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11–1. The simply supported beam is made of timber that

has an allowable bending stress of sallow = 6.5 MPa and an

allowable shear stress of tallow = 500 kPa. Determine its

dimensions if it is to be rectangular and have a height-towidth ratio of 1.25.

8 kN/m

2m

Ix =

1

(b)(1.25b)3 = 0.16276b4

12

Qmax = y¿A¿ = (0.3125b)(0.625b)(b) = 0.1953125b3

Assume bending moment controls:

Mmax = 16 kN # m

sallow =

Mmax c

I

6.5(106) =

16(103)(0.625b)

0.16276b4

b = 0.21143 m = 211 mm

Ans.

h = 1.25b = 264 mm

Ans.

Check shear:

Qmax = 1.846159(10 - 3) m3

I = 0.325248(10 - 3) m4

tmax =

VQmax

16(103)(1.846159)(10 - 3)

= 429 kPa 6 500 kPa‚ OK

=

It

0.325248(10 - 3)(0.21143)

830

4m

2m

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11–2. The brick wall exerts a uniform distributed load

of 1.20 kip>ft on the beam. If the allowable bending stress

is sallow = 22 ksi and the allowable shear stress is

tallow = 12 ksi, select the lightest wide-flange section with

the shortest depth from Appendix B that will safely support

the load.

1.20 kip/

4 ft

10 ft

ft

6 ft

b

Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft. Assuming

bending controls the design and applying the flexure formula.

Sreq d =

=

44.55 (12)

= 24.3 in3

22

W12 * 22

A Sx = 25.4 in3, d = 12.31 in., tw = 0.260 in. B

V

for the W12 * 22 wide tw d

= 6.60 kip.

Shear Stress: Provide a shear stress check using t =

flange section. From the shear diagram, Vmax

tmax =

=

Vmax

tw d

6.60

0.260(12.31)

= 2.06 ksi 6 tallow = 12 ksi (O.K!)

Hence,

Use

9 in.

0.5 in.

Mmax

sallow

Two choices of wide flange section having the weight 22 lb>ft can be made. They

are W12 * 22 and W14 * 22. However, W12 * 22 is the shortest.

Select

0.5 in.

0.5 in.

Ans.

W12 * 22

831

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11–3. The brick wall exerts a uniform distributed load

of 1.20 kip>ft on the beam. If the allowable bending stress

is sallow = 22 ksi, determine the required width b of the

flange to the nearest 14 in.

1.20 kip/

4 ft

10 ft

ft

6 ft

b

0.5 in.

0.5 in.

9 in.

0.5 in.

Section Property:

I =

1

1

(b) A 103 B (b - 0.5) A 93 B = 22.583b + 30.375

12

12

Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft.

sallow =

22 =

Mmax c

I

44.55(12)(5)

22.583b + 30.375

b = 4.04 in.

Use

b = 4.25 in.

Ans.

832

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*11–4. Draw the shear and moment diagrams for the

shaft, and determine its required diameter to the nearest

1

4 in. if sallow = 7 ksi and tallow = 3 ksi. The bearings at A

and D exert only vertical reactions on the shaft. The loading

is applied to the pulleys at B, C, and E.

14 in.

20 in.

15 in.

12 in.

E

A

C

B

D

35 lb

80 lb

110 lb

sallow =

7(103) =

Mmax c

I

1196 c

p 4 ;

4 c

c = 0.601 in.

d = 2c = 1.20 in.

Use d = 1.25 in.

Ans.

Check shear:

2

tmax =

0.625

108(4(0.625)

Vmax Q

3p )(p)( 2 )

= 117 psi 6 3 ksi OK

=

p

4

It

4 (0.625) (1.25)

•11–5.

Select the lightest-weight steel wide-flange beam

from Appendix B that will safely support the machine loading

shown. The allowable bending stress is sallow = 24 ksi and

the allowable shear stress is tallow = 14 ksi.

2 ft

Bending Stress: From the moment diagram, Mmax = 30.0 kip # ft.

Assume bending controls the design. Applying the flexure formula.

Sreq¿d =

=

Select

W12 * 16

Mmax

sallow

30.0(12)

= 15.0 in3

24

A Sx = 17.1 in3, d = 11.99 in., tw = 0.220 in. B

V

for the W12 * 16 wide tw d

= 10.0 kip

Shear Stress: Provide a shear stress check using t =

flange section. From the shear diagram, Vmax

tmax =

=

Vmax

tw d

10.0

0.220(11.99)

= 3.79 ksi 6 tallow = 14 ksi (O.K!)

Hence,

Use

5 kip

5 kip

Ans.

W12 * 16

833

2 ft

5 kip

2 ft

5 kip

2 ft

2 ft

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11–6. The compound beam is made from two sections,

which are pinned together at B. Use Appendix B and select

the lightest-weight wide-flange beam that would be safe for

each section if the allowable bending stress is sallow = 24 ksi

and the allowable shear stress is tallow = 14 ksi. The beam

supports a pipe loading of 1200 lb and 1800 lb as shown.

C

A

B

6 ft

Bending Stress: From the moment diagram, Mmax = 19.2 kip # ft for member AB.

Assuming bending controls the design, applying the flexure formula.

Sreq¿d =

=

Select

Mmax

sallow

19.2(12)

= 9.60 in3

24

A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in. B

W10 * 12

For member BC, Mmax = 8.00 kip # ft.

Sreq¿d =

=

Select

Mmax

sallow

8.00(12)

= 4.00 in3

24

A Sx = 5.56 in3, d = 5.90 in., tw = 0.17 in. B

W6 * 9

V

for the W10 * 12 widetw d

flange section for member AB. From the shear diagram, Vmax = 2.20 kip.

Shear Stress: Provide a shear stress check using t =

tmax =

=

Vmax

tw d

2.20

0.19(9.87)

= 1.17 ksi 6 tallow = 14 ksi (O.K!)

Use

Ans.

W10 * 12

For member BC (W6 * 9), Vmax = 1.00 kip.

tmax =

=

Vmax

tw d

1.00

0.17(5.90)

= 0.997 ksi 6 tallow = 14 ksi (O.K!)

Hence,

Use

1800 lb

1200 lb

W6 * 9

Ans.

834

6 ft

8 ft

10 ft

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Page 835

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11–7. If the bearing pads at A and B support only vertical

forces, determine the greatest magnitude of the uniform

distributed loading w that can be applied to the beam.

sallow = 15 MPa, tallow = 1.5 MPa.

w

A

B

1m

1m

150 mm

25 mm

150 mm

25 mm

The location of c, Fig. b, is

y =

0.1625(0.025)(0.15) + 0.075(0.15)(0.025)

©yA

=

©A

0.025(0.15) + 0.15(0.025)

= 0.11875 m

I =

+

1

(0.025)(0.153) + (0.025)(0.15)(0.04375)2

12

1

(0.15)(0.0253) + 0.15(0.025)(0.04375)2

12

= 21.58203125(10 - 6) m4

Referring to Fig. b,

Qmax = y¿A¿ = 0.059375 (0.11875)(0.025)

= 0.176295313(10 - 4) m3

Referring to the moment diagram, Mmax = 0.28125 w. Applying the Flexure

formula with C = y = 0.11875 m,

sallow =

Mmax c

;

I

15(106) =

0.28125w(0.11875)

21.582(10 - 6)

W = 9.693(103) N>m

Referring to shear diagram, Fig. a, Vmax = 0.75 w.

tallow =

Vallow Qmax

;

It

1.5(106) =

0.75w C 0.17627(10 - 3) D

21.582(10 - 6)(0.025)

W = 6.122(103) N>m

= 6.12 kN>m (Control!)

