6.26 A torque of T A = 525 lb-ft is applied to gear A gear A of the gear train shown in Fig. P6.26. The bearings shown allow the shafts to rotate freely. (a) Determine the torque T D required for equilibrium of the system. (b) Assume shafts (1) and (2) are solid 1.5-in.diameter steel shafts. Determine the magnitude o the maximum shear stresses acting in each shaft. (c) Assume shafts (1) and (2) are solid steel shafts, which have an allowable shear stress of 12,000 psi. Determine Determine the minimum minimum diameter diameter required for each shaft.
Fig. P6.26
Solution (a) Torque T D required for equilibrium: 10 in. 10 in. T D = T A = (525 lb-ft) = 875 lb-ft 6 in. 6 in. (b) Shear stress magnitudes if shafts are solid 1.5-in.-diameter: π 4 π I p = D = (1.5 in.)4 = 0.497010 in in.4 32 32 TR (525 lb-ft)(1.5 in./2)(12 in./ft) ,506.85 ps psi = 9,510 ,510 ps psi τ 1 = 1 1 = = 9,50 I p1 0.497010 in.4 τ 2
=
T2 R2 I p 2
=
(875 lb-ft)(1.5 in./2)(12 in./ft) 0.497010 in.4
,844.7 75 = 15,844.
ps psi = 15,840 ,840 ps psi
Ans. Ans.
torsion formula can be rearranged and rewritten in in (c) Determine minimum diameters: The elastic torsion terms of the unknown diameter D diameter D:: π 4 I p 32 D π 3 T TR D = τ = ∴ = = I p R D/2 16 τ Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T 1 16(525 lb-ft)(12 in./ft) 3 D13 ≥ = = 2.673803 in. π τ allow π (12,000 psi) ∴ D1 ≥ 1.38 .387958 in. =
D23
≥
16 T 2 π τ allow
=
1.388 in.
16(875 lb-ft)(12 in./ft)
1.645607 607 ∴ D2 ≥ 1.6
π (12,000 psi)
in. in. = 1.64 .646 in in.
Ans.
=
4.456338 in.3 Ans.
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6.27 A torque of T D = 1,200 N-m is applied to gear D of D of the gear train shown in Fig. P6.27. The bearings shown allow the shafts to rotate freely. (a) Determine the torque T A required for equilibrium of the system. (b) Assume shafts (1) and (2) are solid 30-mmdiameter steel shafts. Determine the magnitude o the maximum shear stresses acting in each shaft. (c) Assume shafts (1) and (2) are solid steel shafts, which have an allowable shear stress of 60 MPa. Determine the minimum diameter required for each shaft.
Fig. P6.27
Solution (a) Torque T A required for equilibrium: 150 mm 150 mm T A = T D = (1, 200 N-m) = 782.609 N-m 230 mm 230 mm (b) Shear stress magnitudes if shafts are solid 30-mm-diameter: π 4 π I p = D = (30 mm mm)4 = 79,52 ,521.564 mm mm4 32 32 TR (782.609 N-m)(30 N-m)(30 mm/2)(1,000 mm/m) MPa = 147.6 MP MPa τ 1 = 1 1 = = 147.622 MP I p1 79,521.564 mm4 τ 2
=
T2 R2 I p 2
=
(1, 200 N-m)(30 mm/2)(1,000 mm/2)(1,00 0 mm/m) 79,521.564 mm4
=
226.354 MPa = 226 MP MPa
Ans. Ans.
torsion formula can be rearranged and rewritten in in (c) Determine minimum diameters: The elastic torsion terms of the unknown diameter D diameter D:: π 4 D I TR π 3 T p 32 D = τ = ∴ = = I p R D/2 16 τ Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T 1 16(782.609 NN-m)(1,000 mm/m) 3 = = 66,429.915 mm D13 ≥ 2 (60 N/ N/mm ) π τ allow π (60 ∴ D1 ≥
D23
≥
40.500 mm mm = 40.5 mm
16 T 2 π τ allow
∴ D2 ≥
=
Ans.
16(1, 20 200 N-m)(1,000 mm/m) 2
N/mm ) π (60 N/
46.702 mm mm = 46.702 mm mm
= 101,859.164
mm3 Ans.
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6.28 The gear train system shown in Fig. P6.28 includes shafts (1) and (2), which are solid 1.375-in.diameter steel shafts. The allowable shear stress o each shaft is 5,000 psi. The bearings shown allow the shafts to rotate freely. Determine the maximum torque T D that can be applied to the system without exceeding the allowable shear stress in either shaft.
Fig. P6.28
Solution Section properties: π 4 π I p1 = D = (1.375 in.)4 32 32
=
0.350 350922 in.4
=
I p 2
Maximum torque in either shaft: τ allow I p1 (5, (5, 000 psi)( psi)(0.35 0.350922 0922 in.4 ) T max = = = 2,552.16 lb-in. R1 1.375 in./2 Torque relationship: T1 T 2 = or T1 8 in. 5 in.
= T2
8 in. 5 in.
