Mechanics of Materials Solutions Chapter09 Probs18 37
Short Description
Mechanics of Materials Solutions Chapter09 Probs18 37...
Description
9.18 A 50-mm-diameter solid steel shaft supports loads PA = 1.5 kN and PC = 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 in the shaft. (b) the maximum tension bending stress in the shaft.
Fig. P9.18
Solution Section properties: I=
π
D4 =
π
(50 mm) 4
64 64 = 306, 796.158 mm 4 D 3 (50 mm)3 = 12 12 = 10, 416.667 mm3
Q=
Maximum shear force magnitude: Vmax = 1.71 kN (between B and C) Maximum bending moment magnitude: Mmax = 289.29 kN-mm (at C)
(a) Maximum horizontal shear stress: V Q (1, 710 N)(10,416.667 mm3 ) τ= = I t (306, 796.158 mm 4 )(50 mm) = 1.161 MPa
(at neutral axis between B and C )
Ans.
(b) Maximum tension bending stress: My (289.29 kN-mm)( − 50 mm/2)(1, 000 N/kN) =− σx = − I 306, 796.158 mm 4 = 23.574 MPa = 23.6 MPa (T)
(on bottom of shaft at C )
Ans.
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9.19 A 1.25-in.-diameter solid steel shaft supports loads PA = 400 lb and PC = 900 lb, as shown in Fig. P9.19. Assume L1 = 6 in., L2 = 12 in., and L3 = 8 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: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tension bending stress in the shaft.
Fig. P9.19
Solution Section properties: I=
π
D4 =
π
(1.25 in.) 4
64 64 = 0.119842 in.4
D 3 (1.25 in.)3 = 12 12 = 0.162760 in.3
Q=
Maximum shear force magnitude: Vmax = 480 lb (between B and C) Maximum bending moment magnitude: Mmax = 3,360 lb-in. (at C)
(a) Maximum horizontal shear stress: V Q (480 lb)(0.162760 in.3 ) τ= = I t (0.119842 in.4 )(1.25 in.) = 521.519 psi = 522 psi
(at neutral axis between B and C )
Ans.
(b) Maximum tension bending stress: My (3,360 lb-in.)( − 1.25 in./2) σx = − =− I 0.119842 in.4
= 17,523.022 psi = 17,520 psi (T)
(on bottom of shaft at C )
Ans.
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9.20 A 1.00-in.-diameter solid steel shaft supports loads PA = 500 lb and PD = 300 lb, as shown in Fig. P9.20. Assume L1 = 5 in., L2 = 16 in., and L3 = 8 in. The bearing at 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: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tension bending stress in the shaft.
Fig. P9.20
Solution Section properties:
I=
π
D4 =
π
(1.00 in.) 4
64 64 = 0.049087 in.4
D 3 (1.00 in.)3 Q= = 12 12 = 0.083333 in.3 Maximum shear force magnitude: Vmax = 500 lb (between A and B) Maximum bending moment magnitude: Mmax = 2,500 lb-in. (at B)
(a) Maximum horizontal shear stress: V Q (500 lb)(0.083333 in.3 ) τ= = I t (0.049087 in.4 )(1.00 in.) = 848.826 psi = 849 psi
(at neutral axis between A and B )
Ans.
(b) Maximum tension bending stress: My (−2,500 lb-in.)(1.00 in./2) σx = − =− I 0.049087 in.4
= 25, 464.772 psi = 25,500 psi (T)
(on top of shaft at B)
Ans.
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9.21 A 20-mm-diameter solid steel shaft supports loads PA = 1,300 N and PD = 900 N, as shown in Fig. P9.21. Assume L1 = 75 mm, L2 = 300 mm, and L3 = 125 mm. The bearing at 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: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tension bending stress in the shaft.
Fig. P9.21
Solution Section properties:
I=
π
D4 =
π
(20 mm) 4
64 64 = 7,853.982 mm 4 D 3 (20 mm)3 = 12 12 = 666.667 mm3
Q=
Maximum shear force magnitude: Vmax = 1,300 N (between A and B) Maximum bending moment magnitude: Mmax = 112,500 N-mm (at C)
(a) Maximum horizontal shear stress: V Q (1,300 N)(666.667 mm3 ) τ= = I t (7,853.982 mm 4 )(20 mm) = 5.517 MPa = 5.52 MPa
(at neutral axis between A and B )
Ans.
