Explanation Given information: The diameter of the shafts is 25 mm. The applied torque to the gear at E is 500 N ⋅ m . Calculation: Let the reaction at fixed supports A and B be T A and T B , the force in the gears be F. Sketch the free body diagram of the shafts as shown in Figure 1. Apply Equilibrium equation. For the 100 mm gear, Sum of moments at x is equal to 0. Σ M x = 0 T A + F ( 0.1 m ) − 500 N ⋅ m = 0 0.1 F = 500 − T A F = 5 , 000 − 10 T A (1) For the 50 mm gear, Sum of moments at x is equal to 0. Σ M x = 0 T B + F ( 0.05 m ) = 0 0.05 F = T B F = 20 T B (2) Equate Equation (1) and Equation (2) as shown below. 5 , 000 − 10 T A = 20 T B 10 T A + 20 T B = 5 , 000 (3) The angle of twist ( ϕ ) can be expressed as follows: ϕ = T L J G Here, T is the torque applied, L is the length of the shaft, J is the polar moment of inertia, and G is the shear modulus of elasticity of the material. Apply compatibility condition. 0.1 ϕ E = 0.05 ϕ F 0.1 ( T A L A E J G ) = 0.05 ( T B L B F J G ) Substitute 1.5 m for L A E and 0.75 m for L B F . 0.1 ( T A ( 1.5 m ) J G ) = 0.05 ( T B ( 0.75 m ) J G ) 0.15 T A = 0.0375 T B T A = 0.25 T B (4) Substitute 0.25 T B for T A in Equation (3). 10 ( 0.25 T B ) + 20 T B = 5 , 000 22

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Find the angle of twist of the gear at E.

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Chapter 10 Solutions

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Find the angle of twist of the gear at E .

Solution Summary: The author calculates the angle of twist of the gear at E. The reaction at fixed supports A and B is T_A and

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