mechanics of materials solutions chapter06 probs26 41

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6.26 A torque of TA = 525 lb-ft is applied to gear Aof the gear train shown in Fig. P6.26. The bearingsshown allow the shafts to rotate freely. (a) Determine the torque TD required for equilibriumof the system. (b) Assume shafts (1) and (2) are solid 1.5-in.-diameter steel shafts. Determine the magnitude ofthe 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 the minimum diameter required for each shaft.

Fig. P6.26

Solution (a) Torque TD required for equilibrium:

10 in. 10 in.(525 lb-ft) 875 lb-ft6 in. 6 in.D AT T= = =

(b) Shear stress magnitudes if shafts are solid 1.5-in.-diameter:

4 4 4(1.5 in.) 0.497010 in.32 32pI Dπ π= = =

1 11 4

1

(525 lb-ft)(1.5 in./2)(12 in./ft) 9,506.85 psi 9,510 psi0.497010 in.p

T RI

τ = = = = Ans.

2 22 4

2


T RI

τ = = = = Ans.

(c) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter D:

4

332/ 2 16

p

p

DITR TDI R D

ππτ

τ= ∴ = = =

Using this expression, solve for the minimum acceptable diameter of each shaft:

3 311

allow

1

16 16(525 lb-ft)(12 in./ft) 2.673803 in.(12,000 psi)

1.387958 in. 1.388 in.

TD

D

π τ π≥ = =

∴ ≥ = Ans.

3 322

allow

2

16 16(875 lb-ft)(12 in./ft) 4.456338 in.(12,000 psi)

1.645607 in. 1.646 in.

TD

D

π τ π≥ = =

∴ ≥ = Ans.


6.27 A torque of TD = 1,200 N-m is applied to gearD of the gear train shown in Fig. P6.27. The bearingsshown allow the shafts to rotate freely. (a) Determine the torque TA required for equilibriumof the system. (b) Assume shafts (1) and (2) are solid 30-mm-diameter steel shafts. Determine the magnitude ofthe 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 eachshaft. Fig. P6.27

Solution (a) Torque TA required for equilibrium:

150 mm 150 mm(1,200 N-m) 782.609 N-m230 mm 230 mmA DT T= = =

(b) Shear stress magnitudes if shafts are solid 30-mm-diameter:

4 4 4(30 mm) 79,521.564 mm32 32pI Dπ π= = =

1 11 4

1

(782.609 N-m)(30 mm/2)(1,000 mm/m) 147.622 MPa 147.6 MPa79,521.564 mmp

T RI

τ = = = = Ans.

2 22 4

2

(1,200 N-m)(30 mm/2)(1,000 mm/m) 226.354 MPa 226 MPa79,521.564 mmp

T RI

τ = = = = Ans.

(c) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter D:

4

332/ 2 16

p

p

DITR TDI R D

ππτ

τ= ∴ = = =


3 311 2

allow

1

16 16(782.609 N-m)(1,000 mm/m) 66,429.915 mm(60 N/mm )

40.500 mm 40.5 mm

TD

D

π τ π≥ = =

∴ ≥ = Ans.

3 322 2

allow

2

16 16(1,200 N-m)(1,000 mm/m) 101,859.164 mm(60 N/mm )

46.702 mm 46.702 mm

TD

D

π τ π≥ = =

∴ ≥ = Ans.


6.28 The gear train system shown in Fig. P6.28includes shafts (1) and (2), which are solid 1.375-in.-diameter steel shafts. The allowable shear stress ofeach shaft is 5,000 psi. The bearings shown allowthe shafts to rotate freely. Determine the maximumtorque TD that can be applied to the system withoutexceeding the allowable shear stress in either shaft.

Fig. P6.28

Solution Section properties:

4 4 41 2(1.375 in.) 0.350922 in.

32 32p pI D Iπ π= = = =

Maximum torque in either shaft:

4

allow 1max

1

(5,000 psi)(0.350922 in. ) 2,552.16 lb-in.1.375 in./2

pIT

Rτ

= = =

Torque relationship:

1 21 2 2 1

8 in.or 1.60 controls8 in. 5 in. 5 in.T T

T T T T= = = ∴

The torque in shaft (1) controls; therefore, T1 = 2,552.16 lb-in. Consequently, the maximum torque in shaft (2) must be limited to:

2 15 in. 0.625(2,552.16 lb-in.) 1,595.10 lb-in. 132.9 lb-ft8 in.