Ans.

835

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Page 836

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*11–8. The simply supported beam is made of timber that

has an allowable bending stress of sallow = 1.20 ksi and an

allowable shear stress of tallow = 100 psi. Determine its

smallest dimensions to the nearest 18 in. if it is rectangular

and has a height-to-width ratio of 1.5.

12 kip/ft

B

A

3 ft

3 ft

1.5 b

b

The moment of inertia of the beam’s cross-section about the neutral axis is

1

(b)(1.5b)3 = 0.28125b4. Referring to the moment diagram,

I =

12

Mmax = 45.375 kip # ft.

sallow =

Mmax c

;

I

1.2 =

45.375(12)(0.75b)

0.28125b4

b = 10.66 in

Referring to Fig. b, Qmax = y¿A¿ = 0.375b (0.75b)(b) = 0.28125b3. Referring to the

shear diagram, Fig. a, Vmax = 33 kip.

tmax =

Vmax Qmax

;

It

100 =

33(103)(0.28125b3)

0.28125b4(b)

b = 18.17 in (Control!)

Thus, use

b = 18

1

in

4

Ans.

836

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•11–9.

Select the lightest-weight W12 steel wide-flange

beam from Appendix B that will safely support the loading

shown, where P = 6 kip. The allowable bending stress

is sallow = 22 ksi and the allowable shear stress is

tallow = 12 ksi.

P

P

9 ft

From the Moment Diagram, Fig. a, Mmax = 54 kip # ft.

Mmax

sallow

Sreq¿d =

54(12)

22

=

= 29.45 in3

Select W12 * 26

C Sx = 33.4 in3, d = 12.22 in and tw = 0.230 in. D

From the shear diagram, Fig. a, Vmax = 7.5 kip. Provide the shear-stress check

for W 12 * 26,

tmax =

=

Vmax

tw d

7.5

0.230(12.22)

= 2.67 ksi 6 tallow = 12 ksi (O.K!)

Hence

Use

Ans.

W12 * 26

837

6 ft

6 ft

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Page 838

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11–10. Select the lightest-weight W14 steel wide-flange

beam having the shortest height from Appendix B that

will safely support the loading shown, where P = 12 kip.

The allowable bending stress is sallow = 22 ksi and the

allowable shear stress is tallow = 12 ksi.

P

P

9 ft

From the moment diagram, Fig. a, Mmax = 108 kip # ft.

Mmax

sallow

Sreq¿d =

108(12)

22

=

= 58.91 in3

Select W14 * 43

C Sx = 62.7 in3, d = 13.66 in and tw = 0.305 in. D

From the shear diagram, Fig. a, Vmax = 15 kip . Provide the shear-stress check

for W14 * 43 ,

tmax =

=

Vmax

tw d

15

0.305(13.66)

= 3.60 ksi 6 tallow = 12 ksi‚ (O.K!)

Hence,

Use

Ans.

W14 * 43

838

6 ft

6 ft

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11–11. The timber beam is to be loaded as shown. If the ends

support only vertical forces, determine the greatest magnitude

of P that can be applied. sallow = 25 MPa, tallow = 700 kPa.

150 mm

30 mm

120 mm

40 mm

P

4m

A

y =

(0.015)(0.150)(0.03) + (0.09)(0.04)(0.120)

= 0.05371 m

(0.150)(0.03) + (0.04)(0.120)

I =

1

1

(0.150)(0.03)3 + (0.15)(0.03)(0.05371 - 0.015)2 +

(0.04)(0.120)3 +

12

12

B

(0.04)(0.120)(0.09 - 0.05371)2 = 19.162(10 - 6) m4

Maximum moment at center of beam:

Mmax =

P

(4) = 2P

2

Mc

;

I

s =

25(106) =

(2P)(0.15 - 0.05371)

19.162(10 - 6)

P = 2.49 kN

Maximum shear at end of beam:

Vmax =

P

2

VQ

;

t =

It

700(103) =

P 1

C (0.15 - 0.05371)(0.04)(0.15 - 0.05371) D

2 2

19.162(10 - 6)(0.04)

P = 5.79 kN

Thus,

P = 2.49 kN

Ans.

839

4m

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*11–12. Determine the minimum width of the beam to

the nearest 14 in. that will safely support the loading of

P = 8 kip. The allowable bending stress is sallow = 24 ksi

and the allowable shear stress is tallow = 15 ksi.

P

6 ft

6 ft

6 in.

B

A

Beam design: Assume moment controls.

sallow =

Mc

;

I

24 =

48.0(12)(3)

1

3

12 (b)(6 )

b = 4 in.

Ans.

Check shear:

8(1.5)(3)(4)

VQ

= 0.5 ksi 6 15 ksi OK

= 1

3

It

12 (4)(6 )(4)

tmax =

•11–13.

Select the shortest and lightest-weight steel wideflange beam from Appendix B that will safely support the

loading shown.The allowable bending stress is sallow = 22 ksi

and the allowable shear stress is tallow = 12 ksi.

10 kip

6 kip

4 kip

A

B

4 ft

Beam design: Assume bending moment controls.

Sreq¿d =

60.0(12)

Mmax

=

= 32.73 in3

sallow

22

Select a W 12 * 26

Sx = 33.4 in3, d = 12.22 in., tw = 0.230 in.

Check shear:

tavg =

V

10.5

=

= 3.74 ksi 6 12 ksi

Aweb

(12.22)(0.230)

Use W 12 * 26

Ans.

840

4 ft

4 ft

4 ft

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11–14. The beam is used in a railroad yard for loading and

unloading cars. If the maximum anticipated hoist load is

12 kip, select the lightest-weight steel wide-flange section

from Appendix B that will safely support the loading. The

hoist travels along the bottom flange of the beam,

1 ft … x … 25 ft, and has negligible size. Assume the beam

is pinned to the column at B and roller supported at A.

The allowable bending stress is sallow = 24 ksi and

the allowable shear stress is tallow = 12 ksi.

x

27 ft

A

B

12 kip

15 ft

C

Maximum moment occurs when load is in the center of beam.

Mmax = (6 kip)(13.5 ft) = 81 lb # ft

sallow =

M

;

S

24 =

81(12)

Sreq¿d

Sreq¿d = 40.5 in3

Select a W 14 * 30, Sx = 42.0 in3, d = 13.84 in, tw = 0.270 in.

At x = 1 ft, V = 11.56 kip

t =

11.36

V

=

= 3.09 ksi 6 12 ksi

Aweb

(13.84)(0.270)

Use W14 * 30

Ans.

841

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11–15. The simply supported beam is made of timber that

has an allowable bending stress of sallow = 960 psi and an

allowable shear stress of tallow = 75 psi. Determine its

dimensions if it is to be rectangular and have a heightto-width ratio of 1.25.

5 kip/ft

6 ft

1

I =

(b)(1.25b)3 = 0.16276b4

12

Sreq¿d

b

Assume bending moment controls:

Mmax = 60 kip # ft

960 =

Mmax

Sreq¿d

60(103)(12)

0.26042 b3

b = 14.2 in.

