ch08 - mechanics book

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P8.1 During fabrication of a laminated timber arch, one of the 10 in. wide by 1 in. thick Douglas fir [E=

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

Solution

From Eq. (8.3):

1,900 ksi( 0.5 in.) 1.979 ksi

(40 ft)(12 in./f

1.979 k

t

si

)

x

Ey

Ans.


2/128



P8.2 A high-strength steel [E = 200 GPa] tube having an outside diameter of 80 mm and a wall

thickness of 3 mm is bent into a circular curve having a 52-m radius of curvature. Determine themaximum bending stress developed in the tube.

Solution

From Eq. (8.3):

200,000 MPa( 80 mm / 2) 153.846 MPa

(52 m)(1,000 mm/

153.8 MP

m)

axE

y

Ans.


3/128



P8.3A high-strength steel [E= 200 GPa] band saw blade wraps around a pulley that has a diameter o

450 mm. Determine the maximum bending stress developed in the blade. The blade is 12-mm wide and1-mm thick.

Solution

The radius of curvature of the band saw blade is:

450 mm 1 mm225.5 mm

2 2

From Eq. (8.3):

200,000 MPa( 0.5 mm) 443.459 MPa

225.5 mm443 MPa

x

Ey

Ans.


4/128



P8.4The boards for a concrete form are to be bent into a circular shape having an inside radius of 10 m.

What maximum thickness can be used for the boards if the normal stress is not to exceed 7 MPa?Assume that the modulus of elasticity for the wood is 12 GPa.

Solution

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

10,000 mm2

t

From Eq. (8.3):

12,000 MPa7 MPa

210,000 mm

2

x

E ty

t

Solve for t:

12,000 MPa 7 MPa 10,000 mm2 2

6,000 70,000 3.5

5,996.5

11.67 mm

70,000

t t

t t

t

t

Ans.


5/128



P8.5 A beam having a tee-shaped cross section is subjected to equal 12 kN-m bending moments, as

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

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

about thezaxis.

(b) the bending stress at pointH. State whether the normal stress atHis tensionor compression.(c) the maximum bending stress produced in the cross section. State whether the stress is tension or

compression.

FIGURE P8.5a FIGURE P8.5b

Solution

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

Shape AreaAi

yi

(from bottom) yiAi(mm ) (mm) (mm )

top flange 2,500.0 162.5 406,250.0

stem 3,750.0 75.0 281,250.0

6,250.0 mm 687,500.0 mm

3

2

687,500.0 mm

6,250.0 mm110.0 mm

i i

i

y Ay

A

(measured upward from bottom edge of stem) Ans.

Moment of inertia about the zaxis:

Shape IC d=yiy dA IC+ dA

(mm ) (mm) (mm ) (mm )

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

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

Moment of inertia about thezaxis (mm4) = 18,645,833.33418,646,000 mmzI Ans.


6/128



Section moduli:4

3

top

top

43

bot

bot

3

18,645,833.33 mm286,858.974 mm

(175 mm 110 mm)

18,645,833.33 mm169,507.576 mm

110 mm

169,500 mm

z

z

IS

c

IS

c

S

Ans.

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

4

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

18,654,833.33 mm

25.743 MPa 25.7 MPa (C)

x

z

M y

I

Ans.

(c) Maximum bending stress:The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the

cross section is at y= +65 mm, and the bottom of the cross section is at y= 110mm. The largerbending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom

of the cross section.

4

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

18,654,833.33 mm

70.793 MPa 70.8 MPa (T)

x

z

M y

I

Ans.


7/128



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

sectional dimensions of the beam are shown in Figure P8.6b. Determine:(a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus

about thezaxis.

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

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

compression.


Solution

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

Shape AreaAi

yi(from bottom) yiAi

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

left side 8.0 4.0 32.0

top flange 4.0 7.5 30.0

right side 8.0 4.0 32.0

20.0 in. 94.0 in.

3

2

94.0 i4.70

n.

20.0 ii

nn

..

i i

i

y Ay

A

(measured upward from bottom edge of section) Ans.



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

left side 42.667 0.700 3.920 46.587

top flange 0.333 2.800 31.360 31.693

right side 42.667 0.700 3.920 46.587

Moment of inertia about thezaxis (in.4) = 124.867

4124.9 in.zI Ans.


8/128



Section moduli:4

3

top

top

4

3

bot

bot

3

124.867 in.37.8384 in.

(8 in. 4.7 in.)

124.867 in.26.5674 in.

4.7 in

26.6 in.

.

z

z

IS

c

IS

c

S

Ans.

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

4

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

124.867 in.

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

x

z

M y

I

Ans.

(c) Maximum bending stress:

The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of thecross section is aty= +3.30 in., and the bottom of the cross section is at y= 4.7in. The larger bending

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

cross section.

4

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

124.867 in

2,940 psi (

.

2,935. i C)9 ps

x

z

M y

I

Ans.


9/128



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


about thezaxis.

(b) the bending stress at pointH. State whether the normal stress atHis tensionor compression.

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


Solution

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

Shape AreaAi


(mm ) (mm) (mm )

left side 400.0 25.0 10,000.0

bottom flange 272.0 4.0 1,088.0

right side 400.0 25.0 10,000.0

1,072.0 mm 21,088.0 mm

3

2

21,088.0 mm

1,072.0 mm 19.67 mm

i i

i

y A

y A

(measured upward from bottom edge of section) Ans.



(mm ) (mm) (mm ) (mm )

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

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

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

Moment of inertia about thezaxis (mm4) = 257,633.75

4

257,600 mmzI Ans.


10/128



Section moduli:4

3

top

top

43

bot

bot

3

257,633.75 mm8,494.814 mm

(50 mm 19.672 mm)

257,633.75 mm13,096.708 mm

19.672

8,495 mm

mm

z

z

IS

c

IS

c

S

Ans.

