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CCHHAAPPTTEERR 66

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,

reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited

distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it

without permission.

PROBLEM 6.1

Three boards, each of 1.5 3.5-in.× rectangular cross section, are nailed together to form a beam that is subjected to a vertical shear of 250 lb. Knowing that the spacing between each pair of nails is 2.5 in., determine the shearing force in each nail.

SOLUTION

3 3 4

2

1

31

1 1(3.5)(4.5) 26.578 in

12 12

(3.5)(1.5) 5.25 in

1.5 in.

7.875 in

(250)(7.875)74.074 lb/in

26.578

= = =

= ==

= =

= = =

I bh

A

y

Q Ay

VQq

I

nail nail(74.074)(2.5)

22 2

qsqs F F= = = nail 92.6 lbF =




without permission.

PROBLEM 6.2

Three boards, each 2 in. thick, are nailed together to form a beam that is subjected to a vertical shear. Knowing that the allowable shearing force in each nail is 150 lb, determine the allowable shear if the spacing s between the nails is 3 in.

SOLUTION

3 21

3 2 4

3 3 42

43 1

41 2 3

1

121

(6)(2) (6)(2)(3) 112 in121 1

(2)(4) 10.667 in12 12

112 in

234.667 in

I bh Ad

I bh

I I

I I I I

= +

= + =

= = =

= =

= + + =

31 1

nail

(6)(2)(3) 36 in

(1)

(2)

Q A y

qs F

VQq

I

= = ==

=

Dividing Eq. (2) by Eq. (1), nail

1 VQ

s F I=

nail (150)(234.667)

(36)(3)

F IV

Qs= = 326 lbV =




without permission.

PROBLEM 6.3

Three boards are nailed together to form a beam shown, which is subjected to a vertical shear. Knowing that the spacing between the nails is 75mms = and that the allowable shearing force in each nail is 400 N, determine the allowable shear when 120 mm.w =

SOLUTION

Part 2(mm )A (mm)d 2 6 4(10 mm )Ad 6 4(10 mm )I

Top plank 7200 60 25.92 2.16

Middle Plank 12,000 0 0 3.60

Bottom Plank 7200 60 25.92 2.16

Σ 51.84 7.92

2 6 4 6 4

3 3 6 3

nail

nail nail

59.76 10 mm 59.76 10 m

(7200)(60) 432 10 mm 432 10 m

I Ad I

Q

VQq F qs

IF Iq IF

q Vs Q Qs

−

−

= + = × = ×

= = × = ×

= =

= = =

6

6 3

(59.76 10 )(400)

(432 10 )(75 10 )V

−

− −×=

× × 738 NV =




without permission.

PROBLEM 6.4

Solve Prob. 6.3, assuming that the width of the top and bottom boards is changed to 100 mmw = .

PROBLEM 6.3 Three boards are nailed together to form a beam shown, which is subjected to a vertical shear. Knowing that the spacing between the nails is 75mms = and that the allowable shearing force in each nail is 400 N, determine the allowable shear when

120 mm.w =

SOLUTION

Part 2(mm )A (mm)d 2 6 4(10 mm )Ad 6 4(10 mm )I

Top plank 6000 60 21.6 1.80

Middle Plank 12,000 0 0 3.60

Bottom Plank 6000 60 21.6 1.80

Σ 43.2 7.20

2 6 4 6 4

3 3 6 3

nail

nail nail

50.4 10 mm 50.4 10 m

(6000)(60) 360 10 mm 360 10 m

I Ad I

Q

VQq F qs

IF Iq IF

q Vs Q Qs

−

−

= + = × = ×

= = × = ×

= =

= = =

6

6 3

(50.4 10 )(400)

(360 10 )(75 10 )V

−

− −×=

× × 747 NV =




without permission.

PROBLEM 6.5

The American Standard rolled-steel beam shown has been reinforced by attaching to it two 16 200-mm× plates, using 18-mm-diameter bolts spaced longitudinally every 120 mm. Knowing that the average allowable shearing stress in the bolts is 90 MPa, determine the largest permissible vertical shearing force.

SOLUTION

Calculate moment of inertia:

2 2 6 4 6 4

*

*

(mm)(mm ) (10 mm ) (10 mm )Part

Top plate 3200 82.43 0.07160.5

S310 52 6650 0 95.3

Bot. plate 3200 82.43 0.07160.5

164.86 95.44

dA Ad I

×

Σ

2 6 4 6 4

3 3 6 3plate plate

2 3 2 6 2bolt bolt

6 6 3bolt all bolt

* 305 16160.5 mm

2 2

260.3 10 mm 260.3 10 m

(3200)(160.5) 513.6 10 mm 513.6 10 m

(18 10 ) 254.47 10 m4 4

(90 10 )(254.47 10 ) 22.90 10 N

d

I Ad I

Q A d

A d

F A

π π

τ

−

−

− −

−

= + =

= Σ + Σ = × = ×

= = = × = ×

= = × = ×

= = × × = ×3

3boltbolt 3

6 33

6

2 (2)(22.90 10 )2 381.7 10 N/m

120 10

(260.3 10 )(381.7 10 )193.5 10 N

513.6 10

Fqs F q

s

VQ Iqq V

I Q

−

−

−

×= = = = ××

× ×= = = = ××

193.5 kNV =




without permission.

PROBLEM 6.6

Solve Prob. 6.5, assuming that the reinforcing plates are only 12 mm thick.

PROBLEM 6.5 The American Standard rolled-steel beam shown has been reinforced by attaching to it two 16 200-mm× plates, using 18-mm-diameter bolts spaced longitudinally every 120 mm. Knowing that the average allowable shearing stress in the bolts is 90 MPa, determine the largest permissible vertical shearing force.

SOLUTION


2 2 6 4 6 4

*

*

(mm ) (10 mm ) (10 mm )(mm)Part

Top plate 2400 60.29 0.03158.5

S310 52 6650 0 0 95.3

Bot. plate 2400 60.29 0.03158.5

120.58 95.36

A Ad Id

×

Σ

2 6 4 6 4

3 3 6 3plate plate

2 3 2 6 2bolt bolt

6 6bolt all bolt

* 305 12158.5 mm

2 2

215.94 10 mm 215.94 10 m

(200)(12)(158.5) 380.4 10 mm 380.4 10 m

(18 10 ) 254.47 10 m4 4

(90 10 )(254.47 10 ) 22.90

π π

τ

−

−

− −

−

= + =

= Σ + Σ = × = ×

= = = × = ×

= = × = ×

= = × × =

d

I Ad I

Q A d

A d

F A 3

33bolt

bolt 3

6 33

6

2 10 N

2 (2)(22.903 10 )2 381.7 10 N/m

120 10

(215.94 10 )(381.7 10 )217 10 N

380.4 10

−

−

−

×

×= = = = ××

× ×= = = = ××

Fqs F q

s

VQ Iqq V

I Q

217 kNV =




without permission.

PROBLEM 6.7

A columm is fabricated by connecting the rolled-steel members shown by bolts of 3

4-in. diameter spaced longitudinally every 5 in. Determine the

average shearing stress in the bolts caused by a shearing force of 30 kips parallel to the y axis.

SOLUTION

Geometry:

1

( )2

10.00.303 5.303 in.

20.534 in.

5.303 0.534 4.769 in.

w Cs

df t

x

y f x

= +

= + =

== − = − =

Determine moment of inertia.

2 2 4 4(in.)(in ) (in ) (in )Part

8 13.7 4.04 4.769 91.88 1.52

10 25.4 7.4 0 0 123

8 13.7 4.04 4.769 91.88 1.52

183.76 126.04

A Ad Id

C

S

C

×××Σ

2 4

31

bolt

22 2

bolt bolt

183.76 126.04 309.8 in

(4.04)(4.769) 19.267 in

(30)(19.267)1.8658 kip/in

309.8

1 1(1.8658)(5) 4.664 kips

2 2

30.4418 in

4 4 4

I Ad I

Q A y

VQq

I

F qs

A dπ π

= Σ + Σ = + =

= = =

= = =

= = =

= = =

boltbolt

bolt

4.66410.56 ksi

0.4418

F

Aτ = = = bolt 10.56 ksiτ =




without permission.

PROBLEM 6.8

The composite beam shown is fabricated by connecting two W6 20× rolled-steel members, using bolts of 5

8-in. diameter spaced longitudinally every 6 in. Knowing

that the average allowable shearing stress in the bolts is 10.5 ksi, determine the largest allowable vertical shear in the beam.

SOLUTION

2 4W6 20: 5.87 in , 6.20 in., 41.4 in

13.1 in.

2

xA d I

y d

× = = =

= =

Composite: 2

4

3

2[41.4 (5.87)(3.1) ]

195.621 in

(5.87)(3.1) 18.197 in

I

Q Ay

= +

=

= = =

Bolts: all

22

bolt

bolt all bolt

bolt

5 in., 10.5 ksi, 6 in.

8

50.30680 in

4 8

(10.5)(0.30680) 3.2214 kips

2 (2)(3.2214)1.07380 kips/in

6

d s

A

F A

Fq

s

τ

π

τ

= = =

= =

= = =

= = =

Shear: (195.621)(1.0780)

18.197= = =VQ Iq

q VI Q

11.54 kipsV =




without permission.

PROBLEM 6.9

For the beam and loading shown, consider section n-n and determine (a) the largest shearing stress in that section, (b) the shearing stress at point a.

SOLUTION

By symmetry, .A BR R=

0:

15 20 15 0

25 k ips

=

+ − − − == =

y

A B

A B

F

R R

R R

From shear diagram, 30 kipsV = at n-n.

Determine moment of inertia.

(a) 2 4286.74 inI Ad I= + =

Part 2(in )A (in.)y 3(in )Ay

6 4.7 28.2

1.65 2.2 3.63

Σ 31.83

3

maxmax

31.83 in

0.375 in.

(25)(31.83)

(286.74)(0.375)

Q Ay

t

VQ

Itτ

=

==

= =

max 7.40 ksiτ =

Part 2(in )A (in.)d 2 4(in )Ad 4 (in )I

Top Flng 6 4.7 132.54 0.18

Web 3.30 0 0 21.30

Bot. Flng 6 4.7 132.54 0.18

Σ 265.08 21.66




without permission.

PROBLEM 6.9 (Continued)

(b)

Part 2(in )A (in.)y 3(in )Ay

6 4.7 28.2

0.15 4.2 0.63

Σ 28.83

3= 28.83 in

0.375 in.

Q Ay

t

==

(23)(28.83)

(286.74)(0.375)

VQ

Itτ = = 6.70 ksiτ =




without permission.

PROBLEM 6.10


SOLUTION

At section n-n, 10 kN.V =

1 2

3 3 21 1 2 2 2 2

3 3 2

6 6 6

6 4 6 4

4

1 14

12 12

1 1(100)(150) 4 (50)(12) (50)(12)(69)

12 12

28.125 10 4 0.0072 10 2.8566 10

39.58 10 mm 39.58 10 m

I I I

b h b h A d

−

= +

= + +

= + +

= × + × + ×

= × = ×

(a) 1 1 2 2

3 3 6 3

2

(100)(75)(37.5) (2)(50)(12)(69)

364.05 10 mm 364.05 10 m

100 mm = 0.100 m

Q A y A y

t

−

= += +

= × = ×=

3 6

3max 6

(10 10 )(364.05 10 )920 10 Pa

(39.58 10 )(0.100)

VQ

Itτ

−

−× ×= = = ×

× max 920 kPaτ =

(b) 1 1 2 2

3 3 6 3

2

(100)(40)(55) (2)(50)(12)(69)

302.8 10 mm 302.8 10 m

100 mm 0.100 m

Q A y A y

t

−

= += +

= × = ×= =

3 6

36

(10 10 )(302.8 10 )765 10 Pa

(39.58 10 )(0.100)a

VQ

Itτ

−

−× ×= = = ×

× 765 kPaaτ =




without permission.

PROBLEM 6.11


SOLUTION

At section n-n, 80 kNV =

Consider cross section as composed of rectangles of types , , and .

3 2 6 41

3 2 6 42

3 3 43

6 41 2 3

6 4

1(12)(80) (12)(80)(90) 8.288 10 mm

121

(180)(16) (180)(16)(42) 5.14176 10 mm121

(16)(68) 419.24 10 mm12

4 2 2 44.274 10 mm

44.274 10 m−

= + = ×

= + = ×

= = ×

= + + = ×

= ×

I

I

I

I I I I

(a) Calculate Q at neutral axis.

3 41

3 42

3 43

(12)(80)(90) 86.4 10 mm

(180)(16)(42) 120.96 10 mm

(16)(34)(17) 9.248 10 mm

Q

Q

Q

= = ×

= = ×

= = ×

3 3 6 31 2 3

3 66

6 3

2 2 312.256 10 mm 312.256 10 m

(80 10 )(312.256 10 )17.63 10 Pa

(44.274 10 )(2 16 10 )

Q Q Q Q

VQ

Itτ

−

−

− −

= + + = × = ×

× ×= = = ×× × ×

17.63 MPaτ =

(b) At point a, 3 4 6 41 86.4 10 mm 86.4 10 mQ Q −= = × = ×

3 6

66 3

(80 10 )(86.4 10 )13.01 10 Pa

(44.274 10 )(12 10 )

VQ

Itτ

−

− −× ×= = = ×

× × 13.01 MPaτ =




without permission.

PROBLEM 6.12


SOLUTION

By symmetry, .A BR R=

0: 10 10 0

10 kips

y A B

A B

F R R

R R

Σ = + − − =

= =

From the shear diagram, 10 kips at - .V n n=

3 32 2 1 1

3 3 4

1 1

12 121 1

(4)(4) (3)(3) 14.583 in12 12

I b h b h= −

= − =

(a) 31 1 2 2

1 1(3) (1.75) (2) (2)(1) 4.625 in

2 2Q A y A y

= + = + =

1 1

1 in.2 2

t = + =

max(10)(4.625)

(14.583)(1)

VQ

Itτ = = max 3.17 ksiτ =

(b) 31(4) (1.75) 3.5 in

2

1 11 in.

2 2

Q Ay

t

= = =

= + =

(10)(3.5)

(14.583)(1)

VQ

Itτ = = 2.40 ksiaτ =




without permission.

PROBLEM 6.13

For a beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 60 MPa.

SOLUTION

Calculate moment of inertia.

3 4 2

6 4 6 4 6 4

6 4

1 1(50mm)(120mm) 2 (30mm) (30mm 30mm)(35mm)

12 12

7.2 10 mm 2[1.170 10 mm ] 4.86 10 mm

4.86 10 m

I

I−

= − + ×

= × − × = ×

= ×

Assume that mτ occurs at point a.

3 3 6 3

2(10mm) 0.02 m

(10mm 50mm)(55mm) 2[(10mm 30mm)(35mm)]

48.5 10 mm 48.5 10 m

t

Q−

= == × + ×

= × = ×




without permission.


For all 60 MPa,τ = allmVQ

Itτ τ= =

6 3

66 4

(48.5 10 m )60 10 Pa

(4.86 10 m )(0.02m)

V −

−×× =

× 120.3 kNV =

Check τ at neutral axis: 50mm 0.05mt = =

3 3

6 3

6 4

(50 60)(30) (30 30)(35) 58.5 10 mm

(120.3 kN)(58.5 10 m )29.0 MPa 60 MPa OK

(4.86 10 m )(0.05m)

Q

VQ

Itτ

−

−

= × − × = ×

×= = = <×




without permission.

PROBLEM 6.14

For a beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 60 MPa.

SOLUTION

Calculate moment of inertia.

3 3

6 4 6 4

6 4

1 12 (10 mm)(120 mm) (30 mm)(40 mm)

12 12

2[1.440 10 mm ] 0.160 10 mm

3.04 10 mm

I = +

= × + ×

= ×

6 43.04 10 mI −= ×

Assume that mτ occurs at point a.

3 3 6 3

10mm 0.01m

(10mm 40mm)(40mm)

16 10 mm 16 10 m

t

Q

Q −

= == ×

= × = ×

For all 60 MPa,τ = allmVQ

Itτ τ= =

6 3

66 4

(16 10 m )60 10 Pa

(3.04 10 m )(0.01m)

V −

−×× =

× 114.0 kNV =

Check τ at neutral axis:

3 3 6 3

50mm 0.05m

2[(10 60)(30)] (30 20)(10) 42 10 m 42 10 m

t

Q −

= =

= × + × = × = ×

6 3

6 4

(114.0 kN)(42 10 m )31.5 MPa 60 MPa OK

(3.04 10 m )(0.05 m)NA

VQ

Itτ

−

−×= = = <

×




without permission.