(b) Maximum tension bending stress: My (−112,500 N-mm)(20 mm/2) =− σx = − I 7,853.982 mm 4 = 143.239 MPa = 143.2 MPa (T)
(on top of shaft at C )
Ans.
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9.22 A 1.25-in.-diameter solid steel shaft supports loads PA = 600 lb, PC = 1,600 lb, and PE = 400 lb, as shown in Fig. P9.22. Assume L1 = 6 in., L2 = 15 in., L3 = 8 in., and L4 = 10 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: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tension bending stress in the shaft.
Fig. P9.22
Solution Section properties:
I=
π
D4 =
π
(1.25 in.) 4
64 64 = 0.119842 in.4
D 3 (1.25 in.)3 = 12 12 = 0.162760 in.3
Q=
Maximum shear force magnitude: Vmax = 1,060.9 lb (between C and D) Maximum bending moment magnitude: Mmax = 4,487 lb-in. (at C)
(a) Maximum horizontal shear stress: V Q (1, 060.9 lb)(0.162760 in.3 ) τ= = It (0.119842 in.4 )(1.25 in.) = 1,152.632 psi = 1,153 psi
(at neutral axis between C and D )
Ans.
(b) Maximum tension bending stress: My (4, 487 lb-in.)( − 1.25 in./2) σx = − =− I 0.119842 in.4
= 23, 400.309 psi = 23, 400 psi (T)
(on bottom of shaft at C )
Ans.
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9.23 A 25-mm-diameter solid steel shaft supports loads PA = 1,000 N, PC = 3,200 N, and PE = 800 N, as shown in Fig. P9.23. Assume L1 = 80 mm, L2 = 200 mm, L3 = 100 mm, 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 the shaft. (b) the maximum tension bending stress in the shaft.
Fig. P9.23
Solution Section properties: I=
π
D4 =
π
(25 mm) 4
64 64 = 19,174.760 mm 4 D 3 (25 mm)3 = 12 12 = 1,302.083 mm3
Q=
Maximum shear force magnitude: Vmax = 2,200 N (between C and D) Maximum bending moment magnitude: Mmax = 120,000 N-mm (at C)
(a) Maximum horizontal shear stress: V Q (2, 200 N)(1,302.083 mm3 ) τ= = I t (19,174.760 mm 4 )(25 mm) = 5.98 MPa
(at neutral axis between C and D)
Ans.
(b) Maximum tension bending stress: My (120, 000 N-mm)( − 25 mm/2) =− σx = − I 19,174.760 mm 4 = 78.228 MPa = 78.2 MPa (T)
(on bottom of shaft at C )
Ans.
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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:
I=
π
[D4 − d 4 ] =
π
[(3.500 in.) 4 − (3.068 in.)4 ] = 3.017157 in.4
64 64 1 1 Q = [ D 3 − d 3 ] = [(3.500 in.)3 − (3.068 in.)3 ] = 1.166422 in.3 12 12 Maximum shear force magnitude: Vmax = 700 lb Maximum bending moment magnitude: Mmax = (700 lb)(4 ft)(12 in./ft) = 33,600 lb-in.
(a) Maximum horizontal shear stress: VQ (700 lb)(1.166422 in.3 ) τ= = = 626.429 psi = 626 psi I t (3.017157 in.4 )(3.500 in. − 3.068 in.) (b) Maximum tension bending stress: My (−33, 600 lb-in.)(3.500 in./2) σx = − =− = 19, 488.533 psi = 19, 490 psi (T) I 3.017157 in.4
Ans.
Ans.
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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. (b) the maximum tension bending stress in the pipe.
Fig. P9.25a Cantilever beam
Fig. P9.25b Pipe cross section
Solution Section properties:
I=
π
[D4 − d 4 ] =
π
[(170 mm) 4 − (150 mm) 4 ] = 16,147, 786.239 mm 4
64 64 1 1 Q = [ D 3 − d 3 ] = [(170 mm)3 − (150 mm)3 ] = 128,166.667 mm3 12 12 Maximum shear force magnitude: Vmax = 8 kN = 8,000 N Maximum bending moment magnitude: Mmax = (8,000 N)(1.2 m)(1,000 mm/m) = 9,600,000 N-mm
(a) Maximum horizontal shear stress: VQ (8, 000 N)(128,166.667 mm3 ) τ= = = 3.174842 MPa = 3.17 MPa I t (16,147, 786.239 mm 4 )(170 mm − 150 mm) (b) Maximum tension bending stress: My (−9, 600, 000 N-mm)(170 mm/2) σx = − =− = 50.533 MPa = 50.5 MPa (T) I 16,147, 786.239 mm 4
Ans.
