RECONOCIMIENTO DE LA ZONA. 178

Aluminum segment (1) is a tube with an outside diameter of D1 = 4.00 in., a wall thickness of t1= 0.25 in., and a shear modulus of G1 = 4,000 ksi. Steel segment (2) is a tube with an outside diameter of D2= 2.50 in., a wall thickness of t2 = 0.125 in., and a shear modulus of G2 = 12,000 ksi. The compound shaft is subjected to torques applied at B and C, as shown in Fig. P6.20.

(a) Prepare a diagram that shows the internal torque and the maximum shear stress in segments (1), and (2) of the shaft. Use the sign convention presented in Section 6-6.

(b) Determine the rotation angle of B with respect to the support at A.

(c) Determine the rotation angle of C with respect to the support at A.

Fig. P6.20 Solution

Section properties: Polar moments of inertia in the shaft segments will be needed for this calculation.

4 4 4 4 4

650 lb-ft 1,700 lb-ft 0 1,050 lb-ft

(1,050 lb-ft)(4.00 in./2)(12 in./ft)

2, 422.98 psi 2, 420 psi

( 650 lb-ft)(2.50 in./2)(12 in./ft)

7,393.08 psi 7,390 psi 1.3188 in.

p

T R

τ = I = ⁻ = − = − Ans.

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(b) Rotation angle of B with respect to A:

1 _{B A B} 0 _B

φ φ φ= − =φ − = φ Therefore,

2

1 1 1 4

1 1

(1,050 lb-ft)(9 ft)(12 in./ft)

0.032710 rad 0.0327 rad (4,000,000 psi)(10.4004 in. )

B

p

T L

φ = =φ G I = = = Ans.

(c) Rotation angle of C with respect to A:

2 C B C B 2

φ φ φ= − ∴ =φ φ φ+ The angle of twist in shaft (2) is

2 2 2

2 4

2 2

( 650 lb-ft)(6 ft)(12 in./ft)

0.035486 rad (12,000,000 psi)(1.3188 in. )

p

T L

φ =G I = ⁻ = −

and thus, the rotation angle at C is

2 0.032710 rad ( 0.035486 rad) 0.002776 rad 0.00278 rad

C B

φ ⁼φ φ^{+ = + − = − = −} Ans.

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6.21 A solid 1.00-in.-diameter steel [G = 12,000 ksi] shaft is subjected to the torques shown in Fig. P6.21.

(a) Prepare a diagram that shows the internal torque and the maximum shear stress in segments (1), (2), and (3) of the shaft. Use the sign convention presented in Section 6-6.

(a) Determine the rotation angle of pulley C with respect to the support at A.

(b) Determine the rotation angle of pulley D with respect to the support at A.

Fig. P6.21 Solution

Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segments will be needed for this calculation.

4 4 4

240 lb-ft 215 lb-ft 110 lb-ft 0 135 lb-ft

215 lb-ft 110 lb-ft 0 105 lb-ft

( 135 lb-ft)(1.00 in./2)(12 in./ft)

8, 250.57 psi 8, 250 psi

(105 lb-ft)(1.00 in./2)(12 in./ft)

6, 417.11 psi 6, 420 psi

( 110 lb-ft)(1.00 in./2)(12 in./ft)

6,722.69 psi 6,720 psi 0.098175 in.

p

T R

τ = I = ⁻ = − = −

The maximum shear stress in the entire shaft is τ^max = 8,250 psi. Ans.

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(b) Rotation angle of C with respect to A:

The angles of twist in the three shaft segments are:

1 1

1 4

1 1

( 135 lb-ft)(36 in.)(12 in./ft)

0.049503 rad (12,000,000 psi)(0.098175 in. )

p

T L

φ =G I = ⁻ = −

2 2

2 4

2 2

(105 lb-ft)(36 in.)(12 in./ft)

0.038503 rad (12,000,000 psi)(0.098175 in. )

p

T L

φ =G I = =

3 3

3 4

3 3

( 110 lb-ft)(36 in.)(12 in./ft)

0.040336 rad (12,000,000 psi)(0.098175 in. )

p

T L

φ =G I = ⁻ = −

The rotation angle of C with respect to A is found from the sum of the angles of twist in segments (1) and (2):

1 2 0.049503 rad 0.038503 rad 0.011000 rad 0.01100 rad

φC = + = −φ φ + = − = − Ans.

