Mechanics of Materials Solutions Chapter09 Probs18 37

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9.18 A 50-mm-diameter solid steel shaftsupports loads P A = 1.5 kN and P C = 3.0 kN,

as shown in Fig. P9.18. Assume L1 = 150

mm, L2 = 300 mm, and L3 = 225 mm. The bearing at B can be idealized as a roller

support and the bearing at D can be idealized

as a pin support. Determine the magnitude

and location of:

(a) the maximum horizontal shear stress inthe shaft.

(b) the maximum tension bending stress inthe shaft.

Fig. P9.18

Solution

Section properties:

4 4

4

3 3

3

(50 mm)64 64

306,796.158 mm

(50 mm)12 12

10,416.667 mm

I D

DQ

π π = =

=

= =

=

Maximum shear force magnitude:

V max = 1.71 kN (between B and C )

Maximum bending moment magnitude:

M max = 289.29 kN-mm (at C )

(a) Maximum horizontal shear stress:3

4

(1,710 N)(10,416.667 mm )

(306,796.158 mm )(50 mm)

1.161 MPa (at neutral axis between and )

V Q

I t

B C

τ = =

= Ans.

(b) Maximum tension bending stress:

4

(289.29 kN-mm)( 50 mm/2)(1,000 N/kN)

306,796.158 mm

23.574 MPa 23.6 MPa (T) (on bottom of shaft at )

x

M y

I

C

σ −

= − = −

= = Ans.





9.19 A 1.25-in.-diameter solid steel shaftsupports loads P A = 400 lb and P C = 900 lb,

as shown in Fig. P9.19. Assume L1 = 6 in.,

L2 = 12 in., and L3 = 8 in. The bearing at Bcan be idealized as a roller support and the

bearing at D can be idealized as a pin

support. Determine the magnitude and

location of:



Fig. P9.19

Solution

Section properties:

4 4

4

3 3

3

(1.25 in.)64 64

0.119842 in.

(1.25 in.)12 12

0.162760 in.

I D

DQ

π π = =

=

= =

=


V max = 480 lb (between B and C )


M max = 3,360 lb-in. (at C )


4

(480 lb)(0.162760 in. )

(0.119842 in. )(1.25 in.)

521.519 psi 522 psi (at neutral axis between and )

V Q

I t

B C

τ = =

= = Ans.


4

(3,360 lb-in.)( 1.25 in./2)

0.119842 in.

17, 523.022 psi 17, 520 psi (T) (on bottom of shaft at )

x

M y

I

C

σ −

= − = −

= = Ans.





9.20 A 1.00-in.-diameter solid steel shaftsupports loads P A = 500 lb and P D = 300 lb,

as shown in Fig. P9.20. Assume L1 = 5 in.,

L2 = 16 in., and L3 = 8 in. The bearing at Bcan be idealized as a roller support and the

bearing at C can be idealized as a pin support.

Determine the magnitude and location of:

(a) the maximum horizontal shear stress in

the shaft.(b) the maximum tension bending stress in

the shaft.

Fig. P9.20

Solution

Section properties:

4 4

4

3 3

3

(1.00 in.)64 64

0.049087 in.

(1.00 in.)

12 12

0.083333 in.

I D

D

Q

π π = =

=

= =

=


V max = 500 lb (between A and B)


M max = 2,500 lb-in. (at B)


4

(500 lb)(0.083333 in. )

(0.049087 in. )(1.00 in.)

848.826 psi 849 psi (at neutral axis between and )

V Q

I t

A B

τ = =

= = Ans.


4

( 2,500 lb-in.)(1.00 in./2)

0.049087 in.

25, 464.772 psi 25, 500 psi (T) (on top of shaft at )

x

M y

I

B

σ −

= − = −

= = Ans.