Ans.
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9.26 A concentrated load of P = 22 kN is applied to the upper end of a 2-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. (b) the maximum tension bending stress in the pipe.
Fig. P9.26
Solution Section properties:
I=
π
[D4 − d 4 ] =
π
[(330 mm)4 − (300 mm)4 ] = 184,529, 789.364 mm 4
64 64 1 1 Q = [ D 3 − d 3 ] = [(330 mm)3 − (300 mm)3 ] = 744, 750 mm3 12 12 Maximum shear force magnitude: Vmax = 22 kN = 22,000 N Maximum bending moment magnitude: Mmax = (22,000 N)(2 m)(1,000 mm/m) = 44,000,000 N-mm
(a) Maximum vertical shear stress: VQ (22, 000 N)(744, 750 mm3 ) τ= = = 2.959685 MPa = 2.96 MPa I t (184,529, 789.364 mm 4 )(330 mm − 300 mm) (b) Maximum tension bending stress: M c (44, 000, 000 N-mm)(330 mm/2) σx = = = 39.343 MPa = 39.3 MPa (T) I 184,529, 789.364 mm 4
Ans.
Ans.
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9.27 A concentrated load of P = 6 kips is applied to the upper end of a 4-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. (b) the maximum tension bending stress in the pipe.
Fig. P9.27
Solution Section properties:
I=
π
[D4 − d 4 ] =
π
[(8.625 in.)4 − (7.981 in.)4 ] = 72.489241 in.4
64 64 1 1 Q = [ D 3 − d 3 ] = [(8.625 in.)3 − (7.981 in.)3 ] = 11.104874 in.3 12 12 Maximum shear force magnitude: Vmax = 6 kips = 6,000 lb Maximum bending moment magnitude: Mmax = (6,000 lb)(4 ft)(12 in./ft) = 288,000 lb-in.
(a) Maximum vertical shear stress: VQ (6, 000 lb)(11.104874 in.3 ) τ= = = 1, 427.267 psi = 1, 427 psi I t (72.489241 in.4 )(8.625 in. − 7.981 in.) (b) Maximum tension bending stress: M c (288, 000 lb-in.)(8.625 in./2) σx = = = 17,133.564 psi = 17,130 psi (T) I 72.489241 in.4
Ans.
Ans.
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9.28 The internal shear force V at a certain section of an aluminum 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 mm above 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 top flange stem
yi Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) 375.0 72.5 27,187.5 350.0 35.0 12,250.0 39,437.5 mm3 725.0 mm2
Σyi Ai
39, 437.5 mm3 y= = = 54.397 mm ΣAi 725.0 mm 2 = 20.603 mm
(from bottom of shape to centroid) (from top of shape to centroid)
Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 781.250 18.103 122,900.565 stem 142,916.667 −19.397 131,679.177 Moment of inertia about the z axis (mm4) =
IC + d²A (mm4) 123,681.815 274,595.843 398,277.658
(a) Shear stress at H: Q = (5 mm)(30 mm)(39.397 mm) = 5,909.550 mm3
τ=
(8, 000 N)(5,909.550 mm3 ) = 23.7 MPa (398, 277.658 mm 4 )(5 mm)
Ans.
(b) Maximum horizontal shear stress: At neutral axis: Q = (5 mm)(54.397 mm)(27.199 mm) = 7,397.720 mm3
τ=
(8, 000 N)(7,397.720 mm3 ) = 29.7 MPa (398, 277.658 mm 4 )(5 mm)
Ans.
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9.29 The internal shear force V at a certain section of a steel 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 wideflange shape.
Fig. P9.29
Solution Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 59,062.5 97.5 29,944,687.5 web 4,860,000.0 0.0 0.0 bottom flange 59,062.5 −97.5 29,944,687.5 Moment of inertia about the z axis (mm4) =
IC + d²A (mm4) 30,003,750.0 4,860,000.0 30,003,750.0 64,867,500.0
(a) Shear stress at H: Q = (210 mm)(15 mm)(97.5 mm) + (10 mm)(60 mm)(60 mm) = 343,125 mm3
(80, 000 N)(343,125 mm3 ) = 42.3 MPa τ= (64,867,500 mm 4 )(10 mm)
Ans.