PROBLEM 6.15

For the beam and loading shown, determine the minimum required depth h, knowing that for the grade of timber used, all 1750 psiσ = and all 130 psi.τ =

SOLUTION

Total load: 3(750 lb/ft)(16 ft) 12 10 lb= ×

Reaction at A: 36 10 lbA = × ↑R

3max

3 3max

3

6 10 lb

1(8 ft)(6 10 ) 24 10 lb ft

2

288 10 lb in

V

M

= ×

= × = × ⋅

= × ⋅

Bending: 21

6S bh= for rectangular section.

33max

all

288 10164.57 in

1750

6 (6)(164.57)14.05 in.

5

MS

Sh

b

σ×= = =

= = =

Shear: 31

12I bh= for rectangular section.

2

maxmax

1

21

4

1 1 1( )

2 4 8

3

2

A bh

y h

Q Ay b h h bh

VQ V

Ib bhτ

=

=

= = =

= =

3

max

max

3 (3)(6 10 )13.85 in.

2 (2)(5)(130)τ×= = =V

hb

The larger value of h is the minimum required depth. 14.05 in.h =




without permission.

PROBLEM 6.16

For the beam and loading shown, determine the minimum required width b, knowing that for the grade of timber used, all 12 MPaσ = and all 825 kPa.τ =

SOLUTION

0: 3 (2)(2.4) (1)(4.8) 0

3.2 kND A

A

M R

R

= − + + ==

Draw shear and bending moment diagrams.

max max

4.0 kN 4.0 kN mV M= = ⋅

Bending: 3

max6

all

6 3 3 3

4.0 10

12 10

333.33 10 m 333.33 10 mm

MS

σ−

×= =×

= × = ×

For a rectangular cross section,

3

2

3

2 2

2 3

3

3

3 2 3 2

3

1112

1 62

6 (6)(333.33 10 )88.9 mm

1501 1

, 2 4

1 1,

8 123

2

3 3 4.0 10

2 2 825 10

7.2727 10 m 7.2727 10 mm

7.2727 1048.5 mm

150

bhIS bh

c h

Sb

h

A bh y h

Q Ay bh I bh

VQ V

It bh

Vbh

bhb

h

τ

τ−

= = =

×= = =

= =

= = =

= =

×= =×

= × = ×

×= = =

The required value for b is the larger one. 88.9 mmb =




without permission.

PROBLEM 6.17

A timber beam AB of length L and rectangular cross section carries a uniformly distributed load w and is supported as shown. (a) Show that the ratio /m mτ σ of the maximum values of the shearing and normal stresses in the beam is equal to 2 / ,h L where h and L are, respectively, the depth and the length of the beam. (b) Determine the depth h and the width b of the beam, knowing that 5 m,L = 8w = kN/m,

1.08 MPa,mτ = and 12 MPa.mσ =

SOLUTION

2A B

wLR R= =

From shear diagram, | |4m

wLV = (1)

For rectangular section, A bh= (2)

3 3

2 8m

mV wL

A bhτ = = (3)

From bending moment diagram, 2

| |32m

wLM = (4)

For a rectangular cross section,

21

6S bh= (5)

2

2

| | 3

16m

mM wL

S bhσ = = (6)

(a) Dividing Eq. (3) by Eq. (6), 2m

m

h

L

τσ

=

(b) Solving for h:

6

36

(5)(1.08 10 )225 10 m

2 (2)(12 10 )m

m

Lh

τσ

−×= = = ××

225 mmh =

Solving Eq. (3) for b:

3

3 6

3 (3)(8 10 )(5)

8 (8)(225 10 )(1.08 10 )m

wLb

hτ −×= =

× ×

361.7 10 m−= × 61.7 mmb =




without permission.

PROBLEM 6.18

A timber beam AB of length L and rectangular cross section carries a single concentrated load P at its midpoint C. (a) Show that the ratio

/m mτ σ of the maximum values of the shearing and normal stresses in the beam is equal to 2 / ,h L where h and L are, respectively, the depth and the length of the beam. (b) Determine the depth h and the width b of the beam, knowing that 2 m,L = 40 kN,P = 960 kPa,mτ = and

12 MPa.mσ =

SOLUTION

Reactions: /2A BR R P= = ↑

(1) max 2AP

V R= =

(2) A bh= for rectangular section.

(3) max3 3

2 4mV P

A bhτ = = for rectangular section.

(4) max 4

PLM =

(5) 21

6S bh= for rectangular section.

(6) max2

3

2m

M PL

S bhσ = =

(a) 2

m

m

h

L

τσ

=

(b) Solving for h, 3

36

2 (2)(2)(960 10 )320 10 m

12 10m

m

Lh

τσ

−×= = = ××

320 mmh =

Solving Eq. (3) for b,

3

33 3

3 (3)(40 10 )97.7 10 m

4 (4)(320 10 )(960 10 )m

Pb

hτ−

−×= = = ×

× × 97.7 mmb =




without permission.

PROBLEM 6.19

For the wide-flange beam with the loading shown, determine the largest load P that can be applied, knowing that the maximum normal stress is 24 ksi and the largest shearing stress using the approximation web/m V Aτ = is 14.5 ksi.

SOLUTION

0: 15 0

0.6

= − + ==

C A

A

M R qP

R P


max max0.6 0.6

6ft 72 in.

AB

AB

V P M PL

L

= =

= =

Bending. For W 24 104× , 3258 inS =

max

all all

all

0.6

(24)(258)143.3 kips

0.6 (0.6)(72)

AB

AB

M PLS

SP

L

σ σσ

= =

= = =

Shear. web

2

(24.1)(0.500)

12.05 in

wA dt==

=

max

web web

web

0.6

(14.5)(12.05)291 kips

0.6 0.6

V P

A A

AP

τ

τ

= =

= = =

The smaller value of P is the allowable value. 143.3 kipsP =




without permission.

PROBLEM 6.20

For the wide-flange beam with the loading shown, determine the largest load P that can be applied, knowing that the maximum normal stress is 160 MPa and the largest shearing stress using the approximation web/m V Aτ = is 100 MPa.

SOLUTION

0: 3.6 3.0 2.4 1.8 0

2

= − + + + =

= ↑ E A

A

M R P P P

R P


max max

2 , 3

2 3

B AB C D AB

AB

M PL M M PL

V P M PL

= = =

= =

Bending. For W 360 122,× 3 3

6 3

= 2020 10 mm

= 2020 10 m

S−

×

×

max

all all

6 33all

3

(160 10 )(2020 10 )179.6 10 N

3 (3)(0.6)

AB

AB

M PLS

SP

L

σ σ

σ

= =

× ×= = = ×

Shear. web

3 2 3 2

(363)(13.0)

4.719 10 mm 4.719 10 m

wA dt−

= =

= × = ×

max

web web

6 33web

2

(100 10 )(4.719 10 )236 10 N

2 2

V P

A A

AP

τ

τ −

= =

× ×= = = ×

The smaller value of P is the allowable one. 179.6 kNP =




without permission.

PROBLEM 6.21

For the beam and loading shown, consider section n-n and determine the shearing stress at (a) point a, (b) point b.

SOLUTION

Draw the shear diagram. max| | 90 kNV =

Part 2(mm )A (mm)y 3 3(10 mm )Ay d(mm) 2 6 4(10 mm )Ad 6 4(10 mm )I

3200 90 288 25 2.000 0.1067

1600 40 64 −25 1.000 0.8533

1600 40 64 −25 1.000 0.8533

Σ 6400 416 4.000 1.8133

3

2 6 4

6 4 6 4

416 1065 mm

6400

(4.000 1.8133) 10 mm

5.8133 10 mm 5.8133 10 m

AyY

A

I Ad I−

Σ ×= = =Σ

= Σ + Σ = + ×

= × = ×

(a) 2(80)(20) 1600 mmA = =

3 3 6 3

25 mm

40 10 mm 40 10 ma

y

Q Ay −

=

= = × = ×

3 6

66 3

(90 10 )(40 10 )31.0 10 Pa

(5.8133 10 )(20 10 )a

aVQ

Itτ

−

− −× ×= = = ×

× ×

31.0 MPaaτ =

(b) 2

3 3 6 3

(30)(20) 600 mm 65 15 50 mm

30 10 mm 30 10 mb

A y

Q Ay −

= = = − =

= = × = ×

3 6

66 3

(90 10 )(30 10 )23.2 10 Pa

(5.8133 10 )(20 10 )b

bVQ

Itτ

−

− −× ×= = = ×

× ×

23.2 MPabτ =




without permission.

PROBLEM 6.22


SOLUTION

12 kipsA BR R= =

Draw shear diagram.

12 kipsV =

Determine section properties.

Part 2(in )A (in.)y 3(in )Ay d(in.) 2 4(in )Ad 4(in )I

4 4 16 2 16 5.333

8 1 8 −1 8 2.667

Σ 12 24 24 8.000

2 4

242 in.

12

32 in

AyY

A

I Ad I

Σ= = =Σ

= Σ + Σ =

(a) 2 31 in 3.5 in. 3.5 in

1 in.aA y Q Ay

t

= = = ==

(12)(3.5)

(32)(1)a

aVQ

Itτ = = 1.3125 ksiaτ =

(b) 2 32 in 3 in. 6 in

1 in.bA y Q Ay

t

= = = ==

(12)(6)

(32)(1)b

bVQ

Itτ = = 2.25 ksibτ =




without permission.

PROBLEM 6.23

For the beam and loading shown, determine the largest shearing stress in section n-n.

SOLUTION

Draw the shear diagram. max| | 90 kNV =

Part 2(mm )A x′ 3 3(10 mm )Ay d(mm) 2 6 4(10 mm )Ad 6 4(10 mm )I

3200 90 288 25 2.000 0.1067

1600 40 64 −25 1.000 0.8533

1600 40 64 −25 1.000 0.8533

Σ 6400 416 4.000 1.8133

3

2 6 4

6 4 6 4

416 1065 mm

6400

(4.000 1.8133) 10 mm

5.8133 10 mm 5.8133 10 m

AyY

A

I Ad I−

Σ ×= = =Σ

= Σ + Σ = + ×

= × = ×

Part 2(mm )A (mm)y 3 3(10 mm )Ay

3200 25 80

300 7.5 2.25

300 7.5 2.25

Σ 84.5

3 3 6 3

3

3 6

max 6 3

6

84.5 10 mm 84.5 10 m

(2)(20) 40 mm 40 10 m

(90 10 )(84.5 10 )

(5.8133 10 )(40 10 )

32.7 10 Pa

Q Ay

t

VQ

Itτ

−

−

−

− −

= Σ = × = ×

= = = ×

× ×= =× ×

= × 32.7 MPamτ =




without permission.

PROBLEM 6.24


SOLUTION

12 kipsA BR R= =

Draw shear diagram.

12 kipsV =

Determine section properties.


4 4 16 2 16 5.333

8 1 8 −1 8 2.667

Σ 12 24 24 8.000

2 4

242 in.

12

32 in

AyY

A

I Ad I

Σ= = =Σ

= Σ + Σ =

31 1

max

(4)(2) 8 in

1 in.

(12)(8)

(32)(1)τ

= = ==

= =

Q A y

t

VQ

It max 3.00 ksiτ =




without permission.

PROBLEM 6.25

A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the horizontal line along which the shearing stress is maximum, (b) the constant k in the following expression for the maximum shearing stress

maxV

kA

τ =

where A is the cross-sectional area of the beam.

SOLUTION

4 2and4

I c A cπ π= =

For semicircle, 2 4

2 3sc

A c yπ

π= =

2 34 2

2 3 3sc

Q A y c cπ

π= = ⋅ =

(a) maxτ occurs at center where 2 .t c=

(b) 32

3max 4 2

4

4 4

32 3

V cVQ V V

It Ac c cπτπ

⋅= = = =

⋅

41.333

3k = =




without permission.

PROBLEM 6.26


maxV

kA

τ =


SOLUTION

31 1

2 36A bh I bh= =

For a cut at location y,

2

2

1( )

2 2

2 2( )

3 3

( ) ( )3

( )

by byA y y

h h

y y h y

byQ y Ay h y

byt y

h

= =

= −

= = −

=

2

233 331

36

( ) 12 ( ) 12( ) ( )

( )

by

byh

V h yVQ Vy h y Vy hy y

It bh bhbhτ

− −= = = = −

(a) To find location of maximum of ,τ set 0.d

dy

τ =

3

12( 2 ) 0m

d Vh y

dy bh

τ = − = 1

,2my h= i.e., at mid-height

(b) 2

2 23 3

12 12 1 1 3 3( )

2 2 2m m mV V V V

hy y h hbh Abh bh

τ = − = − = =

31.500

2k = =




without permission.

PROBLEM 6.27


maxV

kA

τ =


SOLUTION

For a thin-walled circular section, 2 m mA r tπ=

2 3 312 ,

2m m m m mJ Ar r t I J r tπ π= = = =

For a semicircular arc, 2 mryπ

=

222

s m m

ms m m m m

A r t

rQ A y r t r t

π

ππ

=

= = =

(a) 2 mt t= at neutral axis where maximum occurs.

(b) 2

max 3

(2 ) 2

( )(2 )m m

m mm m m

VQ V r t V V

It r t Ar t tτ

ππ= = = = 2.00k =




without permission.

PROBLEM 6.28


maxV

kA

τ =


SOLUTION

3 31 1 12 2

2 12 6A bh bh I bh bh

= = = =

For a cut at location y, where ,y h≤

2

2 3

1( )

2 2

2( )

3

( )2 3

( )

by byA y y

h h

y y h y

by byQ y Ay

hby

t yh

= =

= −

= = −

=

22 3

3

6( ) 3 2

2 3

VQ h by by V y yy V

It by h bh h hbhτ

= = ⋅ ⋅ − = −

(a) To find location of maximum of ,τ set 0.d

dy

τ =

2

[3 4 ] 0md V y

dy hbh

τ = − = 3

,4

my

h= i.e.,

1

4h± from neutral axis.

(b) 2

3 3 9( ) 3 2 1.125

4 4 8mV V V

ybh bh A

τ = − = =

1.125k =




without permission.

PROBLEM 6.29

The built-up beam shown is made by gluing together five planks. Knowing that in the glued joints the average allowable shearing stress is 350 kPa, determine the largest permissible vertical shear in the beam.

SOLUTION

3

3

6 4

6 4

3 3

6 3

1= (240 mm)(160 mm)

121

(200mm)(80mm)12

73.4 10 mm

= 73.4 10 m

40mm = 0.04m

(100mm 40mm)(60mm)

240 10 mm

240 10 m

I

I

t

Q

Q

−

−

−

= ×

×== ×

= ×

= ×

For 350kPa,mτ =

6 3

36 4

(240 10 m ): 350 10 Pa

(73.4 10 m )0.04 mm

VQ V

Ttτ

−

−×= × =

× 4.28 kNV =




without permission.

PROBLEM 6.30

For the beam of Prob. 6.29, determine the largest permissible horizontal shear.

PROBLEM 6.29 The built-up beam shown is made by gluing together five planks. Knowing that in the glued joints the average allowable shearing stress is 350 kPa, determine the largest permissible vertical shear in the beam.

SOLUTION

3 31 1= 2 (40)(240) (80)(40)

12 12I +

3 3

(40 100)70

280 10 mm

Q = ×

= ×

6 4 6 4= 92.6 10 mm 92.6 10 mI + −× = × 6 3(280 10 ) mQ −= ×

For 350 kPa,τ =

6 3

36 4

(280 10 m ); 350 10 Pa

(92.6 10 m )(0.04 m)τ

−

−×= × =

×VQ V

It

4630 NV = 4.63 kNV =




without permission.

PROBLEM 6.31

Several wooden planks are glued together to form the box beam shown. Knowing that the beam is subjected to a vertical shear of 3 kN, determine the average shearing stress in the glued joint (a) at A, (b) at B.

SOLUTION

3 2 3 2

6 4

3 3 6 4

3 3 6 4

6 4 6 4

1 1(60)(20) (60)(20)(50)

12 12

3.04 10 mm

1 1(60)(20) 0.04 10 mm

12 121 1

(20)(120) 2.88 10 mm12 12

2 2 11.88 10 mm 11.88 10 m

A

B

C

A B C

I bh Ad

I bh

I bh

I I I I −

= + = +

= ×

= = = ×

= = = ×

= + + = × = ×

3 3 6 3

3

(60)(20)(50) 60 10 mm 60 10 m

20 mm 20 mm 40 mm 40 10 m

AQ Ay

t

−

−

= = = × = ×

= + = = ×

(a) 3 6

36 3

(3 10 )(60 10 )379 10 Pa

(11.88 10 )(40 10 )τ

−

− −× ×= = = ×

× ×A

AVQ

It

0BQ = 379 kPaAτ =

(b) 0BB

VQ

Itτ = = 0Bτ =




without permission.

