8.31 A solid steel shaft supports loads P A = 200 lb and P and P D = 300 lb as shown in Fig. P8.31. Assume L1 = 6 in., L2 = 20 in., and L and L3 = 10 in. The bearing at B at B can can be idealized as a roller support and the bearing at C can be idealized as a pin support. If the allowable bending stress is 8 ksi, determine the minimum diameter that can be used for the shaft.
Fig. P8.31
Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = = 3,000 lb-in. Minimum required section modulus M σ x ≥
∴ S ≥
S M σ x
≥
3,000 lb-in. lb-in. 8,000 psi
=
0.375 in.3
Section modulus for solid circular section
S
=
π
32
D3
Minimum shaft diameter π
D 3 32
≥
0.375 in.3
∴ D ≥
1.563 in.
Ans.
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8.32 A solid steel shaft supports loads P A = 500 N and P and P D = 400 N as shown in Fig. P8.32. Assume L1 = 200 mm, L2 = 660 mm, and L3 = 340 mm. The bearing at B at B can can be idealized as a roller support and the bearing at C can be idealized as a pin support. If the allowable bending stress is 25 MPa, determine the minimum diameter that can be used for the shaft.
Fig. P8.32
Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = = 136,000 N-mm Minimum required section modulus M σ x ≥
∴ S ≥
S M σ x
≥
136,000 N-mm N-mm 25 N/mm
2
= 5,440
mm3
Section modulus for solid circular section
S
=
π
32
D3
Minimum shaft diameter π
D3 32
≥ 5,440 ∴ D ≥
mm3
38.1 mm
Ans.
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8.33 A simply supported wood beam (Fig. P8.33a P8.33a) with a span of L = 20 ft supports a uniformly distributed load of w = 800 lb/ft. The allowable bending stress of the wood is 1,400 psi. If the aspect ratio of the solid rectangular wood beam is specified as h/b = h/b = 1.5 (Fig. P8.33b P8.33b), determine the minimum width b that can be used for the beam.
Fig. P8.33a P8.33a
Fig. P8.33b P8.33b
Solution Shear-force and bending-moment diagrams Also, see Example 7-3 for shear-force and bending-moment diagram development. Maximum bending moment wL2 (800 lb/ft)(20 ft) 2 M max = = 8 8 40,000 lblb-fft = 480 480,000 lblb-in in.. = 40,000 Minimum required section modulus M σ x ≥
∴ S ≥
S M σ x
≥
480,000 480, 000 lb-in. lb-in. 1, 400 psi
= 342.8571
in.3
Section modulus for solid rectangular section I bh3 /12 bh2 S = = = c h/2 6
The aspect ratio of the solid rectangular wood beam is specified as h/b h/b = 1.5; therefore, the section modulus can be expressed as: bh 2 b(1.5b)2 2.25b3 3 S= = = = 0.3750b 6 6 6 Minimum allowable beam width 0.3750b3 ≥ 342.8571 in.3 ∴b ≥
9.71 in.
Ans.
The corresponding beam height h is h / b = 1 .5
∴ h = 1.5b = 1.5(9.71
in.) = 14.57 in.
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8.34 A simply supported wood beam (Fig. P8.34a P8.34a) with a span of L = 14 ft supports a uniformly distributed load of w. The beam width is b = 6 in. and the beam height is h = 10 in. (Fig. P8.34b P8.34b). The allowable bending stress of the wood is 900 psi. Determine the magnitude of the maximum load w that may be carried by the beam.
Fig. P8.34a P8.34a
Fig. P8.34b P8.34b
Solution Moment of inertia for rectangular cross section about horizontal centroidal axis bh3 (6 in.)(10 in.)3 4 I = = = 500 in. 12 12 Maximum allowable moment σ x I M c (900 (900 psi)( psi)(500 500 in.4 ) ∴ M = = σ x = I c 5 in.
= 90,000 90,000
lblb-in in.. = 7,500 7,500 lblb-fft
Shear-force and bending-moment diagrams Also, see Example 7-3 for shear-force and bending-moment diagram development. Determine distributed load intensity Equate the moment expression from the bendingmoment diagram to the maximum allowable moment that can be applied to the rectangular cross section: wL2 M max = 7, 500 lb-ft lb-ft = 7,500 8
Solve for the maximum distributed load w that can be applied to the 14-ft simple span: wL2 w(14 ft) 2 7, 500 lb-ft lb-ft = ≤ 7,500 8 8 8(7,500 lb-ft) lb-ft) ∴w ≤ = 306 lb/ft Ans. (14 ft) 2
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8.35 A cantilever timber beam (Fig. P8.35a P8.35a) with a span of L of L = 2.5 m supports a uniformly distributed load of w = 4 kN/m. The allowable bending stress of the wood is 9 MPa. If the aspect ratio of the solid rectangular timber is specified as h/b = h/b = 0.5 (Fig. P8.35b P8.35b), determine the minimum width b that can be used for the beam.
Fig. P8.35a P8.35a
Fig. P8.35b P8.35b
Solution Maximum moment magnitude: The maximum bending moment magnitude in the cantilever beam occurs at support A support A:: wL2 (4 kN/m)(2.5 m) 2 6 M max = = = 12.5 kN-m = 12.5 × 10 N-mm 2 2 Minimum required section modulus M σ x ≥
∴ S ≥
S M σ x
≥
12.5 × 106 N-mm 9 N/mm
2
6
= 1.3889 × 10
mm3
Section modulus for solid rectangular section I bh3 /12 bh2 S = = = c h/2 6
The aspect ratio of the solid rectangular wood beam is specified as h/b h/b = 0.5; therefore, the section modulus can be expressed as: bh 2 b(0.5b) 2 0.25b3 3 S= = = = 0.0416667b 6 6 6 Minimum allowable beam width 0.0416667b3 ≥ 1.3889 × 106 mm3 ∴ b ≥ 321.83
mm = 322 mm
Ans.