= 1.60 T2
∴ T1
control s
The torque in shaft (1) controls; therefore, T 1 = 2,552.16 lb-in. Consequently, the maximum torque in shaft (2) must be limited to: 5 in. T2 = T 1 0.625( 5(2,55 2,552. 2.16 16 lblb-in.) in.) = 1,595.1 ,595.10 0 lblb-in in.. = 132. 132.9 9 lblb-fft = 0.62 Ans. 8 in.
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6.29 The gear train system shown in Fig. P6.29 includes shafts (1) and (2), which are solid 20-mmdiameter steel shafts. The allowable shear stress o each shaft is 30 MPa. The bearings shown allow the shafts to rotate freely. Determine the maximum torque T D that can be applied to the system without exceeding the allowable shear stress in either shaft.
Fig. P6.29
Solution Section properties: π 4 π I p1 = D = (20 mm mm)4 32 32
,707.963 = 15,70
mm mm4
=
I p 2
Maximum torque in either shaft: τ allow I p1 (30 (30 N/mm N/mm2 )(15, (15,70 707. 7.96 963 3 mm mm4 ) T max = = = 47,123.890 N-mm R1 20 mm/2 Torque relationship: T1 T 2
200 mm
=
80 mm
or T1
= T2
200 mm 80 mm
=
2.50 T2
∴ T1
controls
The torque in shaft (1) controls; therefore, T 1 = 47,123.890 N-mm. Consequently, the maximum maximum torque in shaft (2) must be limited to: 80 mm T2 = T 1 0.4(47 47,1 ,123 23.8 .890 90 N-m N-mm) = 18,84 18,849. 9.55 556 6 N-m N-mm = 18.8 18.85 5 N-m = 0.4( Ans. 200 mm
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6.30 In the gear system shown in Fig. P6.30, the motor applies a 160 lb-ft torque to the gear at A at A.. A torque of T C 250 lb-ft is removed from the shaft C = at gear C , and the remaining torque is removed at gear D. D. Segments (1) and (2) are solid 1.5-in.diameter steel [G [G = 12,000 ksi] shafts, and the bearings shown allow free rotation of the shaft. Determine: (a) the maximum shear stress in segments (1) and (2) of the shaft. (b) the rotation angle of gear D gear D relative relative to gear B gear B..
Fig. P6.30
Solution Section properties: π 4 π I p1 = D = (1.5 in.)4 32 32 Torque relationship: T A T B hence T B = 4 in. 10 in.
=
0.497010 in in.4
= TA
10 in. 4 in.
=
=
I p 2
2.50T A
=
2.50(160 lb-ft) = 400 lb-ft
Shaft (1) has the same torque as gear B gear B;; therefore, T 1 = 400 lb-ft. lb-ft. A torque of 250 lb-ft lb-ft is removed from from the shaft at gear C ; therefore, T C equilibrium: C = 250 lb-ft. The torque in shaft (2) must satisfy equilibrium: Σ M x = −T1 + TC + T 2 = 0 ∴ T2 = T1 − T C =
400 lb lb-ft − 250 lb lb-ft = 150 lb lb-ft
To summarize: T1
=
400 lb-ft
T 2
= 150
lb-ft
(a) Maximum shear stress in shafts (1) and (2): TR (400 lb-ft)(1.5 lb-ft)(1.5 in./2)(12 in./ft) ,243.315 psi = 7,240 ,240 psi τ 1 = 1 1 = = 7,243 I p1 0.497010 in.4 τ 2
=
T2 R2 I p 2
=
(150 lb-ft)(1.5 lb-ft)(1.5 in./2)(12 in./ft) 0.497010 in.4
=
2,716 ,716.243 psi = 2,720 ,720 ps psi
Ans. Ans.
(b) Angles of twist in shafts (1) and (2): TL (400 lb-ft)(60 lb-ft)(60 in.)(12 in./ft) φ 1 = 1 1 = = 0.048289 rad G1 I p1 (12,000,000 psi)(0. psi)(0.497 497010 010 in.4 ) φ 2
=
T2 L2 G2 I p 2
=
(150 lb-ft)(40 in.)(12 in./ft) (12,000,000 psi)( psi)(). ).497 497010 010 in.4 )
=
0.012072 rad
Note: There is not enough information given to say whether these twist angles are positive or negative. However, both twist angles will have the same sign. say φ B = 0 rad (since this is our reference) φ D
= φB + φ1 + φ2 =
0 rad + 0.048289 rad + 0.012072 rad = 0.060361 rad = 0.0604 rad
Ans.
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6.31 In the gear system shown in Fig. P6.31, the motor applies a 300 lb-ft torque to the gear at A at A.. A torque of T C D. C = 500 lb-ft is removed from the shaft at gear C , and the remaining torque is removed at gear D. Segments (1) and (2) are solid steel [G [G = 12,000 ksi] shafts, and the bearings shown allow free rotation o the shaft. (a) Determine the minimum permissible diameters for segments (1) and (2) of the shaft i the maximum shear stress must not exceed 6,000 psi. (b) If the same diameter is to be used for both segments (1) and (2), determine the minimum permissible diameter that can be used for the shaft if the maximum shear stress must not exceed 6,000 psi and the rotation angle of gear D gear D relative to gear B gear B must must not exceed 0.10 rad. Fig. P6.31
Solution Torque relationship: The motor applies a torque of 300 lb-ft to gear A gear A.. Since gear B gear B is bigger than gear A gear A,, the torque on gear B gear B will will be increased in proportion to the gear ratio: T A T B 10 in. hence T B = TA = = 2.50T A = 2.50(300 lb-ft) = 750 lb-ft 4 in. 10 in. 4 in. Let the positive x positive x axis axis for shaft BCD shaft BCD extend extend from gear B gear B toward toward gear D gear D.. In order for shaft BCD shaft BCD to to be in equilibrium with the torques as shown on gears C and D, D, the torque at B B must act in the negative x direction. Consequently, T B = −750 lb-ft.
