6.101 A torque of magnitude T = 1.5 kip-in. is applied to each of the bars shown in Fig. P6.101. If the allowable shear stress is specified as τ allow allow = 8 ksi, determine the minimum required dimension b for each bar.
Fig. P6.101
Solution (a) Circular Section Rearrange the elastic torsion formula to group terms with D with D on on the left-hand side:
π
D 4
=
T
which can be simplified simplified to
π D3
=
T
32 ( D / 2) τ 16 τ From this equation, the unknown diameter of the solid shaft can be expressed as D =
3
16T 16T
πτ
To support a torque of T = = 1.5 kip-in. without exceeding the maximum shear stress of 8 ksi, the solid shaft must have a diameter (i.e., dimension b shown in the problem statement) of bmin
≥
3
16T
πτ
=
3
16(1.5 kip-in.)
π (8 ksi)
=
0.985 in.
Ans.
(b) Square Section From Table 6.1,
b
=1 α = 0.208 ⇒ a The maximum shear stress in a rectangular section is given by Eq. (6.22): T
a=b
τ max
=
aspect ratio
α a 2b
For a square section where a = b, T 1.5 kip-in. 3 b3 = = = 0.901442 in. ατ max (0.208)(8 ksi) (c) Rectangular Section From Table 6.1, 2b aspect ratio ⇒ =2 α = 0.246 b For a rectangular section where a = 2b 2b, T 1.5 kip-in. 3 b3 = = = 0.381098 in. 2ατ max 2(0.246)(8 ksi)
∴ bmin =
0.966 in.
∴ bmin =
0.725 in.
Ans.
Ans.
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6.102 A torque of magnitude T = = 270 Nm is applied to each of the bars shown in Fig. P6.102. If the allowable shear stress is specified as τ allow allow = 70 MPa, determine the minimum required dimension b for each bar.
Fig. P6.102
Solution (a) Circular Section Rearrange the elastic torsion formula to group terms with D with D on on the left-hand side:
π
D 4
=
T
which can be simplified simplified to
π D3
=
T
32 ( D / 2) τ 16 τ From this equation, the unknown diameter of the solid shaft can be expressed as D =
3
16T 16T
πτ
To support a torque of T = = 270 N-m without exceeding the maximum shear stress of 70 MPa, the solid shaft must have a diameter (i.e., dimension b shown in the problem statement) of bmin
≥
3
16T
πτ
=
3
16(270 NN-m)(1,000 mm/m)
π (70 N/ N/mm 2 )
=
27.0 mm
Ans.
(b) Square Section From Table 6.1,
b
α = 0.208 =1 ⇒ a The maximum shear stress in a rectangular section is given by Eq. (6.22): T a=b
τ max
=
aspect ratio
α a 2b
For a square section where a = b, T (270 N-m)(1,000 N-m)(1,000 mm/m) 3 = = 18, 543.946 mm b3 = 2 ατ max (0.2 (0.208 08)( )(70 70 N/mm N/mm ) (c) Rectangular Section From Table 6.1, 2b aspect ratio α = 0.246 ⇒ =2 b For a rectangular section where a = 2b 2b, T (270 N-m)(1,000 N-m)(1,000 mm/m) 3 b3 = = = 7, 839.721 mm 2 2ατ max 2(0.246)(70 N/mm )
∴ bmin =
26.5 mm
Ans.
∴ bmin =
19.87 mm
Ans.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Copyright Act without the permission of the copyright copyright owner is unlawful. unlawful.
6.103 The bars shown in Fig. P6.103 have equal cross-sectional areas and they are each subjected to a torque of T = = 160 N-m. Determine: (a) the maximum shear stress in each bar. (b) the rotation angle at the free end i each bar has a length of 300 mm. Assume G = 28 GPa.