Check shear:

tmax =

1.5(15)(103)

1.5V

=

= 88.9 psi 7 75 psi NO

A

(14.2)(1.25)(14.2)

Shear controls:

tallow =

6 ft

1.25 b

I

0.16276b4

=

=

= 0.26042b3

c

0.625b

sallow =

B

A

1.5(15)(103)

1.5V

=

A

(b)(1.25b)

b = 15.5 in.

Ans.

842

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*11–16. The simply supported beam is composed of two

W12 * 22 sections built up as shown. Determine the

maximum uniform loading w the beam will support if

the allowable bending stress is sallow = 22 ksi and the

allowable shear stress is tallow = 14 ksi.

w

Section properties:

24 ft

For W12 * 22 (d = 12.31 in. Ix = 156 in4 tw = 0.260 in. A = 6.48 in2)

I = 2c 156 + 6.48a

S =

12.31 2

b d = 802.98 in4

2

I

802.98

=

= 65.23 in3

c

12.31

Maximum Loading: Assume moment controls.

M = sallowS(72 w)(12) = 22(65.23)

w = 1.66 kip>ft

Check Shear:

tmax =

Ans.

(Neglect area of flanges.)

12(1.66)

Vmax

= 3.11 ksi 6 tallow = 14 ksi OK

=

Aw

2(12.31)(0.26)

•11–17.

The simply supported beam is composed of two

W12 * 22 sections built up as shown. Determine if the beam

will safely support a loading of w = 2 kip>ft. The allowable

bending stress is sallow = 22 ksi and the allowable shear

stress is tallow = 14 ksi.

w

24 ft

Section properties:

For W 12 * 22 (d = 12.31 in.

Ix = 156 in4

tw = 0.260 in.

A = 6.48 in2)

I = 2[156 + 6.48(6.1552)] = 802.98 in4

S =

802.98

I

=

= 65.23 in3

c

12.31

Bending stress:

smax =

144 (12)

Mallow

=

= 26.5 ksi 7 sallow = 22 ksi

S

65.23

No, the beam falls due to bending stress criteria.

Check shear:

tmax =

Ans.

(Neglect area of flanges.)

Vmax

24

=

= 3.75 ksi 6 tallow = 14 ksi OK

Aw

2(12.31)(0.26)

843

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Page 844

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11–18. Determine the smallest diameter rod that will

safely support the loading shown. The allowable bending

stress is sallow = 167 MPa and the allowable shear stress

is tallow = 97 MPa.

25 N/m

15 N/m

15 N/m

1.5 m

Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Assume bending

controls the design. Applying the flexure formula.

sallow =

167 A 10

6

B =

Mmax c

I

24.375

p

4

A d2 B

A d2 B 4

d = 0.01141 m = 11.4 mm

Ans.

Shear Stress: Provide a shear stress check using the shear formula with

I =

p

A 0.0057074 B = 0.8329 A 10 - 9 B m4

4

Qmax =

4(0.005707) 1

c (p) A 0.0057062 B d = 0.1239 A 10 - 6 B m3

3p

2

From the shear diagram, Vmax = 30.0 N.

tmax =

=

Vmax Qmax

It

30.0 C 0.1239(10 - 6) D

0.8329 (10 - 9)(0.01141)

= 0.391 MPa 6 tallow = 97 MPa (O.K!)

844

1.5 m

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Page 845

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11–19. The pipe has an outer diameter of 15 mm.

Determine the smallest inner diameter so that it will safely

support the loading shown. The allowable bending stress

is sallow = 167 MPa and the allowable shear stress is

tallow = 97 MPa.

25 N/m

15 N/m

15 N/m

1.5 m

Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Q. Assume

bending controls the design. Applying the flexure formula.

sallow =

167 A 106 B =

Mmax c

I

24.375(0.0075)

p

4

C 0.00754 - A 2i B 4 D

d

di = 0.01297 m = 13.0 mm

Ans.

Shear Stress: Provide a shear stress check using the shear formula with

I =

p

A 0.00754 - 0.0064864 B = 1.0947 A 10 - 9 B m4

4

Qmax =

4(0.0075) 1

4(0.006486) 1

c (p) A 0.00752 B d c (p) A 0.0064862 B d

3p

2

3p

2

= 99.306 A 10 - 9 B m3

From the shear diagram, Vmax = 30.0 N. Q

tmax =

=

Vmax Qmax

It

30.0 C 99.306(10 - 9) D

1.0947(10 - 9)(0.015 - 0.01297)

= 1.34 MPa 6 tallow = 97 MPa (O.K!)

845

1.5 m

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Page 846

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*11–20. Determine the maximum uniform loading w

the W12 * 14 beam will support if the allowable bending

stress is sallow = 22 ksi and the allowable shear stress is

tallow = 12 ksi.

w

10 ft

10 ft

From the moment diagram, Fig. a, Mmax = 28.125 w. For W12 * 14, Sx = 14.9 in3,

d = 11.91 in and tw = 0.200 in.

sallow =

22 =

Mmax

S

28.125 w (12)

14.9

Ans.

w = 0.9712 kip>ft = 971 lb>ft

From the shear diagram, Fig. a, Vmax = 7.5(0.9712) = 7.284 kip. Provide a shear

stress check on W12 * 14,

tmax =

=

Vmax

tw d

7.284

0.200(11.91)

= 3.06 ksi 6 tallow = 12 ksi (O.K)

846

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•11–21.

Determine if the W14 * 22 beam will safely

support a loading of w = 1.5 kip>ft. The allowable bending

stress is sallow = 22 ksi and the allowable shear stress

is tallow = 12 ksi.

w

10 ft

10 ft

For W14 * 22, Sx = 29.0 in3, d = 13.74 in and tw = 0.23 in. From the moment

diagram, Fig. a, Mmax = 42.1875 kip # ft.

smax =

=

Mmax

S

42.1875(12)

29.0

= 17.46 ksi 6 sallow = 22 ksi (O.K!)

From the shear diagram, Fig. a, Vmax = 11.25 kip.

tmax =

=

Vmax

tw d

11.25

0.23(13.74)

= 3.56 ksi 6 tallow = 12 ksi (O.K!)

Based on the investigated results, we conclude that W14 * 22 can safely support

the loading.

847

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11–22. Determine the minimum depth h of the beam to

the nearest 18 in. that will safely support the loading shown.

The allowable bending stress is sallow = 21 ksi and the

allowable shear stress is tallow = 10 ksi. The beam has a

uniform thickness of 3 in.

4 kip/ft

h

A

B

12 ft

The section modulus of the rectangular cross-section is

S =

I

=

C

1

12

(3)(h3)

h>2

= 0.5 h2

From the moment diagram, Mmax = 72 kip # ft.

Sreq¿d =

Mmax

sallow

0.5h2 =

72(12)

21

h = 9.07 in

Use

h = 9 18 in

Ans.

From the shear diagram, Fig. a, Vmax = 24 kip . Referring to Fig. b,

9.125 9.125

ba

b (3) = 31.22 in3 and

Qmax = y¿A¿ = a

4

2

1

I =

(3) A 9.1253 B = 189.95 in4 . Provide the shear stress check by applying

12

shear formula,

tmax =

=

Vmax Qmax

It

24(31.22)

189.95(3)

= 1.315 ksi 6 tallow = 10 ksi (O.K!)