Draw a free-body diagram that cuts through shaft (1) and includes gear B gear B.. From this FBD, the torque torque in shaft (1) is: Σ M x = −750 lb-ft + T1 = 0 ∴ T 1 = 750 lb-ft From the problem statement, we are told that a torque of 500 lb-ft is removed from the shaft at gear C . In other words, T C 500 lb-ft. Draw a FBD that cuts through shaft C = (2) and includes both gears B gears B and and C . From this FBD, the torque in shaft (2) is: Σ M x = −750 lb-ft + 500 lb-ft + T 2 = 0 ∴ T 2 =
250 lb-ft
To summarize: T1
= 750
lb-ft
T 2
=
250 lb-ft
torsion formula can be rearranged and rewritten in (a) Determine minimum diameters: The elastic torsion terms of the unknown diameter D diameter D:: π 4 D I TR π 3 T p 32 D = τ = ∴ = = I p R D/2 16 τ
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Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T 1 16(750 lb-ft)(12 in./ft) D13 ≥ 7.639437 in.3 = = 7.639437 π τ allow π (6,000 psi) .969490 490 ∴ D1 ≥ 1.96 D23
≥
16 T 2 π τ allow
=
in. in. = 1.969 in in.
16(250 lb-ft)(12 in./ft)
∴ D2 ≥ 1.3 1.36556 5568
π (6,000 psi)
Ans.
=
2.546479 in.3
in. in. = 1.366 in i n.
Ans.
gear D with with respect to gear B gear B is is equal to the sum of the angles of twist in shaft (b) The rotation angle of gear D segments (1) and (2): φ D − φ B = φ1 + φ2 ≤ 0.10 rad Since the same solid steel shaft is to be used for both segments (1) and (2): TL TL 1 0.01 rad ≥ 1 1 + 2 2 = [T1 L1 + T2 L2 ] G1 I p1 G2 I p 2 GI p or rearranging: π 4 I p = D 32 ∴ D ≥
≥
(750 lb-ft lb-ft)(60 )(60 in.)(12 in.)(12 in./ft) in./ft) + (250 lb-ft) lb-ft)(40 (40 in.)( in.)(12 12 in./ft) in./ft) (12, 000, 000 psi)(0.1 rad)
=
0.5500 in.4
1.538 in.
Since the shaft has already been designed for stresses and since the diameter that is required to satisfy the rotation angle requirement is smaller than D than D1 determined in part (a), the minimum diameter required for a constant diameter shaft between gears B gears B and and D D is: is: Dmin
= 1.969
in.
Ans.
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6.32 In the gear system shown in Fig. P6.32, the motor applies a torque of 220 N-m to the gear at A. A. A torque of T C C = 400 N-m is removed from the shaft at gear C , and the remaining torque is removed at gear D gear D.. Segments (1) and (2) are solid 40-mm-diameter steel [G [G = 80 GPa] shafts, and the bearings shown allow free rotation of the shaft. (a) Determine the maximum shear stress in segments (1) and (2) of the shaft. (b) Determine the rotation angle of gear D relative to gear B gear B..
Fig. P6.32
Solution Section properties: π 4 π I p1 = D = (40 mm)4 32 32 Torque relationship: T A T B
100 mm
=
300 mm
=
251,327 ,327.41 mm4
hence T B
= TA
300 mm 100 mm
=
I p 2
=
3T A
=
3(220 N-m) = 660 N-m
Shaft (1) has the same torque as gear B gear B;; therefore, T 1 = 660 N-m. A torque of T C 400 N-m is removed C = from the shaft at gear C . The torque in shaft (2) must satisfy equilibrium: Σ M x = −T1 + TC + T 2 = 0 ∴ T2 = T1 − T C =
660 N-m − 400 N-m = 260 N-m
To summarize: T 1 = 660 N-m T 2
=
260 N-m
(a) Maximum shear stress in shafts (1) and (2): TR (660 N-m)(40 N-m)(40 mm/2)(1,000 mm/m) τ 1 = 1 1 = = 52.521 MPa = 52.5 MPa 251,327.41 mm4 I p1 τ 2
=
T2 R2 I p 2
=
(260 N-m)(40 N-m)(40 mm/2)(1,000 mm/m) 251,327.41 mm4
=
20.690 MPa = 20.7 MP MPa
Ans. Ans.
