Mechanics of Materials Solutions Chapter06 Probs42 66

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:

⎛ 550 lb-ft/s ⎞ ⎟ 1 hp P ⎝ ⎠ T = = = 420.169 lb-ft ω ⎛ 3,500 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎜ min ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

( 280 hp ) ⎜

The minimum diameter required for the shaft can be found from: T (420.169 lb-ft)(12 in./ft) π 3 D ≥ 1.260507 in.3 = = 1.260507 16 4, 000 psi τ allow 858539 539 ∴ D ≥ 1.858

in 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 produced in the shaft if the outside diameter is D 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 ⎞ ⎟ 1 hp P ⎝ ⎠ = 196.954 lb-ft T = = ω ⎛ 4,000 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎜ min ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

(150 hp ) ⎜

The polar moment of inertia of the shaft is: π π ⎡⎣ D 4 − d 4 ⎤⎦ = ⎡⎣(3.000 in I p = in.)4 − (2.750 in in.)4 ⎤⎦ = 2.337403 in.4 32 32 and the maximum shear stress in the shaft is TR (196.954 lb-ft)(3.000 lb-ft)(3.000 in./2)(12 in./ft) τ = = I p 2.337403 in.4

,516.71 psi = = 1,516

1,517 ,517 psi psi

Ans.


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 = 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 = = ω ⎛ 1,700 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞

⎜ ⎝

= 1,263.877

N-m

⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎠⎝ ⎠⎝ ⎠

min

The maximum shear stress in the shaft will be determined from the elastic torsion formula: π TR ⎡⎣ D 4 − d 4 ⎤⎦ where I p = τ = I p 32 Rearrange this equation, grouping the diameter terms on the left-hand side of the equation: 4 4 π ⎡⎣ D − d ⎤⎦ T ≥

32 D / 2 τ Substitute the known values for T , τ , and D and D and and solve for the inside diameter d : 4 π ⎡⎣(75 mm)

32

4

− d

⎤⎦

75 mm / 2 (75 mm)4

≥

(1, 263.877 N-m)(1,000 N-m)(1,000 mm/m) 30 N/mm2

4

− d ≥ 16, 092,181.76

d 4

≤

mm4

(75 mm)4 − 16,092 ,092,181.76 mm4

∴ d ≤

,548,443 ,443.24 = 15,54

mm4

62.7945 mm

The minimum wall thickness for the tubular steel shaft is thus D = d + 2t ∴ t ≥

D − d 2

=

75 mm mm − 62.7945 mm mm 2

=

6.10275 mm = 6.10 mm mm

Ans


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 ⎝ ⎠ = 57.295780 N-m T = = ω ⎛ 25 rev ⎞ ⎛ 2π rad ⎞

⎜ ⎝

s

⎟ ⎜ 1 rev ⎟ ⎠⎝ ⎠

The polar moment of inertia of the shaft is: π 4 π I p = D = (20 mm mm)4 = 15,70 ,707.963 mm mm4 32 32 and the maximum shear stress in the shaft is TR (57.295780 N-m)(20 N-m)(20 mm/2)(1,000 mm/m) MPa = 36.5 MP MPa τ = = = 36.476 MP I p 15,707.963 mm4

Ans.


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 = = ω ⎛ 57 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞

= 1,842.847

N-m

⎜ min ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

The minimum diameter required for the shaft can be found from: T (1,842.847 N-m)(1,000 mm/m) π 3 3 D ≥ = = 30,714.117 mm 2 16 60 N/mm τ allow ∴ D ≥ 53.881 mm =

53.9 mm

Ans.


6.47 A solid steel [G [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: π 4 π I p = D = (3 in.) 4 = 7.952156 in.4 32 32 Angle of twist relationship: From the angle of twist relationship, compute the allowable torque: φ G I p (0.06 rad)(1 rad)(12,000 2,000 ksi)(7 ksi)(7.952156 .952156 in.4 ) T = 29.8205 0585 85 kipkip-in in.. = 2,485.05 2,485.05 lblb-ft = = 29.82 L (16 ft)(12 in./ft)

transmitted at 7 Hz is: is: Power transmission: The maximum power that can be transmitted ⎛ 7 rev ⎞ ⎛ 2π rad ⎞ P = T ω = (2, 485.05 lb-ft) ⎜ ⎟ ⎜ 1 rev ⎟ = 109, 298.21 lb-ft/s s ⎝ ⎠⎝ ⎠ or in units of horsepower, 109,298.21 lb-ft/s P = = = 198.7 hp 550 lb-ft/s

Ans.

1 hp


6.48 A tubular steel [G [G = 80 GPa] shaft with an outside diameter of D 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 shaft can transmit at 335 rpm.

Solution Section properties: The polar moment of inertia of the shaft is: π π ⎡⎣ D 4 − d 4 ⎤⎦ = ⎡⎣(90 mm)4 − (74 mm)4 ⎤⎦ = 3, 49 I p = 497, 32 321.47 mm4 32 32 Angle of twist relationship: From the angle of twist relationship, compute the allowable torque: φ G I p (0.05 (0.05 rad)( rad)(80,0 80,000 00 N/m N/mm m2 )(3,4 )(3,497,3 97,321. 21.47 47 mm mm4 ) T = ,998, 469.4 469.41 1 N-mm N-mm = 1, 998. 998.5 5 N-m N-m = = 1,998, L (7 m)(1,000 mm/m)

transmitted at 335 rpm is: is: Power transmission: The maximum power that can be transmitted ⎛ 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.


6.49 A solid 3-in.-diameter steel [G [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 that 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: π 4 π I p = D = (3 in.) 4 = 7.952156 in.4 32 32 Torque based on allowable shear stress: Compute the allowable allowable torque if the shear stress must must not exceed 8 ksi: τ I p (8 ksi)( ksi)(7.95 7.952156 2156 in. in.4 ) = = 42.4115 kip-in. T ≤ (a) R 1.5 in. Torque based on angle of twist: Compute the allowable torque if the angle of twist must not exceed 0.03 rad: φ G I p (0.03 rad)( rad)(12,000 12,000 ksi)( ksi)(7.952156 7.952156 in.4 ) T ≤ (b) = = 34.0807 kip-in. L (7 ft)(12 in./ft) Controlling torque: From comparison of Eqs. (a) and (b), the maximum torque allowed for this shaft is: T max = 34.0 34.080 807 7 kipkip-in in.. = 2,840 2,840.0 .06 6 lb-f lb-ft (a) Power transmission at 150 rpm: The maximum power that can be transmitted transmitted at 150 rpm is: ⎛ 150 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ P = T ω = (2, 840.06 lb-ft) ⎜ ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ = 44, 611.56 lb-ft/s m i n ⎝ ⎠⎝ ⎠⎝ ⎠ or in units of horsepower, 44,611.56 lb-ft/s P = = = 81.1 hp Ans. 550 lb-ft/s

