.5 in.

below the top surface of the flanged shape.

Fig. P9.33

Solution



(in.4) (in.) (in.

4) (in.

4)

left flange 62.5000 0.0000 0.0000 62.5000

web 0.0885 0.0000 0.0000 0.0885

right flange 62.5000 0.0000 0.0000 62.5000



3

3

4

2(0.75 in.)(2 in.)(4 in.) 12 in.

(75 kips)(12 in. )

(125.0885 in. )(2 0.75 in.)4.80 ksi

H

H

Q

Ans.

(b) Shear stress at K:

3

3

4

2(0.75 in.)(4.5 in.)(2.75 in.) 18.5625 in.

(75 kips)(18.5625 in. )

(125.0885 in. )(2 0.75 in.)7.42 ksi

K

K

Q

Ans.




9.34 Consider a 100-mm-long segment of a simply

supported beam (Fig. P9.34a). The internal bending

moments on the left and right sides of the segment are

75 kN-m and 80 kN-m, respectively. The cross-

sectional dimensions of the flanged shape are shown in

Fig. P9.34b. Determine the maximum horizontal shear

stress in the beam at this location.

Fig. P9.34b Cross-sectional dimensions Fig. P9.34a Beam segment (side view)

Solution


Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

top flange 9,000 300 2,700,000

web 8,400 165 1,386,000

bottom flange 15,000 30 450,000

32,400 mm2

4,536,000 mm

3

3

2

4,536,000 mm140 mm (from bottom of shape to centroid)

32,400 mm

190 mm (from top of shape to centroid)

i i

i

y Ay

A



(mm4) (mm) (mm

4) (mm

4)

top flange 2,700,000 160 230,400,000 233,100,000

web 30,870,000 25 5,250,000 36,120,000

bottom flange 4,500,000 −110 181,500,000 186,000,000


Shear force in beam:

80 kN-m 75 kN-m 5 kN-m

50 kN100 mm 0.1 m

MV

x

Maximum horizontal shear stress:

At neutral axis:

3

max

3

max 4

(250 mm)(60 mm)(110 mm) (40 mm)(80 mm)(40 mm) 1,778,000 m

4.88 MP

m

(50,000 N)(1,778,000 mm )

(455,220,000 mm )(40 mm)a

Q

Ans.




9.35 A simply supported beam supports the loads shown in Fig. P9.35a. The cross-sectional dimensions

of the wide-flange shape are shown in Fig. P9.35b.

(a) Determine the maximum shear force in the beam.

(b) At the section of maximum shear force, determine the shear stress in the cross section at point H,

which is located 100 mm below the neutral axis of the wide-flange shape.

(c) At the section of maximum shear force, determine the maximum horizontal shear stress in the cross

section.

(d) Determine the magnitude of the maximum bending stress in the beam.


Solution

(a) Maximum shear force magnitude:

Vmax = 175 kN (just to the right of B)


Mmax = 156.25 kN-m (between B and C)




Section properties:



(mm4) (mm) (mm

4) (mm

4)

top flange 56,250 142.5 60,918,750 60,975,000

web 16,402,500 0 0 16,402,500

bottom flange 56,250 −142.5 60,918,750 60,975,000


(b) Shear stress at H:

3

3

4

(200 mm)(15 mm)(142.5 mm) (10 mm)(35 mm)(117.5 mm) 468,625 mm

(175,000 N)(468,625 mm )

(138,352,500 mm )(10 mm)59.3 MPa

H

H

Q

Ans.

(c) Maximum horizontal shear stress:

At neutral axis:

3

max

3

max 4

(200 mm)(15 mm)(142.5 mm) (10 mm)(135 mm)(67.5 mm) 518,625 mm

(175,000 N)(518,625 mm )

(138,352,500 mm )(10 mm65.6 MP

)a

Q

Ans.

(d) Maximum tension bending stress:

6

4

(156.25 10 N-mm)(300 mm/2)

138,352,500

169.4 MP

mm

169.40 a a4 MP

x

M c

I

Ans.




9.36 A simply supported beam supports the loads shown in Fig. P9.36a. The cross-sectional dimensions

of the structural tube shape are shown in Fig. P9.36b.

(a) At section a–a, which is located 4 ft to the right of pin support B, determine the bending stress and

the shear stress at point H, which is located 3 in. below the top surface of the tube shape.

(b) Determine the magnitude and the location of the maximum horizontal shear stress in the tube shape

at section a–a.


Solution

Shear force magnitude at a–a:

V = 27.60 kips

Bending moment at a–a:

M = 60.90 kip-ft

Section properties:

3 3

4

(12 in.)(16 in.) (11.25 in.)(15.25 in.)

12 12

771.0830 in.

I

(a) Bending stress at H:

4

(60,900 lb-ft)(5 in.)(12 in./ft)

771.0830 in.

4,738.79 psi 4,740 psi (C)

H

M y

I

Ans.

Shear stress at H:

3

3

4

(12 in.)(0.375 in.)(7.8125 in.) 2(0.375 in.)(2.625 in.)(6.3125 in.) 47.5840 in.

(27,600 lb)(47.5840 in. )

(771.0830 in. )(2 0.375 in.)2,270 psi

H

H

Q

Ans.

(b) Maximum shear force magnitude:

V = 39.60 kips (at pin B)




Maximum horizontal shear stress:

At neutral axis:

3

max

3

max 4

(12 in.)(0.375 in.)(7.8125 in.) 2(0.375 in.)(7.625 in.)(3.8125 in.) 56.9590 in.

(39,600 lb)(56.9590 in. )

(771.0830 in. )(2 0.375 in.)3,900 psi

Q

Ans.




9.37 A cantilever beam supports the loads shown in Fig. P9.37a. The cross-sectional dimensions of the

shape are shown in Fig. P9.37b. Determine:

(a) the maximum horizontal shear stress.

(b) the maximum compression bending stress.

(c) the maximum tension bending stress.


Solution



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

top flange 12.0 0.5 6.00 5.75 34.5000

left stem 0.5 5.5 2.75 2.75 7.5625

right stem 0.5 5.5 2.75 2.75 7.5625

11.50

49.6250

3

2


11.50 in.


i i

i

y Ay

A



(in.4) (in.) (in.

4) (in.

4)

top flange 0.1250 1.4348 12.3519 12.4769

left stem 6.9323 −1.5652 6.7371 13.6694

right stem 6.9323 −1.5652 6.7371 13.6694






V = 5,800 lb

Maximum positive bending moment:

Mpos = 8,850 lb-ft

Maximum negative bending moment:

Mneg = −9,839 lb-ft

(a) Maximum shear stress:

max

3

3

max 4

2(0.5 in.)(4.3152 in.)(4.3152 in./2)

9.3105 in.

(5,800 lb)(9.3105 in. )

(39.8157 in. )(2 0.5 in.)

1,356 psi

Q

Ans.


Check two possibilities. First, check the bending stress created by the largest positive moment at the top

of the cross section:

pos top

4

(8,850 lb-ft)(1.6848 in.)(12 in./ft)4,494 psi

39.8157 in.x

z

M y

I

Next, for the largest negative moment, compute the bending stress at the bottom of the cross section:

neg bot

4

( 9,839 lb-ft)( 4.3152 in.)(12 in./ft)12,796 psi

39.8157 in.x

z

M y

I


comp 12,800 psi (C) Ans.

(c) Maximum tension bending stress:

Check two possibilities. First, check the bending stress created by the largest positive moment at the

bottom of the cross section:

pos bot

4

(8,850 lb-ft)( 4.3152 in.)(12 in./ft)11,510 psi

39.8157 in.x

z

M y

I

Next, for the largest negative moment, compute the bending stress at the top of the cross section:

neg top

4

( 9,839 lb-ft)(1.6848 in.)(12 in./ft)4,996 psi

39.8157 in.x

z

M y

I

Therefore, the maximum tension bending stress is:

tens 11,510 psi (T) Ans.




9.38 A cantilever beam supports the loads shown in Fig.

P9.38a. The cross-sectional dimensions of the shape are

shown in Fig. P9.38b. Determine:

(a) the maximum vertical shear stress.

(b) the maximum compression bending stress.

(c) the maximum tension bending stress.


Solution


Vmax = 5 kN

Maximum positive bending moment:

Mpos = 2.00 kN-m

Maximum negative bending moment:

Mneg = −1.50 kN-m



yi

(from bottom) yi Ai


3)

flange 100 8 800 96 76,800

stem 6 92 552 46 25,392

1,352 102,192




3

2

102,192 mm75.5858 mm (from bottom of shape to centroid)

1,352 mm


i i

i

y Ay

A



(mm4) (mm) (mm

4) (mm

4)

flange 4,266.67 20.4142 333,391.69 337,658.35

stem 389,344.00 -29.5858 483,176.36 872,520.36


(a) Maximum vertical shear stress:

At neutral axis:

3

max

3

max 4

(6 mm)(75.5858 mm)(75.5858 mm/2) 17,139.640 mm

(5,000 N)(17,139.640 mm )

(1,210,178.71 mm )(6 mm)11.80 MPa

Q

Ans.


Check two possibilities. First, check the bending stress created by the largest positive moment at the top

of the cross section:

6

pos top

4

(2.00 10 N-mm)(24.4142 mm)40.348 MPa

1,210,178.71 mmx

z

M y

I

Next, for the largest negative moment, compute the bending stress at the bottom of the cross section:

6

neg bot

4

( 1.50 10 N-mm)( 75.5858 mm)93.688 MPa

1,210,178.71 mmx

z

M y

I


comp 93.7 MPa (C) Ans.

(c) Maximum tension bending stress:

Check two possibilities. First, check the bending stress created by the largest positive moment at the

bottom of the cross section:

6

pos bot

4

(2.00 10 N-mm)( 75.5858 mm)124.917 MPa

1,210,178.71 mmx

z

M y

I

Next, for the largest negative moment, compute the bending stress at the top of the cross section:

6

neg top

4

( 1.50 10 N-mm)(24.4142 mm)30.261 MPa

1,210,178.71 mmx

z

M y

I

Therefore, the maximum tension bending stress is:

tens 124.9 MPa (T) Ans.




9.39 A simply supported beam fabricated from pultruded reinforced plastic supports the loads shown in

Fig. P9.39a. The cross-sectional dimensions of the plastic wide-flange shape are shown in Fig. P9.39b.

(a) Determine the magnitude of the maximum shear force in the beam.

(b) At the section of maximum shear force, determine the shear stress magnitude in the cross section at

point H, which is located 2 in. above the bottom surface of the wide-flange shape.

(c) At the section of maximum shear force, determine the magnitude of the maximum horizontal shear

stress in the cross section.

(d) Determine the magnitude of the maximum compression bending stress in the beam. Where along the

span does this stress occur?


Solution

Section properties:

3 3

4

(4 in.)(8 in.) (3.625 in.)(7.25 in.)

12 12

55.5493 in.

zI

(a) Maximum shear force magnitude:

V = 3,664 lb Ans.

(b) Shear stress magnitude at H:

3

3

4

(4 in.)(0.375 in.)(3.8125 in.)

(0.375 in.)(1.625 in.)(2.8125 in.)

7.4326 in.

(3,664 lb)(7.4326 in. )

(55.5493 in. )(0.375 in.)

1,307 psi

H

H

Q

Ans.




(c) Maximum horizontal shear stress magnitude:

At neutral axis:

3

max

3

max 4

(4 in.)(0.375 in.)(3.8125 in.) (0.375 in.)(3.625 in.)(1.8125 in.) 8.1826 in.

(3,664 lb)(8.1826 in. )

(55.5493 in. )(0.375 in1,439 psi

.)

Q

Ans.

(d) Maximum compression bending stress:

4

(7,719 lb-ft)(4 in.)(12 in./ft)6,669.965 psi

56,670 psi (

5.5493C)

in.H

M y

I Ans.

This stress occurs at 5.86 ft to the right of A.




9.40 A wooden beam is fabricated from one 2 × 10 and two 2 × 4 pieces

of dimension lumber to form the I-beam cross section shown in Fig.

P9.40. The flanges of the beam are fastened to the web with nails that

can safely transmit a force of 120 lb in direct shear. If the beam is

simply supported and carries a 1,000-lb load at the center of a 12-ft

span, determine:

(a) the horizontal force transferred from each flange to the web in a 12-

in. long segment of the beam.

(b) the maximum spacing s (along the length of the beam) required for

the nails.

(c) the maximum horizontal shear stress in the I-beam.

Fig. P9.40

Solution



(in.) (in.) (in.4) (in.) (in.

4) (in.

4)

top flange 4 2 2.667 6.000 288.000 290.667

web 2 10 166.667 0.000 0.000 166.667

bottom flange 4 2 2.667 –6.000 288.000 290.667


Maximum shear force

For P = 1,000 lb, V = P/2 = 500 lb

(a) Horizontal force transferred from each flange

(in a 12-in. length):

3

3

4

(4 in.)(2 in.)(6 in.) 48 in.

(500 lb)(48 in. )32.086 lb/in.

748 in.

(32.086 lb/in.)(12 38 5 li bn.)H

Q

VQq

I

F Ans.

(b) Maximum nail spacing:

(1 nail)(120 lb/nail)

32.086 3.74 in

lb/ n.

i .

f f

f f

q s n V

n Vs

q Ans.

(c) Maximum horizontal shear stress:

3

max

3

max 4

(4 in.)(2 in.)(6 in.) (2 in.)(5 in.)(2.5 in.) 73 in.

(500 lb)(73 in. )

(748 in. )(2 in.)24.4 psi

Q

Ans.




9.41 A wooden beam is fabricated from one 2 × 10 and two 2 × 4 pieces

of dimension lumber to form the I-beam cross section shown in Fig.

P9.41. The I-beam will be used as a simply supported beam to carry a

concentrated load P at the center of a 20-ft span. The wood has an

allowable bending stress of 1,200 psi and an allowable shear stress of 90

psi. The flanges of the beam are fastened to the web with nails that can

safely transmit a force of 120 lb in direct shear.

(a) If the nails are uniformly spaced at an interval of s = 4.5 in. along the

span, what is the maximum concentrated load P that can be supported by

the beam? Demonstrate that the maximum bending and shear stresses

produced by P are acceptable.

(b) Determine the magnitude of load P that produces the allowable

bending stress in the span (i.e., b = 1,200 psi). What nail spacing s is

required to support this load magnitude? Demonstrate that the

maximum horizontal shear stresses produced by P are acceptable.

Fig. P9.41

Solution



(in.) (in.) (in.4) (in.) (in.

4) (in.

4)

top flange 4 2 2.667 6.000 288.000 290.667

web 2 10 166.667 0.000 0.000 166.667

bottom flange 4 2 2.667 –6.000 288.000 290.667


(a) Maximum concentrated load P:

3(4 in.)(2 in.)(6 in.) 48 in.

(1 nail)(120 lb/nail)26.667 lb/in.

4.5 in.

f f

f f

Q

q s n V

n Vq

s

4

3

max

(26.667 lb/in.)(748 in. )415.556 lb

831

48 in

lb

.

2

VQq

I

q IV

Q

PV P Ans.




Check maximum bending and shear stresses:

max 4

3

max

3

maxmax 4

(831.1 lb)(240 in.)/4 (7 in.)( / 4)466.7 psi 1,200 psi OK

748 in.

(2 in.)(5 in.)(2.5 in.) (4 in.)(2 in.)(6 in.) 73 in.

( / 2) (831.1 lb / 2)(73 in. )2

(748 in. )(2 in.)

z z

z z

Mc PL c

I I

Q

VQ P Q

I t I t0.3 psi 90 psi OK

(b) Magnitude of load P that produces the allowable bending stress in the span:

4

max

maxmax 2,1

1,200

40

psi

(1,200 psi)(748 in. )128,228.566 lb-in.

7 in.

4

4 4(128,228.566 lb-in.)2,137.143 lb

(20 ft)(12 in./ft)lb

x

M c

I

M

P LM

MP

L Ans.

Required nail spacing s:

maxmax

3

4

2,137.143 lb1,068.571 lb

2 2

(1,068.571 lb)(48 in. )68.571 lb/in.

748 in.

(1 nail)(1201.7

lb)

68.571 50 in.

lb/in.

f f

f f

PV

VQq

I

q s n V

n Vs

q Ans.

Check maximum shear stresses:

3

maxmax 4

( / 2) (2,137.143 lb / 2)(73 in. )52.143 psi 90 psi OK

(748 in. )(2 in.)z z

VQ P Q

I t I t




9.42 A wooden box beam is fabricated from four boards, which are fastened together with nails, as

shown in Fig. P9.42b. The nails are installed at a spacing of s = 125 mm (Fig. P9.42a), and each nail can

provide a resistance of Vf = 500 N. In service, the box beam will be installed so that bending occurs

about the z axis. Determine the maximum shear force V that can be supported by the box beam based on

the shear capacity of the nailed connections.


Solution

Moment of inertia Iz:

3 3

4(200 mm)(300 mm) (120 mm)(250 mm)293,750,000 mm

12 12zI

First moment of area Q:

3(200 mm)(25 mm)(137.5 mm) 687,500 mmQ

Shear flow q based on nail shear force:

(2 nails)(500 N/nail)

8 N/mm125 mm

f f

f f

q s n V

n Vq

s


4

3

(8 N/mm)(293,750,000 mm )3,418 N

687,500 mm3.42 kNz

z

VQ q Iq V

I Q Ans.




9.43 A wooden box beam is fabricated from four boards, which are fastened together with screws, as

shown in Fig. P9.43b. Each screw can provide a resistance of 800 N. In service, the box beam will be

installed so that bending occurs about the z axis, and the maximum shear force in the beam will be 9 kN.

Determine the maximum permissible spacing interval s for the screws (see Fig. P9.43a).


Solution


3 3

4(190 mm)(250 mm) (140 mm)(150 mm)208,020,833 mm

12 12zI

First moment of area Q:

3(140 mm)(50 mm)(100 mm) 700,000 mmQ

Shear flow q based on beam shear force V:

3

4

(9,000 N)(750,000 mm )30.285 N/mm

208,020,833 mmz

VQq

I

Maximum spacing interval s:

(2 screws)(800 N/screw)

30.2852.8

5 N/m

m m

m

f f

f f

q s n V

n Vs

q Ans.




9.44 A wooden beam is fabricated by nailing together three pieces of dimension lumber, as shown in

Fig. P9.44a. The cross-sectional dimensions of the beam are shown in Fig. P9.44b. The beam must

support an internal shear force of V = 750 lb.

(a) Determine the maximum horizontal shear stress in the cross section for V = 750 lb.

(b) If each nail can provide 100 lb of horizontal resistance, determine the maximum spacing s for the

nails.

(c) If the three boards were connected by glue instead of nails, what minimum shear strength would be

necessary for the glued joints?


Solution



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

left board 2 8 16 4 64

flange board 4 2 8 7 56

right board 2 9 16 4 64

40 184

3

2

184 in.4.6 in. (from bottom of shape to centroid)

40 in.


i i

i

y Ay

A



(in.4) (in.) (in.

4) (in.

4)

left board 85.3333 –0.60 5.7600 91.0933

flange board 2.6667 2.40 46.0800 48.7467

right board 85.3333 –0.60 5.7600 91.0933



At neutral axis:

3

max

3

max 4

2(2 in.)(4.6 in.)(4.6 in./2) 42.32 in.

(750 lb)(42.32 in. )

(230.9333 in. )(4 in.)34.4 psi

Q

Ans.

(b) Shear flow q based on beam shear force V:




3

3

4

(4 in.)(2 in.)(3.4 in. 1 in.) 19.20 in.

(750 lb)(19.20 in. )62.356 lb/in.

230.9333 in.z

Q

VQq

I

Maximum spacing interval s:

(2 nails)(100 lb/nail)

62.356 3.21 in

lb/in..

f f

f f

q s n V

n Vs

q Ans.

(c) Glue joint shear stress:

62.356 lb/in.

2(15.59 psi

2 in.) Ans.




9.45 A wooden beam is fabricated by gluing four dimension lumber

boards, each 40-mm wide and 90-mm deep, to a 32 × 400 plywood

web, as shown in Fig. P9.45. Determine the maximum allowable

shear force and the maximum allowable bending moment that this

section can carry if the allowable bending stress is 6 MPa, the

allowable shear stress in the plywood is 640 kPa, and the allowable

shear stress in the glued joints is 250 kPa.

Fig. P9.45

Solution


3 3

4(112 mm)(400 mm) (80 mm)(220 mm)526,346,667 mm

12 12zI


2 4

max

(6 N/mm )(526,346,667 mm )15,790,4 15.79 kN-m00 N-mm

200 mm

M c

I

IM

c Ans.

Maximum allowable shear force:

Consider maximum shear stress, which occurs at the neutral axis:

3

4

3

(32 mm)(200 mm)(100 mm) 2(40 mm)(90 mm)(200 mm 90 mm/2) 1,756,000 mm

(0.640 N/mm)(526,346,667 mm )(32 mm)6,138.7 N 6.14 kN

1,756,000 mm

Q

VQ I tV

I t Q (a)

Consider shear stress in glue joints:

3(40 mm)(90 mm)(200 mm 90 mm/2) 558,000 mmQ

The shear stress in the glue joints can be found from the shear flow across the glue joint divided by the

width of the glue joint; thus,

glue

glue glue

4glue glue

3

/

(0.250 N/mm)(526,346,667 mm )(90 mm)21,224 N 21.2 kN

558,000 mm

q VQ I

t t

I tV

Q (b)

Compare results (a) and (b) to find that the maximum allowable shear force for the section is:

max 6.14 kNV Ans.




9.46 A wooden beam is fabricated from one 2 × 12 and two 2 × 10

dimension lumber boards to form the double-tee cross section

shown in Fig. P9.46. The beam flange is fastened to the stem with

nails. Each nail can safely transmit a force of 175 lb in direct shear.

The allowable shear stress of the wood is 70 psi.

(a) If the nails are uniformly spaced at an interval of s = 4 in. along

the span, what is the maximum internal shear force V that can be

supported by the double-tee cross section?

(b) What nail spacing s would be necessary to develop the full

strength of the double-tee shape in shear? (Full strength means that

the maximum horizontal shear stress in the double-tee shape equals

the allowable shear stress of the wood.)

Fig. P9.46

Solution



yi

(from bottom) yi Ai

(in.) (in.) (in.2) (in.) (in.

3)

top flange 12 2 24 11 264

left stem 2 10 20 5 100

right stem 2 10 20 5 100

64 464

3

2

464 in.7.25 in. (from bottom of shape to centroid)

64 in.


i i

i

y Ay

A



(in.4) (in.) (in.

4) (in.

4)

left board 8.000 3.750 337.500 345.500

flange board 166.667 –2.250 101.250 267.917

right board 166.667 –2.250 101.250 267.917


(a) Maximum shear force based on capacity of nails at s = 4 in.:

3

4

max 3

(12 in.)(2 in.)(4.75 in. 2 in./2) 90.000 in.

(2 nails)(175 lb/nail)87.5 lb/in.

4 in.

(87.5 lb/in.)(881.333 in. )856.852 lb

90.000 in.

Q

q

V Q q Iq V

I Q

Maximum shear force based on full shear strength of double tee shape:

At neutral axis:

3

max

4

maxmax max 3

2(2 in.)(7.25 in.)(7.25 in./2) 105.125 in.

(70 psi)(881.333 in. )(2 2 in.)2,347.428 lb

90.000 in.

Q

V Q I tV

I t Q




Comparison of two Vmax values:

The smaller of the two values computed above for Vmax gives the limiting shear force:

max 857 lbV Ans.

(b) Nail spacing necessary to develop the full shear strength of the section:

Consider the nailed portion of the beam (i.e., only the top flange) to establish the minimum required nail

spacing for Vmax = 2,347.428 lb, which is the shear force that produces a maximum horizontal shear

stress of 70 psi:

3

4

max

(2,347.428 lb)(90.000 in. )239.715 l

1.46

b/in.881.333 in.

(2 nails)(175 lb/nail)

239.715 lb/i0 n

ni .

.

f f

f f

V Qq

I

q s n V

n Vs

q Ans.




9.47 A box beam is fabricated from two plywood webs that are

secured to dimension lumber boards at its top and bottom flanges

(Fig. P9.47b). The beam supports a concentrated load of P = 5,000 lb

at the center of a 15-ft span (Fig. P9.47a). Bolts (⅜-in. diameter)

connect the plywood webs and the lumber flanges at a spacing of s =

12 in. along the span. Supports A and C can be idealized as a pin and

a roller, respectively. Determine:

(a) the maximum horizontal shear stress in the plywood webs.

(b) the average shear stress in the bolts.

(c) the maximum bending stress in the lumber flanges.


Solution



(in.4) (in.) (in.

4) (in.

4)

left web 576 0 0 576

top flange 16 10 1,200 1,216

bottom flange 16 –10 1,200 1,216

right web 576 0 0 576

Moment of inertia about the z axis (in.4) = 3,584

Maximum shear force:

For P = 5,000 lb, V = P/2 = 2,500 lb


For P = 5,000 lb, M = PL/4 = 18,750 lb-ft = 225,000 lb-in.

(a) Maximum horizontal shear stress (in plywood webs):

max

3

3

max 4

(3 in.)(4 in.)(10 in.)

2(0.5 in.)(12 in.)(6 in.) 192 in.

(2,500 lb)(192 in. )

(3,584 in. )(2133.

0.5 in.)9 psi

Q

VQ

I t Ans.




(b) Bolt shear stress:

Consider the dimension lumber boards that comprise the top flange.

3

flange

3flange

4

(3 in.)(4 in.)(10 in.) 120 in.

(2,500 lb)(120 in. )83.7054 lb/in.

3,584 in.

Q

VQq

I

Determine the force carried by one bolt:

(83.7054 lb/in.)(12 in.)

1,004.4643 lb/bolt1 bolt

f f

f

f

q s n V

q sV

n

The bolt cross-sectional area is:

2 2

bolt (0.375 in.) 0.110447 in.4

A

Each bolt acts in double shear; therefore, the shear stress in each bolt is:

bolt 2

1,004.4643 lb/bolt4,547.284 psi

2(0.110444,550 ps

)i

7 in. Ans.

(c) Maximum bending stress in lumber flanges:

4

(225,000 lb-in.)(12 in.)

3,584 in.753 psi

M c

I Ans.




9.48 A box beam is fabricated from two plywood webs that are secured to

dimension lumber boards at its top and bottom flanges (Fig. P9.48b). The

lumber has an allowable bending stress of 1,500 psi. The plywood has an

allowable shear stress of 300 psi. The ⅜-in. diameter bolts have an

allowable shear stress of 6,000 psi, and they are spaced at intervals of s = 9

in. The beam span is L = 15 ft (Fig. P9.48a). Support A can be assumed to

be pinned and support C can be idealized as a roller.

(a) Determine the maximum load P that can be applied to the beam at

midspan.

(b) Report the bending stress in the lumber, the shear stress in the

plywood, and the average shear stress in the bolts at the load P determined

in part (a).


Solution



(in.4) (in.) (in.

4) (in.

4)

left web 576 0 0 576

top flange 16 10 1,200 1,216

bottom flange 16 –10 1,200 1,216

right web 576 0 0 576

Moment of inertia about the z axis (in.4) = 3,584


V = P/2


M = PL/4

(a) Determine maximum load P:

Consider maximum bending stress:

4

1,500 psi

(1,500 psi)(3,584 in. )448,000 lb-in.

12 in.

M c

I

M

max

448,000 lb-in.4

4(448,000 lb-in.)9,956 lb

(15 ft)(12 in./ft)

PLM

P (a)




Consider maximum horizontal shear stress (in plywood webs):

3

4

max 3

max

max max

(3 in.)(4 in.)(10 in.) 2(0.5 in.)(12 in.)(6 in.) 192 in.

300 psi

(300 psi)(3,584 in. )(2 0.5 in.)5,600 lb

192 in.

2

2 2(5,600 lb) 11,200 lb

Q

VQ

I t

V

PV

P V (b)

Consider bolt shear stress:

The bolt cross-sectional area is:

2 2

bolt (0.375 in.) 0.110447 in.4

A

Each bolt acts in double shear; therefore, the maximum shear force that can be carried by one bolt is:

2

bolt 2(0.110447 in. )(6,000 psi) 1,325.364 lbV

Determine the shear flow that can be allowed based on the bolt shear stress:

(1 bolt)(1,325.364 lb/bolt)

147.263 lb/in.9 in.

f f

f f

q s n V

n Vq

s

Consider the dimension lumber boards that comprise the top flange.

3

flange

flange

4

max 3

flange

max

max max

(3 in.)(4 in.)(10 in.) 120 in.

(147.263 lb/in.)(3,584 in. )4,398.245 lb

120 in.

2

2 2(4,398.245 lb) 8,796 lb

Q

VQq

I

q IV

Q

PV

P V (c)

Compare the three values obtained for Pmax in Eqs. (a), (b), and (c) to find

max 8,796 l 8.80 ipb k sP Ans.




(b) Bending stress in lumber flanges for Pmax:

4

(8,796 lb)(15 ft)32,985 lb-ft

4 4

(32,985 lb-ft)(12 in.)(12 in./ft)

3,584 in.1,325 psi

PLM

M c

I Ans.

Maximum shear stress in plywood webs:

3

4

8,796 lb4,398 lb

2 2

(4,398 lb)(192 in. )

(3,584 in. )(2 0.5 23

in.)6 psi

PV

VQ

I t Ans.

Bolt shear stress:

bolt 6,000 psi Ans.




9.49 A wooden beam is fabricated from three boards, which are fastened together with screws, as shown

in Fig. P9.49b. The screws are uniformly spaced along the span of the beam at intervals of 150 mm (see

Fig. P9.49a). In service, the beam will be positioned so that bending occurs about the z axis. The

maximum bending moment in the beam is Mz = −4.50 kN-m, and the maximum shear force in the beam

is Vy = −2.25 kN. Determine:

(a) the magnitude of the maximum horizontal shear stress in the beam.

(b) the shear force in each screw.

(c) the magnitude of the maximum bending stress in the beam.


Solution



yi

(from bottom) yi Ai


3)

left board 40 180 7,200 90 648,000

bottom board 140 40 5,600 20 112,000

right board 40 180 7,200 90 648,000

20,000 1,408,000

3

2

1,408,000 mm70.4 mm (from bottom of shape to centroid)

20,000 mm


i i

i

y Ay

A



(mm4) (mm) (mm

4) (mm

4)

left board 19,440,000.00 19.60 2,765,952.00 22,205,952.00

bottom board 746,666.67 –50.40 14,224,896.00 14,971,562.67

right board 19,440,000.00 19.60 2,765,952.00 22,205,952.00



At neutral axis:

3

max

3

max 4

2(40 mm)(109.6 mm)(109.6 mm/2) 480,486.4 mm

(2,250 N)(480,486.4 mm )0.2276 MPa

(59,383,466.228 kP

67 mm )(2 0 m )a

4 m

Q

V Q

I t Ans.




(b) Shear force in each screw

Consider bottom board, which is held in place by two screws:

3

3

4

(140 mm)(40 mm)(70.4 mm 40 mm/2) 282,240 mm

(2,250 N)(282,240 mm )10.6939 N/mm

59,383,466.67 mm

(10.6939 N/mm)(150 mm)

2 screw802 N per e

s scr w

f f

f

f

Q

V Qq

I

q s n V

q sV

n Ans.


6

4

( 4.50 10 N-mm)(109.6 mm)

59,383,466.67 8.31 MPa (T)

mm

zx

z

M y

I Ans.




9.50 A wooden beam is fabricated by bolting together three

members, as shown in Fig. P9.50a. The cross-sectional

dimensions are shown in Fig. P9.50b. The 8-mm-diameter

bolts are spaced at intervals of s = 200 mm along the x axis

of the beam. If the internal shear force in the beam is V = 7

kN, determine the shear stress in each bolt.


Solution



yi

(from bottom) yi Ai


3)

left board 40 90 3,600 255 918,000

center board 40 300 12,000 150 1,800,000

right board 40 90 3,600 255 918,000

19,200 3,636,000

3

2


19,200 mm


i i

i

y Ay

A



(mm4) (mm) (mm

4) (mm

4)

left board 2,430,000 65.625 15,503,906.25 17,933,906.25

center board 90,000,000 -39.375 18,604,687.50 108,604,687.50

right board 2,430,000 65.625 15,503,906.25 17,933,906.25





Shear force in each bolt

Consider left board, which is held in place by the bolt:

3

3

4

(40 mm)(90 mm)(110.625 mm 90 mm/2) 236,250 mm

(7,000 N)(236,250 mm )11.4468 N/mm

144,472,500 mm

Q

V Qq

I

Note that this value of q is the shear flow that must be transmitted by one surface of the bolt cross

section. The cross-sectional area of the bolt is:

2 2

bolt (8 mm) 50.2655 mm4

A

Relate the shear flow and the bolt shear stress with Eq. (9.14):

2

(11.4468 N/mm)(200 mm)

(1 bolt surface)(50.45.5 M

2655 mmPa

)

f f f

f

f f

q s n A

q s

n A Ans.




9.51 A wooden beam is fabricated by bolting together three

members, as shown in Fig. P9.51a. The cross-sectional

dimensions are shown in Fig. P9.51b. The allowable shear stress

of the wood is 850 kPa, and the allowable shear stress of the 10-

mm-diameter bolts is 40 MPa. Determine:

(a) the maximum internal shear force V that the cross section can

withstand based on the allowable shear stress in the wood.

(b) the maximum bolt spacing s required to develop the internal

shear force computed in part (a).


Solution



yi

(from bottom) yi Ai


3)

left board 40 90 3,600 255 918,000

center board 40 300 12,000 150 1,800,000

right board 40 90 3,600 255 918,000

19,200 3,636,000

3

2


19,200 mm


i i

i

y Ay

A



(mm4) (mm) (mm

4) (mm

4)

left board 2,430,000 65.625 15,503,906.25 17,933,906.25

center board 90,000,000 -39.375 18,604,687.50 108,604,687.50

right board 2,430,000 65.625 15,503,906.25 17,933,906.25





Consider maximum horizontal shear stress:

3

2 4

max 3

(40 mm)(189.375 mm)(189.375 mm/2) 717,257.813 mm

850 kPa 0.850 MPa

(0.850 N/mm )(144,472,500 mm )(40 mm)6,848.4 N

717,26.85 k

57.8 3 mmN

1

Q

VQ

I t

V Ans.

Maximum bolt spacing

Consider left board, which is held in place by the bolt:

3

3

4

(40 mm)(90 mm)(110.625 mm 90 mm/2) 236,250 mm

(6,848.4 N)(236,250 mm )11.1989 N/mm

144,472,500 mm

Q

V Qq

I

Note that this value of q is the shear flow that must be transmitted by one surface of the bolt cross

section. The cross-sectional area of the bolt is:

2 2

bolt (10 mm) 78.5398 mm4

A


2 2(1 bolt surface)(40 N/mm )(78.5398 mm )

11.1989 N/28

m1 mm

m

f f f

f f f

q s n A

n As

q Ans.




9.52 A cantilever flexural member is fabricated by bolting two identical cold-

rolled steel channels back-to-back, as shown in Fig. P9.52a. The cantilever

beam has a span of L = 1,600 mm and supports a concentrated load of P = 600

N. The cross-sectional dimensions of the built-up shape are shown in Fig.

P9.52b. The effect of the rounded corners can be neglected in determining the

section properties for the built-up shape.

(a) If 4-mm-diameter bolts are installed at intervals of s = 75 mm, determine the

shear stress produced in the bolts.

(b) If the allowable average shear stress in the bolts is 96 MPa, determine the

minimum bolt diameter required if a spacing of s = 400 mm is used.

Fig. P9.52a

Fig. P9.52b

Solution

Centroid location in y direction for the upper channel shown in Figure 9.52b:


yi

(from z axis) yi Ai


3)

left element 3 40 120 20 2400

center element 59 3 177 1.5 265.5

right element 3 40 120 20 2400

417 5065.5

3

2

5,065.5 mm12.1475 mm

417 mm

i i

i

y Ay

A

Note: y is measured from the z axis to the centroid of the upper channel shown in Figure 9.52b.

Moment of inertia (both channels):

3 3

4(3 mm)(40 mm) (65 mm 2(3 mm))(3 mm)2 2 257,062 mm

3 3I

Shear flow:

2 3

3

4

(12.1475 mm)(417 mm ) 5,065.51 mm

(600 N)(5,065.51 mm )11.8232 N/mm

257,062 mm

Q

VQq

I

Bolt area:

2 2

bolt (4 mm) 12.5664 mm4

A




(a) Bolt shear stress:

2

(11.8232 N/mm)(75 mm)

(1 bolt surface)(12.70.6 M

5664 mmPa

)

f f f

f

f f

q s n A

q s

n A Ans.

(b) Minimum bolt diameter for s = 400 mm:

2

2

2 2

bolt

2

bolt

(11.8232 N/mm)(400 mm)49.2633 mm

(1 bolt surface)(96 N/mm )

49.2633 mm4

4(49.2633 m7.92 m

mm

)

f f f

f

f f

q s n A

q sA

n

A D

D Ans.




9.53 A W310 × 60 steel beam (see Appendix B) in an existing

structure is to be strengthened by adding a 250 mm wide by 16

mm thick cover plate to its lower flange, as shown in Fig.

P9.53. The cover plate is attached to the lower flange by pairs

of 24-mm-diameter bolts spaced at intervals of s along the

beam span. Bending occurs about the z centroidal axis.

(a) If the allowable bolt shear stress is 96 MPa, determine the

maximum bolt spacing interval s required to support an internal

shear force in the beam of V = 50 kN.

(b) If the allowable bending stress is 150 MPa, determine the

allowable bending moment for the existing W310 × 60 shape,

the allowable bending moment for the W310 × 60 with the

added cover plate, and the percentage increase in moment

capacity that is gained by adding the cover plate.

Fig. P9.53

Solution



yi

(from bottom) yi Ai


3)

W310 × 60 - - 7,550 167 1,260,850

cover plate 250 16 4,000 8 32,000

11,550 1,292,850

3

2


11,550 mm


i i

i

y Ay

A



(mm4) (mm) (mm

4) (mm

4)

W310 × 60 128,000,000 55.065 22,892,764 150,892,764

cover plate 85,333.33 –103.935 43,209,937 43,295,270


(a) Maximum bolt spacing

Consider the cover plate, which is connected to the W310 × 60 shape with two bolts:

3

3

4

(250 mm)(16 mm)(111.935 mm 16 mm/2) 415,740 mm

(50,000 N)(415,740 mm )107.0457 N/mm

194,188,035 mm

Q

V Qq

I

The cross-sectional area of a 24-mm-diameter bolt is:

2 2

bolt (24 mm) 452.389 mm4

A


2 2(2 bolts)(96 N/mm )(452.389 mm )

107.0457 N/mm811 mm

f f f

f f f

q s n A

n As

q Ans.




(b) Allowable bending moment for W310 × 60 shape (without cover plate):

2 4

allow

150 MPa

(150 N/mm )(128,000,000 mm )127,152, 127318 N-mm

(3.2 kN-m

02 mm/2)

M c

I

M Ans.

Allowable bending moment for W310 × 60 shape (with cover plate):

2 4

allow

150 MPa

(150 N/mm )(194,188,035 mm )141,35 141.4 4,452 N-mm

(206.065 m-

m)kN m

M c

I

M Ans.

Percentage increase in moment capacity:

141,354,452 N-mm 127,152,318 N-mm

% increase (100%)127,152,318 N-mm

11.17% Ans.




9.54 A W410 × 60 steel beam (see Appendix B) is simply supported at its ends and carries a

concentrated load P at the center of a 7-m span. The W410 × 60 shape will be strengthened by adding

two 250 mm wide by 16 mm thick cover plate to its flanges, as shown in Fig. P9.54. Each cover plate is

attached to its flange by pairs of 20-mm-diameter bolts spaced at intervals of s along the beam span. The

allowable bending stress is 150 MPa, the allowable average shear stress in the bolts is 96 MPa, and

bending occurs about the z centroidal axis.

(a) Based on the 150 MPa allowable bending stress, determine the

maximum concentrated load P that may be applied at the center of

a 7-m span for a W410 × 60 steel beam with two cover plates.

(b) For the internal shear force V associated with the concentrated

load P determined in part (a), compute the maximum spacing

interval s required for the bolts that attach the cover plates to the

flanges.

Fig. P9.54

Solution

Moment of inertia about the z axis (with cover plates):


(mm4) (mm) (mm

4) (mm

4)

top cover plate 85,333.33 211 178,084,000 178,169,333

W410 × 60 216,000,000 0 0 216,000,000

bottom cover plate 85,333.33 –211 178,084,000 178,169,333



V = P/2


M = PL/4




Consider W410 × 60 with two cover plates:

150 MPaM c

I

2 3

allow

(150 N/mm )(572,338,666 mm )392,012,785 N-mm 392.013 kN-m

(438 mm/2)M

max

392.013 kN-m4

4(392.013 kN-m)224.007 kN

7 m224 kN

PLM

P Ans.

(b) Maximum bolt spacing

Consider top cover plate, which is held in place by two bolts:

3

3

4

(250 mm)(16 mm)(406 mm/2 16 mm/2) 844,000 mm

(223,418 N/2)(844,000 mm )165.166 N/mm

572,338,666 mm

Q

V Qq

I

Note that this value of q is the shear flow that must be transmitted across two bolt surfaces. The cross-

sectional area of the bolt is:

2 2

bolt (20 mm) 314.159 mm4

A


2 2(2 bolt surfaces)(96 N/mm )(314.159 mm )

165.166 N365 m

m m

/m

f f f

f f f

q s n A

n As

q Ans.




9.55 A W410 × 60 steel beam (see Appendix B) is simply

supported at its ends and carries a concentrated load of P = 420 kN

at the center of a 7-m span. The W410 × 60 shape will be

strengthened by adding two 250 mm wide by 16 mm thick cover

plate to its flanges, as shown in Fig. P9.55. Each cover plate is

attached to its flange by pairs of bolts spaced at intervals of s = 250

mm along the beam span. The allowable average shear stress in the

bolts is 96 MPa, and bending occurs about the z centroidal axis.

Determine the minimum required diameter for the bolts.

Fig. P9.55

Solution

Moment of inertia about the z axis (with cover plates):


(mm4) (mm) (mm

4) (mm

4)

top cover plate 85,333.33 211 178,084,000 178,169,333

W410 × 60 216,000,000 0 0 216,000,000

bottom cover plate 85,333.33 –211 178,084,000 178,169,333



V = P/2 = 420 kN/2 = 210 kN

Minimum bolt diameter

Consider top cover plate, which is held in place by two bolts:

3

3

4

(250 mm)(16 mm)(406 mm/2 16 mm/2) 844,000 mm

(210,000 N)(844,000 mm )309.677 N/mm

572,338,666 mm

Q

V Qq

I




Relate the shear flow and the required bolt area with Eq. (9.14). Note that the shear flow will be

transmitted by means of two fasteners.

2

2

(309.677 N/mm)(250 mm)403.225 mm

2(96 N/mm )

f f f

f

f f

q s n A

q sA

n

Use the minimum required cross-sectional area of the bolt to calculate the minimum bolt diameter:

2 2

bolt

min

403.225 m

22.7 mm

m4

22.658 mm

A D

D Ans.




9.56 A W310 × 60 steel beam (see Appendix B) has a C250 × 45

channel bolted to the top flange, as shown in Fig. P9.56. The beam is

simply supported at its ends and carries a concentrated load of 100 kN at

the center of a 6-m span. Pairs of 24-mm-diameter bolts are spaced at

intervals of s along the beam. If the allowable average shear stress in the

bolts must be limited to 125 MPa, determine the maximum spacing

interval s for the bolts.

Fig. P9.56

Solution


Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

W310 × 60 7,550 151 1,140,050

C250 × 45 5,680 302 + 17.1 – 16.5 = 302.6 1,718,768

13,230 2,858,818

3

2


13,230 mm


i i

i

y Ay

A



(mm4) (mm) (mm

4) (mm

4)

W310 × 60 128,000,000 –65.086 31,983,215 159,983,315

C250 × 45 1,640,000 86.5140 42,512,938 44,152,938



V = P/2 = 100 kN/2 = 50 kN

Shear flow through the bolts

Consider the C250 × 45 shape, which is connected to the

W310 × 60 shape with two bolts:

2

3

3

4

(5,680 mm )(302.6 mm 216.086 mm)

491,399.5 mm

(50,000 N)(491,399.5 mm )

204,136,153 mm

120.361 N/mm

Q

V Qq

I

The cross-sectional area of a single 24-mm-diameter bolt is:




2 2

bolt (24 mm) 452.389 mm4

A

Use Eq. (9.13) to determine the maximum spacing s:

2 2(2 bolts)(125 N/mm )(452.389 mm )

939.653 mm120.361 N/m

940 mmm

f f f

f f f

q s n A

n As

q Ans.




9.57 A W310 × 60 steel beam (see Appendix B) has a C250 × 45

channel bolted to the top flange, as shown in Fig. P9.57. The beam is

simply supported at its ends and carries a concentrated load of 90 kN at

the center of an 8-m span. If pairs of bolts are spaced at 600-mm

intervals along the beam, determine:

(a) the shear force carried by each of the bolts.

(b) the bolt diameter required if the average shear stress in the bolts must

be limited to 75 MPa.

Fig. P9.57

Solution


Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

W310 × 60 7,550 151 1,140,050

C250 × 45 5,680 302 + 17.1 – 16.5 = 302.6 1,718,768

13,230 2,858,818

3

2


13,230 mm


i i

i

y Ay

A



(mm4) (mm) (mm

4) (mm

4)

W310 × 60 128,000,000 –65.086 31,983,215 159,983,315

C250 × 45 1,640,000 86.5140 42,512,938 44,152,938



V = P/2 = 90 kN/2 = 45 kN

(a) Shear force in each bolt

Consider the C250 × 45 shape, which is connected to the

W310 × 60 shape with two bolts:

2

3

3

4

(5,680 mm )(302.6 mm 216.086 mm)

491,399.5 mm

(45,000 N)(491,399.5 mm )

204,136,153 mm

108.325 N/mm

Q

V Qq

I

Relate the shear flow and the bolt shear force with Eq. (9.13):




(108.325 N/mm)(600 mm)

32,497 N2 bolts

32.5 kN per bolt

f f

f

f

q s n V

q sV

n Ans.

(b) Required bolt diameter

2bolt boltbolt 2

bolt allow

2 2

bolt bolt

2

bolt

32,497 N433.299 mm

75 N/mm

433.299

23.5

mm4

4(433.299m

m )m

m

V VA

A

A D

D Ans.




10.1 For the loading shown, use the double-

integration method to determine (a) the

equation of the elastic curve for the

cantilever beam, (b) the deflection at the free

end, and (c) the slope at the free end.

Assume that EI is constant for each beam.

Fig. P10.1

Solution

Integration of moment equation:

2

02( )

d vEI M x M

dx

0 1

dvEI M x C

dx (a)

2

01 2

2

M xEI v C x C (b)

Boundary conditions:

0 at 0

0 at 0

dvx

dx

v x

Evaluate constants:

From Eq. (a), C1 = 0. From Eq. (b), C2 = 0

(a) Elastic curve equation:

0

2

0

2

2 2

M xEI v v

M x

EI Ans.

(b) Deflection at the free end:

2

0

2

0( )

22B

M Lv

E

M L

EI I Ans.

(c) Slope at the free end:

0 0( )B

B

dv M L

dx EI

M L

EI Ans.






equation of the elastic curve for the

cantilever beam, (b) the deflection at the free

end, and (c) the slope at the free end.

Assume that EI is constant for each beam.

Fig. P10.2

Solution


2 2

2( )

2

d v wxEI M x

dx

3

16

dv wxEI C

dx (a)

4

1 224

wxEI v C x C (b)


0 at

0 at

dvx L

dx

v x L

Evaluate constants:

Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1:

3 3

1 1

( )(0)

6 6

w L wLEI C C

Substitute x = L and v = 0 into Eq. (b) to determine C2:

4 4 4 4

1 2 2 2

( )(0) ( )

24 24 6 8

w L wL wL wLEI C L C C C


4 3

4 3 44

24 3

24 6 8 4

wx wL x wLEI v v

wx L x L

EI Ans.


4

4 3 443

(0) 4 (0) 324 24 8

A

w wLv

wL L

EI E

L

EII Ans.


3 3 3(0

6 6

)

6A

A

dv w wL

dx EI EI

wL

EI Ans.






equation of the elastic curve for the cantilever

beam, (b) the deflection at the free end, and

(c) the slope at the free end. Assume that EI is

constant for each beam.