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(b) Angles of twist in shafts (1) and (2): T1 L1 (660 N-m)(1.5 N-m)(1.5 m)(1,000 mm/m)2 φ 1 = = = 0.049239 rad G1 I p1 (80,0 (80,000 00 N/mm N/mm2 )(25 )(251, 1,327 327.4 .41 1 mm mm4 ) φ 2
=
T2 L2 G2 I p 2
=
(260 N-m)(1.0 N-m)(1.0 m)(1,000 mm/m)2 (80, (80,000 000 N/mm N/mm2 )(25 )(251,3 1,327 27.41 .41 mm4 )
=
0.012931 rad
Note: There is not enough information given to say whether these twist angles are positive or negative. However, both twist angles will have the same sign. say φ B φ D
=
0 rad
= φB + φ1 + φ2 =
(since this is our reference)
0 rad + 0.049239 rad + 0.012931 rad = 0.062170 rad = 0.0622 rad
Ans.
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6.33 In the gear system shown in Fig. P6.33, the motor applies a torque of 400 N-m to the gear at A. A. A torque of T C C = 700 N-m is removed from the shaft at gear C , and the remaining torque is removed at gear D gear D.. Segments (1) and (2) are solid steel [G [G = 80 GPa] shafts, and the bearings shown allow free rotation of the shaft. (a) Determine the minimum permissible diameters for segments (1) and (2) of the shaft i the maximum shear stress must not exceed 40 MPa. (b) If the same diameter is to be used for segments (1) and (2), determine the minimum permissible diameter that can be used for the shaft if the maximum shear stress must not exceed 40 MPa and the rotation angle of gear D gear D relative to gear B gear B must must not exceed 3.0°.
Fig. P6.33
Solution Torque relationship: T A T B
100 mm
=
300 mm
hence T B
= TA
300 mm 100 mm
=
3T A
=
3(400 N-m) = 1, 20 200 N-m
Shaft (1) has the same torque as gear B; B; therefore, T 1 = 1,200 N-m. N-m. A torque of T C C = 700 N-m is removed from the shaft at gear C . The torque in shaft shaft (2) must satisfy equilibrium: equilibrium: Σ M x = −T1 + TC + T 2 = 0 ∴ T2 = T1 − T C = 1, 20 200
NN-m − 700 NN-m = 500 NN-m
To summarize: T1
= 1, 200
N-m
T 2
=
500 N-m
(a) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter D diameter D:: π 4 I p 32 D TR π 3 T D = τ = ∴ = = I p R D/2 16 τ
Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T 1 16(1, 20 200 N-m)(1,000 mm/m) 3 = = 152,788.745 mm D13 ≥ 2 (40 N/ N/mm ) π τ allow π (40 ∴ D1 ≥
D23
≥
53.460 mm = 53.5 mm
16 T 2 π τ allow
=
Ans.
16(500 NN-m)(1,000 mm/m)
∴ D2 ≥ 39.929
2
(40 N/mm ) π (40 mm mm = 39.9 mm mm
=
63, 63, 661.977 mm3 Ans.
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gear D with with respect to gear B gear B is is equal to the sum of the angles of twist in shaft (b) The rotation angle of gear D segments (1) and (2): φ D − φB = φ1 + φ2 ≤ (3°)(π / 1 80°) = 0.052360 rad Since the same solid steel shaft is to be used for both segments (1) and (2): TL TL 1 0.052360 rad ≥ 1 1 + 2 2 = [T1 L1 + T2 L2 ] G1 I p1 G2 I p 2 GI p or rearranging: I p
=
π
32
D
∴ D ≥
4
≥
(1, 200 N-m) N-m)(1. (1.5 5 m)(1,0 m)(1,000 00 mm/m mm/m))2 2
+
(500 (500 N-m)( N-m)(1.0 1.0 m)( m)(1,0 1,000 00 mm/m) mm/m)2
(80, 000 N/mm )( ) (0.052360 rad)
= 549,083.270
mm4
48.6 mm
Since the shaft has already been designed for stresses and since the diameter that is required to satisfy the rotation angle requirement is smaller than D than D1 determined in part (a), the minimum diameter required for a constant diameter shaft between gears B gears B and and D D is: is: Dmin
= 53.5
mm
Ans.
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6.34 In the gear system shown in Fig. P6.34, the motor applies a torque of 250 N-m to the gear at A at A.. Shaft (1) is a solid 35-mm-diameter shaft, and shaft (2) is a solid 50-mm-diameter shaft. The bearings shown allow free rotation of the shafts. (a) Determine the torque T E provided by the gear system at gear E. gear E. (b) Determine the maximum shear stresses in in shafts (1) and (2).
Fig. P6.34
Solution (a) Torque relationship: The torque on gear B gear B is: is: 72 T B = TA = 3 T A = 3( 250 N-m) = 750 N-m 24 Shaft (1) has the same torque as gear B gear B;; therefore, T 1 = 750 N-m and T C 750 N-m. The torque on gear C = D is D is found from: 60 T D = TC = 2 T C = 2(750 N-m) = 1, 500 N-m 30 Shaft (2) has the same torque as gear D gear D;; therefore, T 2 = 1,500 N-m and
T E = 1,500 N-m
Ans.