(b) Maximum horizontal shear stress: At neutral axis: Q = (210 mm)(15 mm)(97.5 mm) + (10 mm)(90 mm)(45 mm) = 347, 625 mm3
(80, 000 N)(347, 625 mm3 ) = 42.9 MPa τ= (64,867,500 mm 4 )(10 mm)
Ans.
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9.30 The internal shear force V at a certain section of 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 Centroid location in y direction:
Shape
Area Ai (in.2) 5.0 10.0 8.0 23.0 in.2
top flange (1) web bottom flange
y=
Σyi Ai ΣAi
=
yi (from bottom) (in.) 11.5 6.0 0.5
yi Ai (in.3) 57.5 60.0 4.0 121.5 in.3
121.5 in.3 = 5.2826 in. 23.0 in.2
(from bottom of shape to centroid)
= 6.7174 in.
(from top of shape to centroid)
Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) top flange 0.4167 6.2174 193.2798 web 83.3333 0.7174 5.1465 bottom flange 0.6667 −4.7826 182.9868 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 193.6964 88.4798 183.6534 465.8297
(a) Shear stress at H: Q = (5 in.)(1 in.)(6.2174 in.) + (1 in.)(1 in.)(5.2174 in.) = 36.3044 in.3
(110 kips)(36.3044 in.3 ) = 8.57 ksi τ= (465.8297 in.4 )(1 in.)
Ans.
(b) Maximum horizontal shear stress: At neutral axis: Q = (5 in.)(1 in.)(6.2174 in.) + (1 in.)(5.7174 in.)(2.8587 in.) = 47.4313 in.3
τ=
(110 kips)(47.4313 in.3 ) = 11.20 ksi (465.8297 in.4 )(1 in.)
Ans.
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9.31 The internal shear force V at a certain section 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 the flanged shape.
Fig. P9.31
Solution Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (in. ) (in.) (in.4) left flange 62.5000 0.0000 0.0000 web 0.0885 0.0000 0.0000 right flange 62.5000 0.0000 0.0000 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 62.5000 0.0885 62.5000 125.0885
(a) Shear stress at H: Q = 2(0.75 in.)(2 in.)(4 in.) = 12 in.3
(75 kips)(12 in.3 ) = 4.80 ksi τ= (125.0885 in.4 )(2 × 0.75 in.)
Ans.
(b) Shear stress at K: Q = 2(0.75 in.)(4.5 in.)(2.75 in.) = 18.5625 in.3
τ=
(75 kips)(18.5625 in.3 ) = 7.42 ksi (125.0885 in.4 )(2 × 0.75 in.)
Ans.
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9.32 Consider a 100-mm-long segment of a simply supported 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 crosssectional 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 Centroid location in y direction:
yi Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) top flange 9,000 300 2,700,000 web 8,400 165 1,386,000 bottom flange 15,000 30 450,000 2 4,536,000 mm3 32,400 mm Σyi Ai 4,536, 000 mm3 y= = = 140 mm (from bottom of shape to centroid) ΣAi 32,400 mm 2 Shape
= 190 mm (from top of shape to centroid) Moment of inertia about the z axis: d = yi – y Shape IC + d²A IC d²A 4 (mm ) (mm) (mm4) (mm4) 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: ΔM 80 kN-m − 75 kN-m 5 kN-m V= = = = 50 kN Δx 100 mm 0.1 m Maximum horizontal shear stress: At neutral axis: Q = (250 mm)(60 mm)(110 mm) + (40 mm)(80 mm)(40 mm) = 1, 778, 000 mm3
(50, 000 N)(1, 778, 000 mm3 ) = 4.88 MPa τ= (455, 220, 000 mm 4 )(40 mm)
Ans.
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9.33 A simply supported beam supports the loads shown in Fig. P9.33a. The cross-sectional dimensions of 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: Vmax = 175 kN (just to the right of B) Maximum bending moment magnitude: Mmax = 156.25 kN-m (between B and C)
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Section properties: Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) top flange 56,250 142.5 60,918,750 web 16,402,500 0 0 bottom flange 56,250 −142.5 60,918,750 Moment of inertia about the z axis (mm4) =
IC + d²A (mm4) 60,975,000 16,402,500 60,975,000 138,352,500
(b) Shear stress at H: Q = (200 mm)(15 mm)(142.5 mm) + (10 mm)(35 mm)(117.5 mm) = 468,625 mm3
τ=
(175,000 N)(468,625 mm3 ) = 59.3 MPa (138,352,500 mm 4 )(10 mm)
Ans.