1 hp transmitted at 540 rpm is: (b) Power transmission at 540 rpm: The maximum power that can be transmitted ⎛ 540 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ P = T ω = (2, 840.06 lb-ft) ⎜ ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ = 160, 701.61 lb-ft/s m i n ⎝ ⎠⎝ ⎠⎝ ⎠ or in units of horsepower, 160,701.61 lb-ft/s P = = = 292 hp Ans. 550 lb-ft/s 1 hp


6.50 A 3-m-long hollow steel [G [G = 80 GPa] shaft has an outside diameter of D 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: π π ⎡⎣ D 4 − d 4 ⎤⎦ = ⎡⎣(100 mm)4 − (70 mm)4 ⎤⎦ = 7, 46 I p = 460, 30 300.81 mm4 32 32 Torque based on allowable shear stress: Compute the allowable allowable torque if the shear stress must must not exceed 40 MPa: τ I p (40 (40 N/mm N/mm 2 )(7, )(7,46 460, 0,30 300. 0.81 81 mm mm4 ) ,968, 240. 240.65 65 N-m N-mm = 5,968.2 ,968.24 4 NN-m = = 5,968, T ≤ R 100 mm/2

transmitted at 5 Hz is: (a) Power transmission: The maximum power that can be transmitted ⎛ 5 rev ⎞ ⎛ 2π rad ⎞ P = T ω = (5,968. ,968.24 24 N-m N-m) ⎜ 187,498 8 N-m N-m/s = 187. 187.5 5 kW ⎟ ⎜ 1 rev ⎟ = 187,49 s ⎝ ⎠⎝ ⎠

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 N-m/s = 3,183.10 N-m T = = ω ⎛ 8 rev ⎞ ⎛ 2π rad ⎞

⎜ s ⎟ ⎜ 1 rev ⎟ ⎝ ⎠⎝ ⎠

The corresponding angle of twist is: TL (3,183.10 N-m)(5 m)(1,000 mm/m)2 φ = = = = 0.0267 rad G I p (80,000 80,000 N/mm N/mm2 )(7,460 )(7,460,300 ,300.8 .81 1 mm mm4 )

Ans.


6.51 A tubular aluminum alloy [G [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 shaft when 400 hp is being transmitted Torque from power transmission equation: The torque in the shaft at 1,500 rpm is:

⎛ 550 lb-ft/s ⎞ ⎟ 1 hp P ⎝ ⎠ T = = = 1, 400.56 lb-ft ω ⎛ 1,500 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎜ min ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ (400 hp) ⎜

outside Polar moment of inertia: The inside diameter of the shaft is to be three-fourths of the outside diameter; therefore, d = = 0.75 D. D. From this, the polar moment moment of inertia can be expressed as: 4 4 π π π D πD 4 4 4 4 4 4 4 ⎡⎣ D − d ⎤⎦ = ⎡⎣ D − (0.75D) ⎤⎦ = ⎡⎣1 − (0.75) ⎤⎦ = 0.683594 I p = (a) = 0.067112D 32 32 32 32 must not exceed 6 ksi; thus, from the Diameter based on shear stress: The maximum shear stress must elastic torsion formula: I p T TR τ ≥ ∴ ≥ I p 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 diameter D:: I p 0.067112 D 4 3 = = 0.134224 D R D/2 Solve for the outside diameter D diameter D:: (1, 400.56 lb-ft)(12 lb-ft)(12 in./ft) 3 0.134224 D 3 ≥ (b) = 2.801120 in. ∴ D ≥ 2.753175 in. 6,000 psi is not to exceed 5° in an 8-ft length; thus, from from the Diameter based on twist angle: The angle of twist is angle of twist formula: TL TL φ ≥ ∴ I p ≥ G I p φ G Use the results of Eq. (a) to simplify this equation an d solve for the outside diameter D diameter D:: 2 (1, 400.56 lb-ft)(8 lb-ft)(8 ft)(12 in./ft) 4 0.067112 D 4 ≥ = 4.622180 in. ∴ D ≥ 2.881972 in. (5°)(π /180 180°)(4,000,0 0,000 psi psi)

(c)

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

=

2.88 in.

Ans.


6.52 A tubular steel [G [G = 80 GPa] shaft is being designed to transmit 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: P 150,000 N-m/s N-m/s T = = = 795.7747 N-m ω ⎛ 30 rev ⎞ ⎛ 2π rad ⎞

⎜ ⎝

s

⎟ ⎜ 1 rev ⎟ ⎠⎝ ⎠

diameter to outside diameter for the shaft is to be d/D = d/D = Polar moment of inertia: The ratio of inside diameter 0.80 or in other words d = = 0.80 D. D. From this, the polar moment moment of inertia can be expressed as: 4 4 π π π D πD 4 4 4 4 4 4 4 ⎡⎣ D − d ⎤⎦ = ⎡⎣ D − (0.80 D ) ⎤⎦ = ⎡⎣1 − (0.80) ⎤⎦ = 0.590400 I p = (a) = 0.057962D 32 32 32 32 Diameter based on shear stress: The maximum shear stress stress must not exceed 80 MPa; thus, thus, from the elastic torsion formula: I p T TR τ ≥ ∴ ≥ I p 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 diameter D:: I p 0.057962 D 4 3 = = 0.115925 D R D/2 Solve for the outside diameter D diameter D:: (795.7747 N-m)(1,000 N-m)(1,000 mm/m) 3 0.115925 D 3 ≥ = 9, 947.18375 mm 2 80 N/mm

∴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 ≥ ∴ I p ≥ G I p φ G

Use the results of Eq. (a) to simplify this equation and solve for the outside diameter D diameter D:: 2 (795.7747 N-m)(4 N-m)(4 m)(1,000 mm/m) 4 0.057962 D 4 ≥ = 379, 954.43 mm ∴ 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 diameter D acceptable acceptable in this instance is: Dmin

= 50.6

mm

Ans.


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 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 ⎞ ⎟ 1 hp P ⎝ ⎠ T = = = 225.0906 lb-ft ω ⎛ 4,200 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎜ min ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

(180 hp ) ⎜

for a solid shaft can be found from: (a) Solid driveshaft: The minimum diameter required for T (225.0906 lb-ft)(12 in./ft) π 3 D ≥ 0.540217 in.3 = = 0.540217 16 5, 000 psi τ allow 1.4012 1241 41 in. = ∴ D ≥ 1.40

1.40 1.401 1 in.

Ans.

(b) Hollow driveshaft: From the elastic torsion formula: τ ≥

TR I p

I p

∴

R

=

4 π ⎡⎣ D

32

4

− d

D/2

⎤⎦

≥

T τ

Solve this equation for the maximum inside diameter: 4 π ⎡⎣(2.00 in.)