Fig. P10.3

Solution


2 3

0

2( )

6

d v w xEI M x

dx L

4

01

24

dv w xEI C

dx L (a)

5

01 2

120

w xEI v C x C

L (b)


0 at

0 at

dvx L

dx

v x L

Evaluate constants:


4 3

0 01 1

( )(0)

24 24

w L w LEI C C

L


5 5 3

0 0 01 2 2

4 4 4

0 0 02

( )(0) ( ) ( )

120 120 24

120 24 30

w L w L w LEI C L C L C

L L

w L w L w LC


5 3

0 50

4

0 0 4 5

120 24 305 4

120

wx L x L

L E

w x w L w LEI v

L Ix v Ans.


5 4 50

4

0(0) 5 (0) 412 30 0

A

w L

E

wv L L

L E II Ans.


34 3

0 0 0(0)

24 24 24A

A

dv w w L

dx L E

w L

EI II E Ans.




10.4 For the beam and loading shown in

Fig. P10.4, use the double-integration

method to determine (a) the equation of

the elastic curve for segment AB of the

beam, (b) the deflection at B, and (c) the

slope at A. Assume that EI is constant for

the beam.

Fig. P10.4

Solution


2

2( )

2

d v PEI M x x

dx

2

14

dv PxEI C

dx (a)

3

1 212

PxEI v C x C (b)


0 at 0

0 at2

v x

dv Lx

dx

Evaluate constants:

Substitute x = L/2 and dv/dx = 0 into Eq. (a) to determine C1:

2 2

1 1

( / 2)(0)

4 16

P L PLEI C C

Substitute x = 0 and v = 0 into Eq. (b) to determine C2:

3 2

2 2

(0) (0)(0) 0

12 16

P PLEI C C


3

2

2

23 44

(0 )2812 16

P P xL x

x PL x L

EIEI v v x Ans.

(b) Deflection at B:

2

23( / 2)

3 448 2 48

B

P L L

I

PL

Lv

E EI Ans.

(c) Slope at A:

22 2(0)

4 16 61A

A

dv P PL

dx E

PL

EI II E Ans.







the elastic curve for the beam, (b) the

slope at A, (c) the slope at B, and (d) the

deflection at midspan. Assume that EI is

constant for the beam. Fig. P10.5

Solution

Beam FBD:

0

0 0

0

0

and

y y y

y y

A y

y y

F A B

A B

M B L M

M MB A

L L

Moment equation:

00 0

00

( ) ( ) 0

( )

a a y

MM M x A x M M x x M

L

M xM x M

L


2

002

( )d v M x

EI M x Mdx L

2

00 1

2

dv M xEI M x C

dx L (a)

2 3

0 01 2

2 6

M x M xEI v C x C

L (b)


0 at 0

0 at

v x

v x L

Evaluate constants:


2 3

0 01 2 2

(0) (0)(0) (0) 0

2 6

M MEI C C C

L





2 3

0 01

0 0 01

( ) ( )(0) ( )

2 6

6 2 3

M L M LEI C L

L

M L M L M LC


0 2

3

20

2

0 0

23 2

66 3

M x M x M LxEI v v

L

M xx Lx L

L EI Ans.

(b) Slope at A:

2

0 0 00

(0)(0)

2 33A

A

dv M M LM

dx L EI E

M L

EI I Ans.

(c) Slope at B:

2

0 0 0 0 0( ) ( )6 3 2

2 3 66B

B

dv M L M L M L ML L L

dx EI L EI

M L

EIEI EI Ans.

(d) Deflection at midspan:

20 0

2

/ 2

2( / 2

1

)3 2

6 2 62x L

M L L Lv L

M L

EL

L I IE Ans.




10.6 For the beam and loading shown in Fig.

P10.6, use the double-integration method to

determine (a) the equation of the elastic curve for

the beam, (b) the maximum deflection, and (c) the

slope at A. Assume that EI is constant for the

beam. Fig. P10.6

Solution

Moment equation:

2

2

( ) 02 2

( )2 2

a a

wLx wxM M x

wx wLxM x


2 2

2( )

2 2

d v wx wLxEI M x

dx

3 2

16 4

dv wx wLxEI C

dx (a)

4 3

1 224 12

wx wLxEI v C x C (b)


0 at 0

0 at

v x

v x L

Evaluate constants:


4 3

1 2 2

(0) (0)(0) (0) 0

24 12

w wLEI C C C


4 3 4 4 3

1 1

( ) ( ) ( ) ( )(0) ( )

24 12 24 12 24

w L wL L w L w L wLEI C L C

L L


4 3

3 23

3224 2 2 42 1 4

wx wLx wL wxxEI v v x Lx L

EI Ans.

(b) Maximum deflection: At x = L/2:

3 2 3 32 2

4

max

( / 2)2

5

3824 2 2 48 2 48

w L L L wL L Lv L L L

EI EI

wL

EI Ans.

(c) Slope at A:

3 32 3(0) (0)

6 4 2 24 4A

A

dv w wL wL

dx EI EI

wL

EEI I Ans.







the elastic curve for segment AB of the

beam, (b) the deflection midway between

the two supports, (c) the slope at A, and

(d) the slope at B. Assume that EI is

constant for the beam. Fig. P10.7

Solution

Beam FBD:

0

30

2

3and

2 2

y y y

A y

y y

F A B P

LM B L P

P PB A

Moment equation:

( ) 0 ( )2 2

a a

P PxM M x x M x


2

2( )

2

d v PxEI M x

dx

2

14

dv PxEI C

dx (a)

3

1 212

PxEI v C x C (b)


0 at 0

0 at

v x

v x L

Evaluate constants:


3

1 2 2

(0)(0) (0) 0

12

PEI C C C


3 2

1 1

( )(0) ( )

12 12

P L PLEI C L C




(a) Elastic curve equation for segment AB of the beam:

3

22

2

12 1 22 1

Px PL x Px

II v L xE v

E Ans.

(b) Deflection at midspan:

2

2

2

3

/3

( / 2) 3

12 2 24 4 2x L

P PL

E

L L PL Lv L

I EI IE Ans.

(c) Slope at A:

22 2

12

(0)

4 12A

A

dv P PL

dx EI

PL

EI EI Ans.

(d) Slope at B:

22 2(

1 6

)

4 2B

B

dv P L PL

dx EI

P

EIEI

L Ans.






determine (a) the equation of the elastic curve

for segment BC of the beam, (b) the

deflection midway between B and C, and (c)

the slope at C. Assume that EI is constant for

the beam.

Fig. P10.8

Solution

Beam FBD:

(4 ) (5 ) 0

2 0

B y

y

y y y

y

M PL C L P L

C P

F B C P

B P

Moment equation:

( ) ( ) ( ) ( ) 0

( )

a a yM M x B x P L x M x Px P L x

M x PL


2

2( )

d vEI M x PL

dx

1

dvEI PLx C

dx (a)

2

1 22

PLxEI v C x C (b)


0 at 0

0 at 4

v x

v x L

Evaluate constants:


2

1 2 2

(0)(0) (0) 0

2

PLEI C C C

Substitute x = 4L and v = 0 into Eq. (b) to determine C1:

2

2

1 1

(4 )(0) (4 ) 2

2

PL LEI C L C PL




(a) Elastic curve equation for segment BC of the beam:

2

22 422

PLxEI v PL x

Pxv

LxL

EI Ans.


3

2

(2 )4 ( )

22

2x L

PL Lv L L

EI

PL

EI Ans.

(c) Slope at C:

2 2(4 ) 2 2

C

C

dv PL L PL

dx E

P

I EIEI

L Ans.







segment AB of the beam, (b) the deflection

midway between A and B, and (c) the slope at B.

Assume that EI is constant for the beam.

Fig. P10.9

Solution

Beam FBD:

2 50

2 4

5

2 4

0

2 4

A y

y

y y y

y

wL LM P B L

wL PB

F A B wL P

wL PA

Moment equation:

2 2

2

( ) ( ) 02 2 2 4

( )2 2 4

a a y

wx wx wLx PxM M x A x M x

wx wLx PxM x


2 2

2( )

2 2 4

d v wx wLx PxEI M x

dx

3 2 2

16 4 8

dv wx wLx PxEI C

dx (a)

4 3 3

1 224 12 24

wx wLx PxEI v C x C (b)


0 at 0

0 at

v x

v x L

Evaluate constants:


4 3 3

1 2 2

(0) (0) (0)(0) (0) 0

24 12 24

w wL PEI C C C


4 3 3 3 2

1 1

( ) ( ) ( )(0) ( )

24 12 24 24 24

w L wL L P L wL PLEI C L C




(a) Elastic curve equation for segment AB of the beam:

4 3 3

3 3 2

2

2

3

2

24 12 24 24

224 24

24

wx wLx Px wL x PL xEI v

vwx Px

x Lx L x LEI EI

Ans.


4

3 2 2

3 2

/ 2

3

( / 2) ( / 2)2

24 2 2

5

38 6

24 2

4 4

x L

w L L L P L Lv L L L

EI E

wL PL

EI E

I

I Ans.

(c) Slope at B:

3 2 2 3 2 3 2( ) ( ) ( )

6 4 8 24 24 1224B

B

dv w L wL L P L wL PL

dx EI EI EI E

wL PL

EI EEI II Ans.







segment AC of the beam, (b) the deflection at B,

and (c) the slope at A. Assume that EI is constant

for the beam. Fig. P10.10

Solution

Beam FBD:

3(3 ) (2 ) 0

2

9

4

(3 ) 0

9 33

4 4

A y

y

y y y

y

LM w L C L

wLC

F A C w L

wL wLA wL

Moment equation:

2

3( ) ( ) 0

2 4 2

3( )

2 4

a a y

x wL xM M x A x wx M x x wx

wx wLxM x


2 2

2

3( )

2 4

d v wx wLxEI M x

dx

3 2

1

3

6 8

dv wx wLxEI C

dx (a)

4 3

1 2

3

24 24



0 at 0

0 at 2

v x

v x L

Evaluate constants:


4 3

1 2 2

(0) 3 (0)(0) (0) 0

24 24

w wLEI C C C

Substitute x = 2L and v = 0 into Eq. (b) to determine C1:

4 3

1

3 3 3

1

(2 ) 3 (2 )(0) (2 )

24 24

8 12

24 24 6

w L wL LEI C L

wL wL wLC




(a) Elastic curve equation for segment AC of the beam:

4 3 3

3 2 3

3 2 3

33 4

24 24 6 24

3 424

wx wLx wL x wxEI v x Lx

wxx Lx L

EI

L

v Ans.


3 2 3

4( )( ) ) 4

4 123 (

2B

w Lv L L L

wL

EI

L

EI Ans.

(c) Slope at A:

3 32 3(0) 3 (0)

6 8 6 6A

A

dv w wL wL

dx EI EI

wL

EIEI Ans.




10.11 For the simply supported steel beam [E =

200 GPa; I = 129 × 106 mm

4] shown in Fig.


determine the deflection at B. Assume L = 4 m, P

= 60 kN, and w = 40 kN/m.

Fig. P10.11

Solution

Beam FBD:

( ) 02 2

2 2

( ) 0

2 2

A y

y

y y y

y

L LM wL P C L

wL PC

F A C w L P

wL PA

Moment equation: 2 2

2

( ) ( ) 02 2 2 2

( )2 2 2

a a y

wx wx wLx PxM M x A x M x

wx wLx PxM x


2 2

2( )

2 2 2

d v wx wLx PxEI M x

dx

3 2 2

16 4 4

dv wx wLx PxEI C

dx (a)

4 3 3

1 224 12 12

wx wLx PxEI v C x C (b)


0 at 0

0 at2

v x

dv Lx

dx

Evaluate constants:

Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = L/2 and dv/dx = 0 into

Eq. (b) to determine C1:

3 2 2

1

3 3 2 3 2

1

( / 2) ( / 2) ( / 2)(0)

6 4 4

48 16 16 24 16

w L wL L P LEI C

wL wL PL wL PLC




Elastic curve equation:

4 3 3 3 2

3 2 3 2 2

24 12 12 24 16

2 3 424 48

wx wLx Px wL x PL xEI v

wx Pxv x Lx L L x

EI EI

Deflection at B: At x = L/2:

4 35

384 48B

wL PLv

EI EI

Let E = 200 GPa, I = 129 × 106 mm

4, w = 40 kN/m, P = 60 kN, and L = 4 m.

4 3

2 6 4 2 6 4

5(40 N/mm)(4,000 mm) (60,000 N)(4,000 mm)

384(200,000 N/mm )(129 10 mm ) 48(200,000 N/mm )(129 10 mm )

5.1680 mm 3.1008 mm

8.27 mm

Bv

Ans.




10.12 For the cantilever steel beam [E = 200 GPa;

I = 129 × 106 mm

4] shown in Fig. P10.12, use the

double-integration method to determine the

deflection at A. Assume L = 2.5 m, P = 50 kN, and

w = 30 kN/m.

Fig. P10.12

Solution

Moment equation:

2 2

( ) 0 ( )2 2

a a

wx wxM M x Px M x Px


2 2

2( )

2

d v wxEI M x Px

dx

3 2

16 2

dv wx PxEI C

dx (a)

4 3

1 224 6

wx PxEI v C x C (b)


0 at

0 at

v x L

dvx L

dx

Evaluate constants:


3 2 3 2

1 1

( ) ( )(0)

6 2 6 2

w L P L wL PLEI C C


4 3 3 2 4 4 3 3

2 2

4 3

2

( ) ( )(0) ( ) ( )

24 6 6 2 24 6 6 2

8 3

w L P L wL PL wL wL PL PLEI L L C C

wL PLC


4 3 4 3 2 3

4 3 4 3 2 3

24 6 8 6 2 3

4 3 3 224 6

wx wL x wL Px PL x PLEI v

w Pv x L x L x L x L

EI EI

Deflection at A:

4 3

4 3 4 3 2 3 3(0) 4 (0) 3 (0) 3 (0) 2

24 6 24 3A

w P wL PLv L L L L

EI EI EI EI




Let E = 200 GPa, I = 129 × 106 mm

4, w = 30 kN/m, P = 50 kN, and L = 2.5 m.

4 3

2 6 4 2 6 4

3(30 N/mm)(2,500 mm) (50,000 N)(2,500 mm)

24(200,000 N/mm )(129 10 mm ) 3(200,000 N/mm )(129 10 mm )

5.6777 mm 10.0937 mm

= 15.77 mm

Av

Ans.





I = 129 × 106 mm



deflection at B. Assume L = 3 m, M0 = 70 kN-m,

and w = 15 kN/m.

Fig. P10.13

Solution

Moment equation:

2

0

2

0

( )( ) 0

2

( )( )

2

a a

w L xM M x M

w L xM x M


2 2 2 2

2 2

0 0 02

( )( ) 2

2 2 2 2

d v w L x w wL wxEI M x M L Lx x M wLx M

dx

2 2 3

0 12 2 6

dv wL x wLx wxEI M x C

dx (a)

2 2 3 4 2

01 2

4 6 24 2

wL x wLx wx M xEI v C x C (b)


0 at 0

0 at 0

v x

dvx

dx

Evaluate constants:

Substitute x = 0 and dv/dx = 0 into Eq. (a) to determine C1 = 0. Next, substitute x = 0 and v = 0 into Eq.

(b) to determine C2 = 0.


2 2 3 4 2

0

24 3 2 2 0

4 6 24 2

4 624 2

wL x wLx wx M xEI v

w M xv x Lx L x

EI EI

Deflection at B:

2 4 2

4 3 2 2 0 0( )( ) 4 ( ) 6 ( )

24 2 8 2B

w M L wL M Lv L L L L L

EI EI EI EI




Let E = 200 GPa, I = 129 × 106 mm

4, w = 15 kN/m, M0 = 70 kN-m, and L = 3 m.

4 2

2 6 4 2 6 4

(15 N/mm)(3,000 mm) (70 kN-m)(1,000 N/kN)(1,000 mm/m)(3,000 mm)

8(200,000 N/mm )(129 10 mm ) 2(200,000 N/mm )(129 10 mm )

5.8866 mm 12.2093 mm

= 18.10 mm

Bv

Ans.





I = 129 × 106 mm



deflection at A. Assume L = 2.5 m, P = 50 kN-m,

and w0 = 90 kN/m.

Fig. P10.14

Solution

Moment equation:

0

3

0

( ) ( ) 02 3

( )6

a a

w x xM M x x Px

L

w xM x Px

L


2 3

0

2( )

6

d v w xEI M x Px

dx L

4 2

01

24 2

dv w x PxEI C

dx L (a)

5 3

01 2

120 6

w x PxEI v C x C

L (b)


0 at

0 at

v x L

dvx L

dx

Evaluate constants:


4 2 3 2

0 01 1

( ) ( )(0)

24 2 24 2

w L P L w L PLEI C C

L


5 3 3 2 4 4 3 3

0 0 0 02 2

4 3

02

( ) ( )(0) ( ) ( )

120 6 24 2 120 24 6 2

30 3

w L P L w L PL w L w L PL PLEI L L C C

L

w L PLC


5 3 3 2 4 3

0 0 0

5 4 5 3 2 30

120 6 24 2 30 3

5 4 3 2120 6

w x Px w L PL w L PLEI v x x

L

w Pv x L x L x L x L

L EI EI




Deflection at A: Let E = 200 GPa, I = 129 × 10

6 mm

4, w0 = 90 kN/m, P = 50 kN, and L = 2.5 m.

5 4 5 3 2 30

4 3

0

4 3

2 6 4 2 6 4

(0) 5 (0) 4 (0) 3 (0) 2120 6

30 3

(90 N/mm)(2,500 mm) (50,000 N)(2,500 mm)

30(200,000 N/mm )(129 10 mm ) 3(200,000 N/mm )(129 10 mm )

4.5422 mm 10.0937

1

mm

=

A

w Pv L L L L

L EI EI

w L PL

EI EI

4.64 mm Ans.







the cantilever beam AB, (b) the deflection at the

free end, and (c) the slope at the free end. Assume

that EI is constant for each beam.

Fig. P10.15

Solution

Beam FBD:

0

2

0

0

0

20

2 3

3

02

2

A A

A

y y

y

w L LM M

w LM

w LF A

w LA

Moment equation:

0

2

0 0 0

3 2

0 0 0

( ) ( )2 3

( ) ( ) ( ) 03 2 3 2

( )6 2 3

a a A y

w x xM M x M x A x

L

w L w x x w LM x x x

L

w x w Lx w LM x

L


2 3 2

0 0 0

2( )

6 2 3

d v w x w Lx w LEI M x

dx L

4 2 2

0 0 01

24 4 3

dv w x w Lx w L xEI C

dx L (a)

5 3 2 2

0 0 01 2

120 12 6

w x w Lx w L xEI v C x C

L (b)


0 at 0

0 at 0

v x

dvx

dx

Evaluate constants:

Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq.






5 3 2 2

30

0

3

0

22

0

120 1

10 2012

2 6

0

w x w Lx w L xEI v

L

vw x

x L x LL EI

Ans.


5 2 34

03 20 ( ) 10 ( ) 20 (11

12)

120 0B

wv L L L L L

L EI

w L

EI Ans.


4 2 2 3 3 3

0 0 0 0 0 0

3

0( ) ( ) ( ) 6 8

24 4 3 2 4 84 2 24B

B

dv w L w L w L

EI

L w L L w L w L w L

dx L Ans.







the cantilever beam AB, (b) the deflection at the

free end, and (c) the slope at the free end. Assume

that EI is constant for each beam.

Fig. P10.16

Solution

Beam FBD:

0

2

0

0

0

02 3

6

02

2

A A

A

y y

y

w L LM M

w LM

w LF A

w LA

Moment equation:

3

0 ( )( ) 0

2 3a a

w L xM M x

L

30

3 2 2 30

2 2 3

0 0 0 0

( ) ( )6

( 3 3 )6

6 2 2 6

wM x L x

L

wL L x Lx x

L

w L w Lx w x w x

L


2 3 2 2

0 0 0 0

2( )

6 2 2 6

d v w x w x w Lx w LEI M x

dx L

4 3 2 2

0 0 0 01

24 6 4 6

dv w x w x w Lx w L xEI C

dx L (a)

5 4 3 2 2

0 0 0 01 2

120 24 12 12

w x w x w Lx w L xEI v C x C

L (b)


0 at 0

0 at 0

v x

dvx

dx




Evaluate constants:




5 4 3 2 2

23 2 2 3

0 0 0

0

0

120 24 12

5 10 101

2

2

1

0

w

w xx Lx L x L

L

x w x w Lx

E

w LI v

I

xE

L

v Ans.


5

5 4 2 3 3 20 0

4

04( ) 5 ( ) 10 ( ) 10 ( )

120 3120 0B

w w Lv L L L L L L L

L E

w

IEI EI

L

L Ans.


4 3 2 2

0 0 0 0

3

0( ) ( ) ( ) ( )

24 6 244 6B

B

dv w L w L w L L w L L

dx L EI EI EI EI

w L

EI Ans.







the cantilever beam, (b) the deflection at B, (c) the

deflection at the free end, and (d) the slope at the

free end. Assume that EI is constant for the beam.

Fig. P10.17

Solution

Beam FBD:

2

02 2 4

3

8

02

2

A A

A

y y

y

wL L LM M

wLM

wLF A

wLA

Consider beam segment AB (0 ≤ x ≤ L/2)

Moment equation:

2

2

3( ) ( ) 0

8 2

3( )

8 2

a a A y

wL wLM M x M A x M x x

wL wLxM x


2 2

2

3( )

8 2

d v wL wLxEI M x

dx

2 2

1

3

8 4

dv wL x wLxEI C

dx (a)

2 2 3

1 2

3

16 12

wL x wLxEI v C x C (b)


0 at 0

0 at 0

v x

dvx

dx

Evaluate constants:






Elastic curve equation for beam segment AB:

2 2 3

2

3

16 12

9 4 (0 / 2)48

wL x wLxEI v

wLxv L x x L

EI

Slope at B: Let x = L/2

2 2 33 ( / 2) ( / 2)

8 4 8B

B

dv wL L wL L wL

dx EI EI EI

Deflection at B: Let x = L/2

2 4( / 2) 7

9 448 2 192

B

wL L L wLv L

EI EI

Consider beam segment BC (L/2 ≤ x ≤ L)

Moment equation:

2

22

( )2 2

3( ) 0

8 2 2 2

b b A y

w LM M x M A x x

wL wL w LM x x x

2 2

2 2

3( )

2 2 2 8

2 2

w L wL wLM x x x

wx wLwLx


2 2 2

2( )

2 2

d v wx wLEI M x wLx

dx

3 2 2

36 2 2

dv wx wLx wL xEI C

dx (c)

4 3 2 2

3 424 6 4

wx wLx wL xEI v C x C (d)

Continuity conditions:

4

3

7at

192 2

at8 2

wL Lv x

EI

dv wL Lx

dx EI




Evaluate constants:

Substitute the slope continuity condition into Eq. (c) for x = L/2 and solve for C3:

3 2 2 3

3

3

3

( / 2) ( / 2) ( / 2)

6 2 2 8

48

dv w L wL L wL L wLEI C

dx

wLC

Next, substitute the deflection continuity condition into Eq. (d) for x = L/2 and solve for C4

4 3 2 2 3 4

4

4

4

( / 2) ( / 2) ( / 2) 7( / 2)

24 6 4 48 192

384

w L wL L wL L wL wLEI v L C

wLC

Elastic curve equation for beam segment BC:

4 3 2 2 3 4

4 3 2 2 3 4

24 6 4 48 384

16 64 96 8 ( / 2 )384

wx wLx wL x wL x wLEI v

wv x Lx L x L x L L x L

EI

(a) Elastic curve equations for entire beam:

2

9 4 (0 / 2)48

wLxL x xv L

EI Ans.

4 3 2 2 3 416 64 96 8 ( / 2 )384

wx Lx L x L x L L x L

EIv Ans.


47

192B

wLv

EI Ans.

(c) Deflection at free end of cantilever:

4 3 2 2 3 4

4

16( ) 64 ( ) 96 ( ) 8 ( )3

4

384

1

84C

wv L L L L L L L L

E

L

I

w

EI Ans.

(d) Slope at free end of cantilever:

2

3

3 2 3 38 ( ) 24 ( ) 24 ( ) 7

48 48 48 48 48

7

48C

C

dv w L wL L wL L wL wLEI

dx

dv

EIdx

wL Ans.







the beam, and (b) the deflection at B. Assume that

EI is constant for the beam.

Fig. P10.18

Solution

Beam FBD:

( ) 02 4

8

02

3

8

A y

y

y y y

y

wL LM C L

wLC

wLF A C

wLA

Consider beam segment AB (0 ≤ x ≤ L/2)

Moment equation:

2 2

2

3( ) ( ) 0

2 2 8

3( )

2 8

a a y

wx wx wLM M x A x M x x

wx wLxM x


2 2

2

3( )

2 8

d v wx wLxEI M x

dx

3 2

1

3

6 16

dv wx wLxEI C

dx (a)

4 3

1 224 16



0 at 0v x

Evaluate constants:

Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0.

Slope at B: Let x = L/2 in Eq. (a).

3 2 3 3 3

1 1 1

( / 2) 3 ( / 2) 3 5

6 16 48 64 192B

B

dv w L wL L wL wL wLEI EI C C C

dx (c)




Deflection at B: Let x = L/2 in Eq. (b).

4 3 4 4 4

1 11

( / 2) ( / 2)( / 2)

24 16 384 128 2 192 2B

w L wL L wL wL C L wL C LEI v C L (d)

Consider beam segment BC (L/2 ≤ x ≤ L)

Moment equation:

2

( ) ( ) ( ) ( ) 08

( ) ( )8 8 8

b b y

wLM M x C L x M x L x

wL wL wLxM x L x


2 2

2( )

8 8

d v wLx wLEI M x

dx

2 2

316 8

dv wLx wL xEI C

dx (e)

3 2 2

3 448 16

wLx wL xEI v C x C (f)


0 atv x L

Evaluate constants:

Substitute x = L and v = 0 into Eq. (f) to find

3 2 2

3 4

4

3 4

( ) ( )(0) ( )

48 16

24

wL L wL LEI C L C

wLC L C (g)

Slope at B: Let x = L/2 in Eq. (e).

2 2 3 3 3

3 3 3

( / 2) ( / 2) 3

16 8 64 16 64B

B

dv wL L wL L wL wL wLEI EI C C C

dx (h)

Deflection at B: Let x = L/2 in Eq. (f).

3 2 2 4

33 4 4

( / 2) ( / 2) 5( / 2)

48 16 384 2B

wL L wL L wL C LEI v C L C C (i)


Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h):

3 3

1 3

5 3

192 64

wL wLC C (j)




Further, the deflection at B must be the same for both segments; therefore, equate Eqs. (d) and (i):

4 4

1 34

5

192 2 384 2

wL C L wL C LC (k)

Evaluate constants: Solve Eqs. (g), (j), and (k) simultaneously to determine the values of constants C1,

C3, and C4:

3 3 4

1 3 4

9 17

384 384 384

wL wL wLC C C

(a) Elastic curve equation for beam segment AB:

3

3

4 3

3 216 24 9 (0 / 2)3

9

24 16 3

8

84

4

wx wLx wL xEI v

vwx

x Lx L x LEI

Ans.

(a) Elastic curve equation for beam segment BC:

3 2 2 3

3 2 2 3 417

48 16 38

8 24 17 ( /

4 384

2 )384

wLx wL x wL x wL

wLx Lx L x L L x L

I

E

E

I v

v Ans.


4 44 49 5

192 76

5

7688 768B B

wL wL wLEI v

L

EIv

w Ans.







the entire beam, (b) the deflection at C, and (c) the

slope at B. Assume that EI is constant for the

beam. Fig. P10.19

Solution

Beam FBD:

(3 ) 3 02

7

6

0

6

A y

y

y y y

y

LM B L wL L

wLB

F A B wL

wLA

Consider beam segment AB (0 ≤ x ≤ 3L)

Moment equation:

( ) ( ) 06

( )6

a a y

wLM M x A x M x x

wLxM x


2

2( )

6

d v wLxEI M x

dx

2

112

dv wLxEI C

dx (a)

3

1 236

wLxEI v C x C (b)


0 at 0 and 0 at 3v x v x L

Evaluate constants:

Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 3L and v = 0 into Eq. (b)

and solve for C1:

3 3 3

1 1

(3 ) 9(0) (3 )

36 36 4

wL L wL wLEI C L C

Slope at B: Let x = 3L in Eq. (a).

2 3 3(3 )

12 4 2B

B

dv wL L wL wLEI EI

dx (c)




Consider beam segment BC (3L ≤ x ≤ 4L)

Moment equation:

2( ) (4 ) 02

b b

wM M x L x

2(4 )

( )2

w L xM x


2 2

2

(4 )( )

2

d v w L xEI M x

dx

3

3

(4 )

6

dv w L xEI C

dx (e)

4

3 4

(4 )

24

w L xEI v C x C (f)


0 at 3v x L

Substitute x = 3L and v = 0 into Eq. (f) to find

4 4

3 4 3 4

4

4 3

(4 3 )(0) (3 ) (3 )

24 24

(3 )24

w L L wLEI C L C C L C

wLC L C (g)

Slope at B: Let x = 3L in Eq. (e).

3 3

3 3

(4 3 )

6 6B

B

dv w L L wLEI EI C C

dx (h)


Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h):

3 3 3

3 3

2

2 6 3

wL wL wLC C (i)

Backsubstitute this result into Eq. (g) to determine C4:

4 4 3 4

4 3

2 49(3 ) (3 )

24 24 3 24

wL wL wL wLC L C L




(a) Elastic curve equation for beam segment AB:

3

2

3

2

9

3

9 (0 3 )36

6 36

wL

wL

xx L x L

x wL xEI v

EIv Ans.

Elastic curve equation for beam segment BC:

4 3

4 3 4

4(4 ) 16 49 (3 4 )2

(4 ) 2 49

2

4

4 3 24

w L x

wL x L x

w

L L x LEI

L x wLEI v

v Ans.

(b) Deflection at C:

44 3 4 4 4

4

15(4 4 ) 16 (4 ) 49 64 49

24 24 2

5

8

4C

C

w w wLv L L L L L L L

EI EI EI

wv

L

EI Ans.

(c) Slope at B: Let x = 3L in Eq. (a).

3 3

2 2B B

B B

dv wL dvEI EI

dx dx

wL

EI Ans.







the beam, (b) the location of the maximum

deflection, and (c) the maximum beam deflection.

Assume that EI is constant for the beam. Fig. P10.20

Solution

Beam FBD:

0

0

0

0

20

2 3

3

02

6

A y

y

y y y

y

w L LM B L

w LB

w LF A B

w LA

Moment equation:

2

0

2

0 0

( )2 3

( ) 02 3 6

a a y

w x xM M x A x

L

w x x w LxM x

L

3

0 0( )6 6

w x w LxM x

L


2 3

0 0

2( )

6 6

d v w x w LxEI M x

dx L

4 2

0 01

24 12

dv w x w LxEI C

dx L (a)

5 3

0 01 2

120 36

w x w LxEI v C x C

L (b)


0 at 0

0 at

v x

v x L




Evaluate constants:

Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = L and v = 0 into Eq. (b)

and solve for C1:

5 3 3

0 0 01 1

( ) ( ) 7(0) ( )

120 36 360

w L w L L w LEI C L C

L


5 3 3

4 2 2 400 0 07

120 36 33 10 7

6 3600

ww x w Lx w xx L x L

L EI

L xEI v v

L Ans.

(b) Location of maximum deflection: The maximum deflection occurs where the beam slope is zero.

Therefore, set the beam slope equation [Eq. (a)] equal to zero:

4 2 3

0 0 070

24 12 360

dv w x w Lx w LEI

dx L

Multiply by −360L/w0 to obtain:

4 2 2 415 30 7 0x L x L

Solve this equation numerically to obtain:

= 0.51932962 0.523 1936 3x L L Ans.

(c) Maximum beam deflection:

4 2 2 4

4

0max

440 0 0

(0.51933 )3(0.51933 ) 10 (0.51933 ) 7

360

(0.51933) (0.0065222)4.52118

360

0.00652

w Lv L L L L

L EI

w w LL

EI EI

w L

EI Ans.





Fig. P10.21, integrate the load distribution to

determine (a) the equation of the elastic

curve for the beam, and (b) the maximum

deflection for the beam. Assume that EI is

constant for the beam.

Fig. P10.21

Solution

Integrate the load distribution:

4

0

4

d v w xEI

dx L

3 2

013 2

d v w xEI C

dx L

2 3

01 22 6

d v w xEI C x C

dx L

4 2

0 12 3

24 2

dv w x C xEI C x C

dx L

5 3 2

0 1 23 4

120 6 2

w x C x C xEI v C x C

L

Boundary conditions and evaluate constants:

3

13at 0, 0 0

d vx V EI C

dx

2

22at 0, 0 0

d vx M EI C

dx

4 3

0 03 3

( )at , 0 0

24 24

dv w L w Lx L C C

dx L

5 3 4

0 0 04 4

( ) ( )at , 0 0

120 24 30

w L w L L w Lx L v C C

L


5 3

0 50

4

0 0 4 5

120 24 305 4

120

wx L x L

LE

w x w L x w LE v

L II v Ans.

(b) Maximum deflection:

5 4 50max

4

0(0) 5 (1 30

0) 420

wv L L

LEI

w L

EI Ans.





Fig. P10.22, integrate the load distribution

to determine (a) the equation of the elastic

curve for the beam, and (b) the deflection

midway between the supports. Assume

that EI is constant for the beam. Fig. P10.22

Solution


4

0

4

d v w xEI

dx L

3 2

013 2

d v w xEI C

dx L

2 3

01 22 6

d v w xEI C x C

dx L

4 2

0 12 3

24 2

dv w x C xEI C x C

dx L

5 3 2

0 1 23 4

120 6 2


L


2

22at 0, 0 0

d vx M EI C

dx

2 3

0 01 12

( )at , 0 ( ) 0

6 6

d v w L w Lx L M EI C L C

dx L

4at 0, 0 0x v C

5 3 3

0 0 03 3

( ) 7at , 0 0

120 36 360

w L w Lx w Lx L v C x C

L


5 3 3

5 2 3 400 0 07

120 36 33 10 7

36060

wx L x L x

L

w x w Lx w L xEI v v

L EI Ans.

(b) Deflection midway between the supports:

5 2 3 40/2

4

03( / 2) 10 ( / 2) 7 ( / 2)360

5

768x L

wv L L L L L

LEI

w L

EI Ans.





Fig. P10.23, integrate the load distribution to

determine (a) the equation of the elastic

curve, (b) the deflection at the left end of the

beam, and (c) the support reactions By and

MB. Assume that EI is constant for the beam.

Fig. P10.23

Solution


4 3

0

4 3

d v w xEI

dx L

3 4

013 34

d v w xEI C

dx L

2 5

01 22 320

d v w xEI C x C

dx L

6 2

0 12 33120 2

dv w x C xEI C x C

dx L

7 3 2

0 1 23 43840 6 2


L


3

13at 0, 0 0

d vx V EI C

dx

2

22at 0, 0 0

d vx M EI C

dx

6 3

0 03 33

( )at , 0 0

120 120

dv w L w Lx L C C

dx L

7 3 4

0 0 04 43

( ) ( ) 6at , 0 0

840 120 840

w L w L L w Lx L v C C

L


7 3 4

7 6 70

3

0 0 0

3

7 6

840 840 87 6

84040

wx L x

w x w L x w LEI v

LLL

EIv Ans.

(b) Deflection at left end of beam:

7

7 6 70 0max 3

4

0

3

6(0) 7 (0) 6

840 840 140

ww w Lv L L

L

L

EIEI L EI Ans.

(c) Support reactions By and MB:

3

0

3 4

0 0

3

( )

4 44B y

x L

d v w L w LV EI B

w

dx L

L Ans.

2 5 2

0 0

2 3

2

0( )

20 2

0(cw)

20B B

x L

d v w L w LM EI M

dx L

w L Ans.







curve, (b) the deflection midway between

the supports, and (c) the support reactions

Ay and By. Assume that EI is constant for

the beam. Fig. P10.24

Solution


4 3

0

4 3

d v w xEI

dx L

3 4

013 34

d v w xEI C

dx L

2 5

01 22 320

d v w xEI C x C

dx L

6 2

0 12 33120 2

dv w x C xEI C x C

dx L

7 3 2

0 1 23 43840 6 2


L


2

22at 0, 0 0

d vx M EI C

dx

2 5

0 01 12 3

( )at , 0 ( ) 0

20 20

d v w L w Lx L M EI C L C

dx L

4at 0, 0 0x v C

7 3 3

0 0 03 33

( ) ( ) 6at , 0 ( ) 0

840 120 840

w L w L L w Lx L v C L C

L


7 4 3 60

7 3 3

0 0

3

0

3

6

840 120 847 6

8 00 4

wx L x L x

L

w

E

x w Lx w L xEI

Iv v

L Ans.


7 3

4 60/2 3

4

07 6840 2 2 2

13

5120x L

w L

EI

w L L Lv L L

L EI Ans.

(c) Support reactions Ay and By:

3 4

0 0 0

3 3

0

0(0)

4 20 0 02 2A y

x

d v w w L w LV EI A

dx L

w L Ans.

3 4

0 0 0

3 3

0( ) 4

4 20 2 50B y

x L

d v w L w L w LV EI B

dx L

w L Ans.










Fig. P10.25

Solution


4

04cos

2

d v xEI w

dx L

3

013

2sin

2

d v w L xEI C

dx L

2 2

01 22 2

4cos

2

d v w L xEI C x C

dx L

3 2

0 12 33

8sin

2 2

dv w L x C xEI C x C

dx L

4 3 2

0 1 23 44

16cos

2 6 2

w L x C x C xEI v C x C

L


3

13at 0, 0 0

d vx V EI C

dx

2 2 2

0 02 22 2 2

4 (0) 4at 0, 0 cos 0

2

d v w L w Lx M EI C C

dx L

3 2 3

0 0 03 33 2 3

8 ( ) 4 ( ) 4at , 0 sin 0 (2 )

2

dv w L L w L L w Lx L C C

dx L

4 2 2 3

0 0 044 2 3

4

04 3

16 ( ) 4 ( ) 4 ( )at , 0 cos (2 ) 0

2 2

2(4 )

w L L w L L w L Lx L v C

L

w LC


4 2 2 3 4

0 0 0

4 2 2 2 3 40

0

4 2

4

3 3

16 4 4 2cos (2 ) (4 )

2

32 cos 4 8 (2 ) 4 (4 )2 2

2

w xL L x L x L

w L x w L x w L w LEI v

L

vEI L

Ans.





4 2 2 2 3 40

4

4

4 4

0

0

4

0

4 4

(0)32 cos 4 (0) 8 (0)(2 ) 4 (4 )

2 2

32 4 (4 ) 32 4 (4

0.1089

)2 2

A

wv L L L L

EI L

w w LL

w L

EI

LEI EI

Ans.


3

00 0

3

2 ( ) 2sin

2

2B y

x L

d v w L L w LV EI B

dx L

w L Ans.

2

0

2

2 2 2 2

0 0 0

2 2 2 2

4 ( ) 4 (cw)

4 4cos

2B B

x L

d v w L L w wL w LM EI M

dx L

L Ans.




10.26 For the beam and loading shown

in Fig. P10.26, integrate the load

distribution to determine (a) the equation

of the elastic curve, (b) the deflection

midway between the supports, (c) the

slope at the left end of the beam, and (d)

the support reactions Ay and By. Assume

that EI is constant for the beam.

Fig. P10.26

Solution


4

04sin

d v xEI w

dx L

3

013

cosd v w L x

EI Cdx L

2 2

01 22 2

sind v w L x

EI C x Cdx L

3 2

0 12 33

cos2


dx L

4 3 2

0 1 23 44

sin6 2


L


2

22at 0, 0 0

d vx M EI C

dx

2 2

01 12 2

( )at , 0 sin ( ) 0 0

d v w L Lx L M EI C L C

dx L

4at 0, 0 0x v C

4

03 34

( )at , 0 sin ( ) 0 0

w L Lx L v C L C

L


0

4

44

0

4

ssin inw L x

EI v vL

w L x

EI L Ans.


4

0/2

0

44

4( / 2)sinx L

w L Lv

EI L

w L

EI Ans.

(c) Slope at the left end of the beam:

3 3

0 0

3 3

3

0

3

(0)cosA A

A

dv w L w LEI EI

dx L

w L

EI Ans.




(d) Support reactions Ay and By:

3

00 0

3

0

(0)cosA y

x

d v w L w LV EI A

dx L

w L Ans.

3

00 0

3

( )cosB y

x L

d v w L L w LV EI B

dx L

w L Ans.







curve, (b) the deflection midway between

the supports, (c) the slope at the left end of

the beam, and (d) the support reactions Ay

and By. Assume that EI is constant for the

beam.

Fig. P10.27

Solution


4

04sin

2

d v xEI w

dx L

3

013

2cos

2

d v w L xEI C

dx L

2 2

01 22 2

4sin

2

d v w L xEI C x C

dx L

3 2

0 12 33

8cos

2 2


dx L

4 3 2

0 1 23 44

16sin

2 6 2


L


2

22at 0, 0 0

d vx M EI C

dx

2 2

0 01 12 2 2

4 ( ) 4at , 0 sin ( ) 0

2

d v w L L w Lx L M EI C L C

dx L

4at 0, 0 0x v C

4 3 3

20 0 03 34 2 4

16 ( ) 4 ( ) 2at , 0 sin ( ) 0 (24 )

2 6 3

w L L w L L w Lx L v C L C

L


4

4 2 3 2 30

3 320 0 0

4 2

4

4

16 4 2sin (24 )

224

2 6 3

sin (24 )3 2

w L x w Lx w L xEI v

w xL Lx L

EI

L

v xL

Ans.





3

4 2 2 30/ 2 4

4 2 2

0

4

4

0

4

4

0

4

0

2 ( / 2)24 sin (24 )

3 2 2 2

2 (24 )24sin

3 4 8 2

2(1.2694611)

3

0.008688 0.002 869

x L

w L L Lv L L L

EI L

w L

EI

w L

EI

w L

EI

w L

EI Ans.

(c) Slope at the left end of the beam:

3 2 320 0 0

3 2 4

3 32 3 30 0

0 03 4 3 4

3

0

2

8 (0) 2 (0) 2cos (24 )

2 3

8 2 8 16 2(24 ) 0.0262

0.0262

093 3

A

A

A

dv w L w L w LEI EI

dx L

w L w Lw L

w L

E

w L

I Ans.

(d) Support reactions Ay and By:

3

0 0 0

3 2 2

0

0

2

2 (0) 4 2cos ( 2)

2

2

( 2)

A

x

y

d v w L w L w LV EI

dx

wA

L

L

Ans.

3

0 0 0

3 2

0

2

2

2 ( ) 4 4cos

2

4

B

x L

y

d v w L L w L w LV EI

dx

L

L

Bw

Ans.










Fig. P10.28

Solution


4

04sin

2

d v xEI w

dx L

3

013

2cos

2

d v w L xEI C

dx L

2 2

01 22 2

4sin

2

d v w L xEI C x C

dx L

3 2

0 12 33

8cos

2 2


dx L

4 3 2

0 1 23 44

16sin

2 6 2


L


3

0 01 13

2 (0) 2at 0, 0 cos 0

2

d v w L w Lx V EI C C

dx L

2 2

0 02 22 2

4 (0) 2 (0)at 0, 0 sin 0 0

2

d v w L w Lx M EI C C

dx L

3 2 3

0 0 03 33

8 2 ( )( )at , 0 cos 0

2 2

w L w L L w Ldv Lx L C C

dx L

4 3 3

0 0 044

430

4 4

16 2 ( ) ( )( )at , 0 sin 0

2 6

2(24 )

3

w L w L L w L LLx L v C

L

w LC


4 3 3 430 0 0 0

4

4 3 3 3 3 3 40

4

4

16 2 2sin (

48 sin 3 2(24

24 )2 6

)3 2

3

w L w Lx w L x w LxEI v

L

vw x

L Lx L x LEI L

Ans.





4 3 3 3 3 3 40

4

4

3 40 0

4

4

0

(0)48 sin (0) 3 (0) 2(24 )

3 2

2(24 ) 0

0.04

.047

7

95093

95

A

wv L L

w L

E

L LEI L

w w LL

E

I

I EI

Ans.


3

0

3

0 0 02 ( ) 2 2cos

2

2B y

x L

d v w L L w w LL w LV EI B

dx L Ans.

2

2 2 2 2

0 0 0 0

2

0

2

2 2

2

0

2

4 ( ) 2 ( )

2( 2)

4 2sin

2

2( (cw)2)

B

x L

B

d v w L L w L L w L w LM EI

dx L

LwM

wL Ans.




10.29 For the beam and loading shown, use

discontinuity functions to compute the deflection

of the beam at D. Assume a constant value of EI =

1,750 kip-ft2 for the beam.

Fig. P10.29

Solution

Support reactions: A FBD of the beam is shown to the

right.

(5 kips)(4 ft) (3 kips)(13 ft) (10 ft) 0

5.90 kips

5 kips 3 kips 0

2.10 kips

A y

y

y y y

y

M C

C

F A C

A

Load function w(x):

1 1 1 1

( ) 2.10 kips 0 ft 5 kips 4 ft 5.90 kips 10 ft 3 kips 13 ftw x x x x x


0 0 0 0

( ) 2.10 kips 0 ft 5 kips 4 ft 5.90 kips 10 ft 3 kips 13 ftV x x x x x

1 1 1 1

( ) 2.10 kips 0 ft 5 kips 4 ft 5.90 kips 10 ft 3 kips 13 ftM x x x x x

Equations for beam slope and beam deflection:

From Eq. (10.1), we can write:

2

1 1 1 1

2( ) 2.10 kips 0 ft 5 kips 4 ft 5.90 kips 10 ft 3 kips 13 ft

d vEI M x x x x x

dx

Integrate the moment function to obtain an expression for the beam slope:

2 2

2 2

1

2.10 kips 5 kips0 ft 4 ft

2 2


2 2

dvEI x x

dx

x x C (a)

Integrate again to obtain the beam deflection function:

3 3

3 3

1 2


6 6


6 6

EI v x x

x x C x C (b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection

v or slope dv/dx that are known at particular locations along the beam span. For this beam, the

deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 10 ft). Substitute the

boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain:

2 0C




Next, substitute the boundary condition v = 0 at x = 10 ft into Eq. (b) to obtain:

3 3

1

2

1

2.10 kips 5 kips(10 ft) (6 ft) (10 ft) 0

6 6

17 kip-ft

EI v C

C

The beam slope and elastic curve equations are now complete:

2 2

2 2 2


2 2

5.90 kips 3 kips10 ft 13 ft 17 kip-ft

2 2

dvEI x x

dx

x x (c)

3 3

3 3 2


6 6

5.90 kips 3 kips10 ft 13 ft (17 kip-ft )

6 6

EI v x x

x x x (d)

Beam deflection at D: At the tip of the overhang where x = 13 ft, the beam deflection is:

3 3 3 2

3

3

2

2.10 kips 5 kips 5.90 kips(13 ft) (9 ft) (3 ft) (17 kip-ft )(13 ft)

6 6 6

33.000 kip-ft

33.000 kip-ft0.018857 ft

1,750 kip-f0.226 in.

t

D

D

EI v

v Ans.




10.30 The solid 30-mm-diameter steel [E = 200

GPa] shaft shown in Fig. P10.30 supports two

pulleys. For the loading shown, use discontinuity

functions to compute:

(a) the shaft deflection at pulley B.

(b) the shaft deflection at pulley C.

Fig. P10.30

Solution


right.

800 N 500 N 0

1,300 N

(800 N)(250 mm) (500 N)(600 mm) 0

500,000 N-mm

y y

y

A A

A

F A

A

M M

M

Load function w(x):

2 1

1 1

( ) 500,000 N-mm 0 mm 1,300 N 0 mm

800 N 250 mm 500 N 600 mm

w x x x

x x


1 0

0 0

( ) 500,000 N-mm 0 mm 1,300 N 0 mm

800 N 250 mm 500 N 600 mm

V x x x

x x

0 1

1 1

( ) 500,000 N-mm 0 mm 1,300 N 0 mm

800 N 250 mm 500 N 600 mm

M x x x

x x



20 1

2

1 1

( ) 500,000 N-mm 0 mm 1,300 N 0 mm

800 N 250 mm 500 N 600 mm

d vEI M x x x

dx

x x


1 2

2 2

1

1,300 N500,000 N-mm 0 mm 0 mm

2

800 N 500 N250 mm 600 mm

2 2

dvEI x x

dx

x x C (a)


2 3

3 3

1 2

500,000 N-mm 1,300 N0 mm 0 mm

2 6

800 N 500 N250 mm 600 mm

6 6

EI v x x

x x C x C (b)





v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the

slope dv/dx and the deflection v are known at the fixed support (x = 0 mm). Substitute the boundary

condition dv/dx = 0 at x = 0 mm into Eq. (a) to obtain:

1 0C

Next, substitute the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain:

2 0C


1 2

2 2

1,300 N500,000 N-mm 0 mm 0 mm

2

800 N 500 N250 mm 600 mm

2 2

dvEI x x

dx

x x

2 3

3 3

500,000 N-mm 1,300 N0 mm 0 mm

2 6

800 N 500 N250 mm 600 mm

6 6

EI v x x

x x

Section properties:

4 4 2

9 2

(30 mm) 39,750.782 mm 200 GPa 200,000 N/mm64

7.9522 10 N-mm

I E

EI

(a) Beam deflection at B: The beam deflection at B where x = 250 mm is:

2 3

9 3

9 2

500,000 N-mm 1,300 N(250 mm) (250 mm)

2 6

12.2396 10 N-mm1.5392 mm

7.9522 10 N-m1.539 m

mm

B

B

EI v

v Ans.

