Mechanics of Materials Solutions Chapter08 Probs1 18

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8.1 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 12 ft. Determine the maximum bending stress

developed in the plank.

Solution

From Eq. (8.3):

1,900 ksi( 0.5 in.) 6.597 ksi 6.60 ksi

(12 ft)(12 in./ft) x

E yσ

ρ = − = − ± = ± = Ans.





8.2 A high-strength steel [ E = 200 GPa] tube having an outside diameter of 80 mm and a wall thicknessof 3 mm is bent into a circular curve having a 24-m radius of curvature. Determine the maximum

bending stress developed in the tube.

Solution

From Eq. (8.3):

200,000 MPa( 80 mm / 2) 333.333 MPa 333 MPa

(24 m)(1,000 mm/m) x

E yσ

ρ = − = − ± = ± = Ans.





8.3 A high-strength steel [ E = 200 GPa] band saw blade wraps around a pulley that has a diameter o350 mm. Determine the maximum bending stress developed in the blade. The blade is 12-mm wide and

1-mm thick.

Solution

The radius of curvature of the band saw blade is:

350 mm 1 mm175.5 mm

2 2 ρ = + =

From Eq. (8.3):

200,000 MPa( 0.5 mm) 569.801 MPa 570 MPa

175.5 mm x

E yσ

ρ = − = − ± = ± = Ans.





8.4 The boards for a concrete form are to be bent into a circular shape having an inside radius of 8 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:

8,000 mm2

t ρ = +

From Eq. (8.3):

12,000 MPa7 MPa

28,000 mm

2

x

E t y

t σ

ρ

⎛ ⎞= − = − ≤ ±⎜ ⎟

⎛ ⎞ ⎝ ⎠+⎜ ⎟

⎝ ⎠

Solve for t:

12,000 MPa 7 MPa 8,000 mm2 2

6,000 56,000 3.5

5,996.5 56,000

9.39 mm

t t

t t

t

t

⎛ ⎞ ⎛ ⎞≤ +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

≤ +

≤

∴ ≤ Ans.







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

I S

c

I S

c

S

= = =−

= = =

∴ = Ans.

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

4

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

18,654,833.33 mm

38.615 MPa 38.6 MPa (C)

x

z

M y

I σ = −

= −

= − = 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 = −110 mm. The larger

bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottomof the cross section.

4

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

18,654,833.33 mm

106.190 MPa 106.2 MPa (T)

x

z

M y

I σ = −

−= −

= = Ans.





8.6 A beam is subjected to equal 8.5 kip-ft bending moments, as shown in Fig. P8.6a. The cross-sectional dimensions of the beam are shown in Fig. P8.6b. Determine:

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

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

normal stress at H is tension or compression.

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

compression.

Fig. P8.6a Fig. P8.6b

Solution

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

Shape Area Ai

yi (from bottom) yi Ai

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

3)

left side 8.0 4.0 32.0

top flange 4.0 7.5 30.0

right side 8.0 4.0 32.020.0 in.

2 94.0 in.

3

3

2

94.0 in.4.70 in.

20.0 in.

i i

i

y A y

A

Σ= = =

Σ (measured upward from bottom edge of section) Ans.

Moment of inertia about the z axis:

Shape I C d = yi – d²A I C + d²A

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

4) (in.

4)

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 the z axis (in.4) = 124.867

4124.9 in. z I = Ans.





Section moduli:4

3

top

top

43

bot

bot

3

124.867 in.26.5674 in.

(8 in. 4.7 in.)

124.867 in.37.8384 in.

4.7 in.

26.6 in.

z

z

I S

c

I S

c

S

= = =−

= = =

∴ = Ans.

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

4

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

124.867 in.

1.634 ksi 1,634 psi (C)

x

z

M y

I σ = −

− −= −

= − = 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 at y = +3.30 in., and the bottom of the cross section is at y = −4.7 in. 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

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

124.867 in.

3.839 ksi 3,840 psi (C)

x

z

M y

I σ = −

− −= −

= − = Ans.