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

4

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

257,633.75 m

21.

m

21 3 MPa.293 (T)MPa

x

z

M y

I

Ans.

(c) Maximum bending stress:The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the

cross section is aty= +30.328 mm, and the bottom of the cross section is aty= 19.672mm. The largerbending stress magnitude occurs at the larger magnitude of these two values; in this case, at the top of

the cross section.

4

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

257,633.7

55.3 MPa (

5 mm

55.328 C)MPa

x

z

M y

I

Ans.


11/128



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


about thezaxis.

(b) the bending stress at pointH. State whether the normal stress atHis tensionor compression.

(c) the bending stress at pointK. State whether the normal stress atKis tensionor compression.(d) the maximum bending stress produced in the cross section. State whether the stress is tension or

compression.


Solution

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

Shape AreaAi

yi

(from bottom) yiAi(in. ) (in.) (in. )

top flange 12.0000 13.0000 156.0000

web 20.0000 7.0000 140.0000

bottom flange 20.0000 1.0000 20.0000

52.0000 in. 316.0000 in.

3

2

316.0 in.6.077 in.

52.06.08 in.

in.

i i

i

y Ay

A

(measured upward from bottom edge of bottom

flange) Ans.



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

top flange 4.000 6.923 575.148 579.148

web 166.667 0.923 17.041 183.708bottom flange 6.667 -5.077 515.503 522.170

Moment of inertia about thezaxis (in. ) = 1,285.026

Ans.


12/128



Section Moduli

43

43

6.0769 in.

14 in. 6.0769 in. 7.9231 in.

1,285.026 in.211.460 in.

6.0769 in.

1,285.026 in.162.188 in.

7.9231 in.The controlling section modulus is the smaller of the

bot

top

zbot

bot

ztop

top

c

c

IS

c

IS

c

3

two values; theref

162.2

ore,

in.S Ans.

Bending stress at point H:

From the flexure formula:

4

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

1,285.0256 i968 psi (T)

n.x

z

M y

I

Ans.

Bending stress at point K:

From the flexure formula:

4

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

1,285.026 in666 psi

.(C)x

z

M y

I

Ans.

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

formula:

4

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

1,285.021,295 psi

6 n.(T)

ix

z

M y

I

Ans.

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

modulus:

3

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

162.1,295 ps

1877 in.i

x

M

S Ans.

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


13/128



P8.9The cross-sectional dimensions of a beam are shown

in Figure P8.9.(a) If the bending stress at pointKis 43 MPa (C),

determine the internal bending momentMzacting about

thezcentroidal axis of the beam.

(b) Determine the bending stress at pointH. State whetherthe normal stress atHis tensionor compression.

FIGURE P8.9

Solution

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

Shape AreaAi


(mm ) (mm) (mm )

top flange 375.0 47.5 17,812.5

left stem 225.0 22.5 5,062.5

right stem 225.0 22.5 5,062.5825.0 mm 27,937.5 mm

3

2

27,937.5 mm33.864 mm 33.9 mm

825.0 mm

i i

i

y Ay

A

(measured upward from bottom of section)



(mm ) (mm) (mm ) (mm )

top flange 781.250 13.636 69,731.405 70,512.655

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

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

Moment of inertia about thezaxis (mm4) = 204,559.659

(a) Determine bending moment:

At pointK,y= 50 mm 5 mm 33.864 mm = 11.136 mm. The bending stress atKis x= 43 MPa;therefore, the bending moment magnitude can be determined from the flexure formula:

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

11.136 mm

789,850.765 N- 790 N-mmm

x

z

x z

M y

I

IM

y

Ans.

(b) Bending stress at point H:

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

4

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

2130.8 MPa (

04,559.659 mmT)x

z

M y

I

Ans.


14/128



P8.10The cross-sectional dimensions of the beam shown in Figure

P8.10 are d= 5.0 in., bf = 4.0 in., tf= 0.50 in., and tw= 0.25 in.(a) If the bending stress at pointHis 4,500 psi (T), determine the

internal bending momentMzacting about thezcentroidal axis of

the beam.

(b) Determine the bending stress at pointK. State whether thenormal stress atKis tensionor compression.

FIGURE P8.10

Solution

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

Shape AreaAi

yi


bottom flange 2.0000 0.2500 0.5000

stem 1.1250 2.7500 3.09383.1250 3.5938

3

2

3.5938 in.1.150 in.

3.1250 in.

i i

i

y Ay

A

(measured upward from bottom edge of section)



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

bottom flange 0.0417 0.9000 1.6200 1.6617

stem 1.8984 1.6000 2.8800 4.7784

Moment of inertia about thezaxis (in.4) = 6.4401(a) Determine bending moment:

At point H, y= 1.150 in. The bending stress at K is x= +4,500 psi; therefore, the bending momentmagnitude can be determined from the flexure formula:

4(4,500 psi)(6.4401 in. )25,200.407 lb-in.

1.150 in.2,100 lb-ft

zx

z

x zz

M y

I

IM

y

Ans.

(b) Bending stress at point K:

At pointH,y= 5.00 in.1.150 in. = 3.850 in. The bending stress is computed with the flexure formula:

4

(25,200.407 lb-in.)(3.850 in.)15,065.217 psi

615,070 psi

.4401(

iC)

n.

zx

z

M y

I Ans.


15/128



P8.11The dimensions of the double-box beam cross section

shown in Figure P8.11 are b= 150 mm, d= 50 mm, and t= 4mm. If the maximum allowable bending stress is 17 MPa,

determine the maximum internal bending momentMz

magnitude that can be applied to the beam.

FIGURE P8.11

Solution

Moment of inertia about zaxis:3 3

4(150 mm)(50 mm) (138 mm)(42 mm) 710,488 mm12 12

zI

Maximum internal bending moment Mz:

2 4(17 N/mm )(710,488 mm )483,131.8 N-mm

25 m483

mN-m

zx

z

x z

M c

I

IM

c

Ans.