(c) Maximum horizontal shear stress: At neutral axis: Q = (200 mm)(15 mm)(142.5 mm) + (10 mm)(135 mm)(67.5 mm) = 518,625 mm3
τ=
(175,000 N)(518,625 mm3 ) = 65.6 MPa (138,352,500 mm 4 )(10 mm)
Ans.
(d) Maximum tension bending stress: Mc (156.25 × 106 N-mm)(300 mm/2) σx = =− I 138,352,500 mm 4
= 169.404 MPa = 169.4 MPa
Ans.
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9.34 A simply supported beam supports the loads shown in Fig. P9.34a. The cross-sectional dimensions of 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.
Fig. P9.34a
Fig. P9.34b
Solution Shear force magnitude at a–a: V = 27.60 kips Bending moment at a–a: M = 60.90 kip-ft Section properties: (12 in.)(16 in.)3 (11.25 in.)(15.25 in.)3 I= − 12 12 4 = 771.0830 in. (a) Bending stress at H: My σH = − I (60,900 lb-ft)(5 in.)(12 in./ft) =− 771.0830 in.4 = −4,738.79 psi = 4,740 psi (C)
Ans.
Shear stress at H: Q = (12 in.)(0.375 in.)(7.8125 in.) + 2(0.375 in.)(2.625 in.)(6.3125 in.) = 47.5840 in.3
(27,600 lb)(47.5840 in.3 ) = 2, 270 psi τ= (771.0830 in.4 )(2 × 0.375 in.)
Ans.
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(b) Maximum shear force magnitude: V = 39.60 kips (at pin B) Maximum horizontal shear stress: At neutral axis: Q = (12 in.)(0.375 in.)(7.8125 in.) + 2(0.375 in.)(7.625 in.)(3.8125 in.) = 56.9590 in.3
(39,600 lb)(56.9590 in.3 ) = 3,900 psi τ= (771.0830 in.4 )(2 × 0.375 in.)
Ans.
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9.35 A cantilever beam supports the loads shown in Fig. P9.35a. The cross-sectional dimensions of the shape 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.
Fig. P9.35a
Fig. P9.35b
Solution Centroid location in y direction:
Shape top flange left stem right stem
Width b (in.) 12.0 0.5 0.5
Height h (in.) 0.5 5.5 5.5
Σyi Ai
49.6250 in.3 y= = = 4.3152 in. ΣA i 11.50 in.2 = 1.6848 in.
Area Ai (in.2) 6.00 2.75 2.75 11.50
yi (from bottom) (in.) 5.75 2.75 2.75
yi Ai (in.3) 34.5000 7.5625 7.5625 49.6250
(from bottom of shape to centroid) (from top of shape to centroid)
Moment of inertia about the z axis: d = yi – y Shape d²A IC 4 (in.) (in.4) (in. ) top flange 0.1250 1.4348 12.3519 left stem 6.9323 −1.5652 6.7371 right stem 6.9323 −1.5652 6.7371 Moment of inertia about the z axis (in.4) =
IC + d²A (in.4) 12.4769 13.6694 13.6694 39.8157
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Maximum shear force magnitude: V = 5,800 lb Maximum positive bending moment: Mpos = 8,850 lb-ft Maximum negative bending moment: Mneg = −9,839 lb-ft (a) Maximum shear stress: Q = 2(0.5 in.)(4.3152 in.)(4.3152 in./2) = 9.3105 in.3
τ=
(5,800 lb)(9.3105 in.3 ) (39.8157 in.4 )(2 × 0.5 in.)
= 1,360 psi
Ans.