32

4

− d

⎤⎦

2.00 in. / 2 (2.00 in.)4 − d 4 d 4

≥ ≥

(225.0906 lb-ft)(12 in./ft) 5, 000 psi

0.540217 in.3

5.502605 in.4

≤ 16 ∴

=

in in.4

− 5.502605

d ≤ 1.800 in.

in in.4

= 10.497395

in in.4 Ans.

shafts are proportional proportional to their respective (c) Weight savings: The weights of the solid and hollow shafts cross-sectional areas. The cross-sectional area of of the solid shaft is π Asolid = (1.40 .401 in.)2 = 1.5 1.5416 in in.2 4 and the cross-sectional area of the hollow shaft is π Ahollow = ⎡⎣(2.00 in.)2 − (1.80 in.)2 ⎤⎦ = 0.5969 in.2 4 The weight savings can be determined from ( Asolid − Ahollow ) (1.5416 in in.2 − 0.5929 in in.2 ) weight savings (in percent) = = = 61.5% Ans. Asolid 1.5416 in.2


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 ho llow 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 Hint : The weight of a shaft is proportional to its cross-sectional area.)

Solution Power transmission equation: The torque in the shaft is P 28,000 N-m/s N-m/s T = = = 607.6825 N-m ω ⎛ 440 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞

⎜ min ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

for a solid shaft can be found found from: (a) Solid impeller shaft: The minimum diameter required for T (607.6825 N-m)(1,000 N-m)(1,000 mm/m) π 3 D ≥ 596.0313 mm3 = = 7, 596.0313 2 16 80 N/mm τ allow ∴ D ≥ 33.8209

mm mm = 33.8 in in.

Ans.

(b) Hollow driveshaft: From the elastic torsion formula: τ ≥

TR I p

I p

∴

R

=

4 π ⎡⎣ D

32

4

− d

⎤⎦

D/2

≥

T τ

Solve this equation for the maximum inside diameter: 4 π ⎡⎣(40 mm)

32

4

− d

⎤⎦

40 mm / 2 (40 mm)

4

≥

(607.6825 N-m)(1,000 N-m)(1,000 mm/m) 2

80 N/mm

4

450.779 − d ≥ 1, 547, 45 d 4

≤ ∴

7, 596.0313 596.0313 mm3

mm4

2,56 ,560,000 ,000 mm mm4 d ≤ 31.7 mm

=

− 1,54 ,547,450 ,450.779

mm mm4

=

1,012 ,012,54 ,549.221 mm4 Ans.

shafts are proportional proportional to their respective (c) Weight savings: The weights of the solid and hollow shafts cross-sectional areas. The cross-sectional area of of the solid shaft is π Asolid = (33.8 mm)2 = 897.270 mm mm2 4 and the cross-sectional area of the hollow shaft is π Ahollow = ⎡⎣(40 mm)2 − (31.7 mm)2 ⎤⎦ = 467.398 mm2 4 The weight savings can be determined from ( Asolid − Ahollow ) (897.270 mm2 − 467.398 mm2 ) weight savings (in percent) = = = 4 7 .9 % Ans. Asolid 897.270 mm2


6.55 A motor supplies 200 kW at 6 Hz to flange A flange A of of the shaft shown in Fig. P6.55. Gear B Gear B transfers 125 kW of power to operating machinery in the factory, and the remaining power in the shaft is transferred by gear D gear D.. Shafts (1) and (2) are solid aluminum [G [G = 28 GPa] shafts that have the same diameter and an allowable shear stress of τ = = 40 MPa. Shaft (3) is a solid steel [G [ G = 80 GPa] shaft with an allowable shear stress of τ = = 55 MPa. Determine: (a) the minimum permissible diameter for aluminum shafts (1) and (2). (b) the minimum permissible diameter for steel shaft (3). (c) the rotation angle of gear D gear D with with respect to flange A A if the shafts have the minimum permissible diameters as determined in (a) and (b). Fig. P6.55

Solution flange A is is Power transmission: The torque on flange A P 200,000 N-m/s N-m/s T A = = = 5,305.165 N-m ω ⎛ 6 rev ⎞ ⎛ 2π rad ⎞

⎜ s ⎟ ⎜ 1 rev ⎟ ⎝ ⎠⎝ ⎠

thus, the torque in shaft segment (1) is T 1 = 5,305.165 N-m. The torque transferred transferred by gear B gear B is is P 125,000 N-m/s N-m/s = 3,315.728 N-m T B = = ω ⎛ 6 rev ⎞ ⎛ 2π rad ⎞

⎜ ⎟⎜ ⎟ ⎝ s ⎠ ⎝ 1 rev ⎠

therefore, the remaining torque in segments (2) and (3) of the shaft is T2 = T A − T B = 5,305. ,305.16 165 5 NN-m − 3,315. ,315.72 728 8 NN-m = 1,989. ,989.43 437 7 N-m N-m (a) Minimum diameter for solid aluminum shaft: The minimum diameter required for the solid solid aluminum shaft can be found from: T 1 (5,305.165 (5,305.16 5 N-m)(1,000 mm/m) mm/m) π 3 3 D1 ≥ = = 132,629.125 mm 2 16 40 N/mm τ 1,allow ∴ D1 ≥ 87.7

mm

Ans.

(b) Minimum diameter for solid steel shaft: The minimum diameter diameter required for the solid steel steel shaft can be found from: T 3 (1,989.437 N-m)(1,000 mm/m) π 3 3 = = 36,171.582 mm D3 ≥ 2 16 55 N/mm τ 3,allow ∴ D3 ≥ 56.9

mm

Ans.

(c) Angles of twist: The polar moments of inertia of the aluminum and steel shafts are: π I p1 = I p 2 = (87.7 mm mm)4 = 5,80 ,807,62 ,621 mm4 32 π I p 3 = (56.9 mm)4 = 1,029 ,029,080 ,080 mm4 32 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.

The angles of twist in the three shaft segments are: T1 L1 (5,305.165 N-m)(0.8 N-m)(0.8 m)(1,000 mm/m)2 φ 1 = = G1 I p1 (28, (28,00 000 0 N/m N/mm2 )(5 )(5,807,621 ,807,621 mm mm4 ) φ 2 φ 3

=

=

T2 L2 G2 I p 2 T3 L3 G3 I p 3

=

=

(1,989.437 N-m)(0.4 m)(1,000 mm/m)2 (28, (28,00 000 0 N/mm N/mm2 )(5,807 (5,807,621 ,621 mm mm4 ) (1,989.437 N-m)(0.8 N-m)(0.8 m)(1,000 mm/m)2 (80, (80,00 000 0 N/mm N/mm2 )(1 )(1, 029, 029, 080 mm4 )

=

0.026100 rad

=

0.004894 rad

=

0.019332 rad

at D is is equal to the sum of the angles of twist in the three shaft Rotation angle at D: The rotation angle at D segments: φ D = φ1 + φ2 + φ3 =

0.02 0.0261 6100 00 rad + 0.00 0.0048 4894 94 rad + 0.01 0.0193 9332 32 rad

=

0.0 0.050326 ra rad = 0.05 .0503 rad

Ans.


