Explanation Given information: The allowable bending stress 20 ksi. Calculation: Calculate the reaction for support. Consider the forces acting the x direction. Summation of moment about A . ∑ M A = 0 ( − 400 × 6 ) + ( R B x × 8 ) = 0 − 2 , 400 + 8 R B x = 0 R B x = 400 lb Summation of moment about B . ∑ M B x = 0 400 × 2 − 8 R A x = 0 800 − 8 R A x = 0 R A x = 100 lb Consider the force acting y direction. ∑ F y = 0 ( 400 × 1 ) − R B × 8 = 0 400 − 8 R B = 0 R B y = 50 lb Taking a moment about A . ∑ M y = 0 R A Y = 400 − 50 = 350 lb Sketch the calculated values reaction as shown in Figure 1. Calculate the moment about x axis as follows: Consider a section at point D . Take the left section of portion about point D . M D = 350 × 1 = 350 lb ⋅ ft Calculate at point C . Consider a section at point C Take the left section of portion about point C . M C = 100 × 1 = 100 lb ⋅ ft Calculate the moment about z axis as follows: Calculate point D . M D = 100 × 1 = 100 lb ⋅ ft Calculate point C . M C = 600 × 1 = 600 lb ⋅ ft Calculate the torque at point D as shown below: T D = 50 lb ⋅ ft Calculate the torque at point C as shown below: T C = 50 lb ⋅ ft Sketch the moment and torque diagram as shown in Figure 3. Section just to the left point C is the most critical. Find the moment ( M ) as shown below: M = M Z D 2 + M Z C 2 (1) Here, M Z D is the moment about z axis at D point and M Z D is the z axis at C point. Substitute 600 lb.ft for M Z D and 100lb ft for M Z C in Equation (1). M = 600 2 + 100 2 = 360 , 000 + 10 , 000 = 608.27 lb ⋅ ft Sketch the state of stress point as shown in Figure 3. Section just to the left of point C is the most critical. Both state of stress will yield the same result. Calculate the required diameter using maximum distortion energy. σ a , b = σ 2 ± ( σ 2 ) 2 + τ 2 (2) Here, σ is the stress and τ is the shear stress. Substitute A for σ 2 and B for ( σ 2 ) 2 + τ 2 in Equation (2). σ a = ( A + B ) σ b = ( A − B ) Square on both sides. σ a 2 = ( A + B ) 2 σ b 2 = ( A − B ) 2 σ a σ b = ( A + B ) ( A − B ) = A 2 − B 2 σ a 2 − σ a σ b + σ b 2 = A 2 + B 2 + 2 A B − A 2 + B 2 + A 2 + B 2 − 2 A B = A 2 + 3 B 2 = σ 2 4 + 3 ( σ 2 4 + τ 2 ) = σ 2 + 3 τ 2 σ a 2 − σ a σ b + σ b 2 = σ allow 2 (3) Substitute σ 2 + 3 τ 2 for σ a 2 − σ a σ b + σ b 2 in Equation (3)

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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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 11.4, Problem 11.43P

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The required diameter (d) of the shaft using maximum distortion energy theory.

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Chapter 11.4, Problem 11.42P

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Chapter 11 Solutions

Mechanics of Materials (10th Edition)

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The required diameter (d) of the shaft using maximum distortion energy theory.

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