(b) Maximum compression bending stress: Check two possibilities. First, check the bending stress created by the largest positive moment at the top of the cross section: M y (8,850 lb-ft)(1.6848 in.)(12 in./ft) σ x = − pos top = − = −4, 494 psi Iz 39.8157 in.4 Next, for the largest negative moment, compute the bending stress at the bottom of the cross section: M y (−9,839 lb-ft)( − 4.3152 in.)(12 in./ft) σ x = − neg bot = − = −12,796 psi Iz 39.8157 in.4 Therefore, the maximum compression bending stress is: σ comp = 12,800 psi (C) 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: M y (8,850 lb-ft)( − 4.3152 in.)(12 in./ft) σ x = − pos bot = − = 11,510 psi 39.8157 in.4 Iz Next, for the largest negative moment, compute the bending stress at the top of the cross section: M y (−9,839 lb-ft)(1.6848 in.)(12 in./ft) σ x = − neg top = − = 4,996 psi 39.8157 in.4 Iz Therefore, the maximum tension bending stress is: σ tens = 11,510 psi (T) Ans.
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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. P9.36a
Fig. P9.36b
Solution Maximum shear force magnitude: Vmax = 5 kN Maximum positive bending moment: Mpos = 2.00 kN-m Maximum negative bending moment: Mneg = −1.50 kN-m
Centroid location in y direction:
Shape flange stem
Width b (mm) 100 6
Height h (mm) 8 92
Area Ai (mm2) 800 552 1,352
yi (from bottom) (mm) 96 46
yi Ai (mm3) 76,800 25,392 102,192
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y=
Σyi Ai ΣA i
=
102,192 mm3 = 75.5858 mm 1,352 mm 2 = 24.4142 mm
(from bottom of shape to centroid) (from top of shape to centroid)
Moment of inertia about the z axis: d = yi – y Shape IC d²A 4 (mm ) (mm) (mm4) flange 4,266.67 20.4142 333,391.69 stem 389,344.00 -29.5858 483,176.36 Moment of inertia about the z axis (mm4) =
IC + d²A (mm4) 337,658.35 872,520.36 1,210,178.71
(a) Maximum vertical shear stress: At neutral axis: Q = (6 mm)(75.5858 mm)(75.5858 mm/2) = 17,139.640 mm3
(5,000 N)(17,139.640 mm3 ) τ= = 11.80 MPa (1,210,178.71 mm 4 )(6 mm)
Ans.
(b) Maximum compression bending stress: Check two possibilities. First, check the bending stress created by the largest positive moment at the top of the cross section: M pos ytop (2.00 × 106 N-mm)(24.4142 mm) =− = −40.348 MPa σx = − 1,210,178.71 mm 4 Iz Next, for the largest negative moment, compute the bending stress at the bottom of the cross section: M y (−1.50 × 106 N-mm)( − 75.5858 mm) = −93.688 MPa σ x = − neg bot = − 1,210,178.71 mm 4 Iz Therefore, the maximum compression bending stress is: σ comp = 93.7 MPa (C) 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: M pos ybot (2.00 × 106 N-mm)( − 75.5858 mm) =− = 124.917 MPa σx = − 1,210,178.71 mm 4 Iz Next, for the largest negative moment, compute the bending stress at the top of the cross section: M y (−1.50 × 106 N-mm)(24.4142 mm) = 30.261 MPa σ x = − neg top = − 1,210,178.71 mm 4 Iz Therefore, the maximum tension bending stress is: σ tens = 124.9 MPa (T) Ans.
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9.37 A simply supported beam fabricated from pultruded reinforced plastic supports the loads shown in Fig. 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?
Fig. P9.37a
Fig. P9.37b
Solution Section properties: (4 in.)(8 in.)3 (3.625 in.)(7.25 in.)3 Iz = − 12 12 4 = 55.5493 in. (a) Maximum shear force magnitude: V = 3,644 lb (b) Shear stress at H: Q = (4 in.)(0.375 in.)(3.8125 in.) +(0.375 in.)(1.625 in.)(2.8125 in.) = 7.4326 in.3
τ=
(3,664 lb)(7.4326 in.3 ) (55.5493 in.4 )(0.375 in.)
= 1,307 psi
Ans.
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(c) Maximum horizontal shear stress: At neutral axis: Q = (4 in.)(0.375 in.)(3.8125 in.) + (0.375 in.)(3.625 in.)(1.8125 in.) = 8.1826 in.3
τ=
(3,664 lb)(8.1826 in.3 ) = 1, 439 psi (55.5493 in.4 )(0.375 in.)
(d) Maximum compression bending stress: My (7,719 lb-ft)(4 in.)(12 in./ft) σH = − =− = −6,669.965 psi = 6,670 psi (C) 55.5493 in.4 I
Ans.
Ans.
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