6.56 A motor supplies 150 hp at 520 rpm to flange A flange A of of the shaft shown in Fig. P6.56. Gear B Gear B transfers 90 hp of power to operating machinery machinery in the factory, and the remaining power in the shaft shaft is transferred by gear D gear D.. Shafts (1) and (2) are solid aluminum [G [ G = 4,000 ksi] shafts that have the same diameter and an allowable shear stress of τ = = 6 ksi. Shaft (3) is a solid steel [G [ G = 12,000 ksi] shaft with an allowable shear stress of τ = = 8 ksi. Determine: (a) the minimum permissible diameter for aluminum shafts (1) and (2). (b) the minimum permissible diameter for steel shaft (3). (c) the rotation angle of gear D gear D with with respect to flange A A if the shafts have the minimum permissible diameters as determined in (a) and (b). Fig. P6.56

Solution flange A is is Power transmission: The torque on flange A

⎛ 550 lb-ft/s ⎞ ⎟ 1 hp P ⎝ ⎠ T A = = = 1, 515.033 lb-ft ω ⎛ 520 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎜ min ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

(150 hp ) ⎜

thus, the torque in shaft segment (1) is T 1 = 1,515.033 lb-ft. lb-ft. The torque transferred transferred by gear B gear B is is

⎛ 550 lb-ft/s ⎞ ⎟ 1 hp P ⎝ ⎠ T B = = ω ⎛ 520 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ min ⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠

( 90 hp ) ⎜

= 909.020

lb-ft

therefore, the remaining torque in segments (2) and (3) of the shaft is T2 = T A − T B = 1,515.0 ,515.033 33 lb-ft b-ft − 909. 909.02 020 0 lb-ft b-ft = 606. 606.01 013 3 lblb-ft (a) Minimum diameter for solid aluminum shaft: The minimum diameter required for the solid solid aluminum shaft can be found from: T 1 (1,515.033 lb-ft)(12 in./ft) π 3 3 D1 ≥ = = 3.030066 in. 16 6, 000 psi τ 1,allow ∴ D1 ≥

2.49 in.

Ans.

diameter required for the solid steel steel shaft (b) Minimum diameter for solid steel shaft: The minimum diameter can be found from: T 3 (606.013 lb-ft)(12 in./ft) π 3 = = 0.909020 D3 ≥ 0.909020 in.3 16 8, 000 psi τ 3,allow ∴ D3 ≥ 1.667

in.

Ans.


(c) Angles of twist: The polar moments of inertia of the aluminum and steel shafts are: π I p1 = I p 2 = (2.49 in.)4 = 3.773960 in.4 32 π I p 3 = (1.667 in.) n.)4 = 0.758 758128 in.4 32

The angles of twist in the three shaft segments are: T1 L1 (1,515.033 lb-ft)(2 ft)(12 in./ft) 2 φ 1 = = = 0.028904 rad G1 I p1 (4,000, (4,000,000 000 psi)(3. psi)(3.7739 773960 60 in.4 ) φ 2 φ 3

=

=

T2 L2 G2 I p 2 T3 L3 G3 I p 3

=

=

(606.013 lb-ft)(1 ft)(12 in./ft) 2 (4,000, (4,000,000 000 psi)(3. psi)(3.7739 773960 60 in.4 )

=

(606.013 lb-ft)(2 ft)(12 in./ft) 2 (12,000,000 (12,000,000 psi)(0.758128 psi)(0.758128 in.4 )

0.005781 rad

=

0.019185 rad

at D is is equal to the sum of the angles of twist in the three shaft Rotation angle at D: The rotation angle at D segments: φ D = φ1 + φ2 + φ3 =

0.02 0.0289 8904 04 rad + 0.00 0.0057 5781 81 rad + 0.01 0.0191 9185 85 rad

=

0.0 0.053870 870 ra rad = 0.0539 rad

Ans.


6.57 A motor supplies sufficient power to the system shown in Fig. P6.57 so that gears C and C and D D provide torques of T C C = 700 N-m and T D = 450 N-m, respectively, to machinery in a factory. Power shaft segments (1) and (2) are hollow steel tubes with an outside diameter of D of D = 60 mm and an inside diameter of d = = 50 mm. If the power shaft [i.e., segments (1) and (2)] rotates at 80 rpm, determine: (a) the maximum shear stress in power shaft segments (1) and (2). (b) the power (in kW) that must be provided by the motor as well as the rotation speed (in rpm). (c) the torque applied to gear A gear A by by the motor.

Fig. P6.57

Solution Equilibrium: Σ M x = TD − T 2

=

0

∴ T2 = T D =

450 N-m

Σ M x = TC + TD − T 1 =

0

∴T1 = TC + T D = 700

N-m + 450 N-m

= 1,150

N-m

Section properties: π ⎡⎣(60 mm)4 − (50 mm)4 ⎤⎦ = 658, 75 I p1 = 752.71 mm4 32 (a) Shear stresses: TR (1,150 N-m)(60 mm/2)(1,000 mm/m) τ 1 = 1 1 = I p1 658,752.71 mm4 τ 2

=

T2 R2 I p 2

=

(450 N-m)(60 N-m)(60 mm/2)(1,000 mm/m) 658,752.71 mm4

I p 2

52.372 MP MPa = 52.4 MP MPa

Ans.

20.493 MPa = 20.5 MPa

Ans.

=

=

=

(b) Power transmission: Compute the power in in shaft segment (1): ⎛ 80 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ P = T 1ω = (1,150 N-m) ⎜ ⎟⎜ ⎟⎜ ⎟ = 9, 634.217 N-m/s = 9.63 kW m i n 1 r e v 6 0 s ⎝ ⎠⎝ ⎠⎝ ⎠

This power in the shaft must must be provided by the motor. Therefore: P motor = 9.63 kW

Ans.


The rotation speed of the motor is found with the use of the gear ratio: 72 teeth ω A = ωB = 3ωB = 3(80 rpm) = 240 rpm = ωmotor 24 teeth (c) Motor torque: The motor must provide a torque of N 24 teeth T A = T B A = (1,150 N-m) = 383 N-m N B 72 teeth

Ans.

Ans.


6.58 A motor supplies sufficient power to the system shown in Fig. P6.58 so that gears C and C and D provide D provide torques of T C C = 125 lb-ft and T D = 90 lb-ft, respectively, to machinery in a factory. Power shaft segments (1) and (2) are hollow steel tubes with an outside diameter of D of D = 1.75 in. and an inside diameter of d = = 1.50 in. If the power shaft [i.e., segments (1) and (2)] rotates at 160 rpm, determine: (a) the maximum shear stress in power shaft segments (1) and (2). (b) the power (in hp) that must be provided by the motor as well as the rotation speed (in rpm). (c) the torque applied to gear A gear A by by the motor.