16/128



P8.12The cross-sectional dimensions of a beam are shown in

Figure P8.12. The internal bending moment about thezcentroidal axis isMz= +2.70 kip-ft. Determine:

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

(b) the maximum compression bending stress in the beam.

FIGURE P8.12

Solution

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

Shape AreaAi


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

left stem 2.000 2.000 4.000

top flange 2.500 3.750 9.375

right stem 2.000 2.000 4.000

6.500 in. 17.375 in.

3

217.375 in. 2.673 in.6.500 in.

i i

i

y AyA

(measured upward from bottom edge of section)



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

left stem 2.66667 0.67308 0.90607 3.57273

top flange 0.05208 1.07692 2.89941 2.95149

right stem 2.66667 0.67308 0.90607 3.57273

Moment of inertia about thezaxis (in.4

) = 10.09696

(a) Determine maximum tension bending stress:

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

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

4

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

10.09696 in.8.58 ksi (T)x

z

M y

I

Ans.

(b) Determine maximum compression bending stress:

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

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

1.327 in.):

4

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

10.09694.26 ksi

6 in.(C)x

z

M y

I Ans.


17/128



P8.13The cross-sectional dimensions of a beam are

shown in Figure P8.13.(a) If the bending stress at point Kis 35.0 MPa (T),

determine the bending stress at point H. State

whether the normal stress at H is tension or

compression.

(b) If the allowable bending stress is b= 165 MPa,determine the magnitude of the maximum bending

momentMzthat can be supported by the beam.

FIGURE P8.13

Solution



(mm ) (mm) (mm ) (mm )

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

web 32,518,666.667 0.000 0.000 32,518,666.667bottom flange 540,000.000 160.000 184,320,000.000 184,860,000.000

Moment of inertia about thezaxis (mm4) = 402,238,666.667

(a) At pointK,y= 90 mm, and at pointH,y= 175 mm. The bending stress atKis x= +35 MPa, andthe bending stress is distributed linearly over the depth of the cross section. Therefore, the bending

stress atHcan be found from the ratio:

175 mm(35.0 MPa) 68.056 MPa

906

8.1 MPa (T)

mm

H K

H K

H

H KK

y y

y

y

Ans.

(b) Maximum internal bending moment Mz:

2 4(165 N/mm )(402,238,667 mm )379,253,600 N-mm

175 m3 N

m79 k -m

zx

z

x zz

M c

I

IM

c

Ans.


18/128



P8.14The cross-sectional dimensions of a beam are

shown in Figure P8.14.(a) If the bending stress at point K is 9.0 MPa (T),

determine the bending stress at point H. State

whether the normal stress at H is tension or

compression.

(b) If the allowable bending stress is b= 165 MPa,determine the magnitude of the maximum bending

momentMzthat can be supported by the beam.

FIGURE P8.14

Solution



(mm ) (mm) (mm ) (mm )

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

web 31,680 0 0 31,680

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

Moment of inertia about thezaxis (mm4) = 19,471,680

(a) At pointK,y= 60 mm, and at pointH,y= +90 mm. The bending stress atKis x= +9.0 MPa, andthe bending stress is distributed linearly over the depth of the cross section. Therefore, the bending

stress atHcan be found from the ratio:

90 mm(9.0 MP 13.50a) 13.50 MPa

60 mMPa (C

m)

H K

H K

HH K

K

y y

y

y

Ans.

(b) Maximum bending moment Mz:

2 4(165 N/mm )(19,471,680 mm )35,698,080 N-mm

90 mm35.7 kN-m

zx

z

x z

M c

I

IM

c

Ans.


19/128



P8.15The cross-sectional dimensions of the beam shown in

Figure P8.15 are a= 5.0 in., b= 6.0 in., d= 4.0 in., and t=0.5 in. The internal bending moment about thezcentroidal

axis isMz= 4.25 kip-ft. Determine:


(b) the maximum compression bending stress in the beam.

FIGURE P8.15

Solution

Centroid location in ydirection:

Shape AreaAi

yi


top flange 3.000 3.750 11.250

left web 1.500 2.000 3.000

left bottom flange 2.500 0.250 0.625

right web 1.500 2.000 3.000

right bottom flange 2.500 0.250 0.62511.000 18.500

3

2

18.50 in.1.6818 in.

11.0 in.

i i

i

y Ay

A

(measured upward from bottom edge of bottom flange)



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

top flange 0.0625 2.0682 12.8321 12.8946

left web 1.1250 0.3182 0.1519 1.2769

left bottom flange 0.0521 1.4318 5.1253 5.1773

right web 1.1250 0.3182 0.1519 1.2769

right bottom flange 0.0521 1.4318 5.1253 5.1773

Moment of inertia about thezaxis (in. ) = 25.8030

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

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

4

( 4.25 kip-ft)(4.0 in. 1.6818 in.)(1,000 lb/kip)(12 in./ft)

25.8030 in.

4,581.914 psi 4,580 psi (T)

x

z

M y

I

Ans.

(b) Maximum compression bending stress:The maximum compression bending stress will occur at the bottom surface of the cross section. From

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

4

( 4.25 kip-ft)( 1.6816 in.)(1,000 lb/kip)(12 in./ft)

25.8030 in.

3,324.134 psi 3,320 psi (C)

x

z

M y

I

Ans.


20/128



P8.16 The cross-sectional dimensions of a beam are

shown in Figure P8.16. The internal bending momentabout the z centroidal axis is Mz = +270 lb-ft.

Determine:


(b) the maximum compression bending stress in thebeam.

Solution

FIGURE P8.16


Shape AreaAi

yi


bottom flange 0.40625 0.06250 0.02539

left web 0.28125 1.25000 0.35156

left top flange 0.09375 2.43750 0.22852

right web 0.28125 1.25000 0.35156right top flange 0.09375 2.43750 0.22852

1.15625 in. 1.18555 in.