8.7 A beam is subjected to equal 375 N-m bending moments, as shown in Fig. P8.7a. The cross-sectional dimensions of the beam are shown in Fig. P8.7b. Determine:


about the z axis.(b) the bending stress at point H . State whether the normal stress at H is tension or compression.

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

compression.


Solution

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

Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

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 mm2 21,088.0 mm3

3

2

21,088.0 mm19.67 mm

1,072.0 mm

i i

i

y A y

A

Σ= = =

Σ (measured upward from bottom edge of section) Ans.



(mm4) (mm) (mm

4) (mm

4)

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 the z axis (mm4) = 257,633.75

4257,600 mm z I = Ans.





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 mm

8,495 mm

z

z

I S

c

I S

c

S

= = =−

= = =

∴ = Ans.

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

4

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

257,633.75 mm

16.989 MPa 16.99 MPa (T)

x

z

M y

I σ = −

−= −

= = 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 = +30.328 mm, and the bottom of the cross section is at y = −19.672 mm. The larger

bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the top ofthe cross section.

4

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

257,633.75 mm

44.145 MPa 44.1 MPa (C)

x

z

M y

I σ = −

= −

= − = Ans.





8.8 A beam is subjected to equal 10.5 kip-ft bending moments, as shown in Fig. P8.8a. The cross-sectional dimensions of the beam are shown in Fig. P8.8b. Determine:


about the z axis.(b) the bending stress at point H . State whether the normal stress at H is tension or compression.

(c) the bending stress at point K . State whether the normal stress at K is tension or compression.

(d) the maximum bending stress produced in the cross section. State whether the stress is tension or

compression.


Solution

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

Shape Area Ai

yi

(from bottom) yi Ai

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

3)

top flange 12.0000 11.0000 132.0000

web 16.0000 6.0000 96.0000

bottom flange 20.0000 1.0000 20.0000

48.0000 in.2 248.0000 in.

3

3

2

248.0 in.5.1667 in.

48.0 in.

i i

i

y A y

A

Σ= = =

Σ (measured upward from bottom edge of bottom flange) Ans.



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

4) (in.

4)

top flange 4.0000 5.8333 408.3333 412.3333

web 85.3333 0.8333 11.1111 96.4444

bottom flange 6.6667 –4.1667 347.2222 353.8889


Ans.





Section Moduli

43

43

5.1667 in.

12 in. 5.1667 in. 6.8333 in.

862.6667 in.166.9677 in.

5.1667 in.

862.6667 in.126.2439 in.

6.8333 in.The controlling section modulus is the smaller of the

bot

top

z bot

bot

z top

top

c

c

I S

c

I S

c

=

= − =

= = =

= = =

3

two values; therefore,

126.2439 in.S = Ans.

Bending stress at point H :

From the flexure formula:

4

( 10.5 kip-ft)(6.8333 in. 2 in.)(12 in./ft)0.7059 ksi 706 psi (T)

862.6667 in. x

z

M y

I σ

− −= − = − = = Ans.

Bending stress at point K :

From the flexure formula:

4

( 10.5 kip-ft)( 5.1667 in. 2 in.)(12 in./ft)0.4625 ksi 463 psi (C)

862.6667 in. x

z

M y

I σ

− − += − = − = − = Ans.

Maximum bending stress

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

formula:

4

( 10.5 kip-ft)(6.8333 in.)(12 in./ft)0.9981 ksi 998 psi (T)

862.6667 in. x

z

M y

I σ

−= − = − = = Ans.

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

3

(10.5 kip-ft)(12 in./ft)0.9981 ksi 998 psi

126.2439 in. x

M

S σ = = = = Ans.

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





8.9 The cross-sectional dimensions of a beam areshown in Fig. P8.9.

(a) If the bending stress at point K is 43 MPa (C),

determine the internal bending moment M z actingabout the z centroidal axis of the beam.