Fig. P6.58

Solution Equilibrium: Σ M x = TD − T 2

=

0

∴ T2 = T D = 90

Σ M x = TC + TD − T 1 =

lb-ft

0

∴T1 = TC + T D

125 = 125 =

lb-f b-ft + 90 lb-f b-ft

215 lb-ft

Section properties: π ⎡⎣(1.75 in.)4 I p1 = 32

− (1.50

in in.)4 ⎤⎦ = 0.423762 in.4

(a) Shear stresses: TR (215 lb-ft)(1.75 lb-ft)(1.75 in./2)(12 in./ft) τ 1 = 1 1 = I p1 0.423762 in.4 τ 1

=

T1 R1 I p1

=

(90 lb-ft)(1.75 in./2)(12 in./ft) 0.423762 in.4

=

I p 2

= 5,32 ,327.28

=

ps psi = 5,33 ,330 ps psi

2,230 ,230.03 psi = 2,230 ,230 psi

Ans. Ans.

(b) Power transmission: Compute the power in in shaft segment (1): ⎛ 160 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ P = T 1ω = (215 lb-ft) ⎜ ⎟⎜ ⎟⎜ ⎟ = 3, 602.36 lb-ft/s = 6.55 hp m i n 1 r e v 6 0 s ⎝ ⎠⎝ ⎠⎝ ⎠

This power in the shaft must must be provided by the motor. Therefore: P motor = 6.55 hp

Ans.


The rotation speed of the motor is found with the use of the gear ratio: 72 teeth ω A = ωB = 3ωB = 3(160 rpm) = 480 rpm = ωmotor 24 teeth (c) Motor torque: The motor must provide a torque of N 24 teeth T A = T B A = (215 lb-ft) = 71.7 lb-ft = 860 lb-in. N B 72 teeth

Ans.

Ans.


6.59 A motor supplies 12 hp at 4 Hz to gear A gear A of the drive system shown in Fig. P6.59. Shaft (1) is a solid 2.25-in.-diameter aluminum [G [G = 4,000 ksi] shaft with a length of L1 = 16 in. Shaft (2) is a solid 1.5-in.-diameter steel [G [G = 12,000 ksi] shaft with a length of L2 = 12 in. Shafts (1) and (2) are connected at flange C , and the bearings shown permit free rotation of the shaft. Determine: (a) the maximum shear stress in shafts (1) and (2). (b) the rotation angle of gear D gear D with with respect to gear B gear B..

Fig. P6.59

Solution gear A is: is: Power transmission: The torque acting on gear A

T A

=

P ω

⎛ 550 lb-ft/s ⎞ ⎟ ⎝ 1 hp ⎠ ⎛ 4 rev ⎞ ⎛ 2π rad ⎞ ⎜ s ⎟ ⎜ 1 rev ⎟ ⎝ ⎠⎝ ⎠

(12 hp ) ⎜ =

=

262.606 lb-ft

Equilibrium: The torque applied to gear B gear B is: is: N 60 teeth T B = T A B = (262.606 lb-ft) = 656.515 lb-ft N A 24 teeth

The torque on shaft segments (1) and an d (2) as well as gear D gear D is is equal to the torque applied to gear B gear B.. T1 = T2 = T D = TB = 656.515 lb-ft inertia for aluminum shaft shaft (1) and steel shaft (2) (2) are: Section properties: The polar moments of inertia π 4 π π 4 π I p1 = D1 = (2.25 in.)4 = 2.516112 in.4 I p 2 = D2 = (1.5 in in.)4 = 0.497010 in in.4 32 32 32 32 (a) Maximum shear stresses: TR (656.515 lb-ft)(2.25 in./2)(12 in./ft) τ 1 = 1 1 = I p1 2.516112 in.4 τ 2

=

T2 R2 I p 2

=

(656.515 lb-ft)(1.5 lb-ft)(1.5 in./2)(12 in./ft) 0.497010 in.4

,522.48 = 3,52


Ans.

11,888. 8.36 36 = 11,88

psi psi = 11,89 11,890 0 ps psi

Ans.

(b) Angles of twist: The angles of twist in the two shaft segments are: TL (656.515 lb-ft)(16 lb-ft)(16 in.)(12 in./ft) φ 1 = 1 1 = = 0.012524 rad G1 I p1 (4,000, (4,000,000 000 psi)(2. psi)(2.5161 516112 12 in.4 ) φ 2

=

T2 L2 G2 I p 2

=

(656.515 lb-ft)(12 lb-ft)(12 in.)(12 in./ft) (12,000,000 (12,000,000 psi)(0.497010 psi)(0.497010 in.4 )

=

0.015851 rad

Rotation angle of D relative to B: φ D − φ B

= φ1 + φ2 =

0.012524 ra rad + 0.015851 rad = 0.028375 ra rad = 0.0284 ra rad

Ans.


6.60 A motor supplies 13 kW at 400 rpm to gear A gear A of of the drive system shown in Fig. P6.60. Shaft (1) is a solid 50-mm-diameter aluminum [G = 28 GPa] shaft with a length of L1 = 600 mm. Shaft (2) is a solid 40-mm-diameter steel [G = 80 GPa] shaft with a length of L2 = 400 mm. Shafts (1) and (2) are connected at flange C , and the bearings shown permit free rotation of the shaft. Determine: (a) the maximum shear stress in shafts (1) and (2). (b) the rotation angle of gear D gear D with with respect to gear B gear B..

Fig. P6.60

Solution gear A is: is: Power transmission: The torque acting on gear A P 13,000 N-m/s N-m/s T A = = = 310.35 N-m ω ⎛ 400 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞

⎜ min ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

gear B is: is: Equilibrium: The torque applied to gear B N 60 teeth T B = T A B = (310.35 N-m) = 775.88 N-m N A 24 teeth The torque on shaft segments (1) and an d (2) as well as gear D gear D is is equal to the torque applied to gear B gear B.. T1 = T2 = T D = TB = 775.88 N-m inertia for aluminum shaft shaft (1) and steel shaft (2) (2) are: Section properties: The polar moments of inertia π 4 π π 4 π I p1 = D1 = (50 mm mm)4 = 613,59 ,592.32 mm mm4 I p 2 = D2 = (40 (40 mm)4 = 251,32 ,327.41 mm4 32 32 32 32 (a) Maximum shear stresses: TR (775.88 N-m)(50 N-m)(50 mm/2)(1,000 mm/m) τ 1 = 1 1 = I p1 613,592.32 mm4 τ 2

=

T2 R2 I p 2

=

= 31.612

(775.88 N-m)(40 N-m)(40 mm/2)(1,000 mm/m) 251,327.41 mm4

=

MP MPa = 31.6 MP MPa

61.743 MPa = 61.7 MP MPa

Ans. Ans.