PROBLEM 6.32

The built-up timber beam is subjected to a 1500-lb vertical shear. Knowing that the longitudinal spacing of the nails is 2.5 in.s = and that each nail is 3.5 in. long, determine the shearing force in each nail.

SOLUTION

3 21

4

3 42

1 2

4

1(2)(4) (2)(4)(3)

12

82.6667 in

1(2)(6) 36 in

122 2

237.333 in

I

I

I I I

= +

=

= =

= +

=

31 1 (2)(4)(3) 24 in

(1500)(24)151.685 lb/in

237.333

Q A y

VQq

I

= = =

= = =

nail nail1 1

2 (151.685)(2.5)2 2

F qs F qs = = =

nail 189.6 lbF =




without permission.

PROBLEM 6.33

The built-up wooden beam shown is subjected to a vertical shear of 8 kN. Knowing that the nails are spaced longitudinally every 60 mm at A and every 25 mm at B, determine the shearing force in the nails (a) at A, (b) at B. 9 4(Given: 1.504 10 mm .)xI = ×

SOLUTION

9 4 6 41.504 10 mm 1504 10 m

60 mm 0.060 m

25 mm 0.025 m

x

A

B

I

s

s

−= × = ×= == =

(a) 3 31 1 1

6 3

(50)(100)(150) 750 10 mm

750 10 m

AQ Q A y−

= = = = ×

= ×

3 6

16

(8 10 )(750 10 )(0.060)

1504 10

A A A

A

F q s

VQ s

I

−

−

=

× ×= =×

239 NAF =

(b) 3 32 2 2

3 31 2

6 3

(300)(50)(175) 2625 10 mm

2 4125 10 mm

4125 10 m

B

Q A y

Q Q Q−

= = = ×

= + = ×

= ×

3 6

6

(8 10 )(4125 10 )(0.025)

1504 10B B

B B BVQ s

F q sI

−

−× ×= = =

× 549 NBF =




without permission.

PROBLEM 6.34

Knowing that a vertical shear V of 50 kips is exerted on W14 82× rolled-steel beam, determine the shearing stress (a) at point a, (b) at the centroid C.

SOLUTION

For 4W14 82, 14.3 in., 10.1 in., 0.855 in., 0.510 in., 881 in× = = = = =f f wd b t t I

(a) 2

3

(4.15)(0.855) 3.5482 in

14.3 0.8556.7225 in.

2 2 2 2

23.853 in

0.855 in.

a

fa

a a a

a f

A

tdy

Q A y

t t

= =

= − = − =

= == =

(50)(23.853)

(881)(0.855)a

aa

VQ

Itτ = = 1.583 ksiaτ =

(b)

( )

21

1

22

2

(10.1)(0.855) 8.6355 in

6.7225 in.2 2

(0.510)(6.295) 3.2105 in2

1 16.295 3.1475 in.

2 2 2

= =

= − =

= − = =

= − = =

f f

f

w f

f

A b t

tdy

dA t t

dy t

1 1 2 2

3

(8.6355)(6.7225) (3.2105)(3.1475)

68.157 in

0.510 in.

C

C w

Q A y A y

t t

= + +

== =

(50)(68.157)

(881)(0.510)C

CC

VQ

Itτ = = 7.59 ksiCτ =




without permission.

PROBLEM 6.35

An extruded aluminum beam has the cross section shown. Knowing that the vertical shear in the beam is 150 kN, determine the shearing stress at (a) point a, (b) point b.

SOLUTION

3 3 6 4

6 4

1 1(80)(80) (56)(68) 1.9460 10 mm

12 12

1.946 10 m

I

−

= − = ×

= ×

(a) 1 1 2 2

3 3

6 3

2

(56)(6)(37) (2)(12)(40)(20) 31.632 10 mm

31.632 10 m

(2)(12) 24 mm 0.024 m

a

a

Q A y A y

t

−

= +

= + = ×

= ×= = =

3 6

66

(150 10 )(31.632 10 )101.6 10 Pa

(1.946 10 )(0.024)a

aa

VQ

Itτ

−

−× ×= = = ×

×

101.6 MPaaτ =

(b) 3 31 1

6 3

(56)(6)(37) 12.432 10 mm

12.432 10 m

bQ A y−

= = = ×

= ×

3 66

6

(2)(6) 12 mm 0.012 m

(150 10 )(12.432 10 )79.9 10 Pa

(1.946 10 )(0.012)

b

bb

b

t

VQ

Itτ

−

−

= = =

× ×= = = ××

79.9 MPabτ =




without permission.

PROBLEM 6.36

Knowing that a given vertical shear V causes a maximum shearing stress of 75 MPa in the hat-shaped extrusion shown, determine the corresponding shearing stress (a) at point a, (b) at point b.

SOLUTION

Neutral axis lies 30 mm above bottom.

c a bc a b

a b

VQ VQ VQ

It It Itτ τ τ= = =

a a c b b c

c c a c c b

Q t Q t

Q t Q t

τ ττ τ

= =

3

3

3

(6)(30)(15) (14)(4)(28) 4260 mm

6 mm

(14)(4)(28) 1568 mm

4 mm

(14)(4)(28) 1568 mm

4 mm

c

c

a

a

b

b

Q

t

Q

t

Q

t

= + ==

= ==

= ==

75 MPacτ =

(a) 1568 6

754260 4

a ca c

c a

Q t

Q tτ τ= ⋅ = ⋅ ⋅ 41.4 MPaaτ =

(b) 1568 6

754260 4

b cb c

c b

Q t

Q tτ τ= ⋅ = ⋅ ⋅ 41.4 MPabτ =




without permission.

PROBLEM 6.37

Knowing that a given vertical shear V causes a maximum shearing stress of 75 MPa in an extruded beam having the cross section shown, determine the shearing stress at the three points indicated.

SOLUTION

VQ

Itτ = τ is proportional to Q/t.

Point c: 3 3

2

(30)(10)(75)

22.5 10 mm

10 mm

/ 2250 mm

c

c

c c

Q

t

Q t

=

= ×=

=

Point b: 3 3

2

(20)(50)(55)

77.5 10 mm

20 mm

/ 3875 mm

b c

b

b b

Q Q

t

Q t

= +

= ×=

=

Point a: 3 3

2

2 (120)(30)(15)

209 10 mm

120 mm

/ 1741.67 mm

a b

a

a a

Q Q

t

Q t

= +

= ×=

=

( / )mQ t occurs at b. 75 MPam bτ τ= =

/ / /a b c

a a b b c cQ t Q t Q t

τ τ τ= =

2 2 2

75 MPa

1741.67 mm 3875 mm 2250 mma aτ τ= =

33.7 MPaaτ =

75.0 MPabτ =

43.5 MPacτ =




without permission.

PROBLEM 6.38

An extruded beam has the cross section shown and a uniform wall thickness of 0.20 in. Knowing that a given vertical shear V causes a maximum shearing stress 9 ksi,τ = determine the shearing stress at the four points indicated.

SOLUTION

3

3

3

3

3

(0.2)(0.5)(0.5 0.25) 0.125 in

(0.2)(0.5)(0.3 0.25) 0.055 in

(1.4)(0.2)(0.9) 0.432 in

2 2 (3.0)(0.2)(0.9) 0.900 in.

(0.2)(0.8)(0.4) 0.964 in

a

b

c a b

d a b

m d

Q

Q

Q Q Q

Q Q Q

Q Q

= − =

= + =

= + + =

= + + =

= + =

VQ

Itτ =

Since V, I, and t are constant, τ is proportional to Q.

9

0.125 0.055 0.432 0.900 0.964 0.964a b c d mτ τ τ τ τ= = = = =

1.167 ksi; 0.513 ksi; 4.03 ksi; 8.40 ksia b c dτ τ τ τ= = = =




without permission.

PROBLEM 6.39

Solve Prob. 6.38, assuming that the beam is subjected to a horizontal shear V.

PROBLEM 6.38 An extruded beam has the cross section shown and a uniform wall thickness of 0.20 in. Knowing that a given vertical shear V causes a maximum shearing stress 9 ksi,τ = determine the shearing stress at the four points indicated.

SOLUTION

3

3

3

3

(0.5)(0.2)(1.4) 0.140 in

(0.5)(0.2)(1.4) 0.140 in

(0.2)(1.4)(0.8) 0.504 in

0

0.504 in

= =

= =

= + + ==

= =

a

b

c a b

d

m c

Q

Q

Q Q Q

Q

Q Q

VQ

Itτ = Since V, I, and t are constant, τ is proportional to Q.

9

0.140 0.140 0.504 0 0.504a b c d m

mQ

τ τ τ τ τ= = = = =

2.50 ksi; 2.50 ksi, 9.00 ksi, 0a b c dτ τ τ τ= = = =




without permission.

PROBLEM 6.40

Knowing that a given vertical shear V causes a maximum shearing stress of 50 MPa in a thin-walled member having the cross section shown, determine the corresponding shearing stress (a) at point a, (b) at point b, (c) at point c.

SOLUTION

3 3

3 3

3 3

3 3

(12)(30)(25 10 15) 18 10 mm

(40)(10)(25 5) 12 10 mm

2 (12)(10)(25 5) 45.6 10 mm

25(12)(25) 49.35 10 mm

2

12 mm

10 mm

a

b

c a b

m c

a c m

b

Q

Q

Q Q Q

Q Q

t t t

t

= + + = ×

= + = ×

= + + + = ×

= + = ×

= = ==

50 MPamτ =

(a) 18 12

0.364749.35 12

a a m

m m a

Q t

Q t

ττ

= ⋅ = ⋅ = 18.23 MPaaτ =

(b) 12 12

0.291849.35 10

b b m

m m b

Q t

Q t

ττ

= ⋅ = ⋅ = b 14.59 MPaτ =

(c) 45.6 12

0.924049.35 12

c c m

m m c

Q t

Q t

ττ

= ⋅ = ⋅ = c 46.2 MPaτ =




without permission.

PROBLEM 6.41

The extruded aluminum beam has a uniform wall thickness of 18

in. Knowing that the vertical shear in the beam is 2 kips, determine the corresponding shearing stress at each of the five points indicated.

SOLUTION

3 3 41 1(2.50)(2.50) (2.125)(2.25) 1.2382 in

12 12I = − =

0.125 in.t = at all sections.

2 kipsV =

0aQ = aa

VQ

Itτ = 0aτ =

31.25(0.125)(1.25) 0.09766 in

2bQ = =

(2)(0.09766)

(1.2382)(0.125)b

bVQ


2(1.0625)(0.125)(1.1875) 0.25537in.c bQ Q= + =

(2)(0.25537)

(1.2382)(0.125)c

cVQ

Itτ = = 3.30 ksicτ =

22 (0.125) (1.1875) 0.52929d cQ Q= + =

(2)(0.52929)

(1.2382)(0.125)d

dVQ

Itτ = = 6.84 ksidτ =

1.125

(0.125)(1.125) 0.608392e dQ Q

= + =

(2)(0.60839)

(1.2382)(0.125)eVQ

Itτ = = 7.86 ksieτ =




without permission.

PROBLEM 6.42

The extruded aluminum beam has a uniform wall thickness of 18

in. Knowing that the vertical shear in the beam is 2 kips, determine the corresponding shearing stress at each of the five points indicated.

SOLUTION

3 3 41 1(2.50)(2.50) (2.125)(2.25) 1.2382 in

12 2I = − =

Add symmetric points c’, b’, and a’.

0eQ =

31.125(0.125)(1.125) 0.07910 in

2dQ = =

0.125 in.dt =

2 4(0.125) (1.1875) 0.09765 inc eQ Q= = = 0.25 in.ct =

3(2)(1.0625)(0.125)(1.1875) 0.41308 inb cQ Q= + = 0.25 in.bt =

31.25(2)(0.125)(1.25) 0.60839 in

2a bQ Q = + =

0.25 in.at =

(2)(0.60839)

(1.2382)(0.25)a

aa

VQ

Itτ = = 3.93 ksiaτ =

(2)(0.41308)

(1.2382)(0.25)b

bb

VQ


(2)(0.09765)

(1.2382)(0.25)c

cc

VQ

Itτ = = 0.63 ksicτ =

(2)(0.07910)

(1.2382)(0.25)d

dd

VQ

Itτ = = 1.02 ksidτ =

ee

e

VQ

Itτ = 0eτ =




without permission.

PROBLEM 6.43

Three 1 18-in.× steel plates are bolted to four L6 6 1× × angles to form a beam with the cross section shown. The bolts have a 7

8-in. diameter and are spaced

longitudinally every 5 in. Knowing that the allowable average shearing stress in the bolts is 12 ksi, determine the largest permissible vertical shear in the beam. (Given: 46123 inxI = )

SOLUTION

Flange: 3 2 41(18)(1) (18)(1)(9.5) 1626 in

12fI = + =

Web: 3 41(1)(18) 486 in

12wI = =

Angle: 4 2

2 4

35.5 in , 11.0 in

1.86 in. 9 1.86 7.14 in.

596.18 ina

I A

y d

I I Ad

= == = − =

= + =

42 4 6123 in ,f w aI I I I= + + = which agrees with the given value.

Flange: 3(18)(1)(9.5) 171 infQ = =

Angle: 3

3

(11.0)(7.14) 78.54 in

2 328.08 in

a

f a

Q Ad

Q Q Q

= = =

= + =

22

bolt

bolt bolt bolt

boltall

70.60132 in

4 8

2 (2)(12)(0.60132) 14.4317 kips

14.43172.8863 kip/s

5

A

F A

Fq

s

π

τ

= =

= = =

= = =

allall

(2.8863)(6123)

328.08

VQ q Iq V

I Q= = = all 53.9 kipsV =




without permission.

PROBLEM 6.44

Three planks are connected as shown by bolts of 14-mm diameter spaced every 150 mm along the longitudinal axis of the beam. For a vertical shear of 10 kN, determine the average shearing stress in the bolts.

SOLUTION

Locate neutral axis and compute moment of inertia.

Part 2(mm )A (mm)y 3(mm )Ay (mm)d 2 4(mm )Ad 4(mm )I

12500 200 62.5 10× 37.5 617.5781 10× 610.4167 10×

25000 125 63.125 10× 37.5 635.156 10× 6130.208 10×

12500 200 62.5 10× 37.5 617.5781 10× 610.4167 10×

Σ 50000 68.125 10× 670.312 10× 6151.04 10×

6

3

8.125 10162.5 mm

50 10

AyY

A

Σ ×= = =Σ ×

2 6 4

6 4

221.35 10 mm

221.35 10 m

I Ad I−

= Σ + Σ = ×

= ×

3 3

1 1

6 3

(12500)(37.5) 468.75 10 mm

468.75 10 m

Q A y−

= = = ×

= ×

3 6

6

3

(10 10 )(468.75 10 )

221.35 10

21.177 10 N/m

VQq

I

−

−× ×= =

×= ×

3 3 3bolt

2 2 6 2bolt

36bolt

bolt 6bolt

(21.177 10 )(150 10 ) 3.1765 10 N

(14) 153.938 mm 153.938 10 m4

3.1765 1020.6 10 Pa

153.938 10

F qs

A

F

A

π

τ

−

−

−

= = × × = ×

= = = ×

×= = = ××

bolt 20.6 MPaτ =




without permission.

PROBLEM 6.45

A beam consists of three planks connected as shown by steel bolts with a longitudinal spacing of 225 mm. Knowing that the shear in the beam is vertical and equal to 6 kN and that the allowable average shearing stress in each bolt is 60 MPa, determine the smallest permissible bolt diameter that can be used.

SOLUTION

Part 2(mm )A (mm)y 2 6 4(10 mm )Ay 6 4(10 mm )I

7500 50 18.75 14.06

7500 50 18.75 14.06

15,000 −50 37.50 28.12

Σ 75.00 56.25

2 6 4 6 4

3 31 1

6 3

bolt bolt bolt

131.25 10 mm 131.25 10 m

(7500)(50) 375 10 mm

375 10 m

I Ay I

Q A y

VQsF A qs

Iτ

−

−

= + = × = ×

= = = ×

= ×

= = =

3 66 2

bolt 6 6bolt

2

(6 10 )(375 10 )(0.225)64.286 10 m

(6 10 )(131.25 10 )

64.286 mm

VQsA

Iτ

−−× ×= = = ×

× ×

=

boltbolt

4 (4)(64.286)Ad

π π= = bolt 9.05 mmd =




without permission.