To summarize: T1
=
750 N-m
T 2
= 1, 500
N-m
(b) Section properties: π 4 π I p1 = D1 = (35 mm)4 = 147,32 ,323.51 mm4 32 32 π 4 π I p 2 = D2 = (50 mm mm)4 = 613,59 ,592.32 mm mm4 32 32 Maximum shear stress in shafts (1) and (2): TR (750 N-m)(35 N-m)(35 mm/2)(1,000 mm/m) τ 1 = 1 1 = = 89.090 MPa = 89.1 MPa I p1 147,323.51 mm4 τ 2
=
T2 R2 I p 2
=
(1,500 N-m)(50 mm/2)(1,000 mm/m) 613,592.32 mm4
=
61.115 MPa = 61.1 MPa
Ans. Ans.
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6.35 In the gear system shown in Fig. P6.35, the motor applies a torque of 600 N-m to the gear at A at A.. Shafts (1) and (2) are solid shafts, and the bearings shown allow free rotation of the shafts. (a) Determine the torque T E provided by the gear system at gear E. gear E. (b) If the allowable shear stress in each shaft must be limited to 50 MPa, determine the minimum permissible diameter for each shaft.
Fig. P6.35
Solution (a) Torque relationship: The torque on gear B gear B is: is: 72 T B = TA = 3 T A = 3(600 N-m) = 1, 800 N-m 24 Shaft (1) has the same torque as gear B gear B;; therefore, T 1 = 1,800 N-m and T C 1,800 N-m. The torque on C = gear D gear D is is found from: 60 T D = TC 600 N-m = 2 T C = 2(1, 800 N-m) = 3, 60 30 Shaft (2) has the same torque as gear D gear D;; therefore, T 2 = 3,600 N-m and
T E = 3,600 N-m
Ans.
To summarize: T1
= 1, 800
N-m
T 2
=
3, 600 N-m
(b) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter D diameter D:: π 4 I p 32 D TR π 3 T τ = ∴ = = D = I p R D/2 16 τ
Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T 1 16(1, 80 800 N-m)(1,000 mm/m) = = 183, D13 ≥ 183, 346.494 mm3 2 N/mm ) π τ allow π (50 N/ ∴ D1 ≥
D23
≥
56.810 mm = 56.8 mm
16 T 2 π τ allow
∴ D2 ≥
=
Ans.
16(3, 60 600 NN-m)(1,000 mm/m) 2
N/mm ) π (50 N/
71.576 mm mm = 71.6 mm mm
= 366,692.989
mm3 Ans.
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6.36 In the gear system shown in Fig. P6.36, a torque of T E = 900 lb-ft is delivered at gear E . Shafts (1) and (2) are solid shafts, and the bearings shown allow free rotation of the shafts. (a) Determine the torque provided by the motor to gear A. gear A. (b) If the allowable shear stress in each shaft must be limited to 4,000 psi, determine the minimum permissible diameter for each shaft.
Fig. P6.36
Solution (a) Torque relationship: The torque on gear E creates a torque in shaft (2) of T 2 = 900 lb-ft. Accordingly, the torque applied to gear D gear D is is also equal to 900 lb-ft. The torque on gear C is is found with the gear ratio: 30 TC = TD = 0.5 T D = 0.5(900 lb-ft) = 450 lb-ft 60 Shaft (1) has the same torque as gear C ; therefore, T 1 = 450 lb-ft and T B = 450 N-m. The torque on gear A is A is found from: 24 T C 450 lb-ft = = = 150 lb-ft Ans. T A = T B 72 3 3
To summarize: T1
=
450 lb-ft
T 2
=
900 lb-ft
(b) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter D diameter D:: π 4 I p 32 D TR π 3 T D = τ = ∴ = = I p R D/2 16 τ
Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T 1 16(450 lb-ft)(12 in./ft) 3 D13 ≥ = = 6.875494 in. π τ allow π (4,000 psi) ∴ D1 ≥ 1.9015 in. =
D23
≥
16 T 2 π τ allow
∴ D2 ≥
=
1.90 .902 in.
16(900 lb-ft)(12 in./ft) π (4,000 psi)
2.3958 in. = 2.40 in.
Ans.
13.750987 = 13.750987
in.3 Ans.
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6.37 In the gear system shown in Fig. P6.37, a torque of T E = 720 lb-ft is delivered at gear E . Shaft (1) is a solid 1.50-in.-diameter shaft, and shaft (2) is a solid 2.00-in.-diameter shaft. The bearings shown allow free rotation of the shafts. (a) Determine the torque provided by the motor to gear A. gear A. (b) Determine the maximum shear stresses in shafts (1) and (2).
Fig. P6.37
Solution (a) Torque relationship: The torque on gear E creates a torque in shaft (2) of T 2 = 720 lb-ft. Accordingly, the torque applied to gear D gear D is is also equal to 720 lb-ft. The torque on gear C is is found with the gear ratio: 30 TC = TD = 0.5 T D = 0.5(720 lb-ft) = 360 lb-ft 60 Shaft (1) has the same torque as gear C ; therefore, T 1 = 360 lb-ft and T B = 360 N-m. The torque on gear A is A is found from: 24 T C 360 lb-ft = = = 120 lb-ft Ans. T A = T B 72 3 3
To summarize: T1
= 360
lb-ft
T 2
=
720 lb-ft
(b) 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 Maximum shear stress in shafts (1) and (2): TR (360 lb-ft)(1.50 in./2)(12 in./ft) τ 1 = 1 1 = I p1 0.497010 in.4 τ 2
=
T2 R2 I p 2
=
=
(720 lb-ft)(2.00 lb-ft)(2.00 in./2)(12 in./ft) 1.570796 in.4
6,51 ,519.98 psi = 6,52 ,520 ps psi
=
5,50 ,500.396 ps psi = 5,50 ,500 ps psi
Ans. Ans.