(c) Rotation angle of D with respect to A:

The rotation angle of D with respect to A is found from the sum of the angles of twist in segments (1), (2), and (3):

1 2 3

0.049503 rad 0.038503 rad ( 0.040336 rad) 0.051336 rad 0.0513 rad

φD = + +φ φ φ

= − + + −

= − = − Ans.

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6.22 A compound shaft supports several pulleys as shown in Fig. P6.22. Segments (1) and (4) are solid 25-mm-diameter steel [G = 80 GPa] shafts.

Segments (2) and (3) are solid 50-mm-diameter steel shafts. The bearings shown allow the shaft to turn freely.

(a) Prepare a diagram that shows the internal torque and the maximum shear stress in segments (1), (2), (3), and (4) of the shaft. Use the sign convention presented in Section 6-6.

(b) Determine the rotation angle of pulley D with respect to pulley B.

(c) Determine the rotation angle of pulley E with respect to pulley A.

Fig. P6.22 Solution

Section properties: The polar moments of inertia for the shaft segments will be needed for this calculation.

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4

( 150 N-m)(25 mm/2)(1,000 mm/m)

48.892 MPa 48.9 MPa 38,349.52 mm

(1,700 N-m)(50 mm/2)(1,000 mm/m)

69.264 MPa 69.3 MPa 613,592.32 mm

(380 N-m)(50 mm/2)(1,000 mm/m)

15.48 MPa

( 190 N-m)(25 mm/2)(1,000 mm/m)

61.930 MPa 61.9 MPa 38,349.52 mm

p

T R

τ = I = ⁻ = − = −

The maximum shear stress in the entire shaft is τ^max = 69.3 MPa. Ans.

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Angles of twist:

The angles of twist in the four shaft segments are:

1 1

1 2 4

1 1

( 150 N-m)(750 mm)(1,000 mm/m)

0.036669 rad (80,000 N/mm )(38,349.52 mm )

p

T L

φ =G I = ⁻ = −

2 2 2 2 4

2 2

(1,700 N-m)(500 mm)(1,000 mm/m)

0.017316 rad (80,000 N/mm )(613,592.32 mm )

p

(380 N-m)(625 mm)(1,000 mm/m)

0.004838 rad (80,000 N/mm )(613,592.32 mm )

p

( 190 N-m)(550 mm)(1,000 mm/m)

0.034062 rad (80,000 N/mm )(38,349.52 mm )

p

T L

φ =G I = ⁻ = −

(b) Rotation angle of pulley D with respect to pulley B: The rotation angle of D with respect to B is found from the sum of the angles of twist in segments (2) and (3):

/ 2 3 0.017316 rad 0.004838 rad 0.022154 rad 0.0222 rad

φD B = + =φ φ + = = Ans.

(c) Rotation angle of pulley E with respect to pulley A: The rotation angle of E with respect to A is found from the sum of the angles of twist in all four segments:

1 2 3 4

0.036669 rad 0.017316 rad 0.004838 rad ( 0.034062 rad) 0.048577 rad 0.0486 rad

φE = + + +φ φ φ φ

= − + + + −

= − = − Ans.

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6.23 A solid steel [G = 80 GPa] shaft of variable diameter is subjected to the torques shown in Fig.

P6.23. The diameter of the shaft in segments (1) and (3) is 50 mm, and the diameter of the shaft in segment (2) is 80 mm. The bearings shown allow the shaft to turn freely.

(a) Plot a torque diagram showing the internal torque in segments (1), (2), and (3) of the shaft. Use the sign convention presented in Section 6-6.

(b) Add a plot to the torque diagram that shows the maximum shear stress magnitude in each segment of the shaft.