(b) Angles of twist: The angles of twist in the two shaft segments are: TL (775.88 N-m)(600 N-m)(600 mm)(1,000 mm/m) φ 1 = 1 1 = = 0.027096 rad G1 I p1 (28, (28,000 000 N/mm N/mm2 )(613 )(613,592. ,592.32 32 mm4 ) φ 2

=

T2 L2 G2 I p 2

=

(775.88 N-m)(400 N-m)(400 mm)(1,000 mm/m) (80, (80,000 000 N/mm N/mm2 )(251 )(251,327. ,327.41 41 mm4 )

=

0.015436 rad

Rotation angle of D relative to B: φ D − φ B

= φ1 + φ2 =

0.027096 ra rad + 0.015436 ra rad = 0.042532 ra rad = 0.0425 rad

Ans.


6.61 The motor shown in Fig. P6.61 supplies 60 hp at 700 rpm at A. A. The bearings shown permit free rotation of the shafts. (a) Shaft (1) is a solid 1.75-in.-diameter steel shaft. Determine the maximum shear stress produced in shaft (1). (b) If the shear stress in shaft (2) must be limited to 6,000 psi, determine the minimum acceptable diameter for shaft (2) if a solid shaft is used.

Fig. P6.61

Solution Power transmission: The torque acting in shaft (1) and on gear B gear B is: is:

⎛ 550 lb-ft/s ⎞ ⎟ 1 hp P ⎝ ⎠ T1 = T B = = = 450.181 lb-ft ω ⎛ 700 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎜ min ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

( 60 hp ) ⎜

for shaft (1) (1) is: Section properties: The polar moment of inertia for π 4 π I p1 = D1 = (1.75 in in.)4 = 0.920772 in in.4 32 32 (a) Maximum shear stress in shaft (1): TR (450.181 lb-ft)(1.75 in./2)(12 in./ft) τ 1 = 1 1 = I p1 0.920772 in.4

,133.63 = 5,13


Ans.

is: (b) Equilibrium: The torque applied to gear C is: N 30 teeth TC = T B C = (450.181 lb-ft) = 281.363 lb-ft N B 48 teeth The torque on shaft segment (2) as well as gear D gear D is is equal to the torque applied to gear C . T2 = T D = T C = 281.363 lb-ft The minimum diameter required for solid shaft (2) can be found foun d from: T 2 (281.363 lb-ft)(12 lb-ft)(12 in./ft) π 3 D2 ≥ 0.562726 in.3 = = 0.562726 16 6, 000 psi τ 2,allow 1.420438 438 ∴ D2 ≥ 1.4

in in. = 1.4 1.420 in i n.

Ans.


6.62 The motor shown in Fig. P6.62 supplies 25 kW at 6 Hz at A. A. The bearings shown permit free rotation of the shafts. (a) Shaft (2) is a solid 30-mm-diameter steel shaft. Determine the maximum shear stress produced in shaft (2). (b) If the shear stress in shaft (1) must be limited to 40 MPa, determine the minimum acceptable diameter for shaft (1) if a solid shaft is used.

Fig. P6.62

Solution Power transmission: The torque acting in shaft (1) and on gear B gear B is: is: P 25,000 N-m/s N-m/s = 663.146 N-m T1 = T B = = ω ⎛ 6 rev ⎞ ⎛ 2π rad ⎞

⎜ s ⎟ ⎜ 1 rev ⎟ ⎝ ⎠⎝ ⎠

The torque applied to gear C is: is: N 30 teeth TC = T B C = (663.146 N-m) N B 48 teeth

=

414.466 N-m

The torque on shaft segment (2) as well as gear D gear D is is equal to the torque applied to gear C . T2 = T D = T C = 414.466 N-m for shaft (2) (2) is: Section properties: The polar moment of inertia for π 4 π I p 2 = D2 = (30 mm mm)4 = 79, 52 521.56 mm mm4 32 32 (a) Maximum shear stress in shaft (2): TR (414.466 N-m)(30 N-m)(30 mm/2)(1,000 mm/m) τ 2 = 2 2 = = 78.180 MPa = 78.2 MPa I p 2 79,521.56 mm4

Ans.

(b) Minimum diameter for solid shaft (1): T 1 π 3 (663.146 N-m)(1,000 N-m)(1,000 mm/m) 3 D1 ≥ = = 16,578.65 mm 2 τ 1,allow 16 40 N/mm ∴ D1 ≥

43.8706 mm = 43.9 mm

Ans.


6.63 The motor shown in Fig. P6.63 supplies 4 kW at 2 Hz at A at A.. Shafts (1) and (2) are each solid 30-mm-diameter steel [G [G = 80 GPa] shafts with lengths of L1 = 900 mm and L2 = 1,200 mm, respectively. The bearings shown permit free rotation of the shafts. Determine: (a) the maximum shear stress produced in shafts (1) and (2). (b) the rotation angle of gear D D with respect to flange A flange A..

Fig. P6.63

Solution in shaft (1) and on gear B gear B is: is: Power transmission: The magnitude of the torque acting in P 4,000 4, 000 N-m/s N-m/s T1 = T B = = = 318.310 N-m ω ⎛ 2 rev ⎞ ⎛ 2π rad ⎞

⎜ s ⎟ ⎜ 1 rev ⎟ ⎝ ⎠⎝ ⎠

The magnitude of the torque applied to gear C is: is: N 54 teeth = 477.465 N-m TC = T B C = (318.310 N-m) N B 36 teeth The torque magnitude in shaft segment (2) as well as the magnitude of the torque acting on gear D is equal to the torque applied to gear C . T2 = T D = T C = 477.465 N-m To determine the proper signs for T 1 and T 2, consider equilibrium. First, consider a FBD that cuts through shaft (2) and includes gear D gear D.. Σ M x = TD − T2 = 0 ∴ T2 = T D = +477.465 N-m Since T 2 is positive, the torque in shaft (2) must be negative; therefore, T 1 = −318.310 N-m inertia for shafts (1) and (2) are equal: Section properties: The polar moments of inertia π 4 π I p1 = D1 = (30 mm mm)4 = 79, 52 521.56 mm mm4 = I p 2 32 32 (a) Maximum shear stress magnitudes: TR (318.310 N-m)(30 N-m)(30 mm/2)(1,000 mm/m) MPa = 60.0 MPa τ 1 = 1 1 = = 60.042 MP I p1 79,521.56 mm4 τ 2

=

T2 R2 I p 2

=

(477.465 N-m)(30 N-m)(30 mm/2)(1,000 mm/m) 79,521.56 mm4

= 90.063

MP MPa = 90.1 MP MPa

Ans. Ans.