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6.38 Two solid 30-mm-diameter steel shafts are connected by the gears shown in Fig. P6.38. The shaft lengths are L are L1 = 300 mm and L and L2 = 500 mm. Assume that the shear modulus of both shafts is G = 80 GPa and that the bearings shown allow free rotation of the shafts. If the torque applied at gear D gear D is is T D = 160 N-m, (a) determine the internal torques T 1 and T 2 in the two shafts. (b) determine the angles of twist φ 1 and φ 2. (c) determine the rotation angles φ B and φ C C o gears B gears B and and C . (d) determine the rotation angle of gear D gear D..
Fig. P6.38
Solution (a) Torque relationship: The torque on gear D gear D creates creates a torque in shaft (2) of T 2 = 160 N-m. Σ M x = TD − T2 = 0 ∴ T2 = T D = 160 N-m
Accordingly, the torque applied to gear C is also equal to 160 N-m. The torque on gear B gear B is is found with the gear ratio: 54 teeth = −1.8 T T B = −TC C = −1.8(160 N-m) = − 288 N-m 30 teeth Shaft (1) has the same torque as gear B gear B;; therefore, T 1 = −288 N-m. To summarize: T1
= −288
N-m
T 2
= 160
N-m
Ans.
Section properties: π 4 π I p1 = D1 = (30 mm mm)4 = 79, 52 521.56 mm mm4 = I p 2 32 32 (b) Angles of twist in shafts (1) and (2): TL (−288 N-m N-m)(30 )(300 0 mm)( mm)(1,000 1,000 mm/m) mm/m) φ 1 = 1 1 = = 2 G1 I p1 (80,0 (80,000 00 N/mm N/mm )(79, )(79,52 521.5 1.56 6 mm mm4 ) φ 2
=
T2 L2 G2 I p 2
=
(160 N-m)(500 mm)(1,000 mm/m) (80, (80,000 000 N/mm N/mm2 )(79 )(79,5 ,521. 21.56 56 mm4 )
=
−0.013581 rad
Ans.
0.012575 rad
Ans.
gear B is is found from the angle of twist in shaft (c) Rotation angles of gears B and C: The rotation of gear B (1): φ1
= φ B − φ A
∴φB = φA + φ1 = 0
rad + (−0.013581 rad) =
− 0.01358
rad
Ans.
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: 54 teeth 54 teeth φC = −φ B = −(−0.013581 rad) = 0.024446 rad = 0.0244 rad Ans. 30 teeth 30 teeth The rotation of gear D gear D is is found from: φ2
= φ D − φC
∴ φD = φC + φ2 =
0.024446 ra rad + 0.012575 ra rad = 0.037021 rad = 0.0370 ra rad
Ans.
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6.39 Two solid 1.75-in.-diameter steel shafts are connected by the gears shown in Fig. P6.39. The shaft lengths are L are L1 = 6 ft and L and L2 = 10 ft. Assume that the shear modulus of both shafts is G = 12,000 ksi and that the bearings shown allow free rotation of the shafts. If the torque applied at gear D gear D is is T D = 225 lb-ft, (a) determine the internal torques T 1 and T 2 in the two shafts. (b) determine the angles of twist φ 1 and φ 2. (c) determine the rotation angles φ B and φ C C o gears B gears B and and C . (d) determine the rotation angle of gear D gear D..
Fig. P6.39
Solution (a) Torque relationship: The torque on gear D gear D creates creates a torque in shaft (2) of T 2 = 225 lb-ft. Σ M x = TD − T2 = 0 ∴ T2 = T D = 225 lb-ft
Accordingly, the torque applied to gear C is also equal to 225 lb-ft. lb-ft. The torque on gear B gear B is is found with the gear ratio: 54 teeth = −1.8 T T B = −TC C = −1.8(225 lb-ft) = −405 lb-ft 30 teeth Shaft (1) has the same torque as gear B gear B;; therefore, T 1 = −405 lb-ft. To summarize: T1
= −405
lb-ft
T 2
=
225 lb-ft
Ans.
Section properties: π 4 π I p1 = D1 = (1.75 in in.)4 = 0.920772 in in.4 = I p 2 32 32 (b) Angles of twist in shafts (1) and (2): T1 L1 ( −405 lblb-ft) ft)(6 (6 ft)(1 ft)(12 2 in./ft in./ft)) 2 φ 1 = = = −0.031669 ra rad = G1 I p1 (12,000,000 (12,000,000 psi)(0.9207 psi)(0.920772 72 in. in.4 ) φ 2
=
T2 L2 G2 I p 2
=
(225 lb-ft)(10 lb-ft)(10 ft)(12 in./ft) 2 (12,000,000 (12,000,000 psi)(0.920772 psi)(0.920772 in.4 )
=
− 0.0317
ra rad
0.029323 rad = 0.029 0293 ra rad
Ans. Ans.
gear B is is found from the twist angle in shaft (1): (c) Rotation angles of gears B and C: The rotation of gear B φ1
= φ B − φ A
∴φB = φA + φ1 = 0
rad + (−0.031669 rad) =
− 0.0317
rad
Ans.
