Mechanics of Materials Solutions Chapter08 Probs19 30

7/25/2019 Mechanics of Materials Solutions Chapter08 Probs19 30

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8.19 A WT230 × 26 standard steel shape is used to support the loads shown on the beam in Fig. P8.19 a.The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch o

the cross section (Fig. 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.

Fig. P8.19a Fig. P8.19b

Solution

Section properties

From Appendix B: 6 416.7 10 mm z I = ×

Shear-force and bending-moment diagrams

Maximum bending moments

positive M = 13.61 kN-mnegative M = −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

σ

σ

= −×

=

−= −

×

=

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

σ

σ

−= −

×

=

− −= −×

=

(a) Maximum tension bending stress = 133.9 MPa (T)

(b) Maximum compression bending stress =196.8 MPa (C)

Ans.

Ans.





8.20 A WT305 × 41 standard steel shape is used to support the loads shown on the beam in Fig. P8.20 a.The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch o

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


Solution

Section properties

From Appendix B:6 4

48.7 10 mm z I = ×



positive M = 45.84 kN-m

negative M = −24.00 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

σ

σ

= −×

=

−= −

×

=

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

σ

σ

−= −

×

=

− −

= − ×

=


(b) Maximum compression bending stress =104.0 MPa (C)

Ans.

Ans.







positive M = 100.75 kip-ft

negative M = −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

σ

σ

= −

=

−= −

=

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

σ

σ

−= −

=

− −= −

=

(a) Maximum tension bending stress = 14.73 ksi (T)

(b) Maximum compression bending stress = 9.94 ksi (C)

Ans.

Ans.





8.22 A flanged wooden shape is used to support the loads shown on the beam in Fig. P8.22 a. Thedimensions of the shape are shown in Fig. P8.22b. Consider the entire 18-ft length of the beam and

determine:



SolutionCentroid location in y direction:

Shape Area Ai

yi (from bottom) yi Ai

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

3)

top flange 20.0 13.0 260.0

web 20.0 7.0 140.0

bottom flange 12.0 1.0 12.0

52.0 in.2 412.0 in.

3

3

2412.0 in. 7.9231 in. (from bottom of shape to centroid)52.0 in.

6.0769 in. (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

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

4) (in.

4)

top flange 6.6667 5.0769 515.5030 522.1696

web 166.6667 −0.9231 17.0414 183.7081

bottom flange 4.0000 −6.9231 575.1479 579.1479

Moment of inertia about the z axis (in.4) = 1,285.0256







positive M = 14,851 lb-ft

negative M = −8,400 lb-ft


4

4

(14,851 lb-ft)(6.0769 in.)(12 in./ft)

1,285.0256 in.

843 psi (C)

(14,851 lb-ft)( 7.9231 in.)(12 in./ft)

1,285.0256 in.

1,099 psi (T)

x

x

σ

σ

= −

=

−= −

=


4

4

( 8,400 lb-ft)(6.0769 in.)(12 in./ft)

1,285.0256 in.

477 psi (T)

( 8,400 lb-ft)( 7.9231 in.)(12 in./ft)

1,285.0256 in.

622 psi (C)

x

x

σ

σ

−= −

=

− −= −

=

(a) Maximum tension bending stress = 1,099 psi (T)

(b) Maximum compression bending stress = 843 psi (C)

Ans.

Ans.





8.23 A channel shape is used to support the loads shown on the beam in Fig. P8.23a. The dimensions othe shape are shown in Fig. P8.23b. Consider the entire 12-ft 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.


Solution

Centroid location in y direction:

Shape Area Ai

yi (from bottom) yi Ai

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

3)

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.2 49.625 in.

3

3

2

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

11.500 in.

1.6848 in. (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

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

4) (in.

4)

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

Moment of inertia about the z axis (in.4) = 39.8157







positive M = 8,850 lb-ft

negative M = −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

σ

σ

= −

= =

−= −

= =


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

σ

σ

−= −

= =

− −= −

= =

(a) Maximum tension bending stress = 11.51 ksi (T)

(b) Maximum compression bending stress =12.80 ksi (C)

Ans.

Ans.





8.24 A W360 × 72 standard steel shape is used to support the loads shown on the beam in Fig. P8.24a.The shape is oriented so that bending occurs about the weak axis as shown in Fig. 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 mm z f

I b= × =



positive M = 16.50 kN-mnegative M = −19.12 kN-m

Since the shape is symmetric about the z axis, thelargest 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 at support B, where themoment magnitude is 19.12 kN-m.