9.21 A 20-mm-diameter solid steel shaftsupports loads P A = 1,300 N and P D = 900 N,

as shown in Fig. P9.21. Assume L1 = 75 mm,

L2 = 300 mm, and L3 = 125 mm. The bearingat B can be idealized as a roller support and

the bearing at C can be idealized as a pin

support. Determine the magnitude and

location of:



Fig. P9.21

Solution

Section properties:

4 4

4

3 3

3

(20 mm)64 64

7,853.982 mm

(20 mm)12 12

666.667 mm

I D

DQ

π π = =

=

= =

=


V max = 1,300 N (between A and B)


M max = 112,500 N-mm (at C )


4

(1,300 N)(666.667 mm )

(7,853.982 mm )(20 mm)

5.517 MPa 5.52 MPa (at neutral axis between and )

V Q

I t

A B

τ = =

= = Ans.


4

( 112,500 N-mm)(20 mm/2)

7,853.982 mm

143.239 MPa 143.2 MPa (T) (on top of shaft at )

x

M y

I

C

σ −

= − = −

= = Ans.





9.22 A 1.25-in.-diameter solid steel shaftsupports loads P A = 600 lb, P C = 1,600 lb, and

P E = 400 lb, as shown in Fig. P9.22. Assume

L1 = 6 in., L2 = 15 in., L3 = 8 in., and L4 = 10in. The bearing at B can be idealized as a roller

support and the bearing at D can be idealized

as a pin support. Determine the magnitude and

location of:

(a) the maximum horizontal shear stress in theshaft.

(b) the maximum tension bending stress in theshaft.

Fig. P9.22

Solution

Section properties:

4 4

4

3 3

3

(1.25 in.)64 64

0.119842 in.

(1.25 in.)12 12

0.162760 in.

I D

DQ

π π = =

=

= =

=


V max = 1,060.9 lb (between C and D)


M max = 4,487 lb-in. (at C )


4

(1,060.9 lb)(0.162760 in. )

(0.119842 in. )(1.25 in.)

1,152.632 psi 1,153 psi (at neutral axis between and )

V Q

I t

C D

τ = =

= = Ans.


4

(4,487 lb-in.)( 1.25 in./2)

0.119842 in.

23, 400.309 psi 23, 400 psi (T) (on bottom of shaft at )

x

M y

I

C

σ −

= − = −

= = Ans.





9.23 A 25-mm-diameter solid steel shaftsupports loads P A = 1,000 N, P C = 3,200 N,

and P E = 800 N, as shown in Fig. P9.23.

Assume L1 = 80 mm, L2 = 200 mm, L3 = 100mm, and L4 = 125 mm. The bearing at B can be

idealized as a roller support and the bearing at

D can be idealized as a pin support. Determine

the magnitude and location of:

(a) the maximum horizontal shear stress in theshaft.

(b) the maximum tension bending stress in theshaft.

Fig. P9.23

Solution

Section properties:

4 4

4

3 3

3

(25 mm)64 64

19,174.760 mm

(25 mm)12 12

1,302.083 mm

I D

DQ

π π = =

=

= =

=


V max = 2,200 N (between C and D)


M max = 120,000 N-mm (at C )


4

(2,200 N)(1,302.083 mm )

(19,174.760 mm )(25 mm)

5.98 MPa (at neutral axis between and )

V Q

I t

C D

τ = =

= Ans.


4

(120,000 N-mm)( 25 mm/2)

19,174.760 mm

78.228 MPa 78.2 MPa (T) (on bottom of shaft at )

x

M y

I

C

σ −

= − = −

= = Ans.





9.24 A 3-in. standard steel pipe ( D = 3.500 in.; d = 3.068 in.) supports a concentrated load of P = 700 lb,as shown in Fig. P9.24a. The span length of the cantilever beam is L = 4 ft. Determine the magnitude of:

(a) the maximum horizontal shear stress in the pipe.

(b) the maximum tension bending stress in the pipe.

Fig. P9.24a Cantilever beam Fig. P9.24b Pipe cross section

Solution

Section properties:

4 4 4 4 4

3 3 3 3 3

[ ] [(3.500 in.) (3.068 in.) ] 3.017157 in.

64 641 1

[ ] [(3.500 in.) (3.068 in.) ] 1.166422 in.12 12

I D d

Q D d

π π = − = − =

= − = − =


V max = 700 lb


M max = (700 lb)(4 ft)(12 in./ft) = 33,600 lb-in.

