chapter 9 solution

8/20/2019 Chapter 9 Solution

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0, 0] [ , 0] x y x L y= = = =

PROBLEM 9.12

For the beam and loading shown, (a) express the magnitude and location

of the maximum deflection in terms of w0, L, E , and I . (b) Calculate thevalue of the maximum deflection, assuming that beam AB is a W460 74× rolled shape and that 0 60 kN/m,w = 6 m, L = and 200 GPa. E =

SOLUTION

Using entire beam as a free body,

0

0

10: 0

2 3

16

B A

A

L M R L w L

R w L

Σ = − + =

=

Using AJ as a free body,

20

0

300

1 10: 0

6 2 3

1 1

6 6

J

w x x M w Lx M

L

w M w Lx x

L

Σ = − + + =

= −

230

02

2 400 1

3 500 1 2

1 1

6 6

1 112 24

1 1

36 120

d y w EI w Lx x

Ldx

dy w EI w Lx x C dx L

w EIy w Lx x C x C

L

= −

= − +

= − + +

2 2

34 4 0

0 0 1 1

[ 0, 0]: 0 0 0 0 0

1 1 7[ , 0]: 0 0

36 120 360

x y C C

w L x L y w L w L C L C

= = = − + + ∴ =

= = = − + + ∴ = −

Elastic curve:5

3 30

42 30

1 1 7

36 120 360

1 1 712 24 360

w x y Lx L x

EI L

dy w x Lx Ldx EI L

= − −

= − −

USE DISCONTINUITY FUNCTIONS UP TO PROBLEMS 9.64



PROBLEM 9.12 (Continued)

To find location of maximum deflection, set 0.dy

dx =

2 4 44 2 2 4 2

2 2 2

30 900 42015 30 7 0

30

81 0.2697

15

m m m

m

L L L x L x L x

x L L

− −− + = =

= − =

0.5193m x L=

53 30 1 1 (0.5193 ) 7

(0.5193 ) (0.5193 )36 120 360

m

w L y L L L L

EI L

= − −

4 40 00.00652 or 0.00652

w L w L

EI EI = −

Data: 30 60 kN/m 60 10 N/m 6 mw L= = × =

For 6 4 6 4W460 74, 333 10 mm 333 10 m I −× = × = ×

3 43

9 6

(0.00652)(60 10 )(6)7.61 10 m

(200 10 )(333 10 )m y

−

−

×= = ×

× × 7.61 mmm y =



PROBLEM 9.15

For the beam and loading shown, determine the deflection at

point C. Use 200 GPa. E =

SOLUTION

Reactions: 0 0 / , / A B R M L R M L= ↑ = ↓

0 : x a< <

0

0

0:

0

=

− + =

=

J M

M

x M L

M M x

L

[ 0, 0] [ , 0]

[ , ]

,

x y x L y

x a y y

dy dy x a

dx dx

= = = =

= =

= =

:a x L< <

00

0

0:

0

( )

=

− + + =

= −

K M

M x M M

L

M M x L

L

0 x a< <

20

2

201

1

2

d y M EI x

Ldx

dy M EI x C

dx L

=

= +

(1)

301 2

1

6

M EIy x C x C

L

= + +

(2)

a x L< <

20

2

203

( )

1

2

d y M EI x L

Ldx

dy M EI x Lx C

dx L

= −

= − +

(3)

3 203 4

1 1

6 2

M EIy x Lx C x C

L

= − + +

(4)

[ ] 2 20, 0 Eq. (2): 0 0 0 0 x y C C = = = + + =

2 20 01 3

3 1 0

1 1, Eqs. (1) and (3):

2 2

dy dy M M x a a C a La C

dx dx L L

C C M a

= = + = − +

= +




[ ] 30 1, Eqs. (2) and (4):6

M x a y y a L= = 1C a

+

30 1

6 M a

L= 2 11 (2

La C

− +

0 4

24 0

)