Fig. P6.103
Solution Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22): T
τ max
=
α a 2b
and the angle of twist is given by Eq. (6.23) TL
φ =
β a3bG
Rectangular Section 15 mm by 50 mm From linear interpolation of the values given in Table 6.1, 50 mm aspect ratio = 3.333 ⇒ α = 0.272 β = 0.269 15 mm Shear stress: (160 N-m)(1,000 mm/m) T = = 52.3 MPa τ max = 2 (0.272)((15 mm)2 (50 mm) mm) α a b (0.272) Angle of twist: TL (160 N-m)(300 mm)(1,000 mm/m) = = 0.0378 rad φ = (0.269)((15 mm) mm)3 (50 mm) mm)(28 (28,00 ,000 0 N/mm N/mm2 ) β a3bG (0.269) Rectangular Section 25 mm by 30 mm From Table 6.1, 30 mm aspect ratio = 1.2 α = 0.219 β = 0.166 ⇒ 25 mm Shear stress: T (160 N-m)(1,000 mm/m) τ max = = = 39.0 MPa 2 α a b (0.219) (0.219)(25 (25 mm) mm)2 (30 mm) mm) Angle of twist: TL (160 N-m)(300 mm)(1,000 mm/m) φ = = = 0.0220 rad 3 β a bG (0.166) (0.166)(25 (25 mm) mm)3 (30 mm)(2 mm)(28,00 8,000 0 N/mm N/mm2 )
Ans.
Ans.
Ans.
Ans.
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6.104 The allowable shear stress for each bar shown in Fig. P6.104 is 75 MPa. Determine: (a) the largest torque T that may be applied to each bar. (b) the corresponding rotation angle at the free end if each bar has a length of 300 mm. Assume G = 28 GPa.
Fig. P6.104
Solution Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22): T
τ max
=
α a 2b
and the angle of twist is given by Eq. (6.23) TL
φ =
β a3bG
Rectangular Section 15 mm by 50 mm From linear interpolation of the values given in Table 6.1, 50 mm aspect ratio = 3.333 ⇒ α = 0.272 β = 0.269 15 mm Maximum torque:
T
= τ maxα a
2
Angle of twist: TL
φ =
β a3bG
(0.272) 72)(15 mm)2 (50 mm) = 229,50 29,500 0 N-mm = 230 N-m b = (75 N/mm 2 )(0. =
(229,500 N-mm)(300 mm) (0.269) (0.269)((15 mm) mm)3 (50 mm) mm)(28 (28,00 ,000 0 N/mm N/mm2 )
Rectangular Section 25 mm by 30 mm From Table 6.1, 30 mm aspect ratio α ⇒ = 1.2 25 mm Maximum torque:
T
= τ maxα a
2
Angle of twist: TL
φ =
3
β a bG
=
0.219
=
0.0542 rad
Ans.
β = 0.166
b = (75 N/mm 2 )(0. (0.219) 19)(25 (25 mm)2 (30 mm) = 307,979 ,979 N-mm = 308 308 N-m =
Ans.
(307,979 N-mm)(300 mm) (0.166) (0.166)(25 (25 mm) mm)3 (30 mm)(2 mm)(28,00 8,000 0 N/mm N/mm2 )
=
0.0424 rad
Ans.
Ans.
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6.105 A solid circular rod having diameter D D is to be replaced by a rectangular tube having cross-sectional dimensions D × 2 D D (which are measured to the wall centerlines of the cross section shown in Fig. P6.105). Determine the required minimum thickness t min min of the tube so that the maximum shear stress in the tube will not exceed the maximum shear stress in the solid bar.
Fig. P6.105
Solution For the solid circular rod, the maximum shear stress is given by TR T ( D / 2) 16T = = τ max = π 4 π I p D3 D 32
(a)
For the rectangular tube, the area enclosed by the median line is Am = ( D)(2 D) = 2 D 2 The maximum shear stress for the thin-walled section is given by Eq. (6.25) T T T = = τ max = 2 2 Amt 2(2 D )t 4 D 2 t
(b)
Equate Eqs. (a) and (b) 16T T
π D 3
=
4 D 2t and solve for t : t min
=
π D 64
Ans.
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6.106 A 24 in. wide by 0.100 in. thick by 100 in. long steel sheet is to be formed into a hollow section by bending through 360° and welding (i.e., butt-welding) the long edges together. Assume a crosssectional medial length of 24 in. (no stretching of the sheet due to bending). If the maximum shear stress must be limited to 12 ksi, determine the maximum torque that can be carried by the hollow section if (a) the shape of the section is a circle. (b) the shape of the section is an equilateral triangle. (c) the shape of the section is a square. (d) the shape of the section is an 8 × 4 in. rectangle.
Solution The maximum shear stress for a thin-walled section is given b y Eq. (6.25) T
τ max
=
2 Amt
and thus, the maximum torque that can be carried by the hollow section is Tmax = 2τ max Am t (a) Circle: π D = 24 in.