(c) Add another plot to the torque diagram that shows the rotation angles along the shaft measured with respect to gear A.

Fig. P6.23

Solution

Section properties: The polar moments of inertia for the shaft segments will be needed for this calculation.

(1, 200 N-m)(50 mm/2)(1,000 mm/m)

48.892 MPa 48.9 MPa 613,592.32 mm

( 3,300 N-m)(80 mm/2)(1,000 mm/m)

32.826 MPa 32.8 MPa 4,021,238.60 mm

p

T R

τ = I = ⁻ = − = − Ans.

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3 3

3 4

3

( 500 N-m)(50 mm/2)(1,000 mm/m)

20.372 MPa 20.4 MPa 613,592.32 mm

p

T R

τ = I = ⁻ = − = − Ans.

The maximum shear stress in the entire shaft is τ^max = 48.9 MPa. Ans.

Angles of twist:

The angles of twist in the three shaft segments are:

2

1 1 1 2 4

1 1

(1, 200 N-m)(0.7 m)(1,000 mm/m)

0.017112 rad (80,000 N/mm )(613,592.32 mm )

p

( 3,300 N-m)(1.8 m)(1,000 mm/m)

0.018464 rad (80,000 N/mm )(4,021,238.60 mm )

p

( 500 N-m)(0.7 m)(1,000 mm/m)

0.007130 rad (80,000 N/mm )(613,592.32 mm )

p

0 rad 0.017112 rad 0.017112 rad

0.017112 rad ( 0.018464 rad) 0.001352 rad 0.001352 rad ( 0.007130 rad) 0.008482 rad

A

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6.24 A compound shaft drives three gears, as shown in Fig. P6.24. Segments (1) and (2) of the compound shaft are hollow aluminum [G = 4,000 ksi] tubes, which have an outside diameter of 3.00 in. and a wall thickness of 0.25 in. Segments (3) and (4) are solid 2.00-in.-diameter steel [G = 12,000 ksi] shafts. The bearings shown allow the shaft to turn freely.

(a) Prepare a diagram showing the internal torque and the maximum shear stress in segments (1), (2), (3), and (4) of the shaft.

Also, add a plot of the rotation angles along the shaft measured with respect to A.

(b) Determine the rotation angle of flange C with respect to flange A.

(c) Determine the rotation angle of gear E with respect to flange A.

Fig. P6.24 Solution

Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segments will be needed for this calculation.

4 4 4 4 4

14 kip-in. 42 kip-in. 35 kip-in. 0 7 kip-in.

14 kip-in. 42 kip-in. 0 28 kip-in.

14 kip-in. 42 kip-in. 0 28 kip-in.

Mx T

T

Σ = − − =

∴ = −

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4

( 7 kip-in.)(3.00 in./2)

2.55 ksi

( 28 kip-in.)(3.00 in./2)

10.20 ksi

( 28 kip-in.)(2.00 in./2)

17.83 ksi

(b) Angles of Twist: The angles of twist in the four shaft segments are:

1 1 1 4

1 1

(7 kip-in.)(60 in.)

0.025503 rad (4,000 ksi)(4.117204 in. )

p (4,000 ksi)(4.117204 in. )

p (12,000 ksi)(1.570796 in. )

p (12,000 ksi)(1.570796 in. )

p

0 rad 0.025503 rad 0.025503 rad

0.025503 rad ( 0.010201 rad) 0.015302 rad 0.015302 rad ( 0.026738 rad) 0.011436 rad

0.011436 rad 0.008913 rad 0.002523 rad

A

The rotation angle of flange C with respect to A is 0.01530 rad

φC = Ans.

(c) Rotation angle of E with respect to A:

0.00252 rad

φE = − Ans.

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6.25 A compound shaft drives several pulleys, as shown in Fig. P6.25. Segments (1) and (2) of the compound shaft are hollow aluminum [G = 4,000 ksi]

tubes, which have an outside diameter of 3.00 in. and a wall thickness of 0.125 in. Segments (3) and (4) are solid 1.50-in.-diameter steel [G = 12,000 ksi] shafts.