PROBLEM 6.46

A beam consists of five planks of 1.5 6-in.× cross section connected by steel bolts with a longitudinal spacing of 9 in. Knowing that the shear in the beam is vertical and equal to 2000 lb and that the allowable average shearing stress in each bolt is 7500 psi, determine the smallest permissible bolt diameter that can be used.

SOLUTION

Part 2(in )A 0(in.)y 30(in )Ay (in.)y 2 4(in )Ay 4(in )I

9 5 45 0.8 5.76 27

9 4 36 −0.2 0.36 27

9 3 27 −1.2 12.96 27

9 4 36 −0.2 0.36 27

9 5 45 0.8 5.76 27

Σ 45 189 25.20 135

0

2 4

1894.2 in.

45

160.2 in

AyY

A

I Ad I

Σ= = =Σ

= Σ + Σ =

Between and : 312 1 1 (9)(0.8) 7.2 in= = = =Q Q Ay

Between and : 323 1 2 7.2 (9)( 0.2) 5.4 inQ Q Ay= + = + − =

VQ

qI

=

Maximum q is based on 312 7.2 in .Q =

bolt

(2000)(7.2)89.888 lb/in

160.2(89.888)(9) 809 lb

q

F qs

= =

= = =

2bolt boltbolt bolt

bolt bolt

2 boltbolt bolt bolt

8090.1079 in

7500

4 (4)(0.1079)

4

F FA

A

AA d d

ττ

ππ π

= = = =

= = = bolt 0.371 in.d =




without permission.

PROBLEM 6.47

A plate of 14

-in. thickness is corrugated as shown and then used as a beam. For a vertical shear of 1.2 kips, determine (a) the maximum shearing stress in the section, (b) the shearing stress at point B. Also, sketch the shear flow in the cross section.

SOLUTION

2 2 2(1.2) (1.6) 2.0 in. (0.25)(2.0) 0.5 inBD BDL A= + = = =

Locate neutral axis and compute moment of inertia.

Part 2(in )A (in.)y 3(in )Ay (in.)d 2 4(in )Ad 4(in )I

AB 0.5 0 0 0.4 0.080 neglect

BD 0.5 0.8 0.4 0.4 0.080 *0.1067

DE 0.5 0.8 0.4 0.4 0.080 *0.1067

EF 0.5 0 0 0.4 0.080 neglect

Σ 2.0 0.8 0.320 0.2133

2 2 4

2 4

* 1 1 0.8(0.5)(1.6) 0.1067 in 0.4 in.

12 12 2.0

0.5333 in

BDAy

A h YA

I Ad I

Σ= = = = =Σ

= Σ + Σ =

(a) 3

3

3

(2)(0.25)(0.4) 0.2 in

(0.5)(0.25)(0.2) 0.025 in

0.225 in

m AB BC

AB

BC

m

Q Q Q

Q

Q

Q

= +

= =

= =

=

(1.2)(0.225)

(0.5333)(0.25)m

mVQ

Itτ = = 2.03 ksimτ =

(b) 30.2 inB ABQ Q= =

(1.2)(0.2)

(0.5333)(0.25)B

BVQ

Itτ = = 1.800 ksiBτ =

0Dτ =




without permission.

PROBLEM 6.48

A plate of 4-mm thickness is bent as shown and then used as a beam. For a vertical shear of 12 kN, determine (a) the shearing stress at point A, (b) the maximum shearing stress in the beam. Also, sketch the shear flow in the cross section.

SOLUTION

20

tan 22.6248

α α= = °

Slanted side: 2

3 3 4

(4 sec )(48) 208 mm

1(4 sec )(48) 39.936 10 mm

12

s

s

A

I

α

α

= =

= = ×

Top: 3 2 3 41(50)(4) (50)(4)(24) 115.46 10 mm

12TI = + = ×

Bottom: 3 4115.46 10 mmB TI I= = ×

3 4 9 42 310.8 10 mm 310.8 10 ms T BI I I I −= + + = × = ×

(a) 3 3 6 3

3

(25)(4)(24) 2.4 10 mm 2.4 10 m

4mm 4 10 m

AQ

t

−

−

= = × = ×

= = ×

3 6

69 3

(12 10 )(2.4 10 )23.2 10 Pa

(310.8 10 )(4 10 )A

AVQ

Itτ

−

− −× ×= = = ×

× × 23.2 MPaAτ =

(b) Maximum shearing occurs at point M, 24 mm above the bottom

3 3 3 3

6 3

(4 sec )(24)(12) 2.4 10 1.248 10 3.648 10 mm

3.648 10 m

M AQ Q α−

= + = × + × = ×

= ×

3 6

69 3

(12 10 )(3.648 10 )35.2 10 Pa

(310.8 10 )(4 10 )M

MVQ

Itτ

−

− −× ×= = = ×

× × 35.2 MPaMτ =

23.2 MPaB A B AQ Q τ τ= = =




without permission.

PROBLEM 6.49

A plate of 2-mm thickness is bent as shown and then used as a beam. For a vertical shear of 5 kN, determine the shearing stress at the five points indicated and sketch the shear flow in the cross section.

SOLUTION

3 3 3 2

3 4 9 4

3 9 3

3 9 3

3 3

1 1 12 (2)(48) (2)(52) (20)(2) (20)(2)(25)

12 12 12

133.75 10 mm 133.75 10 mm

(2)(24)(12) 576 mm 576 10 mm

0

(12)(2)(25) 600 mm 600 10 m

(2)(24)(12) 1.176 10 mm 1.176 1

−

−

−

= + + +

= × = ×

= = = ×=

= − = − = − ×

= − = − × = − ×

a

a

c b

d c

I

Q

Q

Q Q

Q Q 6 3

3 9 3

0 m

(2)(26)(13) 600 mm 500 10 m

−

−= + = − = − ×e dQ Q

3 9

69 3

(5 10 )(576 10 )10.77 10 Pa

(133.75 10 )(2 10 )a

aVQ

Itτ

−

− −× ×= = = ×

× × 10.76 MPaaτ =

bb

VQ

Itτ = 0bτ =

3 9

69 3

(5 10 )(600 10 )11.21 10 Pa

(133.75 10 )(2 10 )c

cVQ

Itτ

−

− −× ×= = = ×

× × 11.21 MPacτ =

3 6

69 3

(5 10 )(1.176 10 )22.0 10 Pa

(133.75 10 )(2 10 )d

dVQ

Itτ

−

− −× ×= = = ×

× × 22.0 MPadτ =

3 9

69 3

(5 10 )(500 10 )9.35 10 Pa

(133.75 10 )(2 10 )e

eVQ

Itτ

−

− −× ×= = = ×

× × 9.35 MPaeτ =




without permission.

PROBLEM 6.50

A plate of thickness t is bent as shown and then used as a beam. For a vertical shear of 600 lb, determine (a) the thickness t for which the maximum shearing stress is 300 psi, (b) the corresponding shearing stress at point E. Also, sketch the shear flow in the cross section.

SOLUTION

2 24.8 2 5.2 in.BD EFL L= = + =

Neutral axis lies at 2.4 in. above AB.

Calculate I.

2

2

2

(3 )(2.4) 17.28

1(5.2 )(4.8) 9.984

12

(6 )(2.4) 34.56

9.984

17.28

89.09

AB

BD

DE

EF DB

FG AB

I t t

I t t

I t t

I I t

I I t

I I t

= =

= =

= == == == Σ =

(a) At point C, (3 )(2.4) (2.6 )(1.2) 10.32C AB BCQ Q Q t t t= + = + =

(600)(10.32 )

0.23168 in.(300)(89.09 )

CVQ VQ tt

It I tτ

τ= ∴ = = = 0.232 in.t =

(b) 4(89.09)(0.23168) 20.64 in= =I

30 (3)(0.23168)(2.4) 1.668 in

E EF FGQ Q Q= +

= + =

(600)(1.668)

(20.64)(0.23168)E

EVQ

Itτ = = 209 psiEτ =




without permission.

PROBLEM 6.51

The design of a beam calls for connecting two vertical rectangular 3

84× -in. plates

by welding them to horizontal 1

22× -in. plates as shown. For a vertical shear V,

determine the dimension a for which the shear flow through the welded surface is maximum.

SOLUTION

33 2

2 4

3

2

2

2 2

1 3 1 1 1(2) (4) (2) (2) (2)(2)

12 8 12 2 2

4.041667 2 in

1(2) in

2

Set 0.4.041667 2

(4.041667 2 ) ( )(4 )0

(4.041667 2 )

I a

a

Q a a

VQ Va dqq

I daa

dq a a aV

da a

= + +

= +

= =

= = =+

+ −= = +

22 4.041667a =

1.422 in.a =




without permission.

PROBLEM 6.52

An extruded beam has a uniform wall thickness t. Denoting by V the vertical shear and by A the cross-sectional area of the beam, express the maximum shearing stress as max ( / )k V Aτ = and determine the constant k for each of the two orientations shown.

SOLUTION

(a)

1 2

3

2h a

A A at

=

= =

2 2 31 1

2 2 32 2

31 2

3

41 1 3 1

3 3 4 45

2 42

I A h ath a t

I A h at a a t

I I I a t

= = =

= = =

= + =

( )

21 1

22 2

21 2

2

352

3

2

3

2 4

2 3

3 3

(2 ) 52

6 3 6 3

5 6 5

m

m

Q A h a t

hQ A a t

Q Q Q a t

VQ V a t V

I t ata t t

V V Vk

at A A

τ

= =

= =

= + =

= = =

= = =

6 3

5k = 2.08k =

(b)

1

2

2

1

2

ah

A at

A at

=

=

=

21 1 1

22

3 3 3

1

12 2 2

1 9 7

48 16 12

I I A d

a hath at

a t a t a t

= +

= + +

= + =

( )

33

2

31 2

21

22

31 2

374

352

1 1

3 2 24

54 4

23

2 2 4

1 1

2 4 8

72 2

4

(2 ) (2 )

7 42 21

20 20 6 10

m

aI t a t

I I I a t

a hQ at a t

aQ at a t

Q Q Q a t

V a tVQ

I t a t t

V V V

at at A

τ

= =

= + =

= + =

= =

= + =

⋅= =

= = =

V

kA

= 21

2.1010

k = =




without permission.

PROBLEM 6.53

An extruded beam has a uniform wall thickness t. Denoting by V the vertical shear and by A the cross-sectional area of the beam, express the maximum shearing stress as max ( / )k V Aτ = and determine the constant k for each of the two orientations shown.

SOLUTION

(a) 2

31

33

2

31 2

21

1( )

2 4

1 1

3 2 24

22 4

31

( )2 2

aI at a t

aI t a t

I I I a t

aQ at a t

= =

= =

= + =

= =

( )( )

22

21 2

234

max 323

1 1

2 4 8

32

4

(2 ) (2 )

9 9 9

16 4 4 4

aQ at a t

Q Q Q a t

V a tVQ

I t a t t

V V V

at at AV

kA

τ

= =

= + =

= =

= = =

= 9

2.254

k = =

(b)

2

21 1

3

31

12

2

1 1 2

3 3 2

1

62

43

h a

I A h at a

a t

I I a t

=

= =

=

= =

( )( )

21

21

212

max 323

12

2 4

12 2

2

2

(2 ) (2 )

3 2 3 2

8 2 4

3 2

2

hQ at a t

Q Q a t

V a tVQ

I t a t t

V V

at at

V Vk

A A

τ

= =

= =

= =

= =

= = 3 2

2.122

k = =




without permission.

PROBLEM 6.54

(a) Determine the shearing stress at point P of a thin-walled pipe of the cross section shown caused by a vertical shear V. (b) Show that the maximum shearing stress occurs for

90θ = ° and is equal to 2V/A, where A is the cross-sectional area of the pipe.

SOLUTION

2 3 312 2

2m m m mA r t J Ar r t I J r tπ π π= = = = =

sin

rθ

θ= for a circular arc.

2

2 sinP

P P

A r t

Q A r rt

θθ

== =

(a) ( )3

( )(2 sin )

(2 ) (2 )P

P

m

VQ V rt

I t r t t

θτπ

= = sin

Pm

V

r t

θτπ

=

(b) 22 sin

2mm

V

r t

πτ

π=

2m

V

Aτ =




without permission.

PROBLEM 6.55

For a beam made of two or more materials with different moduli of elasticity, show that Eq. (6.6)

aveVQ

Itτ =

remains valid provided that both Q and I are computed by using the transformed section of the beam (see Sec. 4.6), and provided further that t is the actual width of the beam where aveτ is computed.

SOLUTION

Let refE be a reference modulus of elasticity.

1 21 2

ref ref

, , etc.E E

n nE E

= =

Widths b of actual section are multiplied by n’s to obtain the transformed section. The bending stress distribution in the cross section is given by

xnMy

Iσ = −

where I is the moment of inertia of the transformed cross section and y is measured from the centroid of the transformed section.

The horizontal shearing force over length xΔ is

( ) ( ) ( )

( )xn M y M Q M

H dA dA ny dAI I I

σ Δ Δ ΔΔ = − Δ = = =

Q ny dA= = first moment of transformed section.

Shear flow: H M Q VQ

qx x I I

Δ Δ= = =Δ Δ

q is distributed over actual width t, thus .q

tτ =

VQ

Itτ =




without permission.

PROBLEM 6.56

A steel bar and an aluminum bar are bonded together as shown to form a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the modulus of elasticity is 629 10× psi for the steel and 610.6 10× psi for the aluminum, determine (a) the average stress at the bonded surface, (b) the maximum shearing stress in the beam. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION

1n = in aluminum.

6

6

29 10 psi2.7358

10.6 10 psin

×= =×

in steel.

Part 2(in )nA (in.)y 3(in )nAy (in.)d 2 2(in )nAd 4(in )nI

Steel 8.2074 2.0 16.4148 0.2318 0.4410 2.7358

Alum. 1.5 0.5 0.75 1.2682 2.4125 0.1250

Σ 9.7074 17.1648 2.8535 2.8608

2 4

17.16481.7682 in.

9.7074

5.7143 in

nAyY

A

I nAd nI

Σ= = =Σ

= Σ + Σ =

(a) At the bonded surface, 3(1.5)(1.2682) 1.9023 inQ = =

(4)(1.9023)

(5.7143)(1.5)

VQ

Itτ = = 0.888 ksiτ =

(b) At the neutral axis, 31.2318(2.7358)(1.5)(1.2318) 3.1133 in

2Q

= =

max(4)(3.1133)

(5.7143)(1.5)

VQ





without permission.

PROBLEM 6.57

A steel bar and an aluminum bar are bonded together as shown to form a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the modulus of elasticity is 629 10× psi for the steel and 610.6 10× psi for the aluminum, determine (a) the average stress at the bonded surface, (b) the maximum shearing stress in the beam. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION

1n = in aluminum.

6

6

29 10 psi2.7358

10.6 10 psin

×= =×

in steel.

Part 2(in )nA (in.)y 3(in )nAy (in.)d 2 2(in )nAd 4(in )nI

Alum. 3.0 2.0 6.0 0.8665 2.2525 1.0

Steel 4.1038 0.5 2.0519 0.6335 1.6469 0.3420

Σ 7.1038 8.0519 3.8994 1.3420

2 4

8.05191.1335 in.

7.1038

5.2414 in

nAyY

nA

I nAd nI

Σ= = =Σ

= Σ + Σ =

(a) At the bonded surface, 3(1.5)(2)(0.8665) 2.5995 inQ = =

(4)(2.5995)

(5.2414)(1.5)

VQ

Itτ = = 1.323 ksiτ =

(b) At the neutral axis, 31.8665(1.5)(1.8665) 2.6129 in

2 = =

Q

max(4)(2.6129)

(5.2814)(1.5)

VQ





without permission.