(a) Maximum horizontal shear stress:

3

4

(700 lb)(1.166422 in. ) 626.429 psi 626 psi(3.017157 in. )(3.500 in. 3.068 in.)

V Q

I t τ = = = =

− Ans.


4

( 33,600 lb-in.)(3.500 in./2)19,488.533 psi 19,490 psi (T)

3.017157 in. x

M y

I σ

−= − = − = = Ans.





9.25 A steel pipe ( D = 170 mm; d = 150 mm) supports a concentrated load of P = 8 kN, as shown in Fig.P9.25a. The span length of the cantilever beam is L = 1.2 m. Determine the magnitude of:

(a) the maximum horizontal shear stress in the pipe.


Fig. P9.25a Cantilever beam Fig. P9.25b Pipe cross section

Solution

Section properties:

4 4 4 4 4

3 3 3 3 3

[ ] [(170 mm) (150 mm) ] 16,147,786.239 mm

64 641 1

[ ] [(170 mm) (150 mm) ] 128,166.667 mm12 12

I D d

Q D d

π π = − = − =

= − = − =


V max = 8 kN = 8,000 N


M max = (8,000 N)(1.2 m)(1,000 mm/m) = 9,600,000 N-mm

(a) Maximum horizontal shear stress:

3

4

(8,000 N)(128,166.667 mm ) 3.174842 MPa 3.17 MPa(16,147,786.239 mm )(170 mm 150 mm)

V Q

I t τ = = = =

− Ans.


4

( 9,600,000 N-mm)(170 mm/2)50.533 MPa 50.5 MPa (T)

16,147,786.239 mm x

M y

I σ

−= − = − = = Ans.





9.26 A concentrated load of P = 22 kN is applied to the upper end of a2-m long pipe, as shown in Fig. P9.26. The outside diameter of the pipe

is D = 330 mm and the inside diameter is d = 300 mm. Determine the

magnitude of:(a) the maximum vertical shear stress in the pipe.


Fig. P9.26

Solution

Section properties:

4 4 4 4 4

3 3 3 3 3

[ ] [(330 mm) (300 mm) ] 184,529,789.364 mm64 64

1 1[ ] [(330 mm) (300 mm) ] 744,750 mm

12 12

I D d

Q D d

π π = − = − =

= − = − =


V max = 22 kN = 22,000 N


M max = (22,000 N)(2 m)(1,000 mm/m) = 44,000,000 N-mm

(a) Maximum vertical shear stress:3

4(22,000 N)(744,750 mm ) 2.959685 MPa 2.96 MPa(184,529,789.364 mm )(330 mm 300 mm)

V Q I t

τ = = = =−

Ans.


4

(44,000,000 N-mm)(330 mm/2)39.343 MPa 39.3 MPa (T)

184,529,789.364 mm x

M c

I σ = = = = Ans.





9.27 A concentrated load of P = 6 kips is applied to the upper end of a4-ft long pipe, as shown in Fig. P9.27. The pipe is an 8 in. standard

steel pipe, which has an outside diameter of D = 8.625 in. and an

inside diameter of d = 7.981 in. Determine the magnitude of:(a) the maximum vertical shear stress in the pipe.


Fig. P9.27

Solution

Section properties:

4 4 4 4 4

3 3 3 3 3

[ ] [(8.625 in.) (7.981 in.) ] 72.489241 in.64 64

1 1[ ] [(8.625 in.) (7.981 in.) ] 11.104874 in.

12 12

I D d

Q D d

π π = − = − =

= − = − =


V max = 6 kips = 6,000 lb


M max = (6,000 lb)(4 ft)(12 in./ft) = 288,000 lb-in.

(a) Maximum vertical shear stress:3

4(6,000 lb)(11.104874 in. ) 1, 427.267 psi 1,427 psi(72.489241 in. )(8.625 in. 7.981 in.)

V Q I t

τ = = = =−

Ans.


4

(288,000 lb-in.)(8.625 in./2)17,133.564 psi 17,130 psi (T)

72.489241 in. x

M c

I σ = = = = Ans.