(b) Beam deflection at C: The beam deflection at C where x = 600 mm is:

2 3 3

9 3

9 2

500,000 N-mm 1,300 N 800 N(600 mm) (600 mm) (350 mm)

2 6 6

48.9167 10 N-mm6.1514 mm

7.9522 10 N-mm6.15 mm

C

C

EI v

v Ans.





discontinuity functions to compute (a) the slope of

the beam at C and (b) the deflection of the beam at

C. Assume a constant value of EI = 560×106 N-

mm2 for the beam.

Fig. P10.31

Solution

Support reactions: A FBD of the beam is shown to the right.

(210 N-m)(1,000 mm/m) (1,400 N)(450 mm)

(700 mm) 0

600 N

1,400 N 0

800 N

A

y

y

y y y

y

M

E

E

F A E

A

Load function w(x):

1 2

1 1

( ) 800 N 0 mm 210,000 N-mm 200 mm

1,400 N 450 mm 600 N 700 mm

w x x x

x x


0 1

0 0

( ) 800 N 0 mm 210,000 N-mm 200 mm

1,400 N 450 mm 600 N 700 mm

V x x x

x x

1 0

1 1

( ) 800 N 0 mm 210,000 N-mm 200 mm

1,400 N 450 mm 600 N 700 mm

M x x x

x x



21 0

2

1 1

( ) 800 N 0 mm 210,000 N-mm 200 mm

1,400 N 450 mm 600 N 700 mm

d vEI M x x x

dx

x x


2 1

2 2

1

800 N0 mm 210,000 N-mm 200 mm

2

1,400 N 600 N450 mm 700 mm

2 2

dvEI x x

dx

x x C (a)


3 2

3 3

1 2

800 N 210,000 N-mm0 mm 200 mm

6 2

1,400 N 600 N450 mm 700 mm

6 6

EI v x x

x x C x C (b)






deflection v is known at the pin support (x = 0 mm) and at the roller support (x = 700 mm). Substitute

the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain:

2 0C

Next, substitute the boundary condition v = 0 at x = 700 mm into Eq. (b) to obtain:

3 2 3

1

2

1

800 N 210,000 N-mm 1,400 N0 (700 mm) (500 mm) (250 mm) (700 mm)

6 2 6

22,625,000 N-mm

C

C


2 1

2 2 2

800 N0 mm 210,000 N-mm 200 mm

2

1,400 N 600 N450 mm 700 mm 22,625,000 N-mm

2 2

dvEI x x

dx

x x

3 2

3 3 2

800 N 210,000 N-mm0 mm 200 mm

6 2

1,400 N 600 N450 mm 700 mm (22,625,000 N-mm )

6 6

EI v x x

x x x

(a) Beam slope at C: The beam slope at C is:

2 2

6 2

6 2

800 N(350 mm) (210,000 N-mm)(150 mm) 22,625,000 N-mm

2

5.125 10 N-mm0.009152 rad

50.00915 rad

60 10 N-mm

C

C

dvEI

dx

dv

dx Ans.

(b) Beam deflection at C: The beam deflection at C is:

3 2 2

9 3

9 3

6 2

800 N 210,000 N-mm(350 mm) (150 mm) (22,625,000 N-mm )(350 mm)

6 2

4.564583 10 N-mm

4.564583 10 N-mm8.1510 mm

560 10 N8.15 mm

-mm

C

C

EI v

v Ans.






belt pulleys. Assume that the bearing at A can be

idealized as a pin support and that the bearing at

E can be idealized as a roller support. For the

loading shown, use discontinuity functions to

compute:

(a) the shaft deflection at pulley B.

(b) the shaft deflection at point C.

Fig. P10.32

Solution


right.

(600 N)(300 mm) (800 N)(800 mm)

(1,000 mm) 0

820 N

600 N 800 N 0

580 N

A

y

y

y y y

y

M

E

E

F A E

A

Load function w(x): 1 1 1 1

( ) 580 N 0 mm 600 N 300 mm 800 N 800 mm 820 N 1,000 mmw x x x x x

Shear-force function V(x) and bending-moment function M(x): 0 0 0 0

( ) 580 N 0 mm 600 N 300 mm 800 N 800 mm 820 N 1,000 mmV x x x x x

1 1 1 1( ) 580 N 0 mm 600 N 300 mm 800 N 800 mm 820 N 1,000 mmM x x x x x



21 1

2

1 1

( ) 580 N 0 mm 600 N 300 mm

800 N 800 mm 820 N 1,000 mm

d vEI M x x x

dx

x x


2 2 2

2

1

580 N 600 N 800 N0 mm 300 mm 800 mm

2 2 2

820 N1,000 mm

2

dvEI x x x

dx

x C (a)


3 3 3

3

1 2

580 N 600 N 800 N0 mm 300 mm 800 mm

6 6 6

820 N1,000 mm

6

EI v x x x

x C x C (b)






deflection v is known at the pin support (x = 0 mm) and at the roller support (x = 1,000 mm). Substitute

the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain:

2 0C

Next, substitute the boundary condition v = 0 at x = 1,000 mm into Eq. (b) to obtain:

3 3 3

1

6 2

1

580 N 600 N 800 N0 (1,000 mm) (700 mm) (200 mm) (1,000 mm)

6 6 6

61.3 10 N-mm

C

C


2 2 2

2 6 2

580 N 600 N 800 N0 mm 300 mm 800 mm

2 2 2

820 N1,000 mm 61.3 10 N-mm

2

dvEI x x x

dx

x

3 3 3

3 6 2

580 N 600 N 800 N0 mm 300 mm 800 mm

6 6 6

820 N1,000 mm (61.3 10 N-mm )

6

EI v x x x

x x

Section properties:

4 4 2

9 2

(30 mm) 39,750.782 mm 200 GPa 200,000 N/mm64

7.9522 10 N-mm

I E

EI

(a) Beam deflection at B: The beam deflection at B where x = 300 mm is:

3 6 2

9 3

9 2

580 N(300 mm) (61.3 10 N-mm )(300 mm)

6

15.7800 10 N-mm1.9844 mm

7.9522 10 N-m1.984 m

mm

B

B

EI v

v Ans.

(b) Beam deflection at C: The beam deflection at C where x = 500 mm is:

3 3 6 2

9 3

9 2

580 N 600 N(500 mm) (200 mm) (61.3 10 N-mm )(500 mm)

6 6

19.3667 10 N-mm2.4354 mm

7.9522 10 N-mm2.44 mm

C

C

EI v

v Ans.




10.33 The cantilever beam shown in Fig. P10.33

consists of a W530 × 74 structural steel wide-

flange shape [E = 200 GPa; I = 410 × 106 mm

4].

Use discontinuity functions to compute the

deflection of the beam at C for the loading shown.

Fig. P10.33

Solution


right.

(30 kN/m)(3 m) 40 kN 0

130 kN

(30 kN/m)(3 m)(1.5 m) (40 kN)(5 m) 0

335 kN-m

y y

y

A A

A

F A

A

M M

M

Load function w(x):

1 2

0 0 1

( ) 130 kN 0 m 335 kN-m 0 m

30 kN/m 0 m 30 kN/m 3 m 40 kN 5 m

w x x x

x x x


0 1

1 1 0

( ) 130 kN 0 m 335 kN-m 0 m

30 kN/m 0 m 30 kN/m 3 m 40 kN 5 m

V x x x

x x x

1 0

2 2 1

( ) 130 kN 0 m 335 kN-m 0 m

30 kN/m 30 kN/m0 m 3 m 40 kN 5 m

2 2

M x x x

x x x



21 0

2

2 2 1

( ) 130 kN 0 m 335 kN-m 0 m

30 kN/m 30 kN/m0 m 3 m 40 kN 5 m

2 2

d vEI M x x x

dx

x x x


2 1

3 3 2

1

130 kN0 m 335 kN-m 0 m

2

30 kN/m 30 kN/m 40 kN0 m 3 m 5 m

6 6 2

dvEI x x

dx

x x x C (a)


3 2

4 4 3

1 2

130 kN 335 kN-m0 m 0 m

6 2

30 kN/m 30 kN/m 40 kN0 m 3 m 5 m

24 24 6

EI v x x

x x x C x C (b)






slope dv/dx and the deflection v are known at the fixed support (x = 0 m). Substitute the boundary

condition dv/dx = 0 at x = 0 m into Eq. (a) to obtain:

1 0C

Next, substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain:

2 0C


2 1

3 3 2

130 kN0 m 335 kN-m 0 m

2

30 kN/m 30 kN/m 40 kN0 m 3 m 5 m

6 6 2

dvEI x x

dx

x x x

3 2

4 4 3

130 kN 335 kN-m0 m 0 m

6 2

30 kN/m 30 kN/m 40 kN0 m 3 m 5 m

24 24 6

EI v x x

x x x

Beam deflection at C: For the W530 × 74 structural steel wide-flange shape, EI = 82,000 kN-m2. At

the tip of the overhang where x = 5 m, the beam deflection is:

3 2 4 4

3

3

2

130 kN 335 kN-m 30 kN/m 30 kN/m(5 m) (5 m) (5 m) (2 m)

6 2 24 24

2,240.416667 kN-m

2,240.416667 kN-m0.027322 m

82,000 kN-m27.3 mm

C

C

EI v

v Ans.





consists of a W21 × 50 structural steel wide-flange

shape [E = 29,000 ksi; I = 984 in.4]. Use

discontinuity functions to compute the deflection

of the beam at D for the loading shown.

Fig. P10.34

Solution

Support reactions: A FBD of the beam is shown to

the right.

1

2

1

2

(9 kips)(4 ft)

4 kips/ft 9 ft (13 ft) 0

270.00 kip-ft

(9 kips) 4 kips/ft 9 ft 0

27.00 kips

A

A

A

y y

y

M

M

M

F A

A

Load function w(x):

2 1 1 1

1 0

4 kips/ft( ) 270 kip-ft 0 ft 27 kips 0 ft 9 kips 4 ft 7 ft

9 ft

4 kips/ft16 ft 4 kips/ft 16 ft

9 ft

w x x x x x

x x


1 0 0 2

2 1


2(9 ft)

4 kips/ft16 ft 4 kips/ft 16 ft

2(9 ft)

V x x x x x

x x

0 1 1 3

3 2


6(9 ft)

4 kips/ft 4 kips/ft16 ft 16 ft

6(9 ft) 2

M x x x x x

x x



20 1 1

2

3 3 2

( ) 270 kip-ft 0 ft 27 kips 0 ft 9 kips 4 ft


6(9 ft) 6(9 ft) 2

d vEI M x x x x

dx

x x x





1 2 2

4 4 3

1

27 kips 9 kips270 kip-ft 0 ft 0 ft 4 ft

2 2


24(9 ft) 24(9 ft) 6

dvEI x x x

dx

x x x C

(a)


2 3 3

5 5 4

1 2

270 kip-ft 27 kips 9 kips0 ft 0 ft 4 ft

2 6 6


120(9 ft) 120(9 ft) 24

EI v x x x

x x x C x C (b)



slope dv/dx and the deflection v are known at the fixed support (x = 0 ft). Substitute the boundary

condition dv/dx = 0 at x = 0 ft into Eq. (a) to obtain:

1 0C


2 0C


1 2 2 4

4 3

27 kips 9 kips 4 kips/ft270 kip-ft 0 ft 0 ft 4 ft 7 ft

2 2 24(9 ft)


24(9 ft) 6

dvEI x x x x

dx

x x

2 3 3 5

5 4

270 kip-ft 27 kips 9 kips 4 kips/ft0 ft 0 ft 4 ft 7 ft

2 6 6 120(9 ft)


120(9 ft) 24

EI v x x x x

x x

Beam deflection at D: For the W21 × 50 structural steel wide-flange shape, EI = 198,166.658 kip-ft2.

At the tip of the overhang where x = 16 ft, the beam deflection is:

2 3 3 5

3

3

2

270 kip-ft 27 kips 9 kips 4 kips/ft(16 ft) (16 ft) (12 ft) (9 ft)

2 6 6 120(9 ft)

18,938.7 kip-ft

18,938.7 kip-ft0.095570 ft

198,166.658 kip-f1.147

t in.

D

D

EI v

v Ans.




10.35 The simply supported beam shown in Fig.

P10.35 consists of a W410 × 85 structural steel

wide-flange shape [E = 200 GPa; I = 316 × 106

mm4]. For the loading shown, use discontinuity

functions to compute (a) the slope of the beam at A

and (b) the deflection of the beam at midspan. Fig. P10.35

Solution


right.

(75 kN/m)(2.5 m)(1.25 m)

(75 kN/m)(2.5 m)(6.75 m) (8 m) 0

187.5 kN

(75 kN/m)(2.5 m) (75 kN/m)(2.5 m) 0

187.5 kN

A

y

y

y y y

y

M

D

D

F A D

A

Load function w(x):

1 0 0

0 0 1

( ) 187.5 kN 0 m 75 kN/m 0 m 75 kN/m 2.5 m

75 kN/m 5.5 m 75 kN/m 8 m 187.5 kN 8 m

w x x x x

x x x


0 1 1

1 1 0

( ) 187.5 kN 0 m 75 kN/m 0 m 75 kN/m 2.5 m

75 kN/m 5.5 m 75 kN/m 8 m 187.5 kN 8 m

V x x x x

x x x

1 2 2

2 2 1

75 kN/m 75 kN/m( ) 187.5 kN 0 m 0 m 2.5 m

2 2

75 kN/m 75 kN/m5.5 m 8 m 187.5 kN 8 m

2 2

M x x x x

x x x



21 2 2

2

2 2 1

75 kN/m 75 kN/m( ) 187.5 kN 0 m 0 m 2.5 m

2 2

75 kN/m 75 kN/m5.5 m 8 m 187.5 kN 8 m

2 2

d vEI M x x x x

dx

x x x


2 3 3

3 3 2

1

187.5 kN 75 kN/m 75 kN/m0 m 0 m 2.5 m

2 6 6

75 kN/m 75 kN/m 187.5 kN5.5 m 8 m 8 m

6 6 2

dvEI x x x

dx

x x x C (a)


3 4 4

4 4 3

1 2

187.5 kN 75 kN/m 75 kN/m0 m 0 m 2.5 m

6 24 24

75 kN/m 75 kN/m 187.5 kN5.5 m 8 m 8 m

24 24 6

EI v x x x

x x x C x C (b)






deflection v is known at the pin support (x = 0 m) and at the roller support (x = 8 m). Substitute the

boundary condition v = 0 at x = 0 m into Eq. (b) to obtain:

2 0C


3 4 4 4

1

2

1

187.5 kN 75 kN/m 75 kN/m 75 kN/m0 (8 m) (8 m) (5.5 m) (2.5 m) (8 m)

6 24 24 24

742.1875 kN-m

C

C


2 3 3

3 3 2 2

187.5 kN 75 kN/m 75 kN/m0 m 0 m 2.5 m

2 6 6

75 kN/m 75 kN/m 187.5 kN5.5 m 8 m 8 m 742.1875 kN-m

6 6 2

dvEI x x x

dx

x x x

3 4 4

4 4 3 2

187.5 kN 75 kN/m 75 kN/m0 m 0 m 2.5 m

6 24 24

75 kN/m 75 kN/m 187.5 kN5.5 m 8 m 8 m (742.1875 kN-m )

24 24 6

EI v x x x

x x x x

(a) Beam slope at A: For the W410 × 85 structural steel wide-flange shape, EI = 63,200 kN-m2. The

beam slope at A is:

2

2

2

742.1875 kN-m

742.1875 kN-m0.011743 rad

63,200 kN-0.01174 rad

m

A

A

dvEI

dx

dv

dx Ans.

(b) Beam deflection at midspan: At midspan where x = 4 m, the beam deflection is:

3 4 4 2

midspan

3

3

midspan 2

187.5 kN 75 kN/m 75 kN/m(4 m) (4 m) (1.5 m) (742.1875 kN-m )(4 m)

6 24 24

1,752.929687 kN-m

1,752.929687 kN-m0.027736 m

63,200 kN-27

m.7 mm

EI v

v Ans.






wide-flange shape [E = 29,000 ksi; I = 291 in.4].

For the loading shown, use discontinuity functions

to compute (a) the slope of the beam at A and (b)

the deflection of the beam at midspan.

Fig. P10.36

Solution


right.

(2.5 kips/ft)(12 ft)(12 ft) (24 ft) 0

15 kips

(2.5 kips/ft)(12 ft) 0

15 kips

A y

y

y y y

y

M D

D

F A D

A

Load function w(x):

1 0 0 1

( ) 15 kips 0 ft 2.5 kips/ft 6 ft 2.5 kips/ft 18 ft 15 kips 24 ftw x x x x x


0 1 1 0

( ) 15 kips 0 ft 2.5 kips/ft 6 ft 2.5 kips/ft 18 ft 15 kips 24 ftV x x x x x

1 2 2 12.5 kips/ft 2.5 kips/ft

( ) 15 kips 0 ft 6 ft 18 ft 15 kips 24 ft2 2

M x x x x x



21 2

2

2 1

2.5 kips/ft( ) 15 kips 0 ft 6 ft

2

2.5 kips/ft18 ft 15 kips 24 ft

2

d vEI M x x x

dx

x x


2 3

3 2

1

15 kips 2.5 kips/ft0 ft 6 ft

2 6

2.5 kips/ft 15 kips18 ft 24 ft

6 2

dvEI x x

dx

x x C (a)


3 4

4 3

1 2


6 24

2.5 kips/ft 15 kips18 ft 24 ft

24 6

EI v x x

x x C x C (b)








2 0C


3 4 4

1

2

1

15 kips 2.5 kips/ft 2.5 kips/ft0 (24 ft) (18 ft) (6 ft) (24 ft)

6 24 24

990 kip-ft

C

C


2 3

3 2 2


2 6

2.5 kips/ft 15 kips18 ft 24 ft 990 kip-ft

6 2

dvEI x x

dx

x x

3 4

4 3 2


6 24

2.5 kips/ft 15 kips18 ft 24 ft (990 kip-ft )

24 6

EI v x x

x x x

(a) Beam slope at A: For the W14 × 30 structural steel wide-flange shape, EI = 58,604.164 kip-ft2.

The beam slope at A is:

2

2

2

990 kip-ft

990 kip-ft0.016893 rad

58,604.164 kip-ft0.01689 rad

A

A

dvEI

dx

dv

dx Ans.

(b) Beam deflection at midspan: At midspan where x = 12 ft, the beam deflection is:

3 4 2 3

midspan

3

midspan 2

15 kips 2.5 kips/ft(12 ft) (6 ft) (990 kip-ft )(12 ft) 7,695 kip-ft

6 24

7,695 kip-ft0.131305 ft

58,604.164 kip-f1.5 6 n

t7 i .

EI v

v Ans.








to compute (a) the slope of the beam at A and (b)

the deflection of the beam at B.

Fig. P10.37

Solution


(7 kips/ft)(11 ft)(5.5 ft) (4 kips/ft)(9 ft)(15.5 ft)

(20 ft) 0

49.075 kips


63.925 kips

A

y

y

y y y

y

M

C

C

F A C

A

Load function w(x):

1 0 0

0 0 1

( ) 63.925 kips 0 ft 7 kips/ft 0 ft 7 kips/ft 11 ft

4 kips/ft 11 ft 4 kips/ft 20 ft 49.075 kips 20 ft

w x x x x

x x x


0 1 1

1 1 0

( ) 63.925 kips 0 ft 7 kips/ft 0 ft 7 kips/ft 11 ft


V x x x x

x x x

1 2 2

2 2 1


2 2

4 kips/ft 4 kips/ft11 ft 20 ft 49.075 kips 20 ft

2 2

M x x x x

x x x



21 2 2

2

2 2 1


2 2


2 2

d vEI M x x x x

dx

x x x


2 3 3

3 3 2

1

63.925 kips 7 kips/ft 7 kips/ft0 ft 0 ft 11 ft

2 6 6

4 kips/ft 4 kips/ft 49.075 kips11 ft 20 ft 20 ft

6 6 2

dvEI x x x

dx

x x x C (a)


3 4 4

4 4 3

1 2

63.925 kips 7 kips/ft 7 kips/ft0 ft 0 ft 11 ft

6 24 24


24 24 6

EI v x x x

x x x C x C (b)








2 0C


3 4 4 4

1

2

1

63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft0 (20 ft) (20 ft) (9 ft) (9 ft) (20 ft)

6 24 24 24

1,969.3396 kip-ft

C

C


2 3 3 3

3 2 2

63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft0 ft 0 ft 11 ft 11 ft

2 6 6 6

4 kips/ft 49.075 kips20 ft 20 ft 1,969.3396 kip-ft

6 2

dvEI x x x x

dx

x x

3 4 4 4

4 3 2

63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft0 ft 0 ft 11 ft 11 ft

6 24 24 24

4 kips/ft 49.075 kips20 ft 20 ft (1,969.3396 kip-ft )

24 6

EI v x x x x

x x x

(a) Beam slope at A: For the W21 × 50 structural steel wide-flange shape, EI = 198,166.658 kip-ft2.

The beam slope at A is:

2

2

2

1,969.3396 kip-ft

1,969.3396 kip-ft0.009938 rad

198,166.658 kip0.00994 ra

td

-f

A

A

dvEI

dx

dv

dx Ans.

(b) Beam deflection at B: At midspan where x = 11 ft, the beam deflection is:

3 4 2

3

2

63.925 kips 7 kips/ft(11 ft) (11 ft) (1,969.3396 kip-ft )(11 ft)

6 24

11,752.33123 kip-ft0.059305 ft

198,10.712 in

66.658 ki f.

p- t

B

B

EI v

v Ans.






wide-flange shape [E = 200 GPa; I = 60.8 × 106

mm4]. For the loading shown, use discontinuity

functions to compute (a) the deflection of the beam

at C and (b) the deflection of the beam at F.

Fig. P10.38

Solution


right.

(20 kN)(2 m) (8 kN/m)(6 m)(7 m)

(10 kN)(12 m) (8 m) 0

62 kN

20 kN (8 kN/m)(6 m) 10 kN 0

16 kN

A

y

y

y y y

y

M

D

D

F A D

A

Load function w(x):

1 1 0 1

0 1

( ) 16 kN 0 m 20 kN 2 m 8 kN/m 4 m 62 kN 8 m

8 kN/m 10 m 10 kN 12 m

w x x x x x

x x


0 0 1 0

1 0

( ) 16 kN 0 m 20 kN 2 m 8 kN/m 4 m 62 kN 8 m

8 kN/m 10 m 10 kN 12 m

V x x x x x

x x

1 1 2 1

2 1

8 kN/m( ) 16 kN 0 m 20 kN 2 m 4 m 62 kN 8 m

2

8 kN/m10 m 10 kN 12 m

2

M x x x x x

x x



21 1 2 1

2

2 1

8 kN/m( ) 16 kN 0 m 20 kN 2 m 4 m 62 kN 8 m

2

8 kN/m10 m 10 kN 12 m

2

d vEI M x x x x x

dx

x x


2 2 3 2

3 2

1

16 kN 20 kN 8 kN/m 62 kN0 m 2 m 4 m 8 m

2 2 6 2

8 kN/m 10 kN10 m 12 m

6 2

dvEI x x x x

dx

x x C (a)





3 3 4 3

4 3

1 2

16 kN 20 kN 8 kN/m 62 kN0 m 2 m 4 m 8 m

6 6 24 6

8 kN/m 10 kN10 m 12 m

24 6

EI v x x x x

x x C x C (b)





2 0C


3 3 4

1

2

1

16 kN 20 kN 8 kN/m0 (8 m) (6 m) (4 m) (8 m)

6 6 24

70 kN-m

C

C


2 2 3 2

3 2 2

16 kN 20 kN 8 kN/m 62 kN0 m 2 m 4 m 8 m

2 2 6 2

8 kN/m 10 kN10 m 12 m 70 kN-m

6 2

dvEI x x x x

dx

x x

3 3 4 3

4 3 2

16 kN 20 kN 8 kN/m 62 kN0 m 2 m 4 m 8 m

6 6 24 6

8 kN/m 10 kN10 m 12 m (70 kN-m )

24 6

EI v x x x x

x x x

(a) Beam deflection at C: For the W200 × 59 structural steel wide-flange shape, EI = 12,160 kN-m2.

At C where x = 4 m, the beam deflection is:

3 3 2

C

3

3

2

16 kN 20 kN(4 m) (2 m) (70 kN-m )(4 m)

6 6

136 kN-m

136 kN-m0.011184 m

12,16011.18

mm

kN-mC

EI v

v Ans.

(b) Beam deflection at F: At F where x = 12 m, the beam deflection is:

3 3 4 3

4 2

3

3

2

16 kN 20 kN 8 kN/m 62 kN(12 m) (10 m) (8 m) (4 m)

6 6 24 6

8 kN/m(2 m) (70 kN-m )(12 m)

24

264 kN-m

264 kN-21.7 m

m0.021711 m

12,160 kN-mm

F

F

EI v

v Ans.




10.39 The solid 0.50-in.-diameter steel [E =

30,000 ksi] shaft shown in Fig. P10.39 supports

two belt pulleys. Assume that the bearing at B

can be idealized as a pin support and that the

bearing at D can be idealized as a roller support.

For the loading shown, use discontinuity

functions to compute:

(a) the shaft deflection at pulley A.


Fig. P10.39

Solution


right.

(90 lb)(5 in.) (120 lb)(10 in.) (20 in.) 0

37.5 lb

90 lb 120 lb 0

172.5 lb

B y

y

y y y

y

M D

D

F B D

B

Load function w(x):

1 1 1 1

( ) 90 lb 0 in. 172.5 lb 5 in. 120 lb 15 in. 37.5 lb 25 in.w x x x x x


0 0 0 0

( ) 90 lb 0 in. 172.5 lb 5 in. 120 lb 15 in. 37.5 lb 25 in.V x x x x x

1 1 1 1

( ) 90 lb 0 in. 172.5 lb 5 in. 120 lb 15 in. 37.5 lb 25 in.M x x x x x



2

1 1 1 1

2( ) 90 lb 0 in. 172.5 lb 5 in. 120 lb 15 in. 37.5 lb 25 in.

d vEI M x x x x x

dx


2 2 2 2

1

90 lb 172.5 lb 120 lb 37.5 lb0 in. 5 in. 15 in. 25 in.

2 2 2 2

dvEI x x x x C

dx (a)


3 3 3

3

1 2

90 lb 172.5 lb 120 lb0 in. 5 in. 15 in.

6 6 6

37.5 lb25 in.

6

EI v x x x

x C x C (b)



deflection v is known at the pin support (x = 5 in.) and at the roller support (x = 25 in.). Substitute the

boundary condition v = 0 at x = 5 in. into Eq. (b) to obtain:




3

1 2

3

1 2

90 lb0 (5 in.) (5 in.)

6

(5 in.) 1,875.0 lb-in.

C C

C C (c)

Next, substitute the boundary condition v = 0 at x = 25 in. into Eq. (b) to obtain:

3 3 3

1 2

3

1 2

90 lb 172.5 lb 120 lb0 (25 in.) (20 in.) (10 in.) (25 in.)

6 6 6

(25 in.) 24,375.0 lb-in.

C C

C C (d)

Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2:

2 3

1 21,125 lb-in. and 3,750 lb-in.C C


2 2 2

2 2

90 lb 172.5 lb 120 lb0 in. 5 in. 15 in.

2 2 2

37.5 lb25 in. 1,125 lb-in.

2

dvEI x x x

dx

x

3 3 3

3 2 3

90 lb 172.5 lb 120 lb0 in. 5 in. 15 in.

6 6 6

37.5 lb25 in. (1,125 lb-in. ) 3,750 lb-in.

6

EI v x x x

x x

Section properties:

4 3 4 6

3 2

(0.5 in.) 3.06796 10 in. 30,000 ksi 30 10 psi64

92.0388 10 lb-in.

I E

EI

(a) Beam deflection at A: The beam deflection at A where x = 0 in. is:

3

3

3 2

3,750 lb-in.

3,750 lb-in.0.040744 in.

92.0380.0407 i

8 10 lb- .n

i.

n

A

A

EI v

v Ans.

(b) Beam deflection at C: The beam deflection at C where x = 15 in. is:

3 3 2 3

3

3 2

90 lb 172.5 lb(15 in.) (10 in.) (1,125 lb-in. )(15 in.) 3,750 lb-in.

6 6

8,750 lb-in.0.095069 in.

92.03880.0951 in.

10 lb-in.

C

C

EI v

v Ans.





consists of a W8 × 31 structural steel wide-flange

shape [E = 29,000 ksi; I = 110 in.4]. For the

loading shown, use discontinuity functions to

compute (a) the slope of the beam at A and (b) the

deflection of the beam at A.

Fig. P10.40

Solution


the right.

(3.5 kips/ft)(10 ft) 0

35 kips

75 kip-ft (3.5 kips/ft)(10 ft)(5 ft) 0

100 kip-ft

y y

y

C C

C

F C

C

M M

M

Load function w(x):

2 0 0

1 2

( ) 75 kip-ft 0 ft 3.5 kips/ft 5 ft 3.5 kips/ft 15 ft

35 kips 15 ft 100 kip-ft 15 ft

w x x x x

x x


1 1 1

0 1

( ) 75 kip-ft 0 ft 3.5 kips/ft 5 ft 3.5 kips/ft 15 ft


V x x x x

x x

0 2 2

1 0

3.5 kips/ft 3.5 kips/ft( ) 75 kip-ft 0 ft 5 ft 15 ft

2 2


M x x x x

x x



20 2 2

2

1 0

3.5 kips/ft 3.5 kips/ft( ) 75 kip-ft 0 ft 5 ft 15 ft

2 2


d vEI M x x x x

dx

x x


1 3 3

2 1

1

3.5 kips/ft 3.5 kips/ft75 kip-ft 0 ft 5 ft 15 ft

6 6

35 kips15 ft 100 kip-ft 15 ft

2

dvEI x x x

dx

x x C

(a)


2 4 4

3 2

1 2

75 kip-ft 3.5 kips/ft 3.5 kips/ft0 ft 5 ft 15 ft

2 24 24

35 kips 100 kip-ft15 ft 15 ft

6 2

EI v x x x

x x C x C

(b)








1 3

1

2

1

3.5 kips/ft0 (75 kip-ft)(15 ft) (10 ft)

6

541.666667 kip-ft

C

C


2 4 2

2

3

2

75 kip-ft 3.5 kips/ft0 (15 ft) (10 ft) ( 541.666667 kip-ft )(15 ft)

2 24

1,145.833334 kip-ft

C

C


1 3 3

2 1 2

3.5 kips/ft 3.5 kips/ft75 kip-ft 0 ft 5 ft 15 ft

6 6

35 kips15 ft 100 kip-ft 15 ft 541.666667 kip-ft

2

dvEI x x x

dx

x x

2 4 4

3 2 2 3

75 kip-ft 3.5 kips/ft 3.5 kips/ft0 ft 5 ft 15 ft

2 24 24

35 kips 100 kip-ft15 ft 15 ft (541.666667 kip-ft ) 1,145.833334 kip-ft

6 2

EI v x x x

x x x

(a) Beam slope at A: For the W8 × 31 structural steel wide-flange shape, EI = 22,152.777 kip-ft2. At

the tip of the overhang where x = 0 ft, the beam slope is:

2

2

2

541.666667 kip-ft

541.666667 kip-ft0.024451 rad

22,152.777 ki0.0245 r

p-a

ftd

A

A

dvEI

dx

dv

dx Ans.

(b) Beam deflection at A: At the tip of the overhang where x = 0 ft, the beam deflection is:

3

3

2

1,145.833334 kip-ft

1,145.833334 kip-ft0.051724 ft

22,152.777 kip-f0.621 i .

tn

A

A

EI v

v Ans.








to compute (a) the slope of the beam at E and (b)

the deflection of the beam at C. Fig. P10.41

Solution


1

2

1

2

8 ft(6 kips/ft)(8 ft) (4 kips/ft)(10 ft)(13 ft)

3

(22 ft) 0

20.727 kips


43.273 kips

B

y

y

y y y

y

M

E

E

F B E

B

Load function w(x):

1 1 0 1

0 0 1

6 kips/ft 6 kips/ft( ) 0 ft 8 ft 6 kips/ft 8 ft 43.273 kips 8 ft

8 ft 8 ft


w x x x x x

x x x


2 2 1 0

1 1 0

6 kips/ft 6 kips/ft( ) 0 ft 8 ft 6 kips/ft 8 ft 43.273 kips 8 ft

2(8 ft) 2(8 ft)


V x x x x x

x x x

3 3 2 1

2 2 1

6 kips/ft 6 kips/ft 6 kips/ft( ) 0 ft 8 ft 8 ft 43.273 kips 8 ft

6(8 ft) 6(8 ft) 2


2 2

M x x x x x

x x x



23 3 2

2

1 2 2

1

6 kips/ft 6 kips/ft 6 kips/ft( ) 0 ft 8 ft 8 ft

6(8 ft) 6(8 ft) 2

4 kips/ft 4 kips/ft43.273 kips 8 ft 16 ft 26 ft

2 2

20.727 kips 30 ft

d vEI M x x x x

dx

x x x

x





4 4 3 2

3 3 2

1

6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips0 ft 8 ft 8 ft 8 ft

24(8 ft) 24(8 ft) 6 2


6 6 2

dvEI x x x x

dx

x x x C (a)


5 5 4 3

4 4 3

1 2


120(8 ft) 120(8 ft) 24 6


24 24 6

EI v x x x x

x x x C x C (b)





5

1 2

3

1 2

6 kips/ft0 (8 ft) (8 ft)

120(8 ft)

(8 ft) 204.80 kips-ft

C C

C C (c)


5 5 4 3

4 4

1 2

3

1 2

6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips0 (30 ft) (22 ft) (22 ft) (22 ft)

120(8 ft) 120(8 ft) 24 6

4 kips/ft 4 kips/ft(14 ft) (4 ft) (30 ft)

24 24

(30 ft) 9,334.351 kip-ft

C C

C C (d)


2 3

1 2433.598 kip-ft and 3,673.582 kip-ftC C


4 4 3 2

3 3 2 2


24(8 ft) 24(8 ft) 6 2

4 kips/ft 4 kips/ft 20.727 kips16 ft 26 ft 30 ft 433.598 kip-ft

6 6 2

dvEI x x x x

dx

x x x

5 5 4 3

4 4 3

2 3


120(8 ft) 120(8 ft) 24 6


24 24 6

(433.598 kip-ft ) 3,673.582 kip-ft

EI v x x x x

x x x

x




(a) Beam slope at E: For the W14 × 34 structural steel wide-flange shape, EI = 68,472.219 kip-ft2.

The beam slope at E is:

4 4 3 2

3 3 2

2

2

6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips(30 ft) (22 ft) (22 ft) (22 ft)

24(8 ft) 24(8 ft) 6 2

4 kips/ft 4 kips/ft(14 ft) (4 ft) 433.598 kip-ft

6 6

907.801 kip-ft0.0132

68,472.219 kip-ft

E

E

dvEI

dx

dv

dx0.01358 ra dd 26 ra Ans.

(b) Beam deflection at C: At C where x = 16 ft, the beam deflection is:

5 5 4 3

2 3

3

2

6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips(16 ft) (8 ft) (8 ft) (8 ft)

120(8 ft) 120(8 ft) 24 6

(433.598 kip-ft )(16 ft) 3,673.582 kip-ft

4,896.157 kip-ft0.071506 ft

68,472.219 kip-ft0.858

C

C

EI v

v in. Ans.





discontinuity functions to compute (a) the

deflection of the beam at A and (b) the deflection

of the beam at midspan (i.e., x = 2.5 m). Assume a

constant value of EI = 1,500 kN-m2 for the beam.

Fig. P10.42

Solution


right.

1

2

1

2

9 kN-m (18 kN/m)(3 m)(1 m) (3 m) 0

6 kN

(18 kN/m)(3 m) 0

21 kN

B y

y

y y y

y

M C

C

F B C

B

Load function w(x):

2 1 0

1 1 1

( ) 9 kN-m 0 m 21 kN 1 m 18 kN/m 1 m

18 kN/m 18 kN/m1 m 4 m 6 kN 4 m

3 m 3 m

w x x x x

x x x


1 0 1

2 2 0

( ) 9 kN-m 0 m 21 kN 1 m 18 kN/m 1 m

18 kN/m 18 kN/m1 m 4 m 6 kN 4 m

2(3 m) 2(3 m)

V x x x x

x x x

0 1 2

3 3 1

18 kN/m( ) 9 kN-m 0 m 21 kN 1 m 1 m

2

18 kN/m 18 kN/m1 m 4 m 6 kN 4 m

6(3 m) 6(3 m)

M x x x x

x x x



20 1 2

2

3 3 1

18 kN/m( ) 9 kN-m 0 m 21 kN 1 m 1 m

2

18 kN/m 18 kN/m1 m 4 m 6 kN 4 m

6(3 m) 6(3 m)

d vEI M x x x x

dx

x x x


1 2 3

4 4 2

1

21 kN 18 kN/m9 kN-m 0 m 1 m 1 m

2 6

18 kN/m 18 kN/m 6 kN1 m 4 m 4 m

24(3 m) 24(3 m) 2

dvEI x x x

dx

x x x C (a)





2 3 4

5 5 3

1 2

9 kN-m 21 kN 18 kN/m0 m 1 m 1 m

2 6 24

18 kN/m 18 kN/m 6 kN1 m 4 m 4 m

120(3 m) 120(3 m) 6

EI v x x x

x x x C x C (b)





2

1 2

3

1 2

9 kN-m0 (1 m) (1 m)

2

(1 m) 4.5 kN-m

C C

C C (c)


2 3 4 5

1 2

3

1 2

9 kN-m 21 kN 18 kN/m 18 kN/m0 (4 m) (3 m) (3 m) (3 m) (4 m)

2 6 24 120(3 m)

(4 m) 26.10 kN-m

C C

C C (d)


2 3

1 27.2 kN-m and 2.7 kN-mC C


1 2 3

4 4 2 2

21 kN 18 kN/m9 kN-m 0 m 1 m 1 m

2 6

18 kN/m 18 kN/m 6 kN1 m 4 m 4 m 7.2 kN-m

24(3 m) 24(3 m) 2

dvEI x x x

dx

x x x

2 3 4

5 5 3 2 3

9 kN-m 21 kN 18 kN/m0 m 1 m 1 m

2 6 24

18 kN/m 18 kN/m 6 kN1 m 4 m 4 m (7.2 kN-m ) 2.7 kN-m

120(3 m) 120(3 m) 6

EI v x x x

x x x x

(a) Beam deflection at A: The beam deflection at A is:

3

3

2

2.7 kN-m

2.7 kN-m0.001800 m

1,51.800

00 kN mm

-m

A

A

EI v

v Ans.

(b) Beam deflection at midspan: At x = 2.5 m, the beam deflection is:

2 3 4

midspan

5 2 3

3

3

midspan 2

9 kN-m 21 kN 18 kN/m(2.5 m) (1.5 m) (1.5 m)

2 6 24

18 kN/m(1.5 m) (7.2 kN-m

2.95

)(2.5 m) 2.7 kN-m120(3 m)

4.429688 kN-m

4.429688 kN-m0.002953 m

1,500 kN-m

m m

EI v

v Ans.






the beam at B and (b) the deflection of the beam at

A. Assume a constant value of EI = 133,000 kip-

ft2 for the beam.

Fig. P10.43

Solution


right.

1

2

1

2

(4 kips/ft)(9 ft) 0

18 kips

(4 kips/ft)(9 ft)(11 ft) 0

198 kip-ft

y y

y

C C

C

F C

C

M M

M

Load function w(x):

0 1 1

1 2

4 kips/ft 4 kips/ft( ) 4 kips/ft 0 ft 0 ft 9 ft

9 ft 9 ft


w x x x x

x x


1 2 2

0 1

4 kips/ft 4 kips/ft( ) 4 kips/ft 0 ft 0 ft 9 ft

2(9 ft) 2(9 ft)


V x x x x

x x

2 3 3

1 0


2 6(9 ft) 6(9 ft)


M x x x x

x x



22 3 3

2

1 0


2 6(9 ft) 6(9 ft)


d vEI M x x x x

dx

x x


3 4 4

2 1

1


6 24(9 ft) 24(9 ft)

18 kips14 ft 198 kip-ft 14 ft

2

dvEI x x x

dx

x x C (a)





4 5 5

3 2

1 2


24 120(9 ft) 120(9 ft)

18 kips 198 kip-ft14 ft 14 ft

6 2

EI v x x x

x x C x C (b)





3 4 4

1

2

1

4 kips/ft 4 kips/ft 4 kips/ft0 (14 ft) (14 ft) (5 ft)

6 24(9 ft) 24(9 ft)

1,129.5 kip-ft

C

C


4 5 5 2

2

3

2

4 kips/ft 4 kips/ft 4 kips/ft0 (14 ft) (14 ft) (5 ft) (1,129.5 kip-ft )(14 ft)

24 120(9 ft) 120(9 ft)

11,390.7 kip-ft

C

C


3 4 4

2 1 2


6 24(9 ft) 24(9 ft)

18 kips14 ft 198 kip-ft 14 ft 1,129.5 kip-ft

2

dvEI x x x

dx

x x

4 5 5

3 2 2 3


24 120(9 ft) 120(9 ft)

18 kips 198 kip-ft14 ft 14 ft (1,129.5 kip-ft ) 11,390.7 kip-ft

6 2

EI v x x x

x x x

(a) Beam slope at B: The beam slope at B (i.e., x = 9 ft) is:

3 4 2 2

2

2

4 kips/ft 4 kips/ft(9 ft) (9 ft) 1,129.5 kip-ft 765 kip-ft

6 24(9 ft)

765 kip-ft0.005752 rad

133,000 kip-ft0.00575 rad

B

B

dvEI

dx

dv

dx Ans.

(b) Beam deflection at A: The beam deflection at A is:

3

3

2

11,390.7 kip-ft

11,390.7 kip-ft0.085644 ft

133,000 k1.028 in.

ip-ft

A

A

EI v

v Ans.







C. Assume a constant value of EI = 34×106 lb-ft

2

for the beam.

Fig. P10.44

Solution


right.

1

2

1

2

(7,000 lb/ft)(9 ft)(3 ft) (14 ft) 0

6,750 lbs

(7,000 lb/ft)(9 ft) 0

24,750 lbs

B y

y

y y y

y

M D

D

F B D

B

Load function w(x):

1 0 1

1 1

7,000 lb/ft( ) 24,750 lbs 4 ft 7,000 lb/ft 4 ft 4 ft

9 ft

7,000 lb/ft13 ft 6,750 lbs 18 ft

9 ft

w x x x x

x x


0 1 2

2 0

7,000 lb/ft( ) 24,750 lbs 4 ft 7,000 lb/ft 4 ft 4 ft

2(9 ft)

7,000 lb/ft13 ft 6,750 lbs 18 ft

2(9 ft)

V x x x x

x x

1 2 3

3 1

7,000 lb/ft 7,000 lb/ft( ) 24,750 lbs 4 ft 4 ft 4 ft

2 6(9 ft)

7,000 lb/ft13 ft 6,750 lbs 18 ft

6(9 ft)

M x x x x

x x



21 2 3

2

3 1

7,000 lb/ft 7,000 lb/ft( ) 24,750 lbs 4 ft 4 ft 4 ft

2 6(9 ft)

7,000 lb/ft13 ft 6,750 lbs 18 ft

6(9 ft)

d vEI M x x x x

dx

x x


2 3 4

4 2

1

24,750 lbs 7,000 lb/ft 7,000 lb/ft4 ft 4 ft 4 ft

2 6 24(9 ft)

7,000 lb/ft 6,750 lbs13 ft 18 ft

24(9 ft) 2

dvEI x x x

dx

x x C (a)





3 4 5

5 3

1 2


6 24 120(9 ft)

7,000 lb/ft 6,750 lbs13 ft 18 ft

120(9 ft) 6

EI v x x x

x x C x C

(b)





1 2(4 ft) 0C C (c)


3 4 5 5

1 2

3

1 2

24,750 lbs 7,000 lb/ft 7,000 lb/ft 7,000 lb/ft0 (14 ft) (14 ft) (14 ft) (5 ft)

6 24 120(9 ft) 120(9 ft)

(18 ft)

(18 ft) 3,579,975 lb-ft

C C

C C (d)


2 3

1 2255,712.5 lb-ft and 1,022,850 lb-ftC C


2 3 4

4 2 2


2 6 24(9 ft)

7,000 lb/ft 6,750 lbs13 ft 18 ft 255,712.5 lb-ft

24(9 ft) 2

dvEI x x x

dx

x x

3 4 5

5 3 2 3


6 24 120(9 ft)

7,000 lb/ft 6,750 lbs13 ft 18 ft (255,712.5 lb-ft ) 1,022,850 lb-ft

120(9 ft) 6

EI v x x x

x x x

(a) Beam slope at B: The beam slope at B is:

2

2

6 2

255,712.5 lb-ft

255,712.5 lb-ft0.0075210 rad

34 10.0

0 lb-ft0752 rad

B

B

dvEI

dx

dv

dx Ans.

(b) Beam deflection at C: At C where x = 13 ft, the beam deflection is:

3 4 5

2 3

3

6 2

24,750 lbs 7,000 lb/ft 7,000 lb/ft(9 ft) (9 ft) (9 ft)

6 24 120(9 ft)

(255,712.5 lb-ft )(13 ft) 1,022,850 lb-ft

825,187.5 kip-ft0.0242702 ft

34 10 lb-f0.291 .

tin

C

C

EI v

v Ans.






the beam at A and (b) the deflection of the beam at

B. Assume a constant value of EI = 370,000 kip-

ft2 for the beam.

Fig. P10.45

Solution


right.

1

2

1

2

1

2


(8 kips/ft)(12 ft)(16 ft) (24 ft) 0

48 kips


48 kips

A

y

y

y y y

y

M

C

C

F A C

A

Load function w(x):

1 1 1

0 0 1

1 1

1

8 kips/ft 8 kips/ft( ) 48 kips 0 ft 0 ft 12 ft

12 ft 12 ft

8 kips/ft8 kips/ft 12 ft 8 kips/ft 12 ft 12 ft

12 ft

8 kips/ft24 ft 48 kips 24 ft

12 ft

8 kips/ft48 kips 0 ft 0 f

12 ft

w x x x x

x x x

x x

x x1 1

1 1

2(8 kips/ft)t 12 ft

12 ft


12 ft

x

x x


0 2 2

2 0

8 kips/ft 2(8 kips/ft)( ) 48 kips 0 ft 0 ft 12 ft

2(12 ft) 2(12 ft)


2(12 ft)

V x x x x

x x

1 3 3

3 1


6(12 ft) 6(12 ft)


6(12 ft)

M x x x x

x x






21 3 3

2

3 1


6(12 ft) 6(12 ft)


6(12 ft)

d vEI M x x x x

dx

x x


2 4 4

4 2

1

48 kips 8 kips/ft 2(8 kips/ft)0 ft 0 ft 12 ft

2 24(12 ft) 24(12 ft)

8 kips/ft 48 kips24 ft 24 ft

24(12 ft) 2

dvEI x x x

dx

x x C (a)


3 5 5

5 3

1 2


6 120(12 ft) 120(12 ft)

8 kips/ft 48 kips24 ft 24 ft

120(12 ft) 6

EI v x x x

x x C x C (b)





2 0C


3 5 5

1

2

1

48 kips 8 kips/ft 2(8 kips/ft)0 (24 ft) (24 ft) (12 ft) (24 ft)

6 120(12 ft) 120(12 ft)

2,880 kip-ft

C

C


2 4 4

4 2 2


2 24(12 ft) 24(12 ft)

8 kips/ft 48 kips24 ft 24 ft 2,880 kip-ft

24(12 ft) 2

dvEI x x x

dx

x x

3 5 5

5 3 2


6 120(12 ft) 120(12 ft)

8 kips/ft 48 kips24 ft 24 ft (2,880 kip-ft )

120(12 ft) 6

EI v x x x

x x x

(a) Beam slope at A: The beam slope at A is:

2

2

2

2,880 kip-ft

2,880 kip-ft0.0077838 rad

370,000 kip0.00778 rad

-ft

A

A

dvEI

dx

dv

dx Ans.