Bending stresses at maximum moment2

6 4

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

21.4 10 mm

91.1 MPa (T) and (C)

xσ − ±

= −×

=


(b) Maximum compression bending stress = 91.1 MPa (C)

Ans.

Ans.





8.25 A 1.00-in.-diameter solid steelshaft supports loads P A = 180 lb and P C

= 240 lb as shown in Fig. P8.25.

Assume L1 = 5 in., L2 = 16 in., and L3 =

8 in. The bearing at B can be idealizedas a roller support and the bearing at D

can be idealized as a pin support.

Determine the magnitude and location

of the maximum bending stress in theshaft.

Fig. P8.25

Solution

Section properties

4 4 4(1.00 in.) 0.049087 in.64 64

I Dπ π

= = =



positive M = 980 lb-in.

negative M = −900 lb-in.

Since the circular cross section is symmetric

about the z axis, the largest bending stresseswill occur at the location of the largest moment

magnitude – either positive or negative. In this

case, the largest bending stresses will occur at

C , where the moment magnitude is 980 lb-in.

Bending stresses at maximum moment

4

(980 lb-in.)( 1.00 in./2)

0.049087 in.

9,980 psi

xσ ±

= −

= Ans.





8.26 A 30-mm-diameter solid steelshaft supports loads P A = 1,400 N and

P C = 2,100 N as shown in Fig. P8.26.

Assume L1 = 100 mm, L2 = 200 mm,

and L3 = 150 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 the maximum bending stressin the shaft.

Fig. P8.26

Solution

Section properties

4 4 4(30 mm) 39,760.8 mm64 64

I Dπ π

= = =



positive M = 120,000 N-mm

negative M = −140,000 N-mm





support B, where the moment magnitude is140,000 N-mm.


4

(140,000 N-mm)( 30 mm/2)

39,760.8 mm

52.8 MPa

xσ ±

= −

= Ans.





8.27 A 20-mm-diameter solid steel shaftsupports loads P A = 500 N, P C = 1,750

N, and P E = 500 N as shown in Fig.

P8.27. Assume L1 = 90 mm, L2 = 260

mm, L3 = 140 mm, and L4 = 160 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 themaximum bending stress in the shaft.

Fig. P8.27

Solution

Section properties

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

z I Dπ π

= = =



positive M = 91,500 N-mm

negative M = −80,000 N-mm


about the z axis, the largest bending stresses

will occur at the location of the largest momentmagnitude – either positive or negative. In this


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


4

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

7,853.9816 mm

116.5 MPa

xσ ±

= −

= Ans.





8.28 A 1.75-in.-diameter solid steel shaftsupports loads P A = 250 lb, P C = 600 lb,

and P E = 250 lb as shown in Fig. P8.28.

Assume L1 = 9 in., L2 = 24 in., L3 = 12

in., and L4 = 15 in. 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 the maximum bending stressin the shaft.

Fig. P8.28

Solution

Section properties

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

I Dπ π

= = =



positive M = 1,550 lb-in.

negative M = −3,750 lb-in.





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


4

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

0.460386 in.

7,130 psi

xσ − ±

= −

= Ans.





8.29 A HSS12 × 8 × 1/2 standard steel shapeis used to support the loads shown on the

beam in Fig. P8.29. The shape is oriented so

that bending occurs about the strong axis.

Determine the magnitude and location of themaximum bending stress in the beam.

Fig. P8.29

Solution

Section properties

From Appendix B: 4333 in. 12 in. z I d = =



positive M = 124.59 kip-ft

negative M = −72.00 kip-ft

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 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) 26.9 ksi

333 in. xσ ±= − = Ans.





8.30 A W410 × 60 standard steel shapeis used to support the loads shown on

the beam in Fig. P8.30. The shape is

oriented so that bending occurs aboutthe strong axis. Determine the

magnitude and location of the

maximum bending stress in the beam.

Fig. P8.30

Solution

Section properties

From Appendix B: 6 4216 10 mm 406 mm z I d = × =



positive M = 50 kN-m

negative M = −70 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 between B

and C , where the moment magnitude is 70 kN-

m.

Bending stresses at max moment magnitude

2

6 4

(70 kN-m)( 406 mm/2)(1,000)65.8 MPa

216 10 mm xσ

±= − =

× Ans.

Mechanics of Materials Solutions Chapter08 Probs19 30

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