The corresponding beam height h is h / b = 0 .5 ∴h =
0.5b = 0.5(321.83 mm) = 161 mm
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8.36 A cantilever timber beam (Fig. P8.36a P8.36a) with a span of L = L = 3 m supports a uniformly distributed load of w. The beam width is b = 300 mm and the beam height is h = 200 mm (Fig. P8.36b P8.36b). The allowable bending stress of the wood is 6 MPa. Determine the magnitude of the maximum load w that may be carried by the beam.
Fig. P8.36a P8.36a
Fig. P8.36b P8.36b
Solution Section modulus for solid rectangular section I bh3 / 12 bh2 (300 mm)(200 mm)2 S = = = = c h/2 6 6
=
2 ×106 mm3
Maximum allowable bending moment: M 2 6 3 6 σ x ≥ ∴ M allow ≤ σ x S = (6 N/mm )( 2 ×10 mm ) = 12 ×10 N-mm S Maximum bending moment in cantilever span: The maximum bending moment magnitude in the cantilever beam occurs at support A support A::
M max
=
wL2 2
Maximum distributed load: wL2 ≤ M allow 2 ∴ wallow ≤
2 M allow L2
=
2(12 × 106 N-mm) [(3 m)(1,000 mm/m)]2
=
2.67 N/ N/mm = 2.67 kN kN/m
Ans.
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8.37 The beam shown in Fig. P8.37 will be constructed from a standard steel W-shape using 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.
Fig. P8.37
Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = = 90 kip-ft Minimum required section modulus M σ x ≥
∴ S ≥
S M σ x
≥
(90 kip-ft)(12 in./ft) 24 ksi
=
45 in.3
(a) Acceptable steel W-shapes W10 × 45, S = 49.1 in.3
W12 × 40,
S = 51.5 in.3
W14 × 34,
S = 48.6 in.3
W16 × 31,
S = 47.2 in.3
W18 × 35,
S = 57.6 in.3
(b) Most economical W-shape
W16 × 31
Ans.
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8.38 The beam shown in Fig. P8.38 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 W360, W410, W460, and W530 shapes on the list of possibilities. (b) Select the most economical W shape for this beam.
Fig. P8.38
Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude = 206.630 kN-m M = Minimum required section modulus M σ x ≥
∴ S ≥
S M σ x
≥
(206.63 kN-m)(1,000)2 165 N/mm
2
3
252 × 10 = 1, 25
mm3
(a) Acceptable steel W-shapes W360 × 79, S = 1, 270 × 103 mm3
W 410 × 75,
S = 1, 330 × 103 mm mm3
W 460 × 74,
S = 1, 460 × 103 mm3
W530 × 66,
S = 1, 340 × 103 mm mm3
(b) Most economical W-shape
W530 × 66
Ans.
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8.39 The beam shown in Fig. P8.39 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 W360, W410, W460, and W530 shapes on the list of possibilities. (b) Select the most economical W shape for this beam.
Fig. P8.39
Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = = 238.57 kN-m Minimum required section modulus M σ x ≥
∴ S ≥
S M σ x
≥
(238.57 kN-m)(1,000)2 165 N/mm
2
3
446 × 10 = 1, 44
mm3
(a) Acceptable steel W-shapes W360 × 101, S = 1, 690 × 103 mm3
W 410 × 85,
S = 1, 510 × 103 mm mm3
W 460 × 74,
S = 1, 460 × 103 mm3
W530 × 74,
S = 1, 550 × 103 mm mm3
(b) Most economical W-shape
W 460 × 74 or or W W5 530 × 74
Ans.
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8.40 The beam shown in Fig. 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.
Fig. P8.40
Solution Maximum moment magnitude: The maximum bending moment magnitude occurs at the base of the cantilever beam: 1 1 M max = (15 kN kN)(3.0 m) m) + (40 (40 kN kN/m)(3.0 m) (3.0 m) m) 2 3 = 105.0
kN-m = 105.0 × 106 N-mm
Minimum required section modulus M σ x ≥
∴ S ≥
S M
=
σ x
(105.0 kN-m)(1,000)2 165 N/mm
2
=
636 × 103 mm3
(a) Acceptable steel W-shapes W310 × 60, S = 844 × 103 mm mm3
W360 × 44,
S = 688 × 103 mm mm3
W 410 × 46.1, S = 773× 103 mm3 W 460 × 52,
S = 944 × 103 mm mm3
(b) Most economical W-shape
W360 × 44
Ans.
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8.41 The beam shown in Fig. P8.41 will be constructed from a standard steel HSS-shape using 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.
Fig. P8.41
Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = = 45.56 kip-ft Minimum required section modulus M σ x ≥
∴ S ≥
S M σ x
≥
(45.56 kip-ft)(12 in./ft) 30 ksi
= 18.22
in.3
(a) Acceptable steel HSS shapes HSS8 HSS8 none none are are acce accept ptab able le
HSS10 × 4 × 3 / 8,
S = 20.8 in.3
HSS10 × 6 × 3 / 8,
S = 20.8 in.3
HSS12 × 6 × 3 / 8,
S = 35.9 in.3
HSS12 × 8 × 3 / 8,
S = 43.7 in.3
(b) Most economical HSS shape
HSS10 × 4 × 3 / 8
Ans.
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