9.28 The internal shear force V at a certain section of analuminum beam is 8 kN. If the beam has a cross section

shown in Fig. P9.28, determine:

(a) the shear stress at point H , which is located 30 mmabove the bottom surface of the tee shape.

(b) the maximum horizontal shear stress in the tee shape.

Fig. P9.28

Solution

Centroid location in y direction:

Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm3)

top flange 375.0 72.5 27,187.5

stem 350.0 35.0 12,250.0

725.0 mm2 39,437.5 mm

3

3

2

39,437.5 mm54.397 mm (from bottom of shape to centroid)

725.0 mm

20.603 mm (from top of shape to centroid)

i i

i

y A y

A

Σ= = =

Σ

=

Moment of inertia about the z axis:Shape I C d = yi – d²A I C + d²A

(mm4) (mm) (mm

4) (mm

4)

top flange 781.250 18.103 122,900.565 123,681.815

stem 142,916.667 −19.397 131,679.177 274,595.843

Moment of inertia about the z axis (mm4) = 398,277.658

(a) Shear stress at H :3

3

4

(5 mm)(30 mm)(39.397 mm) 5,909.550 mm

(8,000 N)(5,909.550 mm )23.7 MPa

(398,277.658 mm )(5 mm)

Q

τ

= =

= = Ans.

(b) Maximum horizontal shear stress:

At neutral axis:3

3

4

(5 mm)(54.397 mm)(27.199 mm) 7,397.720 mm

(8,000 N)(7,397.720 mm )29.7 MPa

(398,277.658 mm )(5 mm)

Q

τ

= =

= = Ans.





9.29 The internal shear force V at a certain section of asteel beam is 80 kN. If the beam has a cross section

shown in Fig. P9.29, determine:

(a) the shear stress at point H , which is located 30 mm below the centroid of the wide-flange shape.

(b) the maximum horizontal shear stress in the wide-

flange shape.

Fig. P9.29

Solution

Moment of inertia about the z axis:

Shape I C d = yi – d²A I C + d²A

(mm4

) (mm) (mm4

) (mm4

)top flange 59,062.5 97.5 29,944,687.5 30,003,750.0

web 4,860,000.0 0.0 0.0 4,860,000.0

bottom flange 59,062.5 −97.5 29,944,687.5 30,003,750.0

Moment of inertia about the z axis (mm4) = 64,867,500.0


3

4

(210 mm)(15 mm)(97.5 mm) (10 mm)(60 mm)(60 mm) 343,125 mm

(80,000 N)(343,125 mm )42.3 MPa

(64,867,500 mm )(10 mm)

Q

τ

= + =

= = Ans.


At neutral axis:3

3

4

(210 mm)(15 mm)(97.5 mm) (10 mm)(90 mm)(45 mm) 347,625 mm

(80,000 N)(347,625 mm )42.9 MPa

(64,867,500 mm )(10 mm)

Q

τ

= + =

= = Ans.





9.30 The internal shear force V at a certain sectionof a steel beam is 110 kips. If the beam has a cross

section shown in Fig. P9.30, determine:

(a) the shear stress at point H , which is located 2 in. below the top surface of the flanged shape.

(b) the maximum horizontal shear stress in the

flanged shape.

Fig. P9.30

Solution


Shape Area Ai

yi (from bottom) yi Ai

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

3)

top flange (1) 5.0 11.5 57.5

web 10.0 6.0 60.0

bottom flange 8.0 0.5 4.0

23.0 in.2 121.5 in.

3

3

2

121.5 in.5.2826 in. (from bottom of shape to centroid)

23.0 in.

6.7174 in. (from top of shape to centroid)

i i

i

y A y

A

Σ= = =

Σ

=



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

4) (in.

4)

top flange 0.4167 6.2174 193.2798 193.6964

web 83.3333 0.7174 5.1465 88.4798

bottom flange 0.6667 −4.7826 182.9868 183.6534

Moment of inertia about the z axis (in.4) = 465.8297


3

4

(5 in.)(1 in.)(6.2174 in.) (1 in.)(1 in.)(5.2174 in.) 36.3044 in.