Angles of twist in shafts (1) and (2): TL (−318.310 N-m)(9 N-m)(900 00 mm)( mm)(1,000 1,000 mm/m) mm/m) φ 1 = 1 1 = = −0.045032 rad G1 I p1 (80, (80,000 000 N/mm N/mm2 )(79 )(79,5 ,521. 21.56 56 mm4 ) φ 2

=

T2 L2 G2 I p 2

=

(477.465 N-m)(1,200 N-m)(1,200 mm)(1,000 mm/m) (80,0 (80,000 00 N/mm N/mm2 )(79, )(79,52 521. 1.56 56 mm4 )

=

0.090063 rad

gear B is is found from the angle of twist in shaft (1): Rotation angles of gears B and C: The rotation of gear B φ1 = φ B − φA ∴ φB = φA + φ1 = 0 rad + (−0.045032 rad) = − 0.045032 rad As gear B gear B rotates, rotates, gear C also also rotates but it rotates in the opposite direction. direction. The magnitude of the rotation is dictated by the gear ge ar ratio: N 36 teeth φC = −φ B B = −( −0.045032 rad) = 0.030021 rad N C 54 teeth (b) Rotation angle of gear D: The rotation of gear D gear D with with respect to flange A flange A is is found from: φ2

= φ D − φC

∴ φD = φC + φ2 =

0.030021 rad + 0.090063 ra rad = 0.120084 ra rad = 0.1201 rad

Ans.


6.64 The motor shown in Fig. P6.64 supplies 50 hp at 400 rpm at A. A. Shafts (1) and (2) are each solid 2-in.-diameter steel [G [G = 12,000 ksi] shafts with lengths of L1 = 100 in. and L2 = 120 in., respectively. The bearings shown permit free rotation of the shafts. (a) Determine the maximum shear stress produced in shafts (1) and (2). (b) Determine the rotation angle of gear D gear D with respect to flange A flange A..

Fig. P6.64

Solution Power transmission: The magnitude of the torque acting in in shaft (1) and on gear B gear B is: is:

⎛ 550 lb-ft/s ⎞ ⎟ 1 hp P ⎝ ⎠ T1 = T B = = ω ⎛ 400 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ ⎜ min ⎟ ⎜ 1 rev ⎟ ⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

( 50 hp ) ⎜

= 656.514

lb-ft

The magnitude of the torque applied to gear C is: is: N 54 teeth TC = T B C = (656.514 lb-ft) = 984.771 lb-ft N B 36 teeth The torque magnitude in shaft segment (2) as well as the magnitude of the torque acting on gear D is equal to the torque applied to gear C . T2 = T D = T C = 984.771 lb-ft To determine the proper signs for T 1 and T 2, consider equilibrium. First, consider a FBD that cuts through shaft (2) and includes gear D gear D.. Σ M x = TD − T2 = 0 ∴ T2 = T D = +984.771 lb-ft Since T 2 is positive, the torque in shaft (2) must be negative; therefore, T 1 = −656.514 lb-ft inertia for shafts (1) and (2) are equal: Section properties: The polar moments of inertia π 4 π I p1 = D1 = (2 in in.)4 = 1.570796 in in.4 = I p 2 32 32 (a) Maximum shear stress magnitudes: TR (656.514 lb-ft)(2 lb-ft)(2 in./2)(12 in./ft) τ 1 = 1 1 = I p1 1.570796 in.4 τ 2

=

T2 R2 I p 2

=

(984.771 lb-ft)(2 lb-ft)(2 in./2)(12 in./ 2)(12 in./ft) 1.570796 in.4

,015.40 = 5,015 =

ps psi = 5,020 ps psi

7,52 ,523.10 ps psi = 7,520 ps psi

Ans. Ans.


Angles of twist in shafts (1) and (2): TL (−656.514 lb-ft) lb-ft)(100 (100 in.)(12 in.)(12 in./ft in./ft)) φ 1 = 1 1 = G1 I p1 (12,000,000 (12,000,000 psi)(1.57 psi)(1.570796 0796 in.4 ) φ 2

=

T2 L2 G2 I p 2

=

(984.771 lb-ft)(120 lb-ft)(120 in.)(12 in./ft) (12,000,000 (12,000,000 psi)(1.57079 psi)(1.570796 6 in. in.4 )

= −0.041795

rad

= 0.075231 rad

gear B is is found from the angle of twist in shaft (1): Rotation angles of gears B and C: The rotation of gear B φ1 = φ B − φA ∴ φB = φA + φ1 = 0 rad + (−0.041795 rad) = − 0.041795 rad As gear B gear B rotates, rotates, gear C also also rotates but it rotates in the opposite direction. direction. The magnitude of the rotation is dictated by the gear ge ar ratio: N 36 teeth φC = −φ B B = −( −0.041795 rad) = 0.027863 rad N C 54 teeth (b) Rotation angle of gear D: The rotation of gear D gear D with with respect to flange A flange A is is found from: φ2

= φ D − φC

∴ φD = φC + φ2 =

0.027863 ra rad + 0.075231 rad = 0.103094 ra rad = 0.1031 rad

Ans.


6.65 The gear train shown in Fig. P6.65 transmits power from a motor to a machine at E at E . The motor turns at a frequency of 21 Hz. The diameter of solid shaft (1) is 25 mm, the diameter of solid shaft (2) is 32 mm, and the allowable shear stress for each shaft is 60 MPa. Determine: (a) the maximum power that can be transmitted by the gear train. (b) the torque provided at gear E. gear E. (c) the rotation speed of gear E gear E (in (in Hz).

Fig. P6.65

Solution Section properties: π 4 π I p1 = D1 = (25 mm)4 = 38,34 ,349.52 mm4 32 32 π 4 π I p 2 = D2 = (32 mm mm)4 = 102,94 ,943.71 mm4 32 32 Allowable torques in the shafts: τ allow I p1 (60 (60 N/m N/mm m 2 )(38 )(38,349 ,349.5 .52 2 mm mm4 ) = T 1,allow = R1 25 mm/2

T 2,allow

=

τ allow I p 2

R2

=

= 184,078

(60 (60 N/m N/mm 2 )(10 )(102,94 2,943. 3.71 71 mm4 ) 32 mm/2

=

N-mm N-mm

386,039 N-mm N-mm

magnitudes of the torques acting on the gears are related by: Torque relationships at the gears: The magnitudes of N N T B = TA B TD = TC D (a) N A N C Equilibrium: The magnitudes of magnitudes of the torques in the g ears and shafts are related by: T B = T1 = TC TD = T2 = TE

(b)

Equations (a) and (b) express the relationship between the torque in shafts (1) and (2): N T2 = T 1 D N C

(c)

let T 1 = T 1,allow Controlling shaft torque: In Eq. (c), let 1,allow and compare the result with the allowable torque for shaft (2): N check T1,allow D ≤ T 2,allow N C

⎛ 60 teeth ⎞ ⎟ = 368,156 N-mm ≤ 386, 039 N-m ⎝ 30 teeth ⎠

(184, 078 N-mm) ⎜

O.K.