848

6 ft

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11–23. The box beam has an allowable bending stress

of sallow = 10 MPa and an allowable shear stress of

tallow = 775 kPa. Determine the maximum intensity w of the

distributed loading that it can safely support. Also, determine

the maximum safe nail spacing for each third of the length of

the beam. Each nail can resist a shear force of 200 N.

w

30 mm

250 mm

30 mm

150 mm

30 mm

Section Properties:

I =

1

1

(0.21) A 0.253 B (0.15) A 0.193 B = 0.1877 A 10 - 3 B m4

12

12

QA = y1 ¿A¿ = 0.11(0.03)(0.15) = 0.495 A 10 - 3 B m3

Qmax = ©y¿A¿ = 0.11(0.03)(0.15) + 0.0625(0.125)(0.06)

= 0.96375 A 10 - 3 B m3

Bending Stress: From the moment diagram, Mmax = 4.50w. Assume bending

controls the design. Applying the flexure formula.

sallow =

10 A 106 B =

Mmax c

I

4.50w (0.125)

0.1877 (10 - 3)

w = 3336.9 N>m

Shear Stress: Provide a shear stress check using the shear formula. From the shear

diagram, Vmax = 3.00w = 10.01 kN.

tmax =

=

Vmax Qmax

It

10.01(103) C 0.96375(10 - 3) D

0.1877(10 - 3)(0.06)

= 857 kPa 7 tallow = 775 kPa (No Good!)

Hence, shear stress controls.

tallow =

775 A 103 B =

Vmax Qmax

It

3.00w C 0.96375(10 - 3) D

0.1877(10 - 3)(0.06)

w = 3018.8 N>m = 3.02 kN>m

Ans.

Shear Flow: Since there are two rows of nails, the allowable shear flow is

2(200)

400

=

q =

.

s

s

849

6m

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11–23.

Continued

For 0 … x 6 2 m and 4 m 6 x … 6 m, the design shear force is

V = 3.00w = 9056.3 N.

q =

VQA

I

9056.3 C 0.495(10 - 3) D

400

=

s

0.1877(10 - 3)

s = 0.01675 m = 16.7 mm

Ans.

For 2 m 6 x 6 4 m, the design shear force is V = w = 3018.8 N.

q =

VQA

I

3018.8 C 0.495(10 - 3) D

400

=

s

0.1877(10 - 3)

s = 0.05024 m = 50.2 mm

Ans.

850

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*11–24. The simply supported joist is used in the

construction of a floor for a building. In order to keep the

floor low with respect to the sill beams C and D, the ends of

the joists are notched as shown. If the allowable shear stress

for the wood is tallow = 350 psi and the allowable bending

stress is sallow = 1500 psi, determine the height h that will

cause the beam to reach both allowable stresses at the same

time. Also, what load P causes this to happen? Neglect the

stress concentration at the notch.

P

2 in.

15 ft

B

h

15 ft

D

A

10 in.

C

Bending Stress: From the moment diagram, Mmax = 7.50P. Applying the flexure

formula.

Mmax c

I

salllow =

7.50P(12)(5)

1500 =

1

12

(2)(103)

P = 555.56 lb = 556 lb

Ans.

Shear Stress: From the shear diagram, Vmax = 0.500P = 277.78 lb. The notch is the

critical section. Using the shear formula for a rectangular section.

tallow =

350 =

3Vmax

2A

3(277.78)

2(2) h

h = 0.595 in.

Ans.

11–25. The simply supported joist is used in the

construction of a floor for a building. In order to keep the

floor low with respect to the sill beams C and D, the ends of

the joists are notched as shown. If the allowable shear stress

for the wood is tallow = 350 psi and the allowable bending

stress is sallow = 1700 psi, determine the smallest height h

so that the beam will support a load of P = 600 lb. Also,

will the entire joist safely support the load? Neglect the

stress concentration at the notch.

P

B

tallow =

1.5V

;

A

350 =

D

A

10 in.

600

= 300 lb

2

1.5(300)

(2)(h)

h = 0.643 in.

smax =

Ans.

4500(12)(5)

Mmax c

= 1620 psi 6 1700 psi OK

= 1

3

I

12 (2)(10)

Yes, the joist will safely support the load.

Ans.

851

h

15 ft

C

The reaction at the support is

2 in.

15 ft

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11–26. Select the lightest-weight steel wide-flange beam

from Appendix B that will safely support the loading

shown. The allowable bending stress is sallow = 22 ksi and

the allowable shear stress is tallow = 12 ksi.

5 kip

18 kip ft

B

A

6 ft

From the moment diagram, Fig. a, Mmax = 48 kip # ft.

Sreq¿d =

=

Mmax

sallow

48(12)

22

= 26.18 in3

Select W 14 * 22 C Sx = 29.0 in3, d = 13.74 in. and tw = 0.230 in. D

From the shear diagram, Fig. a, Vmax = 5 kip. Provide the shear stress check for

W 14 * 22,

tmax =

=

Vmax

twd

5

0.230(13.74)

= 1.58 ksi 6 tallow = 12 ksi‚ (O.K!)

Use

Ans.

W14 * 22

W12 * 22 would work also.

852

12 ft

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11–27. The T-beam is made from two plates welded

together as shown. Determine the maximum uniform

distributed load w that can be safely supported on the beam

if the allowable bending stress is sallow = 150 MPa and the

allowable shear stress is tallow = 70 MPa.

w

A

1.5 m

1.5 m

200 mm

20 mm

200 mm

20 mm

The neutral axis passes through centroid c of the beam’s cross-section. The location

of c, Fig. b, is

y =

0.21(0.02)(0.2) + 0.1(0.2)(0.02)

©yA

=

©A

0.02(0.2) + 0.2(0.02)

= 0.155 m

I =

1

(0.2)(0.023) + 0.2(0.02)(0.055)2

12

+

1

(0.02)(0.23) + 0.02(0.2)(0.055)2

12

= 37.667 (10 - 6) m4

Referring to Fig. b,

Qmax = y¿A¿ = 0.0775(0.155)(0.02)

= 0.24025(10 - 3) m3

Referring to the moment diagram, Mmax = -3.375 w. Applying the flexure

formula with C = y = 0.155 m,

sallow =

Mmax c

;

I

150(106) =

3.375 w (0.155)

37.667(10 - 6)

w = 10.80(103) N>m

= 10.8 kN>m (Control!)

Ans.

Referring to the shear diagram, Vmax = 1.5w.

tallow =

Vmax Qmax

;

It

70(106) =

1.5 w C 0.24025(10 - 3) D

37.667(10 - 6)(0.02)

w = 146.33(103) N>m

= 146 kN>m

853

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*11–28. The beam is made of a ceramic material having

an allowable bending stress of sallow = 735 psi and an

allowable shear stress of tallow = 400 psi. Determine the

width b of the beam if the height h = 2b.

15 lb

10 lb

6 lb/in.

2 in.

6 in.

2 in.

h

b

Bending Stress: From the moment diagram, Mmax = 30.0 lb # in. Assume bending

controls the design. Applying the flexure formula.

sallow =

Mmax c

I

30.0

735 =

1

12

A 2b2 B

(b) (2b)3

b = 0.3941 in. = 0.394 in.

Ans.

Shear Stress: Provide a shear stress check using the shear formula for a rectangular

section. From the shear diagram, Vmax = 19.67 lb.

tmax =

=

3Vmax

2A

3(19.67)

2(0.3941)(2)(0.3941)

= 94.95 psi 6 tallow = 400 psi (O.K!)

854

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11–1. The simply supported beam is made of timber that

has an allowable bending stress of sallow = 6.5 MPa and an

allowable shear stress of tallow = 500 kPa. Determine its

dimensions if it is to be rectangular and have a height-towidth ratio of 1.25.