(110 kips)(36.3044 in. )8.57 ksi

(465.8297 in. )(1 in.)

Q

τ

= + =

= = Ans.


At neutral axis:3

3

4

(5 in.)(1 in.)(6.2174 in.) (1 in.)(5.7174 in.)(2.8587 in.) 47.4313 in.

(110 kips)(47.4313 in. )11.20 ksi

(465.8297 in. )(1 in.)

Q

τ

= + =

= = Ans.





9.31 The internal shear force V at a certainsection of a steel beam is 75 kips. If the

beam has a cross section shown in Fig.

P9.31, determine:(a) the shear stress at point H , which is

located 2 in. above the bottom surface of the

flanged shape.

(b) the shear stress at point K , which is

located 4.5 in. below the top surface of theflanged shape.

Fig. P9.31

Solution



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

4) (in.

4)

left flange 62.5000 0.0000 0.0000 62.5000web 0.0885 0.0000 0.0000 0.0885

right flange 62.5000 0.0000 0.0000 62.5000



3

4

2(0.75 in.)(2 in.)(4 in.) 12 in.

(75 kips)(12 in. )4.80 ksi

(125.0885 in. )(2 0.75 in.)

Q

τ

= =

= =×

Ans.

(b) Shear stress at K : 3

3

4

2(0.75 in.)(4.5 in.)(2.75 in.) 18.5625 in.

(75 kips)(18.5625 in. )7.42 ksi

(125.0885 in. )(2 0.75 in.)

Q

τ

= =

= =×

Ans.





9.32 Consider a 100-mm-long segment of a simplysupported beam (Fig. P9.32a). The internal bending

moments on the left and right sides of the segment are

75 kN-m and 80 kN-m, respectively. The cross-sectional dimensions of the flanged shape are shown in

Fig. P9.32b. Determine the maximum horizontal shear

stress in the beam at this location.

Fig. P9.32b Cross-sectional dimensions Fig. P9.32a Beam segment (side view)

Solution


Shape Area Ai

yi (from bottom) yi Ai

(mm2) (mm) (mm

3)

top flange 9,000 300 2,700,000

web 8,400 165 1,386,000

bottom flange 15,000 30 450,000

32,400 mm2 4,536,000 mm

3

3

2

4,536,000 mm140 mm (from bottom of shape to centroid)

32,400 mm

190 mm (from top of shape to centroid)

i i

i

y A y

A

Σ= = =

Σ

=



(mm4) (mm) (mm

4) (mm

4)

top flange 2,700,000 160 230,400,000 233,100,000

web 30,870,000 25 5,250,000 36,120,000

bottom flange 4,500,000 −110 181,500,000 186,000,000

Moment of inertia about the z axis (mm4) = 455,220,000

Shear force in beam:

80 kN-m 75 kN-m 5 kN-m50 kN

100 mm 0.1 m

M V

x

∆ −= = = =

∆

Maximum horizontal shear stress:

At neutral axis:3

3

4

(250 mm)(60 mm)(110 mm) (40 mm)(80 mm)(40 mm) 1,778,000 mm

(50,000 N)(1,778,000 mm )4.88 MPa

(455,220,000 mm )(40 mm)

Q

τ

= + =

= = Ans.





9.33 A simply supported beam supports the loads shown in Fig. P9.33a. The cross-sectional dimensionsof the wide-flange shape are shown in Fig. P9.33b.

(a) Determine the maximum shear force in the beam.

(b) At the section of maximum shear force, determine the shear stress in the cross section at point H ,which is located 100 mm below the neutral axis of the wide-flange shape.

(c) At the section of maximum shear force, determine the maximum horizontal shear stress in the cross

section.

(d) Determine the magnitude of the maximum bending stress in the beam.