T T= = = = Ans.


6.29 The gear train system shown in Fig. P6.29includes shafts (1) and (2), which are solid 20-mm-diameter steel shafts. The allowable shear stress ofeach shaft is 30 MPa. The bearings shown allow theshafts to rotate freely. Determine the maximumtorque TD that can be applied to the system withoutexceeding the allowable shear stress in either shaft.

Fig. P6.29


4 4 41 2(20 mm) 15,707.963 mm

32 32p pI D Iπ π= = = =

Maximum torque in either shaft:

2 4

allow 1max

1

(30 N/mm )(15,707.963 mm ) 47,123.890 N-mm20 mm/2

pIT

Rτ

= = =


1 21 2 2 1

200 mmor 2.50 controls200 mm 80 mm 80 mm

T TT T T T= = = ∴

The torque in shaft (1) controls; therefore, T1 = 47,123.890 N-mm. Consequently, the maximum torque in shaft (2) must be limited to:

2 180 mm 0.4(47,123.890 N-mm) 18,849.556 N-mm 18.85 N-m200 mm

T T= = = = Ans.


6.30 In the gear system shown in Fig. P6.30, themotor applies a 160 lb-ft torque to the gear at A. A torque of TC = 250 lb-ft is removed from the shaftat gear C, and the remaining torque is removed atgear D. Segments (1) and (2) are solid 1.5-in.-diameter steel [G = 12,000 ksi] shafts, and thebearings 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 relative to gear B.

Fig. P6.30


4 4 41 2(1.5 in.) 0.497010 in.

32 32p pI D Iπ π= = = =


10 in.hence 2.50 2.50(160 lb-ft) 400 lb-ft4 in. 10 in. 4 in.

A BB A A

T T T T T= = = = =

Shaft (1) has the same torque as gear B; therefore, T1 = 400 lb-ft. A torque of 250 lb-ft is removed from the shaft at gear C; therefore, TC = 250 lb-ft. The torque in shaft (2) must satisfy equilibrium:

1 2

2 1

0400 lb-ft 250 lb-ft 150 lb-ft

x C

C

M T T TT T T

Σ = − + + =∴ = − = − =

To summarize: 1 2400 lb-ft 150 lb-ftT T= =

(a) Maximum shear stress in shafts (1) and (2):

1 11 4

1


T RI

τ = = = = Ans.

2 22 4

2


T RI

τ = = = = Ans.

(b) Angles of twist in shafts (1) and (2):

1 11 4

1 1

(400 lb-ft)(60 in.)(12 in./ft) 0.048289 rad(12,000,000 psi)(0.497010 in. )p

T LG I

φ = = =

2 22 4

2 2

(150 lb-ft)(40 in.)(12 in./ft) 0.012072 rad(12,000,000 psi)().497010 in. )p

T LG I

φ = = =

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.

1 2

say 0 rad (since this is our reference)

0 rad 0.048289 rad 0.012072 rad 0.060361 rad 0.0604 rad

B

D B

φφ φ φ φ

== + +

= + + = = Ans.


6.31 In the gear system shown in Fig. P6.31, the motor applies a 300 lb-ft torque to the gear at A. A torque of TC = 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 = 12,000 ksi] shafts, and the bearings shown allow free rotation ofthe shaft. (a) Determine the minimum permissiblediameters for segments (1) and (2) of the shaft ifthe maximum shear stress must not exceed 6,000psi. (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 Drelative to gear B must not exceed 0.10 rad.

Fig. P6.31

Solution Torque relationship: The motor applies a torque of 300 lb-ft to gear A. Since gear B is bigger than gear A, the torque on gear B will be increased in proportion to the gear ratio:

10 in.hence 2.50 2.50(300 lb-ft) 750 lb-ft4 in. 10 in. 4 in.

