Explanation The given torque acting on the shaft is 100 N ⋅ m , diameter of sprocket B is 150 mm , the diameter of gear C is 250 mm , diameter of gear F is 125 mm and the pressure angle of the gear is 20 ° . The free body diagram of Gear F is as follows. Figure (1) The free body diagram of the shaft is as follows. Figure (2) Write the tangential force acting on the shaft. F t = T ( c 2 ) (I) Here, the tangential force on the shaft is F t , torque acting on the shaft is T and the diameter of the shaft is c . Write the normal force acting on the shaft. F n = F t tan ϕ (II) Here, F n is the normal force acting on the shaft and ϕ is the pressure angle. Write the torque at gear C . T c = F t × b 2 (III) Here, torque at gear C is T c . Write the load at sprocket B . P = T c ( a 2 ) (IV) Here, the load at sprocket B is P . Write the moment about A in z - direction. ∑ ( M A ) z = 0 ( e + f + g ) R D y − F n ( f + g ) − ( P × e ) = 0 R D y = F n ( f + g ) + ( P × e ) ( e + g + f ) (V) Here, the moment about A in z - direction is ( M A ) z , the reaction at D in y - direction is R D y , the distance between sprocket B and pivot A is e , the distance between sprocket B and gear C is f and the distance between gear C and pivot D is g . Write the moment about A in y - direction. ∑ ( M A ) y = 0 − ( e + f + g ) R D z + F t ( f + g ) = 0 R D z = F t ( f + g ) ( e + g + f ) (VI) Here, the moment about A in y - direction is ( M A ) y and the reaction at D in z direction is R D z . Write the forces in y - direction. ∑ F y = 0 R A y + R D y − F n − P = 0 R A y = F n + P − R D y (VII) Here, the reaction at A in y - direction is R A y . Write the forces in z - direction. ∑ F z = 0 R A z + R D z − F t = 0 R A z = F t − R D z (VIII) Here, the reaction at A in z - direction is R A z . From given figure, the maximum bending moment will be either at B or C . Write the bending moment at B . M B = e R A y 2 + R A z 2 (IX) Here, M B is the bending moment at B . Write the bending moment at C . M C = g R D y 2 + R D z 2 (X) Here, M C is the bending moment at C . Write the bending stress at B . σ B = 32 M B π d 3 (XI) Here, the diameter of the shaft is d and the bending stress at B is σ B . Write the shearing stress at B . τ B = 16 T C π d 3 (XII) Here, the shearing stress at B is τ B . Calculate the maximum principal normal stresses in the shaft. σ max = σ B 2 + ( σ B 2 ) 2 + τ B 2 (XIII) Here, the maximum principal normal stress in the shaft is σ max . Calculate the maximum principal shear stresses in the shaft. τ max = ( σ B 2 ) 2 + τ B 2 (XIV) Here, the maximum principal shear stress in the shaft is τ max . Calculate the factor of safety from maximum-shear-stress theory. n = S y 2 τ max (XV) Here, the maximum yield stress for 1018 CD steel is S y . Calculate the factor of safety from distortion-energy theory. n = S y σ ' (XVI) Here, the Von Mises stress is σ '

Loose Leaf for Shigley's Mechanical Engineering Design Format: LooseLeaf

10th Edition

ISBN: 9780073399652

Author: BUDYNAS

Publisher: Mcgraw Hill Publishers

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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 5, Problem 47P

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The factor of safety for yielding from maximum-shear-stress theory.

The factor of safety for yielding from distortion-energy theory.

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Chapter 5, Problem 46P

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The material of a certain transmission shaft uses AISI 1030 HR, its yield strength is 260 MPa, the transmission torque of this shaft is 160 N·m, and it is subjected to a uniform axial tension of 0.5 kN. Please calculate according to the Distortion energy theory. The minimum shaft diameter with a safety factor n of 2.5 (to 2 decimal places).

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Loose Leaf for Shigley's Mechanical Engineering Design Format: LooseLeaf

Ch. 5 - A ductile hot-rolled steel bar has a minimum yield...Ch. 5 - A ductile hot-rolled steel bar has a minimum yield...Ch. 5 - A ductile hot-rolled steel bar has a minimum yield...Ch. 5 - A ductile hot-rolled steel bar has a minimum yield...Ch. 5 - A ductile hot-rolled steel bar has a minimum yield...Ch. 5 - Prob. 6P Ch. 5 - 5-7 to 5-11 An AISI 1018 steel has a yield...Ch. 5 - 5-7 to 5-11 An AISI 1018 steel has a yield...Ch. 5 - 5-7 to 5-11 An AISI 1018 steel has a yield...Ch. 5 - 5-7 to 5-11 An AISI 1018 steel has a yield...

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The factor of safety for yielding from maximum-shear-stress theory. The factor of safety for yielding from distortion-energy theory.

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The factor of safety for yielding from maximum-shear-stress theory.

The factor of safety for yielding from distortion-energy theory.