(b) Beam deflection at B: At B where x = 12 ft, the beam deflection is:

3 5 2 3

3 3

2

48 kips 8 kips/ft(12 ft) (12 ft) (2,880 kip-ft )(12 ft) 22,118.4 kip-ft

6 120(12 ft)

22,118.4 kip-ft kip-ft0.0597795 ft

370,000 kip0.71 n.

-f7 i

t

B

B

EI v

v Ans.







B. Assume a constant value of EI = 110,000 kN-

m2 for the beam.

Fig. P10.46

Solution


right.

1

2

1

2

(15 kN/m)(4 m) (25 kN/m)(4 m) 0

110 kN

(15 kN/m)(4 m)(2 m)

2(4 m)(25 kN/m)(4 m) 0

3

253.33 kN-m

y y

y

A

A

A

F A

C

M

M

M

Load function w(x):

2 1 0 1

0 1 0

25 kN/m( ) 253.33 kN-m 0 m 110 kN 0 m 15 kN/m 0 m 0 m

4 m

25 kN/m15 kN/m 4 m 4 m 25 kN/m 4 m

4 m

w x x x x x

x x x


1 0 1 2

1 2 1

25 kN/m( ) 253.33 kN-m 0 m 110 kN 0 m 15 kN/m 0 m 0 m

2(4 m)

25 kN/m15 kN/m 4 m 4 m 25 kN/m 4 m

2(4 m)

V x x x x x

x x x

0 1 2 3

2 3 2

15 kN/m 25 kN/m( ) 253.33 kN-m 0 m 110 kN 0 m 0 m 0 m

2 6(4 m)

15 kN/m 25 kN/m 25 kN/m4 m 4 m 4 m

2 6(4 m) 2

M x x x x x

x x x



20 1 2

2

3 2 3 2

15 kN/m( ) 253.33 kN-m 0 m 110 kN 0 m 0 m

2

25 kN/m 15 kN/m 25 kN/m 25 kN/m0 m 4 m 4 m 4 m

6(4 m) 2 6(4 m) 2

d vEI M x x x x

dx

x x x x





1 2 3 4

3 4 3

1

110 kN 15 kN/m 25 kN/m253.33 kN-m 0 m 0 m 0 m 0 m

2 6 24(4 m)

15 kN/m 25 kN/m 25 kN/m4 m 4 m 4 m

6 24(4 m) 6

dvEI x x x x

dx

x x x C (a)


2 3 4 5

4 5 4

1 2

253.33 kN-m 110 kN 15 kN/m 25 kN/m0 m 0 m 0 m 0 m

2 6 24 120(4 m)

15 kN/m 25 kN/m 25 kN/m4 m 4 m 4 m

24 120(4 m) 24

EI v x x x x

x x x C x C (b)





1 0C


2 0C


1 2 3 4

3 4 3

110 kN 15 kN/m 25 kN/m253.33 kN-m 0 m 0 m 0 m 0 m

2 6 24(4 m)

15 kN/m 25 kN/m 25 kN/m4 m 4 m 4 m

6 24(4 m) 6

dvEI x x x x

dx

x x x

2 3 4 5

4 5 4

253.33 kN-m 110 kN 15 kN/m 25 kN/m0 m 0 m 0 m 0 m

2 6 24 120(4 m)

15 kN/m 25 kN/m 25 kN/m4 m 4 m 4 m

24 120(4 m) 24

EI v x x x x

x x x


1 2 3 4

2

2

2

110 kN 15 kN/m 25 kN/m( 253.33 kN-m)(4 m) (4 m) (4 m) (4 m)

2 6 24(4 m)

360 kN-m

2,120 kN-m0.003273 rad

110,000 kN-m0.00327 rad

B

B

dvEI

dx

dv

dx Ans.

(b) Beam deflection at B: The beam deflection at B is:

2 3 4 5

3

3

2

253.33 kN-m 110 kN 15 kN/m 25 kN/m(4 m) (4 m) (4 m) (4 m)

2 6 24 120(4 m)

1,066.67 kN-m

1,066.67 kN-m0.009697 m

110,009.70 m

0 kN-mm

B

B

EI v

v Ans.





discontinuity functions to compute (a) the

deflection of the beam at A and (b) the deflection

of the beam at C. Assume a constant value of EI =

24,000 kN-m2 for the beam.

Fig. P10.47

Solution


right.

1

2

1

2

(35 kN)(2.5 m) (25 kN/m)(4.0 m)(2.0 m)

2(4.0 m)(45 kN/m)(4.0 m) (5.5 m) 0

3

64.09 kN

35 kN (25 kN/m)(4.0 m)

(45 kN/m)(4.0 m) 0

160.91 kN

B

y

y

y y y

y

M

D

D

F B D

B

Load function w(x):

1 1 0

0 1 1

0 1

( ) 35 kN 0 m 160.91 kN 2.5 m 25 kN/m 2.5 m

45 kN/m 45 kN/m25 kN/m 6.5 m 2.5 m 6.5 m

4.0 m 4.0 m

45 kN/m 6.5 m 64.09 kN 8 m

w x x x x

x x x

x x


0 0 1

1 2 2

1 0

( ) 35 kN 0 m 160.91 kN 2.5 m 25 kN/m 2.5 m

45 kN/m 45 kN/m25 kN/m 6.5 m 2.5 m 6.5 m

2(4.0 m) 2(4.0 m)

45 kN/m 6.5 m 64.09 kN 8 m

V x x x x

x x x

x x

1 1 2

2 3 3

2 1

1 1 2

25 kN/m( ) 35 kN 0 m 160.91 kN 2.5 m 2.5 m

2

25 kN/m 45 kN/m 45 kN/m6.5 m 2.5 m 6.5 m

2 6(4.0 m) 6(4.0 m)

45 kN/m6.5 m 64.09 kN 8 m

2

25 kN/m 4535 kN 0 m 160.91 kN 2.5 m 2.5 m

2

M x x x x

x x x

x x

x x x3

2 3 1

kN/m2.5 m

6(4.0 m)

70 kN/m 45 kN/m6.5 m 6.5 m 64.09 kN 8 m

2 6(4.0 m)

x

x x x






21 1 2

2

3 2 3

1

25 kN/m( ) 35 kN 0 m 160.91 kN 2.5 m 2.5 m

2

45 kN/m 70 kN/m 45 kN/m2.5 m 6.5 m 6.5 m

6(4.0 m) 2 6(4.0 m)

64.09 kN 8 m

d vEI M x x x x

dx

x x x

x


2 2 3

4 3 4

2

1

35 kN 160.91 kN 25 kN/m0 m 2.5 m 2.5 m

2 2 6

45 kN/m 70 kN/m 45 kN/m2.5 m 6.5 m 6.5 m

24(4.0 m) 6 24(4.0 m)

64.09 kN8 m

2

dvEI x x x

dx

x x x

x C (a)


3 3 4

5 4 5

3

1 2

35 kN 160.91 kN 25 kN/m0 m 2.5 m 2.5 m

6 6 24

45 kN/m 70 kN/m 45 kN/m2.5 m 6.5 m 6.5 m

120(4.0 m) 24 120(4.0 m)

64.09 kN8 m

6

EI v x x x

x x x

x C x C (b)



deflection v is known at the pin support (x = 2.5 m) and at the roller support (x = 8 m). Substitute the

boundary condition v = 0 at x = 2.5 m into Eq. (b) to obtain:

3

1 2

3

1 2

35 kN0 (2.5 m) (2.5 m)

6

(2.5 m) 91.145833 kN-m

C C

C C (c)


3 3 4 5

4 5

1 2

3

1 2

35 kN 160.91 kN 25 kN/m 45 kN/m0 (8 m) (5.5 m) (5.5 m) (5.5 m)

6 6 24 120(4.0 m)

70 kN/m 45 kN/m(1.5 m) (1.5 m) (8 m)

24 120(4.0 m)

(8 m) 65.666667 kN-m

C C

C C (d)


2 3

1 228.511 kN-m and 162.424 kN-mC C





2 2 3

4 3 4

2 2

35 kN 160.91 kN 25 kN/m0 m 2.5 m 2.5 m

2 2 6

45 kN/m 70 kN/m 45 kN/m2.5 m 6.5 m 6.5 m

24(4.0 m) 6 24(4.0 m)

64.09 kN8 m 28.511 kN-m

2

dvEI x x x

dx

x x x

x

3 3 4

5 4 5

3 2 3

35 kN 160.91 kN 25 kN/m0 m 2.5 m 2.5 m

6 6 24

45 kN/m 70 kN/m 45 kN/m2.5 m 6.5 m 6.5 m

120(4.0 m) 24 120(4.0 m)

64.09 kN8 m (28.511 kN-m ) 162.424 kN-m

6

EI v x x x

x x x

x x

(a) Beam deflection at A: The beam deflection at A is:

3

3

2

162.424 kN-m

162.424 kN-m0.006768 m

24,000 kN-m6.77 mm

A

A

EI v

v Ans.

(b) Beam deflection at C: At x = 6.5 m, the beam deflection is:

3 3 4

5 2 3

3

3

2

35 kN 160.91 kN 25 kN/m(6.5 m) (4.0 m) (4.0 m)

6 6 24

45 kN/m(4.0 m) (28.511 kN-m )(6.5 m) 162.424 kN-m

120(4.0 m)

271.1797 kN-m

271.1797 kN-m0.0011299 m

24,000 k11.3 m

N-m0 m

C

C

EI v

v Ans.







A. Assume a constant value of EI = 54,000 kN-m2

for the beam.

Fig. P10.48

Solution


the right.

1

2

1

2

(20 kN/m)(3 m) (30 kN/m)(3 m) 0

105 kN

(20 kN/m)(3 m)(2.5 m)

(30 kN/m)(3 m)(2 m) 0

240 kN-m

y y

y

C

C

C

F C

C

M

M

M

Load function w(x):

0 1 0 1

0 1 2


3 m 3 m

30 kN/m 3 m 105 kN 4 m 240 kN-m 4 m

w x x x x x

x x x


1 2 1 2

1 0 1


2(3 m) 2(3 m)

30 kN/m 3 m 105 kN 4 m 240 kN-m 4 m

V x x x x x

x x x

2 3 2 3

2 1 0

20 kN/m 30 kN/m 20 kN/m 30 kN/m( ) 0 m 0 m 3 m 3 m

2 6(3 m) 2 6(3 m)

30 kN/m3 m 105 kN 4 m 240 kN-m 4 m

2

M x x x x x

x x x



22 3 2

2

3 2 1 0

20 kN/m 30 kN/m 20 kN/m( ) 0 m 0 m 3 m

2 6(3 m) 2

30 kN/m 30 kN/m3 m 3 m 105 kN 4 m 240 kN-m 4 m

6(3 m) 2

d vEI M x x x x

dx

x x x x


3 4 3 4

3 2 1

1


6 24(3 m) 6 24(3 m)

30 kN/m 105 kN3 m 4 m 240 kN-m 4 m

6 2

dvEI x x x x

dx

x x x C (a)





4 5 4 5

4 3 2

1 2


24 120(3 m) 24 120(3 m)

30 kN/m 105 kN 240 kN-m3 m 4 m 4 m

24 6 2

EI v x x x x

x x x C x C (b)





3 4 3 4 3

1

2

1

20 kN/m 30 kN/m 20 kN/m 30 kN/m 30 kN/m0 (4 m) (4 m) (1 m) (1 m) (1 m)

6 24(3 m) 6 24(3 m) 6

311.25 kN-m

C

C


4 5 4 5

4 2

2

3

2

20 kN/m 30 kN/m 20 kN/m 30 kN/m0 (4 m) (4 m) (1 m) (1 m)

24 120(3 m) 24 120(3 m)

30 kN/m(1 m) (311.25 kN-m )(4 m)

24

948.50 kN-m

C

C


3 4 3 4

3 2 1 2


6 24(3 m) 6 24(3 m)

30 kN/m 105 kN3 m 4 m 240 kN-m 4 m 311.25 kN-m

6 2

dvEI x x x x

dx

x x x

4 5 4 5

4 3 2

3


24 120(3 m) 24 120(3 m)

30 kN/m 105 kN 240 kN-m3 m 4 m 4 m

24 6 2

(311.25 kN-m) 948.50 kN-m

EI v x x x x

x x x

x


3 4 2

2

2

2

20 kN/m 30 kN/m(3 m) (3 m) 311.25 kN-m

6 24(3 m)

187.5 kN-m

187.5 kN-m0.003472 rad

54,000 kN0.00347 d

-mra

B

B

dvEI

dx

dv

dx Ans.

(b) Beam deflection at A: The beam deflection at A is:

3

3

2

948.50 kN-m

948.50 kN-m0.017565 m

54,000 kN-m17.56 mm

A

A

EI v

v Ans.




10.49a For the beams and loadings shown

below, determine the beam deflection at

point H. Assume that EI = 8 × 104 kN-m

2 is


Fig. P10.49a

Solution

Determine beam slope at A.

[Appendix C, SS beam with concentrated moment.]

Relevant equation from Appendix C:

6

A

ML

EI (slope magnitude)

Values:

M = 150 kN-m, L = 8 m, EI = 8 × 104 kN-m

2

Computation:

4 2

(150 kN-m)(8 m)0.00250 rad

6 6(8 10 kN-m )A

ML

EI

Determine beam deflection at H. [Skill 1]

(3 m)(0.00250 rad) 0.00750 7.50m mmHv Ans.




10.49b For the beams and loadings shown

below, determine the beam deflection at point

H. Assume that EI = 8 × 104 kN-m

2 is constant

for each beam.

Fig. P10.49b

Solution

Determine beam deflection at A. [Appendix C, Cantilever beam with uniformly distributed load.]


4

8A

wLv

EI

Values:

w = 6 kN/m, L = 4 m, EI = 8 × 104 kN-m

2

Computation:

4 4

4 2

(6 kN/m)(4 m)0.00240 m

8 8(8 10 kN-m )A

wLv

EI

Determine beam slope at A. [Appendix C, Cantilever beam with uniformly distributed load.]


3

6A

wL


Values:

w = 6 kN/m, L = 4 m, EI = 8 × 104 kN-m

2

Computation:

3 3

4 2

(6 kN/m)(4 m)0.00080 rad

6 6(8 10 kN-m )A

wL

EI


0.00240 m (2 m)(0.00080 rad) 0.00400 m 4.00 mmHv Ans.




10.49c For the beams and loadings shown



2 is


Fig. P10.49c

Solution


[Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )6

H

Pbxv L b x

LEI (elastic curve)

Values:

P = 30 kN-m, L = 12 m, b = 4 m, x = 4 m,

EI = 8 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(30 kN)(4 m)(4 m)(12 m) (4 m) (4 m)

6(12 m)(8 10 kN-m )

0.0 9.33 mm0933 m

H

Pbxv L b x

LEI

Ans.




10.49d For the beams and loadings shown



2 is


Fig. P10.49d

Solution

Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated load.]


3

,cant3

H

PLv

EI (assuming fixed support at B)

Values:

P = 15 kN, L = 4 m, EI = 8 × 104 kN-m

2

Computation:

3 3

,cant 4 2

(15 kN)(4 m)0.004000 m

3 3(8 10 kN-m )H

PLv

EI

Determine beam slope at B. [Appendix C, SS beam with concentrated moment.]


3

B

ML


Values:

M = (15 kN)(4 m) = 60 kN-m, L = 8 m,

EI = 8 × 104 kN-m

2

Computation:

4 2

(60 kN-m)(8 m)0.002000 rad

3 3(8 10 kN-m )B

ML

EI


0.00400 m (4 m)(0.00200 rad) 0.01200 m 12.00 mmHv Ans.






point H. Assume that EI = 1.2 × 107 kip-in.

2

is constant for each beam.

Fig. P10.50a

Solution

Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated moment.]


2

2B

MLv

EI

Values:

M = 40 kip-ft, L = 9 ft, EI = 1.2 × 107 kip-in.

2

Computation:

2 2 3

7 2

(40 kip-ft)(9 ft) (12 in./ft)0.23328 in.

2 2(1.2 10 kip-in. )B

MLv

EI

Determine beam slope at B. [Appendix C, Cantilever beam with concentrated moment.]


B

ML


Values:


2

Computation:

2

7 2

(40 kip-ft)(9 ft)(12 in./ft)0.004320 rad

(1.2 10 kip-in. )B

ML

EI


0.23328 in. (6 ft)(12 in./ft)(0.004320 rad) 0.54432 in 0.. 544 in.Hv Ans.






point H. Assume that EI = 1.2 × 107 kip-

in.2 is constant for each beam.

Fig. P10.50b

Solution

Determine beam slope at C. [Appendix C, SS beam with concentrated load not at midspan.]


2 2( )

6C

Pa L a

LEI

(slope magnitude)

Values:

P = 25 kips, L = 18 ft, a = 12 ft,

EI = 1.2 × 107 kip-in.

2

Computation:

2 2

2 2 2

7 2

( )

6

(25 kips)(12 ft)(18 ft) (12 ft) (12 in./ft) 0.00600 rad

6(18 ft)(1.2 10 kip-in. )

C

Pa L a

LEI


(7 ft)(12 in./ft)(0.00600 rad) 0.5040 0.504 in.in.Hv Ans.






H. Assume that EI = 1.2 × 107 kip-in.

2 is


Fig. P10.50c

Solution


[Appendix C, Cantilever beam with uniformly distributed load.]


2

2 2(6 4 )24

H

wxv L Lx x

EI (elastic curve)

Values:

w = 2.5 kips/ft, L = 15 ft, x = 9 ft,

EI = 1.2 × 107 kip-in.

2

Computation:

2

2 2

2

2 2 3

7 2

(6 4 )24

(2.5 kips/ft)(9 ft)6(15 ft) 4(15 ft)(9 ft) (9 ft) (12 in./ft)

24(1.2 10 ki

1.0

p-in. )

1.082565 83 in.in.

H

wxv L Lx x

EI

Ans.




10.50d For the beams and loadings

shown below, determine the beam

deflection at point H. Assume that EI =

1.2 × 107 kip-in.

2 is constant for each

beam.

Fig. P10.50d

Solution

Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.]


4

,cant8

H

wLv

EI (assuming fixed support at A)

Values:

w = 5 kips/ft, L = 8 ft, EI = 1.2 × 107 kip-in.

2

Computation:

4 4 3

,cant 7 2

(5 kips/ft)(8 ft) (12 in./ft)0.36864 in.

8 8(1.2 10 kip-in. )H

wLv

EI

Determine beam slope at A. [Appendix C, SS beam with concentrated moment.]


3

A

ML


Values:

M = (5 kips/ft)(8 ft)(4 ft) = 160 kip-ft, L = 22 ft,

EI = 1.2 × 107 kip-in.

2

Computation:

2

7 2

(160 kip-ft)(22 ft)(12 in./ft)0.014080 rad

3 3(1.2 10 kip-in. )A

ML

EI









2


Fig. P10.51a

Solution




2 2(2 3 )6

H

M xv L Lx x

LEI (elastic curve)

Values:

M = −60 kN-m, L = 12 m, x = 6 m,

EI = 6 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 60 kN-m)(6 m)2(12 m) 3(12 m)(6 m) (6 m)

6(12 m)(6 10 kN-

9.

m )

0.009 0000 mmm 0

H

M xv L Lx x

LEI

Ans.







2 is constant

for each beam.

Fig. P10.51b

Solution

Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.]


4

,cant8

H

wLv


Values:

w = 7.5 kN/m, L = 3 m, EI = 6 × 104 kN-m

2

Computation:

4 4

,cant 4 2

(7.5 kN/m)(3 m)0.00126563 m

8 8(6 10 kN-m )H

wLv

EI



3

A

ML


Values:

M = (7.5 kN/m)(3 m)(1.5 m) = 33.75 kN-m,

L = 6 m, EI = 6 × 104 kN-m

2

Computation:

4 2

(33.75 kN-m)(6 m)0.001125 rad

3 3(6 10 kN-m )A

ML

EI


0.00126563 m (3 m)(0.001125 rad) 0.00464063 m 4.64 mmHv Ans.







2 is


Fig. P10.51c

Solution

Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated load.]


3

3B

PLv

EI

Values:

P = 30 kN, L = 3 m, EI = 6 × 104 kN-m

2

Computation:

3 3

4 2

(30 kN)(3 m)0.004500 m

3 3(6 10 kN-m )B

PLv

EI

Determine beam slope at B. [Appendix C, Cantilever beam with concentrated load.]


2

2B

PL


Values:

P = 30 kN, L = 3 m, EI = 6 × 104 kN-m

2

Computation:

2 2

4 2

(30 kN)(3 m)0.002250 rad

2 2(6 10 kN-m )B

PL

EI


0.004500 m (3 m)(0.002250 rad) 0.01125 m 11.25 mmHv Ans.

Alternative solution for beam deflection at B. [Appendix C, Cantilever beam with concentrated load at midspan.]

Relevant equation from Appendix C: 35

48H

PLv

EI

Values: P = 30 kN, L = 6 m, EI = 6 × 104 kN-m

2

Computation: 3 3

4 2

5 5(30 kN)(6 m)0.011250 m 11.25 mm

48 48(6 10 kN-m )H

PLv

EI







2 is


Fig. P10.51d

Solution

Determine beam slope at C. [Appendix C, SS beam with uniformly distributed load over a portion of the span.]


2

2(2 )24

C

waL a

LEI (slope magnitude)

Values:

w = 5 kN/m, L = 9 m, a = 6 m,

EI = 6 × 104 kN-m

2

Computation:

2 2

22

4 2

(5 kN/m)(6 m)(2 ) 2(9 m) (6 m) 0.00200 rad

24 24(9 m)(6 10 kN-m )C

waL a

LEI


(3 m)(0.00200 rad) 0.00600 6.00m mmHv Ans.






point H. Assume that EI = 3.0 × 106 kip-

in.2 is constant for each beam.

Fig. P10.52a

Solution

Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated moment.]


2

,cant2

H

MLv


Values:


2

Computation:

2 2 3

,cant 6 2

(50 kip-ft)(6 ft) (12 in./ft)0.51840 in.

2 2(3.0 10 kip-in. )H

MLv

EI



3

A

ML


Values:


2

Computation:

2

6 2

(50 kip-ft)(18 ft)(12 in./ft)0.01440 rad

3 3(3.0 10 kip-in. )A

ML

EI









2 is


Fig. P10.52b

Solution


[Appendix C, Cantilever beam with concentrated load.]


2

(3 )6

H

Pxv L x

EI (elastic curve)

Values:

P = 10 kips, L = 10 ft, x = 7 ft,

EI = 3.0 × 106 kip-in.

2

Computation:

2

2 3

6 2

(3 )6

(10 kips)(7 ft) (12 in./ft)[3(10 ft) (7 ft)] 1.081920 in.

6(3.0 101.082 in.

kip-in. )

H

Pxv L x

EI

Ans.







2 is


Fig. P10.52c

Solution



6

A

ML


Values:

M = (2 kips/ft)(8 ft)(4 ft) = 64 kip-ft,

L = 18 ft, EI = 3.0 × 106 kip-in.

2

Computation:

2

6 2

(64 kip-ft)(18 ft)(12 in./ft)0.009216 rad

6 6(3.0 10 kip-in. )A

ML

EI


(6 ft)(12 in./ft)(0.009216 rad) 0.663552 0.664 inn. . iHv Ans.







2 is


Fig. P10.52d

Solution

Determine beam deflection at B. [Appendix C, Cantilever beam with uniformly distributed load.]


4

8B

wLv

EI

Values:

w = 1.5 kips/ft, L = 10 ft, EI = 3.0 × 106 kip-in.

2

Computation:

4 4 3

6 2

(1.5 kips/ft)(10 ft) (12 in./ft)1.0800 in.

8 8(3.0 10 kip-in. )B

wLv

EI

Determine beam slope at B. [Appendix C, Cantilever beam with uniformly distributed load.]


3

6B

wL


Values:

w = 1.5 kips/ft, L = 10 ft, EI = 3.0 × 106 kip-in.

2

Computation:

3 3 2

6 2

(1.5 kips/ft)(10 ft) (12 in./ft)0.01200 rad

6 6(3.0 10 kip-in. )B

wL

EI






10.53 The simply supported beam shown

in Fig. P10.53 consists of a W24 × 94

structural steel wide-flange shape [E =

29,000 ksi; I = 2,700 in.4]. For the loading

shown, determine the beam deflection at

point C.

Fig. P10.53

Solution

Consider distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.]


3

2 2(4 7 3 )24

C

wav L aL a

LEI

Values:

w = 3.2 kips/ft, L = 28 ft, a = 21 ft,

EI = 7.830 × 107 kip-in.

2

Computation:

32 2

3 32 2

7 2

(4 7 3 )24

(3.2 kips/ft)(21 ft) (12 in./ft)4(28 ft) 7(21 ft)(28 ft) 3(21 ft) 0.333822 in.

24(28 ft)(7.830 10 kip-in. )

C

wav L aL a

LEI

Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.]


2 2(3 4 )48

C

Pxv L x

EI (elastic curve)

Values:

P = 36 kips, L = 28 ft, x = 7 ft,

EI = 7.830 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(3 4 )48

(36 kips)(7 ft)(12 in./ft)3(28 ft) 4(7 ft) 0.249799 in.

48(7.830 10 kip-in. )

C

Pxv L x

EI

Beam deflection at C

0.333822 in. 0.249799 in. 0.583620 in. 0.584 in.Cv Ans.






structural steel wide-flange shape [E = 200

GPa; I = 370 × 106 mm

4]. For the loading


point C.

Fig. P10.54

Solution

Consider distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.]


3

2 2(4 7 3 )24

C

wav L aL a

LEI

Values:

w = 26 kN/m, L = 8 m, a = 6 m,

EI = 7.4 × 104 kN-m

2

Computation:

32 2

32 2

4 2

(4 7 3 )24

(26 kN/m)(6 m)4(8 m) 7(6 m)(8 m) 3(6 m) 0.011068 m

24(8 m)(7.40 10 kN-m )

C

wav L aL a

LEI

Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )

6C

Pbxv L b x

LEI (elastic curve)

Values:

P = 60 kN, L = 8 m, b = 3 m, x = 2 m,

EI = 7.4 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(60 kN)(3 m)(2 m)(8 m) (3 m) (2 m) 0.005169 m

6(8 m)(7.40 10 kN-m )

C

Pbxv L b x

LEI


0.011068 m 0.005169 m 0.016237 m 16.24 mmCv Ans.






structural steel wide-flange shape [E = 200

GPa; I = 216 × 106 mm

4]. For the loading


point B.

Fig. P10.55

Solution



2 2 2( )6

B

Pabv L a b

LEI

Values:

P = 60 kN, L = 9 m, a = 3 m, b = 6 m,

EI = 4.32 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(60 kN)(3 m)(6 m)(9 m) (3 m) (6 m) 0.016667 m

6(9 m)(4.32 10 kN-m )

B

Pabv L a b

LEI

Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )

6B

M xv L Lx x

LEI (elastic curve)

Values:

M = −45 kN-m, L = 9 m, x = 6 m,

EI = 4.32 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 45 kN-m)(6 m)2(9 m) 3(9 m)(6 m) (6 m) 0.004167 m

6(9 m)(4.32 10 kN-m )

B

M xv L Lx x

LEI

Beam deflection at B

0.016667 m 0.004167 m 0.012500 m 12.50 mmBv Ans.







29,000 ksi; I = 843 in.4]. For the loading


point B.

Fig. P10.56

Solution

Consider uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load over a portion of the span.]


3

2 2(4 7 3 )24

B

wav L aL a

LEI

Values:

w = 5 kips/ft, L = 24 ft, a = 16 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:

32 2

3 32 2

7 2

(4 7 3 )24

(5 kips/ft)(16 ft) (12 in./ft)4(24 ft) 7(16 ft)(24 ft) 3(16 ft) 0.965066 in.

24(24 ft)(2.4447 10 kip-in. )

B

wav L aL a

LEI

Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )

6B

M xv L Lx x

LEI (elastic curve)

Values:

M = −200 kip-ft, L = 24 ft, x = 8 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6

( 200 kip-ft)(8 ft)(12 in./ft)2(24 ft) 3(24 ft)(8 ft) (8 ft) 0.502638 in.

6(24 ft)(2.4447 10 kip-in. )

B

M xv L Lx x

LEI


0.965066 in. 0.502638 in. 0.462428 in. 0.462 in.Bv Ans.




10.57 The cantilever beam shown in Fig.

P10.57 consists of a rectangular

structural steel tube shape [E = 29,000

ksi; I = 476 in.4]. For the loading shown,

determine:

(a) the beam deflection at point B.

(b) the beam deflection at point C.

Fig. P10.57

Solution

(a) Beam deflection at point B




4

8B

wLv

EI

Values:


2

Computation:

4 4 3

7 2

(2 kips/ft)(6 ft) (12 in./ft)0.040559 in.

8 8(1.3804 10 kip-in. )B

wLv

EI

Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.]


2

(3 )6

B

Pxv L x

EI (elastic curve)

Values:

P = 12 kips, L = 10 ft, x = 6 ft,

EI = 1.3804 × 107 kip-in.

2

Computation:

2 2 3

7 2

(12 kips)(6 ft) (12 in./ft)(3 ) 3(10 ft) (6 ft) 0.216313 in.

6 6(1.3804 10 kip-in. )B

Pxv L x

EI


0.040559 in. 0.216313 in. 0.256872 in. 0.257 in.Bv Ans.




(b) Beam deflection at point C




3

6B

wL

EI

Values:


2

Computation:

3 3 26

7 2

6

(2 kips/ft)(6 ft) (12 in./ft)751.0866 10 rad

6 6(1.3804 10 kip-in. )

0.040559 in. (4 ft)(12 in./ft)(751.0866 10 rad) 0.076611 in.

B

C

wL

EI

v



3

3C

PLv

EI

Values:

P = 12 kips, L = 10 ft, EI = 1.3804 × 107 kip-in.

2

Computation:

3 3 3

7 2

(12 kips)(10 ft) (12 in./ft)0.500724 in.

3 3(1.3804 10 kip-in. )C

PLv

EI


0.076611 in. 0.500724 in. 0.577336 in. 0.577 in.Cv Ans.





P10.58 consists of a rectangular structural

steel tube shape [E = 200 GPa; I = 400 ×

106 mm

4]. For the loading shown,

determine:

(a) the beam deflection at point A.

(b) the beam deflection at point B.

Fig. P10.58

Solution

(a) Beam deflection at point A

Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]


4

8A

wLv

EI

Values:

w = 25 kN/m, L = 4 m, EI = 8.0 × 104 kN-m

2

Computation:

4 4

4 2

(25 kN/m)(4 m)0.010000 m

8 8(8.0 10 kN-m )A

wLv

EI


Relevant equations from Appendix C:

3 2

and3 2

B B

PL PLv

EI EI

Values:

P = 55 kN, L = 2.5 m, EI = 8.0 × 104 kN-m

2

Computation:

3 3

4 2

2 2

4 2

(55 kN)(2.5 m)0.003581 m

3 3(8.0 10 kN-m )

(55 kN)(2.5 m)0.002148 rad

2 2(8.0 10 kN-m )

0.003581 m (1.5 m)(0.002148 rad) 0.006803 m

B

B

A

PLv

EI

PL

EI

v

Beam deflection at A

0.010000 m 0.006803 m 0.016803 m 16.80 mmAv Ans.




(b) Beam deflection at point B



2

2 2(6 4 )24

B

wxv L Lx x

EI (elastic curve)

Values:

w = 25 kN/m, L = 4 m, x = 2.5 m,

EI = 8.0 × 104 kN-m

2

Computation:

22 2

22 2

4 2

(6 4 )24

(25 kN/m)(2.5 m)6(4.0 m) 4(4.0 m)(2.5 m) (2.5 m) 0.005066 m

24(8.0 10 kN-m )

B

wxv L Lx x

EI



3

3B

PLv

EI

Values:

P = 55 kN, L = 2.5 m, EI = 8.0 × 104 kN-m

2

Computation:

3 3

4 2

(55 kN)(2.5 m)0.003581 m

3 3(8.0 10 kN-m )B

PLv

EI


0.005066 m 0.003581 m 0.008647 m 8.65 mmBv Ans.





29,000 ksi] shaft shown in Fig. P10.59

supports two pulleys. For the loading

shown, determine:

(a) the shaft deflection at point B.

(b) the shaft deflection at point C.

Fig. P10.59

Solution

Section properties:

4 4(1.25 in.) 0.119842 in.64

I

(a) Shaft deflection at point B

Consider concentrated load at pulley B. [Appendix C, Cantilever beam with concentrated load at tip.]


3

3B

PLv

EI

Values:

P = 200 lb, L = 10 in., EI = 3.47543 × 106 lb-in.

2

Computation:

3 3

6 2

(200 lb)(10 in.)0.019182 in.

3 3(3.47543 10 lb-in. )B

PLv

EI

Consider concentrated load at pulley C. [Appendix C, Cantilever beam with concentrated load at tip.]


2

(3 )6

B

Pxv L x

EI (elastic curve)

Values:

P = 120 lb, L = 25 in., x = 10 in.,

EI = 3.47543 × 106 lb-in.

2

Computation:

2 2

6 2

(120 lb)(10 in.)(3 ) 3(25 in.) (10 in.) 0.037405 in.

6 6(3.47543 10 lb-in. )B

Pxv L x

EI

Shaft deflection at B

0.019182 in. 0.037405 in. 0.056588 0.0566 in.in.Bv Ans.




(b) Shaft deflection at point C

Consider concentrated load at pulley B. [Appendix C, Cantilever beam with concentrated load at tip.]


3 2

and3 2

B B

PL PLv

EI EI (magnitude)

Values:

P = 200 lb, L = 10 in., EI = 3.47543 × 106 lb-in.

2

Computation:

3 3

6 2

(200 lb)(10 in.)0.019182 in.

3 3(3.47543 10 lb-in. )B

PLv

EI

2 2

6 2

(200 lb)(10 in.)0.0028773 rad

2 2(3.47543 10 lb-in. )

0.019182 in. (15 in.)(0.0028773 rad) 0.062342 in.

B

C

PL

EI

v

Consider concentrated load at pulley C. [Appendix C, Cantilever beam with concentrated load at tip.]


3

3C

PLv

EI

Values:

P = 120 lb, L = 25 in.,

EI = 3.47543 × 106 lb-in.

2

Computation:

3 3

6 2

(120 lb)(25 in.)0.179834 in.

3 3(3.47543 10 lb-in. )C

PLv

EI

Shaft deflection at C

0.062342 in. 0.179834 in. 0.242176 in. 0.242 in.Cv Ans.





P10.60 consists of a rectangular structural

steel tube shape [E = 29,000 ksi; I = 1,710

in.4]. For the loading shown, determine:



Fig. P10.60

Solution


Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.]


2

2A

MLv

EI

Values:

M = −200 kip-ft, L = 15 ft,

EI = 4.959 × 107 kip-in.

2

Computation:

2 2 3

7 2

( 200 kip-ft)(15 ft) (12 in./ft)0.784029 in.

2 2(4.959 10 kip-in. )A

MLv

EI



3 2

and3 2

B B

PL PLv

EI EI (slope magnitude)

Values:

P = −18 kips, L = 9 ft, EI = 4.959 × 107 kip-in.

2

Computation:

3 3 3

7 2

2 2 2

7 2

( 18 kips)(9 ft) (12 in./ft)0.152415 in.

3 3(4.959 10 kip-in. )

(18 kips)(9 ft) (12 in./ft)0.0021169 rad

2 2(4.959 10 kip-in. )

0.152415 in. (6 ft)(12 in./ft)(0.0021169 rad) 0.3

B

B

A

PLv

EI

PL

EI

v

04830 in.


0.784029 in. 0.304830 in. 1.088860 1.in. 089 in.Av Ans.





Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.]


2

2B

Mxv

EI (elastic curve)

Values:

M = −200 kip-ft, L = 15 ft, x = 9 ft,

EI = 4.959 × 107 kip-in.

2

Computation:

2 2 3

7 2

( 200 kip-ft)(9 ft) (12 in./ft)0.282250 in.

2 2(4.959 10 kip-in. )B

Mxv

EI



3

3B

PLv

EI

Values:

P = −18 kips, L = 9 ft,

EI = 4.959 × 107 kip-in.

2

Computation:

3 3 3

7 2

( 18 kips)(9 ft) (12 in./ft)0.152415 in.

3 3(4.959 10 kip-in. )B

PLv

EI


0.282250 in. 0.152415 in. 0.434665 0.in. 435 in.Bv Ans.




10.61 The simply supported beam shown in

Fig. P10.61 consists of a W21 × 44



shown, determine:



Fig. P10.61

Solution


Determine cantilever deflection due to uniformly distributed load on overhang. [Appendix C, Cantilever beam with uniform load.]


4

8A

wLv


Values:


2

Computation:

4 4 3

7 2

(4 kips/ft)(8 ft) (12 in./ft)0.144760 in.

8 8(2.4447 10 kip-in. )A

wLv

EI

Consider deflection at A resulting from rotation at B caused by distributed load on overhang.



3

B

ML


Values:


L = 22 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2

7 2

(128 kip-ft)(22 ft)(12 in./ft)0.0055290 rad

3 3(2.4447 10 kip-in. )

(8 ft)(12 in./ft)(0.0055290 rad) 0.530786 in.

B

A

ML

EI

v






2

16B

PL


Values:


2

Computation:

2 2 2

7 2

(45 kips)(22 ft) (12 in./ft)0.0080182 rad

16 16(2.4447 10 kip-in. )

(8 ft)(12 in./ft)(0.0080182 rad) 0.769744 in.

B

A

PL

EI

v


0.144760 in. 0.530786 in. 0.769744 in. 0.094198 in. 0.0942 in.Av Ans.


Consider distributed load on overhang.



2 2(2 3 )

6C

M xv L Lx x

LEI (elastic curve)

Values:

M = (4 kips/ft)(8 ft)(4 ft) = −128 kip-ft,

L = 22 ft, x = 11 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6


6(22 ft)(2.4447 10 kip-in. )

C

M xv L Lx x

LEI



3

48C

PLv

EI

Values:


2

Computation:

3 3 3

7 2

(45 kips)(22 ft) (12 in./ft)0.705598 in.

48 48(2.4447 10 kip-in. )C

PLv

EI


0.273687 in. 0.705598 in. 0.431912 0.432 in.in.Cv Ans.







200 GPa; I = 351 × 106 mm

4]. For the

loading shown, determine:

(a) the beam deflection at point B.

(b) the beam deflection at point D. Fig. P10.62

Solution

(a) Beam deflection at point B

Consider distributed load between supports. [Appendix C, SS beam with uniformly distributed load.]


45

384B

wLv

EI

Values:

w = 55 kN/m, L = 7.2 m, EI = 7.02 × 104 kN-m

2

Computation:

4 4

4 2

5 5(55 kN/m)(7.2 m)0.027415 m

384 384(7.02 10 kN-m )B

wLv

EI

Consider distributed load on overhang. [Appendix C, SS beam with concentrated moment.]


2 2(2 3 )

6B

M xv L Lx x

LEI (elastic curve)

Values:

M = (55 kN/m)(2.8 m)(1.4 m) = −215.6 kN-m,

L = 7.2 m, x = 3.6 m, EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

7 2

(2 3 )6

( 215.6 kN-m)(3.6 m)2(7.2 m) 3(7.2 m)(3.6 m) (3.6 m) 0.009951 m

6(7.2 m)(7.02 10 kN-m )

B

M xv L Lx x

LEI


0.027415 m 0.009951 m 0.017464 m 17.46 mmBv Ans.




(b) Beam deflection at point D

Consider distributed load between supports. [Appendix C, SS beam with uniformly distributed load.]


3

24C

wL


Values:

w = 55 kN/m, L = 7.2 m, EI = 7.02 × 104 kN-m

2

Computation:

3 3

4 2

(55 kN/m)(7.2 m)0.0121846 rad

24 24(7.02 10 kN-m )

(2.8 m)(0.0121846 rad) 0.034117 m

C

D

wL

EI

v

Consider deflection at D resulting from rotation at C caused by distributed load on overhang.



3

C

ML


Values:

M = (55 kN/m)(2.8 m)(1.4 m) = 215.6 kN-m,

L = 7.2 m, EI = 7.02 × 104 kN-m

2

Computation:

7 2

(215.6 kN-m)(7.2 m)0.0073709 rad

3 3(7.02 10 kN-m )

(2.8 m)(0.0073709 rad) 0.020639 m

C

D

ML

EI

v

Determine cantilever deflection due to uniformly distributed load on overhang. [Appendix C, Cantilever beam with uniform load.]


4

8D

wLv

EI (assuming fixed support at C)

Values:

w = 55 kN/m, L = 2.8 m, EI = 7.02 × 104 kN-m

2

Computation:

4 4

4 2

(55 kN/m)(2.8 m)0.006020 m

8 8(7.02 10 kN-m )D

wLv

EI

Beam deflection at D

0.034117 m 0.020639 m 0.006020 m 0.007459 7m .46 mmDv Ans.







For a loading of w = 6 kips/ft, determine:


(b) the beam deflection at point C. Fig. P10.63

Solution


Determine cantilever deflection due to linearly distributed load on overhang. [Appendix C, Cantilever beam with linear load.]


4

0

30A

w Lv


Values:

w0 = −6 kips/ft, L = 12 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:

4 4 3

0

7 2

( 6 kips/ft)(12 ft) (12 in./ft)0.293139 in.

30 30(2.4447 10 kip-in. )A

w Lv

EI

Consider deflection at A resulting from rotation at B caused by linear load on overhang.



3

B

ML


Values:

M = ½(6 kips/ft)(12 ft)(4 ft) = 144 kip-ft,

L = 18 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2

7 2

(144 kip-ft)(18 ft)(12 in./ft)0.0050892 rad

3 3(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0050892 rad) 0.732847 in.

B

A

ML

EI

v

Consider uniformly distributed loads between supports. [Appendix C, SS beam with uniformly distributed load.]


3

24B

wL


Values:

w = 6 kips/ft, L = 18 ft,

EI = 2.4447 × 107 kip-in.

2




Computation:

3 3 2

7 2

(6 kips/ft)(18 ft) (12 in./ft)0.0085880 rad

24 24(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0085880 rad) 1.236679 in.

B

A

wL

EI

v

Consider deflection at A resulting from rotation at B caused by uniform load on overhang DE.



6

B

ML


Values:


L = 18 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2

7 2

(108 kip-ft)(18 ft)(12 in./ft)0.0019085 rad

6 6(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0019085 rad) 0.274818 in.

B

A

ML

EI

v


0.293139 in. 0.732847 in. 1.236679 in. 0.274818 in. 0.064124 in 0.0641 in. .Av Ans.





Consider deflection at C from moment caused by linear load on overhang.



2 2(2 3 )6

C

M xv L Lx x

LEI (elastic curve)

Values:


L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6

(144 kip-ft)(9 ft)(12 in./ft)2(18 ft) 3(18 ft)(9 ft) (9 ft) 0.206112 in.

6(18 ft)(2.4447 10 kip-in. )

C

M xv L Lx x

LEI



45

384C

wLv

EI

Values:

w = −6 kips/ft, L = 18 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:

4 4 2

7 2

5 5( 6 kips/ft)(18 ft) (12 in./ft)0.579693 in.

384 384(2.4447 10 kip-in. )C

wLv

EI

Consider deflection at C resulting from moment caused by uniform load on overhang DE.



2 2(2 3 )

6C

M xv L Lx x

LEI (elastic curve)

Values:


L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6


6(18 ft)(2.4447 10 kip-in. )

C

M xv L Lx x

LEI


0.206112 in. 0.579693 in. 0.154585 in. 0.218995 in 0.219 in. .Cv Ans.







For a loading of w = 8 kips/ft, determine:

(a) the beam deflection at point C.

(b) the beam deflection at point E. Fig. P10.64

Solution

(a) Beam deflection at point C

Consider deflection at C from moment caused by linear load on overhang.



2 2(2 3 )6

C

M xv L Lx x

LEI (elastic curve)

Values:


L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2 2

3

2 2

7 2

(2 3 )6


6(18 ft)(2.4447 10 kip-in. )

C

M xv L Lx x

LEI



45

384C

wLv

EI

Values:

w = −8 kips/ft, L = 18 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:

4 4 2

7 2

5 5( 8 kips/ft)(18 ft) (12 in./ft)0.772924 in.

384 384(2.4447 10 kip-in. )C

wLv

EI

Consider deflection at C resulting from moment caused by uniform load on overhang DE.



2 2(2 3 )

6C

M xv L Lx x

LEI (elastic curve)

Values:


L = 18 ft, x = 9 ft, EI = 2.4447 × 107 kip-in.

2

Computation:




2 2

3

2 2

7 2

(2 3 )6


6(18 ft)(2.4447 10 kip-in. )

C

M xv L Lx x

LEI


0.274816 in. 0.772924 in. 0.206113 in. 0.291995 in 0.292 in. .Cv Ans.

(b) Beam deflection at point E

Consider deflection at E resulting from rotation at D caused by linear load on overhang.



6

D

ML


Values:


L = 18 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2

7 2

(192 kip-ft)(18 ft)(12 in./ft)0.0033928 rad

6 6(2.4447 10 kip-in. )

(6 ft)(12 in./ft)(0.0033928 rad) 0.244282 in.

D

E

ML

EI

v



3

24D

wL


Values:

w = 8 kips/ft, L = 18 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:

3 3 2

7 2


24 24(2.4447 10 kip-in. )

(6 ft)(12 in./ft)(0.0114507 rad) 0.824448 in.

D

E

wL

EI

v

Consider deflection at E resulting from rotation at D caused by uniform load on overhang DE.



3

D

ML


Values:


L = 18 ft, EI = 2.4447 × 107 kip-in.

2




Computation:

2

7 2

(144 kip-ft)(18 ft)(12 in./ft)0.0050892 rad

3 3(2.4447 10 kip-in. )

(6 ft)(12 in./ft)(0.0050892 rad) 0.366422 in.

D

E

ML

EI

v

Determine cantilever deflection due to uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with uniformly distributed load.]


4

8E

wLv

EI (assuming fixed support at D)

Values:

w = −8 kips/ft, L = 6 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

4 4 3

7 2

( 8 kips/ft)(6 ft) (12 in./ft)0.091605 in.

8 8(2.4447 10 kip-in. )E

wLv

EI

Beam deflection at E

0.244282 in. 0.824448 in. 0.366422 in. 0.091605 in.

0.122139 i 0n. .1221 in.

Ev

Ans.






belt pulleys. Assume that the bearing at B can

be idealized as a roller support and that the

bearing at D can be idealized as a pin support.

For the loading shown, determine:

(a) the shaft deflection at pulley A.


Fig. P10.65

Solution

Section properties:

4 4(30 mm) 39,760.78 mm64

I

(a) Shaft deflection at pulley A

Determine cantilever deflection due to pulley A load. [Appendix C, Cantilever beam with concentrated load.]


3

3A

PLv


Values:

P = 700 N, L = 500 mm,

EI = 7.95216 × 109 N-mm

2

Computation:

3 3

9 2

(700 N)(500 mm)3.6678 mm

3 3(7.95216 10 N-mm )A

PLv

EI

Consider deflection at A resulting from rotation at B caused by pulley A load.



3

B

ML


Values:

M = (700 N)(500 mm) = 350,000 N-mm,

L = 1,800 mm, EI = 7.95216 × 109 N-mm

2

Computation:

9 2

(350,000 N-mm)(1,800 mm)0.0264079 rad

3 3(7.95216 10 N-mm )

(500 mm)(0.0264079 rad) 13.2040 mm

B

A

ML

EI

v

Consider deflection at A resulting from rotation at B caused by pulley C load.

[Appendix C, SS beam with concentrated load.]





2

16B

PL


Values:

P = 1,000 N, L = 1,800 mm,

EI = 7.95216 × 109 N-mm

2

Computation:

2 2

9 2

(1,000 N)(1,800 mm)0.0254648 rad

16 16(7.95216 10 N-mm )

(500 mm)(0.0254648 rad) 12.7324 mm

B

A

PL

EI

v

Shaft deflection at A

3.6678 mm 13.2040 mm 12.7324 mm 4.1393 mm 4.14 mmAv Ans.

(b) Shaft deflection at pulley C

Consider pulley A load. [Appendix C, SS beam with concentrated moment.]


2 2(2 3 )

6C

M xv L Lx x

LEI (elastic curve)

Values:

M = (700 N)(500 mm) = −350,000 N-mm,

L = 1,800 mm, x = 900 mm,

EI = 7.95216 × 109 N-mm

2

Computation:

2 2

2 2

9 2

(2 3 )6

( 350,000 N-mm)(900 mm)2(1,800 mm) 3(1,800 mm)(900 mm) (900 mm)

6(1,800 mm)(7.95216 10 N-mm )

8.9127 mm

C

M xv L Lx x

LEI

Consider pulley C load. [Appendix C, SS beam with concentrated load.]