A BB A A

T T T T T= = = = =

Let the positive x axis for shaft BCD extend from gear B toward gear D. In order for shaft BCD to be in equilibrium with the torques as shown on gears C and D, the torque at B must act in the negative x direction. Consequently, TB = −750 lb-ft. Draw a free-body diagram that cuts through shaft (1) and includes gear B. From this FBD, the torque in shaft (1) is: 1 1750 lb-ft 0 750 lb-ftxM T TΣ = − + = ∴ = 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, TC = 500 lb-ft. Draw a FBD that cuts through shaft (2) and includes both gears B and C. From this FBD, the torque in shaft (2) is:

2

2

750 lb-ft 500 lb-ft 0250 lb-ft

xM TT

Σ = − + + =∴ =

To summarize: 1 2750 lb-ft 250 lb-ftT T= = (a) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter D:

4

332/ 2 16

p

p

DITR TDI R D

ππτ

τ= ∴ = = =



3 311

allow

1

16 16(750 lb-ft)(12 in./ft) 7.639437 in.(6,000 psi)

1.969490 in. 1.969 in.

TD

D

π τ π≥ = =

∴ ≥ = Ans.

3 322

allow

2

16 16(250 lb-ft)(12 in./ft) 2.546479 in.(6,000 psi)

1.365568 in. 1.366 in.

TD

D

π τ π≥ = =

∴ ≥ = Ans. (b) The rotation angle of gear D with respect to gear B is equal to the sum of the angles of twist in shaft segments (1) and (2): 1 2 0.10 radD Bφ φ φ φ− = + ≤ Since the same solid steel shaft is to be used for both segments (1) and (2):

[ ]1 1 2 21 1 2 2

1 1 2 2

10.01 radp p p

T L T L T L T LG I G I GI

≥ + = +

or rearranging:

4 4(750 lb-ft)(60 in.)(12 in./ft) (250 lb-ft)(40 in.)(12 in./ft) 0.5500 in.32 (12,000,000 psi)(0.1 rad)

1.538 in.

pI D

D

π += ≥ =

∴ ≥ 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 D1 determined in part (a), the minimum diameter required for a constant diameter shaft between gears B and D is: min 1.969 in.D = Ans.


6.32 In the gear system shown in Fig. P6.32, themotor applies a torque of 220 N-m to the gear atA. A torque of TC = 400 N-m is removed from theshaft at gear C, and the remaining torque isremoved at gear D. Segments (1) and (2) are solid40-mm-diameter steel [G = 80 GPa] shafts, andthe bearings shown allow free rotation of theshaft. (a) Determine the maximum shear stress insegments (1) and (2) of the shaft. (b) Determine the rotation angle of gear Drelative to gear B.

Fig. P6.32


4 4 41 2(40 mm) 251,327.41 mm

32 32p pI D Iπ π= = = =


300 mmhence 3 3(220 N-m) 660 N-m100 mm 300 mm 100 mm

A BB A A

T TT T T= = = = =

Shaft (1) has the same torque as gear B; therefore, T1 = 660 N-m. A torque of TC = 400 N-m is removed from the shaft at gear C. The torque in shaft (2) must satisfy equilibrium:

1 2

2 1

0660 N-m 400 N-m 260 N-m

x C

C

M T T TT T T

Σ = − + + =∴ = − = − =

To summarize:

1

2

660 N-m260 N-m

TT

==

(a) Maximum shear stress in shafts (1) and (2):

1 11 4

1

(660 N-m)(40 mm/2)(1,000 mm/m) 52.521 MPa 52.5 MPa251,327.41 mmp

T RI

τ = = = = Ans.

2 22 4

2


T RI

τ = = = = Ans.



2

1 11 2 4

1 1

(660 N-m)(1.5 m)(1,000 mm/m) 0.049239 rad(80,000 N/mm )(251,327.41 mm )p

T LG I

φ = = =

2

2 22 2 4

2 2

(260 N-m)(1.0 m)(1,000 mm/m) 0.012931 rad(80,000 N/mm )(251,327.41 mm )p

T LG I

φ = = =

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.

1 2

say 0 rad (since this is our reference)

0 rad 0.049239 rad 0.012931 rad 0.062170 rad 0.0622 rad

B

D B

φφ φ φ φ

== + +

= + + = = Ans.