1

2

M a a C

C M a

+ +

= −

[ ] 3 3 201 0 0

2 201

1 1 1, 0 Eq. (4): ( ) 0

6 2 2

1 1

3 2

M x L y L L C M a L M a

L

M C L a aL

L

= = − + + − =

= + −

Elastic curve for 0 : x a< < 3 2 20 1 1 1

6 3 2

M y x L a aL x

EIL

= + + −

Make . x a= 3 2 3 2 3 2 20 01 1 1 2 1

6 3 2 3 3C

M M y a L a a a L a L a La

EIL EIL

= + + − = + −

Data: 9200 10 Pa, E = × 6 4 6 434.4 10 mm 34.4 10 m , I −= × = × 30 60 10 N m M = × ⋅

1.2 m, 4.8 ma L= =

3 3 2 2

3

9 6

(60 10 ) (2)(1.2) / 3 (4.8) (1.2) / 3 (4.8)(1.2)6.28 10 m

(200 10 )(34.4 10 )(4.8)C y −

−

× + − = = ×× ×

6.28 mmC y = ↑



PROBLEM 9.45

For the beam and loading shown, determine (a) the slope at

end A, (b) the deflection at point C . Use 200 GPa. E =

SOLUTION

Units: Forces in kN, lengths in m

0:

1.6 (9.6)(0.8) (20)(0.4) 0

9.8 kN

D

A

A

M

R

R

=

− + + =

=

0 0

0 0

( ) 12 0.4 12 1.2 kN/m

( ) 12 0.4 12 1.2 kN/m

w x x x

dV w x x x

dx

= − − −

= − = − − + −

1 1 0

22 2 1

2

2 3 3 2 2

1

3 4 4 3 31 2

9.8 12 0.4 12 1.2 20 1.2 kN

9.8 6 0.4 6 1.2 20 1.2 kN m

4.9 2 0.4 2 1.2 10 1.2 kN m

1 1 101.63333 0.4 1.2 1.2 kN m

2 2 3

dM V x x x

dx

d y EI M x x x x

dx

dy

EI x x x x C dx

EIy x x x x C x C

= = − − + − − −

= = − − + − − − ⋅

= −

−

+

−

−

−

+ ⋅

= − − + − − − + + ⋅

2 2[ 0, 0] : 0 0 0 0 0 0 0 x y C C = = − + − + + = =

3 4 4 31

1 1 10[ 1.6, 0] : (1.63333)(1.6) (1.2) (0.4) (0.4) (1.6) 0 0

2 2 3 x y C = = − + − + + =

21 3.4080 kN mC = − ⋅

Data: 9 6 4 6 4

4 6 6 2 2

200 10 Pa, 6.83 10 mm 6.83 10 mm

(200 10 )(6.83 10 ) 1.366 10 N m 1366 kN m

E I

EI

−

−

= × = × = ×

= × × = × ⋅ = ⋅

(a) Slope at A: at 0dy

xdx

=

20 0 0 0 3.4080 kN mdy

EI dx

= − + − − ⋅

33.40802.49 10 rad

1366 Aθ

−= − = − × 32.49 10 rad Aθ −= ×




(b) Deflection at C : ( at 1.2 m) y x =

3 4

3

1(1.63333)(1.2) (0.8) 0 0 (3.4080)(1.2) 0

2

1.4720 kN m

C EIy = − + − − +

= − ⋅

31.47201.078 10 m

1366C y −= − = − × 1.078 mmC y = ↓



SOLUTION

Units: Forces in kN; lengths in

0: 4.8 (

( B A

M R= − +

+

6 4

9

2

212 10 mm

(200 10 )(212

42400 kN m

I

EI

= ×

= ×

= ⋅

( ) 30 30 2.4w x x= − −

30 30dV

w

dx

= − = − +

79 30dM

V xdx

= = − +

2

79 15d y

EI M x xdx

= = −

2 3795 5

2

dy EI x x

dx= − +

3 479 5 5

6 4 4 EIy x x= − +

[ 0, 0] 0

7[ 4.8, 0]

x y

x y

= =

= =

PROBLEM 9.64

The rigid bar DEF is welded at point D to

For the loading shown, determine (a) thedeflection at midpoint C of the beam. Use

meters.

