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6.42 The driveshaft of an automobile is being designed to transmit 280 hp at 3,500 rpm. Determine the minimum diameter required for a solid steel shaft if the allowable shear stress in the shaft is not to exceed 4,000 psi.

Solution The torque in the driveshaft is: ( 280 hp) ⎛⎜ 550 lb-ft/s ⎞⎟ P ⎝ 1 hp ⎠ = 420.169 lb-ft T= = ω ⎛ 3,500 rev ⎞⎛ 2 π rad ⎞⎛ 1 min ⎞ ⎜ min ⎟⎜ 1 rev ⎟⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ The minimum diameter required for the shaft can be found from: T (420.169 lb-ft)(12 in./ft) π 3 = = 1.260507 in.3 D ≥ 16 4,000 psi τallow

∴ D ≥ 1.858539 in. = 1.859 in.

Ans.

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6.43 A tubular steel shaft transmits 150 hp at 4,000 rpm. Determine the maximum shear stress produce d in the shaft if the outside diameter is D = 3.000 in. and the wall thickness is t = 0.125 in.

Solution The torque in the tubular steel shaft is: ⎛ 550 lb-ft/s ⎞ ( 150 hp) ⎜ ⎟ P ⎝ 1 hp ⎠ = 196.954 lb-ft T= = ω ⎛ 4,000 rev ⎞⎛ 2 π rad ⎞⎛1 min ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ min ⎠⎝ 1 rev ⎠⎝ 60 s ⎠ The polar moment of inertia of the shaft is:

Ip =

π

π

⎡ D 4 − d 4 ⎤⎦ = ⎡⎣ (3.000 in.) 4 − (2.750 in.) 4 ⎤⎦ = 2.337403 in.4 32 ⎣ 32

and the maximum shear stress in the shaft is TR (196.954 lb-ft)(3.000 in./2)(12 in./ft) = = 1,516.71 psi = 1,517 psi τ= Ip 2.337403 in.4

Ans.

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6.44 A tubular steel shaft is being designed to transmit 225 kW at 1,700 rpm. The maximum shear stress in the shaft must not exceed 30 MPa. If the outside diameter of the shaft is D = 75 mm, determine the minimum wall thickness for the shaft.

Solution The torque in the tubular steel shaft is: ⎛ 1,000 N-m/s ⎞ 225 kW ) ⎜ ( ⎟ P ⎝ 1 kW ⎠ = T= = 1, 263.877 N-m π ω ⎛ 1,700 rev ⎞⎛ 2 rad ⎞⎛ 1 min ⎞ ⎜ min ⎟⎜ 1 rev ⎟⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ The maximum shear stress in the shaft will be determined from the elastic torsion formula: π TR ⎡⎣ D4 − d 4 ⎤⎦ τ= where I p = Ip 32 Rearrange this equation, grouping the diameter terms on the left-hand side of the equation: 4 4 π ⎡⎣ D − d ⎤⎦ T ≥ τ 32 D / 2 Substitute the known values for T, τ, and D and solve for the inside diameter d: 4 4 π ⎡⎣ (75 mm) − d ⎤⎦ (1, 263.877 N-m)(1,000 mm/m) ≥ 2 32 75 mm / 2 30 N/mm (75 mm)4 − d4 ≥ 16,092,181.76 mm4 d 4 ≤ (75 mm)4 −16,092,181.76 mm 4 = 15,548,443.24 mm 4 ∴ d ≤ 62.7945 mm The minimum wall thickness for the tubular steel shaft is thus D = d + 2t D − d 75 mm − 62.7945 mm ∴t ≥ = = 6.10275 mm = 6.10 mm 2 2

Ans

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6.45 A solid 20-mm-diameter bronze shaft transmits 9 kW at 25 Hz to the propeller of a small sailboat. Determine the maximum shear stress produced in the shaft.