6.33 In the gear system shown in Fig. P6.33, themotor applies a torque of 400 N-m to the gear atA. A torque of TC = 700 N-m is removed from theshaft at gear C, and the remaining torque isremoved at gear D. Segments (1) and (2) are solidsteel [G = 80 GPa] shafts, and the bearings shownallow free rotation of the shaft. (a) Determine the minimum permissiblediameters for segments (1) and (2) of the shaft ifthe maximum shear stress must not exceed 40MPa. (b) If the same diameter is to be used forsegments (1) and (2), determine the minimumpermissible diameter that can be used for theshaft if the maximum shear stress must notexceed 40 MPa and the rotation angle of gear Drelative to gear B must not exceed 3.0°. Fig. P6.33

Solution Torque relationship:

300 mmhence 3 3(400 N-m) 1,200 N-m100 mm 300 mm 100 mm

A BB A A

T TT T T= = = = =

Shaft (1) has the same torque as gear B; therefore, T1 = 1,200 N-m. A torque of TC = 700 N-m is removed from the shaft at gear C. The torque in shaft (2) must satisfy equilibrium:

1 2

2 1

01,200 N-m 700 N-m 500 N-m

x C

C

M T T TT T T

Σ = − + + =∴ = − = − =

To summarize: 1 21, 200 N-m 500 N-mT T= =

(a) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter D:

4

332/ 2 16

p

p

DITR TDI R D

ππτ

τ= ∴ = = =


3 311 2

allow

1

16 16(1,200 N-m)(1,000 mm/m) 152,788.745 mm(40 N/mm )

53.460 mm 53.5 mm

TD

D

π τ π≥ = =

∴ ≥ = Ans.

3 322 2

allow

2

16 16(500 N-m)(1,000 mm/m) 63,661.977 mm(40 N/mm )

39.929 mm 39.9 mm

TD

D

π τ π≥ = =

∴ ≥ = Ans.


(b) The rotation angle of gear D with respect to gear B is equal to the sum of the angles of twist in shaft segments (1) and (2): 1 2 (3 )( /180 ) 0.052360 radD Bφ φ φ φ π− = + ≤ ° ° = Since the same solid steel shaft is to be used for both segments (1) and (2):

[ ]1 1 2 21 1 2 2

1 1 2 2

10.052360 radp p p

T L T L T L T LG I G I GI

≥ + = +

or rearranging:

2 24 4

2

(1,200 N-m)(1.5 m)(1,000 mm/m) (500 N-m)(1.0 m)(1,000 mm/m) 549,083.270 mm32 (80,000 N/mm )(0.052360 rad)

48.6 mm

pI D

D

π += ≥ =

∴ ≥ 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 D1 determined in part (a), the minimum diameter required for a constant diameter shaft between gears B and D is: min 53.5 mmD = Ans.


6.34 In the gear system shown in Fig. P6.34, themotor applies a torque of 250 N-m to the gear at A. Shaft (1) is a solid 35-mm-diameter shaft, andshaft (2) is a solid 50-mm-diameter shaft. Thebearings shown allow free rotation of the shafts. (a) Determine the torque TE provided by the gearsystem at gear E. (b) Determine the maximum shear stresses inshafts (1) and (2).

Fig. P6.34

Solution (a) Torque relationship: The torque on gear B is:

72 3 3(250 N-m) 750 N-m24B A AT T T= = = =

Shaft (1) has the same torque as gear B; therefore, T1 = 750 N-m and TC = 750 N-m. The torque on gear D is found from:

60 2 2(750 N-m) 1,500 N-m30D C CT T T= = = =

Shaft (2) has the same torque as gear D; therefore, T2 = 1,500 N-m and 1,500 N-mET = Ans. To summarize: 1 2750 N-m 1,500 N-mT T= = (b) Section properties:

4 4 41 1

4 4 42 2

(35 mm) 147,323.51 mm32 32

(50 mm) 613,592.32 mm32 32

p

p

I D

I D

π π

π π

= = =

= = =

Maximum shear stress in shafts (1) and (2):

1 11 4

1


T RI

τ = = = = Ans.

2 22 4

2

(1,500 N-m)(50 mm/2)(1,000 mm/m) 61.115 MPa 61.1 MPa613,592.32 mmp

T RI

τ = = = = Ans.