PROBLEM 6.58

A composite beam is made by attaching the timber and steel portions shown with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of 4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress at the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION

Let ref 200 GPa

10 GPa 11

200 GPa 20

s

ws w

s

E E

En n

E

= =

= = = =

Widths of transformed section:

3 2 3

6 6 6

6 4 6 4

1150 mm (150) 7.5 mm

20

1 12 (150)(12) (150)(12)(125 6) (7.5)(250)

12 12

2[0.0216 10 30.890 10 ] 9.766 10

71.589 10 mm 71.589 10 m

s wb b

I

−

= = =

= + + + = × + × + ×

= × = ×

(a) 3 3 6 31 (150)(12)(125 6) 235.8 10 mm 235.8 10 mQ −= + = × = ×

3 631

6

3 3 3bolt

2 2 2 6 2bolt bolt

36bolt

bolt 6bolt

(4 10 )(235.8 10 )13.175 10 N/m

71.589 10

(23.187 10 )(200 10 ) 2.635 10 N

(12) 113.1 mm 113.1 10 m4 4

2.635 1023.3 10 Pa

113.1 10

VQq

I

F qs

A d

F

A

π π

τ

−

−

−

−

−

× ×= = = ××

= = × × = ×

= = = = ×

×= = = ××

bolt 23.3 MPaτ =

(b) 3 3 3 3 6 32 1 (7.5)(125)(62.5) 235.8 10 58.594 10 294.4 10 mm 294.4 10 mQ Q −= + = × + × = × = ×

3

3 632

6 3

150 mm 150 10 m

(4 10 )(294.4 10 )109.7 10 Pa

(71.589 10 )(150 10 )c

t

VQ

Itτ

−

−

− −

= = ×

× ×= = = ×× ×

109.7 kPacτ =




without permission.

PROBLEM 6.59

A composite beam is made by attaching the timber and steel portions shown with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of 4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress at the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION

Let steel be the reference material.

10 GPa

1.0 0.05200 GPa

= = = =ws w

s

En n

E

Depth of section: 90 84 90 264 mmd = + + =

For steel portion: 3 3 6 41 12 (2) (6)(264) 18.400 10 mm

12 12sI bd = = = ×

For the wooden portion: ( )3 3 3 3 6 41 2

1 1(140)(264 84 ) 207.75 10 mm

12 12wI b d d= − = − = ×

For the transformed section: s s w wI n I n I= +

6 6 6 4 6 4(1.0)(18.400 10 ) (0.05)(207.75 10 ) 28.787 10 mm 28.787 10 mI −= × + × = × = ×

(a) Shearing stress in the bolts.

For the upper wooden portion 6 3(90)(140)(42 45) 1.0962 10 mmwQ = + = ×

For the transformed wooden portion

6 3 3 6 3(0.05)(1.0962 10 ) 54.81 10 mm 54.81 10 mw wQ n Q −= = × = × = ×

Shear flow on upper wooden portion.

6

6

(4000)(54.81 10 )7616 N/m

28.787 10

VQq

I

−

−×= = =

×

bolt (7616)(0.200) 1523.2 NF qs= = =

2 2 2 6 2bolt bolt (12) 113.1 mm 113.1 10 m

4 4A d

π π −= = = = ×

Double shear: boltbolt 6

bolt

1523.2

2 (2)(113.1 10 )

F

Aτ −= =

×

66.73 10 Pa= × bolt 6.73 MPaτ =




without permission.


(b) Shearing stress at the center of the cross section.

For two steel plates 3 3 6 3(2)(6)(90 42)(90 42) 76.032 10 mm 76.032 10 msQ −= + − = × = ×

For the neutral axis 6 6 6 354.81 10 76.032 10 130.842 10 mQ − − −= × + × = ×

Shear flow across the neutral axis

6

36

(4000)(130.842 10 )18.181 10 N/m

28.787 10

VQq

I

−

−×= = = ×

×

Double thickness 2 12 mm 0.012 mt = =

Shearing stress 3

618.181 101.515 10 Pa

2 0.012τ ×= = = ×q

t

1.515 MPaτ =




without permission.

PROBLEM 6.60

Consider the cantilever beam AB discussed in Sec. 6.8 and the portion ACKJ of the beam that is located to the left of the transverse section CC′ and above the horizontal plane JK, where K is a point at a distance Yy y< above the neutral axis. (See Figure). (a) Recalling that x Yσ σ= between C and E and ( / )x Y Yy yσ σ= between E and K, show that the magnitude of the horizontal shearing force H exerted on the lower face of the portion of beam ACKJ is

21

22 Y Y

Y

yH b c y

yσ

= − −

(b) Observing that the shearing stress at K is

0 0

1 1lim limxyA x

H H H

A b x b x

∂τ∂Δ → Δ →

Δ Δ= = =Δ Δ

and recalling that Yy is a function of x defined by Eq. (6.14), derive Eq. (6.15).

SOLUTION

Point K is located a distance y above the neutral axis.

The stress distribution is given by

for 0 and for .Y Y Y YY

yy y y y c

yσ σ σ σ= ≤ < = ≤ ≤




without permission.


For equilibrium of horizontal forces acting on ACKJ,

2 2

( )2

Y

Y

y cY

Yy yY

Y Yy Y

Y

ybH dA dy bdy

y

b y yb c y

y

σσ σ

σ σ

= = +

−= + −

21

22 Y Y

Y

yH b c y

yσ

= − −

(a)

Note that Yy is a function of x.

2

2

2

2

1 1

2

11

2

Y Yxy Y

Y

YY

Y

y dyH y

b x x y dx

dyy

y dx

∂∂τ σ∂ ∂

σ

= = − +

= − −

But 2

2

3 11

2 3Y

yy

M Px Mc

= = −

Differentiating, 2

3 2

2 3Y Y

Yy dydM

P Mdx dxc

= = −

2 2

223

3

2Y

Y Y Y YY Y

dy Pc Pc P

dx y M byy bc σσ= − = − = −

Then 2 2

2 21 3 31 1

2 2 4xy YY Y YY Y

y P P y

y b by yτ σ

σ σ

= − = −

(b)




without permission.

PROBLEM 6.61

Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION

23 3

2 3 3

3 3 3

1 3 7( )

12 2 3

1(2 ) (2 ) 2

121 2 28

(2 )12 3 3

AB FG

DB EF

DE

aI I ta ta ta

I I at a a t a t

I t a ta I I ta

= = + =

= = + ≈

= = = Σ =

Part AB:

2 2

2 2

2(2 );

31

(2 ) (2 )2

1(4 )

2

(4 )2

a yA t a y y

Q Ay t a y a y

t a y

VQ Va y

It Iτ

+= − =

= = − +

= −

= = −

22 2

1

23 3 32

3

(4 )2

(2) 14 (4)(2) (4)(1)

2 3 2 3 3

5 5

6 56

a

a

a

a

VF dA a y tdy

I

Vt y Vtaa y

I I

VtaV

I

τ= = −

= − = − − +

= =




without permission.


Part DB: 3

( )2

3

2

3

2

aQ ta txa

ata x

VQ Va ax

It Iτ

= +

= +

= = +

2 2

20 0

22 3 2

0

3

3 3

2 2

3 (3)(2) (2)

2 2 2 2

155

28

a a

a

Va a Vta aF dA x tdx x dx

I I

Vta ax x Vta

I I

VtaV

I

τ = = + = +

= + = +

= =

HMΣ = :HMΣ 2 1(2 ) 2 (2 )

30 20 5

28 56 7

Ve F a F a

Va Va Va

= −

= − =

5

0.7147

e a a= =




without permission.

PROBLEM 6.62


SOLUTION

23 3

23 2

3 3

3

3 1 9

2 12 4

1 1

2 12 4

1 9(3 )

12 429

4

AB HJ

DE FG

AH

aI I at at ta

aI I at at ta

I t a ta

I I ta

= = + ≈

= = + ≈

= =

= Σ =

Part AB:

32

3 2294

1 2 20 0

3 3

2 2

6

29

6 6 3

2929 29

a a

aA tx y Q atx

V atxVQ Vx

It ta t a t

Vx VF dA tdx x dx V

a t a

τ

τ

= = =

⋅= = =

= = = =

Part DE:

12

3 2294

1

2 2

2

29

aA tx y Q atx

V atxVQ Vx

It ta t a tτ

= = =

⋅= = =

2 2 20 0

2 2 1

2929 29

a aVx VF dA t dx xdx V

a t aτ= = = =

KMΣ = :KMΣ 1 2(3 ) ( )

9 1 10

29 29 29

Ve F a F a

Va Va Va

= +

= + =

10

29e a= 0.345e a=




without permission.

PROBLEM 6.63

An extruded beam has the cross section shown. Determine (a) the location of the shear center O, (b) the distribution of the shearing stresses caused by the vertical shearing force V shown applied at O.

SOLUTION

2

3 3

6 4 6 4

1 192 12 (72) (12) (72) (12) (6)(192)

12 2 12

19.4849 10 mm 19.4849 10 m

I

−

= + +

= × = ×

Part AB: 192

12 (12 ) 11522

1152

A x Q Ay x x

VQ Vxq

I I

= = = =

= =

0x = at point A. 72 mmABx l= = at point B.

2

2

6

1152 1152 (72)721 0 2

(576) (72)0.153246

19.4849 10

= = =

= =×

B

A

xx

Vx VF qdx dx

I I

V V

CMΣ = : (0.153246 ) (192)CM Ve V=

(a) 29.423 mme = 29.4 mme =

(b) Point : 0 0, 0A x Q q= = = 0Aτ =

Point B in part AB: 72 mmx =

3 3 6 3

3 6

6

(1152) (72) 82.944 10 mm 82.944 10 m

12 mm 0.012 m

(110 10 )(82.944 10 )

(19.4849 10 )(0.012)B

Q

t

VQ

Itτ

−

−

= = × − ×= =

× ×= =×

639.0 10 Pa= × 39.0 MPa in B ABτ =





without permission.

Part BD:

Point B : 3 3 6 3

3 6

6

96 mm 82.944 10 mm 82.944 10 m

6 mm 0.006 m

(110 10 ) (82.944 10 )

(19.4849 10 ) (0.006)B

y Q

t

VQ

Itτ

−

−

−

= = × = ×= =

× ×= =×

678.0 10 Pa= × 78.0 MPa in B BDτ =

Point C : 0, 6 mm 0.006 my t= = =

3 3 3 6 39682.944 10 (6) (96) 110.592 10 mm 110.592 10 m

2Q − = × + = × = ×

3 6

66

(110.10 ) (110.592 10 )104.1 10 Pa

(19.4849 10 ) (0.006)

VQ

Itτ

−

−×= = = ×

× 104.1 MPaCτ =




without permission.

PROBLEM 6.64


SOLUTION

2

3 3

6 4 6 4

1 192 12 (72)(6) (72)(6) (12) (192)

12 2 12

15.0431 10 mm 15.0431 10 m−

= + +

= × = ×

I

Part AB: 192

6 (6 ) 5762

576

A x Q Ay x x

VQ Vxq

I I

= = = =

= =

0x = at point A. 72 mmABx l= = at point B.

272

10

2

6

576 576 (72)

2

(288) (72)0.099247

15.0431 10

B

A

x

x

Vx VF qdx dx

I I

V V

= = =

= =×

CM = : (0.099247) (192)CM Ve V=

(a) 19.0555 mme = 19.06 mme =

(b) Point : 0 0, 0A x Q q= = = 0Aτ =

Point B in part AB: 72 mmx =

3 3 6 3

3 6

6

(576) (72) 41.472 10 mm 41.472 10 m

6 mm 0.006 m

(110 10 )(41.472 10 )

(15.0431 10 )(0.006)B

Q

t

VQ

Itτ

−

−

−

= = × = ×= =

× ×= =×

650.5 10 Pa= × 50.5 MPaBτ =




without permission.


Part BD:

Point B : 3 3 6 3

3 6

6

96 mm 41.472 10 mm 41.472 10 m

12 mm 0.012 m

(110 10 ) (41.472 10 )

(15.0431 10 )(0.012)B

y Q

t

VQ

Itτ

−

−

−

= = × = ×= =

× ×= =×

625.271 10 Pa= × 25.3 MPaBτ =

Point C :

3 3 3 6 3

0, 0.012 m

9641.472 10 (12) (96) 96.768 10 mm 96.768 10 m

2

y t

Q −

= =

= × + = × = ×

3 6

66

(110 10 ) (96.768 10 )58.967 10 Pa

(15.0431 10 ) (0.012)

VQ

Itτ

−

−× ×= = = ×

× 59.0 MPaCτ =




without permission.

PROBLEM 6.65


SOLUTION

Part A (in2) (in.)d 2 4(in )Ad 4(in )I

BD 0.50 3 4.50 0≈

ABEG 1.25 0 0 10.417

EF 0.50 3 4.50 0≈

Σ 2.25 9.00 10.417

2 419.417 inI Ad I= Σ + Σ =

(a) Part BD: ( ) 3

( )( ) (3 )

Q x tx

VQ x Vq x tx

I I

=

= =

4

0

3 3 24(8)BD

Vt Vt VtF xdx

I I I= = =

Its moment about H : 72

( ) 3BD H BDVt

M FI

= =

Part EF: By same method, 24 72

( )EF EF HVt Vt

F MI I

= =

Moments of , , and AB BE EGF F F about H are zero.

72 72 144

HVt Vt Vt

Ve MI I I

= Σ = + =

144 (144)(0.125)

19.417

te

I= = 0.927 in.e =




without permission.


(b) At , , , and : 0A D F G Q = 0A D F Gτ τ τ τ= = = =

Just above B : 1 (2 )(4) 8ABQ Q t t= = =

11

(2.75) (8 )

(19.417)

VQ t

It tτ = = 1 1.133 ksiτ =

Just to the right of B : 2 (3) (4) 12BDQ Q t t= = =

22

(2.75)(12 )

(19.417)

VQ t

It tτ = = 2 1.700 ksiτ =

Just below B : 3 1 2 20Q Q Q t= + =

33

(2.75)(20 )

(19.417)

VQ t

It tτ = = 3 2.83 ksiτ =

At H (neutral axis): 3 20 (3)(1.5) 24.5H BHQ Q Q t t t= + = + =

(2.75) (24.5 )

(19.417)H

HVQ t

It tτ = = 3.47 ksiHτ =

By symmetry: 4 3 2.83 ksiτ τ= =

5 2 1.700 ksiτ τ= =

6 1 1.133 ksiτ τ= =




without permission.

PROBLEM 6.66


SOLUTION

3 4

3 2 4

3 4

4

4

4

1(0.125)(3) 1.125 in

31

(4) (0.125) (4)(0.125)(3) 4.50065 in121

(0.125) (6) 2.25 in12

4.50065 in

1.125 in

13.50 in

AB

BD

DE

EF BD

FG AB

I

I

I

I I

I I

I I

= =

= + =

= =

= =

= =

= Σ =

(a) Part AB: 2

2

3 32

0 0

( ) 0.52( ) 0.5

( )

0.5( ) 4.5

= =

= =

= = = ↑ AB

yQ y ty ty

VQ y Vtq y y

I IVt Vt

F q y dy y dyI I

Its moment about H is 4 18 ABVt

FI

=

2(0.5) ( )(3) 4.5BQ t t= =

Part BD:

4 4

0 0

( ) (3) (4.5 3 )

( )( ) (4.5 3 )

( ) (4.5 3 ) 42

B

BD

Q x Q xt x t

Vq x Vtq x x

I It Vt

F q x dx x dxI I

= + = +

= = +

∇= = + = ←

Its moment about : 3 126BDVt

H FI

=

[4.5 (3)(4)] 16.5DQ t t= + =




without permission.


Part EF: By symmetry with part BD, 42 EFVt

FI

= →

Its moment about H is 3 126EFVt

FI

=

Part FG: By symmetry with part AB, 4.5FGVT

FI

= ↑

Its moment about H is 4. 18 FGVt

FI

=

Moment about H of force in part DE is zero.

144(18 126 0 126 18)

144 (2.88)(0.125)

13.50

HVt Vt

Ve MI I

te

I

= Σ = + + + + =

= = 2.67 in.e =

(b) 0A GQ Q= = 0A Gτ τ= =

4.5B FQ Q t= =

(2.75) (4.5 )

13.50B

B FVQ t

It tτ τ= = = 0.917 ksiB Fτ τ= =

16.5D EQ Q t= =

0 (2.75) (16.5 )

13.50D EVQ t

It tτ τ= = = 3.36 ksiD Eτ τ= =

At H (neutral axis): (3)(1.5) 21H DQ Q t t= + =

(2.75) (21 )

13.50H

HVQ t

It tτ = = 4.28 ksiHτ =




without permission.