8 kN/m

2m

Ix =

1

(b)(1.25b)3 = 0.16276b4

12

Qmax = y¿A¿ = (0.3125b)(0.625b)(b) = 0.1953125b3

Assume bending moment controls:

Mmax = 16 kN # m

sallow =

Mmax c

I

6.5(106) =

16(103)(0.625b)

0.16276b4

b = 0.21143 m = 211 mm

Ans.

h = 1.25b = 264 mm

Ans.

Check shear:

Qmax = 1.846159(10 - 3) m3

I = 0.325248(10 - 3) m4

tmax =

VQmax

16(103)(1.846159)(10 - 3)

= 429 kPa 6 500 kPa‚ OK

=

It

0.325248(10 - 3)(0.21143)

830

4m

2m

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11–2. The brick wall exerts a uniform distributed load

of 1.20 kip>ft on the beam. If the allowable bending stress

is sallow = 22 ksi and the allowable shear stress is

tallow = 12 ksi, select the lightest wide-flange section with

the shortest depth from Appendix B that will safely support

the load.

1.20 kip/

4 ft

10 ft

ft

6 ft

b

Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft. Assuming

bending controls the design and applying the flexure formula.

Sreq d =

=

44.55 (12)

= 24.3 in3

22

W12 * 22

A Sx = 25.4 in3, d = 12.31 in., tw = 0.260 in. B

V

for the W12 * 22 wide tw d

= 6.60 kip.

Shear Stress: Provide a shear stress check using t =

flange section. From the shear diagram, Vmax

tmax =

=

Vmax

tw d

6.60

0.260(12.31)

= 2.06 ksi 6 tallow = 12 ksi (O.K!)

Hence,

Use

9 in.

0.5 in.

Mmax

sallow

Two choices of wide flange section having the weight 22 lb>ft can be made. They

are W12 * 22 and W14 * 22. However, W12 * 22 is the shortest.

Select

0.5 in.

0.5 in.

Ans.

W12 * 22

831

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11–3. The brick wall exerts a uniform distributed load

of 1.20 kip>ft on the beam. If the allowable bending stress

is sallow = 22 ksi, determine the required width b of the

flange to the nearest 14 in.

1.20 kip/

4 ft

10 ft

ft

6 ft

b

0.5 in.

0.5 in.

9 in.

0.5 in.

Section Property:

I =

1

1

(b) A 103 B (b - 0.5) A 93 B = 22.583b + 30.375

12

12

Bending Stress: From the moment diagram, Mmax = 44.55 kip # ft.

sallow =

22 =

Mmax c

I

44.55(12)(5)

22.583b + 30.375

b = 4.04 in.

Use

b = 4.25 in.

Ans.

832

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*11–4. Draw the shear and moment diagrams for the

shaft, and determine its required diameter to the nearest

1

4 in. if sallow = 7 ksi and tallow = 3 ksi. The bearings at A

and D exert only vertical reactions on the shaft. The loading

is applied to the pulleys at B, C, and E.

14 in.

20 in.

15 in.

12 in.

E

A

C

B

D

35 lb

80 lb

110 lb

sallow =

7(103) =

Mmax c

I

1196 c

p 4 ;

4 c

c = 0.601 in.

d = 2c = 1.20 in.

Use d = 1.25 in.

Ans.

Check shear:

2

tmax =

0.625

108(4(0.625)

Vmax Q

3p )(p)( 2 )

= 117 psi 6 3 ksi OK

=

p

4

It

4 (0.625) (1.25)

•11–5.

Select the lightest-weight steel wide-flange beam

from Appendix B that will safely support the machine loading

shown. The allowable bending stress is sallow = 24 ksi and

the allowable shear stress is tallow = 14 ksi.

2 ft

Bending Stress: From the moment diagram, Mmax = 30.0 kip # ft.

Assume bending controls the design. Applying the flexure formula.

Sreq¿d =

=

Select

W12 * 16

Mmax

sallow

30.0(12)

= 15.0 in3

24

A Sx = 17.1 in3, d = 11.99 in., tw = 0.220 in. B

V

for the W12 * 16 wide tw d

= 10.0 kip

Shear Stress: Provide a shear stress check using t =

flange section. From the shear diagram, Vmax

tmax =

=

Vmax

tw d

10.0

0.220(11.99)

= 3.79 ksi 6 tallow = 14 ksi (O.K!)

Hence,

Use

5 kip

5 kip

Ans.

W12 * 16

833

2 ft

5 kip

2 ft

5 kip

2 ft

2 ft

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11–6. The compound beam is made from two sections,

which are pinned together at B. Use Appendix B and select

the lightest-weight wide-flange beam that would be safe for

each section if the allowable bending stress is sallow = 24 ksi

and the allowable shear stress is tallow = 14 ksi. The beam

supports a pipe loading of 1200 lb and 1800 lb as shown.

C

A

B

6 ft

Bending Stress: From the moment diagram, Mmax = 19.2 kip # ft for member AB.

Assuming bending controls the design, applying the flexure formula.

Sreq¿d =

=

Select

Mmax

sallow

19.2(12)

= 9.60 in3

24

A Sx = 10.9 in3, d = 9.87 in., tw = 0.19 in. B

W10 * 12

For member BC, Mmax = 8.00 kip # ft.

Sreq¿d =

=

Select

Mmax

sallow

8.00(12)

= 4.00 in3

24

A Sx = 5.56 in3, d = 5.90 in., tw = 0.17 in. B

W6 * 9

V

for the W10 * 12 widetw d

flange section for member AB. From the shear diagram, Vmax = 2.20 kip.

Shear Stress: Provide a shear stress check using t =

tmax =

=

Vmax

tw d

2.20

0.19(9.87)

= 1.17 ksi 6 tallow = 14 ksi (O.K!)

Use

Ans.

W10 * 12

For member BC (W6 * 9), Vmax = 1.00 kip.

tmax =

=

Vmax

tw d

1.00

0.17(5.90)

= 0.997 ksi 6 tallow = 14 ksi (O.K!)

Hence,

Use

1800 lb

1200 lb

W6 * 9

Ans.

834

6 ft

8 ft

10 ft

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Page 835

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11–7. If the bearing pads at A and B support only vertical

forces, determine the greatest magnitude of the uniform

distributed loading w that can be applied to the beam.

sallow = 15 MPa, tallow = 1.5 MPa.

w

A

B

1m

1m

150 mm

25 mm

150 mm

25 mm

The location of c, Fig. b, is

y =

0.1625(0.025)(0.15) + 0.075(0.15)(0.025)

©yA

=

©A

0.025(0.15) + 0.15(0.025)

= 0.11875 m

I =

+

1

(0.025)(0.153) + (0.025)(0.15)(0.04375)2

12

1

(0.15)(0.0253) + 0.15(0.025)(0.04375)2

12

= 21.58203125(10 - 6) m4

Referring to Fig. b,

Qmax = y¿A¿ = 0.059375 (0.11875)(0.025)

= 0.176295313(10 - 4) m3

Referring to the moment diagram, Mmax = 0.28125 w. Applying the Flexure

formula with C = y = 0.11875 m,

sallow =

Mmax c

;

I

15(106) =

0.28125w(0.11875)

21.582(10 - 6)

W = 9.693(103) N>m

Referring to shear diagram, Fig. a, Vmax = 0.75 w.

tallow =

Vallow Qmax

;

It

1.5(106) =

0.75w C 0.17627(10 - 3) D

21.582(10 - 6)(0.025)

W = 6.122(103) N>m

= 6.12 kN>m (Control!)