6.35 In the gear system shown in Fig. P6.35, themotor applies a torque of 600 N-m to the gear at A. Shafts (1) and (2) are solid shafts, and the bearingsshown allow free rotation of the shafts. (a) Determine the torque TE provided by the gear system at gear E. (b) If the allowable shear stress in each shaft mustbe limited to 50 MPa, determine the minimumpermissible diameter for each shaft.

Fig. P6.35

Solution (a) Torque relationship: The torque on gear B is:

72 3 3(600 N-m) 1,800 N-m24B A AT T T= = = =

Shaft (1) has the same torque as gear B; therefore, T1 = 1,800 N-m and TC = 1,800 N-m. The torque on gear D is found from:

60 2 2(1,800 N-m) 3,600 N-m30D C CT T T= = = =

Shaft (2) has the same torque as gear D; therefore, T2 = 3,600 N-m and 3,600 N-mET = Ans. To summarize: 1 21,800 N-m 3,600 N-mT T= = (b) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter D:

4

332/ 2 16

p

p

DITR TDI R D

ππτ

τ= ∴ = = =


3 311 2

allow

1

16 16(1,800 N-m)(1,000 mm/m) 183,346.494 mm(50 N/mm )

56.810 mm 56.8 mm

TD

D

π τ π≥ = =

∴ ≥ = Ans.

3 322 2

allow

2

16 16(3,600 N-m)(1,000 mm/m) 366,692.989 mm(50 N/mm )

71.576 mm 71.6 mm

TD

D

π τ π≥ = =

∴ ≥ = Ans.


6.36 In the gear system shown in Fig. P6.36, atorque of TE = 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 togear A. (b) If the allowable shear stress in each shaft mustbe limited to 4,000 psi, determine the minimumpermissible diameter for each shaft.

Fig. P6.36

Solution (a) Torque relationship: The torque on gear E creates a torque in shaft (2) of T2 = 900 lb-ft. Accordingly, the torque applied to gear D is also equal to 900 lb-ft. The torque on gear C is found with the gear ratio:

30 0.5 0.5(900 lb-ft) 450 lb-ft60C D DT T T= = = =

Shaft (1) has the same torque as gear C; therefore, T1 = 450 lb-ft and TB = 450 N-m. The torque on gear A is found from:

24 450 lb-ft 150 lb-ft72 3 3

CA B

TT T= = = = Ans.

To summarize: 1 2450 lb-ft 900 lb-ftT T= = (b) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter D:

4

332/ 2 16

p

p

DITR TDI R D

ππτ

τ= ∴ = = =


3 311

allow

1

16 16(450 lb-ft)(12 in./ft) 6.875494 in.(4,000 psi)

1.9015 in. 1.902 in.

TD

D

π τ π≥ = =

∴ ≥ = Ans.

3 322

allow

2

16 16(900 lb-ft)(12 in./ft) 13.750987 in.(4,000 psi)

2.3958 in. 2.40 in.

TD

D

π τ π≥ = =

∴ ≥ = Ans.


6.37 In the gear system shown in Fig. P6.37, atorque of TE = 720 lb-ft is delivered at gear E. Shaft (1) is a solid 1.50-in.-diameter shaft, andshaft (2) is a solid 2.00-in.-diameter shaft. Thebearings shown allow free rotation of the shafts. (a) Determine the torque provided by the motor togear A. (b) Determine the maximum shear stresses inshafts (1) and (2).

Fig. P6.37

Solution (a) Torque relationship: The torque on gear E creates a torque in shaft (2) of T2 = 720 lb-ft. Accordingly, the torque applied to gear D is also equal to 720 lb-ft. The torque on gear C is found with the gear ratio:

30 0.5 0.5(720 lb-ft) 360 lb-ft60C D DT T T= = = =

Shaft (1) has the same torque as gear C; therefore, T1 = 360 lb-ft and TB = 360 N-m. The torque on gear A is found from:

24 360 lb-ft 120 lb-ft72 3 3

CA B

TT T= = = = Ans.

