Explanation Given information: The power delivered by the motor is 300 kW , power required at gear B is 120 kW , power required at gear C is 180 kW , allowable shear stress is 50 MPa , angle of twist between motor and gear C is 4 ° , and the frequency of the shaft is 32 Hz . Write the expression for the power supplied by the motor. P = 2 π N T A ....... (I) Here, the torque applied at point A is T A , and speed of the shaft is N . Write the expression for the power required at gear B . P B = 2 π N T B ....... (II) Here, the torque required at point B is T B . Write the expression for the power required at gear C . P C = 2 π N T C ....... (III) Here, the torque required at point C is T C . Write the expression for the torque in the segment AB. T A B = T A ....... (IV) Write the expression for the torque in the segment BC. T B C = T A − T B ....... (V) Write the expression for the allowable shear stress. τ max = 16 T max π d 3 ....... (VI) Here, diameter of the shaft is d , and maximum torque applied on the shaft is T max . Write the expression for the angle of twist in the segment AB . ϕ A B = 32 T A B L A B G π d 4 ....... (VII) Here, modulus of rigidity is G , and length of the shaft between A and B is L A B . Write the expression for the angle of twist in the segment BC . ϕ B C = 32 T B C L B C G π d 4 ....... (VIII) Here, the length of the shaft between B and C is L B C . Write the expression for the total angle of twist in the shaft. ϕ = ϕ A B + ϕ B C ....... (IX) Calculation: Substitute 300 kW for P , 32 Hz for N in Equation (I). 300 kW = 2 π ( 32 Hz ) T A ( 300 kW ) ( 1000 W 1 kW ) = 2 π ( 32 Hz ) T A T A = 1492.07 N ⋅ m Substitute 120 kW for P B , 32 Hz for N in Equation (II). 120 kW = 2 π ( 32 Hz ) T B ( 120 kW ) ( 1000 W 1 kW ) = 2 π ( 32 Hz ) T B T B = 596.83 N ⋅ m Substitute 180 kW for P C , 32 Hz for N in Equation (III). 180 kW = 2 π ( 32 Hz ) T C ( 180 kW ) ( 1000 W 1 kW ) = 2 π ( 32 Hz ) T C T C ≈ 895.24 N ⋅ m Substitute 1492.07 N ⋅ m for T A in Equation (IV). T A B = 1492.07 N ⋅ m Substitute 1492.07 N ⋅ m for T A , and 7878.17 lb ⋅ in for T B in Equation (V). T B C = 1492.07 N ⋅ m − 596.83 N ⋅ m = 895.24 N ⋅ m Maximum torque will be selected for maximum shear stress so T max = 1492.07 N ⋅ m . Substitute 50 MPa for τ a l l o w , and 1492.07 N ⋅ m for T max in Equation (VI). 50 MPa = 16 ( 1492

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN: 9781337093347

Author: Barry J. Goodno, James M. Gere

Publisher: Cengage Learning

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Design of Shafts and Shaft Components

The shaft is basically defined as the machine element that rotates, is in circular cross-section, it is majorly used to transfer the power from one part of the system to another, or from a power generating machine to a power absorbing machine. The shaft …

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Chapter 3, Problem 3.7.12P

To determine

The minimum shaft diameter.

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Chapter 3 Solutions

Mechanics of Materials (MindTap Course List)

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The minimum shaft diameter.

Solution Summary: The author explains the minimum shaft diameter, allowable shear stress, angle of twist between motor and gear C, and frequency of the shaft.

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The minimum shaft diameter.

Chapter 3 Solutions