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PROBLEMS ON TORSION

PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

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Page 1: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

PROBLEMS ON TORSION

Page 2: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

SIGN CONVENTION (REVIEW)

Page 3: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

FREE-BODY DIAGRAMS (REVIEW)

Page 4: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

TORQUE DIAGRAM (REVIEW)

Page 5: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

Example: The gears attached to the fixed-end steel shaft are subjected to the torques shown in figure. If the shear modulus of elasticity is 80 GPa and the shaft has a diameter of 14 mm, determine the displacement of the tooth P on gear A. The shaft turns freely within the bearing at B.

Soln.

Internal Torque: By inspection, the torques in segments, AC, CD, and DE are different yet constant throughout each segment. Free-body diagrams of appropriate segments of the shaft are shown in Figure (b):

Page 6: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

mNTor

T

givesM

DE

DE

.170,

040280150

0

But, from the right hand rule and the established sign convention, we get

TDE = - 170 N.m

Similarly,

-150 + 280 – TCD = 0; OR, TCD = 130 N.m

But, from the right hand rule and the established sign convention, we getTCD = - 130 N.m

And, -150 + TAC = 0, OR, TAC = 150 N.m and is POSITIVE

Page 7: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

The polar moment of inertia for the shaft is

J = π/2(0.007 m)4 = 3.77 (10-9) m4. Applying the formula for the angle of twist for each segment and adding the results algebraically, we have:

rad

GJ

TLAE

212.0)10(80)10(77.3

)5.0)(170(

)10(80)10(77.3

)3.0)(130(

)10(80)10(77.3

)4.0)(150(

99

99

99

Since the answer is negative, by the right-hand rule the thumb is directed toward the end E of the shaft, and therefore, gear A will rotate as shown in figure (d)

Page 8: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

The displacement of tooth P on gear A is

mmmmradrS AEP 2.21)100)(212.0(

Remember that this analysis is valid only if the shear stress does not exceed the proportional limit of the material.

Page 9: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

Problem 1: A hollow member AC has the outer diameter of 60 mm and the inner diameter of 40 mm. If the member is loaded as shown, determine:

(a) the maximum shear stress

(b) the angle of twist of end C relative to the fixed end, and

(c) the angle of twist of end C relative to point B

Use G = 80 GPa.

Page 10: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

Problem 2: A shaft is designed such that it is solid in section AB with a diameter of 50 mm and hollow in section BC with the inner diameter of 45 mm and outer diameter of 60 mm. If the shaft is loaded as shown, determine:

(a) the maximum shear stress on the shaft,

(b) the angle of twist of end A relative to the fixed support C, and

(c) the angle of twist of end A relative to point B.

Take G = 75 GPa.

Page 11: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

Problem 3: A vertical shaft is designed such that it is solid in section AB with a diameter of 60 mm and hollow in section BC with the inner diameter of 40 mm and outer diameter of 60 mm. If the shaft is loaded as shown, determine:

(a) the maximum shear stress on the shaft,

(b) the angle of twist of end D relative to the fixed support A, and

(c) the angle of twist of end C relative to point B.

Use G = 80 GPa for all sections.

Page 12: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

Problem 4: Determine the power that can be transmitted by a turbine driven solid circular shaft of 35 mm in diameter if it is to rotate at 500 rpm and the maximum allowable shear stress on the shaft is limited to 60 MPa.

Page 13: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

Problem 5: Determine the diameter of a solid steel shaft to transmit 30 MW at a speed of 1500 rev/min if the angle of twist is limited to 10 for every twenty diameters of length.

Take G = 80 GN/m2.

Problem 6: A torque of 50 kN m is to be transmitted by a hollow shaft of internal diameter half the external diameter. If the maximum shear stress is not to exceed 80 MN/m2, calculate the outside diameter of the shaft. What would be the angle of twist over a length of 3 m of the shaft under the above torque?Take G = 80 GN/m2.

Problem 7: A hollow shaft is 50 mm outside diameter and 30 mm inside diameter. An applied torque of 1.6 kN m is found to produce an angular twist of 0.40, measured on a length of 0.2 m of the shaft. Calculate:(a) the value of the modulus of rigidity, and(b) the maximum power that can be transmitted by the shaft at 2000 rev/min if the maximum allowable shear stress is 65 kN/m2.

Page 14: PROBLEMS ON TORSION. SIGN CONVENTION (REVIEW) FREE-BODY DIAGRAMS (REVIEW)

Problem 8: A shaft 50 mm in diameter and 0.75 m long has a concentric hole drilled for a portion of its length. Find the maximum length and diameter of the hole so that when the shaft is subjected to a torque of 1.67 kN m, the maximum shearing stress will not exceed 75 kN/m2 and the total angle of twist will not exceed 1.50 deg.

Take G = 80 GPa.