To summarize: 1 2360 lb-ft 720 lb-ftT T= = (b) Section properties:

4 4 41 1

4 4 42 2

(1.50 in.) 0.497010 in.32 32

(2.00 in.) 1.570796 in.32 32

p

p

I D

I D

π π

π π

= = =

= = =

Maximum shear stress in shafts (1) and (2):

1 11 4

1


T RI

τ = = = = Ans.

2 22 4

2


T RI

τ = = = = Ans.


6.38 Two solid 30-mm-diameter steel shafts areconnected by the gears shown in Fig. P6.38. Theshaft lengths are L1 = 300 mm and L2 = 500 mm.Assume that the shear modulus of both shafts isG = 80 GPa and that the bearings shown allowfree rotation of the shafts. If the torque applied atgear D is TD = 160 N-m, (a) determine the internal torques T1 and T2 in the two shafts. (b) determine the angles of twist φ1 and φ2. (c) determine the rotation angles φB and φC of gears B and C. (d) determine the rotation angle of gear D.

Fig. P6.38

Solution (a) Torque relationship: The torque on gear D creates a torque in shaft (2) of T2 = 160 N-m. 2 20 160 N-mx D DM T T T TΣ = − = ∴ = = Accordingly, the torque applied to gear C is also equal to 160 N-m. The torque on gear B is found with the gear ratio:

54 teeth 1.8 1.8(160 N-m) 288 N-m30 teethB C CT T T= − = − = − = −

Shaft (1) has the same torque as gear B; therefore, T1 = −288 N-m. To summarize: 1 2288 N-m 160 N-mT T= − = Ans. Section properties:

4 4 41 1 2(30 mm) 79,521.56 mm

32 32p pI D Iπ π= = = =


1 11 2 4

1 1

( 288 N-m)(300 mm)(1,000 mm/m) 0.013581 rad(80,000 N/mm )(79,521.56 mm )p

T LG I

φ −= = = − Ans.

2 22 2 4

2 2

(160 N-m)(500 mm)(1,000 mm/m) 0.012575 rad(80,000 N/mm )(79,521.56 mm )p

T LG I

φ = = = Ans.

(c) Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft (1): 1 1 0 rad ( 0.013581 rad) 0.01358 radB A B Aφ φ φ φ φ φ= − ∴ = + = + − = − Ans. As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the rotation is dictated by the gear ratio:

54 teeth 54 teeth( 0.013581 rad) 0.024446 rad 0.0244 rad30 teeth 30 teethC Bφ φ= − = − − = = Ans.

The rotation of gear D is found from: 2 2 0.024446 rad 0.012575 rad 0.037021 rad 0.0370 radD C D Cφ φ φ φ φ φ= − ∴ = + = + = = Ans.


6.39 Two solid 1.75-in.-diameter steel shafts areconnected by the gears shown in Fig. P6.39. Theshaft lengths are L1 = 6 ft and L2 = 10 ft. Assumethat the shear modulus of both shafts is G = 12,000 ksi and that the bearings shown allowfree rotation of the shafts. If the torque applied atgear D is TD = 225 lb-ft, (a) determine the internal torques T1 and T2 in the two shafts. (b) determine the angles of twist φ1 and φ2. (c) determine the rotation angles φB and φC of gears B and C. (d) determine the rotation angle of gear D.

Fig. P6.39

Solution (a) Torque relationship: The torque on gear D creates a torque in shaft (2) of T2 = 225 lb-ft. 2 20 225 lb-ftx D DM T T T TΣ = − = ∴ = = Accordingly, the torque applied to gear C is also equal to 225 lb-ft. The torque on gear B is found with the gear ratio:

54 teeth 1.8 1.8(225 lb-ft) 405 lb-ft30 teethB C CT T T= − = − = − = −

Shaft (1) has the same torque as gear B; therefore, T1 = −405 lb-ft. To summarize: 1 2405 lb-ft 225 lb-ftT T= − = Ans. Section properties:

4 4 41 1 2(1.75 in.) 0.920772 in.

32 32p pI D Iπ π= = = =


2

1 11 4

1 1

( 405 lb-ft)(6 ft)(12 in./ft) 0.031669 rad 0.0317 rad(12,000,000 psi)(0.920772 in. )p

T LG I

φ −= = = − = − Ans.