Ans.

835

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*11–8. The simply supported beam is made of timber that

has an allowable bending stress of sallow = 1.20 ksi and an

allowable shear stress of tallow = 100 psi. Determine its

smallest dimensions to the nearest 18 in. if it is rectangular

and has a height-to-width ratio of 1.5.

12 kip/ft

B

A

3 ft

3 ft

1.5 b

b

The moment of inertia of the beam’s cross-section about the neutral axis is

1

(b)(1.5b)3 = 0.28125b4. Referring to the moment diagram,

I =

12

Mmax = 45.375 kip # ft.

sallow =

Mmax c

;

I

1.2 =

45.375(12)(0.75b)

0.28125b4

b = 10.66 in

Referring to Fig. b, Qmax = y¿A¿ = 0.375b (0.75b)(b) = 0.28125b3. Referring to the

shear diagram, Fig. a, Vmax = 33 kip.

tmax =

Vmax Qmax

;

It

100 =

33(103)(0.28125b3)

0.28125b4(b)

b = 18.17 in (Control!)

Thus, use

b = 18

1

in

4

Ans.

836

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•11–9.

Select the lightest-weight W12 steel wide-flange

beam from Appendix B that will safely support the loading

shown, where P = 6 kip. The allowable bending stress

is sallow = 22 ksi and the allowable shear stress is

tallow = 12 ksi.

P

P

9 ft

From the Moment Diagram, Fig. a, Mmax = 54 kip # ft.

Mmax

sallow

Sreq¿d =

54(12)

22

=

= 29.45 in3

Select W12 * 26

C Sx = 33.4 in3, d = 12.22 in and tw = 0.230 in. D

From the shear diagram, Fig. a, Vmax = 7.5 kip. Provide the shear-stress check

for W 12 * 26,

tmax =

=

Vmax

tw d

7.5

0.230(12.22)

= 2.67 ksi 6 tallow = 12 ksi (O.K!)

Hence

Use

Ans.

W12 * 26

837

6 ft

6 ft

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11–10. Select the lightest-weight W14 steel wide-flange

beam having the shortest height from Appendix B that

will safely support the loading shown, where P = 12 kip.

The allowable bending stress is sallow = 22 ksi and the

allowable shear stress is tallow = 12 ksi.

P

P

9 ft

From the moment diagram, Fig. a, Mmax = 108 kip # ft.

Mmax

sallow

Sreq¿d =

108(12)

22

=

= 58.91 in3

Select W14 * 43

C Sx = 62.7 in3, d = 13.66 in and tw = 0.305 in. D

From the shear diagram, Fig. a, Vmax = 15 kip . Provide the shear-stress check

for W14 * 43 ,

tmax =

=

Vmax

tw d

15

0.305(13.66)

= 3.60 ksi 6 tallow = 12 ksi‚ (O.K!)

Hence,

Use

Ans.

W14 * 43

838

6 ft

6 ft

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11–11. The timber beam is to be loaded as shown. If the ends

support only vertical forces, determine the greatest magnitude

of P that can be applied. sallow = 25 MPa, tallow = 700 kPa.

150 mm

30 mm

120 mm

40 mm

P

4m

A

y =

(0.015)(0.150)(0.03) + (0.09)(0.04)(0.120)

= 0.05371 m

(0.150)(0.03) + (0.04)(0.120)

I =

1

1

(0.150)(0.03)3 + (0.15)(0.03)(0.05371 - 0.015)2 +

(0.04)(0.120)3 +

12

12

B

(0.04)(0.120)(0.09 - 0.05371)2 = 19.162(10 - 6) m4

Maximum moment at center of beam:

Mmax =

P

(4) = 2P

2

Mc

;

I

s =

25(106) =

(2P)(0.15 - 0.05371)

19.162(10 - 6)

P = 2.49 kN

Maximum shear at end of beam:

Vmax =

P

2

VQ

;

t =

It

700(103) =

P 1

C (0.15 - 0.05371)(0.04)(0.15 - 0.05371) D

2 2

19.162(10 - 6)(0.04)

P = 5.79 kN

Thus,

P = 2.49 kN

Ans.

839

4m

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Page 840

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*11–12. Determine the minimum width of the beam to

the nearest 14 in. that will safely support the loading of

P = 8 kip. The allowable bending stress is sallow = 24 ksi

and the allowable shear stress is tallow = 15 ksi.

P

6 ft

6 ft

6 in.

B

A

Beam design: Assume moment controls.

sallow =

Mc

;

I

24 =

48.0(12)(3)

1

3

12 (b)(6 )

b = 4 in.

Ans.

Check shear:

8(1.5)(3)(4)

VQ

= 0.5 ksi 6 15 ksi OK

= 1

3

It

12 (4)(6 )(4)

tmax =

•11–13.

Select the shortest and lightest-weight steel wideflange beam from Appendix B that will safely support the

loading shown.The allowable bending stress is sallow = 22 ksi

and the allowable shear stress is tallow = 12 ksi.

10 kip

6 kip

4 kip

A

B

4 ft

Beam design: Assume bending moment controls.

Sreq¿d =

60.0(12)

Mmax

=

= 32.73 in3

sallow

22

Select a W 12 * 26

Sx = 33.4 in3, d = 12.22 in., tw = 0.230 in.

Check shear:

tavg =

V

10.5

=

= 3.74 ksi 6 12 ksi

Aweb

(12.22)(0.230)

Use W 12 * 26

Ans.

840

4 ft

4 ft

4 ft

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Page 841

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11–14. The beam is used in a railroad yard for loading and

unloading cars. If the maximum anticipated hoist load is

12 kip, select the lightest-weight steel wide-flange section

from Appendix B that will safely support the loading. The

hoist travels along the bottom flange of the beam,

1 ft … x … 25 ft, and has negligible size. Assume the beam

is pinned to the column at B and roller supported at A.

The allowable bending stress is sallow = 24 ksi and

the allowable shear stress is tallow = 12 ksi.

x

27 ft

A

B

12 kip

15 ft

C

Maximum moment occurs when load is in the center of beam.

Mmax = (6 kip)(13.5 ft) = 81 lb # ft

sallow =

M

;

S

24 =

81(12)

Sreq¿d

Sreq¿d = 40.5 in3

Select a W 14 * 30, Sx = 42.0 in3, d = 13.84 in, tw = 0.270 in.

At x = 1 ft, V = 11.56 kip

t =

11.36

V

=

= 3.09 ksi 6 12 ksi

Aweb

(13.84)(0.270)

Use W14 * 30

Ans.

841

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11–15. The simply supported beam is made of timber that

has an allowable bending stress of sallow = 960 psi and an

allowable shear stress of tallow = 75 psi. Determine its

dimensions if it is to be rectangular and have a heightto-width ratio of 1.25.

5 kip/ft

6 ft

1

I =

(b)(1.25b)3 = 0.16276b4

12

Sreq¿d

b

Assume bending moment controls:

Mmax = 60 kip # ft

960 =

Mmax

Sreq¿d

60(103)(12)

0.26042 b3

b = 14.2 in.

Check shear:

tmax =

1.5(15)(103)

1.5V

=

= 88.9 psi 7 75 psi NO

A

(14.2)(1.25)(14.2)

Shear controls:

tallow =

6 ft

1.25 b

I

0.16276b4

=

=

= 0.26042b3

c

0.625b

sallow =

B

A

1.5(15)(103)

1.5V

=

A

(b)(1.25b)

b = 15.5 in.