(b) Determine the bending stress at point H . State

whether the normal stress at H is tension or

compression.

Fig. P8.9

Solution

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

Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

top flange 375.0 47.5 17,812.5left stem 225.0 22.5 5,062.5

right stem 225.0 22.5 5,062.5

825.0 mm2 27,937.5 mm

3

3

2

27,937.5 mm33.864 mm 33.9 mm

825.0 mm

i i

i

y A y

A

Σ= = = =

Σ (measured upward from bottom of section)



(mm4) (mm) (mm

4) (mm

4)

top flange 781.250 13.636 69,731.405 70,512.655

left stem 37,968.750−

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

Moment of inertia about the z axis (mm4) = 204,559.659

(a) Determine bending moment:

At point K , y = 50 mm − 5 mm − 33.864 mm = 11.136 mm. The bending stress at K is σ 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-mm 790 N-m

x

z

x z

M y

I

I M

y

σ

σ

= −

−∴ = − = −

= = Ans.

(b) Bending stress at point H :

At point H , 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 130.8 MPa (T)

204,559.659 mm x

z

M y

I σ

−= − = − = = Ans.






(a) If the bending stress at point K is 2,600 psi (T),

determine the internal bending moment M z actingabout the z centroidal axis of the beam.

(b) Determine the bending stress at point H . State

whether the normal stress at H is tension or

compression.

Fig. P8.10

Solution

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

Shape Area Ai


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

3)

bottom flange 0.56250 0.12500 0.07031

stem 0.56250 1.37500 0.773441.12500 in.

2 0.84375 in.

3

3

2

0.84375 in.0.750 in.

1.1250 in.

i i

i

y A y

A

Σ= = =

Σ (measured upward from bottom edge of section)



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

4) (in.

4)

bottom flange 0.00293 −0.62500 0.21973 0.22266

stem 0.23730 0.62500 0.21973 0.45703


(a) Determine bending moment:

At point K , y = 2.50 in. − 0.75 in. = 1.750 in. The bending stress at K is σ x = +2,600 psi; therefore, the bending moment magnitude can be determined from the flexure formula:

4(2,600 psi)(0.67967 in. )

1.750 in.

1,009.820 lb-in. 1,010 lb-in. 84.2 lb-ft

x

z

x z

M y

I

I M

y

σ

σ

= −

∴ = − = −

= − = − = − Ans.

(b) Bending stress at point H :

At point H , y = −0.75 in. The bending stress is computed with the flexure formula:

4

( 1,009.820 lb-in.)( 0.75 in.)1,114.286 psi 1,114 psi (C)

0.67969 in. x

z

M y

I σ

− −= − = − = − = Ans.





8.11 The cross-sectional dimensions of a box-shaped beam are shown in Fig. P8.11. If the

maximum allowable bending stress is σ b =15,000 psi, determine the maximum internal

bending moment M z magnitude that can be

applied to the beam.

Fig. P8.11

Solution

Moment of inertia about z axis:3 3

4(3 in.)(2 in.) (2.5 in.)(1 in.)1.791667 in.

12 12 z I = − =

Maximum internal bending moment M z :

4(15,000 psi)(1.791667 in. )26,875 lb-in. 2,240 lb-ft

1 in.

z x

z

x z

M c

I I

M c

σ

σ

=

∴ = = = = Ans.





8.12 The cross-sectional dimensions of a beam areshown in Fig. P8.12. The internal bending moment

about the z centroidal axis is M z = +2.70 kip-ft.

Determine:(a) the maximum tension bending stress in the

beam.

(b) the maximum compression bending stress in the

beam.

Fig. P8.12

Solution

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

Shape Area Ai


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

3)

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.2 17.375 in.33

2

17.375 in.2.673 in.

6.500 in.

i i

i

y A y

A

Σ= = =

Σ

(measured upward from bottom edge of section)



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

4) (in.