2

2 22 4

2 2

(225 lb-ft)(10 ft)(12 in./ft) 0.029323 rad 0.0293 rad(12,000,000 psi)(0.920772 in. )p

T LG I

φ = = = = Ans.

(c) Rotation angles of gears B and C: The rotation of gear B is found from the twist angle in shaft (1): 1 1 0 rad ( 0.031669 rad) 0.0317 radB A B Aφ φ φ φ φ φ= − ∴ = + = + − = − Ans. As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the rotation is dictated by the gear ratio:

54 teeth 54 teeth( 0.031669 rad) 0.057004 rad 0.0570 rad30 teeth 30 teethC Bφ φ= − = − − = = Ans.

The rotation of gear D is found from: 2 2 0.057004 rad 0.029323 rad 0.086327 rad 0.0863 radD C D Cφ φ φ φ φ φ= − ∴ = + = + = = Ans.


6.40 Two solid steel shafts are connected by thegears shown in Fig. P6.40. The designrequirements for the system require (1) that bothshafts must have the same diameter, (2) that themaximum shear stress in each shaft must be lessthan 6,000 psi, and (3) that the rotation angle ofgear D must not exceed 3°. Determine theminimum required diameter of the shafts if thetorque applied at gear D is TD = 1,000 lb-ft. The shaft lengths are L1 = 10 ft and L2 = 8 ft.Assume that the shear modulus of both shafts isG = 12,000 ksi and that the bearings shownallow free rotation of the shafts. Fig. P6.40

Solution Torque relationship: The torque on gear D creates a torque in shaft (2) of T2 = 1,000 lb-ft. 2 20 1,000 lb-ft 12,000 lb-in.x D DM T T T TΣ = − = ∴ = = = Accordingly, the torque applied to gear C is also 12,000 lb-in. The torque on gear B is found with the gear ratio:

48 teeth 0.6667 0.6667(12,000 lb-in.) 8,000 lb-in.72 teethB C CT T T= − = − = − = −

Shaft (1) has the same torque as gear B; therefore, T1 = −8,000 lb-in. Diameters based on allowable shear stresses: The elastic torsion formula gives the relationship between shear stress and torque in a shaft.

p

TRI

τ =

The torques in both shafts have been 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:

pI TR τ

=

Express the left-hand side of this equation in terms of the shaft diameter D:

4

332/ 2 16

D TDD

ππ

τ= =

Solve for the minimum acceptable diameter in shaft (1):

3 311 2

1

16 16(8,000 lb-in.) 6.790614 in.(6,000 lb/in. )

1.894 in.

TD

Dπ τ π

≥ = =

∴ ≥



3 322 2

2

16 16(12,000 lb-in.) 10.185916 in.(6,000 lb/in. )

2.17 in.

TD

Dπ τ π

≥ = =

∴ ≥ Of these two 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:

1 1 2 21 2

1 1 2 2p p

T L T LG I G I

φ φ= =

The rotations of gears B and C are related since the arclengths turned by the two gears have the same magnitude. The gears turn in opposite directions; therefore, a negative sign is introduced. C C B BR Rφ φ= − The rotation angle of gear B is equal to the angle of twist in shaft (1): φB = φ1. Therefore, the rotation angle of gear C can be expressed in terms of T1.

1 1 1 11

1 1 1 1

B B B BC B

C C C p C p

R R R T L N T LR R R G I N G I

φ φ φ= − = − = − = −

The rotation angle of gear D is equal to the rotation angle of gear C plus the twist that occurs in shaft (2): 2D Cφ φ φ= + and so the rotation angle of gear D can be expressed in terms of the torques T1 and T2:

1 1 2 2

1 1 2 2

3BD

C p p

N T L T LN G I G I

φ = − + ≤ °

Shafts having the same diameters and the same shear moduli are required for this system; therefore, Ip1 = Ip2 = Ip and G1 = G2 = G. Factor these terms out to obtain:

1 1 2 2 1 1 2 21 13

(3 )B B

pp C C

N NT L T L I T L T LG I N G N

⎡ ⎤ ⎡ ⎤− + ≤ ° ∴ ≥ − +⎢ ⎥ ⎢ ⎥°⎣ ⎦ ⎣ ⎦

Express the polar moment of inertia in terms of diameter to obtain the following relationship:

( )

41 1 2 2

4

32(3 )

32 48 teeth ( 8,000 lb-in.)(120 in.) (12,000 lb-in.)(96 in.) rad 72 teeth(12,000,000 psi)(3 ) 180

29.050810 in.2.32 in.