PROBLEM 6.67


SOLUTION

3 2 6 4

3 2 6 4

1(30)(6) (30)(6)(45) 0.365 10 mm

121

(30)(4) (30)(4)(15) 0.02716 10 mm12

AB HJ

DE FG

I I

I I

= = + = ×

= = + = ×

3 6 4

6 4

1(6)(90) 0.3645 10 mm

12

1.14882 10 mm

AHI

I I

= = ×

= Σ = ×

(a) For a typical flange, ( )A s ts=

2

0

( )

( )( )

( )2

b

Q s yts

VQ s Vytsq s

I I

VytbF q s ds

I

=

= =

= =

Flange AB: 2

6

(45)(6)(30 )0.10576

(2)(1.14882 10 )AB

VF V−= = ←

×

Flange DE: 2

6

(15)(4)(30)0.023502

(2)(1.14882 10 )DE

VF V= = ←

×

Flange FG: 0.023502FGF V= →

Flange HJ: 0.10576HJF V= →

KMΣ = : 45 15 15

45 10.223K AB DE FG

HJ

M Ve F F F

F V

Σ = + ++ =

Dividing by V, 10.22 mme =




without permission.


(b) Calculation of shearing stresses.

3 6 435 10 N 1.14882 10 mV I −= × = ×

At B, E, G, and J , 0τ =

At A and H ,

3 3 6 3(30)(6)(45) 8.1 10 mm 8.1 10 mQ −= = × = ×

3

3 66

6 3

6 10 m

(35 10 )(8.1 10 )41.1 10 Pa

(1.14882 10 )(6 10 )

t

VQ

Itτ

−

−

− −

= ×

× ×= = = ×× ×

41.1 MPaτ =

Just above D and just below F :

3 3 3 6 3

3

8.1 10 (6)(30)(30) 13.5 10 mm 13.5 10 m

6 10 m

Q

t

−

−

= × + = × = ×

= ×

3 6

66 3

(35 10 )(13.5 10 )68.5 10 Pa

(1.14882 10 )(6 10 )

VQ

Itτ

−

− −× ×= = = ×

× × 68.5 MPaτ =

Just to right of D and just to the right of F :

3 3 6 3 3(30)(4)(15) 1.8 10 mm 1.8 10 m 4 10 mQ t− −= = × = × = ×

3 6

66 3

(35 10 )(1.8 10 )13.71 10 Pa

(1.14882 10 )(4 10 )

VQ

Itτ

−

− −× ×= = = ×

× × 13.71 MPaτ =

Just below D and just above F :

3 3 3 3 6 3

3

13.5 10 1.8 10 15.3 10 mm 15.3 10 m

6 10 m

Q

t

−

−

= × + × = × = ×

= ×

3 6

66 3

(35 10 )(15.3 10 )77.7 10 Pa

(1.14882 10 )(6 10 )

VQ

Itτ

−

− −× ×= = = ×

× × 77.7 MPaτ =

At K , 3 3 3 6 315.3 10 (6)(15)(7.5) 15.975 10 mm 15.975 10 mQ −= × + = × = ×

3 6

66 3

(35 10 )(15.975 10 )81.1 10 Pa

(1.14882 10 )(6 10 )

VQ

Itτ

−

− −× ×= = = ×

× × 81.1 MPaτ =




without permission.

PROBLEM 6.68


SOLUTION

3 2 6 4

3 2 6 4

1(30)(4) (30)(4)(45) 0.24316 10 mm

121

(30)(6) (30)(6)(15) 0.04104 10 mm12

AB HJ

DE FG

I I

I I

= = + = ×

= = + = ×

3 6 4

6 4

1(6)(90) 0.3645 10 mm

12

0.9329 10 mm

= = ×

= Σ = ×

AHI

I I

(a) For a typical flange, ( )A s ts=

2

0

( )

( )( )

( )2

b

Q s yts

VQ s Vytsq s

I I

VytbF q s ds

I

=

= =

= =

Flange AB: 2

6

(45)(4)(30)0.086826

(2)(0.9329 10 )AB

VF V= = ←

×

Flange DE: 2

6

(15)(6)(30)0.043413

(2)(0.9329 10 )DE

VF V= = ←

×

Flange FG: 0.043413 FGF V= →

Flange HJ: 0.086826 HJF V= →

KMΣ = : 45 15 15

45 9.1167 K AB DE FG

HJ

M Ve F F F

F V

Σ = + ++ =





without permission.


(b) Calculation of shearing stresses.

3 6 435 10 N 0.9329 10 mV I −= × = ×

At B, E, G, and J , 0τ =

At A and H , 3 3 6 3

3 63

6

(30)(4)(45) 5.4 10 mm 5.4 10 m

(35 10 )(5.4 10 )202.59 10 N/m

0.9329 10

Q

VQq

I

−

−

−

= = × = ×

× ×= = = ××

Just to the right of A and H : 34 10 mt −= ×

3

63

202.59 1050.6 10 Pa

4 10

q

tτ −

×= = = ××

50.6 MPaτ =

Just below A and just above H : 36 10 mt −= ×

3

63

202.59 1033.8 10 Pa

6 10

q

tτ −

×= = = ××

33.8 MPaτ =

Just above D and just below F : 36 10 mt −= ×

3 3 3 6 35.4 10 (6)(30)(30) 10.8 10 mm 10.8 10 mQ −= × + = × = ×

3 6

66 6

(35 10 )(10.8 10 )67.5 10 Pa

(0.9329 10 )(6 10 )

VQ

Itτ

−

− −× ×= = = ×

× × 67.5 MPaτ =

Just to the right of D and E : 36 10 mt −= ×

3 2 6 3(30)(6)(15) 2.7 10 mm 2.7 10 mQ −= = × = ×

3 6

66 3

(35 10 )(2.7 10 )16.88 10 Pa

(0.9329 10 )(6 10 )

VQ

Itτ

−

− −× ×= = = ×

× × 16.88 MPaτ =

Just below D and just above F : 36 10t −= ×

3 3 3 3 6 310.8 10 2.7 10 13.5 10 mm 13.5 10 mQ −= × + × = × = ×

3 6

66 6

(35 10 )(13.5 10 )84.4 10 Pa

(0.9329 10 )(6 10 )

VQ

Itτ

−

− −× ×= = = ×

× × 84.4 MPaτ =

At K , 3 3 3 6 313.5 10 (6)(15)(7.5) 14.175 10 mm 14.175 10 mQ −= × + = × = ×

3 6

66 3

(35 10 )(14.175 10 )88.6 10 Pa

(0.9329 10 )(6 10 )

VQ

Itτ

−

− −× ×= = = ×

× × 88.6 MPaτ =




without permission.

PROBLEM 6.69


SOLUTION

2 2

2 2 2 2 4

3 4

4

4 3 5 in. 5

1 1(5 )(3) (5 )(4) 83.75 in

12 121

( )(5) 10.417 in12

2 177.917 in

= + = =

= + = + =

= =

= + =

AB AB

AB AB AB

BD

AB BD

L A t

I A h A d t t t

I t t

I I I t

In part BD, AB BYQ Q Q= +

2 2

1(5 )(4) (2.5 ) (2.5 )

2

1 120 3.125 23.125

2 2

Q t y t y

t t ty y t

= + − +

= + − = −

212.52

2.5

2.52.5

2 3

2.52.5

3

(23.125 )

1 123.125 23.125

2 6

(2.5) (110.417)2 (23.125)(2.5) 0.62061

6 177.917

BD

V y tVQF dA tdy

It It

Vt Vty dy y y

I I

Vt VtV

I t

τ τ−

− −

−= = = ⋅

= − = −

= ⋅ − = =

KMΣ = :KMΣ 10 10(0.62061 )

3 3V e V − − = −

10

[1 0.62061]3

e = − 1.265 in.e =

Note that the lines of action of ABF and DEF pass through point K. Thus, these forces have zero moment

about point K.




without permission.

PROBLEM 6.70


SOLUTION

3 3 4

2

2 2 3 4

3 3 3 4

1(6)(35) 85.75 10 mm

3

70 mm (70)(6) 420 mm

1 1(420)(35) 171.5 10 mm

3 3

(2)(85.75 10 ) (2)(171.5 10 ) 514.5 10 mm

DB

AB AB

AB AB

I

L A

I A h

I

= = ×

= = =

= = = ×

= × + × = ×

Part AB:

2

2

6

1 1sin 30

2 43

2

3

A ts s

y s s

Q Ay s

VQ Vs

It Itτ

= =

= ° =

= =

= =

270 70

21

0 0

3

3 3

2

(3)(70) 1

(2)(3) 3

Vs VF dA tds s ds

It I

V VI

τ= = =

= =

DMΣ = 1: 2[( cos 60°)(70 sin 60°)]

20.2DM Ve F

V

Σ ==





without permission.

PROBLEM 6.71


SOLUTION

2 3

2 2

2 2 3

3

(40 ) (60) 144 10

80 60 100 mm 100

1 1(100 )(60) 120 10

3 3

2 2 528 10

AB

DB DB

DB DB

AB DB

I t t

L A t

I A h t t

I I I t

= = ×

= + = =

= = = ×

= + = ×

Part AB: 3

40 40

10 0

302 2

30

60 mm

60 mm

(60 ) 60

60 60

60 (60) (30)0.051136

2 (2)(528 10 )

A tx y

Q A y tx

VQ V tx Vx

It It IVx Vt

F dA tdx xdxI I

Vt x VtV

I t

τ

τ

= =

= =

= = =

= = =

= = =×

DMΣ = : (0.051136 ) (120)DM Ve VΣ = 6.14 mme =




without permission.

PROBLEM 6.72


SOLUTION

3 2 4

2 2 2

2 2 4

4

1(1.5) (0.1) (1.5) (0.1)(2) 0.600125 in

12

1.5 2 2.5 in. (2.5) (0.1) 0.25 in

1 1(0.25)(2) 0.33333 in

3 3

2 2 1.86692 in

AB

BD BD

BD BD

AB BD

I

L A

I A h

I I I

= + =

= + = = =

= = =

= + =

Part AB: 3

1.5 1.5

10 0

2

( ) 0.1 , 2 in.

( ) ( ) 0.2 in

( ) 0.2( )

0.2( )

(0.2)(1.5)0.225

2

A x tx x y

Q x A x y x

VQ x Vxq x

I IV

F q x dx xdxI

V V

I I

= = =

= =

= =

= =

= =

Likewise, by symmetry in part EF: 1 0.225V

FI

=

DMΣ = 1: 4 0.9 0.482DV

M Ve F VI

Σ = = =

Dividing by V, 0.482 in.e =




without permission.

PROBLEM 6.73


SOLUTION

For a thin-walled hollow circular cross section, 2A atπ=

2 3 312

2J a A a t I J a tπ π= = = =

For the half-pipe section, 3

2I a t

π=

Use polar coordinate θ for partial cross section.

2

2 2

arc length

sin where

2sin cos

cos

sin cos(2 sin cos )

sin 2 sin

A st a t s

r a

y r a

Q Ay a t a a t

a t a t

θα θα

αα ααα

α αθ α αα

α θ

= = =

= =

= =

= = =

= =

2

2 4

00

4

sin

sin cos

42

H

VQ Va

It I

Va Va tM a dA a tad

I I

Va tVa

I

ππ

τ θ

τ θ θ θ

π

= =

= = = −

= =

But ,HM Ve= hence 4

1.273e a aπ

= =




without permission.

PROBLEM 6.74


SOLUTION

For whole cross section, 2A atπ=

2 3 312

2J Aa a t I J a tπ π= = = =

Use polar coordinate θ for partial cross section.

2

22 2

2 2 2

arc length

sin 1 where

2

sin sin

sin2 sin

2sin (1 cos )2

A st a t s

r a

y r a

Q Ay a t a a t

a t a t

θα α θ

ααα

ααθ α

αθ θ

= = =

= =

= =

= = =

= = −

2

23 42

00

4

3

(1 cos )

(1 cos ) ( sin )

22

C

VQ Va

It I

Va Va tM a dA tad

I I

Va taV

a t

ππ

τ θ

τ θ θ θ θ

ππ

= = −

= = − = −

= =

But , henceCM Ve= 2e a=




without permission.

PROBLEM 6.75

A thin-walled beam has the cross section shown. Determine the location of the shear center O of the cross section.

SOLUTION

3 31 1 2 2

1 1

12 12I t h t h= +

Right flange: 2 2

2 2

2 2 2

2 22 2

1

2

1 1

2 2

1 1 1

2 2 2

1 1

2 4

A h y t

y h y t

Q Ay

h y h y t

h y t

= − = +

=

= − + = −

2 22 2

2 2

22

1

2 4

2

VQ Vh y t

It It

VtF dA

I t

τ

τ

= = −

= =2

2

/22 22

/22

1

4

h

hh y t

−

−

2

2

/2322

2 2/2

3 3 3 32 22 2 2 2 2 2 2 2 22 2 3 3

1 1 2 2

1

2 4 3

1 1 1 1

2 4 2 3 2 4 2 3 2 12

h

h

Vt ydy h y

I

Vt h h h h Vt h V t hh h

I I t h t h

= −

= − + − = = +

HM =3

2 22 3 3

1 1 2 2

:Ht h b

M Ve F b Vt h t h

− = − = −+

3 3

2 23 3 3 3

1 1 2 2

(0.75)(6) (8)

(0.75) (8) (0.75) (6)

t h be

t h t h= =

+ + 2.37 in.e =




without permission.

PROBLEM 6.76

A thin-walled beam has the cross section shown. Determine the location of the shear center O of the cross section.

SOLUTION

Let 3 3 31 2 3 1 2 3

1, , and . ( )

12h AB h DE h FG I t h h h= = = = + +

Part AB: 1 1

2 21 1 1

1 1 1

2 2 2

1 1 1 1 1

2 2 2 2 4

A h y t y h y

Q Ay t h y h y t h y

= − = +

= = − + = −

1

12

112

2 21

13122 2 2

1 1 1 112

3 3 32 1 1 11 1 3 3 3

1 2 3

1

2 4

1 1

2 4 2 4 3

1 1 1

4 2 3 2 12

hh

h h

VQ Vh y

It I

V Vt yF dA h y tdy h y

I I

h Vt h h VVth h

I I h h h

τ

τ−

= = −

= = − = −

= − = = + +

Likewise, for part DE, 32

2 3 3 31 2 3

h VF

h h h=

+ +

and for part FG, 33

3 3 3 31 2 3

h VF

h h h=

+ +

HMΣ = :HMΣ 3 3

2 2 3 32 2 3 3 3 3 3

1 2

a h a hVe a F a F V

h h h

+= + =+ +

Dividing by V, 3 3 3 3

2 2 3 33 3 2 3 3 31 2 3

(3)(5) (5)(4)

5 5 4

a h a he

h h h

+ += =+ + + +

2.21 in.e =




without permission.

PROBLEM 6.77

A thin-walled beam of uniform thickness has the cross section shown. Determine the dimension b for which the shear center O of the cross section is located at the point indicated.

SOLUTION

Part AB:

2

2

0

1( ) ( )

21

( ) ( ) ( )2

( ) 1( )

2

( )AB

A

A

A

l

AB

A s ts y s y s

Q s A s y s ty s ts

VQ s Vtq s y s s

I I

F q s ds

= = −

= = −

= = −

=

2 3

2 6A AB ABy l lVt

I

= − ↓

At B: 21

2B A AB ABQ ty l t l= −

By symmetry, FG ABF F=

Part BD:

2

0

( )

( ) ( )

( )( ) ( )

1( )

2

B B B B

B B

b

BD B B

A x tx

Q x Q y A x Q t y x

V Q x Vq x Q ty x

I IV

F q x dx Q b t y bI

== + = +

= = +

= = + →

By symmetry, EF BDF F=

DEF is not required, since its moment about O is zero.




without permission.


2 32

2 3 2 2 2

0 : ( ) 0

2 2 0

12 2

2 6 2

2 1 1 2 1 10

2 6 2 2

O AB FG B BD F EF

AB B BD

A AB ABB B B

A AB AB A AB AB B B

M b F F y F y F

b F y F

y l lVt Vb y Q b ty b

I I

Vt Vty l l b y l l y b y b

I I

Σ = + − + =− =

⋅ − − + − − − − =

Dividing by 2Vt

I and substituting numerical data,

2 3 2 2 21 1 1 1(90) (60) (60) (90)(60) (60) (30) (30) 0

2 6 2 2b b b

− − − + =

3 3 2126 10 108 10 450 0b b b× − × + =

3 218 10 450 0b b× − = 0 and 40.0 mmb b= =




without permission.