3

48C

PLv

EI

Values:

P = 1,000 N, L = 1,800 mm,

EI = 7.95216 × 109 N-mm

2

Computation:

3 3

9 2

(1,000 N)(1,800 mm)15.2789 mm

48 48(7.95216 10 N-mm )C

PLv

EI

Shaft deflection at C

8.9127 mm 15.2789 mm 6.3662 mm 6.37 mmCv Ans.






flange shape [E = 200 GPa; I = 552 × 106 mm

4].




Fig. P10.66

Solution


Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span.



4

8A

wLv

EI

Values:

w = −85 kN/m, L = 4 m, EI = 1.104 × 105 kN-m

2

Computation:

4 4

5 2

( 85 kN/m)(4 m)0.024638 m

8 8(1.104 10 kN-m )A

wLv

EI

Consider a downward 85 kN/m uniformly distributed load acting over span BC.



4 3

and8 6

B B

wL wLv

EI EI (magnitude)

Values:

w = 85 kN/m, L = 2.5 m, EI = 1.104 × 105 kN-m

2

Computation:

4 4

5 2

3 3

5 2

(85 kN)(2.5 m)0.003759 m

8 8(1.104 10 kN-m )

(85 kN)(2.5 m)0.0020050 rad

6 6(1.104 10 kN-m )

0.003759 m (1.5 m)(0.0020050 rad) 0.006767 m

B

B

A

wLv

EI

wL

EI

v


0.024638 m 0.006767 m 0.017871 17 m .87 mmAv Ans.








2

2 2(6 4 )24

B

wxv L Lx x

EI (elastic curve)

Values:

w = −85 kN/m, L = 4 m, x = 2.5 m,

EI = 1.104 × 105 kN-m

2

Computation:

22 2

22 2

5 2

(6 4 )24

( 85 kN/m)(2.5 m)6(4 m) 4(4 m)(2.5 m) (2.5 m) 0.012481 m

24(1.104 10 kN-m )

B

wxv L Lx x

EI

Consider a downward 85 kN/m uniformly distributed load acting over span BC.



4

8B

wLv

EI

Values:

w = 85 kN/m, L = 2.5 m, EI = 1.104 × 105 kN-m

2

Computation:

4 4

5 2

(85 kN)(2.5 m)0.003759 m

8 8(1.104 10 kN-m )B

wLv

EI


0.012481 m 0.003759 m 0.008722 m 8.72 mmBv Ans.




10.67 The solid 30-mm-diameter steel [E

= 200 GPa] shaft shown in Fig. P10.67

supports two belt pulleys. Assume that

the bearing at A can be idealized as a pin

support and that the bearing at E can be

idealized as a roller support. For the

loading shown, determine the shaft

deflection at pulley B.

Fig. P10.67

Solution

Section properties:

4 4(30 mm) 39,760.78 mm64

I

Shaft deflection at pulley B

Consider pulley B load. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )

6B

Pabv L a b

LEI

Values:

P = 750 N, L = 1,000 mm, a = 300 mm,

b = 700 mm, EI = 7.95216 × 109 N-mm

2

Computation:

2 2 2

2 2 2

9 2

( )6

(750 N)(300 mm)(700 mm)(1,000 mm) (300 mm) (700 mm)

6(1,000 mm)(7.95216 10 N-mm )

1.38642 mm

B

Pabv L a b

LEI

Consider pulley D load. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )

6B

Pbxv L b x

LEI (elastic curve)

Values:

P = 500 N, L = 1,000 mm, x = 300 mm,

b = 200 mm, EI = 7.95216 × 109 N-mm

2

Computation:

2 2 2

2 2 2

9 2

( )6

(500 N)(200 mm)(300 mm)(1,000 mm) (200 mm) (300 mm)

6(1,000 mm)(7.95216 10 N-mm )

0.54702 mm

B

Pbxv L b x

LEI

Shaft deflection at B

1.38642 mm 0.54702 mm 1.93344 mm 1.933 mmBv Ans.




10.68 The solid 30-mm-diameter steel [E

= 200 GPa] shaft shown in Fig. P10.68

supports two belt pulleys. Assume that


support and that the bearing at E can be

idealized as a roller support. For the

loading shown, determine the shaft

deflection at pulley D.

Fig. P10.68

Solution

Section properties:

4 4(30 mm) 39,760.78 mm64

I

Shaft deflection at pulley D

Consider pulley B load. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )6

D

Pbxv L b x

LEI (elastic curve)

Values:

P = 750 N, L = 1,000 mm, x = 200 mm,

b = 300 mm, EI = 7.95216 × 109 N-mm

2

Computation:

2 2 2

2 2 2

9 2

( )6

(750 N)(300 mm)(200 mm)(1,000 mm) (300 mm) (200 mm)

6(1,000 mm)(7.95216 10 N-mm )

0.82053 mm

D

Pbxv L b x

LEI

Consider pulley D load. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )

6D

Pabv L a b

LEI

Values:

P = 500 N, L = 1,000 mm, a = 800 mm,

b = 200 mm, EI = 7.95216 × 109 N-mm

2

Computation:

2 2 2

2 2 2

9 2

( )6

(500 N)(800 mm)(200 mm)(1,000 mm) (800 mm) (200 mm)

6(1,000 mm)(7.95216 10 N-mm )

0.53654 mm

D

Pabv L a b

LEI

Shaft deflection at D

0.82053 mm 0.53654 mm 1.35707 mm 1.357 mmDv Ans.







mm4]. For the loading shown, determine the

beam deflection at point B.

Fig. P10.69

Solution

Beam deflection at point B Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

B

M xv L Lx x

LEI (elastic curve)

Values:

M = −180 kN-m, L = 6 m, x = 1.5 m,

EI = 4.32 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 180 kN-m)(1.5 m)2(6 m) 3(6 m)(1.5 m) (1.5 m) 0.008203 m

6(6 m)(4.32 10 kN-m )

B

M xv L Lx x

LEI



2 2 2( )

6B

Pabv L a b

LEI

Values:

P = 70 kN, L = 6 m, a = 1.5 m, b = 4.5 m,

EI = 4.32 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(70 kN)(1.5 m)(4.5 m)(6 m) (1.5 m) (4.5 m) 0.004102 m

6(6 m)(4.32 10 kN-m )

B

Pabv L a b

LEI




2

3 2 2 2 2(2 6 4 )24

B

wav x Lx a x L x a L

LEI

Values:

w = 80 kN/m, L = 6 m, a = 3 m, x = 4.5 m,

EI = 4.32 × 104 kN-m

2

Computation:




23 2 2 2 2

23 2 2 2 2

4 2

(2 6 4 )24

(80 kN/m)(3 m)2(4.5) 6(6)(4.5) (3) (4.5) 4(6) (4.5) (3) (6)

24(6.0 m)(4.32 10 kN-m )

0.010156 m

B


LEI


0.008203 m 0.004102 m 0.010156 m 0.006055 m 6.06 mmBv Ans.







mm4]. For the loading shown, determine the

beam deflection at point C.

Fig. P10.70

Solution

Beam deflection at point C Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

C

M xv L Lx x

LEI (elastic curve)

Values:

M = −180 kN-m, L = 6 m, x = 3.0 m,

EI = 4.32 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 180 kN-m)(3.0 m)2(6 m) 3(6 m)(3.0 m) (3.0 m) 0.009375 m

6(6 m)(4.32 10 kN-m )

C

M xv L Lx x

LEI



2 2 2( )

6C

Pbxv L b x

LEI (elastic curve)

Values:

P = 70 kN, L = 6 m, x = 3.0 m, b = 1.5 m,

EI = 4.32 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(70 kN)(1.5 m)(3.0 m)(6 m) (1.5 m) (3.0 m) 0.005013 m

6(6 m)(4.32 10 kN-m )

C

Pbxv L b x

LEI







3

2 2(4 7 3 )24

C

wav L aL a

LEI

Values:

w = 80 kN/m, L = 6 m, a = 3 m,

EI = 4.32 × 104 kN-m

2

Computation:

32 2

32 2

4 2

(4 7 3 )24

(80 kN/m)(3 m)4(6 m) 7(3 m)(6 m) 3(3 m) 0.015625 m

24(6.0 m)(4.32 10 kN-m )

C

wav L aL a

LEI


0.009375 m 0.005013 m 0.015625 m 0.011263 11.26 mm mCv Ans.







mm4]. If w = 80 kN/m, determine



Fig. P10.71

Solution


Determine cantilever deflection due to concentrated load on overhang AB. [Appendix C, Cantilever beam with concentrated load.]


3

3A

PLv


Values:

P = 35 kN, L = 4 m, EI = 7.02 × 104 kN-m

2

Computation:

3 3

4 2

(35 kN)(4 m)0.0106363 m

3 3(7.02 10 kN-m )A

PLv

EI

Consider deflection at A resulting from rotation at B caused by concentrated load on overhang

AB. [Appendix C, SS beam with concentrated moment.]


3

B

ML


Values:

M = (35 kN)(4 m) = 140 kN-m, L = 8 m,

EI = 7.02 × 104 kN-m

2

Computation:

4 2

(140 kN-m)(8 m)0.0053181 rad

3 3(7.02 10 kN-m )

(4 m)(0.0053181 rad) 0.0212726 m

B

A

ML

EI

v

Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.]


2

2 2(2 )24

B

waL a


Values:

w = 80 kN/m, L = 8 m, a = 4 m,

EI = 7.02 × 104 kN-m

2




Computation:

2 2

2 2 2 2

4 2

(80 kN/m)(4 m)(2 ) 2(8 m) (4 m) 0.0106363 rad

24 24(8 m)(7.02 10 kN-m )

(4 m)(0.0106363 rad) 0.0425451 m

B

A

waL a

LEI

v

Consider deflection at A resulting from rotation at B caused by uniform load on overhang DE.



6

B

ML


Values:

M = (80 kN/m)(2 m)(1 m) = 160 kN-m,

L = 8 m, EI = 7.02 × 104 kN-m

2

Computation:

4 2

(160 kN-m)(8 m)0.0030389 rad

6 6(7.02 10 kN-m )

(4 m)(0.0030389 rad) 0.0121557 m

B

A

ML

EI

v


0.0106363 m 0.0212726 m 0.0425451 m 0.0121557 m

0.0015195 m 1.520 mm

Av

Ans.


Consider concentrated moment from overhang AB.



2 2(2 3 )

6C

M xv L Lx x

LEI(elastic curve)

Values:

M = (35 kN)(4 m) = −140 kN-m, L = 8 m,

x = 4 m, EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 140 kN-m)(4 m)2(8 m) 3(8 m)(4 m) (4 m) 0.0079772 m

6(8 m)(7.02 10 kN-m )

C

M xv L Lx x

LEI






3

2 2(4 7 3 )24

C

wav L aL a

LEI

Values:

w = 80 kN/m, L = 8 m, a = 4 m,

EI = 7.02 × 104 kN-m

2

Computation:

32 2

32 2

4 2

(4 7 3 )24

(80 kN/m)(4 m)4(8 m) 7(4 m)(8 m) 3(4 m) 0.0303894 m

24(8 m)(7.02 10 kN-m )

C

wav L aL a

LEI

Consider concentrated moment from overhang DE.



2 2(2 3 )

6C

M xv L Lx x

LEI (elastic curve)

Values:

M = (80 kN/m)(2 m)(1 m) = 160 kN-m,

L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 160 kN-m)(4 m)2(8 m) 3(8 m)(4 m) (4 m) 0.0091168 m

6(8 m)(7.02 10 kN-m )

C

M xv L Lx x

LEI


0.0079772 m 0.0303894 m 0.0091168 m 0.0132954 1m 3.30 mmCv Ans.







mm4]. If w = 90 kN/m, determine:


(b) the beam deflection at point E.

Fig. P10.72

Solution


Consider concentrated moment from overhang AB.



2 2(2 3 )6

C

M xv L Lx x

LEI(elastic curve)

Values:

M = (35 kN)(4 m) = −140 kN-m, L = 8 m,

x = 4 m, EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 140 kN-m)(4 m)2(8 m) 3(8 m)(4 m) (4 m) 0.0079772 m

6(8 m)(7.02 10 kN-m )

C

M xv L Lx x

LEI



3

2 2(4 7 3 )24

C

wav L aL a

LEI

Values:

w = 90 kN/m, L = 8 m, a = 4 m,

EI = 7.02 × 104 kN-m

2

Computation:

32 2

3

2 2

4 2

(4 7 3 )24

(90 kN/m)(4 m)4(8 m) 7(4 m)(8 m) 3(4 m) 0.0341881 m

24(8 m)(7.02 10 kN-m )

C

wav L aL a

LEI




Consider concentrated moment from overhang DE.



2 2(2 3 )6

C

M xv L Lx x

LEI (elastic curve)

Values:

M = (90 kN/m)(2 m)(1 m) = 180 kN-m,

L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 180 kN-m)(4 m)2(8 m) 3(8 m)(4 m) (4 m) 0.0102564 m

6(8 m)(7.02 10 kN-m )

C

M xv L Lx x

LEI


0.0079772 m 0.0341881 m 0.0102564 m 0.0159545 1m 5.95 mmCv Ans.


Consider deflection at E resulting from rotation at D caused by concentrated load on overhang

AB. [Appendix C, SS beam with concentrated moment.]


6

D

ML


Values:

M = (35 kN)(4 m) = 140 kN-m, L = 8 m,

EI = 7.02 × 104 kN-m

2

Computation:

4 2

(140 kN-m)(8 m)0.0026591 rad

6 6(7.02 10 kN-m )

(2 m)(0.0026591 rad) 0.0053181 m

D

E

ML

EI

v



2

2(2 )24

D

waL a


Values:

w = 90 kN/m, L = 8 m, a = 4 m,

EI = 7.02 × 104 kN-m

2

Computation:

2 2

22

4 2

(90 kN/m)(4 m)(2 ) 2(8 m) (4 m) 0.0153846 rad

24 24(8 m)(7.02 10 kN-m )

(2 m)(0.0153846 rad) 0.0307692 m

D

E

waL a

LEI

v




Consider deflection at E resulting from rotation at D caused by uniform load on overhang DE.



3

D

ML


Values:

M = (90 kN/m)(2 m)(1 m) = 180 kN-m,

L = 8 m, EI = 7.02 × 104 kN-m

2

Computation:

4 2

(180 kN-m)(8 m)0.0068376 rad

3 3(7.02 10 kN-m )

(2 m)(0.0068376 rad) 0.0136753 m

D

E

ML

EI

v

Determine cantilever deflection due to uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with distributed load.]


4

8E

wLv


Values:

w = 90 kN/m, L = 2 m, EI = 7.02 × 104 kN-m

2

Computation: 4 4

4 2

(90 kN/m)(2 m)0.0025641 m

8 8(7.02 10 kN-m )E

wLv

EI


0.0053181 m 0.0307692 m 0.0136753 m 0.0025641 m

0.0092117 9.21 mmm

Ev

Ans.





Fig. P10.73 consists of a W16 × 40



shown, determine:


(b) the beam deflection at point F.

Fig. P10.73

Solution


Consider 40-kip concentrated load at B. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )6

C

Pbxv L b x

LEI (elastic curve)

Values:

P = 40 kips, L = 18 ft, b = 4 ft, x = 10 ft,

EI = 1.5022 × 107 kip-in.

2

Computation:

2 2 2

32 2 2

7 2

( )6

(40 kips)(4 ft)(10 ft)(12 in./ft)(18 ft) (4 ft) (10 ft) 0.354467 in.

6(18 ft)(1.5022 10 kip-in. )

C

Pbxv L b x

LEI

Consider 30-kip concentrated load at D. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )

6C

Pbxv L b x

LEI (elastic curve)

Values:

P = 30 kips, L = 18 ft, b = 6 ft, x = 8 ft,

EI = 1.5022 × 107 kip-in.

2

Computation:

2 2 2

32 2 2

7 2

( )6


6(18 ft)(1.5022 10 kip-in. )

C

Pbxv L b x

LEI




Consider 20-kip concentrated load at F. [Appendix C, SS beam with concentrated moment.]


2 2(2 3 )6

C

M xv L Lx x

LEI (elastic curve)

Values:

M = −(20 kips)(6 ft) = −120 kip-ft, L = 18 ft,

x = 10 ft, EI = 1.5022 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6


6(18 ft)(1.5022 10 kip-in. )

C

M xv L Lx x

LEI


0.354467 in. 0.343560 in. 0.265850 in. 0.432177 in. 0.432 in.Cv Ans.

(b) Beam deflection at point F

Consider 40-kip concentrated load at B. [Appendix C, SS beam with concentrated load not at midspan.]


2 2( )

6E

Pa L a


Values:

P = 40 kips, L = 18 ft, a = 4 ft,

EI = 1.5022 × 107 kip-in.

2

Computation:

2 2 22 2

7 2

( ) (40 kips)(4 ft)(12 in./ft)(18 ft) (4 ft) 0.0043740 rad

6 6(18 ft)(1.5022 10 kip-in. )

(6 ft)(12 in./ft)(0.0043740 rad) 0.314930 in.

E

F

Pa L a

LEI

v

Consider 30-kip concentrated load at D. [Appendix C, SS beam with concentrated load not at midspan.]


2 2( )

6E

Pa L a


Values:

P = 30 kips, L = 18 ft, x = 8 ft, a = 12 ft,

EI = 1.5022 × 107 kip-in.

2

Computation:

2 2 22 2

7 2


6 6(18 ft)(1.5022 10 kip-in. )

(6 ft)(12 in./ft)(0.0057516 rad) 0.414113 in.

E

F

Pa L a

LEI

v




Consider deflection at F resulting from rotation at E caused by 20-kip load on overhang EF.



3

E

ML


Values:

M = (20 kips)(6 ft) = 120 kip-ft, L = 18 ft,

EI = 1.5022 × 107 kip-in.

2

Computation:

2

7 2

(120 kip-ft)(18 ft)(12 in./ft)0.0069019 rad

3 3(1.5022 10 kip-in. )

(6 ft)(12 in./ft)(0.0069019 rad) 0.496935 in.

E

F

ML

EI

v

Determine cantilever deflection due to concentrated load on overhang EF. [Appendix C, Cantilever beam with concentrated load.]


3

3F

PLv

EI (assuming fixed support at E)

Values:


2

Computation: 3 3 3

7 2

(20 kips)(6 ft) (12 in./ft)0.165645 in.

3 3(1.5022 10 kip-in. )F

PLv

EI

Beam deflection at F

0.314930 in. 0.414113 in. 0.496935 in. 0.165645 in.

0.066463 in 0.0665 in. .

Fv

Ans.





consists of a rectangular structural steel tube

shape [E = 200 GPa; I = 170 × 106 mm

4]. For

the loading shown, determine:



Fig. P10.74

Solution




4

8A

wLv

EI

Values:

w = −65 kN/m, L = 6 m, EI = 3.4 × 104 kN-m

2

Computation:

4 4

4 2

( 65 kN/m)(6 m)0.309706 m

8 8(3.4 10 kN-m )A

wLv

EI

Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.]


3

3A

PLv

EI

Values:

P = 90 kN, L = 6 m, EI = 3.4 × 104 kN-m

2

Computation:

3 3

4 2

(90 kN)(6 m)0.190588 m

3 3(3.4 10 kN-m )A

PLv

EI

Consider 30-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.]


3 2

and3 2

B B

PL PLv

EI EI (magnitude)

Values:

P = 30 kN, L = 3.5 m, EI = 3.4 × 104 kN-m

2

Computation:

3 3

4 2

(30 kN)(3.5 m)0.012610 m

3 3(3.4 10 kN-m )B

PLv

EI (a)




2 2

4 2

(30 kN)(3.5 m)0.0054044 rad

2 2(3.4 10 kN-m )

0.012610 m (2.5 m)(0.0054044 rad) 0.026121 m

B

A

PL

EI

v

Consider 225 kN-m concentrated moment at B. [Appendix C, Cantilever beam with concentrated moment at tip.]


2

and2

B B

ML MLv


Values:

M = 225 kN-m, L = 3.5 m, EI = 3.4 × 104 kN-m

2

Computation:

2 2

4 2

(225 kN-m)(3.5 m)0.040533 m

2 2(3.4 10 kN-m )B

MLv

EI (b)

4 2

(225 kN-m)(3.5 m)0.0231618 rad

(3.4 10 kN-m )

0.040533 m (2.5 m)(0.0231618 rad) 0.098438 m

B

A

ML

EI

v


0.309706 m 0.190588 m 0.026121 m 0.098438 m 0.005441 m 5.44 mmAv Ans.




2

2 2(6 4 )24

B

wxv L Lx x

EI (elastic curve)

Values:

w = −65 kN/m, L = 6 m, x = 3.5 m,

EI = 3.4 × 104 kN-m

2

Computation:

22 2

22 2

4 2

(6 4 )24

( 65 kN/m)(3.5 m)6(6 m) 4(6 m)(3.5 m) (3.5 m) 0.140759 m

24(3.4 10 kN-m )

B

wxv L Lx x

EI




Consider 90-kN concentrated load at A. [Appendix C, Cantilever beam with concentrated load at tip.]


2

(3 )6

B

Pxv L x

EI (elastic curve)

Values:

P = 90 kN, L = 6 m, x = 3.5 m,

EI = 3.4 × 104 kN-m

2

Computation:

2 2

4 2

(90 kN)(3.5 m)(3 ) 3(6 m) (3.5 m) 0.078364 m

6 6(3.4 10 kN-m )B

Pxv L x

EI

Consider 30-kN concentrated load at B. Previously calculated in Eq. (a).

Consider 225 kN-m concentrated moment at B. Previously calculated in Eq. (b).


0.140759 m 0.078364 m 0.012610 m 0.040533 m 0.009252 m 9.25 mmBv Ans.





P10.75 consists of a rectangular structural steel

tube shape [E = 200 GPa; I = 350 × 106 mm

4].



(b) the beam deflection at point E.

Fig. P10.75

Solution


Consider 315 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )

6C

M xv L Lx x

LEI (elastic curve)

Values:

M = −315 kN-m, L = 9 m, x = 6 m,

EI = 7.0 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 315 kN-m)(6 m)2(9 m) 3(9 m)(6 m) (6 m) 0.018000 m

6(9 m)(7.0 10 kN-m )

C

M xv L Lx x

LEI

Consider 120 kN/m uniformly distributed load.



2

3 2 2 2 2(2 6 4 )24

C


LEI

Values:

w = 120 kN/m, L = 9 m, a = 3 m, x = 6 m,

EI = 7.0 × 104 kN-m

2

Computation: 2

3 2 2 2 2

23 2 2 2 2

4 2

(2 6 4 )24

(120 kN/m)(3 m)2(6 m) 6(9 m)(6 m) (3 m) (6 m) 4(9 m) (6 m) (3 m) (9 m)

24(9 m)(7.0 10 kN-m )

0.028929 m

C


LEI




Consider 100-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]


2 2 2( )6

C

Pabv L a b

LEI

Values:

P = 100 kN, L = 9 m, a = 6 m, b = 3 m,

EI = 7.0 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(100 kN)(6 m)(3 m)(9 m) (6 m) (3 m) 0.017143 m

6(9 m)(7.0 10 kN-m )

C

Pabv L a b

LEI

Consider 60 kN/m uniformly distributed load on overhang DE. [Appendix C, SS beam with concentrated moment.]


2 2(2 3 )

6C

M xv L Lx x

LEI (elastic curve)

Values:

M = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m,

L = 9 m, x = 3 m, EI = 7.0 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 270 kN-m)(3 m)2(9 m) 3(9 m)(3 m) (3 m) 0.019286 m

6(9 m)(7.0 10 kN-m )

C

M xv L Lx x

LEI


0.018000 m 0.028929 m 0.017143 m 0.019286 m 0.008786 m 8.79 mmCv Ans.




6

D

ML


Values:

M = −315 kN-m, L = 9 m,

EI = 7.0 × 104 kN-m

2

Computation:

4 2

( 315 kN-m)(9 m)0.0067500 rad

6 6(7.0 10 kN-m )

(3 m)( 0.0067500 rad) 0.020250 m

D

E

ML

EI

v




Consider 120 kN/m uniformly distributed load.



2

2 2(2 )24

D

waL a


Values:

w = 120 kN/m, L = 9 m, a = 3 m,

EI = 7.0 × 104 kN-m

2

Computation:

2 22 2 2 2

4 2

(120 kN/m)(3 m)(2 ) 2(9 m) (3 m) 0.0109286 rad

24 24(9 m)(7.0 10 kN-m )

(3 m)(0.0109286 rad) 0.032786 m

D

E

waL a

LEI

v



2 2( )

6D

Pa L a


Values:

P = 100 kN, L = 9 m, a = 6 m,

EI = 7.0 × 104 kN-m

2

Computation:

2 22 2

4 2

( ) (100 kN)(6 m)(9 m) (6 m) 0.0071429 rad

6 6(9 m)(7.0 10 kN-m )

(3 m)(0.0071429 rad) 0.021429 m

D

E

Pa L a

LEI

v

Consider 60 kN/m uniformly distributed load on overhang DE. [Appendix C, SS beam with concentrated moment.]


3

D

ML


Values:

M = −(60 kN/m)(3 m)(1.5 m) = −270 kN-m,

L = 9 m, EI = 7.0 × 104 kN-m

2

Computation:

4 2

(270 kN-m)(9 m)0.0115714 rad

3 3(7.0 10 kN-m )

(3 m)(0.0115714 rad) 0.034714 m

D

E

ML

EI

v




Determine cantilever deflection due to 60 kN/m uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with concentrated load.]


4

8E

wLv


Values:

w = 60 kN/m, L = 3 m, EI = 7.0 × 104 kN-m

2

Computation: 4 4

4 2

(60 kN-m)(3 m)0.008679 m

8 8(7.0 10 kN-m )E

wLv

EI


0.020250 m 0.032786 m 0.021429 m 0.034714 m 0.008679 m

0.009429 m 9.43 mm

Ev

Ans.






shape [E = 200 GPa; I = 95 × 106 mm

4]. For the

loading shown, determine the beam deflection at

point B.

Fig. P10.76

Solution

Consider the downward 50 kN/m uniformly distributed load acting over span AB.



4

8B

wLv

EI

Values:

w = 50 kN/m, L = 2 m, EI = 1.9 × 104 kN-m

2

Computation:

4 4

4 2

(50 kN/m)(2 m)0.0052632 m

8 8(1.9 10 kN-m )B

wLv

EI




2

2 2(6 4 )24

B

wxv L Lx x

EI (elastic curve)

Values:

w = −25 kN/m, L = 5 m, x = 2 m,

EI = 1.9 × 104 kN-m

2

Computation:

22 2

22 2

4 2

(6 4 )24

( 25 kN/m)(2 m)6(5 m) 4(5 m)(2 m) (2 m) 0.0250000 m

24(1.9 10 kN-m )

B

wxv L Lx x

EI

Consider a downward 25 kN/m uniformly distributed load acting over span AB.



4

8B

wLv

EI

Values:

w = 25 kN/m, L = 2 m, EI = 1.9 × 104 kN-m

2

Computation:

4 4

4 2

(25 kN/m)(2 m)0.0026316 m

8 8(1.9 10 kN-m )B

wLv

EI






3

3B

PLv

EI

Values:

P = −20 kN, L = 2 m, EI = 1.9 × 104 kN-m

2

Computation:

3 3

4 2

( 20 kN)(2 m)0.0028070 m

3 3(1.9 10 kN-m )B

PLv

EI

Consider 50-kN concentrated load at C. [Appendix C, Cantilever beam with concentrated load at tip.]


2

(3 )6

B

Pxv L x

EI (elastic curve)

Values:

P = 50 kN, L = 5 m, x = 2 m, EI = 1.9 × 104 kN-m

2

Computation:

2 2

4 2

(50 kN)(2 m)(3 ) 3(5 m) (2 m) 0.0228070 m

6 6(1.9 10 kN-m )B

Pxv L x

EI


0.0052632 m 0.0250000 m 0.0026316 m 0.0028070 m 0.0228070 m

0.0028947 m 2.89 mm

Bv

Ans.






shape [E = 200 GPa; I = 95 × 106 mm

4]. For the

loading shown, determine the beam deflection at

point C.

Fig. P10.77

Solution

Consider the downward 50 kN/m uniformly distributed load acting over span AB.



4

8B

wLv

EI and

3

6B

wL


Values:

w = 50 kN/m, L = 2 m, EI = 1.9 × 104 kN-m

2

Computation:

4 4

4 2

(50 kN/m)(2 m)0.0052632 m

8 8(1.9 10 kN-m )B

wLv

EI

3 3

4 2

(50 kN/m)(2 m)0.0035088 rad

6 6(1.9 10 kN-m )

0.0052632 m (3 m)(0.0035088 rad) 0.0157895 m

B

C

wL

EI

v




4

8C

wLv

EI

Values:

w = −25 kN/m, L = 5 m, EI = 1.9 × 104 kN-m

2

Computation:

4 4

4 2

( 25 kN/m)(5 m)0.1027961 m

8 8(1.9 10 kN-m )C

wLv

EI

Consider a downward 25 kN/m uniformly distributed load acting over span AB.



4

8B

wLv

EI and

3

6B

wL


Values:

w = 25 kN/m, L = 2 m, EI = 1.9 × 104 kN-m

2

Computation:




4 4

4 2

(25 kN/m)(2 m)0.0026316 m

8 8(1.9 10 kN-m )B

wLv

EI

3 3

4 2

(25 kN)(2 m)0.0017544 rad

6 6(1.9 10 kN-m )

0.0026316 m (3 m)(0.0017544 rad) 0.0078948 m

B

C

wL

EI

v



3

3B

PLv

EI and

2

2B

PL


Values:

P = −20 kN, L = 2 m, EI = 1.9 × 104 kN-m

2

Computation:

3 3

4 2

( 20 kN)(2 m)0.0028070 m

3 3(1.9 10 kN-m )B

PLv

EI

2 2

4 2

(20 kN)(2 m)0.0021053 rad

2 2(1.9 10 kN-m )

0.0028070 m (3 m)(0.0021053 rad) 0.0091228 m

B

C

PL

EI

v

Consider 50-kN concentrated load at C. [Appendix C, Cantilever beam with concentrated load at tip.]


3

3C

PLv

EI

Values:

P = 50 kN, L = 5 m, EI = 1.9 × 104 kN-m

2

Computation:

3 3

4 2

(50 kN)(5 m)0.1096491 m

3 3(1.9 10 kN-m )C

PLv

EI


0.0157895 m 0.1027961 m 0.0078948 m 0.0091228 m 0.1096491 m

0.0214145 m 21.4 mm

Cv

Ans.







If w = 5 kips/ft, determine:



Fig. P10.78

Solution


Consider cantilever beam deflection of 85 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at one end.]


2

2A

MLv

EI

Values:


2

Computation:

2 2 3

6 2

(85 kip-ft)(3 ft) (12 in./ft)0.134069 in.

2 2(4.93 10 kip-in. )A

MLv

EI

Consider rotation at B caused by 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


3

B

ML


Values:


2

Computation:

2

6 2

(85 kip-ft)(15 ft)(12 in./ft)0.0124138 rad

3 3(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0124138 rad) 0.446897 in.

B

A

ML

EI

v

Consider cantilever beam deflection of 5 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]


4

8A

wLv

EI

Values:


2




Computation:

4 4 3

6 2

(5 kips/ft)(3 ft) (12 in./ft)0.017744 in.

8 8(4.93 10 kip-in. )A

wLv

EI

Consider rotation at B caused by 5 kips/ft uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.]


3

B

ML


Values:

M = (5 kips/ft)(3 ft)(1.5 ft) = 22.5 kip-ft,

L = 15 ft, EI = 4.93 × 106 kip-in.

2

Computation:

2

6 2

(22.5 kip-ft)(15 ft)(12 in./ft)0.0032860 rad

3 3(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0032860 rad) 0.118296 in.

B

A

ML

EI

v

Consider 5 kips/ft uniformly distributed load on segment BC.



2

2(2 )24

B

waL a


Values:

w = 5 kips/ft, L = 15 ft, a = 5 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

2 2 222

6 2

(5 kips/ft)(5 ft) (12 in./ft)(2 ) 2(15 ft) (5 ft) 0.0063387 rad

24 24(15 ft)(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0063387 rad) 0.228195 in.

B

A

waL a

LEI

v

Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.]


2 2( )

6B

Pb L b


Values:

P = 25 kips, L = 15 ft, b = 5 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

2 2 22 2

6 2


6 6(15 ft)(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0081136 rad) 0.292089 in.

B

A

Pb L b

LEI

v





0.134069 in. 0.446897 in. 0.017744 in. 0.118296 in. 0.228195 in. 0.292089 in.

0.196722 in. 0.1967 in.

Av

Ans.


Consider 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

C

M xv L Lx x

LEI (elastic curve)

Values:

M = −85 kip-ft, L = 15 ft, x = 5 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

2 2

32 2

6 2

(2 3 )6


6(15 ft)(4.93 10 kip-in. )

C

M xv L Lx x

LEI

Consider moment at B caused by 5 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )

6C

M xv L Lx x

LEI (elastic curve)

Values:

M = −(5 kips/ft)(3 ft)(1.5 ft) = −22.5 kip-ft,

L = 15 ft, x = 5 ft, EI = 4.93 × 106 kip-in.

2

Computation:

2 2

32 2

6 2

(2 3 )6

( 22.5 kip-ft)(5 ft)(12 in./ft)2(15 ft) 3(15 ft)(5 ft) (5 ft) 0.109533 in.

6(15 ft)(4.93 10 kip-in. )

C

M xv L Lx x

LEI




3

2 2(4 7 3 )24

C

wav L aL a

LEI

Values:

w = 5 kips/ft, L = 15 ft, a = 5 ft,

EI = 4.93 × 106 kip-in.

2




Computation:

32 2

3 32 2

6 2

(4 7 3 )24


24(15 ft)(4.93 10 kip-in. )

C

wav L aL a

LEI



2 2 2( )6

C

Pbxv L b x

LEI (elastic curve)

Values:

P = 25 kips, L = 15 ft, b = 5 ft, x = 5 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

2 2 2

32 2 2

6 2

( )6


6(15 ft)(4.93 10 kip-in. )

C

Pbxv L b x

LEI


0.413793 in. 0.109533 in. 0.273834 in. 0.425963 in.

0.176471 i 0n. .1765 in.

Cv

Ans.







If w = 9 kips/ft, determine:


(b) the beam deflection at point D.

Fig. P10.79

Solution


Consider cantilever beam deflection of 85 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at one end.]


2

2A

MLv

EI

Values:


2

Computation:

2 2 3

6 2

(85 kip-ft)(3 ft) (12 in./ft)0.134069 in.

2 2(4.93 10 kip-in. )A

MLv

EI

Consider rotation at B caused by 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


3

B

ML


Values:


2

Computation:

2

6 2

(85 kip-ft)(15 ft)(12 in./ft)0.0124138 rad

3 3(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0124138 rad) 0.446897 in.

B

A

ML

EI

v

Consider cantilever beam deflection of 9 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]


4

8A

wLv

EI

Values:


2




Computation:

4 4 3

6 2

(9 kips/ft)(3 ft) (12 in./ft)0.031939 in.

8 8(4.93 10 kip-in. )A

wLv

EI

Consider rotation at B caused by 9 kips/ft uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.]


3

B

ML


Values:

M = (9 kips/ft)(3 ft)(1.5 ft) = 40.5 kip-ft,

L = 15 ft, EI = 4.93 × 106 kip-in.

2

Computation:

2

6 2

(40.5 kip-ft)(15 ft)(12 in./ft)0.0059148 rad

3 3(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0059148 rad) 0.212933 in.

B

A

ML

EI

v




2

2(2 )24

B

waL a


Values:

w = 9 kips/ft, L = 15 ft, a = 5 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

2 2 222

6 2


24 24(15 ft)(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0114097 rad) 0.410748 in.

B

A

waL a

LEI

v



2 2( )

6B

Pb L b


Values:

P = 25 kips, L = 15 ft, b = 5 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

2 2 22 2

6 2


6 6(15 ft)(4.93 10 kip-in. )

(3 ft)(12 in./ft)(0.0081136 rad) 0.292089 in.

B

A

Pb L b

LEI

v





0.134069 in. 0.446897 in. 0.031939 in. 0.212933 in. 0.410748 in. 0.292089 in.

0.123001 in. 0.1230 in.

Av

Ans.

(b) Beam deflection at point D

Consider 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

D

M xv L Lx x

LEI (elastic curve)

Values:

M = −85 kip-ft, L = 15 ft, x = 10 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

2 2

32 2

6 2

(2 3 )6


6(15 ft)(4.93 10 kip-in. )

D

M xv L Lx x

LEI

Consider moment at B caused by 9 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )

6D

M xv L Lx x

LEI (elastic curve)

Values:

M = −(9 kips/ft)(3 ft)(1.5 ft) = −40.5 kip-ft,

L = 15 ft, x = 10 ft, EI = 4.93 × 106 kip-in.

2

Computation:

2 2

3

2 2

6 2

(2 3 )6

( 40.5 kip-ft)(10 ft)(12 in./ft)2(15 ft) 3(15 ft)(10 ft) (10 ft) 0.157729 in.

6(15 ft)(4.93 10 kip-in. )

D

M xv L Lx x

LEI




2

3 2 2 2 2(2 6 4 )24

D


LEI

Values:

w = 9 kips/ft, L = 15 ft, a = 5 ft, x = 10 ft,

EI = 4.93 × 106 kip-in.

2




Computation: 2

3 2 2 2 2

2 3

3 2 2 2 2

6 2

(2 6 4 )24

(9 kips/ft)(5 ft) (12 in./ft)2(10 ft) 6(15 ft)(10 ft) (5 ft) (10 ft) 4(15 ft) (10 ft) (5 ft) (15 ft)

24(15 ft)(4.93 10 kip-in. )

0.410751 in.

D


LEI



2 2 2( )6

D

Pabv L a b

LEI

Values:

P = 25 kips, L = 15 ft, a = 10 ft, b = 5 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

2 2 2

32 2 2

6 2

( )6


6(15 ft)(4.93 10 kip-in. )

D

Pabv L a b

LEI

Beam deflection at D

0.331034 in. 0.157729 in. 0.410751 in. 0.486815 in.

0.408803 in. 0.409 in.

Dv

Ans.





Fig. P10.80 consists of a W10 × 30 structural

steel wide-flange shape [E = 29,000 ksi; I =

170 in.4]. For the loading shown, determine:



Fig. P10.80

Solution


Consider cantilever beam deflection of linearly distributed load on overhang AB. [Appendix C, Cantilever beam with linearly distributed load.]


4

0

30A

w Lv

EI

Values:

w0 = 8 kips/ft, L = 9 ft, EI = 4.93 × 106 kip-in.

2

Computation:

4 4 3

0

6 2

(8 kips/ft)(9 ft) (12 in./ft)0.613247 in.

30 30(4.93 10 kip-in. )A

w Lv

EI

Consider rotation at B caused by linearly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]


3

B

ML


Values:


L = 18 ft, EI = 4.93 × 106 kip-in.

2

Computation:

2

6 2

(108 kip-ft)(18 ft)(12 in./ft)0.0189274 rad

3 3(4.93 10 kip-in. )

(9 ft)(12 in./ft)(0.0189274 rad) 2.044157 in.

B

A

ML

EI

v




Consider linearly distributed load from 8 kips/ft to 0 kips/ft over span BD. [Appendix C, SS beam with linearly distributed load.]


3

0

45B

w L


Values:

w0 = 8 kips/ft, L = 18 ft, EI = 4.93 × 106 kip-in.

2

Computation:

3 3 2

0

6 2


45 45(4.93 10 kip-in. )

(9 ft)(12 in./ft)(0.0302838 rad) 3.270652 in.

B

A

w L

EI

v

Consider 4 kips/ft uniformly distributed load on segment CD.



2

2 2(2 )24

B

waL a


Values:

w = 4 kips/ft, L = 18 ft, a = 9 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

22 2

2 22 2

6 2

(2 )24

(4 kips/ft)(9 ft) (12 in./ft)2(18 ft) (9 ft) 0.0124211 rad

24(18 ft)(4.93 10 kip-in. )

(9 ft)(12 in./ft)(0.0124211 rad) 1.341478 in.

B

A

waL a

LEI

v


0.613247 in. 2.044157 in. 3.270652 in. 1.341478 in. 1.954726 in. 1.955 in.Av Ans.


Consider moment at B caused by linearly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )

6C

M xv L Lx x

LEI

Values:

M = −½(8 kips/ft)(9 ft)(3 ft) = −108 kip-ft,

L = 18 ft, x = 9 ft, EI = 4.93 × 106 kip-in.

2




Computation:

2 2

32 2

6 2

(2 3 )6


6(18 ft)(4.93 10 kip-in. )

C

M xv L Lx x

LEI

Consider linearly distributed load from 8 kips/ft to 0 kips/ft over span BD. [Appendix C, SS beam with linearly distributed load.]


4 2 2 40 (7 10 3 )360

C

w xv L L x x

LEI

Values:

w0 = 8 kips/ft, L = 18 ft, x = 9 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

4 2 2 40

34 2 2 4

6 2

(7 10 3 )360

(8 kips/ft)(9 ft)(12 in./ft)7(18 ft) 10(18 ft) (9 ft) 3(9 ft) 1.916398 in.

360(18 ft)(4.93 10 kip-in. )

C

w xv L L x x

LEI

Consider 4 kips/ft uniformly distributed load on segment CD.



3

2 2(4 7 3 )24

C

wav L aL a

LEI

Values:

w = 4 kips/ft, L = 18 ft, a = 9 ft,

EI = 4.93 × 106 kip-in.

2

Computation:

32 2

3 32 2

6 2

(4 7 3 )24


24(18 ft)(4.93 10 kip-in. )

C

wav L aL a

LEI










shown, determine:



Fig. P10.81

Solution


Consider cantilever beam deflection of downward 4 kips/ft uniform load over AB. [Appendix C, Cantilever beam with uniformly distributed load.]


4

8A

wLv

EI

Values:


2

Computation:

4 4 3

7 2

(4 kips/ft)(12 ft) (12 in./ft)0.732847 in.

8 8(2.4447 10 kip-in. )A

wLv

EI

Consider cantilever beam deflection of upward 4 kips/ft uniform load over 6-ft segment. [Appendix C, Cantilever beam with uniformly distributed load.]


4 3

and8 6

wL wLv

EI EI(slope magnitude)

Values:

w = −4 kips/ft, L = 6 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

4 4 3

7 2

3 3 2

7 2

( 4 kips/ft)(6 ft) (12 in./ft)0.045803 in.

8 8(2.4447 10 kip-in. )


6 6(2.4447 10 kip-in. )

0.045803 in. (6 ft)(12 in./ft)(0.0008482 rad)A

wLv

EI

wL

EI

v 0.106873 in.




Consider rotation at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.]


3

B

ML


Values:


L = 24 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2

7 2

(216 kip-ft)(24 ft)(12 in./ft)0.0101784 rad

3 3(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0101784 rad) 1.465693 in.

B

A

ML

EI

v



2 2( )

6B

Pb L b


Values:

P = 42 kips, L = 24 ft, b = 18 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:

2 2 22 2

7 2


6 6(24 ft)(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0077929 rad) 1.122172 in.

B

A

Pb L b

LEI

v

Consider 4 kips/ft uniformly distributed load on 6-ft segment near D.



2

2 2(2 )24

B

waL a


Values:

w = 4 kips/ft, L = 24 ft, a = 6 ft,

EI = 2.4447 × 107 kip-in.

2

Computation:

2 2 22 2 2 2

7 2


24 24(24 ft)(2.4447 10 kip-in. )

(12 ft)(12 in./ft)(0.0016434 rad) 0.236648 in.

B

A

waL a

LEI

v


0.732847 in. 0.106873 in. 1.465693 in. 1.122172 in. 0.236648 in.

0.732847 in. 0.733 in.

Av

Ans.





Consider moment at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.]


2 2(2 3 )6

C

M xv L Lx x

LEI (elastic curve)

Values:

M = −(4 kips/ft)(6 ft)(9 ft) = −216 kip-ft,

L = 24 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2 2

32 2

7 2

(2 3 )6


6(24 ft)(2.4447 10 kip-in. )

C

M xv L Lx x

LEI



2 2 2( )

6C

Pbxv L b x

LEI (elastic curve)

Values:

P = 42 kips, L = 24 ft, b = 6 ft,

x = 12 ft, EI = 2.4447 × 107 kip-in.

2

Computation:

2 2 2

32 2 2

7 2

( )6


6(24 ft)(2.4447 10 kip-in. )

C

Pbxv L b x

LEI

Consider 4 kips/ft uniformly distributed load on 6-ft segment near D.



3 2 2 2 2(2 6 4 )24

C


LEI

(elastic curve)

Values:

w = 4 kips/ft, L = 24 ft, a = 6 ft, x = 12 ft,

EI = 2.4447 × 107 kip-in.

2




Computation: 2

3 2 2 2 2

3 2 22 3

7 2 2 2

(2 6 4 )24

2(12 ft) 6(24 ft)(12 ft) (6 ft) (12 ft)(4 kips/ft)(6 ft) (12 in./ft)

24(24 ft)(2.4447 10 kip-in. ) 4(24 ft) (12 ft) (6 ft) (24 ft)

0.175578 in.

C


LEI


0.549635 in. 0.587804 in. 0.175578 in. 0.213747 in 0.. 214 in.Cv Ans.







mm4]. If w = 85 kN/m, determine the beam

deflection at point B.

Fig. P10.82

Solution

Beam deflection at point B



2 2(2 3 )

6B

M xv L Lx x

LEI (elastic curve)

Values:

M = −300 kN-m, L = 9 m, x = 4 m,

EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 300 kN-m)(4 m)2(9 m) 3(9 m)(4 m) (4 m) 0.022159 m

6(9 m)(7.02 10 kN-m )

B

M xv L Lx x

LEI

Consider 85 kN/m uniformly distributed load on segment AB.



3

2 2(4 7 3 )24

B

wav L aL a

LEI

Values:

w = 85 kN/m, L = 9 m, a = 4 m,

EI = 7.02 × 104 kN-m

2

Computation:

32 2

32 2

4 2

(4 7 3 )24

(85 kN/m)(4 m)4(9 m) 7(4 m)(9 m) 3(4 m) 0.043052 m

24(9 m)(7.02 10 kN-m )

B

wav L aL a

LEI






2 2 2( )6

B

Pbxv L b x

LEI (elastic curve)

Values:

P = 140 kN, L = 9 m, b = 3 m, x = 4 m,

EI = 7.02 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(140 kN)(3 m)(4 m)(9 m) (3 m) (4 m) 0.024818 m

6(9 m)(7.02 10 kN-m )

B

Pbxv L b x

LEI



2 2(2 3 )

6B

M xv L Lx x

LEI (elastic curve)

Values:

M = −175 kN-m, L = 9 m, x = 5 m,

EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 175 kN-m)(5 m)2(9 m) 3(9 m)(5 m) (5 m) 0.012003 m

6(9 m)(7.02 10 kN-m )

B

M xv L Lx x

LEI


0.022159 m 0.043052 m 0.024818 m 0.012003 m 0.033708 m 33.7 mmBv Ans.







mm4]. If w = 115 kN/m, determine the beam

deflection at point C.

Fig. P10.83

Solution




2 2(2 3 )

6C

M xv L Lx x

LEI (elastic curve)

Values:

M = −300 kN-m, L = 9 m, x = 6 m,

EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 300 kN-m)(6 m)2(9 m) 3(9 m)(6 m) (6 m) 0.017094 m

6(9 m)(7.02 10 kN-m )

C

M xv L Lx x

LEI

Consider 115 kN/m uniformly distributed load on segment AB.



2

3 2 2 2 2(2 6 4 )24

C


LEI

Values:

w = 115 kN/m, L = 9 m, a = 4 m, x = 6 m,

EI = 7.02 × 104 kN-m

2

Computation: 2

3 2 2 2 2

2

3 2 2 2 2

4 2

(2 6 4 )24

(115 kN/m)(4 m)2(6 m) 6(9 m)(6 m) (4 m) (6 m) 4(9 m) (6 m) (4 m) (9 m)

24(9 m)(7.02 10 kN-m )

0.046597 m

C


LEI






2 2 2( )6

C

Pabv L a b

LEI

Values:

P = 140 kN, L = 9 m, a = 6 m, b = 3 m,

EI = 7.02 × 104 kN-m

2

Computation:

2 2 2

2 2 2

4 2

( )6

(140 kN)(6 m)(3 m)(9 m) (6 m) (3 m) 0.023932 m

6(9 m)(7.02 10 kN-m )

C

Pabv L a b

LEI



2 2(2 3 )

6C

M xv L Lx x

LEI (elastic curve)

Values:

M = −175 kN-m, L = 9 m, x = 3 m,

EI = 7.02 × 104 kN-m

2

Computation:

2 2

2 2

4 2

(2 3 )6

( 175 kN-m)(3 m)2(9 m) 3(9 m)(3 m) (3 m) 0.012464 m

6(9 m)(7.02 10 kN-m )

C

M xv L Lx x

LEI


0.017094 m 0.046597 m 0.023932 m 0.012464 m 0.040971 m 41.0 mmCv Ans.