30)(2.4)(3.6)

0)(2.4) 0

79 kN A R

=

= ↑

6 4

6 6 2

212 10 m

10 ) 42.4 10 N m

−

−

×

× = × ⋅

0

02.4 x −

1 030 2.4 50 3.6 x x − − −

2 1 015 2.4 50 3.6 60 3.6 x x x+ − − − − −

3 2 112.4 25 3.6 60 3.6 x x x C − − − − − +

4 3 21

252.4 3.6 30 3.6

3 x x x C x − − − − − + +

( )

2 2

43 4

3 21

0 0 0 0 0 0 0

9 5 5(4.8) 4.8 (2.4)

4 4

25(1.2) (30)(1.2) 4.8 0

3

C C

C

+ − − + + = =

− +

− − + =

1 161.7C = −

the rolled-steel beam AB.

slope at point A, (b) the200 GPa. E =

kN/m

kN

kN m⋅

2kN m⋅

2 3kN m⋅

26 kN m⋅




(a) Slope at point A: at 0dy

xdx

=

2

3

0 0 0 0 0 161.76

161.76 kN m

161.763.82 10

42400

A

A

dy EI

dx

dy

dx

−

= − + − − −

= − ⋅

− = = − ×

33.82 10 rad. Aθ

−= ×

(b) Deflection at midpoint C : ( at 2.4) y x =

3 4

3

3

79 5

(2.4) (2.4) 0 0 0 (161.76)(2.4) 06 4

247.68 kN m

247.685.84 10 m

42400

C

C

EIy

y −

= − + − − − +

= − ⋅

−= = − × 5.84 mmC y = ↓



PROBLEM 9.77

For the beam and loading shown, determine (a) the slope at

end A, (b) the deflection at point C . Use 200 GPa. E =

SOLUTION

Units: Forces in kN; lengths in m.

Loading I: Moment at B.

Case 7 of Appendix D: 80 kN m, 5.0 m, 2.5 m M L x= ⋅ = =

3 2 3 2

(80)(5.0) 66.667

6 6

80 125( ) [2.5 (5.0) (2.5)]

6 6 (5.0)

A

C

ML

EI EI EI

M y x L x

EIL EI EI

θ = = =

= − − = − − =

Loading II: Moment at A: (Case 7 of Appendix D.)

80 kN m, 5.0 m, 2.5 m

(80)(5.0) 133.333

3 3 A

M L x

ML

EI EI EI θ

= ⋅ = =

= = =

125C y

EI = (Same as loading I.)

Loading III: 140 kN concentrated load at C : 140 kNP =

2 2

3 3

(140)(5.0) 218.75

16 16

(140)(5.0) 364.583

48 48

A

C

PL

EI EI EI

PL y

EI EI EI

θ = − = − = −

= − = − = −

Data: 9 6 4 6 4200 10 Pa, 156 10 mm 156 10 m E I −= × = × = ×

9 6 6 2 2(200 10 )(156 10 ) 31.2 10 N m 31200 kN m EI −= × × = × ⋅ = ⋅

(a) Slope at A: 367.667 133.333 218.750.601 10 rad

31200 Aθ

−+ −= = − ×

30.601 10 rad Aθ −= ×

(b) Deflection at C : 3125 125 364.5833.67 10 m

31200C y

−+ −= = − × 3.67 mmC y = ↓



PROBLEM 9.86

The two beams shown have the same cross section and

are joined by a hinge at C . For the loading shown,determine (a) the slope at point A, (b) the deflection at

point B. Use 200 GPa. E =

SOLUTION

Using free body ABC ,

3 3 4

6 12 2

0: 0.45 (0.3)(3.5) 0 2.33 kN

200 GPa

1 1

(30)(30) 67500 mm12 12

(200 10 )(67500 10 ) 13.5 kNm

A C C M R R

E

I bh

EI −

Σ = − = =

=

= = =

= × × =

Using cantilever beam CD with load ,C R

Case 1 of Appendix D:

3 3(2.33)(0.3)0.001555 mm

3 (3)(13.5)

C CDC

R L y

EI = − = − =

Calculation of Aθ ′ and B y′ assuming that point C does not move.