Am
=
Tmax
π 4
=
D2
∴D =
=
π 4
7.639437 in.
(7.6 7.639437 437 in. in.))2
=
45.836 8366 in in.2
2τ max Am t = 2(12 ksi)(45.8366 in.2 )(0.100 in.) = 110.0 kip-in.
Ans.
(b) Equilateral triangle: tria triangl nglee side sidess are are each each 24 in./ in./3 3 = 8 in. in.
Am
=
Tmax
1 2
=
bh =
1 2
(8 in. in.))(8 in. in.)si )sin n 60° = 27.7 7.7128 in in.2
2τ max Am t = 2(12 ksi)(27.7128 in.2 )(0.100 in.) = 66.5 kip-in.
Ans.
(c) Square: side sidess of the the squ squar aree are are each each 24 in./ in./4 4 = 6 in.
Am
= bh = (6
Tmax
=
in.) n.)(6 in. in.)) = 36 in in.2
2τ max Am t = 2(12 ksi)(36 in.2 )(0.100 in.) = 86.4 kip-in.
Ans.
(d) 8 × 4 in. rectangle: Am = bh = (8 in. in.))(4 in in.) = 32 in. in.2
Tmax
=
2τ max Am t = 2(12 ksi)(32 in.2 )(0.100 in.) = 76.8 kip-in.
Ans.
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6.107 A 500 mm wide by 3 mm thick by 2 m long aluminum sheet is to be formed into a hollow section by bending through 360° and welding (i.e., butt-welding) the long edges together. Assume a crosssectional medial length of 500 mm (no stretching of the sheet due to bending). If the maximum shear stress must be limited to 75 MPa, determine the maximum torque that can be carried by the hollow section if (a) the shape of the section is a circle. (b) the shape of the section is an equilateral triangle. (c) the shape of the section is a square. (d) the shape of the section is a 150 × 100 mm rectangle.
Solution The maximum shear stress for a thin-walled section is given b y Eq. (6.25) T
τ max
=
2 Amt
and thus, the maximum torque that can be carried by the hollow section is Tmax = 2τ max Am t (a) Circle: π D = 500 mm
Am
=
Tmax
π 4
=
D2
=
π 4
∴ D = 159.155
(159. 159.15 155 5 mm)2
mm
19,894. 4.38 382 2 = 19,89
mm2
2τ max Am t = 2(75 N/mm2 )(19, 894.382 mm2 )(3 mm) = 8, 95 952, 47 472 N-mm = 8.95 kN-m
Ans.
(b) Equilateral triangle: triangl trianglee sides sides are each each 500 mm/3 mm/3 = 166.667 166.667 mm mm
Am
=
Tmax
1 2
=
bh =
1 2
(166.6 166.667 67 mm)( mm)(16 166.6 6.667 67 mm) mm) sin sin 60° = 12, 12, 028.1 028.131 31 mm2
2τ max Am t = 2(75 N/mm2 )(12, 02 028.131 mm2 )(3 mm) = 5, 41 412, 65 659 N-mm = 5.41 kN-m
Ans.
(c) Square: side sidess of the the squ squar aree are are each each 500 500 mm/4 mm/4 = 125 125 mm
Am
2
mm)(125 125 mm) mm) = 15, 15, 625 mm = bh = (125 mm)(
Tmax
=
2τ max Am t = 2(75 N/mm2 )(15, 62 625 mm2 )(3 mm) = 7, 03 031, 25 250 N-mm = 7.03 kN-m
Ans.
(d) 150 × 100 mm rectangle: Am = bh = (150 150 mm) mm)(100 (100 mm) mm) = 15,000 15,000 mm mm2
Tmax
=
2τ max Am t = 2(75 N/mm2 )(15, 00 000 mm2 )(3 mm) = 6, 75 750, 00 000 N-mm = 6.75 kN-m
Ans.
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6.108 A torque of T = = 150 kip-in. will be applied to the hollow, thin-walled aluminum alloy section shown in Fig. P6.108. If the maximum shear stress must be limited to 10 ksi, determine the minimum thickness required for the section. ( Note: Note: The dimensions shown are measured to the wall centerline.)
