Explanation Given information: The load act at the first pulley ( P 1 ) is 50 lb . The load act at the second pulley ( P 2 ) is 80 lb . The flexural rigidity of the shaft is E I . The length of the shaft ( L ) is 60 in . Calculation: Show the free body diagram of the shaft as in Figure 1. Refer to Figure 1. Calculate the support reactions A and B ( R A and R B ) : Calculate the moment about point A by applying the equation of equilibrium: Sum of moments about point A is equal to 0. ∑ M A = 0 ( R B × 48 ) − ( 80 × 24 ) + ( 50 × 12 ) = 0 48 R B − 1 , 320 = 0 R B = 27.5 lb Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 R A + R B − 50 − 80 = 0 R A + 27.5 = 130 R A = 102.5 lb Calculate the moment function with respect of the x distance M ( x ) : M ( x ) = − 50 ( x − 0 ) − ( − 102.5 ) 〈 x − 12 〉 − 80 〈 x − 36 〉 = − 50 x + 102.5 〈 x − 12 〉 − 80 〈 x − 36 〉 Elastic Curve and slope is as follows: E I d 2 υ d x 2 = M ( x ) Substitute [ − 50 x + 102.5 〈 x − 12 〉 − 80 〈 x − 36 〉 ] for M ( x ) . E I d 2 υ d x 2 = − 50 x + 102.5 〈 x − 12 〉 − 80 〈 x − 36 〉 Integrate both sides of the Equation. E I d υ d x = − 50 x 2 2 + 102.5 2 〈 x − 12 〉 2 − 80 2 〈 x − 36 〉 2 + C 1 = − 25 x 2 + 102.5 2 〈 x − 12 〉 2 − 40 〈 x − 36 〉 2 + C 1 (1) Integrate both sides of the Equation (1). E I υ = − 25 x 3 3 + 102.5 3 〈 x − 12 〉 3 − 40 3 〈 x − 36 〉 3 + C 1 x + C 2 = [ − 25 3 x 3 + 102.5 6 〈 x − 12 〉 3 − 40 3 〈 x − 36 〉 3 + C 1 x + C 2 ] (2) Boundary conditions are as follows: υ = 0 at x = 12 in . υ = 0 at x = 60 in . Apply the first boundary condition in Equation (2). E I ( 0 ) = [ − 25 3 ( 12 ) 3 + 0 − 0 + 12 C 1 + C 2 ] 0 = − 14 , 400 + 12 C 1 + C 2 12 C 1 + C 2 = 14 , 400 (3) Apply the second boundary condition in Equation (2)

Statics and Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (5th Edition)

5th Edition

ISBN: 9780134301006

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 16.3, Problem 42P

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Find the slope at A(θA).

Find the slope at B(θB).

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Chapter 16.3, Problem 41P

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Chapter 16 Solutions

Statics and Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (5th Edition)

Ch. 16.2 - In each ease, determine the internal bending...Ch. 16.2 - Prob. 1FP Ch. 16.2 - Determine the slope and deflection of end A of the...Ch. 16.2 - Prob. 3FP Ch. 16.2 - Prob. 4FP Ch. 16.2 - Determine the maximum deflection of the simply...Ch. 16.2 - Prob. 6FP Ch. 16.2 - An L2 steel strap having a thickness of 0.125 in....Ch. 16.2 - The L2 steel blade of the band saw wraps around...Ch. 16.2 - A picture is taken of a man performing a pole...

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Find the slope at A ( θ A ) . Find the slope at B ( θ B ) .

Solution Summary: The author shows the free body diagram of the shaft as in Figure 1.

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Find the slope at A(θA).

Find the slope at B(θB).

Want to see the full answer?

Chapter 16 Solutions

Find the slope at A ( θ A ) . Find the slope at B ( θ B ) .

Solution Summary: The author shows the free body diagram of the shaft as in Figure 1.

Concept explainers

Find the slope at A(θA). Find the slope at B(θB).

Want to see the full answer?

Chapter 16 Solutions

Find the slope at A(θA).

Find the slope at B(θB).