This calculation shows that the torque in shaft (1) controls the capacity of the gear train. Using this fact, the torques throughout the gear train can be determined: T1 = 184, 078 N-mm T 2 = 368,156 N-mm T B

= TC = 184, 078

T A

= T B

N A N B

=

N-mm

TD

= TE =

368,156 N-m m

⎛ 24 teeth ⎞ ⎟ = 61, 359 N-mm ⎝ 72 teeth ⎠

(184, 078 N-mm) ⎜

(a) Power transmission: The maximum power that can be transmitted by the gear train can be found from the torque and rotation speed at A at A:: ⎛ 21 rev ⎞ ⎛ 2π rad ⎞ P = T Aω A = (61.359 N-m) ⎜ Ans. ⎟⎜ ⎟ = 8, 096.13 N-m/s = 8.10 kW s 1 r e v ⎝ ⎠⎝ ⎠ (b) Torque at gear E : Equations (a) and (b) can be combined to express the relationship between the torque supplied by the motor at A at A and and the torque available at gear E gear E :

T E

= TD = TC

N D N C

= TB

ND NC

⎛

NB ⎞ ND

⎝

N A ⎠ NC

= ⎜ TA

⎟

= TA

⎛ N B ⎞ ⎛ N D ⎞ ⎟ ⎜ ⎟ ⎜ N ⎝ A ⎠ ⎝ N C ⎠

(d)

The torque available at gear E gear E is is thus: T E

⎛ N B ⎞ ⎛ N D ⎞ teeth ⎞ ⎛ 72 teeth ⎞ ⎛ 60 te ⎟ = (61.359 N-m) ⎜ ⎟⎜ ⎟ ⎜ ⎟ = 368.154 N-m = 368 N-m N N 2 4 t e e t h 3 0 t e e t h ⎝ ⎠⎝ ⎠ ⎝ A ⎠ ⎝ C ⎠

= T A ⎜

Ans.

This result could also be found from the torque in shaft (2), as calculated p reviously. (c) Rotation speed of gear E : There are a couple of ways to compute this speed. Using the gear ratios, the speed of gear E gear E can can be found from: ω E

⎛ N A ⎞ ⎛ N C ⎞ ⎛ 24 teeth ⎞ ⎛ 30 teeth ⎞ ⎟⎜ ⎟ = (21 Hz) ⎜ ⎟⎜ ⎟ = 3.5 Hz N N 7 2 t e e t h 6 0 t e e t h ⎝ ⎠ ⎝ ⎠ ⎝ B ⎠ ⎝ D ⎠

= ω A ⎜

Ans.


6.66 The gear train shown in Fig. P6.66 transmits power from a motor to a machine at E . The motor turns at 750 rpm. The diameter of solid shaft (1) is 1.50 in., the diameter o solid shaft (2) is 2.00 in., and the allowable shear stress for each shaft is 6 ksi. Determine: (a) the maximum power that can be transmitted by the gear train. (b) the torque provided at gear E. gear E. (c) the rotation speed of gear E gear E (in (in Hz).

Fig. P6.66

Solution Section properties: π 4 π I p1 = D1 = (1.50 in in.)4 = 0.497010 in in.4 32 32 π 4 π I p 2 = D2 = (2. (2.00 in in.)4 = 1.570796 in in.4 32 32 Allowable torques in the shafts: τ allow I p1 (6,000 psi)( psi)(0.49 0.497010 7010 in.4 ) = = 3,976.1 lb-in. T 1,allow = R1 1.50 in./2

T 2,allow

=

τ allow I p 2

R2

=

(6,000 psi)( psi)(1.57079 1.570796 6 in.4 ) 2.00 in./2

= 9, 424.8

lb-in. lb-in.

magnitudes of the torques acting on the gears are related by: Torque relationships at the gears: The magnitudes of N N T B = TA B TD = TC D (a) N A N C Equilibrium: The magnitudes of magnitudes of the torques in the gears and shafts are related by: T B = T1 = TC TD = T2 = TE

Equations (a) and (b) express the relationship between the torque in shafts (1) and (2): N T2 = T 1 D N C

(b)

(c)

let T 1 = T 1,allow Controlling shaft torque: In Eq. (c), let 1,allow and compare the result with the allowable torque for shaft (2): N check T1,allow D ≤ T 2,allow N C

⎛ 60 teeth ⎞ 424.8 lb-in. ⎟ = 7, 952.2 lb-in. ≤ 9, 42 ⎝ 30 teeth ⎠

(3, 97 976.1 lb-in.) ⎜

O.K.


This calculation shows that the torque in shaft (1) controls the capacity of the gear train. Using this fact, the torques throughout the gear train can be determined: T1 = 3, 976.1 lb-in. T 2 = 7, 952.2 lb-in. T B

= TC = 3, 976.1 lb-in.

T A

= T B

N A N B

=

TD

= TE =

7, 952.2 lb-i n.

⎛ 24 teeth ⎞ ⎟ = 1, 325.4 lb-in. = 110.45 lb-ft ⎝ 72 teeth ⎠

(3, 976.1 lb-in.) ⎜

(a) Power transmission: The maximum power that can be transmitted by the gear train can be found from the torque and rotation speed at A at A:: ⎛ 750 rev ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞ P = T Aω A = (110.45 lb-ft) ⎜ Ans. ⎟⎜ ⎟⎜ ⎟ = 8, 675.723 lb-ft/s = 15.77 hp m i n 1 r e v 6 0 s ⎝ ⎠⎝ ⎠⎝ ⎠ (b) Torque at gear E : Equations (a) and (b) can be combined to express the relationship between the torque supplied by the motor at A at A and and the torque available at gear E gear E :

T E

= TD = TC

N D N C

= TB

ND NC

⎛

NB ⎞ ND

⎝

N A ⎠ NC

= ⎜ TA

⎟

= TA

⎛ N B ⎞ ⎛ N D ⎞ ⎟ ⎜ ⎟ ⎜ N ⎝ A ⎠ ⎝ N C ⎠

(d)

The torque available at gear E gear E is is thus: T E

⎛ N B ⎞ ⎛ N D ⎞ ⎛ 72 teeth ⎞ ⎛ 60 teeth ⎞ ⎟⎜ ⎟ = (110.45 lb-ft) ⎜ ⎟⎜ ⎟ = 662.7 lb-ft = 663 lb-ft N N 2 4 t e e t h 3 0 t e e t h ⎝ ⎠ ⎝ ⎠ ⎝ A ⎠ ⎝ C ⎠

= T A ⎜

Ans.

This result could also be found from the torque in shaft (2), as calculated p reviously. (c) Rotation speed of gear E : There are a couple of ways to compute this speed. Using the gear ratios, the speed of gear E gear E can can be found from: ω E

⎛ N A ⎞ ⎛ N C ⎞ ⎛ 24 teeth ⎞ ⎛ 30 teeth ⎞ ⎟⎜ ⎟ = (750 rpm) ⎜ ⎟⎜ ⎟ = 125 rpm N N 7 2 t e e t h 6 0 t e e t h ⎝ ⎠ ⎝ ⎠ ⎝ B ⎠ ⎝ D ⎠

= ω A ⎜

Ans.


Mechanics of Materials Solutions Chapter06 Probs42 66

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