B

C

ND T L T LG N

D

π

ππ

⎡ ⎤≥ − +⎢ ⎥° ⎣ ⎦

⎡ ⎤= − − +⎢ ⎥⎣ ⎦° °=∴ ≥

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 ≥ 2.32 in. Ans.


6.41 Two solid steel shafts are connected by thegears shown in Fig. P6.41. The designrequirements for the system require (1) that bothshafts must have the same diameter, (2) that themaximum shear stress in each shaft must be lessthan 40 MPa, and (3) that the rotation angle ofgear D must not exceed 3°. Determine theminimum required diameter of the shafts if thetorque applied at gear D is TD = 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 isG = 80 GPa and that the bearings shown allowfree rotation of the shafts. Fig. P6.41

Solution Torque relationship: The torque on gear D creates a torque in shaft (2) of T2 = 1,200 N-m. 2 20 1,200 N-mx D DM T T T TΣ = − = ∴ = = Accordingly, the torque applied to gear C is also 1,200 N-m. The torque on gear B is found with the gear ratio:

48 teeth 0.6667 0.6667(1,200 N-m) 800 N-m72 teethB C CT T T= − = − = − = −

Shaft (1) has the same torque as gear B; therefore, T1 = −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:

3

16p

TR TDI

πττ

= ⇒ =


3 311 2

1

16 16(800 N-m)(1,000 mm/m) 101,859.16 mm(40 N/mm )

46.70 mm

TD

Dπ τ π

≥ = =

∴ ≥ Solve for the minimum acceptable diameter in shaft (2):

3 322 2

2

16 16(1, 200 N-m)(1,000 mm/m) 152,788.75 mm(40 N/mm )

53.46 mm

TD

Dπ τ π

≥ = =

∴ ≥ 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:

1 1 2 21 2

1 1 2 2p p

T L T LG I G I

φ φ= =


The rotations of gears B and C are related since the arclengths turned by the two gears have the same magnitude. The gears turn in opposite directions; therefore, a negative sign is introduced. C C B BR Rφ φ= − The rotation angle of gear B is equal to the angle of twist in shaft (1): φB = φ1. Therefore, the rotation angle of gear C can be expressed in terms of T1.

1 1 1 11

1 1 1 1

B B B BC B

C C C p C p

R R R T L N T LR R R G I N G I

φ φ φ= − = − = − = −

The rotation angle of gear D is equal to the rotation angle of gear C plus the twist that occurs in shaft (2): 2D Cφ φ φ= + and so the rotation angle of gear D can be expressed in terms of the torques T1 and T2:

1 1 2 2

1 1 2 2

3BD

C p p

N T L T LN G I G I

φ = − + ≤ °

Shafts having the same diameters and the same shear moduli are required for this system; therefore, Ip1 = Ip2 = Ip and G1 = G2 = G. Factor these terms out to obtain:

1 1 2 2 1 1 2 21 13

(3 )B B

pp C C

N NT L T L I T L T LG I N G N

⎡ ⎤ ⎡ ⎤− + ≤ ° ∴ ≥ − +⎢ ⎥ ⎢ ⎥°⎣ ⎦ ⎣ ⎦

Express the polar moment of inertia in terms of diameter to obtain the following relationship:

41 1 2 2

2

2

4

32(3 )

48 teeth32 ( 800 N-m)(2.5 m) (1,200 N-m)(2.0 m) (1,000 mm/m)72 teeth

(80,000 N/mm )(3 )( /180 )9,078,378.05 mm

54.89 mm

B

C

ND T L T LG N

D

π

π π

⎡ ⎤≥ − +⎢ ⎥° ⎣ ⎦

⎡ ⎤− − +⎢ ⎥⎣ ⎦=° °

=∴ ≥

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 ≥ 54.9 mm. Ans.

mechanics of materials solutions chapter06 probs26 41

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