The bearings shown allow the shaft to turn freely.

(a) Prepare a diagram showing the internal torque and the maximum shear stress in segments (1), (2), (3), and (4) of the shaft. Also, add a plot of the rotation angles along the shaft measured with respect to A.

(b) Determine the rotation angle of flange C with respect to pulley A.

(c) Determine the rotation angle of pulley E with respect to pulley A.

Fig. P6.25

70 lb-ft 200 lb-ft 0 130 lb-ft

Mx T T

Σ = − − = ∴ = −

Note: The internal torque in shaft segment (2) is the same as in segment (3); therefore, T2 = –130 lb-ft.

1 480 lb-ft 0 1 480 lb-ft

Mx T T

Σ = − = ∴ =

Polar moments of inertia in the shaft segments will be needed for this calculation. Segments (1) and (2) are hollow tubes with an outside diameter of 3.00 in. and an inside diameter of 3.00 in. – 2(0.125 in.) = 2.75 in. The polar moment of inertia in segments (1) and (2) is:

4 4 4 4 4

Segments (3) and (4) are solid 1.50-in.-diameter shafts, which have a polar moment of inertia of:

4 4 4

3 3 (1.50 in.) 0.49701 in. 4

32 32

p p

I ₌ π D ₌ π _{= =}I

Shear stresses: The shear stresses in each segment can be calculated by the elastic torsion formula:

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1 1 1 4

1

(480 lb-ft)(1.50 in.)(12 in./ft)

3,696.4 psi 3,700 psi 2.33740 in.

( 130 lb-ft)(1.50 in.)(12 in./ft)

1,001.1 psi 1,001 psi 2.33740 in.

( 130 lb-ft)(0.75 in.)(12 in./ft)

2,354.1 psi 2,350 psi 0.49701 in.

(70 lb-ft)(0.75 in.)(12 in./ft)

1, 267.6 psi 1, 268 psi

(480 lb-ft)(64 in.)(12 in./ft)

0.039428 rad (4,000,000 lb/in. )(2.33740 in. )

p

T L

φ =G I = =

2 2 2 2 4

2 2

( 130 lb-ft)(32 in.)(12 in./ft)

0.005339 rad (4,000,000 lb/in. )(2.33740 in. )

p

( 130 lb-ft)(48 in.)(12 in./ft)

0.012555 rad (12,000,000 lb/in. )(0.49701 in. )

p

T L

φ =G I = ⁻ = −

4 4 4 2 4

4 4

(70 lb-ft)(48 in.)(12 in./ft)

0.006760 rad (12,000,000 lb/in. )(0.49701 in. )

p

T L

φ =G I = =

Rotation angles: The angles of twist can be defined in terms of the rotation angles at the ends of each segment:

1 B A 2 C B 3 D C 4 E D

φ φ φ= − φ φ φ= − φ φ φ= − φ φ φ= −

The origin of the coordinate system is located at pulley A. We will arbitrarily define the rotation angle at pulley A to be zero (φ^A = 0). The rotation angle at B can be calculated from the angle of twist in segment (1):

1

1 0 0.039428 rad 0.039428 rad

B A

Similarly, the rotation angle at C is determined from the angle of twist in segment (2) and the rotation angle of pulley B:

2

2 0.039428 rad ( 0.005339 rad) 0.034089 rad

C B

The rotation angle at D is:

3

3 0.034089 rad ( 0.012555 rad) 0.021534 rad

D C

and the rotation angle at E is:

4

4 0.021534 rad 0.006760 rad 0.028294 rad

E D

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Rotation angle of flange C with respect to pulley A: Using the rotation angles determined for the system, the rotation angle of flange C with respect to pulley A is simply:

0.034089 rad 0.0341 rad

φC = = Ans.

Rotation angle of pulley E with respect to pulley A: Using the rotation angles determined for the system, the rotation angle of flange C with respect to pulley A is simply:

0.028294 rad 0.0283 rad

φE = = Ans.

In document Proyecto de carreteras des5642008/01/08 (página 164-169)