3

2

1.18555 in.1.0253 in.

1.15625 in.

i i

i

y Ay

A

(measured upward from bottom edge of bottom flange)



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

bottom flange 0.000529 0.962838 0.376617 0.377146

left web 0.118652 0.224662 0.014196 0.132848left top flange 0.000122 1.412162 0.186956 0.187079

right web 0.118652 0.224662 0.014196 0.132848

right top flange 0.000122 1.412162 0.186956 0.187079


(a) Maximum tension bending stress:

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

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

4

(270 lb-ft)( 1.0253 in.)(12 in./ft)

3,26 3,270 psi6.446 psi1.016999 (i )n. Tx z

M y

I

Ans.

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

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

section is:

4

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

1.016999 in.(C)x

z

M y

I Ans.


21/128



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

section shown in Figure P8.17b. Determine the maximum tension and compression bending stressesproduced in segmentBCof the beam.


Solution


Shape AreaAi

yi

(from bottom) yiAi(mm ) (mm) (mm )

top flange 3,000.0 167.5 502,500.0

stem 1,440.0 80.0 115,200.0

4,440 mm 617,700 mm

3

2

617,700 mm139.1216 mm

4,440 mm

i i

i

y Ay

A

(measured upward from bottom edge of stem)


Shape IC d=yiy dA IC+ dA(mm ) (mm) (mm ) (mm )

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

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

Moment of inertia about thezaxis (mm ) 10,577,574.32


22/128



Shear-force and bending-moment diagrams:

The maximum moment occurs betweenBand C. The moment magnitude is 12 kN-m.

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

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

6 4(12 kN-m)( 139.1216 mm)(1,000 N/kN)(1,000 mm/m)10.5776 10 m157.8 MPa )

m(Tx

z

M yI

Ans.

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

6 4

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

10.5776 10 mm

40.7 MPa 40.7 MPa (C)

x

z

M y

I

Ans.


23/128



P8.18Two vertical forces ofP= 240 lb are applied to a simply supported beam (Figure P8.18a) having

the cross section shown in Figure P8.18b. Using a= 30 in.,L= 84 in., b= 3.0 in., d= 4.0 in., and t= 0.5in., calculate the maximum tension and compression bending stresses produced in segmentBCof the

beam.


Solution


Shape AreaAi


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

left stem 2.000 2.000 4.000bottom flange 1.000 0.250 0.250

right stem 2.000 2.000 4.000

5.000 8.250

3

2

8.250 in.1.65 in.

5.000 in.

i i

i

y Ay

A

(measured upward from bottom edge of stem)



(in. ) (in.) (in. ) (in. )left stem 2.66667 0.35000 0.24500 2.91167

bottom flange 0.02083 1.40000 1.96000 1.98083

right stem 2.66667 0.35000 0.24500 2.91167

Moment of inertia about thezaxis (in. ) 7.80417


24/128



Shear-force and bending-moment diagrams:

The maximum moment occurs betweenBand C. The moment magnitude is 7,200 lb-in.

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

cross section, wherey= 2.350 in.:

4

( 7,200 lb-in.)(2.350 in.)2,168.073 psi

7.82,170 psi (T)

0417 in.x

z

M y

I

Ans.

Maximum compression bending stress:For a negative bending moment, the maximum compression bending stress will occur at the bottomsurface of this cross section at y= 1.650 in. From the flexure formula, the bending stress at the bottom

of the U shape is:

4

( 7,200 lb-in.)( 1.650 in.)1,522.264 psi

7.81,522 psi

0417 in.(C)x

z

M y

I

Ans.

P8.19A WT230 26 standard steel shape is used to support the loads shown on the beam in Figure

P8.19a. The dimensions from the top and bottom of the shape to the centroidal axis are shown on the

sketch of the cross section (Figure P8.19b). Consider the entire 4-m length of the beam and determine:(a) the maximum tension bending stress at any location along the beam, and

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


25/128




Solution

Section properties

From Appendix B: 6 416.7 10 mmzI

Shear-force and bending-moment diagrams

Maximum bending momentspositiveM= 13.61 kN-m

negativeM= 20.00 kN-m

Bending stresses at max positive moment2

6 4

2

6 4

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

16.7 10 mm

49.5 MPa (C)

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

16.7 10 mm

133.9 MPa (T)

x

x

Bending stresses at max negative moment2

6 4

2

6 4

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

16.7 10 mm

72.7 MPa (T)

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

16.7 10 mm

196.8 MPa (C)

x

x

(a) Maximum tension bending stress

(b) Maximum compression bending stress

133.9 MPa (T)

196.8 MPa (C)

Ans.

Ans.


26/128



P8.20A WT305 41 standard steel shape is used to support the loads shown on the beam in Figure

P8.20a. The dimensions from the top and bottom of the shape to the centroidal axis are shown on thesketch of the cross section (Figure P8.19b). Consider the entire 10-m length of the beam and determine:

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



Solution

Section properties

From Appendix B:6 4

48.7 10 mmzI


Maximum bending momentspositiveM= 45.84 kN-m


Bending stresses at max positive moment2

6 4

2

6 4

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

48.7 10 mm

83.7 MPa (C)(45.84 kN-m)( 211.1 mm)(1,000)

48.7 10 mm

198.7 MPa (T)

x

x

Bending stresses at max negative moment2

6 4

2

6 4

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

48.7 10 mm

43.8 MPa (T)

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

48.7 10 mm

104.0 MPa (C)

x

x



198.7 MPa (T)

104.0 MPa (C)

Ans.

Ans.


27/128



P8.21 A steel tee shape is used to support the loads shown on the beam in Figure P8.21a. The

dimensions of the shape are shown in Figure P8.21b. Consider the entire 24-ft length of the beam anddetermine:




SolutionCentroid location in ydirection:

Shape AreaAi

yi


top flange 24.0000 19.2500 462.0000

stem 13.8750 9.2500 128.3438

37.875 in. 590.3438 in.

