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 '

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

10th Edition

ISBN: 9780073398204

Author: Richard G Budynas, Keith J Nisbett

Publisher: McGraw-Hill Education

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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

Chapter 5, Problem 48P

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The figure (attached) shows a belt pulley mechanism which is loaded statically. The shaft is made of AISI 1030 steel with the yield strength of 480 MPa. Using distortion energy theory (DET), determine the diameter of the shaft with a factor of safety of 2.

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Figure 2 shows a solid shaft consisting of two prismatic circular parts in equilibrium under the torques applied to the pulleys fastened to it. Give that the diameter for part AB and BC are 200 mm and 120 mm, respectively. (a) Determine the torque, TR on the shaft. (b) State the segment at which the maximum shearing stress occurs. Verify your answer by shear stress analysis. d- 200 mm d- 120 mm TA= 36 kN.m Te=14 kN.m Ta C Figure 2: Solid shaft with pulleys

Chapter 5 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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.