Fig. P9.33a Fig. P9.33b

Solution

(a) Maximum shear force magnitude:

V max = 175 kN (just to the right of B)


M max = 156.25 kN-m (between B and C )





Section properties:



(mm4) (mm) (mm

4) (mm

4)

top flange 56,250 142.5 60,918,750 60,975,000

web 16,402,500 0 0 16,402,500

bottom flange 56,250 −142.5 60,918,750 60,975,000

Moment of inertia about the z axis (mm

4

) = 138,352,500

(b) Shear stress at H :3

3

4

(200 mm)(15 mm)(142.5 mm) (10 mm)(35 mm)(117.5 mm) 468,625 mm

(175,000 N)(468,625 mm )59.3 MPa

(138,352,500 mm )(10 mm)

Q

τ

= + =

= = Ans.

(c) Maximum horizontal shear stress:

At neutral axis:3

3

4

(200 mm)(15 mm)(142.5 mm) (10 mm)(135 mm)(67.5 mm) 518,625 mm

(175,000 N)(518,625 mm ) 65.6 MPa(138,352,500 mm )(10 mm)

Q

τ

= + =

= = Ans.

(d) Maximum tension bending stress:6

4

(156.25 10 N-mm)(300 mm/2)

138,352,500 mm

169.404 MPa 169.4 MPa

x

M c

I σ

×= = −

= = Ans.





9.34 A simply supported beam supports the loads shown in Fig. P9.34a. The cross-sectional dimensionsof the structural tube shape are shown in Fig. P9.34b.

(a) At section a–a, which is located 4 ft to the right of pin support B, determine the bending stress and

the shear stress at point H , which is located 3 in. below the top surface of the tube shape.(b) Determine the magnitude and the location of the maximum horizontal shear stress in the tube shape

at section a–a.


Solution

Shear force magnitude at a–a:

V = 27.60 kips

Bending moment at a–a:

M = 60.90 kip-ft

Section properties:3 3

4

(12 in.)(16 in.) (11.25 in.)(15.25 in.)

12 12771.0830 in.

I = −

=

(a) Bending stress at H :

4

(60,900 lb-ft)(5 in.)(12 in./ft)

771.0830 in.

4,738.79 psi 4,740 psi (C)

H

M y

I σ = −

= −

= − = Ans.

Shear stress at H :3

3

4

(12 in.)(0.375 in.)(7.8125 in.) 2(0.375 in.)(2.625 in.)(6.3125 in.) 47.5840 in.

(27,600 lb)(47.5840 in. )2,270 psi

(771.0830 in. )(2 0.375 in.)

Q

τ

= + =

= =×

Ans.





(b) Maximum shear force magnitude:

V = 39.60 kips (at pin B)

Maximum horizontal shear stress:

At neutral axis:3

3

4

(12 in.)(0.375 in.)(7.8125 in.) 2(0.375 in.)(7.625 in.)(3.8125 in.) 56.9590 in.

(39,600 lb)(56.9590 in. )3,900 psi

(771.0830 in. )(2 0.375 in.)

Q

τ

= + =

= =

×

Ans.





9.35 A cantilever beam supports the loads shown in Fig. P9.35a. The cross-sectional dimensions of theshape are shown in Fig. P9.35b. Determine:

(a) the maximum horizontal shear stress.

(b) the maximum compression bending stress.(c) the maximum tension bending stress.


Solution


Shape Width b Height h Area Ai

yi

(from bottom) yi Ai (in.) (in.) (in.

2) (in.) (in.

3)

top flange 12.0 0.5 6.00 5.75 34.5000

left stem 0.5 5.5 2.75 2.75 7.5625

right stem 0.5 5.5 2.75 2.75 7.5625

11.50 49.6250

3

2

49.6250 in.4.3152 in. (from bottom of shape to centroid)

11.50 in.

1.6848 in. (from top of shape to centroid)

i i

i

y A y

A

Σ= = =

Σ

=



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

4) (in.

4)

top flange 0.1250 1.4348 12.3519 12.4769

left stem 6.9323 −1.5652 6.7371 13.6694

right stem 6.9323 −1.5652 6.7371 13.6694







V = 5,800 lb

Maximum positive bending moment:

M pos = 8,850 lb-ft

Maximum negative bending moment:

M neg = −9,839 lb-ft

(a) Maximum shear stress:

3

3

4

2(0.5 in.)(4.3152 in.)(4.3152 in./2)

9.3105 in.