4)

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


(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 point K (i.e., y = −2.673 in.):

4

(2.70 kip-ft)( 2.673 in.)(12 in./ft)8.578 ksi 8.58 ksi (T)

10.09696 in. x

z

M y

I σ

−= − = − = = 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. − 2.673 in. =1.327 in.):

4

(2.70 kip-ft)(1.327 in.)(12 in./ft)4.258 ksi 4.26 ksi (C)

10.09696 in. x

z

M y

I σ = − = − = − = Ans.






(a) If the bending stress at point K is 35.0 MPa (T),

determine the bending stress at point H . Statewhether the normal stress at H is tension or

compression.

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

moment M z that can be supported by the beam.

Fig. P8.13

Solution


Shape I C d = yi – y d²A I C + d²A

(mm4) (mm) (mm

4) (mm

4)

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

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

Moment of inertia about the z axis (mm4) = 402,238,666.667

(a) At point K , y = −90 mm, and at point H , y = −175 mm. The bending stress at K is σ x = +35 MPa, andthe bending stress is distributed linearly over the depth of the cross section. Therefore, the bending

stress at H can be found from the ratio:

175 mm(35.0 MPa) 68.056 MPa 68.1 MPa (T)

90 mm

H K

H K

H

H K K

y y

y

y

σ σ

σ σ

=

−∴ = = = =

−

Ans.

(b) Maximum internal bending moment M z :

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

175 mm

z x

z

x z

M c

I

I M

c

σ

σ

=

∴ = = = = Ans.






(a) If the bending stress at point K is 9.0 MPa (T),

determine the bending stress at point H . Statewhether the normal stress at H is tension or

compression.

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

moment M z that can be supported by the beam.

Fig. P8.14

Solution



(mm4) (mm) (mm

4) (mm

4)

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 the z axis (mm4) = 19,471,680

(a) At point K , y = −60 mm, and at point H , y = +90 mm. The bending stress at K is σ x = +9.0 MPa, andthe bending stress is distributed linearly over the depth of the cross section. Therefore, the bending

stress at H can be found from the ratio:

90 mm(9.0 MPa) 13.50 MPa 13.50 MPa (C)

60 mm

H K

H K

H H K

K

y y

y

y

σ σ

σ σ

=

∴ = = = − =−

Ans.

(b) Maximum bending moment M z :

2 4(165 N/mm )(19,471,680 mm )35,698,080 N-mm 35.7 kN-m

90 mm

z x

z

x z

M c

I

I M

c

σ

σ

=

∴ = = = = Ans.






about the z centroidal axis is M z = −1.55 kip-ft.

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


beam.

Fig. P8.15

Solution

Centroid location in y direction:

Shape Area Ai


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

3)

top flange 8.0 4.5 36.0

left web 3.0 2.5 7.5

left bottom flange 3.0 0.5 1.5

right web 3.0 2.5 7.5

right bottom flange 3.0 0.5 1.5

20.0 in.2 54.0 in.

3

3

2

54.0 in.2.70 in.

20.0 in.

i i

i

y A y

A

Σ= = =

Σ (measured upward from bottom edge of bottom flange)



(in.

4

) (in.) (in.

4

) (in.

4

)top flange 0.6667 1.8000 25.9200 26.5867

left web 2.2500 −0.2000 0.1200 2.3700

left bottom flange 0.2500 −2.2000 14.5200 14.7700

right web 2.2500 −0.2000 0.1200 2.3700

right bottom flange 0.2500 −2.2000 14.5200 14.7700


(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; that is, point H for this shape. From the flexure formula, the bending stress at point H is:

4( 1.55 kip-ft)(5.0 in. 2.70 in.)(12 in./ft) 0.7028 ksi 703 psi (T)

60.8667 in. x

z

M y I

σ − −= − = − = = Ans.

(b) Maximum compression bending stress:

The maximum compression bending stress will occur at the bottom surface of the cross section, which is point K for this shape. From the flexure formula, the bending stress at point K is:

4

( 1.55 kip-ft)( 2.70 in.)(12 in./ft)0.8251 ksi 825 psi (C)

60.8667 in. x

z

M y

I σ

− −= − = − = − = Ans.






about the z centroidal axis is M z = +270 lb-ft.