PROBLEM 6.78


SOLUTION

Part AB: 3

60 mm

60 mm

60

A tx y

Q Ay tx

VQ Vx

It Iτ

= =

= =

= =

30 30

10 0

302 23

0

60 60

60 (60)(30)27 10

2 2

Vx VtF dA t dx x dx

I I

Vt x Vt Vt

I I I

τ= = =

= = = ×

Part DE: 45 mm

45

45

A tx y

Q Ay tx

VQ Vx

It Iτ

= == =

= =

2

245 45 45

2b b

O O

Vx Vt b VtF dA t dx xdx

I I Iτ= = = =

OM = : OM 2 10 (2)(45) (2)(60)F F= −

2 2 3(45) (2)(60)(27 10 ) 0Vt

bI

− × =

3

2 22

(2)(60)(27 10 )1600 mm

45b

×= = 40 mmb =

Note that the pair of 1F forces form a couple. Likewise, the pair of 2F forces form a couple. The lines of action of the forces in BDOGK pass through point O.




without permission.

PROBLEM 6.79

For the angle shape and loading of Sample Prob. 6.6, check that 0qdz = along the horizontal leg of the

angle and qdy P= along its vertical leg.

SOLUTION

Refer to Sample Prob. 6.6.

Along horizontal leg: 2 23 3

3 ( )( 3 ) 3( 4 3 )

4 4f

P a z a z Pa az z

ta taτ − −= = − +

2 230 0

2 22

30

3( 4 3 )

4

34 3

2 34

a a

f

a

Pqdz t dz a az z dz

a

P z za z a

a

τ= = − +

= − +

3 3 33

3( 2 ) 0

4

Pa a a

a= − + =

Along vertical leg: 2 23 3

3 ( )( 5 ) 3( 4 5 )

4 4e

P a y a y Pa ay y

ta taτ − += = + −

2 230 0

3( 4 5 )

4

a a

eP

qdy t dy a ay y dya

τ= = + −

2 32

30

3 3 3 33 3

34 5

2 34

3 5 3 42

3 34 4

aP y y

a y aa

P Pa a a a P

a a

= + −

= + − = ⋅ =




without permission.

PROBLEM 6.80

For the angle shape and loading of Sample Prob. 6.6, (a) determine the points where the shearing stress is maximum and the corresponding values of the stress, (b) verify that the points obtained are located on the neutral axis corresponding to the given loading.

SOLUTION

Refer to Sample Prob. 6.6.

(a) Along vertical leg: 2 23 3

3 ( )( 5 ) 3( 4 5 )

4 4e

P a y a y Pa ay y

ta taτ − += = + −

3

3(4 10 ) 0

4ed P

a ydy ta

τ= − =

2

5y a=

2

2 23 3

3 2 2 3 9(4 ) (5)

5 5 54 4m

P Pa a a a a

ta taτ

= + − =

27

20mP

taτ =

Along horizontal leg: 2 23 3

3 ( )( 3 ) 3( 4 3 )

4 4f

P a z a z Pa az z

ta taτ − −= = − +

3

3( 4 6 ) 0

4

fd Pa z

dz ta

τ= − + =

2

3z a=

2

2 2m 3 3

3 2 2 3 5(4 ) (3)

3 3 34 4

P Pa a a a a

ta taτ

= − + = −

m1

4

P

taτ = −

At the corner: 0, 0,y z= = 3

4

P

taτ =

(b) 3 31 145

3 12y zI ta I ta θ′ ′= = = °

1

tan tan 14.0364

z

y

I

Iϕ θ ϕ′

′= = = °

45 14.036 30.964

( /2) 1

2 4( /2) 1

2 4

Ay at ay a

A atAz at a

z aA at

θ ϕ− = − = °Σ= = =Σ

Σ= = =Σ




without permission.


Neutral axis intersects vertical leg at

tan 30.964y y z= + °

1 1

tan 30.964 0.4004 4

a a = + ° =

2

5y a=

Neutral axis intersects horizontal leg at

tan (45 + )

1 1tan 59.036 0.667

4 4

z z y

a a

ϕ= + °

= + ° =

2

3z a=




without permission.

PROBLEM 6.81*

A steel plate, 160 mm wide and 8 mm thick, is bent to form the channel shown. Knowing that the vertical load P acts at a point in the midplane of the web of the channel, determine (a) the torque T that would cause the channel to twist in the same way that it does under the load P, (b) the maximum shearing stress in the channel caused by the load P.

SOLUTION

Use results of Example 6.06 with 30 mm,b = 100 mm, and 8 mm.h t= =

3100

3 (3)(30)

2 2 6 4 6 4

3

309.6429 mm 9.6429 10 m

2 2

1 1(6 ) (8)(100) [(16)(30) 100] 1.86667 10 mm 1.86667 10 m

12 12

15 10 N

hb

be

I t h b h

V

−

−

= = = = ×+ +

= + = + = × = ×

= ×

(a) 3 3(15 10 )(9.6429 10 )T Ve −= = × × 144.64 N mT = ⋅

Stress at neutral axis due to V:

[ ] 3 3 6 3

3

1( 4 )

2 2 4 8

1(8)(100) 100 (4)(30) 22 10 mm 22 10 m

8

8 10 m

h h hQ bt t th h b

t

−

−

= + = +

= + = × = ×

= ×

3 6

66 3

(15 10 )(22 10 )22.10 10 Pa 22.10 MPa

(1.86667 10 )(8 10 )V

VQ

Itτ

−

− −× ×= = = × =

× ×

Stress due to T : 2 160 mm 0.160 ma b h= + = =

1

62 3 2

1

1 1 81 0.630 1 (0.630) 0.3228

3 3 160

144.6443.76 10 Pa 43.76 MPa

(0.3228)(0.160)(8 10 )τ −

= − = − =

= = = × =×V

tc

a

T

c at

(b) By superposition, max V Tτ τ τ= + max 65.9 MPaτ =




without permission.

PROBLEM 6.82*

Solve Prob. 6.81, assuming that a 6-mm-thick plate is bent to form the channel shown.

PROBLEM 6.81* A steel plate, 160 mm wide and 8 mm thick, is bent to form the channel shown. Knowing that the vertical load P acts at a point in the midplane of the web of the channel, determine (a) the torque T that would cause the channel to twist in the same way that it does under the load P, (b) the maximum shearing stress in the channel caused by the load P.

SOLUTION

Use results of Example 6.06 with 30 mm,b = 100 mm, and 6 mm.h t= =

3100

3 (3)(30)

2 2 6 4 6 4

3

309.6429 mm 9.6429 10 m

2 2

1 1(6 ) (6)(100) [(6)(30) 100] 1.400 10 mm 1.400 10 m

12 12

15 10 N

hb

be

I t h b h

V

−

−

= = = = ×+ +

= + = + = × = ×

= ×

(a) 3 3(15 10 )(9.6429 10 )T Ve −= = × × 144.64 N mT = ⋅

Stress at neutral axis due to V:

[ ] 3 3 6 3

3

1( 4 )

2 2 4 8

1(6)(100) 100 (4)(30) 16.5 10 mm 16.5 10 m

8

6 10 m

h h hQ bt t th h b

t

−

−

= + = +

= + = × = ×

= ×

3 6

66 6

(15 10 )(16.5 10 )29.46 10 Pa 29.46 MPa

(1.400 10 )(6 10 )V

VQ

Itτ

−

− −× ×= = = × =

× ×

Stress due to T : 2 160 mm 0.160 ma b h= + = =

1

62 3 2

1

1 1 61 0.630 1 (0.630) 0.32546

3 3 160

144.6477.16 10 Pa 77.16 MPa

(0.32546)(0.160)(6 10 )τ −

= − = − =

= = = × =×V

tc

a

T

c a t

(b) By superposition, max V Tτ τ τ= + max 106.6 MPaτ =




without permission.

PROBLEM 6.83∗

The cantilever beam AB, consisting of half of a thin-walled pipe of 1.25-in. mean radius and 3

8 -in. wall thickness, is subjected to a 500-lb vertical load. Knowing that the line of action of the load passes through the centroid C of the cross section of the beam, determine (a) the equivalent force-couple system at the shear center of the cross section, (b) the maximum shearing stress in the beam. (Hint: The shear center O of this cross section was shown in Prob. 6.73 to be located twice as far from its vertical diameter as its centroid C.)

SOLUTION

From the solution to Prob. 6.73,

3 2

2max

sin 24

I a t Q a t

e a Q a t

π θ

π

= =

= =

For a half-pipe section, the distance from the center of the semi-circle to the centroid is

2

x aπ

=

At each section of the beam, the shearing force V is equal to P. Its line of action passes through the centroid C. The moment arm of its moment about the shear center O is

4 2 2

d e x a a aπ π π

= − = − =

(a) Equivalent force-couple system at O.

02

PaV P M Vdπ

= = =

Data: 500 lb = 1.25 in.P a=

500 lbV =

02

(500)(1.25)Mπ

=

0 398 lb inM = ⋅




without permission.

PROBLEM 6.83* (Continued)

(b) Shearing stresses.

(1) Due to max: VVQ

VIt

τ =

2

3

( )( ) 2 (2)(500)679 psi

(1.25)(0.375)( )

2

VP a t P

ata t t

τπ π π

= = = =

(2) Due to the torque. 0 ,M

For a long rectangular section of length l and width t, the shearing stress due to torque 0M is

012

1

1where 1 0.630

3τ = = −

M

M tc

lc lt

Data: 1(1.25) 3.927 in. 0.375 in. 0.31328π π= = = = =l a t c

2

397.92300 psi

(0.31328)(3.927)(0.375)Mτ = =

By superposition, 679 psi 2300 psiV Mτ τ τ= + = + 2980 psiτ =




without permission.

PROBLEM 6.84∗

Solve Prob. 6.83, assuming that the thickness of the beam is reduced to 1

4 in.

PROBLEM 6.83∗ The cantilever beam AB, consisting of half of a thin-walled pipe of 1.25-in. mean radius and 3

8 -in. wall thickness, is subjected to a 500-lb vertical load. Knowing that the line of action of the load passes through the centroid C of the cross section of the beam, determine (a) the equivalent force-couple system at the shear center of the cross section, (b) the maximum shearing stress in the beam. (Hint: The shear center O of this cross section was shown in Prob. 6.73 to be located twice as far from its vertical diameter as its centroid C.)

SOLUTION

From the solution to Prob. 6.73,

3 2

2max

sin

4

I a t Q a t

e a Q a t

π θ

π

= =

= =

For a half-pipe section, the distance from the center of the semi-circle to the centroid is

2

x aπ

=

At each section of the beam, the shearing force V is equal to P. Its line of action passes through the centroid C. The moment arm of its moment about the shear center O is

4 2 2

d e x a a aπ π π

= − = − =

(a) Equivalent force-couple system at O.

02

PaV P M Vdπ

= = =

Data: 500 lb 1.25 in. P a= =

500 lbV =

0 398 lb inM = ⋅




without permission.


(b) Shearing stresses.

(1) Due to max, VVQ

VIt

τ =

2

3

( )( ) 2 (2)(500)1019 psi

(1.25)(0.250)( )

2

VP a t P

ata t t

τπ π π

= = = =

(2) Due to the torque. 0 ,M

For a long rectangular section of length l and width t, the shearing stress due to torque 0M is

012

1

1where = 1 0.630

3τ = −

M

M tc

lc lt

Data: 1

2

(1.25) 3.927 in. 0.250 in. 0.31996

397.95067 psi

(0.31996)(3.927)(0.250)

π π

τ

= = = = =

= =M

l a t c

By superposition 1019 psi 5067 psiV Mτ τ τ= + = + 6090 psiτ =




without permission.

PROBLEM 6.85∗

The cantilever beam shown consists of a Z-shape of 1

4 -in. thickness. For the given loading, determine the distribution of the shearing stresses along line A B′ ′ in the upper horizontal leg of the Z-shape. The x′ and y′ axes are the principal centroidal axes of the cross section and the corresponding moments of inertia are 4166.3 inxI ′ = and

413.61 in .yI ′ =

SOLUTION

3 kips 22.5

sin cos x y

V

V V V V

ββ β′ ′

= = °= =

In upper horizontal leg, use coordinate : ( 6 in 0)x x− � �

1(6 ) in.

41

( 6 ) in.26 in.

cos + sin

cos sin

A x

x x

y

x x y

y y x

β ββ β

= +

= − +

=′ =′ = −

Due to :xV ′ 1x

y

V Ax

I tτ ′ ′

=

1

1 1( sin ) (6 ) ( 6 )cos 6sin

4 21

(13.61)4

0.084353(6 )( 0.47554 0.46194 )

V x x

x x

β β βτ

+ − + + =

= + − +

Due to :yV ′ 2

1 1( cos ) (6 ) 6 cos ( 6 )sin

4 21

(166.3)4

0.0166665(6 )[6.69132 0.19134 ]

y

x

V x xV Ay

I t

x x

β β βτ ′

+ − − + ′ = =

= + −

Total: 1 2 (6 )[ 0.07141 0.035396 ]x xτ τ+ = + − +

(in)x 6− 5− 4− 3− 2− 1− 0

(ksi)τ 0 0.105− 0.140− 0.104− 0.003 0.180 0.428




without permission.

PROBLEM 6.86∗

For the cantilever beam and loading of Prob. 6.85, determine the distribution of the shearing stress along line B D′ ′ in the vertical web of the Z-shape.

PROBLEM 6.85∗ The cantilever beam shown consists of a Z-shape of 1

4 -in. thickness. For the given loading, determine the distribution of the shearing stresses along line A B′ ′ in the upper horizontal leg of the Z-shape. The x′ and y′ axes are the principal centroidal axes of the cross section and the corresponding moments of inertia are

4166.3 inxI ′ = and 413.61 in .yI ′ =

SOLUTION

3 kips 22.5

sin cos x y

V

V V V V

ββ β′ ′

= = °= =

For part AB’ 21(6) 1.5 in

4

3 in., 6 in.

A

x y

= =

= − =

For part B′Y, 1

(6 )4

10 (6 )

2cos sin

cos sin

A y

x y y

x x y

y y x

β ββ β

= −

= = +

′ = +′ = −

Due to :xV ′

( )

1

1 14 2

1 14

22

( )

( sin )[(1.5)( 3cos 6sin ) (6 ) (6 ) sin ]

(13.61)

( sin )[ 0.7133 1.7221 0.047835 ] 0.3404 0.01614

3.4025

x AB AB BY BY

y

V A x A x

I t

V y y

V yy

τ

β β β βτ

β

′

′

′+=

− + + − +=

− + −= = −




without permission.


Due to :yV ′

( )

( )

2

1 14 2

2 14

22

14

( )

( cos )[(1.5)(6 cos 3 sin ) (6 ) (6 )cos ]

(166.3)

( cos )[10.037 4.1575 0.11548 ]0.9463 0.00770

(166.3)

y AB AB BY

x

V A y A y

I t

V y y

V yy

τ

β β β βτ

β

′

′

′ ′+=

+ + − +=

+ −= = −

Total: 21 2 1.2867 0.02384 yτ τ+ = −

(in.)y 0 ± 2 4± 6±

(ksi)τ 1.287 1.191 0.905 0.428




without permission.

PROBLEM 6.87∗

Determine the distribution of the shearing stresses along line D B′ ′ in the horizontal leg of the angle shape for the loading shown. The x′ and y′ axes are the principal centroidal axes of the cross section.

SOLUTION

15.8 cos sin

1( ) (2 ) (2 ), 0

2

x yV P V P

A y a y t y a y x

β β β′ ′= ° = = −

= − = + =

Coordinate transformation. 2 1

cos sin3 6

1 2cos sin

6 3

y y a x a

x x a y a

β β

β β

′ = − − − ′ = − + −

In particular, 2 1

cos sin3 6

1 1 1cos sin

2 3 6

0.48111 0.36612

y y a x a

y a a

y a

β β

β β

′ = − − − = + − −

= +

1 2cos sin

6 3

1 1 1cos sin

6 2 3

0.13614 0.06961

x x a y a

a y a

y a

β β

β β

′ = − + − = − + +

= −




without permission.

PROBLEM 6.87∗ (Continued)

yx

y x

V AyV Ax

I t I tτ ′′

′ ′

′′+ → = +

3

3

3

( cos )(2 )( )(0.13614 0.06961 )

(0.1557 )( )

( sin )(2 )(0.48111 0.36612 )

(1.428 )( )

(2 )(0.750 0.500 )

P a y t y a

ta t

P a y y a

a t t

P a y y a

ta

β

β

− −=

− − ++

− −=

( )y a 0 1

3

2

3 1

4

3

5

3 2

P

atτ −

1.000− 0.417− 0 0.250 0.333 0.250 0




without permission.

PROBLEM 6.88∗

For the angle shape and loading of Prob. 6.87, determine the distribution of the shearing stresses along line D A′ ′ in the vertical leg.

PROBLEM 6.87* Determine the distribution of the shearing stresses along line D B′ ′ in the horizontal leg of the angle shape for the loading shown. The x′ and y′ axes are the principal centroidal axes of the cross section.