Ans.

842

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Page 843

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*11–16. The simply supported beam is composed of two

W12 * 22 sections built up as shown. Determine the

maximum uniform loading w the beam will support if

the allowable bending stress is sallow = 22 ksi and the

allowable shear stress is tallow = 14 ksi.

w

Section properties:

24 ft

For W12 * 22 (d = 12.31 in. Ix = 156 in4 tw = 0.260 in. A = 6.48 in2)

I = 2c 156 + 6.48a

S =

12.31 2

b d = 802.98 in4

2

I

802.98

=

= 65.23 in3

c

12.31

Maximum Loading: Assume moment controls.

M = sallowS(72 w)(12) = 22(65.23)

w = 1.66 kip>ft

Check Shear:

tmax =

Ans.

(Neglect area of flanges.)

12(1.66)

Vmax

= 3.11 ksi 6 tallow = 14 ksi OK

=

Aw

2(12.31)(0.26)

•11–17.

The simply supported beam is composed of two

W12 * 22 sections built up as shown. Determine if the beam

will safely support a loading of w = 2 kip>ft. The allowable

bending stress is sallow = 22 ksi and the allowable shear

stress is tallow = 14 ksi.

w

24 ft

Section properties:

For W 12 * 22 (d = 12.31 in.

Ix = 156 in4

tw = 0.260 in.

A = 6.48 in2)

I = 2[156 + 6.48(6.1552)] = 802.98 in4

S =

802.98

I

=

= 65.23 in3

c

12.31

Bending stress:

smax =

144 (12)

Mallow

=

= 26.5 ksi 7 sallow = 22 ksi

S

65.23

No, the beam falls due to bending stress criteria.

Check shear:

tmax =

Ans.

(Neglect area of flanges.)

Vmax

24

=

= 3.75 ksi 6 tallow = 14 ksi OK

Aw

2(12.31)(0.26)

843

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Page 844

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11–18. Determine the smallest diameter rod that will

safely support the loading shown. The allowable bending

stress is sallow = 167 MPa and the allowable shear stress

is tallow = 97 MPa.

25 N/m

15 N/m

15 N/m

1.5 m

Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Assume bending

controls the design. Applying the flexure formula.

sallow =

167 A 10

6

B =

Mmax c

I

24.375

p

4

A d2 B

A d2 B 4

d = 0.01141 m = 11.4 mm

Ans.

Shear Stress: Provide a shear stress check using the shear formula with

I =

p

A 0.0057074 B = 0.8329 A 10 - 9 B m4

4

Qmax =

4(0.005707) 1

c (p) A 0.0057062 B d = 0.1239 A 10 - 6 B m3

3p

2

From the shear diagram, Vmax = 30.0 N.

tmax =

=

Vmax Qmax

It

30.0 C 0.1239(10 - 6) D

0.8329 (10 - 9)(0.01141)

= 0.391 MPa 6 tallow = 97 MPa (O.K!)

844

1.5 m

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Page 845

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11–19. The pipe has an outer diameter of 15 mm.

Determine the smallest inner diameter so that it will safely

support the loading shown. The allowable bending stress

is sallow = 167 MPa and the allowable shear stress is

tallow = 97 MPa.

25 N/m

15 N/m

15 N/m

1.5 m

Bending Stress: From the moment diagram, Mmax = 24.375 N # m. Q. Assume

bending controls the design. Applying the flexure formula.

sallow =

167 A 106 B =

Mmax c

I

24.375(0.0075)

p

4

C 0.00754 - A 2i B 4 D

d

di = 0.01297 m = 13.0 mm

Ans.

Shear Stress: Provide a shear stress check using the shear formula with

I =

p

A 0.00754 - 0.0064864 B = 1.0947 A 10 - 9 B m4

4

Qmax =

4(0.0075) 1

4(0.006486) 1

c (p) A 0.00752 B d c (p) A 0.0064862 B d

3p

2

3p

2

= 99.306 A 10 - 9 B m3

From the shear diagram, Vmax = 30.0 N. Q

tmax =

=

Vmax Qmax

It

30.0 C 99.306(10 - 9) D

1.0947(10 - 9)(0.015 - 0.01297)

= 1.34 MPa 6 tallow = 97 MPa (O.K!)

845

1.5 m

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Page 846

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*11–20. Determine the maximum uniform loading w

the W12 * 14 beam will support if the allowable bending

stress is sallow = 22 ksi and the allowable shear stress is

tallow = 12 ksi.

w

10 ft

10 ft

From the moment diagram, Fig. a, Mmax = 28.125 w. For W12 * 14, Sx = 14.9 in3,

d = 11.91 in and tw = 0.200 in.

sallow =

22 =

Mmax

S

28.125 w (12)

14.9

Ans.

w = 0.9712 kip>ft = 971 lb>ft

From the shear diagram, Fig. a, Vmax = 7.5(0.9712) = 7.284 kip. Provide a shear

stress check on W12 * 14,

tmax =

=

Vmax

tw d

7.284

0.200(11.91)

= 3.06 ksi 6 tallow = 12 ksi (O.K)

846

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Page 847

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•11–21.

Determine if the W14 * 22 beam will safely

support a loading of w = 1.5 kip>ft. The allowable bending

stress is sallow = 22 ksi and the allowable shear stress

is tallow = 12 ksi.

w

10 ft

10 ft

For W14 * 22, Sx = 29.0 in3, d = 13.74 in and tw = 0.23 in. From the moment

diagram, Fig. a, Mmax = 42.1875 kip # ft.

smax =

=

Mmax

S

42.1875(12)

29.0

= 17.46 ksi 6 sallow = 22 ksi (O.K!)

From the shear diagram, Fig. a, Vmax = 11.25 kip.

tmax =

=

Vmax

tw d

11.25

0.23(13.74)

= 3.56 ksi 6 tallow = 12 ksi (O.K!)

Based on the investigated results, we conclude that W14 * 22 can safely support

the loading.

847

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Page 848

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11–22. Determine the minimum depth h of the beam to

the nearest 18 in. that will safely support the loading shown.

The allowable bending stress is sallow = 21 ksi and the

allowable shear stress is tallow = 10 ksi. The beam has a

uniform thickness of 3 in.

4 kip/ft

h

A

B

12 ft

The section modulus of the rectangular cross-section is

S =

I

=

C

1

12

(3)(h3)

h>2

= 0.5 h2

From the moment diagram, Mmax = 72 kip # ft.

Sreq¿d =

Mmax

sallow

0.5h2 =

72(12)

21

h = 9.07 in

Use

h = 9 18 in

Ans.

From the shear diagram, Fig. a, Vmax = 24 kip . Referring to Fig. b,

9.125 9.125

ba

b (3) = 31.22 in3 and

Qmax = y¿A¿ = a

4

2

1

I =

(3) A 9.1253 B = 189.95 in4 . Provide the shear stress check by applying

12

shear formula,

tmax =

=

Vmax Qmax

It

24(31.22)

189.95(3)

= 1.315 ksi 6 tallow = 10 ksi (O.K!)