3

2

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

37.8750 in.

4.4134 in. (from top of shape to centroid)

i i

i

y Ay

A



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

top flange 4.5000 3.6634 322.0861 326.5861

stem 395.7266 6.3366 557.1219 952.8484

Moment of inertia about thezaxis (in. ) = 1,279.4345


28/128




Maximum bending momentspositiveM= 100.75 kip-ft

negativeM= 68.00 kip-ft

Bending stresses at max positive moment

4

4

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

1,279.4345 in.

4.17 ksi (C)

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

1,279.4345 in.

14.73 ksi (T)

x

x

Bending stresses at max negative moment

4

4

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

1,279.4345 in.

2.81 ksi (T)

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

1,279.4345 in.

9.94 ksi (C)

x

x



14.73 ksi (T)

9.94 ksi (C)

Ans.

Ans.


29/128



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





Solution


Shape AreaAi


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

top flange 20.0 11.0 220.0

web 16.0 6.0 96.0

bottom flange 12.0 1.0 12.0

48.0 in. 328.0 in.

3

2


48.0 in.


i i

i

y Ay

A



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

top flange 6.667 4.167 347.222 353.889

web 85.333 0.833 11.111 96.444

bottom flange 4.000 5.833 408.333 412.333



30/128




Maximum bending momentspositiveM= 10,580 lb-ft

negativeM= 8,400 lb-ft


4

4

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

862.667 in.

760.4 psi (C)

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

862.667 in.

1,005.6 psi (T)

x

x


4

4

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

862.667 in.

603.7 psi (T)

( 8,400 lb-ft)( 6.8333 in.)(12 in./ft)862.667 in.

798.5 psi (C)

x

x


(b) Maximum compression bending stres

1,006 psi (T)

799 psi (C)s

Ans.

Ans.


31/128



P8.23 A channel shape is used to support the loads shown on the beam in Figure P8.23a. The





Solution


Shape AreaAi

yi

(from bottom) yiAi

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

left stem 3.000 3.000 9.000

top flange 5.500 5.750 31.625

right stem 3.000 3.000 9.000

11.500 in. 49.625 in.

3

2


11.500 in.


i i

i

y Ay

A



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

left stem 9.0000 1.3152 5.1894 14.1894

top flange 0.1146 1.4348 11.3223 11.4369

right stem 9.0000 1.3152 5.1894 14.1894



32/128




Maximum bending momentspositiveM= 8,850 lb-ft

negativeM= 9,839 lb-ft


4

4

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

39.8157 in.

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

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

39.8157 in.

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

x

x


4

4

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

39.8157 in.

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

( 9,839 lb-ft)( 4.3152 in.)(12 in./ft)39.8157 in.

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

x

x



11.51 ksi (T)

12.80 ksi (C)

Ans.

Ans.


33/128



P8.24A W360 72 standard steel shape is used to support the loads shown on the beam in Figure

P8.24a. The shape is oriented so that bending occurs about the weak axis as shown in Figure P8.24b.Consider the entire 6-m length of the beam and determine:




Solution

Section properties

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

Shear-force and bending-moment diagramsMaximum bending moments

positiveM= 31.50 kN-m


Since the shape is symmetric about the z axis,

the largest bending stresses will occur at the

location of the largest moment magnitude either positive or negative. In this case, the

largest bending stresses will occur where the

moment magnitude is 31.50 kN-m.

Bending stresses at maximum moment2

6 4

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

21.4 10 mm

150.1 MPa (T) and (C)

x



150.1 MPa (T)

150.1 MPa (C)

Ans.

Ans.


34/128



P8.25 A 20-mm-diameter solid steel

shaft supports loads PA = 500 N, PC =1,750 N, and PE = 500 N as shown in

Figure P8.25/26. AssumeL1= 90 mm,L2

= 260 mm, L3= 140 mm, and L4= 160

mm. The bearing atBcan be idealized asa roller support and the bearing at Dcan

be idealized as a pin support. Determine

the magnitude and location of themaximum bending stress in the shaft.

FIGURE P8.25/26

Solution

Section properties

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

zI D


Maximum bending momentspositiveM= 91,500 N-mm

negativeM= 80,000 N-mm

Since the circular cross section is symmetricabout the z axis, the largest bending stresses

will occur at the location of the largest moment

magnitudeeither positive or negative. In thiscase, the largest bending stresses will occur at

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

mm.

Bending stresses at maximum moment

4

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

7,853.9816

1

mm

16.5 MPa

x

Ans.


35/128



P8.26 A 1.75-in.-diameter solid steel

shaft supports loads PA = 250 lb, PC =600 lb, and PE= 250 lb as shown in

Figure P8.28. AssumeL1= 9 in.,L2= 24

in., L3 = 12 in., and L4= 15 in. The

bearing at B can be idealized as a rollersupport and the bearing at D can be

idealized as a pin support. Determine the

magnitude and location of the maximumbending stress in the shaft.

FIGURE P8.25/26

Solution

Section properties

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

I D


Maximum bending moments

positiveM= 1,550 lb-in.

negativeM= 3,750 lb-in.

Since the circular cross section is symmetricabout the z axis, the largest bending stresseswill occur at the location of the largest moment

magnitudeeither positive or negative. In this

case, the largest bending stresses will occur at

support D, where the moment magnitude is3,750 lb-in.

Bending stresses at maximum moment

4

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

0.460386 in.

7,130 psi

x

Ans.


36/128



P8.27 The steel beam in Figure P8.27a/28ahas the cross section shown in Figure P8.27b/28b. The

beam length isL= 6.0 m, and the cross-sectional dimensions are d= 350 mm, bf= 205 mm, tf= 14 mm,and tw= 8 mm. Calculate the largest intensity of distributed load w0that can be supported by this beam

if the allowable bending stress is 200 MPa.