Case 5 of Appendix D:

2 2 2 2

2 2 2 2

3.5 kN, 0.45 m, 0.3 m, 0.15 m

( ) (3.5)(0.15)(0.45 0.15 )0.00259 rad

6 (6)(13.5)(0.45)

(3.5)(0.15) (0.3)0.000389 m

3 (3)(13.5)(0.45)

A

B

P L a b

Pb L b

EIL

Pb a y

EIL

θ

= = = =− −

′ = − = − = −

′ = − = − = −

Additional slope and deflection due to movement of point C .

0.0015550.003456 rad

0.45

(0.3)(0.001555)0.0010367 mm

0.45

C A

AC

B C

y

L

a y y

L

θ ′′ = = − = −

′′ = = − = −

(a) Slope at A: 0.00259 0.003456 A A Aθ θ θ ′ ′′= + = − −

0.006046 rad 0.00605 rad− =

(b) Deflection at B: 0.000389 0.0010367 B B B y y y′ ′′= + = − −

31.4257 10 m 1.43 mm−= − × =



SOLUTION

Using portion AB and applying

3

1

2

1

1 1

3

(13( )

3(1

( )2

( ) ( )

1.56 10

B

B

C B

PL y

EI PL

EI

y y L

θ

−

= =

= =

= +

= ×

Due to load P at point C : (C

( y

Total deflection at point C :

PROBLEM 9.87

Beam DE rests on the cantilever beam

that a square rod of side 10 mm is usedthe deflection at end C if the 25-N m⋅ c

E of beam DE , (b) to end C of beam AC .

9

3 4

2

200 10 Pa

1(10)(10) 833.33 mm

12

166.667 N m

E

I

EI

= ×

= = =

= ⋅

(a) Couple applied to beam DE .

Free body DE . 0: 0.180 25 0 M PΣ = − =

Loads on cantilever beam ABC are P at point

as shown.

Due to P at point B.

ase 1 of Appendix D,

33

23

3 31

8.889)(0.120)0.480 10 m

(3)(166.667)8.889)(0.120)

6.00 10(2)(166.667)

( ) 0.480 10 (0.180)(6.00 10 )

m

C Bθ

−

−

− −

= ×

= ×

= × + ×

se 1 of App. D applied to ABC .)

3 3

2

(138.889)(0.120 0.180)) 7.50 1

3 (3)(166.667)

PL

EI

+= − = − = − ×

31 2( ) ( ) 5.94 10 mC C C y y y −

= + = − ×

(b) Couple applied to beam AC :

Case 3 of Appendix D.

2 23(25)(0.300)

6.75 10 m2 (2)(166.667)

C

ML y

EI

−= − = − = − ×

AC as shown. Knowing

for each beam, determineuple is applied (a) to endUse 200 GPa. E =

12 433.33 10 m−×

138.889 NP =

B and P at point C

30 m−

5.94 mmC y =

6.75 mmC y =



PROBLEM 9.91

Before the load P was applied, a gap, 0 0.5mm,δ = existed

between the cantilever beam AC and the support at B.Knowing that 200 GPa, E = determine the magnitude of P for which the deflection at C is 1 mm.

SOLUTION

Let length 0.5 m AB L= =

length 0.2 m BC a= =

Consider portion AB of beam ABC .