10.84 A 25-ft-long soldier beam is used as a key

component of an earth retention system at an

excavation site. The soldier beam is subjected to

a soil loading that is linearly distributed from

520 lb/ft to 260 lb/ft, as shown in Fig. P10.84.

The soldier beam can be idealized as a

cantilever with a fixed support at A. Added

support is supplied by a tieback anchor at B,

which exerts a force of 5,000 lb on the soldier

beam. Determine the horizontal deflection of the

soldier beam at point C. Assume EI = 5 × 108

lb-

in.2.

Fig. P10.84

Solution

Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]


4

8C

wLv

EI

Values:

w = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.

2

Computation:

4 4 3

8 2

(260 lb/ft)(25 ft) (12 in./ft)43.875 in.

8 8(5.0 10 lb-in. )C

wLv

EI

Consider a linearly distributed load that varies from 260 lb/ft at A to 0 lb/ft at C. [Appendix C, Cantilever beam with linearly distributed load.]


4

0

30C

w Lv

EI

Values:

w0 = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.

2

Computation:

4 4 3

0

8 2

(260 lb/ft)(25 ft) (12 in./ft)11.700 in.

30 30(5.0 10 lb-in. )C

w Lv

EI




Consider 5,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.]


3 2

and3 2

B B

PL PLv


Values:

P = 5,000 lb, L = 18 ft, EI = 5.0 × 108 lb-in.

2

Computation:

3 3 3

8 2

2 2 2

8 2

(5,000 lb)(18 ft) (12 in./ft)33.592320 in.

3 3(5.0 10 lb-in. )

(5,000 lb)(18 ft) (12 in./ft)0.2332800 rad

2 2(5.0 10 lb-in. )

33.592320 in. (7 ft)(12 in./ft)(0.2332800 rad) 53.187

B

B

C

PLv

EI

PL

EI

v 840 in.






10.85 A 25-ft-long soldier beam is used as a key

component of an earth retention system at an

excavation site. The soldier beam is subjected to a

uniformly distributed soil loading of 260 lb/ft, as

shown in Fig. P10.85. The soldier beam can be

idealized as a cantilever with a fixed support at A.

Added support is supplied by a tieback anchor at B,

which exerts a force of 4,000 lb on the soldier beam.

Determine the horizontal deflection of the soldier

beam at point C. Assume EI = 5 × 108

lb-in.2.

Fig. P10.85

Solution

Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.]


4

8C

wLv

EI

Values:

w = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.

2

Computation:

4 4 3

8 2

(260 lb/ft)(25 ft) (12 in./ft)43.875 in.

8 8(5.0 10 lb-in. )C

wLv

EI

Consider 4,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.]


3 2

and3 2

B B

PL PLv


Values:

P = 4,000 lb, L = 18 ft, EI = 5.0 × 108 lb-in.

2

Computation:

3 3 3

8 2

2 2 2

8 2

(4,000 lb)(18 ft) (12 in./ft)26.873856 in.

3 3(5.0 10 lb-in. )

(4,000 lb)(18 ft) (12 in./ft)0.1866240 rad

2 2(5.0 10 lb-in. )

26.873856 in. (7 ft)(12 in./ft)(0.1866240 rad) 42.550

B

B

C

PLv

EI

PL

EI

v 272 in.


43.875 in. 42.550272 in. 1.324728 in. 1.325 in.Cv Ans.




11.1 A beam is loaded and supported as

shown in Fig. P11.1. Use the double-

integration method to determine the

magnitude of the moment M0 required to

make the slope at the left end of the beam

zero.

Fig. P11.1

Solution

Moment equation:

0

2

0

( ) 02

( )2

a a

xM M x wx M

wxM x M

Integration:

2 2

02( )

2

d v wxEI M x M

dx

3

0 16

dv wxEI M x C

dx

2 4

01 2

2 24

M x wxEI v C x C


3

0 1

3

1 0

( )at , 0 ( ) 0

6

6

dv w Lx L M L C

dx

wLC M L

Beam slope equation:

3 3

0 06 6

dv wx wLEI M x M L

dx

Constraint:

At x = 0, the slope of the beam is to be zero; therefore,

3

2

3

0 0

0

(0)(0) 0

6 6

6

A

dv w wLEI M M L

dx

MwL

Ans.




11.2 When moment M0 is applied to the left

end of the cantilever beam shown in Fig.

P11.2, the slope of the beam is zero. Use the


magnitude of the moment M0.

Fig. P11.2

Solution

Moment equation:

0

0

( ) 0

( )

a aM M x Px M

M x Px M

Integration:

2

02( )

d vEI M x Px M

dx

2

0 12

dv PxEI M x C

dx

3 2

01 2

6 2

Px M xEI v C x C


2

0 1

2

1 0

( )at , 0 ( ) 0

2

2

dv P Lx L M L C

dx

PLC M L

Beam slope equation:

2 2

0 02 2

dv Px PLEI M x M L

dx

Constraint:

At x = 0, the slope of the beam is to be zero; therefore,

2 2

0 0

0

(0)(0) 0

2 2

2

A

dv P PLEI M M

MP

Ldx

L Ans.




11.3 When the load P is applied to the right

end of the cantilever beam shown in Fig.

P11.3, the deflection at the right end of the

beam is zero. Use the double-integration

method to determine the magnitude of the

load P.

Fig. P11.3

Solution

Moment equation:

( ) ( ) ( ) 02

a a

L xM w L x P L x M x

2( ) ( ) ( )2

wM x L x P L x

Integration:

2

2

2( ) ( ) ( )

2

d v wEI M x L x P L x

dx

3 2

1( ) ( )6 2

dv w PEI L x L x C

dx

4 3

1 2( ) ( )24 6

w PEI v L x L x C x C


3 2

1

3 2

1

at 0, 0 ( 0) ( 0) 06 2

6 2

dv w Px L L C

dx

wL PLC

4 3

1 2

4 3

2

at 0, 0 ( 0) ( 0) (0) 024 6

24 6

w Px v L L C C

wL PLC

Beam elastic curve equation:

3 2 4 34 3

3 4 2 34 3

( ) ( )24 6 6 2 24 6

( ) ( )24 6 24 6 2 6

w P wLx PL x wL PLEI v L x L x

w wLx wL P PL x PLL x L x




Constraint:

At x = L, the deflection of the beam is to be zero; therefore,

3 4 2 3

4 3( ) ( )( ) ( ) 0

24 6 24 6 2 6B

w wL L wL P PL L PLEI v L L L L

which simplifies to

4 4 3 3 4 3

06 24 2 6 8 3

B

wL wL PL PL wL PLEI v

Therefore, the magnitude of P is

3

8P

wL Ans.







reactions at supports A and B.

Fig. P11.4

Solution

Beam FBD:

0

0

0

y y y

A A y

F A B

M M B L M

Moment equation:

0 0( ) ( ) 0 ( ) ( )a a y yM M x M B L x M x B L x M

Integration:

2

02( ) ( )y

d vEI M x B L x M

dx

2

0 1( )2

yBdvEI L x M x C

dx

2

3 01 2( )

6 2

yB M xEI v L x C x C


2

2

0 1 1at 0, 0 ( 0) (0) 02 2

y yB B Ldvx L M C C

dx

32

3 0

1 2 2

(0)at 0, 0 ( 0) (0) 0

6 2 6

y yB B LMx v L C C C

2 32

3 0

3 2

0 00

( )at , 0 ( ) ( ) 0

6 2 2 6

3

3 2 2

3

2

y y y

y

y

B B L B LM Lx L v L L L

B L M M M

L

LB

L Ans.

Backsubstitute into equilibrium equations:

0 030

2

3

2y y y y y y

MMF A B A B A

L L Ans.

0

0 0 0

00

30 0

2

(cw)22

A A y A y

A

MM M B L M M B L

M

M L ML

MM Ans.






(a) Use the double-integration method to

determine the reactions at supports A and B.

(b) Draw the shear-force and bending-

moment diagrams for the beam.

Fig. P11.5

Solution

Beam FBD:

0

02

y y y

A B y

F A B wL

LM M B L wL

Moment equation:

2

( ) 0 ( )2 2

a a y y

x wxM M x wx A x M x A x

Integration:

2 2

2( )

2y

d v wxEI M x A x

dx

23

16 2

yA xdv wxEI C

dx

34

1 224 6

yA xwxEI v C x C


34

1 2 2

(0)(0)at 0, 0 (0) 0 0

24 6

yAwx v C C C

2 23 3

1 1

( )( )at , 0 0

6 2 6 2

y yA L A Ldv w L wLx L C C

dx (a)

3 24 3

1 1

( )( )at , 0 ( ) 0

24 6 24 6

y yA L A Lw L wLx L v C L C (b)

Solve Eqs. (a) and (b) simultaneously to find:

3

1

3and

48 8

3

8y

wL wLC A

wL Ans.


3 5

05

88 8y y y y y y

wL wLF A B wL B wL

wLA wL B Ans.

2

2 2 2 2

2

50

2

(cw)

2 2 8 8

8 8

A B y B y

B

L wL wL wL wL

wL

M M B L wL M B L

wLM Ans.








Fig. P11.6

Solution

Beam FBD:

0

0

02

20

2 3

y y y

A B y

w LF A B

w L LM M B L

Moment equation:

2

0

3

0

( ) 02 3

( )6

a a y

y

w x xM M x A x

L

w xM x A x

L

Integration:

2 3

0

2( )

6y

d v w xEI M x A x

dx L

24

01

24 2

yA xdv w xEI C

dx L

35

01 2

120 6

yA xw xEI v C x C

L


35

01 2 2

(0)(0)at 0, 0 (0) 0 0

120 6

yAwx v C C C

L

2 24 3

0 01 1

( )( )at , 0 0

24 2 24 2

y yA L A Ldv w L w Lx L C C

dx L (a)

3 25 3

0 0

1 1

( )( )at , 0 ( ) 0

120 6 120 6

y yA L A Lw L w Lx L v C L C

L (b)




Solve Eqs. (a) and (b) simultaneously to find:

3

0 0 01 and

120 110 0y

w L w LC A

w L Ans.


0 0 0 00 040

2 2 2 10 1 50

2y y y y y y

w L w L w L w L w LF A B B A B

w L Ans.

2 2 2 2

0 0 0 0 0

2 2

0 0

2 20

2 3 3 3 5 15

(cw15

)1 5

A B y B y

B

w L L w L w L w L w LM M B L M B L

w Lw LM Ans.





shown in Fig. P11.7. Use the fourth-order

integration method to determine the reaction

at roller support B.

Fig. P11.7

Solution


4 2

0

4 2

d v w xEI

dx L

3 3

013 23

d v w xEI C

dx L

2 4

01 22 212

d v w xEI C x C

dx L

5 2

0 12 3260 2

dv w x C xEI C x C

dx L

6 3 2

0 1 23 42360 6 2


L


6 3 2

0 1 23 4 42

(0) (0) (0)at 0, 0 (0) 0 0

360 6 2

w C Cx v C C C

L

5 2

0 12 3 32

(0) (0)at 0, 0 (0) 0 0

60 2

dv w Cx C C C

dx L

6 3 2 2

0 1 2 01 22

( ) ( ) ( )at , 0 0 3

360 6 2 60

w L C L C L w Lx L v C L C

L (a)

2 4 2

0 01 2 1 22 2

( )at , 0 ( ) 0

12 12

d v w L w Lx L M EI C L C C L C

dx L (b)

Solve Eqs. (a) and (b) simultaneously to obtain:

2 2 2 2

0 0 0 02 2

42

60 12 60 30

w L w L w L w LC C

2 2 2

0 0 0 01 1

7 7

12 30 60 60

w L w L w L w LC L C

Roller reaction at B:

3 3

0 00 0 0 0

3 2

( ) 7 20 7 13

3 60 60 60 60

13

60B y

x L

d v w L w L w L w L w LV EI B

dx L

w L Ans.






integration method to determine the reaction

at roller support A.

Fig. P11.8

Solution


4 2

0

4 2

d v w xEI

dx L

3 3

013 23

d v w xEI C

dx L

2 4

01 22 212

d v w xEI C x C

dx L

5 2

0 12 3260 2

dv w x C xEI C x C

dx L

6 3 2

0 1 23 42360 6 2


L


6 3 2

0 1 23 4 42

(0) (0) (0)at 0, 0 (0) 0 0

360 6 2

w C Cx v C C C

L

2 4

01 2 22 2

(0)at 0, 0 (0) 0 0

12

d v wx M EI C C C

dx L

5 2 3

20 1 03 1 32

( ) ( )at , 0 0 2

60 2 30

dv w L C L w Lx L C C L C

dx L (a)

6 3 3

20 1 03 1 32

( ) ( )at , 0 ( ) 0 6

360 6 60

w L C L w Lx L v C L C L C

L (b)


3 3 3 3

0 0 0 03 34

30 60 60 240

w L w L w L w LC C

3 3 3

2 0 0 0 0 01 1

5 5

30 120 120 120 24

w L w L w L w L w LC L C

Roller reaction at A:

3 3

0 0 0

3 2

0

0(0)

3 24 4 42 2A y

x

d v w w L w LV EI A

dx L

w L Ans.







reaction at roller support A.

Fig. P11.9

Solution


4

04sin

2

d v xEI w

dx L

3

013

2cos

2

d v w L xEI C

dx L

2 2

01 22 2

4sin

2

d v w L xEI C x C

dx L

3 2

0 12 33

8cos

2 2


dx L

4 3 2

0 1 23 44

16sin

2 6 2


L


4 3 2

0 1 23 4 44

16 (0) (0) (0)at 0, 0 sin (0) 0 0

2 6 2

w L C Cx v C C C

L

2 2

01 2 22 2

4 (0)at 0, 0 sin (0) 0 0

2

d v w Lx M EI C C C

dx L

3 2

20 13 1 33

8 ( ) ( )at , 0 cos 0 2 0

2 2

dv w L L C Lx L C C L C

dx L (a)

4 3 3

20 1 03 1 34 4

16 ( ) ( ) 96at , 0 sin ( ) 0 6

2 6

w L L C L w Lx L v C L C L C

L (b)


3 3

0 03 34 4

96 244

w L w LC C

3

2 0 01 14 4

24 482

w L w LC L C

Roller reaction at A:

3

0 0 0

3 4

0

4

0

2 (0) 48cos

2

2 48A y

x

d v w L w LV EI

w L

x

LA

w

d L Ans.








Fig. P11.10

Solution


4

04cos

2

d v xEI w

dx L

3

013

2sin

2

d v w L xEI C

dx L

2 2

01 22 2

4cos

2

d v w L xEI C x C

dx L

3 2

0 12 33

8sin

2 2


dx L

4 3 2

0 1 23 44

16cos

2 6 2


L


3 2

0 12 3 33

8 (0) (0)at 0, 0 sin (0) 0 0

2 2

dv w L Cx C C C

dx L

4 3 2 4

0 1 2 04 44 4

16 (0) (0) (0) 16at 0, 0 cos 0

2 6 2

w L C C w Lx v C C

L

3 2 2

0 1 02 1 23 3

8 ( ) ( ) 16at , 0 sin ( ) 0 2

2 2

dv w L L C L w Lx L C L C L C

dx L(a)

4 3 2 4 2

0 1 2 0 01 24 4 4

16 ( ) ( ) ( ) 16 96at , 0 cos 0 3

2 6 2

w L L C L C L w L w Lx L v C L C

L(b)


2 2 2

0 0 02 23 4 4

16 96 166

w L w L w LC C

2 2

0 0 01 13 4 4

48 192 484

w L w L w LC L C

Reactions at supports A and B

3

0 0 0

3 4 4

0

0

4

2 (0) 48 48sin 4 4

4

2

84

A

x

y

d v w L w L

w

w

L

LV EI

dx L

A Ans.




330 0

4

4

30

0

3 4

2 ( ) 48 2si

296

n 4 24 9

4

62

2

B

x L

y

d v w L L w L w LV EI

dx L

w LB Ans.

2 2 2

0 0 0

2 2 4 4

22

0

2 2

0 0

2 4

0

4

4 (0) 48 (0) 16cos 4

424 4 (c

62

4 1

)

66

w

A

x

A

d v w L w L w LM EI

dx L

w L w L

Mw L

Ans.

2 2 2

0 0 0

2 2 4 4

2 2 2 2

0 0 0 0

4 4 4

2

0

4

4

4 ( ) 48 (

323 (c

) 16cos 4 6

2

48 16

cw

16 164 6 3 12 6 2 6

)

B

x L

B

d v w L L w L L w LM EI

dx L

w L w L

L

w L w L

wM Ans.








Fig. P11.11

Solution


4

04sin

d v xEI w

dx L

3

013

cosd v w L x

EI Cdx L

2 2

01 22 2

sind v w L x

EI C x Cdx L

3 2

0 12 33

cos2


dx L

4 3 2

0 1 23 44

sin6 2


L


3 2 3

0 1 02 3 33 3

(0) (0)at 0, 0 cos (0) 0

2

dv w L C w Lx C C C

dx L

4 3 2

0 1 23 4 44

(0) (0) (0)at 0, 0 sin (0) 0 0

6 2

w L C Cx v C C C

L

3 2 3 2

0 1 0 02 1 23 3 3

( ) ( ) 4at , 0 cos ( ) 0 2

2

dv w L L C L w L w Lx L C L C L C

dx L (a)

4 3 2 3 2

0 1 2 0 01 24 3 3

( ) ( ) ( ) 6at , 0 sin ( ) 0 3

6 2

w L L C L C L w L w Lx L v L C L C

L (b)


2 2 2

0 0 02 23 3 3

6 4 2w L w L w LC C

2 2

0 01 13 3

4 22 0

w L w LC L C

Reactions at supports A and B

3

0

0 0

3

0

(0)cosA

x

y

d v w L w LV EI

dx L

Aw L

Ans.




3

0

3

0 0( )cosB

x L

y

d v w L L w LV EI

dx L

Bw L

Ans.

2 2 2 2

0 0 0

2 2 3 3

0

2

0

3

(0) 2

2 )

sin

(cw

2A

x

A

d v w L w L w LM

w

Idx L

L

E

M Ans.

2 2 2 2

0 0 0

2 2 3 3

2

0

3

( ) 2

2 (ccw)

2sinB

x L

B

d v w L L w L w

w

LM E

L

Idx L

M Ans.







determine the reactions at supports A and C.



(c) Determine the deflection in the middle

of the span. Fig. P11.12

Solution

Beam FBD:

from symmetry,

2

and

y y

A C

PA C

M M

Moment equation:

( ) 0 ( )2 2

a a A A

P PxM M x M x M x M

Integration:

2

2( )

2A

d v PxEI M x M

dx

2

14

A

dv PxEI M x C

dx

3 2

1 212 2

APx M xEI v C x C


2

1 1

(0)at 0, 0 (0) 0 0

4A

dv Px M C C

dx

3 2

2 2

(0) (0)at 0, 0 0 0

12 2

AP Mx v C C

(a) Beam reaction forces:

2

y yAP

C Ans.




(a) Beam reaction moments:

2

(ccw) (cw)8

( / 2)at , 0 0

2 4 2

8 8

A

A C

L dv P L Lx M

dx

PL PL PM M

L Ans.


3 2 3 2 2

2

3 412 2 12 16 48

3 448

APx M x Px PLx PxEI v L x

Pxv L x

EI

(c) Midspan deflection:

2 3( / 2)

3 4(192

/ 2)48

B

P Lv L L

E

PL

EII Ans.







determine the reactions at supports A and B.




of the span.

Fig. P11.13

Solution

Beam FBD:

from symmetry,

2

and

y y

A B

wLA B

M M

Moment equation:

2

( ) 02 2

( )2 2

a a A

A

x wLM M x M wx x

wx wLxM x M

Integration:

2 2

2( )

2 2A

d v wx wLxEI M x M

dx

3 2

16 4

A

dv wx wLxEI M x C

dx

4 3 2

1 224 12 2

Awx wLx M xEI v C x C


3 2

1 1

(0) (0)at 0, 0 (0) 0 0

6 4A

dv w wLx M C C

dx

4 3 2

2 2

(0) (0) (0)at 0, 0 0 0

24 12 2

Aw wL Mx v C C


2

y yA BwL

Ans.





3

2

2

2 22

(ccw) (cw)

( / 2) ( / 2)at , 0 0

2

12 12

6 4 2

12 12

A

A B

L dv w L wL L Lx M

dx

wL wLM M

wL wL Ans.


4 3 2 4 3 2 2 2 22 2 2

22

2 ( )24 12 2 24 12 24 24 24

( )24

Awx wLx M x wx wLx wL x wx wxEI v x Lx L x L

wxv x L

EI


22

2

4

/

( / 2)

4 2 3)

842x L

w L Lv L

EI

wL

EI Ans.








(b) Determine the deflection in the middle

of the span.

Fig. P11.14

Solution

Beam FBD:

0

from symmetry,

2

and

y y

A C

w LA C

M M

Moment equation:

2

0 0

3

0 0

( ) 02 3 2

( )6 2

a a A

A

w x x w LM M x M x

L

w x w LxM x M

L

Integration:

2 3

0 0

2( )

6 2A

d v w x w LxEI M x M

dx L

4 2

0 01

24 4A

dv w x w LxEI M x C

dx L

5 3 2

0 01 2

120 12 2

Aw x w Lx M xEI v C x C

L


4 2

0 01 1

(0) (0)at 0, 0 (0) 0 0

24 4A

dv w w Lx M C C

dx L

5 3 2

0 02 2

(0) (0) (0)at 0, 0 0 0

120 12 2

Aw w L Mx v C C

L


0

2y yA C

w L Ans.





2

4 2

0 0

2 2

0 0

2

0 0

( ) ( )at , 0 ( ) 0

24 4

5 5 5 (ccw) (cw)

24

5

24 24 24

A

A C

d

w

v w L w L Lx L M L

dx L

w L w LL w LM M Ans.


5 3 2 5 3 2 2

0 0 0 0 0

5 2 3 3 2 23 2 30 0 0 0

23 2 30

5

120 12 2 120 12 48

2 20 252 20 25

240 240 240 240

2 20 25240

Aw x w Lx M x w x w Lx w L xEI v

L L

w x w L x w L x w xx L x L

L L L L

w xv x L x L

L EI


2

3 2 34

0 0( )2( ) 20 ( ) 25

2

7

0 24 40B

w Lv L L L L

w

L EI

L

EI Ans.











of the span. Fig. P11.15

Solution

Beam FBD:

0

02

y y y

A C y

F A C P

LM M C L P

Moment equation:

( ) 0

( ) 02

a a y

y

M M x A x

LM x A x x

( ) 02

( )2 2

b b y

y

LM M x A x P x

PL LM x A x Px x L

Integration:

For beam segment AB:

2

2( ) y

d vEI M x A x

dx

2

12

yA xdvEI C

dx

3

1 26

yA xEI v C x C

For beam segment BC:

2

2( )

2y

d v PLEI M x A x Px

dx

2 2

32 2 2

yA xdv Px PLxEI C

dx

3 3 2

3 46 6 4

yA x Px PLxEI v C x C


3

1 2 2

(0)at 0, 0 (0) 0 0

6

yAx v C C C

2 22

3 3

( ) ( ) ( )at , 0 0

2 2 2 2

y yA L A Ldv P L PL Lx L C C

dx

3 2 33 2 3

4 4

( ) ( ) ( )at , 0 ( ) 0

6 6 4 2 3 12

y y yA L A L A LP L PL L PLx L v L C C




Slope continuity condition at x = L/2:

2 2 22

1

22

1

at ,2

( / 2) ( / 2) ( / 2) ( / 2)

2 2 2 2 2

8 2

AB BC

y y y

y

L dv dvx

dx dx

A L A L A LP L PL LC

A LPLC

Deflection continuity condition at x = L/2:

3 2 3 2 32 3 2 3

at ,2

6 8 2 6 6 4 2 3 12

B BAB BC

y y y y y

Lx v v

A x A L x A x A L x A LPL x Px PLx PL

eliminate terms and rearrange:

3 3 2 2 3

3 6 4 8 12

yA L Px PLx PL x PL

Substitute x = L/2 to obtain:

3 3 2 2 3 3( / 2) ( / 2) ( / 2) 5

3 6 4 8 12 48

5

16

y

y

A L P L PL L PL L PL PL

PA


5 11

16 16y yA C

P P Ans.

(a) Beam reaction moment:

11 3

(cw)16

3 3

2 2 16 16 16C y C

L PL PL PL PLM P C L

PM

L Ans.

Elastic curve equation for beam segment AB:

3 22 3 2 2 3 2

2 2

5 5 5 3

6 8 2 96 8 32 96 96

5 396

y yA x A L xPL x Px PL x PL x Px PL xEI v

Pxv x L

EI


2 32( / 2)

5 396 2

7

768B

P L PL

EI

Lv L

EI Ans.










Fig. P11.16

Solution

Beam FBD:

0

0

0

y y y y y

A A y

F A C C A

M M C L M

Moment equation:

( ) 0

( ) 02

a a y A

y A

M M x A x M

LM x A x M x

0

0

( ) 0

( )2

b b y A

y A

M M x A x M M

LM x A x M M x L

Integration:


2

2( ) y A

d vEI M x A x M

dx

2

12

y

A

A xdvEI M x C

dx

3 2

1 26 2

y AA x M x

EI v C x C


2

02( ) y A

d vEI M x A x M M

dx

2

0 32

y

A

A xdvEI M x M x C

dx

3 2 2

03 4

6 2 2

y AA x M x M x

EI v C x C

Boundary conditions and evaluate constants for segment AB:

3 2

1 2 2

(0) (0)at 0, 0 (0) 0 0

6 2

y AA M

x v C C C

2

1 1

(0)at 0, 0 (0) 0 0

2

y

A

Advx M C C

dx





2 2

0 3

03

at ,2

2 2

2

AB BC

y y

A A

L dv dvx

dx dx

A x A xM x M x M x C

M LC


3 32 2 2

0 04

2

04

at ,2

6 2 6 2 2 2

8

B BAB BC

y yA A

Lx v v

A x A xM x M x M x M LxC

M LC

Boundary condition for segment BC:

3 2 2 2

0 0 0 0( ) ( ) ( ) 3

at , 0 ( ) 0 36 2 2 2 8 4

y Ay A

A L M L M L M L M L Mx L v L A L M

Also, the beam moment equilibrium equation can be written as:

0y AA L M M

(a) Beam Reactions: Solve these two equations simultaneously to obtain:

0 0 0 00 9 9 (cw)

8 88 8

9

8A y y

M MM A C

L

M M M

L L Ans.




11.17 A beam is loaded and supported as shown

in Fig. P11.17.



(b) Draw the shear-force and bending-moment

diagrams for the beam.

Fig. P11.17

Solution

Beam FBD:

02

02 4

y y y

A C y

wLF A C

wL LM M C L

Moment equation:

2

( ) 02

( ) 02 2

a a y

y

xM M x wx A x

wx LM x A x x

( ) 02 4

( )2 4 2

b b y

y

wL LM M x x A x

wL L LM x x A x x L

Integration:


2 2

2( )

2y

d v wxEI M x A x

dx

23

16 2

yA xdv wxEI C

dx

34

1 224 6

yA xwxEI v C x C


2

2( )

2 4y

d v wL LEI M x x A x

dx

2 2

34 4 2

yA xdv wL LEI x C

dx

3 3

3 412 4 6

yA xwL LEI v x C x C

Boundary conditions and evaluate constants for segment AB:

34

1 2 2

(0)(0)at 0, 0 (0) 0 0

24 6

yAwx v C C C





Equate the slope expressions for the two beam segments:

22 23

1 36 2 4 4 2

y yA x A xwx wL LC x C

Set x = L/2 and solve for the constant C1:

2 23 3 3 3

1 3 3 3

3

1 3

( / 2)

6 4 4 6 4 2 4 48 64

192

wx wL L w L wL L L wL wLC C x C C

wLC C


Equate the deflection expressions for the two beam segments:

33 34

1 3 424 6 12 4 6

y yA x A xwx wL LC x x C x C

Set x = L/2 and solve for the constant C4:

34 3

3 3 4

4 4 4

3 3 4

4

4

( / 2)

24 192 2 12 2 4 2

384 2 384 768 2

768

w L wL L wL L L LC C C

wL L wL wL LC C C

wLC

Boundary conditions and evaluate constants for segment BC:

2 2 23

3 3

( ) 9at , 0 0

4 4 2 64 2

y yA L A Ldv wL L wLx L L C C

dx

at x = L, v = 0

3 3 4

3

3 24 3 4

3 34 4 4

3 3 4 4

3 4

( )( ) 0

12 4 6 768

27 9( ) 0

768 6 64 2 768

27 90

768 6 64 2 768

3 26 108

6 6 768 768

82 41

3 768 128

y

y y

y y

y y

y

y

A LwL L wLL C L

A L A LwL wL wLL

A L A LwL wL wL

A L A L wL wL

A L wL wLA




Solve for C3:

3 3 3

3

9 41 5

64 256 256

wL wL wLC

and for C1:

3 3 3

1

5 11

256 192 768

wL wL wLC

(a) Beam force reactions:

41

128y

wLA Ans.

41 23

2 2 128

23

12 281 8y y y

wL wL wL wLC

wA C

L Ans.

Beam moment reaction:

2 2 22 2 223 7 7

8 8 1

7 (cw)

12828 128 128C y C

wL wL wL wL wLM

wLC L M Ans.




11.18 A propped cantilever beam is loaded as shown

in Fig. P11.18. Assume EI = 200,000 kN-m2. Use

discontinuity functions to determine:

(a) the reactions at A and C.

(b) the beam deflection at B.

Fig. P11.18

Solution

Moment equation:

(a) Support reactions:

150 kN 0

150 kN

y y y

y y

F A C

A C (a)

(150 kN)(7 m) (12 m) 0

(12 m) 1,050 kN-m

A y A

A y

M C M

M C (b)

Discontinuity expressions:

2 1 1 1

1 0 0 0

0 1 1 1

20 1 1

2

0 m 0 m 150 kN 7 m 12 m

( ) 0 m 0 m 150 kN 7 m 12 m

( ) 0 m 0 m 150 kN 7 m 12 m

( ) 0 m 0 m 150 kN 7 m 1

A y y

A y y

A y y

A y y

w x M x A x x C x

V x w x dx M x A x x C x

M x V x dx M x A x x C x

d vEI M x M x A x x C x

dx

12 m

1 2 2 2

1

150 kN0 m 0 m 7 m 12 m

2 2 2

y y

A

A CdvEI M x x x x C

dx (c)

2 3 3 3

1 2

150 kN0 m 0 m 7 m 12 m

2 6 6 6

y yAA CM

EI v x x x x C x C (d)


1at 0 m, 0 0dv

x Cdx

2at 0 m, 0 0x v C

2 3 3

2 3 3

150 kNat 12 m, 0 0 (12 m) (12 m) (5 m)

2 6 6

(72 m ) (288 m ) 3,125 kN-m

yA

A y

AMx v

M A (e)

(a) Solve for Ay, Cy, and MA:

Solve equations (a), (b), and (c) simultaneously to obtain the results:

88.3 kN 61.7 kN

310 kN-

88.3247

m

kN 61.6753 kN

309.8958 kN-m (ccw)

y y

A

A C

M Ans.




(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 7 m) is computed as follows:

2 3

3

3

2

309.8958 kN-m 88.3247 kN(7 m) (7 m)

2 6

2,543.2219 kN-m

2,543.2219 kN-m0.0127161 m

200,12.72

000 kmm

N-m

B

B

EI v

v Ans.




11.19 A propped cantilever beam is loaded as

shown in Fig. P11.19. Assume EI = 200,000 kN-

m2. Use discontinuity functions to determine:

(a) the reactions at A and B.

(b) the beam deflection at C.

Fig. P11.19

Solution

Moment equation:


0y y y

y y

F A B

A B (a)

750 kN-m (5 m) 0

(5 m) 750 kN-m

A y A

A y

M B M

M B (b)


2 1 1 2

1 0 0 1

0 1 1 0

20 1 1

2

0 m 0 m 5 m 750 kN-m 7.5 m

( ) 0 m 0 m 5 m 750 kN-m 7.5 m

( ) 0 m 0 m 5 m 750 kN-m 7.5 m

( ) 0 m 0 m 5 m

A y y

A y y

A y y

A y y

w x M x A x B x x

V x w x dx M x A x B x x

M x V x dx M x A x B x x

d vEI M x M x A x B x

dx

0750 kN-m 7.5 mx

1 2 2 1

10 m 0 m 5 m 750 kN-m 7.5 m2 2

y y

A

A BdvEI M x x x x C

dx (c)

2 3 3 2

1 2

750 kN-m0 m 0 m 5 m 7.5 m

2 6 6 2

y yAA BM



1at 0 m, 0 0dv

x Cdx

2at 0 m, 0 0x v C

2 3

2 3

at 5 m, 0 0 (5 m) (5 m)2 6

(12.5 m ) (20.83333 m ) 0

yA

A y

AMx v

M A (e)

(a) Solve for Ay, By, and MA:

Solve equations (a), (b), and (c) simultaneously to obtain the results:

225 kN 225 kN225.000 kN 225.000 kN

375.000 kN-m 375 kN-m (cw)

y y

A

A B

M Ans.




(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 7.5 m) is computed as follows:

2 3 3

3

3

2

(7.5 m) (7.5 m) (2.5 m)2 6 6

4,687.500 kN-m

4,687.500 kN-m0.023438 m

200,000 kN-m23.4 mm

y yAC

C

A BMEI v

v Ans.




11.20 A propped cantilever beam is loaded as shown

in Fig. P11.20. Assume EI = 100,000 kip-ft2. Use


(a) the reactions at A and E.


Fig. P11.20

Solution

Moment equation:


20 kips 30 kips 20 kips 0

70 kips

y y y

y y

F A E

A E (a)

(28 ft) 20 kips(21 ft)

30 kips(14 ft) 20 kips(7 ft) 0

(28 ft) 980 kip-ft

E y

E

y E

M A

M

A M (b)


1 1 1 1

2 1

0 0 0 0

1 0

1

0 ft 20 kips 7 ft 30 kips 14 ft 20 kips 21 ft

28 ft 28 ft

( ) 0 ft 20 kips 7 ft 30 kips 14 ft 20 kips 21 ft

28 ft 28 ft

( ) 0 ft 20

y

E y

y

E y

y

w x A x x x x

M x E x

V x w x dx A x x x x

M x E x

M x V x dx A x1 1 1

0 1

21 1 1 1

2

0 1

kips 7 ft 30 kips 14 ft 20 kips 21 ft

28 ft 28 ft

( ) 0 ft 20 kips 7 ft 30 kips 14 ft 20 kips 21 ft

28 ft 28 ft

E y

y

E y

x x x

M x E x

d vEI M x A x x x x

dx

M x E x

2 2 2 2

1 2

1

20 kips 30 kips 20 kips0 ft 7 ft 14 ft 21 ft

2 2 2 2

28 ft 28 ft2

y

y

E

AdvEI x x x x

dx

EM x x C (c)

3 3 3 3

2 3

1 2

20 kips 30 kips 20 kips0 ft 7 ft 14 ft 21 ft

6 6 6 6

28 ft 28 ft2 6

y

yE

AEI v x x x x

EMx x C x C (d)


2at 0 ft, 0 0x v C




3 3 3 3

1

at 28 ft, 0

20 kips 30 kips 20 kips(28 ft) (21 ft) (14 ft) (7 ft) (28 ft) 0

6 6 6 6

y

x v

AC (e)

2 2 2 2

1

at 28 ft, 0

20 kips 30 kips 20 kips(28 ft) (21 ft) (14 ft) (7 ft) 0

2 2 2 2

y

dvx

dx

AC (f)

(a) Solve for Ay, Ey, and ME:

Solve equations (e) and (f) simultaneously to obtain:

2

1 1,470.000 kip-ft

23.7500 ki 23.8 kips psy

C

A Ans.

With the value of Ay, calculate Ey and ME from equations (a) and (b), respectively.

46.2500 kips

315.000 kip-f

46.3 kips

315 kip-ft ( wt c )

y

E

E

M Ans.

(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 14 ft) is computed as follows:

3 3 2

3

3

2

23.75 kips 20 kips(14 ft) (7 ft) (1,470 kip-ft )(14 ft)

6 6

10,861.6667 kip-ft

10,861.6667 kip-ft0.108617 ft 1.3034 in.

100,000 ki1.303 in.

p-ft

C

C

EI v

v Ans.





shown in Fig. P11.21. Assume EI = 100,000 kip-

ft2. Use discontinuity functions to determine:


(b) the beam deflection at x = 7 ft.

Fig. P11.21

Solution

Moment equation:


1(12 kips/ft)(16 ft) 0

2

96 kips

y y y

y y

F A B

A B (a)

1 2(16 ft)(16 ft) (12 kips/ft)(16 ft)

2 3

0

(16 ft) 1,024 kip-ft

B y

B

y B

M A

M

A M (b)


1 0 1

2 1

0 1 2

1 0

1

12 kips/ft0 ft 12 kips/ft 0 ft 0 ft

16 ft

16 ft 16 ft

12 kips/ft( ) 0 ft 12 kips/ft 0 ft 0 ft

2(16 ft)

16 ft 16 ft

12 kips/ft( ) 0 ft 0

2

y

B y

y

B y

y

w x A x x x

M x B x

V x w x dx A x x x

M x B x

M x V x dx A x x2 3

0 1

21 2 3

2

0 1

12 kips/ft ft 0 ft

6(16 ft)

16 ft 16 ft

12 kips/ft 12 kips/ft( ) 0 ft 0 ft 0 ft

2 6(16 ft)

16 ft 16 ft

B y

y

B y

x

M x B x

d vEI M x A x x x

dx

M x B x

2 3 4

1 2

1

12 kips/ft 12 kips/ft0 ft 0 ft 0 ft

2 6 24(16 ft)

16 ft 16 ft2

y

y

B

AdvEI x x x

dx

BM x x C

(c)

3 4 5

2 3

1 2


6 24 120(16 ft)

16 ft 16 ft2 6

y

yB

AEI v x x x

BMx x C x C (d)





2at 0 ft, 0 0x v C

3 4 5

1

at 16 ft, 0

12 kips/ft 12 kips/ft(16 ft) (16 ft) (16 ft) (16 ft) 0

6 24 120(16 ft)

y

x v

AC (e)

2 3 4

1

at 16 ft, 0

12 kips/ft 12 kips/ft(16 ft) (16 ft) (16 ft) 0

2 6 24(16 ft)

y

dvx

dx

AC (f)

(a) Solve for Ay, By, and MB:

Solve equations (e) and (f) simultaneously to obtain:

2

1 614.4000 kip-ft

52.8000 ki 52ps .8 kipsy

C

A Ans.

With the value of Ay, calculate By and MB from equations (a) and (b), respectively.

43.200 kips

179.200 kip-ft

43.2 kips

179.2 kip-ft (cw)

y

B

B

M Ans.

(b) Beam deflection at x = 7 ft: From Eq. (d), the beam deflection at x = 7 ft is computed as follows:

3 4 5 2

3

3

2

52.8 kips 12 kips/ft 12 kips/ft(7 ft) (7 ft) (7 ft) (614.400 kip-ft )(7 ft)

6 24 120(16 ft)

2,377.85625 kip-ft

2,377.85625 kip-ft0.023779 ft 0.2853 in.

100,000 ki0.285

p- i

fn

t.

EI v

v Ans.





shown in Fig. P11.22. Assume EI = 200,000 kN-

m2. Use discontinuity functions to determine:



Fig. P11.22

Solution

Moment equation:


1(120 kN/m)(8 m) 0

2

480 kN

y y y

y y

F A B

A B (a)

1 2(8 m)(120 kN/m)(8 m) (6 m)

2 3

0

(6 m) 2,560 kN-m

A y

A

y A

M B

M

B M (b)


2 1 1 1

1 0 2 0

0 1 3 1

20 1

2

120 kN/m0 m 0 m 0 m 6 m

8 m

120 kN/m( ) 0 m 0 m 0 m 6 m

2(8 m)

120 kN/m( ) 0 m 0 m 0 m 6 m

6(8 m)

1( ) 0 m 0 m

A y y

A y y

A y y

A y

w x M x A x x B x

V x w x dx M x A x x B x

M x V x dx M x A x x B x

d vEI M x M x A x

dx

3 120 kN/m0 m 6 m

6(8 m)yx B x

1 2 4 2

1

120 kN/m0 m 0 m 0 m 6 m

2 24(8 m) 2

y y

A

A BdvEI M x x x x C

dx (c)

2 3 5 3

1 2

120 kN/m0 m 0 m 0 m 6 m

2 6 120(8 m) 6

y yAA BM



1at 0 m, 0 0dv

x Cdx

2at 0 m, 0 0x v C

2 3 5120 kN/m

at 6 m, 0 (6 m) (6 m) (6 m) 02 6 120(8 m)

yAAM

x v (e)




(a) Solve for Ay, By, and MA:

Solve equations (a), (b), and (e) simultaneously to obtain:

66.5 kN 414 kN66.5000 kN

79.0 kN-m

413.5000 kN

79.0000 kN-m (ccw)

y y

A

A B

M Ans.

(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 8 m) is computed as follows:

2 3 5 3

3

3

2

79.0 kN-m 66.5 kN 120 kN/m 413.5 kN(8 m) (8 m) (8 m) (2 m)

2 6 120(8 m) 6

398.0000 kN-m

398.0000 kN-m0.001990 m 1.990 mm

200,1.990

000 kN-mm

m

EI v

v Ans.




11.23 For the beam shown in Fig. P11.23, assume

EI = 200,000 kN-m2 and use discontinuity

functions to determine:

(a) the reactions at A, C, and D.


Fig. P11.23

Solution

Moment equation:


(120 kN/m)(6 m) 0

720 kN

y y y y

y y y

F A C D

A C D (a)

(120 kN/m)(6 m)(3 m) (6 m)

(10 m) 0

(6 m) (10 m) 2,160 kN-m

A y

y

y y

M C

D

C D (b)


1 0 0

1 1

0 1 1

0 0

1 2 2

1

0 m 120 kN/m 0 m 120 kN/m 6 m

6 m 10 m

( ) 0 m 120 kN/m 0 m 120 kN/m 6 m

6 m 10 m

120 kN/m 120 kN/m( ) 0 m 0 m 6 m

2 2

6 m 10 m

y

y y

y

y y

y

y y

w x A x x x

C x D x

V x w x dx A x x x

C x D x

M x V x dx A x x x

C x D x1

21 2 2

2

1 1

120 kN/m 120 kN/m( ) 0 m 0 m 6 m

2 2

6 m 10 m

y

y y

d vEI M x A x x x

dx

C x D x

2 3 3

2 2

1

120 kN/m 120 kN/m0 m 0 m 6 m

2 6 6

6 m 10 m2 2

y

y y

AdvEI x x x

dx

C Dx x C (c)

3 4 4

3 3

1 2

120 kN/m 120 kN/m0 m 0 m 6 m

6 24 24

6 m 10 m6 6

y

y y

AEI v x x x

C Dx x C x C (d)


2at 0 m, 0 0x v C

3 4

1

120 kN/mat 6 m, 0 (6 m) (6 m) (6 m) 0

6 24

yAx v C (e)




3 4 4 3

1

at 10 m, 0

120 kN/m 120 kN/m(10 m) (10 m) (4 m) (4 m) (10 m) 0

6 24 24 6

y y

x v

A CC (f)

(a) Solve for Ay, Cy, and Dy:

Solve equations (a), (b), (e), and (f) simultaneously to obtain:

2

1 756.000 kN-m

306.0000 kN

4

306 kN

495 kN95.0000 kN

81. 81.0 k0000 N N k

y

y

y

C

A

C

D Ans.

(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 3 m) is computed as follows:

3 4 2

3

3

2

306.00 kN 120 kN/m(3 m) (3 m) (756.000 kN-m )(3 m)

6 24

1,296.0000 kN-m

1,296.0000 kN-m0.006480 m 6.48 mm

200,006.

0 48

k

N-m

mm

B

B

EI v

v Ans.





EI = 100,000 kip-ft2 and use discontinuity


(a) the reactions at A, C, and D.


Fig. P11.24

Solution

Moment equation:



187 kips

y y y y

y y y

F A C D

A C D (a)

(75 kips)(8 ft) (7 kips/ft)(16 ft)(24 ft)

(16 ft) (32 ft) 0

(16 ft) (32 ft) 3,288 kip-ft

A

y y

y y

M

C D

C D (b)


1 1 1

0 0 1

0 0 0

1 1 0

0 ft 75 kips 8 ft 16 ft

7 kips/ft 16 ft 7 kips/ft 32 ft 32 ft

( ) 0 ft 75 kips 8 ft 16 ft

7 kips/ft 16 ft 7 kips/ft 32 ft 32 ft

( ) 0 f

y y

y

y y

y

y

w x A x x C x

x x D x

V x w x dx A x x C x

x x D x

M x V x dx A x1 1 1

2 2 1

21 1 1

2

2 2 1

t 75 kips 8 ft 16 ft


2 2

( ) 0 ft 75 kips 8 ft 16 ft


2 2

y

y

y y

y

x C x

x x D x

d vEI M x A x x C x

dx

x x D x

2 2 2

3 3 2

1

75 kips0 ft 8 ft 16 ft

2 2 2


6 6 2

y y

y

A CdvEI x x x

dx

Dx x x C (c)

3 3 3

4 4 3

1 2

75 kips0 ft 8 ft 16 ft

6 6 6


24 24 6

y y

y

A CEI v x x x

Dx x x C x C (d)


2at 0 ft, 0 0x v C




3 3

1

75 kipsat 16 ft, 0 (16 ft) (8 ft) (16 ft) 0

6 6

yAx v C (e)

3 3 3 4

1

at 32 ft, 0

75 kips 7 kips/ft(32 ft) (24 ft) (16 ft) (16 ft) (32 ft) 0

6 6 6 24

y y

x v

A CC (f)

(a) Solve for Ay, Cy, and Dy:

Solve equations (a), (b), (e), and (f) simultaneously to obtain:

2

1 601.3333 kip-ft

23.4688 kips

12

23.5 kips

121.6 kip1.5625 kips

41.9688 kips

s

42.0 kips

y

y

y

C

A

C

D Ans.

(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 8 ft) is computed as follows:

3 2

3

3

2

23.4688 kips(8 ft) (601.3333 kip-ft )(8 ft)

6

2,808.0000 k

0.337 in

ip-ft

2,808.0000 kip-ft0.028080 ft 0.3370 in.

100,000 ki.

p-ft

B

B

EI v

v Ans.




11.25 For the propped cantilever beam shown in

Fig. P11.25, assume EI = 100,000 kip-ft2 and use


(a) the reactions at B and D.


Fig. P11.25

Solution

Moment equation:


1(5 kips/ft)(10 ft)

2


85 kips

y y y

y y

F B D

B D (a)

1 10 ft(5 kips/ft)(10 ft) 20 ft

2 3


(20 ft) 0

(20 ft) 1,423.3333 kip-ft

D

y D

y D

M

B M

B M (b)


1 1 0

1 0 0

2 1

1 1


10 ft 10 ft

10 ft 5 kips/ft 10 ft 5 kips/ft 22 ft

30 ft 30 ft

5 kips/ft 5 kips/ft0 ft 10 ft 5 kips/ft 22 f

10 ft 10 ft

y

D y

w x x x x

B x x x

M x D x

x x x0

2 1

2 2 0

1 1 0

3

t

30 ft 30 ft


2(10 ft) 2(10 ft)

5 kips/ft 22 ft 30 ft 30 ft

5 kips/ft 5 kips/ft( ) 0 ft 10 f

6(10 ft) 6(10 ft)

D y

y

D y

M x D x

V x w x dx x x B x

x M x D x

M x V x dx x x3 1

2 0 1

23 3 1

2

2 0 1

t 10 ft

5 kips/ft22 ft 30 ft 30 ft

2


6(10 ft) 6(10 ft)

5 kips/ft22 ft 30 ft 30 ft

2

y

D y

y

D y

B x

x M x D x

d vEI M x x x B x

dx

x M x D x




4 4 2 3

1 2

1

5 kips/ft 5 kips/ft 5 kips/ft0 ft 10 ft 10 ft 22 ft

24(10 ft) 24(10 ft) 2 6

30 ft 30 ft2

y

y

D

BdvEI x x x x

dx

DM x x C (c)

5 5 3 4

2 3

1 2

5 kips/ft 5 kips/ft 5 kips/ft0 ft 10 ft 10 ft 22 ft

120(10 ft) 120(10 ft) 6 24

30 ft 30 ft2 6

y

yD

BEI v x x x x

DMx x C x C (d)


5

1 2

at 10 ft, 0

5 kips/ft(10 ft) (10 ft) 0

120(10 ft)

x v

C C (e)

5 5 3

4

1 2

at 30 ft, 0

5 kips/ft 5 kips/ft(30 ft) (20 ft) (20 ft)

120(10 ft) 120(10 ft) 6

5 kips/ft(8 ft) (30 ft) 0

24

y

x v

B

C C (f)

4 4 2 3

1

at 30 ft, 0

5 kips/ft 5 kips/ft 5 kips/ft(30 ft) (20 ft) (20 ft) (8 ft) 0

24(10 ft) 24(10 ft) 2 6

y

dvx

dx

BC (g)

(a) Solve for By, Dy, and MD:

Solve equations (a), (b), (e), (f), and (g) simultaneously to obtain:

2

1

3

2

65.9 kip

59.0000 k

s

19.13 kips

105

ip-ft

1,006.6667 kip-ft

65.8700 kips

19.13

.9 kip-

00 kips

105.93 ft (cw33 )kip-ft

y

y

D

C

C

B

D

M Ans.