FIGURE P8.27a/28a FIGURE P8.27b/28b

Solution


Shape AreaAi

yi

(from bottom) IC d=yiy dA IC+ dA

(mm2

) (mm) (mm4

) (mm) (mm4

) (mm4

)top flange 2,870 343 46,876.67 168.00 81,002,880.00 81,049,756.67

web 2,576 175 22,257,498.67 0.00 0.00 22,257,498.67

bottom flange 2,870 7 46,876.67 -168.00 81,002,880.00 81,049,756.67

Moment of inertia about thezaxis (mm4) = 184,357,012.00

Allowable bending moment: Based on the allowable bending stress, the maximum moment that can be

applied to this beam is2 4

6allowallow

(200 N/mm )(184,357,012 mm )210.694 10 N-mm

350 mm / 2

zIMc

Moment in the simply supported beam:From a FBD of the beam, determine the reactionforce atA:

0

0

10

2 2

4

C y

y

LM w L A L

w LA

The maximum moment will occur in the center ofthe span atB. From the FBD shown, determine

the bending momentM:

0 0

2

0

04 6 4 2

12

w L L w L LM M

w LM


37/128



Largest intensity of distributed load w0: Equate the allowable moment with the moment produced at

midspan for this beam.2

0allow

6

allow0 2 2

12

12 12(210.694 10 N-mm)70.231 N/mm

(6,000 mm70.2 k m

)N/

w LM

Mw

L

Ans.


38/128



P8.28 The steel beam in Figure P8.27a/28ahas the cross section shown in Figure P8.27b/28b. The

beam length isL= 22 ft, and the cross-sectional dimensions are d= 16.3 in., bf= 10.0 in., tf= 0.665 in.,and tw= 0.395 in. Calculate the maximum bending stress in the beam if w0= 6 kips/ft.


Solution


Shape AreaAi

yi

(from bottom) IC d=yiy dA IC+ dA

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

top flange 6.6500 15.9675 0.2451 7.8175 406.4035 406.6486web 5.9131 8.1500 110.4285 0.0000 0.0000 110.4285

bottom flange 6.6500 0.3325 0.2451 -7.8175 406.4035 406.6486

Moment of inertia about thezaxis (in.4) = 923.7256

Moment in the simply supported beam:

From a FBD of the beam, determine the reaction

force atA:

0

0

10

2 2

4

C y

y

LM w L A L

w LA

The maximum moment will occur in the center of

the span atB. From the FBD shown, determine

the bending momentM:

0 0

2

0

2

04 6 4 2

12

(6 kips/ft)(22 ft)12

242 kip-ft 2,904 kip-in.

w L L w L LM M

w LM

Maximum Bending Stress: The maximum bending stress in the beam occurs at midspan:

4

(2,904 kip-in.)(16.3 in. / 2)25.622 ksi

923.725625.6 ksi

in.x

z

Mc

I Ans.


39/128



P8.29 A HSS12 8 1/2 standard steel

shape is used to support the loads shown onthe beam in Figure P8.29. The shape is

oriented so that bending occurs about the

strong axis. Determine the magnitude and

location of the maximum bending stress inthe beam.

FIGURE P8.29

Solution

Section properties

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


Maximum bending momentspositiveM= 124.59 kip-ft

negativeM= 72.00 kip-ft

Since the shape is symmetric about the zaxis,


location of the largest moment magnitude

either positive or negative. In this case, thelargest bending stresses will occur at C, where

the moment magnitude is 124.59 kip-ft.

Bending stresses at max moment magnitude

4

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

333 in.

26.9 ksix

Ans.


40/128



P8.30 A W410 60 standard steel

shape is used to support the loadsshown on the beam in Figure P8.30.

The shape is oriented so that bending

occurs about the strong axis.

Determine the magnitude and locationof the maximum bending stress in the

beam.

FIGURE P8.30

Solution

Section properties

From Appendix B: 6 4216 10 mm 406 mmzI d


Maximum bending momentspositiveM= 50 kN-m

negativeM= 70 kN-m

Since the shape is symmetric about the zaxis,


location of the largest moment magnitude either positive or negative. In this case, the

largest bending stresses will occur between B

and C, where the moment magnitude is 70 kN-m.

Bending stresses at max moment magnitude2

6 4

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

216 10 m

65.8 MPa

m

x


41/128



P8.31 A solid steel shaft supports

loads PA= 200 lb and PD= 300 lb asshown in Figure P8.31. AssumeL1= 6

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

bearing atBcan be idealized as a roller

support and the bearing at C can beidealized as a pin support. If the

allowable bending stress is 8 ksi,

determine the minimum diameter thatcan be used for the shaft.

FIGURE P8.31

Solution


Maximum bending moment magnitudeM= 3,000 lb-in.

Minimum required section modulus

33,000 lb-in. 0.375 in.8,000 psi

x

x

M

S

MS

Section modulus for solid circular section3

32

dS

Minimum shaft diameter3

30.375 in.

1.

32

563 in.

d

d

Ans.


42/128



P8.32A solid steel shaft supports

loadsPA= 250 N andPC= 620 N asshown in Figure P8.32. AssumeL1=

500 mm,L2= 700 mm, andL3= 600

mm. The bearing atBcan be idealized

as a roller support and the bearing atDcan be idealized as a pin support. If the

allowable bending stress is 105 MPa,

determine the minimum diameter thatcan be used for the shaft.

FIGURE P8.32

Solution


Maximum bending moment magnitudeM= 142,615.4 N-mm


3

2

142,615.4 N-mm1,358.242 mm

105 N/mm

x

x

M

S

MS

Section modulus for solid circular section3

32

dS

Minimum shaft diameter3

31,358.242 mm

24.0 mm

32

d

d

Ans.