848

6 ft

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Page 849

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11–23. The box beam has an allowable bending stress

of sallow = 10 MPa and an allowable shear stress of

tallow = 775 kPa. Determine the maximum intensity w of the

distributed loading that it can safely support. Also, determine

the maximum safe nail spacing for each third of the length of

the beam. Each nail can resist a shear force of 200 N.

w

30 mm

250 mm

30 mm

150 mm

30 mm

Section Properties:

I =

1

1

(0.21) A 0.253 B (0.15) A 0.193 B = 0.1877 A 10 - 3 B m4

12

12

QA = y1 ¿A¿ = 0.11(0.03)(0.15) = 0.495 A 10 - 3 B m3

Qmax = ©y¿A¿ = 0.11(0.03)(0.15) + 0.0625(0.125)(0.06)

= 0.96375 A 10 - 3 B m3

Bending Stress: From the moment diagram, Mmax = 4.50w. Assume bending

controls the design. Applying the flexure formula.

sallow =

10 A 106 B =

Mmax c

I

4.50w (0.125)

0.1877 (10 - 3)

w = 3336.9 N>m

Shear Stress: Provide a shear stress check using the shear formula. From the shear

diagram, Vmax = 3.00w = 10.01 kN.

tmax =

=

Vmax Qmax

It

10.01(103) C 0.96375(10 - 3) D

0.1877(10 - 3)(0.06)

= 857 kPa 7 tallow = 775 kPa (No Good!)

Hence, shear stress controls.

tallow =

775 A 103 B =

Vmax Qmax

It

3.00w C 0.96375(10 - 3) D

0.1877(10 - 3)(0.06)

w = 3018.8 N>m = 3.02 kN>m

Ans.

Shear Flow: Since there are two rows of nails, the allowable shear flow is

2(200)

400

=

q =

.

s

s

849

6m

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11–23.

Continued

For 0 … x 6 2 m and 4 m 6 x … 6 m, the design shear force is

V = 3.00w = 9056.3 N.

q =

VQA

I

9056.3 C 0.495(10 - 3) D

400

=

s

0.1877(10 - 3)

s = 0.01675 m = 16.7 mm

Ans.

For 2 m 6 x 6 4 m, the design shear force is V = w = 3018.8 N.

q =

VQA

I

3018.8 C 0.495(10 - 3) D

400

=

s

0.1877(10 - 3)

s = 0.05024 m = 50.2 mm

Ans.

850

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*11–24. The simply supported joist is used in the

construction of a floor for a building. In order to keep the

floor low with respect to the sill beams C and D, the ends of

the joists are notched as shown. If the allowable shear stress

for the wood is tallow = 350 psi and the allowable bending

stress is sallow = 1500 psi, determine the height h that will

cause the beam to reach both allowable stresses at the same

time. Also, what load P causes this to happen? Neglect the

stress concentration at the notch.

P

2 in.

15 ft

B

h

15 ft

D

A

10 in.

C

Bending Stress: From the moment diagram, Mmax = 7.50P. Applying the flexure

formula.

Mmax c

I

salllow =

7.50P(12)(5)

1500 =

1

12

(2)(103)

P = 555.56 lb = 556 lb

Ans.

Shear Stress: From the shear diagram, Vmax = 0.500P = 277.78 lb. The notch is the

critical section. Using the shear formula for a rectangular section.

tallow =

350 =

3Vmax

2A

3(277.78)

2(2) h

h = 0.595 in.

Ans.

11–25. The simply supported joist is used in the

construction of a floor for a building. In order to keep the

floor low with respect to the sill beams C and D, the ends of

the joists are notched as shown. If the allowable shear stress

for the wood is tallow = 350 psi and the allowable bending

stress is sallow = 1700 psi, determine the smallest height h

so that the beam will support a load of P = 600 lb. Also,

will the entire joist safely support the load? Neglect the

stress concentration at the notch.

P

B

tallow =

1.5V

;

A

350 =

D

A

10 in.

600

= 300 lb

2

1.5(300)

(2)(h)

h = 0.643 in.

smax =

Ans.

4500(12)(5)

Mmax c

= 1620 psi 6 1700 psi OK

= 1

3

I

12 (2)(10)

Yes, the joist will safely support the load.

Ans.

851

h

15 ft

C

The reaction at the support is

2 in.

15 ft

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

11–26. Select the lightest-weight steel wide-flange beam

from Appendix B that will safely support the loading

shown. The allowable bending stress is sallow = 22 ksi and

the allowable shear stress is tallow = 12 ksi.

5 kip

18 kip ft

B

A

6 ft

From the moment diagram, Fig. a, Mmax = 48 kip # ft.

Sreq¿d =

=

Mmax

sallow

48(12)

22

= 26.18 in3

Select W 14 * 22 C Sx = 29.0 in3, d = 13.74 in. and tw = 0.230 in. D

From the shear diagram, Fig. a, Vmax = 5 kip. Provide the shear stress check for

W 14 * 22,

tmax =

=

Vmax

twd

5

0.230(13.74)

= 1.58 ksi 6 tallow = 12 ksi‚ (O.K!)

Use

Ans.

W14 * 22

W12 * 22 would work also.

852

12 ft

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11–27. The T-beam is made from two plates welded

together as shown. Determine the maximum uniform

distributed load w that can be safely supported on the beam

if the allowable bending stress is sallow = 150 MPa and the

allowable shear stress is tallow = 70 MPa.

w

A

1.5 m

1.5 m

200 mm

20 mm

200 mm

20 mm

The neutral axis passes through centroid c of the beam’s cross-section. The location

of c, Fig. b, is

y =

0.21(0.02)(0.2) + 0.1(0.2)(0.02)

©yA

=

©A

0.02(0.2) + 0.2(0.02)

= 0.155 m

I =

1

(0.2)(0.023) + 0.2(0.02)(0.055)2

12

+

1

(0.02)(0.23) + 0.02(0.2)(0.055)2

12

= 37.667 (10 - 6) m4

Referring to Fig. b,

Qmax = y¿A¿ = 0.0775(0.155)(0.02)

= 0.24025(10 - 3) m3

Referring to the moment diagram, Mmax = -3.375 w. Applying the flexure

formula with C = y = 0.155 m,

sallow =

Mmax c

;

I

150(106) =

3.375 w (0.155)

37.667(10 - 6)

w = 10.80(103) N>m

= 10.8 kN>m (Control!)

Ans.

Referring to the shear diagram, Vmax = 1.5w.

tallow =

Vmax Qmax

;

It

70(106) =

1.5 w C 0.24025(10 - 3) D

37.667(10 - 6)(0.02)

w = 146.33(103) N>m

= 146 kN>m

853

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*11–28. The beam is made of a ceramic material having

an allowable bending stress of sallow = 735 psi and an

allowable shear stress of tallow = 400 psi. Determine the

width b of the beam if the height h = 2b.

15 lb

10 lb

6 lb/in.

2 in.

6 in.

2 in.

h

b

Bending Stress: From the moment diagram, Mmax = 30.0 lb # in. Assume bending

controls the design. Applying the flexure formula.

sallow =

Mmax c

I

30.0

735 =

1

12

A 2b2 B

(b) (2b)3

b = 0.3941 in. = 0.394 in.

Ans.

Shear Stress: Provide a shear stress check using the shear formula for a rectangular

section. From the shear diagram, Vmax = 19.67 lb.

tmax =

=

3Vmax

2A

3(19.67)

2(0.3941)(2)(0.3941)

= 94.95 psi 6 tallow = 400 psi (O.K!)

854

## Mechanics of Materials, Seventh Edition

## Ebook mechanics of materials (7th edition) part 2

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