(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 22 ft) is computed as follows:

5 5 3

2 3

3

3

2

5 kips/ft 5 kips/ft 65.8700 kips(22 ft) (12 ft) (12 ft)

120(10 ft) 120(10 ft) 6

(59.0000 kip-ft )(22 ft) 1,006.6667 kip-ft

1,757.4400 kip-ft

1,757.4400 kip-ft0.017574 ft

100,000 kip-ft

C

C

EI v

v 0.2109 in. 0.211 in. Ans.







(a) the reactions at B, C, and D.

(b) the beam deflection at A.

Fig. P11.26

Solution

Moment equation:


120 kN (60 kN/m)(12 m) 0

840 kN

y y y y

y y y

F B C D

B C D (a)

(120 kN)(3 m) (60 kN/m)(12 m)(6 m)

(6 m) (12 m) 0

(6 m) (12 m) 3,960 kN-m

B

y y

y y

M

C D

C D (b)


1 1 0

1 1

0 0 1

0 0

1 1 2

1 1

2

120 kN 0 m 3 m 60 kN/m 3 m

9 m 15 m

( ) 120 kN 0 m 3 m 60 kN/m 3 m

9 m 15 m

60 kN/m( ) 120 kN 0 m 3 m 3 m

2

9 m 15 m

y

y y

y

y y

y

y y

w x x B x x

C x D x

V x w x dx x B x x

C x D x

M x V x dx x B x x

C x D x

d vEI

1 1 2

2

1 1

60 kN/m( ) 120 kN 0 m 3 m 3 m

2

9 m 15 m

y

y y

M x x B x xdx

C x D x

2 2 3

2 2

1

120 kN 60 kN/m0 m 3 m 3 m

2 2 6

9 m 15 m2 2

y

y y

BdvEI x x x

dx

C Dx x C (c)

3 3 4

3 3

1 2

120 kN 60 kN/m0 m 3 m 3 m

6 6 24

9 m 15 m6 6

y

y y

BEI v x x x

C Dx x C x C (d)


3

1 2

at 3 m, 0

120 kN(3 m) (3 m) 0

6

x v

C C (e)




3 3 4

1 2

at 9 m, 0

120 kN 60 kN/m(9 m) (6 m) (6 m) (9 m) 0

6 6 24

y

x v

BC C (f)

3 3 4 3

1 2

at 15 m, 0

120 kN 60 kN/m(15 m) (12 m) (12 m) (6 m) (15 m) 0

6 6 24 6

y y

x v

B CC C (g)

(a) Solve for By, Cy, and Dy:


2

1

3

2

330 kN

900.0000

360 kN

150.0 kN

kN-m

2,160.0000 kN-m

330.0000 kN

360.0000 kN

150.0000 kN

y

y

y

C

C

B

C

D Ans.

(b) Beam deflection at A: From Eq. (d), the beam deflection at A (x = 0 m) is computed as follows:

3

3

2

2,160.0000 kN-m

2,160.0000 kN-m0.010800 m 10.80 mm

200,000 kN-m10.80 mm

A

A

EI v

v Ans.







(a) the reactions at B, C, and D.


Fig. P11.27

Solution

Moment equation:


(60 kN/m)(6 m) 0

360 kN

y y y y

y y y

F B C D

B C D (a)

420 kN-m (60 kN/m)(6 m)(3 m)

(6 m) (12 m) 0

(6 m) (12 m) 660 kN-m

B

y y

y y

M

C D

C D (b)


2 1 0

1 0 1

1 0 1

0 1 0

0 1

420 kN-m 0 m 3 m 60 kN/m 3 m

9 m 60 kN/m 9 m 15 m

( ) 420 kN-m 0 m 3 m 60 kN/m 3 m

9 m 60 kN/m 9 m 15 m

60 ( ) 420 kN-m 0 m 3 m

y

y y

y

y y

y

w x x B x x

C x x D x

V x w x dx x B x x

C x x D x

M x V x dx x B x2

1 2 1

20 1 2

2

1 2 1

kN/m3 m

2

60 kN/m9 m 9 m 15 m

2

60 kN/m( ) 420 kN-m 0 m 3 m 3 m

2

60 kN/m9 m 9 m 15 m

2

y y

y

y y

x

C x x D x

d vEI M x x B x x

dx

C x x D x

1 2 3

2 3 2

1

60 kN/m420 kN-m 0 m 3 m 3 m

2 6

60 kN/m9 m 9 m 15 m

2 6 2

y

y y

BdvEI x x x

dx

C Dx x x C (c)

2 3 4

3 4 3

1 2

420 kN-m 60 kN/m0 m 3 m 3 m

2 6 24

60 kN/m9 m 9 m 15 m

6 24 6

y

y y

BEI v x x x

C Dx x x C x C (d)





2

1 2

at 3 m, 0

420 kN-m(3 m) (3 m) 0

2

x v

C C (e)

2 3 4

1 2

at 9 m, 0

420 kN-m 60 kN/m(9 m) (6 m) (6 m) (9 m) 0

2 6 24

y

x v

BC C (f)

2 3 4

3 4

1 2

at 15 m, 0

420 kN-m 60 kN/m(15 m) (12 m) (12 m)

2 6 24

60 kN/m(6 m) (6 m) (15 m) 0

6 24

y

y

x v

B

CC C (g)

(a) Solve for By, Cy, and Dy:


2

1

3

2

1,590.0000 kN-m

2,880.0000 kN-m

245.0000 kN

120.0000 kN

245 kN

120 kN

5.00 kN5.0000 kN

y

y

y

C

C

B

C

D Ans.

(b) Beam deflection at A: From Eq. (d), the beam deflection at A (x = 0 m) is computed as follows:

3

3

2

2,880.0000 kN-m

2,880.0000 kN-m0.014400 m 14.40 mm

200,000 kN-m14.40 mm

A

A

EI v

v Ans.





below, assume that EI = 3.0 × 104 kN-m

2 is


(a) For the beam in Fig. P11.28a, determine

the concentrated upward force P required to

make the total beam deflection at B equal to

zero (i.e., vB = 0).

Fig. P11.28a

Solution

Downward deflection at B due to 15 kN/m uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load over portion of span.]


3

2 2(4 7 3 )24

B

wav L aL a

LEI

Values:

w = 15 kN/m, L = 7 m, a = 3.5 m,

EI = 3.0 × 104 kN-m

2

Computation:

32 2

3 32 2

(4 7 3 )24

(15 kN/m)(3.5 m) 234.472656 kN-m4(7 m) 7(3.5 m)(7 m) 3(3.5 m)

24(7 m)

B

wav L aL a

LEI

EI EI

Upward deflection at B due to concentrated load P.

[Appendix C, SS beam with concentrated load at midspan.]


3

48B

PLv

EI

Values:

L = 7 m, EI = 3.0 × 104 kN-m

2

Computation:

3 3 3(7 m) (7.145833 m )

48 48B

PL P Pv

EI EI EI

Compatibility equation at B:

3 3

3

3

234.472656 kN-m (7.145833 m )0

234.472656 kN-m32.8125 kN

7.14532.8 kN

833 m

P

EI EI

P Ans.






2 is


(b) For the beam in Fig. P11.28b, determine

the concentrated moment M required to

make the total beam slope at A equal to zero

(i.e., A = 0).

Fig. P11.28b

Solution

Slope at A due to 32 kN concentrated load.

[Appendix C, Cantilever beam with concentrated load at tip.]


2

2A

PL


Values:

P = 32 kN, L = 4 m, EI = 3.0 × 104 kN-m

2

Computation:

2 2 2(32 kN)(4 m) 256 kN-m

(negative slope by inspection)2 2

A

PL

EI EI EI

Slope at A due to concentrated moment M.

[Appendix C, Cantilever beam with concentrated moment at tip.]


A

ML


Values:

L = 4 m, EI = 3.0 × 104 kN-m

2

Computation:

(4 m) (4 m)

(positive slope by inspection)A

ML M M

EI EI EI

Compatibility equation at A:

2

2

64.

256 kN-m (4 m)0

256 kN-m0 kN-m

4 m

M

EI EI

M Ans.





below, assume that EI = 5.0 × 106 kip-in.

2 is


(a) For the beam in Fig. P11.29a, determine

the concentrated upward force P required to

make the total beam deflection at B equal to

zero (i.e., vB = 0).

Fig. P11.29a

Solution

Downward deflection at B due to 4 kips/ft uniformly distributed load.



4

8B

wLv

EI

Values:


2

Computation:

4 4 3(4 kips/ft)(13 ft) 14,280.5 kip-ft

8 8B

wLv

EI EI EI

Upward deflection at B due to concentrated load P.



3

3B

PLv

EI

Values:

L = 13 ft, EI = 5.0 × 106 kip-in.

2

Computation:

3 3 3(13 ft) (732.333333 ft )

3 3B

PL P Pv

EI EI EI


3 3

3

319.50 kip

14,280.5 kip-ft (732.333333 ft )0

14,280.5 kip-ft

732.333333 s

ft

P

EI EI

P Ans.






2 is




make the total beam slope at C equal to zero

(i.e., C = 0).

Fig. P11.29b

Solution

Slope at C due to 40-kip concentrated load.



2

16C

PL


Values:


2

Computation:

2 2 2(40 kips)(18 ft) 810 kip-ft

(negative slope by inspection)16 16

C

PL

EI EI EI

Slope at C due to concentrated moment M.

[Appendix C, SS beam with concentrated moment at one end.]


3

C

ML


Values:

L = 18 ft, EI = 5.0 × 106 kip-in.

2

Computation:

(18 ft) (6 ft)

(positive slope by inspection)3 3

C

ML M M

EI EI EI

Compatibility equation at C:

2

2

810 kip-ft (6 ft)0

810 kip-f135.0

t

6 kip-f

ftt

M

EI EI

M Ans.






2


(a) For the beam in Fig. P11.30a,

determine the concentrated downward

force P required to make the total beam

deflection at B equal to zero (i.e., vB = 0).

Fig. P11.30a

Solution

Upward deflection at B due to 105 kN-m concentrated moment.



2 2(2 3 )6

B

M xv L Lx x

LEI (elastic curve)

Values:

M = −105 kN-m, L = 8 m, x = 4 m,

EI = 5.0 × 104 kN-m

2

Computation:

2 2

32 2

(2 3 )6

( 105 kN-m)(4 m) 420 kN-m2(8 m) 3(8 m)(4 m) (4 m)

6(8 m)

B

M xv L Lx x

LEI

EI EI

Downward deflection at B due to concentrated load P.



3

48B

PLv

EI

Values:

L = 8 m, EI = 5.0 × 104 kN-m

2

Computation:

3 3 3(8 m) (10.666667 m )

48 48B

PL P Pv

EI EI EI


3 3

3

3

420 kN-m (10.666667 m )0

420 kN-m39.375 kN

10.66666739.4 kN

m

P

EI EI

P Ans.






2 is




make the total beam slope at A equal to zero

(i.e., A = 0).

Fig. P11.30b

Solution

Slope at A due to 6 kN/m uniformly distributed load.



3

6A

wL


Values:

w = 6 kN/m, L = 5 m, EI = 5.0 × 104 kN-m

2

Computation:

3 3 2(6 kN/m)(5 m) 125 kN-m


A

wL

EI EI EI




A

ML


Values:

L = 5 m, EI = 5.0 × 104 kN-m

2

Computation:

(5 m)

(negative slope by inspection)A

ML M

EI EI


2

2

25.0 kN-

125 kN-m (5 m)0

125 kN-m

mm

5

M

EI EI

M Ans.






2


(a) For the beam in Fig. P11.31a,

determine the concentrated downward

force P required to make the total beam

deflection at B equal to zero (i.e., vB = 0).

Fig. P11.31a

Solution

Upward deflection at B due to 125 kip-ft concentrated moment.



2

2B

MLv

EI

Values:

M = −125 kip-ft, L = 15 ft, EI = 8.0 × 106 kip-in.

2

Computation:

2 2 3( 125 kip-ft)(15 ft) 14,062.5 kip-ft

2 2B

MLv

EI EI EI

Downward deflection at B due to concentrated load P.



3

3B

PLv

EI

Values:

L = 15 ft, EI = 8.0 × 106 kip-in.

2

Computation:

3 3 3(15 ft) (1,125 ft )

3 3B

PL P Pv

EI EI EI


3 3

3

3

14,062.5 kip-ft (1,125 ft )0

14,062.5 kip-ft

1,125 12.50 kips

ft

P

EI EI

P Ans.






2


(b) For the beam in Fig. P11.31b,

determine the concentrated moment M

required to make the total beam slope at A

equal to zero (i.e., A = 0). Fig. P11.31b

Solution

Slope at A due to 7 kips/ft uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load over portion of span.]


2

2 2(2 )24

A

waL a


Values:

w = 7 kips/ft, L = 23 ft, a = 15 ft,

EI = 8.0 × 106 kip-in.

2

Computation:

22 2

22 2

2

(2 )24

(7 kips/ft)(15 ft)2(23 ft) (15 ft)

24(23 ft)

2,376.766304 kip-ft(negative slope by inspection)

A

waL a

LEI

EI

EI




3

A

ML


Values:

L = 23 ft, EI = 8.0 × 106 kip-in.

2

Computation:

(23 ft) (7.666667 ft)


A

ML M M

EI EI EI


2

2

310 kip-f

2,376.766304 kip-ft (7.666667 ft)0

2,376.766304 kip-ft

7.666667 ftt

M

EI EI

M Ans.





below, derive an expression for the reactions

at supports A and B. Assume that EI is


Fig. P11.32

Solution

Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever.

Consider downward deflection of cantilever beam at B due to concentrated moment M0.



2 2

0

2 2B

ML M Lv

EI EI

Consider upward deflection of cantilever beam at B due to concentrated load By.



33

3 3

y

B

B LPLv

EI EI

Compatibility equation for deflection at B:

32

0 002

3

3 2

y

y

B LM LB

EI E

M

I L Ans.

Equilibrium equations

for entire beam:

0 03

2

3

20y y y y y

MF A B A B

L

M

L Ans.

0 0A A yM M M B L

0 00 0 0

03 3( )

2 (c )

22wA y

M MM B L M L M

L

MM Ans.








Fig. P11.33

Solution


Consider downward deflection of cantilever beam at B due to linearly distributed load.

[Appendix C, Cantilever beam with linearly distributed load.]


4

0

30B

w Lv

EI




33

3 3

y

B

B LPLv

EI EI


3

0

4

0 030 3 10

y

y

B Lw LB

EI E

w L

I Ans.

Equilibrium equations for entire beam:

0 0 0 00 40

2 2 10

2

510y y y y

w L w L w L w LF A B A

w L Ans.

0 02 3

A A y

w L LM M B L

2 2 2

0 0 0 0

2

0( )6 1

(ccw)150 6 15

A y

w L w L w L w LM B

LL L

w Ans.




11.34 For the beam and loading shown below,

derive an expression for the reactions at

supports A and B. Assume that EI is constant

for the beam.

Fig. P11.34

Solution

Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever.

Consider downward deflection of cantilever beam at A due to concentrated load P.



2

(3 )6

A

Pxv L x

EI (elastic curve)

2 3

3Let ,

2

( ) 3 73

6 2 12A

Lx L L

P L L PLv L

EI EI

Consider upward deflection of cantilever beam at A due to concentrated load Ay.



33

3 3

y

A

A LPLv

EI EI

Compatibility equation for deflection at A:

3370

12 3

7

4

y

y

A LPLA

EI EI

P Ans.


0

7 3

4 4

3

4

y y y

y

F A B P

P PB P

P Ans.

3

02

B B y

LM M A L P

3 7 3

( (cc42

w)4 4

)2

B y

L P PL PLM A L P L

PL Ans.




11.35 For the beam and loading shown below,

derive an expression for the reactions at

supports A and B. Assume that EI is constant

for the beam.

Fig. P11.35

Solution


Consider downward deflection of cantilever beam at A due to uniformly distributed load.



2

2 2(6 4 )24

A

wxv L Lx x

EI (elastic curve)

22 42

3Let ,

2

( ) 3 3 176 4 ( ) ( )

24 2 2 48A

Lx L L

w L L L wLv L L

EI EI




33

3 3

y

A

A LPLv

EI EI


34170

48 3

17

16

y

y

A LwL

EI

LA

E

w

I Ans.





3 3 17 7 7

11 60

2 2 16 6y y y y

wL wL wL wLF A B B

wL Ans.

3 30

2 4B B y

wL LM M A L

2 22 29 17 9

( )8 16 8

6

(cw)161

B y

wL wL wL w wLLM A L L Ans.








Fig. P11.36

Solution


Consider downward deflection of cantilever beam at A due to concentrated load P.

[Appendix C, Cantilever beam with concentrated load at midspan.]


35

48A

PLv

EI

3 3

Let 2

5 (2 ) 5

48 6A

L L

P L PLv

EI EI




3

3A

PLv

EI

3 3

Let 2 ,

( )(2 ) 8

3 3

y

y y

A

L L P A

A L A Lv

EI EI


33 850

6 3

5

16

y

y

A LPLA

EI EI

P Ans.


0

5 11 11

11 16 66

y y y

y

F A B P

P PB

PP Ans.

(2 ) ( ) 0B B yM M A L P L

3

(c5 3

(2 ) ( )8 8

w)8

B y

PL PLM A L P

PLL PL Ans.





below, derive an expression for the

reactions at supports A and C. Assume that


Fig. P11.37

Solution

Choose the reaction force at C as the redundant; therefore, the released beam is a cantilever.

Consider downward deflection of cantilever beam at C due to concentrated moment M0.

[Appendix C, Cantilever beam with concentrated moment.]


2

and2

B B

ML MLv

EI EI

2 2

0 0 03( )

2 2C

M L M L M Lv L

EI EI EI

Consider upward deflection of cantilever beam at C due to concentrated load Cy.



3

3C

PLv

EI

3 3

Let 2 ,

( )(2 ) 8

3 3

y

y y

C

L L P C

C L C Lv

EI EI

Compatibility equation for deflection at C:

32

0083

02 3

9

16

y

y

C LM L MC

E EI LI Ans.


0 0

0

9

16

9

16

y y y

y

M

F A

L

C

MA

L Ans.

0 (2 ) 0A A yM M M C L

0 00 0

09(2 ) (2 ) (c )

6w

8 81A y

M MM C L M M

ML

L Ans.





below, derive an expression for the

reactions at supports A and C. Assume that


Fig. P11.38

Solution


Consider downward deflection of cantilever beam at C due to uniformly distributed load.



4 3

and8 6

B B

wL wLv

EI EI

4 3 47

( )8 6 24

C

wL wL wLv L

EI EI EI




3

3C

PLv

EI

3 3

Let 2 ,

( )(2 ) 8

3 3

y

y y

C

L L P C

C L C Lv

EI EI


34 870

24 3

7

64

y

y

C LwL

EI I

LC

w

E Ans.


57

64

0

7

64

y y y

y

F A C wL

wLA wL

wL Ans.

(2 ) 02

A A y

LM M wL C L

2 22 2 27 18 9

(2 ) (2 )2 64 2 64

9 (ccw)

3 322A y

wL wL wL wLwL wLM C L L Ans.





below, derive an expression for the reaction

forces at A, C, and D. Assume that EI is

constant for the beam. (Reminder: The

roller symbol implies that both upward and

downward displacement is restrained.)

Fig. P11.39

Solution

Choose the reaction force at C as the redundant; therefore, the released beam is simply supported.

Consider downward deflection of simply supported beam at C due to P.



2 2 2( ) (elastic curve)

6C

Pbxv L b x

LEI

2 2 2

32

Let 3 , ,

( )( )(3 ) ( ) ( )

6(3 )

77

18 18

C

L L b L x L

P L Lv L L L

L EI

PL PLL

EI EI

Consider upward deflection of simply supported beam at C due to concentrated load Cy.



2 2 2( )

6C

Pabv L a b

LEI

2 2 2

3

2

Let 3 , 2 , ,

( )(2 )( )(3 ) (2 ) ( )

6(3 )

2 84

18 18

y

y

C

y y

L L a L b L P C

C L Lv L L L

L EI

C L C LL

EI EI


33 870

18 1

7

8 8

y

y

C LPLC

EI EI

P Ans.





(2 ) (3 ) 0A y yM PL C L D L

7 3(3 ) (2 ) (2 )

8 4

4 4

y y

y

P PLD L PL C L PL L

PD

P Ans.

0

7 5 3

8 4 8 8

3

8

y y y y

y y y

F A C D P

P PP P PA P C D P P Ans.






force at B. Assume that EI is constant for

the beam. (Reminder: The roller symbol

implies that both upward and downward

displacement is restrained.)

Fig. P11.40

Solution

Choose the reaction force at B as the redundant; therefore, the released beam is simply supported.

Consider upward deflection of simply supported beam at B due to M0.



2 2(2 3 ) (elastic curve)

6B

M xv L Lx x

LEI

0

22 2 20 0 0

Let 2 , ,

( )( )2(2 ) 3(2 )( ) ( ) 3

6(2 ) 12 4B

L L x L M M

M L M M Lv L L L L L

L EI EI EI

Consider upward deflection of simply supported beam at B due to concentrated load By.



3

48B

PLv

EI

3 3

Let 2 ,

( )(2 )

48 6

y

y y

B

L L P B

B L B Lv

EI EI


32

0 0 030

4 6 2

3

2

y

y

B LM L MB

EI EI

M

LL Ans.







the beam.

Fig. P11.41

Solution


Consider downward deflection of simply supported beam at B due to uniformly distributed load.



3

2 2(4 7 3 )24

B

wav L aL a

LEI

3 2 42 2 2

Let 3 , 2

(2 ) 24(3 ) 7(2 )(3 ) 3(2 ) 6

24(3 ) 9 3B

L L a L

w L wL wLv L L L L L

L EI EI EI




2 2 2( )

6B

Pabv L a b

LEI

3

2 2 2 2

Let 3 , , 2 ,

( )( )(2 ) 4(3 ) ( ) (2 ) 4

6(3 ) 9 9

y

y y y

B

L L a L b L P B

B L L B L B Lv L L L L

L EI EI EI


34 42 30

3 9

3

22

y

y

B LwL wLB

EI EI

wL Ans.







the beam. (Reminder: The roller symbol

implies that both upward and downward

displacement is restrained.)

Fig. P11.42

Solution


Consider upward deflection of simply supported beam at B due to uniformly distributed load.




6B

M xv L Lx x

LEI

2

2

2 42 2 2

Let 2 , , 2 4 8

( )8

2(2 ) 3(2 )( ) ( ) 36(2 ) 96 32

B

L L wLL L x L M w

wLL

wL wLv L L L L L

L EI EI EI




3

48B

PLv

EI

3 3

Let 2 ,

( )(2 )

48 6

y

y y

B

L L P B

B L B Lv

EI EI


34 30

32 6 16

3

16

y

y

B LwL wLB

EI EI

wL Ans.







the beam.

Fig. P11.43

Solution


Consider downward deflection of simply supported beam at B due to one concentrated load P.




6B

Pbxv L b x

LEI

32 2 2 2

Let 4 , , 2

( )(2 ) 11(4 ) ( ) (2 ) 11

6(4 ) 12 12B

L L b L x L

P L L PL PLv L L L L

L EI EI EI

The second concentrated load will cause an additional deflection at B of the same magnitude.




3

48B

PLv

EI

3 3 3

Let 4 ,

( )(4 ) 64 16

48 48 12

y

y y y

B

L L P B

B L B L B Lv

EI EI EI


33 3 1611 110

12 1 82

11

2 1

y

y

B LPL PLB

I EI

P

E EI Ans.







the beam.

Fig. P11.44

Solution





3

2 2(4 7 3 )24

B

wav L aL a

LEI

3 2 42 2 2

Let 5 , 3

(3 ) 27 994(5 ) 7(3 )(5 ) 3(3 ) 22

24(5 ) 120 20B

L L a L

w L wL wLv L L L L L

L EI EI EI

Consider downward deflection of simply supported beam at B due to concentrated load.




6B

Pbxv L b x

LEI

2 42 2 2 2

Let 5 , , 3 , 3

( )(3 )3

(5 ) ( ) (3 ) 156(5 ) 30 2

B

wLL L b L x L P

wLL L

wL wLv L L L L

L EI EI EI







2 2 2( )6

B

Pabv L a b

LEI

3

2 2 2 2

Let 5 , 3 , 2 ,

( )(3 )(2 ) 12(5 ) (3 ) (2 ) 12

6(5 ) 5 5

y

y y y

B

L L a L b L P B

B L L B L B Lv L L L L

L EI EI EI


34 4

2.21299 109

020 2 5 48

70833y

y

B LwL wL wLB

EI EI EIwL Ans.




11.45 The beam shown in Fig. P11.45 consists of a

W360 × 79 structural steel wide-flange shape [E =

200 GPa; I = 225 × 106 mm

4]. For the loading

shown, determine:

(a) the reactions at A, B, and C.

(b) the magnitude of the maximum bending stress

in the beam.

Fig. P11.45

Solution

(a) Reactions at A, B, and C. Choose the reaction force at B as the redundant; therefore, the released

beam is simply supported between A and C.




3

2 2(4 7 3 )24

B

wav L aL a

LEI

Values:

w = 90 kN/m, L = 9 m, a = 6 m

Calculation:

32 2

3 32 2

(4 7 3 )24

(90 kN/m)(6 m) 4,860 kN-m4(9 m) 7(6 m)(9 m) 3(6 m)

24(9 m)

B

wav L aL a

LEI

EI EI

Consider downward deflection of simply supported beam at B due to concentrated moment.

[Appendix C, SS beam with concentrated moment at one end of span.]



6B

M xv L Lx x

LEI

Values:

M = 180 kN-m, L = 9 m, x = 3 m

Calculation:

2 2

32 2

(2 3 )6

(180 kN-m)(3 m) 900 kN-m2(9 m) 3(9 m)(3 m) (3 m)

6(9 m)

B

M xv L Lx x

LEI

EI EI







2 2 2( )6

B

Pabv L a b

LEI

Values:

P = −By, L = 9 m, a = 3 m, b = 6 m

Calculation:

2 2 2

3

2 2 2

( )6

( )(3 m)(6 m) (12 m )(9 m) (3 m) (6 m)

6(9 m)

B

y y

Pabv L a b

LEI

B B

EI EI


33 3

3

3

(12 m )4,860 kN-m 900 kN-m0

5,760 kN-m480 480 kkN

12 N

m

y

y

B

EI EI EI

B Ans.

Equilibrium equations for entire beam: (3 m) (9 m) 180 kN-m (90 kN/m)(6 m)(6 m) 0A y yM B C

180 kN-m (90 kN/m)(6 m)(6 m) (480 kN)(3 m)

9 m

220.0 kN 220 kN

yC

Ans.

(90 kN/m)(6 m) 0y y y yF A B C

(90 kN/m)(6 m) 480 kN 220 kN 160.0 kN 160.0 kNyA Ans.





(b) Magnitude of maximum bending stress:

Section properties (from Appendix B):

6 4

3 3

225 10 mm 353 mm

1,270 10 mm

I d

S


Mmax = 300 kN-m


2

6 4

(300 kN-m)(353 mm/2)(1,000)

225 10

235 MPa

mmx

Ans.

or using the tabulated section modulus value:

2

3 3

(300 kN-m)(1,000)

1,270 10 m

23 M

m

6 Pa

x

Ans.





W610 × 140 structural steel wide-flange shape [E

= 200 GPa; I = 1,120 × 106 mm

4]. For the loading

shown, determine:

(a) the reactions at A, B, and D.

(b) the magnitude of the maximum bending stress

in the beam.

Fig. P11.46

Solution

(a) Reactions at A, B, and D. Choose the reaction force at B as the redundant; therefore, the released

beam is simply supported between A and D.




3 2 3( 2 ) (elastic curve)24

B

wxv L Lx x

EI

Values:

w = 60 kN/m, L = 7.5 m, x = 1.5 m

Calculation:

3 2 3

3

3 2 3

( 2 )24

(60 kN/m)(1.5 m) 1,468.125 kN-m(7.5 m) 2(7.5 m)(1.5 m) (1.5 m)

24

B

wxv L Lx x

EI

EI EI

Consider downward deflection of simply supported beam at B due to concentrated load.




6B

Pbxv L b x

LEI

Values:

P = 125 kN, L = 7.5 m, b = 2.5 m, x = 1.5 m

Calculation:

2 2 2

32 2 2

( )6

(125 kN)(2.5 m)(1.5 m) 497.396 kN-m(7.5 m) (2.5 m) (1.5 m)

6(7.5 m)

B

Pbxv L b x

LEI

EI EI







2 2 2( )6

B

Pabv L a b

LEI

Values:

P = −By, L = 7.5 m, a = 1.5 m, b = 6 m

Calculation:

2 2 2

3

2 2 2

( )6

( )(1.5 m)(6 m) (3.6 m )(7.5 m) (1.5 m) (6 m)

6(7.5 m)

B

y y

Pabv L a b

LEI

B B

EI EI


33 3

3

3

(3.6 m )1,468.125 kN-m 497.396 kN-m0

1,965.521 kN-m545.978 kN

3.6 546 k

mN

y

y

B

EI EI EI

B Ans.

Equilibrium equations for entire beam: (1.5 m) (7.5 m) (60 kN/m)(7.5 m)(3.75 m) (125 kN)(5 m) 0A y yM B D

(60 kN/m)(7.5 m)(3.75 m) (125 kN)(5 m) (545.978 kN)(1.5 m)

7.5 m

199.138 kN 199.1 kN

yD

Ans.

(60 kN/m)(7.5 m) 125 kN 0y y y yF A B D

(60 kN/m)(7.5 m) 125 kN 545.978 kN 199.138 kN

170.116 170.1 kNkN

yA

Ans.







6 4

3 3

1,120 10 mm 617 mm

3,640 10 mm

I d

S


Mmax = 322.67 kN-m


2

6 4

(322.67 kN-m)(617 mm/2)(1,000)

1,120 10 mm

88.9 MPa

x

Ans.

or using the tabulated section modulus value:

2

3 3

(322.67 kN-m)(1,000

88.6

)

3,640 10 m

M a

m

P

x

Ans.





shown in Fig. P11.47. Determine the reactions at A

and D for the beam. Assume EI = 12.8 × 106 lb-in.

2.

Fig. P11.47

Solution

Choose the reaction force at D as the redundant; therefore, the released beam is a cantilever.

Consider downward deflection of cantilever beam at D due to uniformly distributed load.



4 3

and8 6

C C

wL wLv

EI EI

Values:

w = 25 lb/in., L = 72 in.

Calculation:

4 4 3

3 3 2

3 2 3

(25 lb/in.)(48 in.) 16,588,800 lb-in.

8 8

(25 lb/in.)(48 in.) 460,800 lb-in.

6 6

16,588,800 lb-in. 460,800 lb-in. 27,648,000 lb-in.(24 in.)

C

C

D

wLv

EI EI EI

wL

EI EI EI

vEI EI EI

Consider downward deflection of cantilever beam at D due to the 360-lb concentrated load.



3 2

and3 2

B B

PL PLv

EI EI

Values:

P = 360 lb, L = 24 in.

Calculation:

3 3 3

2 2 2

3 2 3

(360 lb)(24 in.) 1,658,880 lb-in.

3 3

(360 lb)(24 in.) 103,680 lb-in.

2 2

1,658,880 lb-in. 103,680 lb-in. 6,635,520 lb-in.(48 in.)

B

B

D

PLv

EI EI EI

PL

EI EI EI

vEI EI EI




Consider upward deflection of cantilever beam at D due to concentrated load Dy.



3

3D

PLv

EI

Values:

P = −Dy, L = 72 in.

Calculation:

3 33 ( )(72 in.) (124,416 in. )

3 3

y y

D

D DPLv

EI EI EI

Compatibility equation for deflection at D:

33 3

3

3

(124,416 in. )27,648,000 lb-in. 6,635,520 lb-in.0

34,283,520 lb-in.275.556 lb

124,416 276 l

in.b

y

y

D

EI EI EI

D Ans.



(25 lb/in.)(48 in.) 360 lb 0

(25 lb/in.)(48 in.) 360 lb 275.556 lb 1,28 1,4 284 l.4 b44 lb

y y y

y

F A D

A Ans.

(25 lb/in.)(48 in.)(24 in.) (360 lb)(24 in.) (72 in.) 0A A yM M D

(275.556 lb)(72 in.) (25 lb/in.)(48 in.)(24 in.) (360 lb)(24 in.)

17,600 17,600 lb-in. (clb-in. cw)

AM

Ans.





shown in Fig. P11.48. Assume EI = 24 × 106

kip-in.2. Determine:

(a) the reactions at B and C for the beam.


Fig. P11.48

Solution


Consider downward deflection of cantilever beam at B due to uniformly distributed load.



4

8B

wLv

EI

Values:

w = 8 kips/ft, L = 24 ft

Calculation:

4 4 3(8 kips/ft)(24 ft) 331,776 kip-ft

8 8B

wLv

EI EI EI (a)

Consider downward deflection of cantilever beam at B due to the 40-kip concentrated load.



2

(3 ) (elastic curve)6

B

Pxv L x

EI

Values:

P = 40 kips, L = 36 ft, x = 24 ft

Calculation:

2 2 3(40 kips)(24 ft) 322,560 kip-ft

(3 ) 3(36 ft) (24 ft)6 6

B

Pxv L x

EI EI EI







3

3B

PLv

EI

Values:

P = −By, L = 24 ft

Calculation:

3 33 ( )(24 ft) (4,608 ft )

3 3

y y

B

B BPLv

EI EI EI (b)


33 3

3

3

(4,608

142.0

ft )331,776 kip-ft 322,560 kip-ft0

654,336 kip-ft142.0 kips

4,608 ftkips

y

y

B

EI EI EI

B Ans.



40 kips (8 kips/ft)(24 ft) 142. 90.0 k0 kip ss ip

y y y

y

F B C

C Ans.

(40 kips)(36 ft) (8 kips/ft)(24 ft)(12 ft) (24 ft) 0C C yM M B

(142.0 kips)(24 ft) (40 kips)(36 ft) (8 kips/ft)(24 ft)(12 ft)

336. 30 kip-f 36 kip-ft t (cw)

CM

Ans.

(b) Beam deflection at A:

Consider downward deflection of cantilever beam at A due to uniformly distributed load.



4 3

and (slope magnitude)8 6

B B

wL wLv

EI EI

Values:

w = 8 kips/ft, L = 24 ft, EI = 24 × 106 kip-in.

2




Calculation:

3

3 3 2

3 2 3

331,776 kip-ftcalculated previously in Eq. (a)

(8 kips/ft)(24 ft) 18,432 kip-ft

6 6

331,776 kip-ft 18,432 kip-ft 552,960 kip-ft(12 ft)

B

B

A

vEI

wL

EI EI EI

vEI EI EI

Consider downward deflection of cantilever beam at A due to the 40-kip concentrated load.



3

3A

PLv

EI

Values:

P = 40 kips, L = 36 ft, EI = 24 × 106 kip-in.

2

Calculation:

3 3 3(40 kips)(36 ft) 622,080 kip-ft

3 3A

PLv

EI EI EI




3 2


B B

PL PLv

EI EI

Values:

P = −By = 142 kips, L = 24 ft, EI = 24 × 106 kip-in.

2

Calculation:

3 3

2 2 2

3 2 3

(4,608 ft )(142.0 kips) 654,336 kip-ftusing the results from Eq. (b)

(142.0 kips)(24 ft) 40,896 kip-ft

2 2

654,336 kip-ft 40,896 kip-ft 1,145,088 kip-ft(12 ft)

B

B

A

vEI EI

PL

EI EI EI

vEI EI EI

Beam deflection at A.

3 3 3

3 3 3

6 2

552,960 kip-ft 622,080 kip-ft 1,145,088 kip-ft

29,952 kip-ft (29,952 kip-ft )(12 in./ft)2.156544 in.

24 10 kip-i2.16 i

n.n.

AvEI EI EI

EI Ans.




11.49 A propped cantilever beam is loaded

as shown in Fig. P11.49. Assume EI = 86.4

× 106 N-mm

2. Determine:

(a) the reactions at A and C for the beam.


Fig. P11.49

Solution


Consider downward deflection of cantilever beam at C due to uniformly distributed load.



2

2 2(6 4 ) (elastic curve)24

C

wxv L Lx x

EI

Values:

w = 25 N/mm, L = 400 mm, x = 300 mm,

EI = 86.4 × 106 N-mm

2

Calculation:

22 2

22 2

9 3

(6 4 )24

(25 N/mm)(300 mm)6(400 mm) 4(400 mm)(300 mm) (300 mm)

24

53.43750 10 N-mm

C

wxv L Lx x

EI

EI

EI

Consider downward deflection of cantilever beam at C due to the 4,000-N concentrated load.



3 2


B B

PL PLv

EI EI

Values:

P = 4,000 N, L = 120 mm, EI = 86.4 × 106 N-mm

2

Calculation:

3 3 9 3

2 2 7 2

9 3 7 2 9 3

(4,000 N)(120 mm) 2.304 10 N-mm

3 3

(4,000 N)(120 mm) 2.88 10 N-mm

2 2

2.304 10 N-mm 2.88 10 N-mm 7.488 10 N-mm(180 mm)

B

B

C

PLv

EI EI EI

PL

EI EI EI

vEI EI EI







3

3C

PLv

EI

Values:

P = −Cy, L = 300 mm, EI = 86.4 × 106 N-mm

2

Calculation:

3 6 33 ( )(300 mm) (9 10 mm )

3 3

y y

C

C CPLv

EI EI EI


6 39 3 9 3

9 3

6 3

(9 10 mm )53.43

6,

750 10 N-mm 7.488 10 N-mm0

60.9255 10770 N

N-mm6,769.5 N

9 10 mm

y

y

C

EI EI EI

C Ans.


4,000 N (25 N/mm)(400 mm) 0

4,000 N (25 N/mm)(400 mm) 6,769.5 7,230 NN 7,230.5 N

y y y

y

F A C

A Ans.

(4,000 N)(120 mm) (25 N/mm)(400 mm)(200 mm) (300 mm) 0A A yM M C

(6,769.5 N)(300 mm) (4,000 N)(120 mm) (25 N/mm)(400 mm)(200 mm)

449,15 449,000 N-mm (c0 N-mm cw)

AM

Ans.

(b) Beam deflection at B:

Consider downward deflection of cantilever beam at B due to uniformly distributed load.



2


B

wxv L Lx x

EI

Values:

w = 25 N/mm, L = 400 mm, x = 120 mm,

EI = 86.4 × 106 N-mm

2




Calculation:

22 2

22 2

9 3

(6 4 )24

(25 N/mm)(120 mm)6(400 mm) 4(400 mm)(120 mm) (120 mm)

24

11.736 10 N-mm

B

wxv L Lx x

EI

EI

EI

Consider downward deflection of cantilever beam at B due to the 4,000-N concentrated load.



3

3B

PLv

EI

Values:

P = 4,000 N, L = 120 mm, EI = 86.4 × 106 N-mm

2

Calculation:

3 3 9 3(4,000 N)(120 mm) 2.304 10 N-mm

3 3B

PLv

EI EI EI

Consider upward deflection of cantilever beam at B due to concentrated load Cy.



2

(3 )6

B

Pxv L x

EI

Values:

P = −Cy = −6,769.5 N, L = 300 mm, x = 120 mm,

EI = 86.4 × 106 N-mm

2

Calculation:

2 2

9 3

( 6,769.5 N)(120 mm)(3 ) 3(300 mm) (120 mm)

6 6

12.6725 10 N-mm

B

Pxv L x

EI EI

EI

Beam deflection at B.

9 3 9 3 9 3

9 3 9 3

6 2

11.736 10 N-mm 2.304 10 N-mm 12.6725 10 N-mm

1.367496 10 N-mm 1.367496 10 N-mm15.8275 mm

86.4 1015.83 m

mm

N-m

BvEI EI EI

EI Ans.





W610 × 82 structural steel wide-flange shape [E =

200 GPa; I = 562 × 106 mm

4]. For the loading shown,

determine:

(a) the reaction force at C.

(b) the beam deflection at A. Fig. P11.50

Solution

(a) Reaction force at C. Choose the reaction force at C as the redundant; therefore, the released beam

is simply supported between B and D.

Consider upward deflection of simply supported beam at C due to uniformly distributed load on

overhang AB. [Appendix C, SS beam with concentrated moment at one end.]



6C

M xv L Lx x

LEI

Values:

M = −(105 kN/m)(3 m)(1.5 m) = −472.5 kN-m,

L = 14 m, x = 7 m

Calculation:

2 2

32 2

(2 3 )6

( 472.5 kN-m)(7 m) 5,788.125 kN-m2(14 m) 3(14 m)(7 m) (7 m)

6(14 m)

C

M xv L Lx x

LEI

EI EI

Consider downward deflection of simply supported beam at C due to uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load.]


45

384C

wLv

EI

Values:

w = 105 kN/m, L = 14 m

Calculation:

4 4 35 5(105 kN/m)(14 m) 52,521.875 kN-m

384 384C

wLv

EI EI EI







3

48C

PLv

EI

Values:

P = −Cy, L = 14 m

Calculation:

3 33 ( )(14 m) (57.1667 m )

48 48

y y

C

C CPLv

EI EI EI


33 3

3

3

(57.1667

818

m )5,788.125 kN-m 52,521.875 kN-m0

46,733.750 kN-m817.5 kN

57.1667 mkN

y

y

C

EI EI EI

C Ans.

(b) Beam deflection at A.

Consider downward cantilever beam deflection caused by uniformly distributed load on overhang

AB. [Appendix C, Cantilever beam with uniformly distributed load.]


4

8A

wLv

EI

Values:

w = 105 kN/m, L = 3 m

Calculation:

4 4 3(105 kN/m)(3 m) 1,063.125 kN-m

8 8A

wLv

EI EI EI

Consider downward deflection at A resulting from rotation at B caused by concentrated load on

overhang AB. [Appendix C, SS beam with concentrated moment at one end.]


3

B

ML


Values:

M = (105 kN/m)(3 m)(1.5 m) = 472.5 kN-m,

L = 14 m

Computation:

2

2 3

(472.5 kN-m)(14 m) 2,205 kN-m

3 3

2,205 kN-m 6,615 kN-m(3 m)

B

A

ML

EI EI EI

vEI EI




Consider upward deflection at A due to uniformly distributed load between B and D.



3

(slope magnitude)24

B

wL

EI

Values:

w = 105 kN/m, L = 14 m

Calculation:

3 3 2

2 3

(105 kN/m)(14 m) 12,005 kN-m

24 24

12,005 kN-m 36,015 kN-m(3 m)

B

A

wL

EI EI EI

vEI EI

Consider downward deflection at A due to concentrated load Cy.



2

(slope magnitude)16

B

PL

EI

Values:

P = −Cy = −817.5 kN, L = 14 m

Calculation:

2 2 2

2 3

(817.5 kN)(14 m) 10,014.375 kN-m

16 16

10,014.375 kN-m 30,043.125 kN-m(3 m)

B

A

PL

EI EI EI

vEI EI

Beam deflection at A.

3 3 3 3

3 3

2

1,063.125 kN-m 6,615 kN-m 36,015 kN-m 30,043.125 kN-m

1,706.25 kN-m 1,706.25 kN-m0.015180 m

112,400 kN15.18

-mm

m

AvEI EI EI EI

EI Ans.




11.51 The beam shown in Fig. P11.51


flange shape [E = 29,000 ksi; I = 48 in.4].



(b) the magnitude of the maximum bending

stress in the beam.

(Reminder: The roller symbol implies that

both upward and downward displacement is

restrained.)

Fig. P11.51

Solution


Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam

span. [Appendix C, Cantilever beam with uniformly distributed load.]


2


B

wxv L Lx x

EI

Values:

w = −80 lb/in., L = 150 in., x = 100 in.

Calculation:

22 2

22 2

9 3

(6 4 )24

( 80 lb/in.)(100 in.)6(150 in.) 4(150 in.)(100 in.) (100 in.)

24

2.8333 10 lb-in.

B

wxv L Lx x

EI

EI

EI

Consider deflection of cantilever beam at B due to the force caused by the linear portion of the

distributed load. [Appendix C, Cantilever beam with concentrated load at tip.]


3

3B

PLv

EI

Values:

P = −½(50 in.)(60 lb/in.) = −1,500 lb, L = 100 in.

Calculation:

3 3 6 3( 1,500 lb)(100 in.) 500 10 lb-in.

3 3B

PLv

EI EI EI




Consider deflection of cantilever beam at B due to the moment caused by the linear portion of the

distributed load. [Appendix C, Cantilever beam with concentrated moment at tip.]


2

2B

MLv

EI

Values:

M = −½(50 in.)(60 lb/in.)[⅔(50 in.)]

= −50,000 lb-in., L = 100 in.

Calculation:

2 2 6 3( 50,000 lb-in.)(100 in.) 250 10 lb-in.

2 2B

MLv

EI EI EI

Consider deflection of cantilever beam at B due to concentrated load By.



3

3B

PLv

EI

Values:

P = −By, L = 100 in.

Calculation:

3 33 ( )(100 in.) (333.3333 in. )

3 3

y y

B

B BPLv

EI EI EI


3 39 3 6 3 6 3

9 3

3 3

(333.3333 10 in. )2.8333 10 lb-in. 500 10 lb-in. 250 10 lb-in.0

3.5833 10 lb-in.10,750 lb

333.3333 10 in.10,750 lb

y

y

B

EI EI EI EI

B Ans.


1(80 lb/in.)(150 in.) (140 lb/in. 80 lb/in.)(50 in.) 0

2

(60 lb/in.)(50 in.)(80

2,

lb/in.)(150 in.) ( 10,750 lb)2

2,750 l 75 bb 0 l

y y y

y

F A B

A

Ans.




(80 lb/in.)(150 in.)(75 in.)

(140 lb/in. 80 lb/in.)(50 in.) 2100 in. (50 in.) (100 in.) 0

2 3

A A

y

M M

B

(60 lb/in.)(50 in.)(80 lb/in.)(150 in.)(75 in.) (133.3333 in.) ( 10,750 in.)(100 in.)

2

25,000 lb-in 25,000 l. b-in. (cw)

AM

Ans.



4 348 in. 8.11 in. 11.8 in.I d S


Mmax = 150,000 lb-in. (at B)


4

(150,000 lb-in.)(8.11 in./2)12,671.875 psi

4812

in.,670 psix Ans.

or, using the tabulated value for the section modulus:

3

12,710 ps150,000 lb-in.

12,711.864 psi11.8 in.

ix Ans.




11.52 The beam shown in Fig. P11.52


flange shape [E = 29,000 ksi; I = 2,700 in.4].


(a) the reactions at A and D.


stress in the beam.

Fig. P11.52

Solution


Consider downward deflection of cantilever beam at A due to the 50-kip concentrated load.



3 2


B B

PL PLv

EI EI

Values:

P = 50 kips, L = 20 ft

Calculation:

3 3 3

2 2 3

3 3 3

(50 kips)(20 ft) 133,333.333 kip-ft

3 3

(50 kips)(20 ft) 10,000 kip-ft

2 2

133,333.333 kip-ft 10,000 kip-ft 193,333.333 kip-ft(6 ft)

B

B

A

PLv

EI EI EI

PL

EI EI EI

vEI EI EI

Consider downward deflection of cantilever beam at A due to the uniformly distributed load.