Fig. P6.108
Solution The maximum shear stress for a thin-walled section is given b y Eq. (6.25) T
τ max
=
2 Amt
For the aluminum alloy section, Am
=
(6 in.)(8 in.) +
π 4
(6 in.)2
=
76.274 in.2
The minimum thickness required for the section if the maximum shear stress must be limited to 10 ksi is thus: T 150 kip-in. t min = = = 0.0983 in. Ans. 2τ max Am 2(10 ksi)(76.274 in.2 )
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6.109 A torque of T = = 2.5 kN-m will be applied to the hollow, thin-walled aluminum alloy section shown in Fig. P6.109. I the maximum shear stress must be limited to 50 MPa, determine the minimum thickness required for the section. ( Note: Note: The dimensions shown are measured to the wall centerline.)
Fig. P6.109
Solution The maximum shear stress for a thin-walled section is given b y Eq. (6.25) T
τ max
=
2 Amt
For the aluminum alloy section, 1 Am = π (100 mm)2 = 7,85 ,853.982 mm2 4 The minimum thickness required for the section if the maximum shear stress must be limited to 50 MPa is thus: T (2.5 kN-m)(1,000 N/kN)(1,000 mm/m) mm/m) t min = = = 3.18 mm Ans. 2 2τ max Am 2(50 N/mm )(7, 853.982 mm2 )
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6.110 A torque of T = = 100 kip-in. will be applied to the hollow, thin-walled aluminum alloy section shown in Fig. P6.110. If the section has a uniform thickness of 0.100 in., determine the magnitude o the maximum shear stress developed in the section. ( Note: Note: The dimensions shown are measured to the wall centerline.)
Fig. P6.110
Solution The maximum shear stress for a thin-walled section is given b y Eq. (6.25) T
τ max
=
2 Amt
For the aluminum alloy section, Am
=
π 8
(10 in in.)2
=
39.2699 in in.2
The maximum shear stress developed in the section is: T 100 kip-in. = = 12.73 ksi τ max = 2 Amt 2(39.2699 in in.2 )(0.100 in in.)
Ans.
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6.111 A torque of T = 2.75 kN-m will be applied to the hollow, thin-walled aluminum alloy section shown in Fig. P6.111. If the section has a uniform thickness of 4 mm, determine the magnitude of the maximum shear stress developed in the section. ( Note: Note: The dimensions shown are measured to the wall centerline.)
Fig. P6.111
Solution The maximum shear stress for a thin-walled section is given b y Eq. (6.25) T
τ max
=
2 Amt
For the aluminum alloy section, Am
=
(150 mm mm)(50 mm) +
π 2
(25 mm)2
= 8, 48 481.748
mm2
The maximum shear stress developed in the section is: T (2.75 kN-m)(1,000 N/kN)(1,000 mm/m) mm/m)
τ max
=
2 Amt
=
2(8, 481.748 mm2 )(4 mm)
=
40.5 MPa
Ans.
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6.112 A cross section of the leading edge of an airplane wing is shown in Fig. P6.112. The enclosed 2 area is 82 in. . Sheet thicknesses are shown on the diagram. For an applied torque of T = 100 kip-in., determine the magnitude of the maximum shear stress developed in the section. ( Note: Note: The dimensions shown are measured to the wall centerline.)
Fig. P6.112
Solution The maximum shear stress for a thin-walled section is given b y Eq. (6.25) T
τ max
=
2 Amt
The maximum shear stress developed in the section is: T 100 kip-in. τ max = = = 12.20 ksi 2 Amt 2(82 in.2 )(0.050 in.)
Ans.
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6.113 A cross section of an airplane fuselage made of aluminum alloy is shown in Fig. P6.113. For an applied torque of T = = 1,250 kip-in. and an allowable shear stress of τ = = 7.5 ksi, determine the minimum thickness of the sheet (which must be constant for the entire periphery) required to resist the torque. ( Note: Note: The dimensions shown are measured to the wall centerline.)
Fig. P6.113
Solution The maximum shear stress for a thin-walled section is given b y Eq. (6.25) T
τ max
=
2 Amt
For the fuselage, Am = (30 in in.)(20 in in.) + π (15 in in.)2
,306.858 = 1,30
in in.2
The minimum thickness required for the sheet if the maximum shear stress must be limited to 7.5 ksi is thus: T 1, 250 kip-in. t min = = = 0.0638 in. Ans. 2τ max Am 2(7.5 ksi)(1, 306.858 in.2 )
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