Solution The torque in the tubular steel shaft is: ⎛ 1,000 N-m/s ⎞ 9 kW )⎜ ( ⎟ P ⎝ 1 kW ⎠= T= = 57.295780 N-m ω ⎛ 25 rev ⎞⎛ 2π rad ⎞ ⎜ ⎟⎜ ⎟ ⎝ s ⎠⎝ 1 rev ⎠ The polar moment of inertia of the shaft is: Ip =

π 32

D4 =

π 32

(20 mm) 4 =15, 707.963 mm 4

and the maximum shear stress in the shaft is TR (57.295780 N-m)(20 mm/2)(1,000 mm/m) = = 36.476 MPa = 36.5 MPa τ= Ip 15, 707.963 mm 4

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 Act without the permission of the copyright owner is unlawful.

6.46 A steel propeller for a windmill transmits 11 kW at 57 rpm. If the allowable shear stress in the shaft must be limited to 60 MPa, determine the minimum diameter required for a solid shaft.

Solution The torque in the tubular steel shaft is: ⎛ 1,000 N-m/s ⎞ 11 kW ) ⎜ ( ⎟ P ⎝ 1 kW ⎠ = T= = 1,842.847 N-m ω ⎛ 57 rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ min ⎠⎝ 1 rev ⎠⎝ 60 s ⎠ The minimum diameter required for the shaft can be found from: π 3 T (1,842.847 N-m)(1,000 mm/m) D ≥ = = 30, 714.117 mm 3 2 τ 16 60 N/mm allow ∴ D ≥ 53.881 mm = 53.9 mm

Ans.

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6.47 A solid steel [G = 12,000 ksi] shaft with a 3-in. diameter must not twist more than 0.06 rad in a 16ft length. Determine the maximum horsepower that the shaft can transmit at 7 Hz.

Solution Section properties: The polar moment of inertia of the shaft is:

Ip =

π 32

D4 =

π 32

(3 in.) 4 = 7.952156 in. 4

Angle of twist relationship: From the angle of twist relationship, compute the allowable torque: φ G I p (0.06 rad)(12,000 ksi)(7.952156 in.4 ) T= = = 29.820585 kip-in. = 2, 485.05 lb-ft (16 ft)(12 in./ft) L Power transmission: The maximum power that can be transmitted at 7 Hz is: ⎛ 7 rev ⎞⎛ 2π rad ⎞ P = T ω = (2, 485.05 lb-ft) ⎜ ⎟⎜ ⎟ = 109,298.21 lb-ft/s ⎝ s ⎠⎝ 1 rev ⎠ or in units of horsepower, 109,298.21 lb-ft/s P= = 198.7 hp 550 lb-ft/s 1 hp

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 Act without the permission of the copyright owner is unlawful.

6.48 A tubular steel [G = 80 GPa] shaft with an outside diameter of D = 90 mm and a wall thickness of t = 8 mm must not twist more than 0.05 rad in a 7-m length. Determine the maximum power that the shaf t can transmit at 335 rpm.

Solution Section properties: The polar moment of inertia of the shaft is:

Ip =

π

π

⎡⎣ D 4 − d 4 ⎤⎦ = ⎡⎣ (90 mm) 4 − (74 mm) 4 ⎦⎤ = 3, 497,321.47 mm 4 32 32

Angle of twist relationship: From the angle of twist relationship, compute the allowable torque: φ G I p (0.05 rad)(80,000 N/mm 2 )(3,497,321.47 mm 4 ) T= = = 1,998, 469.41 N-mm = 1,998.5 N-m L (7 m)(1,000 mm/m) Power transmission: The maximum power that can be transmitted at 335 rpm is: ⎛ 335 rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞ P = T ω = (1998.5 N-m) ⎜ ⎟⎜ ⎟⎜ ⎟ = 70,109.6 N-m/s = 70.1 kW ⎝ min ⎠⎝ 1 rev ⎠⎝ 60 s ⎠

Ans.

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6.49 A solid 3-in.-diameter steel [G = 12,000 ksi] shaft is 7-ft long. The allowable shear stress in the shaft is 8 ksi and the angle of twist must not exceed 0.03 rad. Determine the maximum horsepower tha t this shaft can deliver (a) when rotating at 150 rpm. (b) when rotating at 540 rpm.