(5,800 lb)(9.3105 in. )

(39.8157 in. )(2 0.5 in.)

1,360 psi

Q

τ

=

=

=×

= Ans.

(b) Maximum compression bending stress:

Check two possibilities. First, check the bending stress created by the largest positive moment at the topof the cross section:

pos top

4

(8,850 lb-ft)(1.6848 in.)(12 in./ft)4,494 psi

39.8157 in. x

z

M y

I σ = − = − = −

Next, for the largest negative moment, compute the bending stress at the bottom of the cross section:

neg bot

4( 9,839 lb-ft)( 4.3152 in.)(12 in./ft) 12,796 psi

39.8157 in. x

z

M y I

σ − −= − = − = −

Therefore, the maximum compression bending stress is:

comp 12,800 psi (C)σ = Ans.

(c) Maximum tension bending stress:

Check two possibilities. First, check the bending stress created by the largest positive moment at the

bottom of the cross section:

pos bot

4

(8,850 lb-ft)( 4.3152 in.)(12 in./ft)11,510 psi

39.8157 in. x

z

M y

I σ

−= − = − =

Next, for the largest negative moment, compute the bending stress at the top of the cross section:

neg top

4

( 9,839 lb-ft)(1.6848 in.)(12 in./ft)4,996 psi

39.8157 in. x

z

M y

I σ

−= − = − =

Therefore, the maximum tension bending stress is:

tens 11,510 psi (T)σ = Ans.





9.36 A cantilever beam supports the loads shown in Fig.P9.36a. The cross-sectional dimensions of the shape are

shown in Fig. P9.36b. Determine:

(a) the maximum vertical shear stress.(b) the maximum compression bending stress.

(c) the maximum tension bending stress.

Fig. P9.36b Fig. P9.36a

Solution

Maximum shear force magnitude:V max = 5 kN

Maximum positive bending moment:

M pos = 2.00 kN-m

Maximum negative bending moment:

M neg = −1.50 kN-m


Shape Width b Height h Area Ai

yi

(from bottom) yi Ai

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

3)

flange 100 8 800 96 76,800

stem 6 92 552 46 25,392

1,352 102,192







9.37 A simply supported beam fabricated from pultruded reinforced plastic supports the loads shown inFig. P9.37a. The cross-sectional dimensions of the plastic wide-flange shape are shown in Fig. P9.37b.

(a) Determine the maximum shear force in the beam.

(b) At the section of maximum shear force, determine the shear stress in the cross section at point H ,which is located 2 in. above the bottom surface of the wide-flange shape.

(c) At the section of maximum shear force, determine the maximum horizontal shear stress in the cross

section.

(d) Determine the magnitude of maximum compression bending stress in the beam. Where along the

span does this stress occur?


Solution

Section properties:3 3

4

(4 in.)(8 in.) (3.625 in.)(7.25 in.)

12 12

55.5493 in.

z I = −

=

(a) Maximum shear force magnitude:V = 3,644 lb

(b) Shear stress at H :

3

3

4

(4 in.)(0.375 in.)(3.8125 in.)

(0.375 in.)(1.625 in.)(2.8125 in.)

7.4326 in.

(3,664 lb)(7.4326 in. )

(55.5493 in. )(0.375 in.)

1,307 psi

Q

τ

=

+

=

=

= Ans.



(c) Maximum horizontal shear stress:

At neutral axis:3

3

4

(4 in.)(0.375 in.)(3.8125 in.) (0.375 in.)(3.625 in.)(1.8125 in.) 8.1826 in.

(3,664 lb)(8.1826 in. )1,439 psi

(55.5493 in. )(0.375 in.)

Q

τ

= + =

= = Ans.

(d) Maximum compression bending stress:

4

(7,719 lb-ft)(4 in.)(12 in./ft) 6,669.965 psi 6,670 psi (C)55.5493 in.

H M y

I σ = − = − = − = Ans.

Mechanics of Materials Solutions Chapter09 Probs18 37

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