The loading becomes forces P and B R at B plus the couple Pa.The deflection at B is 0.δ Using Cases 1 and 3 of Appendix D ,

3 2

0

3 2 3

0

( )

3 2

3 2 3

B

B

P R L PaL

EI EI

L L a LP R EI

δ

δ

−= +

+ − =

(1)

The deflection at C depends on the deformation of beam ABC

subjected to loads P and . B R For loading I, using Case 1 of

Appendix D,

3

1

( )( )

3C

P L a

EI δ +=

For loading II, using Case 1 of Appendix D,

3 2

3 2

B B B B

R L R L y

EI EI θ = =

Portion BC remains straight.

3 2

3 2

BC B B

L L a R y y a

EI θ

= + = +

By superposition, the downward deflection at C is3 3 2( )

3 3 2

BC

P L a L L a R

EI EI δ

+= − +

3 3 2( )

3 3 2 B C

L a L L aP R EI δ

+− + =

(2)




Data:9

200 10 Pa E = × 3 6 4 6 4

3 2

3 30

1(60)(60) 1.08 10 mm 1.08 10 m12

216 10 N m

0.5 10 m 1.0 10 mC

I

EI

δ δ

−

− −

= = × = ×

= × ⋅

= × = ×

Using the data, eqs (1) and (2) become

0.06667 0.04167 108 BP R− = (1)′

0.11433 0.06667 216 BP R− = (2)′

Solving simultaneously,

3

5.63 10 NP = ×

5.63 kNP = ↓

36.42 10 N B R = ×



PROBLEM 9.93

A 22-mm-diameter rod BC is attached to the lever AB and

to the fixed support at C . Lever AB has a uniform crosssection 10 mm thick and 25 mm deep. For the loadingshown, determine the deflection of point A. Use

200 GPa E = and 77 GPaG =

SOLUTION

Deformation of rod BC : (Torsion)

4 4

9 12

1 1(22) 11 mm

2 2

22998 mm2

(350)(0.25) 87.5 N m

0.5 m

(87.5)(0.5)

(77 10 )(22998 10 )

0.0247 rad

B

C d

J C

T Pa

L

TL

GJ

π

ϕ −

= = =

= =

= = = ⋅

=

= =× ×

=

Deflection of point A assuming lever AB to be rigid:

1( ) (0.25)(0.0247) B B y aϕ = =

0.006175 m=

Additional deflection due to bending of lever AB.

Refer to Case 1 of Appendix D.

3 4

3 3

9 12

1(10)(25) 13021 mm

12

(350)(0.25)( )

3 (3)(200 10 )(13021 10 ) A

I

PL y

EI −

= =

= =× ×

32.1 10 m−= ×

Total deflection at point A:

1 2( ) ( ) 8.28 mm A A A y y y= + =



PROBLEM 9.94

A 16-mm-diameter rod has been bent into the shape shown.

Determine the deflection of end C after the 200-N force is applied.Use 200 GPa E = and 80 GPa.G =

SOLUTION

Let 200 N .P=

Consider torsion of rod AB.

2

3

( ) B

C B

TL PL L PL

JG JG JG

PL y L

JG

φ

φ

= = =

′ = − = −

Consider bending of AB. (Case 1 of Appendix D.)

3

3C B

PL y y

EI ′′ = = −

Consider bending of BC . (Case 1 of Appendix D.)

3

3C

PL y EI

′′′ = −

Superposition:

3 3 3 3 2

3 3 3

C C C C y y y y

PL PL PL PL EI

JG EI EI EI JG

′ ′′ ′′′= + +

= − − − = − +

Data:

9 4 9 4

9 9 4

2 2

33

180(10 ) Pa (0.008) 6.4340(10 ) m

2

1200(10 ) Pa 3.2170(10 ) m2

643.40 N m 514.72 N m

(200)(0.25) 643.40 29.3093(10 ) m

643.40 514.72 3C

G J

E I J

EI JG

y

π −

−

−

= = =

= = =

= ⋅ = ⋅

= − + = −

9.31 mmC y = ↓

chapter 9 solution

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