Solution Section properties: The polar moment of inertia of the shaft is:

Ip =

π 32

D4 =

π 32

(3 in.) 4 = 7.952156 in. 4

Torque based on allowable shear stress: Compute the allowable torque if the shear stress must not exceed 8 ksi: τ I p (8 ksi)(7.952156 in.4 ) = = 42.4115 kip-in. (a) T≤ 1.5 in. R Torque based on angle of twist: Compute the allowable torque if the angle of twist must not exceed 0.03 rad: φG I p (0.03 rad)(12,000 ksi)(7.952156 in. 4 ) = = 34.0807 kip-in. (b) T≤ L (7 ft)(12 in./ft) Controlling torque: From comparison of Eqs. (a) and (b), the maximum torque allowed for this shaft is: Tmax = 34.0807 kip-in. = 2,840.06 lb-ft (a) Power transmission at 150 rpm: The maximum power that can be transmitted at 150 rpm is: ⎛ 150 rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞ P = T ω = (2,840.06 lb-ft) ⎜ ⎟⎜ ⎟⎜ ⎟ = 44,611.56 lb-ft/s ⎝ min ⎠⎝ 1 rev ⎠⎝ 60 s ⎠ or in units of horsepower, 44,611.56 lb-ft/s P= = 81.1 hp Ans. 550 lb-ft/s 1 hp (b) Power transmission at 540 rpm: The maximum power that can be transmitted at 540 rpm is: ⎛ 540 rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞ P = T ω = (2,840.06 lb-ft) ⎜ ⎟⎜ ⎟⎜ ⎟ = 160, 701.61 lb-ft/s ⎝ min ⎠⎝ 1 rev ⎠⎝ 60 s ⎠ or in units of horsepower, 160,701.61 lb-ft/s P= = 292 hp Ans. 550 lb-ft/s 1 hp

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6.50 A 3-m-long hollow steel [G = 80 GPa] shaft has an outside diameter of D = 100 mm and a wall thickness of t = 15 mm. The maximum shear stress in the shaft must be limited to 40 MPa. Determine: (a) the maximum power that can be transmitted by the shaft if the rotation speed must be limited to 5 Hz. (b) the magnitude of the angle of twist in a 5-m length of the shaft when 160 kW is being transmitted at 8 Hz.

Solution Section properties: The polar moment of inertia of the shaft is:

Ip =

π

π

⎡ D 4 − d 4 ⎤⎦ = ⎡⎣ (100 mm) 4 − (70 mm) 4 ⎤⎦ = 7, 460,300.81 mm 4 32 ⎣ 32

Torque based on allowable shear stress: Compute the allowable torque if the shear stress must not exceed 40 MPa: τ I p (40 N/mm 2 )(7,460,300.81 mm 4 ) = = 5,968, 240.65 N-mm = 5,968.24 N-m T≤ R 100 mm/2 (a) Power transmission: The maximum power that can be transmitted at 5 Hz is: 5 rev ⎞⎛ 2π rad ⎞ P = T ω = (5,968.24 N-m) ⎛⎜ ⎟⎜ ⎟ = 187, 498 N-m/s = 187.5 kW ⎝ s ⎠⎝ 1 rev ⎠

Ans.

(b) Angle of twist: The torque in the shaft when 160 kW is being transmitted at 8 Hz is: P 160,000 N-m/s = 3,183.10 N-m T= = ω ⎛ 8 rev ⎞⎛ 2π rad ⎞ ⎜ ⎟⎜ ⎟ ⎝ s ⎠⎝ 1 rev ⎠

The corresponding angle of twist is: (3,183.10 N-m)(5 m)(1,000 mm/m)2 TL φ= = = 0.0267 rad G I p (80,000 N/mm2 )(7, 460,300.81 mm 4 )

Ans.

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6.51 A tubular aluminum alloy [G = 4,000 ksi] shaft is being designed to transmit 400 hp at 1,500 rpm. The maximum shear stress in the shaft must not exceed 6 ksi and the angle of twist is not to exceed 5° in an 8-ft length. Determine the minimum permissible outside diameter if the inside diameter is to be threefourths of the outside diameter.