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: 54 teeth 54 teeth = −( −0.031669 rad) = 0.057004 rad = 0.0570 rad Ans. φC = −φ B 30 teeth 30 teeth The rotation of gear D gear D is is found from: φ2
= φ D − φC
∴ φD = φC + φ2 =
0.057004 ra rad + 0.029323 rad = 0.086327 ra rad = 0.0863 rad
Ans.
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6.40 Two solid steel shafts are connected by the gears shown in Fig. P6.40. The design requirements for the system require (1) that both shafts must have the same diameter, (2) that the maximum shear stress in each shaft must be less than 6,000 psi, and (3) that the rotation angle o gear D D must not exceed 3°. Determine the minimum required diameter of the shafts if the torque applied at gear D gear D is is T D = 1,000 lb-ft. The shaft lengths are L1 = 10 ft and L2 = 8 ft. Assume that the shear modulus of both shafts is G = 12,000 ksi and that the bearings shown allow free rotation of the shafts.
Fig. P6.40
Solution Torque relationship: The torque on gear D gear D creates creates a torque in shaft (2) of T 2 = 1,000 lb-ft. Σ M x = TD − T2 = 0 ∴ T2 = T D = 1, 000 lb-ft = 12, 000 lb-in.
Accordingly, the torque applied to gear C is is also 12,000 lb-in. lb-in. The torque on gear B gear B is is found with the gear ratio: 48 teeth T B = −TC 000 lb lb-in.) = − 8, 00 000 lb lb-in. = −0.6667 T C = −0.6667 (12, 00 72 teeth Shaft (1) has the same torque as gear B gear B;; therefore, T 1 = −8,000 lb-in. Diameters based on allowable shear stresses: between shear stress and torque in a shaft. TR τ = I p
The elastic torsion formula gives the relationship
The torques in both shafts have been determined, determined, and the allowable shear stress is specified. Rearrange the elastic torsion formula, putting the known terms on the right-hand side of the equation: I p T =
R τ Express the left-hand side of this equation in terms of the shaft diameter D diameter D:: π 4 D π 3 T 32 D = = D / 2 16 τ Solve for the minimum acceptable acceptab le diameter in shaft (1): 16 T 1 16(8, 00 000 lb-in.) 3 D13 ≥ = = 6.790614 in. 2 (6, 000 000 lb/i lb/in. n. ) π τ π (6, ∴ D1 ≥ 1.894
in.
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Solve for the minimum acceptable acceptab le diameter in shaft (2): 16 T 2 16(12, 00 000 lb-in.) 3 D23 ≥ = = 10.185916 in. 2 (6, 000 000 lb/i lb/in. n. ) π τ π (6, ∴ D2 ≥
2.17 in.
Of these two values, D values, D2 controls. Therefore, both shafts could have a diameter of 2.17 in. or more and the shear stress constraint would be satisfied. The angles of twists in shafts (1) and (2) can be expressed as: TL TL φ1 = 1 1 φ 2 = 2 2 G1 I p1 G2 I p 2 The rotations of gears B and C are are related since the arclengths turned by the two gears have the same magnitude. The gears turn in opposite directions; directions; therefore, a negative sign sign is introduced. RCφC = − RBφ B The rotation angle of gear B gear B is is equal to the angle of twist in shaft (1): φ B = φ 1. Therefore, the rotation angle of gear C can can be expressed in terms of T 1. R R R TL N TL φC = − B φB = − B φ 1 = − B 1 1 = − B 1 1 RC RC RC G1 I p1 N C G1 I p1 The rotation angle of gear D gear D is is equal to the rotation angle of gear C plus plus the twist that occurs in shaft (2): φ D = φC + φ 2 and so the rotation angle of gear D gear D can can be expressed in terms of the torques T 1 and T 2: N T L TL φ D = − B 1 1 + 2 2 ≤ 3° N C G1 I p1 G2 I p 2 Shafts having the same diameters and the same shear moduli are required for this system; therefore, I therefore, I p1 p1 = I p2 = I p and G1 = G2 = G. Factor these terms out to obtain: p2 = I 1 ⎡ N B
⎢−
⎤
T1 L1 + T2 L2 ⎥ ≤ 3°
∴ I p ≥
⎡ N B ⎤ T L T L − + 1 1 2 2⎥ ⎢ G (3°) ⎣ N C ⎦ 1
G I p ⎣ N C ⎦ Express the polar moment of inertia in terms of diameter to obtain the following relationship: D 4
≥
=
=
⎡ N B ⎤ T L T L − + 1 1 2 2⎥ ⎢ π G (3°) ⎣ N C ⎦ 32
32
(
12, 000,000 000,000 psi)( psi)(3 3° ) π rad π (12,
180°
)
⎡ 48 teeth ⎤ ( lb -in. n.)( )(120 120 in.) in .) 12 , 000 00 0 lbl b-in in.) .)(9 (96 6 in. i n.) ) − −8, 000 lb-i + (12, ⎢⎣ 72 teeth ⎥⎦
29.050810 in.4
∴ D ≥
2.32 in.