4 3


C C

wL wLv

EI EI

Values:

w = 4 kips/ft, L = 14 ft

Calculation:

4 4 3

3 3 3

3 3 3

(4 kips/ft)(14 ft) 19,208 kip-ft

8 8

(4 kips/ft)(14 ft) 1,829.333 kip-ft

6 6

19,208 kip-ft 1,829.333 kip-ft 41,160 kip-ft(12 ft)

C

C

A

wLv

EI EI EI

wL

EI EI EI

vEI EI EI







3

3A

PLv

EI

Values:

P = −Ay, L = 26 ft

Calculation:

3 33 ( )(26 ft) (5,858.667 ft )

3 3

y y

A

A APLv

EI EI EI (b)


33 3

3

3

(5,858.667 ft )193,333.333 kip-ft 41,160 kip-ft0

234,493.333 kip-ft40.025 kips

5,858.667 40.0 kips

ft

y

y

A

EI EI EI

A Ans.



50 kips (4 kips/ft)(14 ft) 40.025 kips 65.975 kips 66.0 kips

y y y

y

F A D

D Ans.

(50 kips)(20 ft) (4 kips/ft)(14 ft)(7 ft) (26 ft) 0D D yM M A

(40.025 kips)(26 ft) (50 kips)(20 ft) (4 kips/ft)(14 ft)(7 ft)

351.350 kip- 351 kip-ft (cw)ft

DM

Ans.







4

3

2,700 in.

24.3 in.

222 in.

I

d

S


Mmax = 351.350 kip-ft


4

(351.350 kip-ft)(24.3 in./2)(12 in./ft)

2,700

18.97 ksi

in.x

Ans.

or, using the tabulated section modulus:

3

(351.350 kip-ft)(12 in./ft)

222 in

18.99 ksi

.x

Ans.






belt pulleys. Assume that the bearing at A can be

idealized as a pin support and that the bearings at

C and E can be idealized as roller supports. For


(a) the reaction forces at bearings A, C, and E.


stress in the shaft.

Fig. P11.53

Solution

(a) Reaction forces at A, C, and E. Choose the reaction force at C as the redundant; therefore, the

released beam is simply supported between A and E.

Consider downward deflection of simply supported beam at C due to pulley B load.




6C

Pbxv L b x

LEI

Values:

P = 850 N, L = 2,000 mm, b = 600 mm,

x = 1,000 mm

Calculation:

2 2 2

2 2 2

9 3

( )6

(850 N)(600 mm)(1,000 mm)(2,000 mm) (600 mm) (1,000 mm)

6(2,000 mm)

112.2 10 N-mm

C

Pbxv L b x

LEI

EI

EI

Consider downward deflection of simply supported beam at C due to pulley D load.




6C

Pbxv L b x

LEI

Values:

P = 600 N, L = 2,000 mm, b = 400 mm,

x = 1,000 mm




Calculation:

2 2 2

2 2 2

9 3

( )6

(600 N)(400 mm)(1,000 mm)(2,000 mm) (400 mm) (1,000 mm)

6(2,000 mm)

56.8 10 N-mm

C

Pbxv L b x

LEI

EI

EI




3

48C

PLv

EI

Values:

P = −Cy, L = 2,000 mm

Calculation:

3 6 33 ( )(2,000 mm) (166.6667 10 mm )

48 48

y y

C

C CPLv

EI EI EI


6 39 3 9 3

9 3

6 3

(166.6667 10 mm )112.2 10 N-mm 56.8 10 N-mm0

169.0 10 N-mm1,014 N

166.661,014 N

67 10 mm

y

y

C

EI EI EI

C Ans.

Equilibrium equations for entire beam: (850 N)(600 mm) (600 N)(1,600 mm) (1,000 mm) (2,000 mm) 0A y yM C E

(850 N)(600 mm) (600 N)(1,600 mm) (1,014 N)(1,000 mm)

2,000 mm228 NyE Ans.

850 N 600 N 0

850 N 600 N 1,014 N 228 N 208 N

y y y y

y

F A C E

A Ans.






Section properties:

4 4(20 mm) 7,853.9816 mm64

I


Mmax = 132,000 N-mm


4

(132,000 N-mm)(20 mm/2)

7,853.9816 m

168.1 MPa

mx

Ans.




11.54 The solid 1.00-in.-diameter steel [E

= 29,000 ksi] shaft shown in Fig. P11.54

supports three belt pulleys. Assume that


support and that the bearings at C and E

can be idealized as roller supports. For the


(a) the reaction forces at bearings A, C,

and E.

(b) the magnitude of the maximum

bending stress in the shaft.

Fig. P11.54

Solution

(a) Reaction forces at A, C, and E. Choose the reaction force at C as the redundant; therefore, the

released beam is simply supported between A and E.

Consider downward deflection of simply supported beam at C due to pulley B load.




6C

Pbxv L b x

LEI

Values:

P = 200 lb, L = 60 in., b = 15 in., x = 30 in.

Calculation:

2 2 2

2 2 2

3

( )6

(200 lb)(15 in.)(30 in.)(60 in.) (15 in.) (30 in.)

6(60 in.)

618,750 lb-in.

C

Pbxv L b x

LEI

EI

EI





6C

Pbxv L b x

LEI

Values:

P = 280 lb, L = 60 in., b = 15 in., x = 30 in.

Calculation:

2 2 2

2 2 2

3

( )6

(280 lb)(15 in.)(30 in.)(60 in.) (15 in.) (30 in.)

6(60 in.)

866,250 lb-in.

C

Pbxv L b x

LEI

EI

EI




Consider upward deflection of simply supported beam at C due to pulley F load.




C

M xv L Lx x

LEI

Values:

M = −(120 lb)(10 in.) = −1,200 lb-in.,

L = 60 in., x = 30 in.

Calculation:

2 2

2 2

3

(2 3 )6

( 1,200 lb-in.)(30 in.)2(60 in.) 3(60 in.)(30 in.) (30 in.)

6(60 in.)

270,000 lb-in.

C

M xv L Lx x

LEI

EI

EI




3

48C

PLv

EI

Values:

P = −Cy, L = 60 in.

Calculation:

3 33 ( )(60 in.) (4,500 in. )

48 48

y y

C

C CPLv

EI EI EI


33 3 3

3

3

(4,500 in. )618,750 lb-in. 866,250 lb-in. 270,000 lb-in.0

1,215,000 lb-in.270 lb

4,500 in270 lb

.

y

y

C

EI EI EI EI

C Ans.

Equilibrium equations for entire beam: (200 lb)(15 in.) (280 lb)(45 in.) (120 lb)(70 in.) (30 in.) (60 in.) 0A y yM C E




(200 lb)(15 in.) (280 lb)(45 in.) (120 lb)(70 in.) (215 lb)(30 in.)

60 in.

265.0 lb 265 lb

yE

Ans.

200 lb 280 lb 120 lb 0

200 lb 280 lb 120 lb 270 lb 2 65.0 65 l lbb

y y y y

y

F A C E

A Ans.



Section properties:

4 4(1.00 in.) 0.0490874 in.64

I


Mmax = 1,200 lb-in.


4

(1,200 lb-in.)(1.00 in./2)

0.0490874 in.

12,223.1 psi

12,220 psi

x

Ans.





29,000 ksi] shaft shown in Fig. P11.55

supports two belt pulleys. Assume that the

bearing at E can be idealized as a pin

support and that the bearings at B and C can

be idealized as roller supports. For the


(a) the reaction forces at bearings B, C, and

E.


stress in the shaft.

Fig. P11.55

Solution

(a) Reaction forces at B, C, and E. Choose the reaction force at C as the redundant; therefore, the

released beam is simply supported between B and E.

Consider upward deflection of simply supported beam at C due to pulley A load.




6C

M xv L Lx x

LEI

Values:

M = −(90 lb)(7 in.) = −630 lb-in.,

L = 45 in., x = 15 in.

Calculation:

2 2

2 2

3

(2 3 )6

( 630 lb-in.)(15 in.)2(45 in.) 3(45 in.)(15 in.) (15 in.)

6(45 in.)

78,750 lb-in.

C

M xv L Lx x

LEI

EI

EI





6C

Pbxv L b x

LEI

Values:

P = 240 lb, L = 45 in., b = 15 in., x = 15 in.




Calculation:

2 2 2

2 2 2

3

( )6

(240 lb)(15 in.)(15 in.)(45 in.) (15 in.) (15 in.)

6(45 in.)

315,000 lb-in.

C

Pbxv L b x

LEI

EI

EI




2 2 2( )6

C

Pabv L a b

LEI

Values:

P = −Cy, L = 45 in., a = 15 in., b = 30 in.

Calculation:

2 2 2

3

2 2 2

( )6

( )(15 in.)(30 in.) (1,500 in. )(45 in.) (15 in.) (30 in.)

6(45 in.)

C

y y

Pabv L a b

LEI

C C

EI EI


33 3

3

3

(1,500 i

157.

n. )78,750 lb-in. 315,000 lb-in.0

236,250 lb-in.157.50 lb

1,500 inl

.5 b

y

y

C

EI EI EI

C Ans.

Equilibrium equations for entire beam: (240 lb)(15 in.) (90 lb)(52 in.) (45 in.) (30 in.) 0E y yM B C

(240 lb)(15 in.) (90 lb)(52 in.) (157.5 lb)(30 in.)

45 in.

79.0 lb 79.0 lb

yB

Ans.

90 lb 240 lb 0

90 lb 240 lb 157.5 lb 79.0 lb 93.5 lb 93.5 lb

y y y y

y

F B C E

E Ans.






Section properties:

4 4(1.00 in.) 0.0490874 in.64

I


Mmax = 1,402.5 lb-in.


4

(1,402.5 lb-in.)(1.00 in./2)

0.0490874 in.

14,285.74 psi

14,290 psi

x

Ans.




11.56 The beam shown in Fig. P11.56 consists

of a W360 × 101 structural steel wide-flange

shape [E = 200 GPa; I = 301 × 106 mm

4]. For




stress in the beam.

Fig. P11.56

Solution


Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam

span. [Appendix C, Cantilever beam with uniformly distributed load.]


2


B

wxv L Lx x

EI

Values:

w = 30 kN/m, L = 8 m, x = 5.5 m

Calculation:

22 2

22 2

3

(6 4 )24

(30 kN/m)(5.5 m)6(8 m) 4(8 m)(5.5 m) (5.5 m)

24

9,008.828125 kN-m

B

wxv L Lx x

EI

EI

EI

Consider deflection of cantilever beam at B due a linearly distributed load.

[Appendix C, Cantilever beam with linearly distributed load.]


3 2 2 30 (10 10 5 )120

B

w xv L L x Lx x

LEI (elastic curve)

Values:

w0 = 60 kN/m, L = 8 m, x = 5.5 m

Calculation:

23 2 2 30

23 2 2 3

3

(10 10 5 )120

(60 kN/m)(5.5 m)10(8 m) 10(8 m) (5.5 m) 5(8 m)(5.5 m) (5.5 m)

120(8 m)

4,998.103516 kN-m

B

w xv L L x Lx x

LEI

EI

EI




Consider deflection of cantilever beam at B due to concentrated load By.



3

3B

PLv

EI

Values:

P = −By, L = 5.5 m

Calculation:

3 33 ( )(5.5 m) (55.458333 m )

3 3

y y

B

B BPLv

EI EI EI


33 3

3

3

(55.458333

25

m )9,008.828125 kN-m 4,998.103516 kN-m0

14,006.93164 kN-m252.56676 kN

55.458333 N

m3 k

y

y

B

EI EI EI

B Ans.


1(30 kN/m)(8 m) (90 kN/m 30 kN/m)(8 m) 0

2

1(30 kN/m)(8 m) (90 kN/m 30 kN/m)(8 m) 252.56676 kN

227

2

227.43324 kN kN

y y y

y

F A B

A

Ans.

(90 kN/m 30 kN/m)(8 m) 8 m

(30 kN/m)(8 m)(4 m) (5.5 m) 02 3

A A yM M B

960 kN-m 640 kN-m (252.56676 kN)(5.5 m)

210.88281 211 kN-m (ccw) kN-m

AM

Ans.







6 4

3 3

301 10 mm

356 mm

1,690 10 mm

I

d

S


Mmax = 210.88281 kN-m


2

6 4

(210.88281 kN-m)(356 mm/2)(1,000)

301 10

124.7 MPa

mmx

or, using the tabulated value for the section

modulus:

2

3 3

(210.88281 kN-m)(1,000)

1,690 10 mm

124.8 MPa

x

Ans.




11.57 A W530 × 92 structural steel wide-

flange shape [E = 200 GPa; I = 554 × 106

mm4] is loaded and supported as shown in

Fig. P11.57. Determine:

(a) the force and moment reactions at

supports A and C.

(b) the maximum bending stress in the

beam.

(c) the deflection of the beam at B.

Fig. P11.57

Solution

(a) Reactions at A and C. Choose the moment reactions at A and C as the redundants. This will leave a

simply supported beam between A and C as the released beam.

Determine the slopes at A and C caused by the 150-kN concentrated load.


Relevant equations from Appendix C: 2 2 2 2( ) ( )

and6 6

A C

Pb L b Pa L a

LEI LEI

Values:

P = 150 kN, L = 10 m, a = 6 m, b = 4 m

Calculation:

2 2 22 2

2 2 22 2

( ) (150 kN)(4 m) 840 kN-m(10 m) (4 m)

6 6(10 m)

( ) (150 kN)(6 m) 960 kN-m(10 m) (6 m)

6 6(10 m)

A

C

Pb L b

LEI EI EI

Pa L a

LEI EI EI

Determine the slopes at A and C caused by moment reaction MA.



and3 6

A C

ML ML

EI EI

Values:

M = MA, L = 10 m

Calculation:

(10 m) (3.333333 m)

3 3

(10 m) (1.666667 m)

6 6

A AA

A AC

ML M M

EI EI EI

ML M M

EI EI EI




Determine the slopes at A and C caused by moment reaction MC.



and6 3

A C

ML ML

EI EI

Values:

M = MC, L = 10 m

Calculation:

(10 m) (1.666667 m)

6 6

(10 m) (3.333333 m)

3 3

C CA

C CC

ML M M

EI EI EI

ML M M

EI EI EI

Compatibility equation for slope at A:

2840 kN-m (3.333333 m) (1.666667 m)

0A CM M

EI EI EI (a)

Compatibility equation for slope at C:

2960 kN-m (1.666667 m) (3.333333 m)

0A CM M

EI EI EI (b)

Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as:

2

2

(3.333333 m) (1.666667 m) 840 kN-m

(1.666667 m) (3.333333 m) 960 kN-m

A C

A C

M M

M M

and solved simultaneously for MA and MC:

144 kN-m (ccw144 kN-m )AM Ans.

2216 kN-m 16 kN-m (cw)CM Ans.

Equilibrium equations for entire beam: (150 kN)(6 m) (10 m) 0A A C yM M M C

(150 kN)(6 m) ( 144 kN-m) ( 216 kN-m)

10 m

97 97.2 kN.2 kN

yC

Ans.

150 kN 0y y yF A C

150 kN 97.2 kN 52.8 k 52.8 kNNyA Ans.







6 4

3 3

554 10 mm

533 mm

2,080 10 mm

I

d

S


Mmax = 216 kN-m (at C)


2

6 4

(216 kN-m)(533 mm/2)(1,000)

554 10 m

103.

m

9 MPa

x

Ans.

or using the tabulated value for the section modulus:

2

3 3

(216 kN-m)(1,000)

2,080 10

103.8 M a

mm

P

x

Ans.

(c) Beam deflection at B:

Determine the deflection at B caused by the 150-kN concentrated load.



2 2 2( )

6B

Pabv L a b

LEI

Values:

P = 150 kN, L = 10 m, a = 6 m,

b = 4 m

Calculation:

3

2 2 2 2 2 2(150 kN)(6 m)(4 m) 2,880 kN-m( ) (10 m) (6 m) (4 m)

6 6(10 m)B

Pabv L a b

LEI EI EI

Determine the deflection at B caused by MA.



2 2(2 3 )

6B

M xv L Lx x

LEI

Values:

M = −144 kN-m, L = 10 m, x = 6 m




Calculation:

2 2

32 2

(2 3 )6

( 144 kN-m)(6 m) 806.4 kN-m2(10 m) 3(10 m)(6 m) (6 m)

6(10 m)

B

M xv L Lx x

LEI

EI EI

Determine the deflection at B caused by MC.



2 2(2 3 )6

B

M xv L Lx x

LEI

Values:

M = −216 kN-m, L = 10 m, x = 4 m

Calculation:

2 2

32 2

(2 3 )6

( 216 kN-m)(4 m) 1,382.4 kN-m2(10 m) 3(10 m)(4 m) (4 m)

6(10 m)

B

M xv L Lx x

LEI

EI EI

Beam deflection vB:

3 3 3

3 3

2

2,880 kN-m 806.4 kN-m 1,382.4 kN-m

691.2 kN-m 691.2 kN-m0.006238 m

110,806.

0 24 mm

kN-m

BvEI EI EI

EI

Ans.




11.58 A W530 × 92 structural steel wide-

flange shape [E = 200 GPa; I = 554 × 106

mm4] is loaded and supported as shown in

Fig. P11.58. Determine:

(a) the force and moment reactions at

supports A and C.

(b) the maximum bending stress in the beam.


Fig. P11.58

Solution

(a) Reactions at A and C. Choose the moment reactions at A and C as the redundants. This will leave a

simply supported beam between A and C as the released beam.

Determine the slopes at A and C caused by the 80 kN/m uniformly distributed load.



22

22 2

(2 ) and24

(2 )24

A

C

waL a

LEI

waL a

LEI

Values:

w = 80 kN/m, L = 9 m, a = 4.5 m

Calculation:

2 2 2

22

2 2 22 2 2 2

(80 kN/m)(4.5 m) 1,366.875 kN-m(2 ) 2(9 m) (4.5 m)

24 24(9 m)

(80 kN/m)(4.5 m) 1,063.125 kN-m(2 ) 2(9 m) (4.5 m)

24 24(9 m)

A

C

waL a

LEI EI EI

waL a

LEI EI EI

Determine the slopes at A and C caused by moment reaction MA.



and3 6

A C

ML ML

EI EI

Values:

M = MA, L = 9 m

Calculation:

(9 m) (3 m)

3 3

(9 m) (1.5 m)

6 6

A AA

A AC

ML M M

EI EI EI

ML M M

EI EI EI

Determine the slopes at A and C caused by moment reaction MC.






and6 3

A C

ML ML

EI EI

Values:

M = MC, L = 9 m

Calculation:

(9 m) (1.5 m)

6 6

(9 m) (3 m)

3 3

C CA

C CC

ML M M

EI EI EI

ML M M

EI EI EI

Compatibility equation for slope at A:

21,366.875 kN-m (3 m) (1.5 m)

0A CM M

EI EI EI (a)

Compatibility equation for slope at C:

21,063.125 kN-m (1.5 m) (3 m)

0A CM M

EI EI EI (b)

Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as:

2

2

(3 m) (1.5 m) 1,366.875 kN-m

(1.5 m) (3 m) 1,063.125 kN-m

A C

A C

M M

M M

and solved simultaneously for MA and MC:

371.25 kN-m (ccw371.25 kN-m )AM Ans.

168.75 kN- 168.8 kN-m (cw)mCM Ans.

Equilibrium equations for entire beam: (80 kN/m)(4.5 m)(2.25 m) (9 m) 0A A C yM M M C

(80 kN/m)(4.5 m)(2.25 m) ( 371.25 kN-m) ( 168.75 kN-m)

9 m

67.5 kN 67.5 kN

yC

Ans.

(80 kN/m)(4.5 m) 0y y yF A C

(80 kN/m)(4.5 m) 67.5 kN 292.5 kN 293 kNyA Ans.







6 4

3 3

554 10 mm

533 mm

2,080 10 mm

I

d

S


Mmax = 371.25 kN-m (at A)


2

6 4

(371.25 kN-m)(533 mm/2)(1,000)

554 10

178.6 MPa

mmx

Ans.

or using the tabulated value for the section modulus:

2

3 3

(371.25 kN-m)(1,000)

2,080 10 mm

178.5 MPa

x

Ans.

(c) Beam deflection at B:

Determine the deflection at B caused by the 80 kN/m uniformly distributed load.



3

2 2(4 7 3 )24

B

wav L aL a

LEI

Values:

w = 80 kN/m, L = 9 m, a = 4.5 m

Calculation:

32 2

3 32 2

(4 7 3 )24

(80 kN/m)(4.5 m) 3,417.1875 kN-m4(9 m) 7(4.5 m)(9 m) 3(4.5 m)

24(9 m)

B

wav L aL a

LEI

EI EI




Determine the deflection at B caused by MA.



2 2(2 3 )6

B

M xv L Lx x

LEI

Values:

M = −371.25 kN-m, L = 9 m,

x = 4.5 m

Calculation:

2 2

32 2

(2 3 )6

( 371.25 kN-m)(4.5 m) 1,879.4531 kN-m2(9 m) 3(9 m)(4.5 m) (4.5 m)

6(9 m)

B

M xv L Lx x

LEI

EI EI

Determine the deflection at B caused by MC.



2 2(2 3 )

6B

M xv L Lx x

LEI

Values:

M = −168.75 kN-m, L = 9 m,

x = 4.5 m

Calculation:

2 2

32 2

(2 3 )6

( 168.75 kN-m)(4.5 m) 854.2969 kN-m2(9 m) 3(9 m)(4.5 m) (4.5 m)

6(9 m)

B

M xv L Lx x

LEI

EI EI

Beam deflection vB:

3 3 3

3 3

2

3,417.1875 kN-m 1,879.4531 kN-m 854.2969 kN-m

683.4375 kN-m 683.4375 kN-m0.006168 m

110,800 k6.

N-17 mm

m

BvEI EI EI

EI

Ans.




11.59 A timber [E = 1,800 ksi] beam is

loaded and supported as shown in Fig.

P11.59. The cross section of the timber beam

is 4-in. wide and 8-in. deep. The beam is

supported at B by a ½-in.-diameter steel [E =

30,000 ksi] rod, which has no load before the

distributed load is applied to the beam. After

a distributed load of 900 lb/ft is applied to

the beam, determine:

(a) the force carried by the steel rod.


timber beam.


Fig. P11.59

Solution

Section properties:

34

beam beam

2 2

1 1

(4 in.)(8 in.)Beam: 170.6667 in. 1,800,000 psi

12

Rod (1): (0.50 in.) 0.1963495 in. 30,000,000 psi4

I E

A E

(a) Force carried by the steel rod.

The reaction force from rod (1) will be taken as the

redundant, leaving a simply supported beam

between A and C as the released beam.

For this analysis, a tension force is assumed to exist

in axial member (1).

Beam free-body diagram

Downward deflection of wood beam at B due to 900 lb/ft uniformly distributed load.



45

384B

wLv

EI

Values:

w = 900 lb/ft, L = 14 ft, EI = 307.2 ×106 lb-in.

2

Calculation:

4 4 3

6 2

5 5(900 lb/ft)(14 ft) (12 in./ft)2.532305 in.

384 384(307.2 10 lb-in. )B

wLv

EI




Upward deflection of wood beam at B due to force F1 in rod (1).



3

48B

PLv

EI

Values:

P = −F1, L = 14 ft, EI = 307.2 ×106 lb-in.

2

Calculation:

3 3 3

6116 2

( )(14 ft) (12 in./ft)(321.5625 10 in./lb)

48 48(307.2 10 lb-in. )B

PL Fv F

EI

Elongation of steel rod (1) due to force F1.

61 1

1 1 12 6

1 1

(16 ft)(12 in./ft)(32.59493 10 in./lb)

(0.1963495 in. )(30 10 psi)

F LF F

A E

The deflection of the wood beam at B will not equal zero in this instance because the rod that supports

the beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore,

6

1 1(32.59493 10 in./lb)Bv F


The sum of the downward deflection caused by the uniformly distributed load and the upward deflection

caused by the force in rod (1) must equal the elongation of the steel rod. Elongation of the steel rod will

produce a downward (i.e., negative) deflection of the wood beam at B.

6 6

1 1

1 6

2.532305 in. (321.5625 10 in./lb) (32.59493 10 in./lb)

2.532305 in.7,150.22 lb

354.1574 10 in.7,150 lb (T)

/lb

F F

F

Ans.




(b) Determine maximum bending stress in wood

beam:


Mmax = 4,125 lb-ft = 49,500 lb-in.

Bending stress at maximum moment

4

(49,500 lb-in.)(8 in./2)

170.6667 in1,160 psi

.x Ans.

(c) Beam deflection at B.

The beam deflection at B is equal to the elongation of the steel rod:

1 11 2 6

1 1

(7,150.22 lb)(16 ft)(12 in./ft)0.233 in.

(0.1963495 in. )(30 10 p0.233

s )i

in.B

F Lv

A E

Ans.




11.60 A W360 × 72 structural steel [E = 200

GPa] wide-flange shape is loaded and

supported as shown in Fig. P11.60. The beam

is supported at B by 20-mm-diameter solid

aluminum [E = 70 GPa] rod. After a

concentrated load of 40 kN is applied to the tip

of the cantilever, determine:

(a) the force produced in the aluminum rod.



Fig. P11.60

Solution

Section properties:

6 4

beam beam

3 3

beam

2 2

1 1

W360 × 72: 201 10 mm 351 mm 200,000 MPa

1,150 10 mm

Rod (1): (20 mm) 314.159266 mm 70,000 MPa4

I d E

S

A E

(a) Force in the aluminum rod.


redundant, leaving a cantilever beam as the released

beam.

For this analysis, a tension force is assumed to exist

in axial member (1).

Downward deflection of W360 × 72 beam at B due to 40-kN concentrated load.



2

(3 ) (elastic curve)6

B

Pxv L x

EI

Values:

P = 40 kN, L = 5 m, x = 3.6 m,

EI = 40,200 kN-m2

Calculation:

2 2

3

2

(40 kN)(3.6 m)(3 ) 3(5 m) (3.6 m) 24.50149 10 m

6 6(40,200 kN-m )B

Pxv L x

EI




Upward deflection of W360 × 72 beam at B due to force F1 in rod (1).



3

3B

PLv

EI

Values:

P = −F1, L = 3.6 m, EI = 40,200 kN-m2

Calculation:

3 3

6112

( )(3.6 m)(386.8657 10 m/kN)

3 3(40,200 kN-m )B

PL Fv F

EI

Elongation of aluminum rod (1) due to force F1.

61 1

1 1 12 2

1 1

(3 m)(136.4185 10 m/kN)

(314.159266 mm )(70,000 N/mm )(1 kN/1,000 N)

F LF F

A E

The deflection of the W360 × 72 beam at B will not equal zero in this instance because the rod that

supports the beam at B will elongate, thus permitting the W360 × 72 beam to deflect downward.

Therefore,

6

1 1(136.4185 10 m/kN)Bv F


The sum of the downward deflection caused by the 40-kN concentrated load and the upward deflection

caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the

aluminum rod will produce a downward (i.e., negative) deflection of the W360 × 72 beam at B.

3 6 6

1 1

3

1 6

24.50149 10 m (386.8657 10 m/kN) (136.4185 10 m/kN)

24.50149 10 m46.82254 46.8 kN (T) kN

523.2842 10 m/kN

F F

F

Ans.




(b) Determine maximum bending stress in the

W360 × 72 beam:


Mmax = 56 kN-m (at B)


2

6 4

(56 kN-m)(351 mm/2)(1,000)

201 10 m

48.9 MPa

mx

Ans.

or using the tabulated value for the section modulus

of the W360 × 72 beam:

2

3 3

(56 kN-m)(1,000)

1,150 10 mm48.7 MPax

Ans.


The beam deflection at B is equal to the elongation of the aluminum rod:

1 11 2 2

1 1

(46,822.54 N)(3,000 mm)6.38746 mm

(314.159266 mm )(70,000 N/6.39 mm

mm )B

F Lv

A E Ans.




11.61 A W18 × 55 structural steel [E = 29,000

ksi] wide-flange shape is loaded and supported

as shown in Fig. P11.61. The beam is supported

at C by a ¾-in.-diameter aluminum [E = 10,000

ksi] rod, which has no load before the

distributed load is applied to the beam. After a

distributed load of 4 kips/ft is applied to the

beam, determine:

(a) the force carried by the aluminum rod.

(b) the maximum bending stress in the steel

beam.

(c) the deflection of the beam at C. Fig. P11.61

Solution

Section properties:

4

beam beam

3

beam

2 2

1 1

Beam: 890 in. 18.1 in. 29,000 ksi

98.3 in.

Rod (1): (0.75 in.) 0.4417865 in. 10,000 ksi4

I d E

S

A E

(a) Force in the aluminum rod.

The reaction force from rod (1) will be taken as

the redundant, leaving a cantilever beam as the

released beam.

For this analysis, a tension force is assumed to

exist in axial member (1).

Downward deflection of W18 × 55 beam at C due to 4 kips/ft uniformly distributed load.



4 3

and8 6

B B

wL wLv

EI EI

Values:

w = 4 kips/ft, L = 11 ft, EI = 25.81×106 kip-in.

2

Calculation:

4 4 3

6 2

3 3 23

6 2

(4 kips/ft)(11 ft) (12 in./ft)0.490113 in.

8 8(25.81 10 kip-in. )

(4 kips/ft)(11 ft) (12 in./ft)4.950639 10 rad

6 6(25.81 10 kip-in. )

0.490113 in. (5 ft)(12 in./ft)( 4.9

B

B

C

wLv

EI

wL

EI

v

350639 10 rad) 0.787152 in.




Upward deflection of W18 × 55 beam at C due to force F1 in rod (1).



3

3C

PLv

EI

Values:

P = −F1, L = 16 ft, EI = 25.81×106 kip-in.

2

Calculation:

3 3 3

3116 2

( )(16 ft) (12 in./ft)(91.41015 10 in./kip)

3 3(25.81 10 kip-in. )C

PL Fv F

EI


31 1

1 1 12

1 1

(14 ft)(12 in./ft)(38.02742 10 in./kip)

(0.4417865 in. )(10,000 ksi)

F LF F

A E

The deflection of the W18 × 55 beam at C will not equal zero in this instance because the rod that

supports the beam at C elongates, thus permitting the W18 × 55 beam to deflect downward. Therefore,

3

1 1(38.02742 10 in./kip)Cv F




aluminum rod will produce a downward (i.e., negative) deflection of the W18 × 55 beam at C.

3 3

1 1

1 3

0.787152 in. (91.41015 10 in./kip) (38.02742 10 in./kip)

0.787152 in.6.081323 kips

129.4376 106.08 kips (T

in./ p)

ki

F F

F

Ans.

(b) Determine maximum bending stress in W18

× 55 beam:


Mmax = 144.70 kip-ft (at A)


4

(144.70 kip-ft)(18.1 in./2)(12 in./ft)

890

17.66 ksi

in.x

Ans.

or using the tabulated value for the section

modulus of the W18 × 55 beam:

3

(144.70 kip-ft)(12 in./ft)

98.3 in

17.66 ksi

.x

Ans.




(c) Beam deflection at C.

The beam deflection at C is equal to the elongation of the aluminum rod:

1 11 2

1 1

(6.081323 kips)(14 ft)(12 in./ft)0.231257 in.

(0.4417865 in. )(10,000 ksi)0.231 in.C

F Lv

A E Ans.




11.62 A W250 × 32.7 structural steel [E =

200 GPa] wide-flange shape is loaded and

supported as shown in Fig. P11.62. A

uniformly distributed load of 16 kN/m is

applied to the beam, causing the roller

support at B to settle downward (i.e.,

displace downward) by 15 mm. Determine:

(a) the reactions at supports A, B, and C.


beam.

Fig. P11.62

Solution

Section properties:

6 4 3 3250 32.7 : 49.1 10 mm 259 mm 380 10 mmW I d S

(a) Reactions at A, B, and C. Choose the reaction force at B as the redundant; therefore, the released

beam is simply supported between A and C.





B

wxv L Lx x

EI

Values:

w = 16 kN/m, L = 10 m, x = 4 m Calculation:

3 2 3

33 2 3

( 2 )24

(16 kN/m)(4 m) 1,984 kN-m(10 m) 2(10 m)(4 m) (4 m)

24

B

wxv L Lx x

EI

EI EI




2 2 2( )

6B

Pabv L a b

LEI

Values:

P = −By, L = 10 m, a = 4 m, b = 6 m

Calculation:

2 2 2

3

2 2 2

( )6

( )(4 m)(6 m) (19.2 m )(10 m) (4 m) (6 m)

6(10 m)

B

y y

Pabv L a b

LEI

B B

EI EI





2 6 4 12 2 3 2

33

3 2 3

3

(200,000 N/mm )(49.1 10 mm ) 9.82 10 N-mm 9.82 10 kN-m

(19.2 m )1,984 kN-m0.015 m

( 0.015 m)(9.82 10 kN-m ) 1,984 kN-m95.6615 kN

1995.7 kN

.2 m

y

y

EI

B

EI EI

B

Ans.

Equilibrium equations for entire beam: (4 m) (10 m) (16 kN/m)(10 m)(5 m) 0A y yM B C

(16 kN/m)(10 m)(5 m) (95.6615 kN)(4 m)

10 m

41.7354 kN 41.7 kN

yC

Ans.

(16 kN/m)(10 m) 0y y y yF A B C

(16 kN/m)(10 m) 95.6615 kN 41.7354 kN

22.6031 k 22.6 kNN

yA

Ans.


(b) Determine maximum bending stress in W250 ×

32.7 beam:


Mmax = 54.4327 kN-m


2

6 4

(54.4327 kN-m)(259 mm/2)(1,000)

49.1 10

143.6 MPa

mmx

Ans.


of the W250 × 32.7 beam:

2

3 3

(54.4327 kN-m)(1,000)

380 10

143.2 M a

mm

P

x

Ans.




11.63 A W10 × 22 structural steel [E =

29,000 ksi] wide-flange shape is loaded

and supported as shown in Fig. P11.63.

The beam is supported at C by a timber

[E = 1,800 ksi] post having a cross-

sectional area of 16 in.2. After a

concentrated load of 10 kips is applied to

the beam, determine:

(a) the reactions at supports A and C.


beam.

(c) the deflection of the beam at C. Fig. P11.63

Solution

Section properties:

4

beam beam

3

beam

1 1

W10 22: 118 in. 10.2 in. 29,000 ksi

23.2 in.

Post (1): 16 in. 1,800 ksi

I d E

S

A E

(a) Reactions at supports A and C.

The reaction force from post (1) will be taken as the

redundant, leaving a cantilever beam as the released

beam. To be consistent with earlier sign

conventions (e.g., Chapter 5), we will assume that

the force in the axial member is tension (even

though intuitively we recognize that the post must be

in compression).

Beam free-body diagram

Downward deflection of W10 × 22 beam at C due to 10-kip concentrated load.



3 2

and3 2

B B

PL PLv

EI EI

Values:

P = 10 kips, L = 14 ft, EI = 3,422,000 kip-in.2

Calculation:

3 3 3

2

2 2 2

2

(10 kips)(14 ft) (12 in./ft)4.618773 in.

3 3(3,422,000 kip-in. )

(10 kips)(14 ft) (12 in./ft)0.0412390 rad

2 2(3,422,000 kip-in. )

4.618773 in. (0.0412390 rad)(6 ft)(12 in./f

B

B

C

PLv

EI

PL

EI

v

t) 7.587984 in.

Since we are assuming that a tension force exists in post (1), the tension force from the post will

cause a downward deflection of W10 × 22 beam at C.






3

3B

PLv

EI

Values:

P = F1, L = 20 ft, EI = 3,422,000 kip-in.2

Calculation:

3 3 3

112

( )(20 ft) (12 in./ft)(1.346581 in./kip)

3 3(3,422,000 kip-in. )B

PL Fv F

EI

Axial deformation of post (1) due to force F1.

1 11 1 12

1 1

(12 ft)(12 in./ft)(0.005 in./kip)

(16 in. )(1,800 ksi)

F LF F

A E

The deflection of the steel beam will not equal zero at C in this instance because the post deforms. If we

are consistent and assume that there is tension in the post, then the steel beam must deflect upward at C.

Therefore,

1 1(0.005 in./kip)Cv F


The sum of the downward deflection caused by the 10-kip concentrated load and the upward deflection

caused by tension in post (1) must equal the elongation of the post.

1

1

7.587984 in. (1.346581 in./kip) (0.005 in./kip)

7.587984 in.5.614154 kips

1.351582 in./k5.61 kips (C)

ipy

F F

F C

Ans.

Equilibrium equations for entire beam: 1(20 ft) (10 kips)(14 ft) 0A AM M F

( 5.614154 kips)(20 ft) (10 kips)(14 ft) 27.7169 kip 27.7 kip-ft (-ft ccw)AM Ans.

1 (10 kips) 0y yF A F

10 kips ( 5.614154 kips) 4.3858 kips 4.39 kipsyA Ans.





beam:


Mmax = 33.685 kip-ft = 404.218 kip-in. (at B)


4

(404.218 kip-in.)(10.2 in./2)

118 in

17.47 ksi

.x

Ans.


modulus of the W10 × 22 beam:

3

404.218 kip-in.

23.2 in.17.42 ksix Ans.

(c) Beam deflection at C.

The beam deflection at C is equal to the deformation of the post:

1 11 2

1 1

( 5.614154 kips)(12 ft)(12 in./ft)0.0281 in.

(16 in. )(1,800 k0.0281

)i

in.

sC

F Lv

A E

Ans.




11.64 A timber [E = 12 GPa] beam is loaded

and supported as shown in Fig. P11.64. The

cross section of the timber beam is 100-mm.

wide and 300-mm deep. The beam is

supported at B by a 12-mm-diameter steel [E

= 200 GPa] rod, which has no load before the

distributed load is applied to the beam. After

a distributed load of 7 kN/m is applied to the

beam, determine:

(a) the force carried by the steel rod.

(b) the maximum bending stress in the timber

beam.


Fig. P11.64

Solution

Section properties:

36 4

beam beam

2 2

1 1

(100 mm)(300 mm)Beam: 225 10 mm 12,000 MPa

12

Rod (1): (12 mm) 113.097336 mm 200,000 MPa4

I E

A E

(a) Force carried by the steel rod.


redundant, leaving a simply supported beam between

A and C as the released beam.

For this analysis, a tension force is assumed to exist in

axial member (1).

Downward deflection of wood beam at B due to 7 kN/m uniformly distributed load.




B

wxv L Lx x

EI

Values:

w = 7 kN/m, L = 6 m, x = 4 m, EI = 2,700 kN-m2

Calculation:

3 2 3

3 2 3 3

2

( 2 )24

(7 kN/m)(4 m)(6 m) 2(6 m)(4 m) (4 m) 38.02469 10 m

24(2,700 kN-m )

B

wxv L Lx x

EI







2 2 2( )6

B

Pabv L a b

LEI

Values:

P = −F1, L = 6 m, a = 4 m, b = 2 m,

EI = 2,700 kN-m2

Calculation:

2 2 2

2 2 2 3112

( )6

( )(4 m)(2 m)(6 m) (4 m) (2 m) (1.316872 10 m/kN)

6(6 m)(2,700 kN-m )

B

Pabv L a b

LEI

FF

Elongation of steel rod (1) due to force F1.

61 1

1 1 12 2

1 1

(5 m)(221.0485 10 m/kN)

(113.097336 mm )(200,000 N/mm )(1 kN/1,000 N)

F LF F

A E



6

1 1(221.0485 10 m/kN)Bv F



caused by the force in rod (1) must equal the elongation of the steel rod. Elongation of the steel rod will

produce a downward (i.e., negative) deflection of the wood beam at B.

3 3 6

1 1

3

1 3

38.02469 10 m (1.316872 10 m/kN) (221.0485 10 m/kN)

38.02469 10 m24.72474 24.7 kN (T) kN

1.537921 10 m/kN

F F

F

Ans.





beam:


Mmax = 11.6270 kN-m


2

6 4

(11.6270 kN-m)(300 mm/2)(1,000)

225 10 mm

7.75 MPa

x

Ans.


The beam deflection at B is equal to the elongation of the steel rod:

1 1

1

1 1

2 2

(24.72474 kN)(5 m)(1,000 N/kN)(1,000 mm/m)

(113.097336 mm )(200,000 N/mm )

5.4653 5.47 mm67 mm

B

F Lv

A E

Ans.






supported as shown in Fig. P11.65. The

beam is supported at B by a timber [E = 12

GPa] post having a cross-sectional area of

20,000 mm2. After a uniformly distributed

load of 50 kN/m is applied to the beam,

determine:

(a) the reactions at supports A, B, and C.


beam.


Fig. P11.65

Solution

Section properties:

6 4

beam beam

3 3

beam

2

1 1

W360 × 72: 201 10 mm 351 mm 200,000 MPa

1,150 10 mm

Post (1): 20,000 mm 12,000 MPa

I d E

S

A E

(a) Reactions at supports A, B, and C.

The reaction force from post (1) will be taken as the

redundant, leaving a simply supported beam as the

released beam. To be consistent with earlier sign

conventions (e.g., Chapter 5), we will assume that the

force in the axial member is tension (even though

intuitively we recognize that the post must be in

compression).

Downward deflection of W360 × 72 beam at B due to 50 kN/m uniformly distributed load.



3

2 2(4 7 3 )24

B

wav L aL a

LEI

Values:

P = 50 kN/m, L = 13 m, a = 6 m,

EI = 40,200 kN-m2

Calculation:

32 2

32 2 3

2

(4 7 3 )24

(50 kN/m)(6 m)4(13 m) 7(6 m)(13 m) 3(6 m) 204.937 10 m

24(13 m)(40,200 kN-m )

B

wav L aL a

LEI

Since we are assuming that a tension force exists in post (1), the tension force from the post will

cause a downward deflection of W360 × 72 beam at B.






2 2 2( )6

B

Pabv L a b

LEI

Values:

P = F1, L = 13 m, a = 6 m, b = 7 m,

EI = 40,200 kN-m2

Calculation:

2 2 2

2 2 2 3112

( )6

( )(6 m)(7 m)(13 m) (6 m) (7 m) (1.12514 10 m/kN)

6(13 m)(40,200 kN-m )

B

Pabv L a b

LEI

FF

Deformation of wood post (1) due to force F1.

61 1

1 1 12 2

1 1

(5 m)(20.8333 10 m/kN)

(20,000 mm )(12,000 N/mm )(1 kN/1,000 N)

F LF F

A E

The deflection of the W360 × 72 beam at B will not equal zero in this instance because the post deforms.

If we are consistent and assume that there is tension in the post, then the steel beam must deflect upward

at B. Therefore,

6

1 1(20.8333 10 m/kN)Bv F


The sum of the downward deflection caused by the 50 kN/m uniformly distributed load and the

downward deflection caused by the force in post (1) must equal the deformation of the wood post.

3 3 6

1 1

3

1 3178.8 kN

204.937 10 m (1.12514 10 m/kN) (20.8333 10 m/kN)

204.937 10 m178.832 kN

1.145(C)

98 10 m/kNy

F F

F B

Ans.

Equilibrium equations for entire beam: 1(6 m) (13 m) (50 kN/m)(6 m)(3 m) 0A yM F C

(50 kN/m)(6 m)(3 m) ( 178.832 kN)(6 m)

13.3071 kN13 m

13.31 kNyC

Ans.

1 (50 kN/m)(6 m) 0y y yF A F C

(50 kN/m)(6 m) ( 178.832 kN) ( 13.3071 kN)

134.475 kN 134.5 kN

yA

Ans.





W360 × 72 beam:


Mmax = 180.836 kN-m


2

6 4

(180.836 kN-m)(351 mm/2)(1,000)

201 10

157.9 MPa

mmx

Ans.


of the W360 × 72 beam:

2

3 3

(180.836 kN-m)(1,000

157.2

)

1,150 10 mm

MPa

x

Ans.


The beam deflection at B is equal to the deformation of the wood post:

1 11 2 2

1 1

( 178.832 kN)(5 m)(1,000 N/kN)(1,000 mm/m)

(20,000 mm )(12,000 N/

3

mm )

3.7257 mm .73 mm

B

F Lv

A E

Ans.




11.66 A timber [E = 1,800 ksi] beam is loaded

and supported as shown in Fig. P11.66. The

cross section of the timber beam is 4-in. wide

and 8-in. deep. The beam is supported at B by

a ¾-in.-diameter aluminum [E = 10,000 ksi]

rod, which has no load before the distributed

load is applied to the beam. After a distributed

load of 800 lb/ft is applied to the beam,

determine:

(a) the force carried by the aluminum rod.

(b) the maximum bending stress in the timber

beam.


Fig. P11.66

Solution

Section properties:

34

beam beam

2 2

1 1

(4 in.)(8 in.)Beam: 170.6667 in. 1,800,000 psi

12

Rod (1): (0.75 in.) 0.441786 in. 10,000,000 psi4

I E

A E

(a) Force carried by the aluminum rod.


redundant, leaving a cantilever beam between A and C

as the released beam.

For this analysis, a tension force is assumed to exist in

axial member (1).

Downward deflection of wood beam at B due to 800 lb/ft uniformly distributed load.



2


B

wxv L Lx x

EI

Values:

w = 800 lb/ft, L = 16 ft, x = 12 ft,

EI = 307.2×106 lb-in.

2

Calculation:

22 2

2 32 2

6 2

(6 4 )24

(800 lb/ft)(12 ft) (12 in./ft)6(16 ft) 4(16 ft)(12 ft) (12 ft) 24.624 in.

24(307.2 10 lb-in. )

B

wxv L Lx x

EI





[Appendix C, SS beam with concentrated load at tip.]


3

3B

PLv

EI

Values:

P = −F1, L = 12 ft, EI = 307.2×106 lb-in.

2

Calculation:

3

3 331

16 2

3

( )(12 ft) (12 in./ft)(3.24 10 in./lb)

3(307.2 10 lb-in. )

B

PLv

EI

FF


61 1

1 1 12 6

1 1

(14 ft)(12 in./ft)(38.02746 10 in./lb)

(0.441786 in. )(10 10 psi)

F LF F

A E



6

1 1(38.02746 10 in./lb)Bv F




aluminum rod will produce a downward (i.e., negative) deflection of the wood beam at B.

3 6

1 1

1 3

24.624 in. (3.24 10 in./lb) (38.02746 10 in./lb)

24.624 in.7,511.835 lb

3.278027 10 in./l7,510 lb (

bT)

F F

F

Ans.





beam:


Mmax = 12,258 lb-ft


4

(12,258 lb-ft)(8 in./2)(12 in./ft)

170.6667 in.

3,447.56 psi 3,450 psi

x

Ans.


The beam deflection at B is equal to the elongation of the aluminum rod:

1 1

1

1 1

2 6

(7,511.835 lb)(14 ft)(12 in./ft)

(0.441786 in. )(10 10 psi)

0.28566 0.286 in. in.

B

F Lv

A E

Ans.






supported as shown in Fig. P11.67. A

uniformly distributed load of 70 kN/m is

applied to the beam, causing the roller support

at B to settle downward (i.e., displace

downward) by 10 mm. Determine:

(a) the reactions at supports A and B.


Fig. P11.67

Solution

Section properties:

6 4 3 3W530 66: 351 10 mm 526 mm 1,340 10 mmI d S

(a) Reactions at supports A and B.

The reaction force at B will be taken as the redundant, leaving a cantilever beam between A and C as the

released beam.

Downward deflection of beam at B due to 70 kN/m uniformly distributed load.



2


B

wxv L Lx x

EI

Values:

w = 70 kN/m, L = 6 m, x = 4.5 m,

EI = 70,200 kN-m2

Calculation:

22 2

22 2

2

(6 4 )24

(70 kN/m)(4.5 m)6(6 m) 4(6 m)(4.5 m) (4.5 m) 0.107903 m

24(70,200 kN-m )

B

wxv L Lx x

EI

Upward deflection of beam at B due to reaction force By.

[Appendix C, SS beam with concentrated load at tip.]


3

3B

PLv

EI

Values:

P = −By, L = 4.5 m, EI = 70,200 kN-m2

Calculation:

3

3

6

2

3

( )(4.5 m)(432.6923 10 m/kN)

3(70,200 kN-m )

B

y

y

PLv

EI

BB






caused by the reaction force By must equal the support settlement:

6

3

6

0.107903 m (432.6923 10 m/kN) 0.010 m

97.90264 10 m226.2639 kN

432.6923 10 m/k226 k

NN

y

y

B

B

Ans.

Equilibrium equations for entire beam: (4.5 m) (70 kN/m)(6 m)(3 m) 0A A yM M B

(226.2639 kN)(4.5 m) (70 kN/m)(6 m)(3 m)

241.8125 242 kN-m (cc-m )kN w

AM

Ans.

(70 kN/m)(6 m) 0y y yF A B

(70 kN/m)(6 m) 226.2639 kN 193.7361 kN 193.7 kNyA Ans.

(b) Determine maximum bending stress in

beam:


Mmax = 241.8125 kN-m (at A)


2

6 4

(241.8125 kN-m)(526 mm/2)(1,000)

351 10 mm

181.187 MPa 181.2 MPa

x

Ans.


modulus:

2

3 3

(241.8125 kN-m)(1,000)

1,340 10 mm

180.457 MPa 180.5 MPa

x

Ans.

philpot mom 2nd ch07-11 ism

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