Solution Torque from power transmission equation: The torque in the shaft when 400 hp is being transmitted at 1,500 rpm is: ⎛ 550 lb-ft/s ⎞ (400 hp)⎜ ⎟ P ⎝ 1 hp ⎠ T= = = 1, 400.56 lb-ft ω ⎛ 1,500 rev ⎞⎛ 2π rad ⎞⎛1 min ⎞ ⎜ min ⎟⎜ 1 rev ⎟⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ Polar moment of inertia: The inside diameter of the shaft is to be three-fourths of the outside diameter; therefore, d = 0.75D. From this, the polar moment of inertia can be expressed as: π π π D4 4 π D4 ⎡⎣1 − (0.75)4 ⎤⎦ = 0.683594 = 0.067112 D4 I p = ⎡⎣ D4 − d4 ⎤⎦ = ⎡⎣ D4 − (0.75 D)4 ⎤⎦ = (a) 32 32 32 32 Diameter based on shear stress: The maximum shear stress must not exceed 6 ksi; thus, from the elastic torsion formula: Ip T TR ∴ ≥ τ≥ Ip R τ

Use the results of Eq. (a) to simplify the left-hand side of this equation, giving an expression in terms of the outside diameter D: I p 0.067112D 4 = = 0.134224 D 3 R D/2 Solve for the outside diameter D: (1, 400.56 lb-ft)(12 in./ft) 0.134224 D 3 ≥ = 2.801120 in.3 ∴ D ≥ 2.753175 in. (b) 6,000 psi Diameter based on twist angle: The angle of twist is not to exceed 5° in an 8-ft length; thus, from the angle of twist formula: TL TL ∴ Ip ≥ φ≥ φG GI p

Use the results of Eq. (a) to simplify this equation and solve for the outside diameter D: (1, 400.56 lb-ft)(8 ft)(12 in./ft) 2 = 4.622180 in.4 ∴ D ≥ 2.881972 in. 0.067112 D4 ≥ (5°)( π /180 °)(4,000,000 psi)

(c)

From the results of Eqs. (b) and (c), the minimum outside diameter D acceptable in this instance is: Dmin = 2.88 in. Ans.

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6.52 A tubular steel [G = 80 GPa] shaft is being designed to transmit 150 kW at 30 Hz. The maximum shear stress in the shaft must not exceed 80 MPa and the angle of twist is not to exceed 6° in a 4- m length. Determine the minimum permissible outside diameter if the ratio of the inside diameter to the outside diameter is 0.80.

Solution Torque from power transmission equation: The torque in the shaft when 150 kW is being transmitted at 1,500 rpm is: 150,000 N-m/s P T= = = 795.7747 N-m ω ⎛ 30 rev ⎞⎛ 2π rad ⎞ ⎜ s ⎟⎜ 1 rev ⎟ ⎝ ⎠⎝ ⎠ Polar moment of inertia: The ratio of inside diameter to outside diameter for the shaft is to be d/D = 0.80 or in other words d = 0.80D. From this, the polar moment of inertia can be expressed as: π π π D4 4 π D4 4 4 4 4 ⎡⎣1 − (0.80)4 ⎤⎦ = 0.590400 = 0.057962 D4 I p = ⎡⎣ D − d ⎤⎦ = ⎡⎣ D − (0.80 D) ⎤⎦ = (a) 32 32 32 32 Diameter based on shear stress: The maximum shear stress must not exceed 80 MPa; thus, from the elastic torsion formula: Ip T TR ∴ ≥ τ≥ Ip R τ

Use the results of Eq. (a) to simplify the left-hand side of this equation, giving an expression in terms of the outside diameter D: I p 0.057962D 4 = = 0.115925D 3 R D/2 Solve for the outside diameter D: (795.7747 N-m)(1,000 mm/m) 0.115925 D 3 ≥ = 9,947.18375 mm 3 80 N/mm 2

∴ D ≥ 44.1070 mm

(b)

Diameter based on twist angle: The angle of twist is not to exceed 6° in an 4-m length; thus, from the angle of twist formula: TL TL φ≥ ∴ Ip ≥ GI p φG

Use the results of Eq. (a) to simplify this equation and solve for the outside diameter D: 2 (795.7747 N-m)(4 m)(1,000 mm/m) 0.057962 D4 ≥ = 379,954.43 mm 4 ∴ D ≥ 50.5996 mm 2 (6° )(π /180° )(80,000 N/mm )

(c)

From the results of Eqs. (b) and (c), the minimum outside diameter D acceptable in this instance is: Dmin = 50.6 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 Act without the permission of the copyright owner is unlawful.