43/128



P8.33A simply supported wood beam (Figure P8.33a/34a) with a span ofL= 15 ft supports a uniformly

distributed load of w0= 320 lb/ft. The allowable bending stress of the wood is 1,200 psi. If the aspectratio of the solid rectangular wood beam is specified as h/b= 2.0 (Figure P8.33b/34b), calculate the

minimum width bthat can be used for the beam.


Solution


Maximum bending moment magnitudeM= 7,111 lb-ft = 85,332 lb-in.


385,332 lb-in. 71.110 in.1,200 psi

x

x

MS

MS

Section modulus for solid rectangular

section3 2/12

/ 2 6

I bh bhS

c h

The aspect ratio of the solid rectangular wood beam is specified as h/b = 2.0; therefore, the sectionmodulus can be expressed as:

2 2 33(2.0 ) 4 0.6667

6 6 6

bh b b bS b

Minimum allowable beam width3 30.6667 71.110

4.74 in

in

.

.b

b

Ans.

The corresponding beam height his

/ 2.0 2.0 2.0(4.74 in.) 9.48 in.h b h b


44/128



P8.34A simply supported wood beam (Figure P8.33a/34a) with a span ofL= 5 m supports a uniformly

distributed load of w0. The beam width is b= 140 mm and the beam height is h= 260 mm (FigureP8.33b/34b). The allowable bending stress of the wood is 9.5 MPa. Calculate the magnitude of the

maximum load w0that may be carried by the beam.


Solution

Moment of inertia for rectangular cross section about horizontal centroidal axis3 3

4(140 mm)(260 mm) 205,053,333 mm12 12

bhI

Maximum allowable moment

2 4(9.5 N/mm )(205,053,333 mm )14,984,667 N-mm 14,984.667 N-m

130 mm

x

x

Mc

I

IM

c

Equilibrium

Determine the reaction forces on the beam interms of the distributed load intensity w0:

0

0

0

0

2 20

3 6

29

2 20

3 3

4

9

A y

y

C y

y

L LM C L w

C w L

L LM A L w

A w L


45/128



Construct the shear-force diagram for the beam

with its loading. Calculate the area under theshear-force diagram to determine an expression

for the maximum bending moment.

2

max 0 0

1 4 4 8

2 9 9 81M w L L w L

Determine distributed load intensity

Equate the moment expression to themaximum allowable moment that can be

applied to the rectangular cross section:

2

max 0

814,984.667 N-m

81M w L

Solve for the distributed load w0:2

2 00

0 2

8 8 (5 m)14,984.667 N-m

81 81

81(14,984.667 N-m)6,068.79 N/m

8(5 m)6.07 kN/m

ww L

w

Ans.


46/128



P8.35A cantilever timber beam (Figure P8.35a/36a) with a span ofL= 3.6 m supports a linearly

distributed load with maximum intensity of w0. The beam width is b= 240 mm and the beam height is h= 180 mm (Figure P8.35b/36b). The allowable bending stress of the wood is 7.6 MPa. Calculate the

magnitude of the maximum load w0that may be carried by the beam.


Solution

Moment of inertia for rectangular cross section about horizontal centroidal axis3 3

4(240 mm)(180 mm) 116,640,000 mm12 12

bhI

Maximum allowable moment

2 4(7.6 N/mm )(116,640,000 mm )9,849,600 N-mm 9,849.6 N-m

90 mm

x

x

McI

IM

c

Maximum moment magnitude:

The maximum bending moment magnitude in the cantilever beam occurs at supportA:2

0 0max

2 3 6

w L L w LM

Solve for the distributed load w0:2 2

0 0

0 2

(3.6 m)9,849.6 N-m

6 6

6(9,849.6 N-m)4,560 N/m

(3.6 m)4.56 kN/m

w L w

w

Ans.


47/128



P8.36A cantilever timber beam (Figure P8.35a/36a) with a span ofL= 15 ft supports a linearly

distributed load with maximum intensity of w0= 420 lb/ft. The allowable bending stress of the wood is1,400 psi. If the aspect ratio of the solid rectangular timber is specified as h/b= 0.75 (Figure

P8.35b/36b), determine the minimum width bthat can be used for the beam.


Solution

Maximum moment magnitude:The maximum bending moment magnitude in the cantilever beam occurs at supportA:

2 2

0 0max

(420 lb/ft)(15 ft)15,750 lb-ft 189,000 lb-in.

2 3 6 6

w L L w LM

Minimum required section modulus3189,000 lb-in. 135.0 in.

1,400 psix

x

M MS

S

Section modulus for solid rectangular section3 2/12

/ 2 6

I bh bhS

c h

The aspect ratio of the solid rectangular wood beam is specified as h/b= 0.75; therefore, the section

modulus can be expressed as:2 2 3

3(0.75 ) 0.5625 0.093756 6 6

bh b b bS b

Minimum allowable beam width3 3

11.29

0.09375 135.0 in.

11.2924 in .. in

b

b

Ans.

The corresponding beam height his

/ 0.75

0.75 0.75(11.2924 in.) 8.47 in.

h b

h b


48/128



P8.37 The beam shown in Figure P8.37 will

be constructed from a standard steel W-shapeusing an allowable bending stress of 24 ksi.

(a) Develop a list of five acceptable shapes

that could be used for this beam. On this

list, include the most economical W10,W12, W14, W16, and W18 shapes.

(b) Select the most economical W shape for

this beam.FIGURE P8.37

Solution


Maximum bending moment magnitude

M= 90 kip-ft


3(90 kip-ft)(12 in./ft) 45 in.24 ksi

x

x

M

S

MS

(a) Acceptable steel W-shapes3

3

3

3

3

W10 45, 49.1 in.

W12 40, 51.5 in.

W14 34, 48.6 in.

W16 31, 47.2 in.

W18 35, 57.6 in.

S

S

S

S

S

(b) Most economical W-shape

W16 31 Ans.