Since this minimum diameter is greater than the diameter needed to satisfy the shear stress requirements, the rotation angle constraint controls. Therefore, the minimum shaft diameter that satisfies all requirements is D is D ≥ 2.32 in.
Ans.
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6.41 Two solid steel shafts are connected by the gears shown in Fig. P6.41. The design requirements for the system require (1) that both shafts must have the same diameter, (2) that the maximum shear stress in each shaft must be less than 40 MPa, and (3) that the rotation angle o gear D D must not exceed 3°. Determine the minimum required diameter of the shafts if the torque applied at gear D gear D is is T D = 1,200 N-m. The shaft lengths are L1 = 2.5 m and L2 = 2.0 m. Assume that the shear modulus of both shafts is G = 80 GPa and that the bearings shown allow free rotation of the shafts.
Fig. P6.41
Solution gear D creates creates a torque in shaft (2) of Torque relationship: The torque on gear D T 2 = 1,200 N-m. Σ M x = TD − T2 = 0 ∴ T2 = T D = 1, 200 N-m Accordingly, the torque applied to gear C is is also 1,200 N-m. The torque on gear B gear B is is found with the gear ratio: 48 teeth T B = −TC 200 N-m) = − 800 N-m = −0.6667 T C = −0.6667(1, 20 72 teeth Shaft (1) has the same torque as gear B gear B;; therefore, T 1 = −800 N-m. Diameters based on allowable shear stresses: The elastic torsion formula gives the relationship between shear stress and torque in a shaft. This equation can be rearranged in terms of the outside diameter: TR π 3 T D = τ = ⇒ I p 16 τ
Solve for the minimum acceptable acceptab le diameter in shaft (1): 16 T 1 16(800 NN-m)(1,000 mm/m) 3 D13 ≥ = = 101,859.16 mm 2 (40 N/ N/mm ) π τ π (40 ∴ D1 ≥
46.70 mm
Solve for the minimum acceptable acceptab le diameter in shaft (2): 16 T 2 16(1,200 ,200 NN-m)(1,000 mm mm/m) 3 D23 ≥ = = 152,788.75 mm 2 (40 N/ N/mm ) π τ π (40 ∴ D2 ≥ 53.46
mm
Of these two values, D2 controls. Therefore, both shafts could have a diameter of 53.46 mm or more and the shear stress constraint would be satisfied. The angles of twists in shafts (1) and (2) can be expressed as: TL TL φ1 = 1 1 φ 2 = 2 2 G1 I p1 G2 I p 2 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 rotations of gears B and C are are related since the arclengths turned by the two gears have the same magnitude. The gears turn in opposite directions; directions; therefore, a negative sign sign is introduced. RCφC = − RBφ B The rotation angle of gear B gear B is is equal to the angle of twist in shaft (1): φ B = φ 1. Therefore, the rotation angle of gear C can can be expressed in terms of T 1. R R R TL N TL φC = − B φB = − B φ 1 = − B 1 1 = − B 1 1 RC RC RC G1 I p1 N C G1 I p1 The rotation angle of gear D gear D is is equal to the rotation angle of gear C plus plus the twist that occurs in shaft (2): φ D = φC + φ 2 and so the rotation angle of gear D gear D can can be expressed in terms of the torques T 1 and T 2: N T L TL φ D = − B 1 1 + 2 2 ≤ 3° N C G1 I p1 G2 I p 2 Shafts having the same diameters and the same shear moduli are required for this system; therefore, I therefore, I p1 p1 = I p2 = I p and G1 = G2 = G. Factor these terms out to obtain: p2 = I 1 ⎡ N B
⎢−
⎤
T1 L1 + T2 L2 ⎥ ≤ 3°
∴ I p ≥
⎡ N B ⎤ T L T L − + 1 1 2 2 ⎢ ⎥ G (3°) ⎣ N C ⎦ 1
G I p ⎣ N C ⎦ Express the polar moment of inertia in terms of diameter to obtain the following relationship: D 4
⎡ N B ⎤ T1 L1 + T2 L2 ⎥ ⎢− π G (3°) ⎣ N C ⎦ ⎡ 48 teeth ⎤ 32 ⎢ − (−800 N-m)(2.5 m) + (1, 20 200 N-m)(2.0 m)⎥ (1,000 mm/m)2 ⎣ 72 teeth ⎦ =
≥
32
2
000 N/ N/mm )(3°)(π / 18 180°) π (80, 00
= 9,078,378.05 ∴ D ≥ 54.89
mm4
mm
Since this minimum diameter is greater than the diameter needed to satisfy the shear stress requirements, the rotation angle constraint controls. Therefore, the minimum shaft diameter that satisfies all requirements is D is D ≥ 54.9 mm.
Ans.
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