6.53 An automobile engine supplies 180 hp at 4,200 rpm to a driveshaft. If the allowable shear stress in the driveshaft must be limited to 5 ksi, determine: (a) the minimum diameter required for a solid driveshaft. (b) the maximum inside diameter permitted for a hollow driveshaft if the outside diameter is 2.00 in. (c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The weight of a shaft is proportional to its cross-sectional area.)

Solution Power transmission equation: The torque in the shaft is ⎛ 550 lb-ft/s ⎞ ( 180 hp) ⎜ ⎟ P 1 hp ⎠ ⎝ T= = = 225.0906 lb-ft ω ⎛ 4,200 rev ⎞⎛ 2 π rad ⎞⎛1 min ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ min ⎠⎝ 1 rev ⎠⎝ 60 s ⎠ (a) Solid driveshaft: The minimum diameter required for a solid shaft can be found from: π 3 T (225.0906 lb-ft)(12 in./ft) D ≥ = = 0.540217 in. 3 τ allow 16 5, 000 psi

∴ D ≥ 1.401241 in. = 1.401 in.

Ans.

(b) Hollow driveshaft: From the elastic torsion formula: 4 4 I p π ⎡⎣ D − d ⎤⎦ T TR ∴ = ≥ τ≥ Ip R 32 D / 2 τ

Solve this equation for the maximum inside diameter: 4 4 π ⎡⎣ (2.00 in.) − d ⎤⎦ (225.0906 lb-ft)(12 in./ft) ≥ = 0.540217 in.3 32 2.00 in./ 2 5, 000 psi (2.00 in.)4 − d 4 ≥ 5.502605 in.4 4 4 4 4 d ≤ 16 in. − 5.502605 in. = 10.497395 in.

∴ d ≤ 1.800 in.

Ans.

(c) Weight savings: The weights of the solid and hollow shafts are proportional to their respective cross-sectional areas. The cross-sectional area of the solid shaft is

Asolid =

π

(1.401 in.) 2 = 1.5416 in.2

4 and the cross-sectional area of the hollow shaft is

π

⎡ (2.00 in.) 2 − (1.80 in.) 2 ⎤⎦ = 0.5969 in.2 4⎣ The weight savings can be determined from ( A − Ahollow ) (1.5416 in. 2 − 0.5929 in. 2) weight savings (in percent) = solid = = 61.5% 2 Asolid 1.5416 in.

Ahollow =

Ans.

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6.54 The impeller shaft of a fluid agitator transmits 28 kW at 440 rpm. If the allowable shear stress in the impeller shaft must be limited to 80 MPa, determine: (a) the minimum diameter required for a solid impeller shaft. (b) the maximum inside diameter permitted for a hollow impeller shaft if the outside diameter is 40 mm. (c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The weight of a shaft is proportional to its cross-sectional area.)

Solution Power transmission equation: The torque in the shaft is 28, 000 N-m/s P T= = = 607.6825 N-m ω ⎛ 440 rev ⎞⎛ 2π rad ⎞⎛ 1 min ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ min ⎠⎝ 1 rev ⎠⎝ 60 s ⎠ (a) Solid impeller shaft: The minimum diameter required for a solid shaft can be found from: π 3 T (607.6825 N-m)(1,000 mm/m) D ≥ = = 7,596.0313 mm 3 2 τ allow 16 80 N/mm

∴ D ≥ 33.8209 mm = 33.8 in.

Ans.

(b) Hollow driveshaft: From the elastic torsion formula: I p π ⎡⎣ D 4 − d 4 ⎤⎦ T TR ∴ = ≥ τ≥ Ip R 32 D / 2 τ

Solve this equatio...