49/128



P8.38The beam shown in Figure P8.38 will be

constructed from a standard steel W-shape usingan allowable bending stress of 165 MPa.

(a) Develop a list of four acceptable shapes that

could be used for this beam. Include the most

economical W360, W410, W460, and W530shapes on the list of possibilities.

(b) Select the most economical W shape for this

beam.FIGURE P8.38

Solution



M= 206.630 kN-m


23 3

2

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

165 N/mm

x

x

M

S

MS

(a) Acceptable steel W-shapes3 3

3 3

3 3

3 3

W360 79, 1,270 10 mm

W410 75, 1,330 10 mm

W460 74, 1,460 10 mm

W530 66, 1,340 10 mm

S

S

S

S


W530 66 Ans.


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P8.39The beam shown in Figure P8.39 will be

constructed from a standard steel W-shape usingan allowable bending stress of 165 MPa.

(a) Develop a list of four acceptable shapes that

could be used for this beam. Include the

most economical W360, W410, W460, andW530 shapes on the list of possibilities.

(b) Select the most economical W shape for this

beam.FIGURE P8.39

Solution



M= 238.57 kN-m


23 3

2

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

165 N/mm

x

x

M

S

MS


3 3

3 3

3 3

W360 101, 1,690 10 mm

W410 85, 1,510 10 mm

W460 74, 1,460 10 mm

W530 74, 1,550 10 mm

S

S

S

S


orW460 74 W530 74 Ans.


51/128



P8.40 The beam shown in Figure P8.40 will be constructed from a

standard steel W-shape using an allowable bending stress of 165 MPa.(a) Develop a list of four acceptable shapes that could be used for this

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

shapes on the list of possibilities.

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

FIGURE P8.40

Solution

Maximum moment magnitude:

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

max

6

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

2 3

105.0 kN-m 105.0 10 N-mm

M


23 3

2

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

165 N/mm

x

x

M

S

MS


3 3

3 3

3 3

W310 60, 844 10 mm

W360 44, 688 10 mm

W410 46.1, 773 10 mm

W460 52, 944 10 mm

S

S

S

S


W360 44 Ans.


52/128



P8.41 The beam shown in Figure P8.41 will be

constructed from a standard steel HSS-shapeusing an allowable bending stress of 30 ksi.

(a) Develop a list of three acceptable shapes that

could be used for this beam. On this list,

include the most economical HSS8, HSS10,and HSS12 shapes.

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

beam.FIGURE P8.41

Solution



M= 45.56 kip-ft


3(45.56 kip-ft)(12 in./ft) 18.22 in.30 ksi

x

x

M

S

MS

(a) Acceptable steel HSS shapes

3

3

3

3

HSS8 none are acceptable

HSS10 4 3 / 8, 20.8 in.

HSS10 6 3 / 8, 27.4 in.

HSS12 6 3 / 8, 35.9 in.

HSS12 8 3 / 8, 43.7 in.

S

S

S

S

(b) Most economical HSS shape

HSS10 4 3 / 8 Ans.


53/128



P8.42A composite beam is fabricated by bolting two 3 in. wide 12 in. deep timber planks to the sidesof a 0.50 in. 12 in. steel plate (Figure P8.42b). The moduli of elasticity of the timber and the steel are

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

two concentrated loadsP, which are applied at the quarter points of the span (Figure P8.42a).

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

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

respectively. Determine the largest acceptable magnitude for concentrated loads P. (You may neglectthe weight of the beam in your calculations.)

FIGURE P8.42a

FIGURE P8.42b

Solution

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

2

1

30,000 ksi16.6667

1,800 ksi

En

E

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

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

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

Moment of inertia about the horizontal centroidal axis


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

timber (1) 864 0 0 864

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

Moment of inertia about thezaxis = 2,064 in.

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

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

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


54/128



1 4

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

2,523

064 in.psi

My

I

Ans.

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

maximum bending stress in steel plate (2) is:

2 4(180 kip-in.)( 6 in.)(16.6667) 8.7209 ksi2,0648,720

np

.s

iiMyn

I

Ans.

Determine maximum P

If the allowable bending stress in the timber is 1,200 psi, then the maximum bending moment that maybe supported by the beam is:

4

11 max

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

6 in.

IMyM

I y

If the allowable bending stress in the steel is 165 MPa, then the maximum bending moment that may besupported by the beam is:

4

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

(16.667)(6 in.)IMyn M

I n y

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

magnitude is of interest here.

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

maximum concentrated load magnitudePthat can be supported is found from:

max

max

(5 ft)

412.80 kip-in.

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

M P

MP

Ans.


55/128



P8.43 The cross section of a composite beam that

consists of 4-mm-thick fiberglass faces bonded to a 20-mm-thick particleboard core is shown in Figure P8.43.

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

acting about the z axis. The elastic moduli for the

fiberglass and the particleboard are 30 GPa and 10 GPa,respectively. Determine:

(a) the maximum bending stresses in the fiberglass

faces and the particleboard core.(b) the stress in the fiberglass at the joint where the two

materials are bonded together.

FIGURE P8.43

Solution

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

2

1

30 GPa3

10 GPa

En

E

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

the modular ratio: b2, trans= 3(50 mm) = 150 mm. Thus, for calculation purposes, the 50 mm 4 mmfiberglass faces are replaced by particleboard faces that are 150-mm wide and 4-mm thick.

Moment of inertia about the horizontal centroidal axis


(mm ) (mm) (mm ) (mm )

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

particleboard core 33,333.33 0 0 33,333.33

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

Moment of inertia about thezaxis = 207,733.33 mm

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

1 4

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

207,733.332.65 MPa

mm

My

I

Ans.

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

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

2 4

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

207,733.33 mm11.12 MPa

Myn

I

Ans.

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

2 4

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

207,733.33 m7.94

mMPa

Myn

I

Ans.


56/128



P8.44A composite bea

ch08 - mechanics book

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