SOLUTION

15.8 cos

sin ( ) ( )

1( ), 0

2

x

x

V P

V P A x a x t

x a x y

β ββ

′

′

= ° == − − −

= + =

Coordinate transformation.

2 1

cos sin3 6

1 2cos sin

6 3

y y a x a

x x a y a

β β

β β

′ = − − − ′ = − + −

In particular,

2 1cos sin

3 6

2 1 1cos sin

3 2 3

0.13614 0.73224

1 2cos sin

6 3

1 1 2cos sin

2 3 3

0.48111 0.13922

y y a x a

a x a

x a

x x a y a

x a a

x a

β β

β β

β β

β β

′ = − − − = − − +

= − −

′ = − + − = + + −

= +




without permission.


3

3

3

( )( )

( cos )( )( )(0.48111 0.13922 )

(0.1557 )( )

( sin )( )( )( 0.13614 0.73224 )

(1.428 )( )

( )(3.00 1.000 )

yx

y x

V A x yV A x x

I t I t

P a x t x a

ta t

P a x t x a

ta t

P a x x a

ta

τ

β

β

′′

′

′′= −

′

− +=

− − − −+

− +=

( )x a 0 1

6

1

3

1

2

2

3

5

6 1

P

atτ ↓

1.000 1.250 1.333 1.250 1.000 0.583 0




without permission.

PROBLEM 6.89

A square box beam is made of two 20 80-mm× planks and two 20 120-mm× planks nailed together as shown. Knowing that the spacing between the nails is 30s = mm and that the vertical shear in the beam is 1200 N,V = determine (a) the shearing force in each nail, (b) the maximum shearing stress in the beam.

SOLUTION

3 32 2 1 1

3 3 6 4

6 4

1 1

12 121 1

(120)(120) (80)(80) 13.8667 10 mm12 12

13.8667 10 m

I b h b h

−

= −

= − = ×

= ×

(a) 21

1

3 3 6 31 1 1

(120)(20) 2400 mm

50 mm

120 10 mm 120 10 m

A

y

Q A y −

= ==

= = × = ×

63

6

nail

(1200)(120 10 )10.385 10 N/m

13.8667 102

VQq

Iqs F

−

−×= = = ×

×=

3 3

nail(10.385 10 )(30 10 )

2 2

qsF

−× ×= = nail 155.8 NF =

(b) 1

3 3 3 3

6 3

(2)(20)(40)(20)

120 10 32 10 152 10 mm

152 10 m

Q Q

−

= +

= × + × = ×

= ×

6

max 6 3

(1200)(152 10 )

(13.8667 10 )(2 20 10 )

VQ

Itτ

−

− −×= =

× × ×

3329 10 Pa= × max 329 kPaτ =




without permission.

PROBLEM 6.90

The beam shown is fabricated by connecting two channel shapes and two plates, using bolts of 3

4-in. diameter spaced longitudinally every 7.5 in.

Determine the average shearing stress in the bolts caused by a shearing force of 25 kips parallel to the y-axis.

SOLUTION

4C12 20.7: 12.00 in., 129 inxd I× = =

For top plate,

32 4

12.00 1 16.25 in.

2 2 2

1 1 1(16) (16) (6.25) 312.667 in

12 2 2t

y

I

= + =

= + =

For bottom plate, 4312.667 inbI =

Moment of inertia of fabricated beam:

4

3plate plate

bolt

22 2

bolt bolt

(2)(129) 312.667 312.667

883.33 in

1 (16) (6.25) 50 in

2

(25)(50)1.41510 kips/in

883.331 1

(1.41510)(7.5) 5.3066 kips2 2

3( ) 0.44179 in

4 4 4

I

Q A y

VQq

I

F qs

A dπ π

τ

= + +

=

= = =

= = =

= = =

= = =

boltbolt

bolt

5.306612.01 ksi

0.44179

F

A= = = bolt 12.01 ksiτ =




without permission.

PROBLEM 6.91


SOLUTION

0: 2.3 (1.5)(72) 0

46.957 kN

BM AΣ = − + =

= ↑A

At section n-n, 46.957 kNV A= =


3 3 3

6 4 6 4

1 1 12 (15)(40) 2 (15)(80) (30)(120 )

12 12 12

5.76 10 mm 5.76 10 m

I

−

= + + = × = ×

At a, 3 3

6 3

3 6

6

6

30 mm 0.030 m

(30 20)(50) 30 10 mm

30 10 m

(46.957 10 )(30 10 )

(5.76 10 )(0.030)

8.15 10 Pa = 8.15 MPa

a

a

aa

a

t

Q

VQ

Itτ

−

−

−

= =

= × = ×

= ×

× ×= =×

= ×

At b, 3 3 3 3 6 4

3 66

6

60 mm 0.060 m

(60 20)(30) 30 10 36 10 66 10 mm 66 10 m

(46.957 10 )(66 10 )8.97 10 Pa 8.97 MPa

(5.76 10 )(0.060)

b

b a

bb

b

t

Q Q

VQ

Itτ

−

−

−

= =

= + × = × + × = × = ×

× ×= = = × =×

At NA, NA

3 3 3 3 6 3NA

3 66NA

NA 6NA

90 mm 0.090 m

(90 20)(10) 66 10 18 10 84 10 mm 84 10 m

(46.957 10 )(84 10 )7.61 10 Pa 7.61 MPa

(5.76 10 )(0.090)

b

t

Q Q

VQ

Itτ

−

−

−

= =

= + × = × + × = × = ×

× ×= = = × =×

(a) maxτ occurs at b. max 8.97 MPaτ =

(b) 8.15 MPaaτ =




without permission.

PROBLEM 6.92

For the beam and loading shown, determine the minimum required width b, knowing that for the grade of timber used, all 12 MPaσ = and all 825 kPa.τ =

SOLUTION

0: 3 (2)(2.4) (1)(4.8) (0.5)(7.2) 0

2 kN

DM AΣ = − + + − =

= ↑A

Draw the shear and bending moment diagrams.

3

max

3max

| | 7.2 kN 7.2 10 N

| | 3.6 kN m 3.6 10 N m

V

M

= = ×

= ⋅ = × ⋅

Bending: M

Sσ =

maxmin

3

6

6 3

3 3

| |

3.6 10

12 10300 10 m

300 10 mm

MS

σ

−

=

×=×

= ×= ×

For a rectangular section, 21

6S bh=

3

2 2

6 (6)(300 10 )80 mm

(150)

Sb

h

×= = =

Shear: Maximum shearing stress occurs at the neutral axis of bending for a rectangular section.

2

3

218

3112

33

3 3

1 1 1, ,

2 4 81

12( ) 3

2( )( )

3 (3)(7.2 10 )87.3 10 m

2 (2)(150 10 )(825 10 )

τ

τ−

−

= = = =

= =

= = =

×= = = ×× ×

A bh y h Q Ay bh

I bh t b

V bhVQ V

It bhbh b

Vb

h

The required value of b is the larger one. 87.3 mmb =




without permission.

PROBLEM 6.93


SOLUTION

25 kipsA BR R= =

At section n-n, 25 kipsV =

Locate centroid and calculate moment of inertia.


4.875 6.875 33.52 2.244 24.55 0.23

10.875 3.625 39.42 1.006 11.01 47.68

Σ 15.75 72.94 35.56 47.86

2 4

72.944.631 in.

15.75

35.56 47.86 83.42 in

AyY

A

I Ad I

Σ= = =Σ

= Σ + Σ = + =

(a) 33(1.5)(4.631 0.75) 4.366 in

4

30.75 in.

4

aQ Ay

t

= = − =

= =

(25)(4.366)

(83.42)(0.75)aVQ

Itτ = = 1.745 ksiaτ =

(b) 33(3)(4.631 1.5) 7.045 in

4

0.75 in.

bQ Ay

t

= = − =

=

(25)(7.045)

(83.42)(0.75)bVQ





without permission.

PROBLEM 6.94


SOLUTION

25 kipsA BR R= =

At section n-n, 25 kips.V =

Locate centroid and calculate moment of inertia.


4.875 6.875 33.52 2.244 24.55 0.23

10.875 3.625 39.42 1.006 11.01 47.68

Σ 15.75 72.94 35.56 47.86

2 4

72.944.631 in.

15.75

35.56 47.86 83.42 in

AyY

A

I Ad I

Σ= = =Σ

= Σ + Σ = + =

Largest shearing stress occurs on section through centroid of entire cross section.

33 4.631(4.631) 8.042 in

4 2

30.75 in.

4(25)(8.042)

(83.42)(0.75)

Q Ay

t

VQ

Itτ

= = =

= =

= = 3.21 ksimτ =




without permission.

PROBLEM 6.95

The composite beam shown is made by welding C200 17.1× rolled-steel channels to the flanges of a W250 80× wide-flange rolled-steel shape. Knowing that the beam is subjected to a vertical shear of 200 kN, determine (a) the horizontal shearing force per meter at each weld, (b) the shearing stress at point a of the flange of the wide-flange shape.

SOLUTION

For 6 4W250 80, 257 mm, 15.6 mm, 126 10 mm× = = = ×f xd t I

For 2C200 17.1, 2170mm , 57.4 mm, 9.91 mmf fA b t× = = =

6 40.545 10 mm , 14.5 mmyI x= × =

For the channel in the composite beam,

257

57.4 14.5 171.4 mm2cy = + − =

For the composite beam,

6 6 2

6 4 6 4

126 10 +2 0.545 10 (2170)(171.4)

254.59 10 mm 254.59 10 m

I

−

= × × +

= × = ×

(a) For the two welds,

3 3 6 3(2170) (171.4) 371.94 10 mm 371.94 10 mw cQ Ay −= = = × = ×

3 6

36

(200 10 ) (371.94 10 )292.2 10 N/m

254.59 10

VQq

I

−

−× ×= = = ×

×

For one weld, 3146.1 10 N m2

q = ×

Shearing force per meter of weld: 146.1 kN m

(b) For cuts at a and 'a together,

2

3 3 3 6 3

257 15.62(112) (15.6) 3494.4 mm 120.7 mm

2 2

371.94 10 (3494.4)(120.7) 793.71 10 mm 793.71 10 m

a a

a

A y

Q −

= = = − =

= × + = × = ×

Since there are cuts at a and ',a 2 (2)(15.6) 31.2 mm 0.0312 m.ft t= = = =

3 66

6

(200 10 )(793.71 10 )19.99 10 Pa

(254.59 10 )(0.0312)a

aVQ

Itτ

−× ×= = = ×× 19.99 MPaaτ =




without permission.

PROBLEM 6.96

An extruded beam has the cross section shown and a uniform wall thickness of 3 mm. For a vertical shear of 10 kN, determine (a) the shearing stress at point A, (b) the maximum shearing stress in the beam. Also, sketch the shear flow in the cross section.

SOLUTION

16

tan 28.0730

α α= = °

Side: 2

3 3 4

(3 sec )(30) 102 mm

1(3 sec )(30) 7.6498 10 mm

12

A

I

α

α

= =

= = ×

Part 2(mm )A 0 (mm)y 3 3(10 mm )Ay (mm)d 2 3 4(10 mm )Ad 3 4(10 mm )I

Top 180 30 5.4 11.932 25.627 neglect

Side 102 15 1.53 3.077 0.966 7.6498

Side 102 15 1.53 3.077 0.966 7.6498

Bot 84 0 0 18.077 27.449 neglect

Σ 468 8.46 55.008 15.2996

3

0

2 3 4 9 4

8.46 1018.077 mm

468

70.31 10 mm 70.31 10 m

AyY

A

I Ad I −

Σ ×= = =Σ

= Σ + Σ = × = ×

(a) 3 3 6 3(180)(11.932) 2.14776 10 mm 2.14776 10 mAQ −= = × = ×

3 3

3 66

9 6

(2)(3 10 ) 6 10 m

(10 10 )(2.14776 10 )50.9 10 Pa

(70.31 10 )(6 10 )A

t

VQ

Itτ

− −

−

− −

= × = ×

× ×= = = ×× ×

50.9 MPaτ =A

(b)

3 3 3

6 3

3

1(2)(3 sec )(11.932) 11.932

2

2.14776 10 484.06 2.6318 10 mm

2.6318 10 m

6 10 m

m AQ Q

t

α

−

−

= + ×

= × + = ×

= ×

= ×

3 6

69 3

(10 10 )(2.6318 10 )62.4 10 Pa

(70.31 10 )(6 10 )m

mVQ

Itτ

−

− −× ×= = = ×

× × 62.4 MPamτ =




without permission.


3 3

3

3

(28)(3)(18.077) 1.51847 10 mm

1.51847 10(50.9)

2.14776 10

36.0 MPa

B

BB A

A

Q

Q

Qτ τ

= = ×

×= =×

=

Multiply shearing stresses by (3 mm 0.003 m)t = to get shear flow.




without permission.

PROBLEM 6.97

The design of a beam requires welding four horizontal plates to a vertical 0.5 5-in.× plate as shown. For a vertical shear V, determine the dimension h for which the shear flow through the welded surface is maximum.

SOLUTION

Horizontal plate:

3 2

2

1(4.5)(0.5) (4.5)(0.5)

12

0.046875 2.25

hI h

h

= +

= +

Vertical plate: 3 41(0.5)(5) 5.2083 in

12vI = =

Whole section: 2 44 9 5.39583 inh vI I I h= + = +

For one horizontal plate, 3(4.5)(0.5) 2.25 inQ h h= =

2

2.25

9 5.39583

VQ Vhq

I h= =

+

To maximize q, set 0.dq

dh=

2 2

2 2

(9 5.39583) 182.25 0

(9 5.39583)

h hV

h

+ − =+

0.774 in.h =




without permission.

PROBLEM 6.98


SOLUTION

23 2

3 2

1 1( ) ( ) ( )

2 12 4

1 1(6 6 )

12 12

AB EF

DG

hI I a b t a b t t a b h

I th I I t a b h h

= = + + + ≈ +

= = Σ = + +

Part AD:

10 0

2 2

0

1

2 2

2

2 2

2 2 4

a a

a

hQ tx thx

VQ Vhx

It IVhx Vht

F dA tdx xdxI I

Vht x Vhta

I I

τ

τ

= =

= =

= = =

= =

Part BD: 1

2 2

2

hQ tx thx

VQ Vhx

It Iτ

= =

= =

20 0

2 2

0

2 2

2 2 4

b b

b

Vhx VhtF dA tdx xdx

I I

Vht x Vhtb

I I

τ= = =

= =

HMΣ = :HMΣ

2 2 2

2 1

2 2 2 2 2

2112

( )

4

( ) 3 ( )

6 64 (6 6 )

Vh t b aVe F h F h

I

Vh t b a V b a

a b ht a b h h

−= − =

− −= =+ +⋅ + +

2 23( )

6( )

b ae

a b h

−=+ +




without permission.

PROBLEM 6.99


SOLUTION

3 4

2

2 2 4

4

1 1(1.5) 0.28125 in

3 4

13 in. (3) 0.75 in

4

1 1(0.75)(1.5) 0.5625 in

3 3

(2)(0.28125) (2) (0.5625) 1.6875 in

AB

BD BD

BD BD

I

L A

I A h

I

= =

= = =

= = =

= + =

Part AB: 2

2 2

21.5

10

1.53 3

0

1 1 1

4 2 8

(8) (1.6875) (0.25) 3.375

(0.25 )3.375

(0.25) (0.25) (1.5)

3.375 3 (3.375) (3)

0.08333

A y y y Q Ay y

VQ Vy Vy

It

VyF dA dy

V y

V

τ

τ

= = = =

= = =

= =

= =

=

DM = 1: 2 (3 sin 60 )

(2) (0.08333) (3 sin 60 )DM Ve F

Ve V

= °= °

(2)(0.08333)(3 sin 60 )e = ° 0.433 in.e =




without permission.

PROBLEM 6.100


SOLUTION

Part AB:

60

10

23

, 100 mm, 100

100

100

(100)(60)180 10

2

B

A

x

x

A tx y Q tx

VQ Vtxq

I IVt

F qdx x dxI

Vt Vt

I I

= = =

= =

= =

= = ×

Part DE:

20

2

, 80 mm, 80

80

80

(40 )

D

x b

x

A tx y Q tx

VQ Vtxq

I IVt

F qdx xdxI

Vtb

I

= = =

= =

= =

=

3 20 0: (200)(180 10 ) (160)(40 ) 0

Vt VtM b

I IΣ = × − =

3

2 2(200)(180 10 )5625 mm

(160)(40)b

×= = 75.0 mmb =

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