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


beam.

SolutionFig. P8.16


Shape Area Ai


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

3)

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 0.35156

right top flange 0.09375 2.43750 0.228521.15625 in.

2 1.18555 in.

3

3

2

1.18555 in.1.0253 in.

1.15625 in.

i i

i

y A y

A

Σ= = =

Σ (measured upward from bottom edge of bottom flange)



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

4) (in.

4)

bottom flange 0.000529 -0.962838 0.376617 0.377146

left web 0.118652 0.224662 0.014196 0.132848

left top flange 0.000122 1.412162 0.186956 0.187079right 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 of M z = +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 stress

at the bottom of the cross section is:

4

(270 lb-ft)( 1.0253 in.)(12 in./ft)3, 266.446 psi 3,270 psi (T)

1.016999 in. x

z

M y

I σ

−= − = − = = 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)4,698.164 psi 4,700 psi (C)

1.016999 in. x

z

M y

I σ = − = − = − = Ans.





8.17 Two vertical forces are applied to a simply supported beam (Fig. P8.17a) having the cross sectionshown in Fig. P8.17b. Determine the maximum tension and compression bending stresses produced in

segment BC of the beam.


Solution


Shape Area Ai

yi

(from bottom) yi Ai

(mm2) (mm) (mm

3)

top flange 3,000.0 167.5 502,500.0

stem 1,440.0 80.0 115,200.0

4,440 mm2 617,700 mm

3

3

2

617,700 mm139.1216 mm

4,440 mm

i i

i

y A y

A

Σ= = =

Σ (measured upward from bottom edge of stem)


Shape I C d = yi – d²A I C + d²A (mm

4) (mm) (mm

4) (mm

4)

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 the z axis (mm4) 10,577,574.32





Shear-force and bending-moment diagrams:

The maximum moment occurs between B and 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 ofthis 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)157.8 MPa (T)10.5776 10 mm

x

z

M y

I σ

−

= − = − =× 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 σ = −

−= −

×

= − = Ans.





8.18 Two vertical forces are applied to a simply supported beam (Fig. P8.18a) having the cross sectionshown in Fig. P8.18b. Determine the maximum tension and compression bending stresses produced in

segment BC of the beam.


Solution


Shape Area Ai


(in.2

) (in.) (in.3

)left stem 0.7500 1.5000 1.1250

bottom flange 0.5000 0.1250 0.0625

right stem 0.7500 1.5000 1.1250

2.000 in.2 2.3125 in.

3

3

2

2.3125 in.1.1563 in.

2.000 in.

i i

i

y A y

A

Σ= = =

Σ (measured upward from bottom edge of stem)


Shape I C d = yi – d²A I C + d²A (in.

4) (in.) (in.

4) (in.

4)

left stem 0.56250 0.34375 0.08862 0.65112

bottom flange 0.00260 -1.03125 0.53174 0.53434

right stem 0.56250 0.34375 0.08862 0.65112

Moment of inertia about the z axis (in.4) 1.83659



Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only

Shear-force and bending-moment diagrams:

The maximum moment occurs between B and C . The moment magnitude is 600 lb-ft.

Maximum tension bending stress:

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

this cross section at y = −1.1563 in. From the flexure formula, the bending stress at the bottom of the Ushape is:

4

(600 lb-ft)( 1.1563 in.)(12 in./ft)

4,533.053 psi 4,530 psi (T)1.83659 in. x z

M y

I σ

−= − = − = =

Ans.

Maximum compression bending stress:

The maximum compression bending stress will occur at the top of the U shape, where y = 1.8438 in.:

4

(600 lb-ft)(1.8438 in.)(12 in./ft)7, 228.265 psi 7,230 psi (C)

1.83659 in. x

z

M y

I σ = − = − = − = Ans.

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Mechanics of Materials Solutions Chapter08 Probs1 18