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 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 + R B = 130 (1) 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 48 R B = 1 , 320 R B = 1 , 320 48 R B = − 27.5 lb R B = 27.5 lb ( ↓ ) Substitute 27.5 lb for R B in Equation (1). R A + 27.5 = 130 R A = 130 − 27.5 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 ) (2) Substitute [ − 50 x + 102.5 ( x − 12 ) − 80 ( x − 36 ) ] for M ( x ) in Equation (2). E I d 2 υ d x 2 = − 50 x + 102.5 ( x − 12 ) − 80 ( x − 36 ) (3) Integrate both sides of the Equation (3). 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 (4) Integrate both sides of the Equation (4). 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 ] (5) Boundary conditions are as follows: υ = 0 at x = 12 in . υ = 0 at x = 60 in

Mechanics of Materials, Student Value Edition Plus Mastering Engineering with Pearson eText -- Access Card Package (10th Edition)

10th Edition

ISBN: 9780134326054

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 …

Question

Chapter 12.3, Problem 12.42P

To determine

The slop at A(θA).

The slop at B(θB).

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Chapter 12.3, Problem 12.41P

Chapter 12.3, Problem 12.43P

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Mechanics of Materials, Student Value Edition Plus Mastering Engineering with Pearson eText -- Access Card Package (10th Edition)

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The slop at A ( θ A ) . The slop at B ( θ B ) .

Solution Summary: The author calculates the slop at A and B by applying the equation of equilibrium: Sum of moments about point A is equal to 0.

Concept explainers

The slop at A(θA).

The slop at B(θB).

Want to see the full answer?

Chapter 12 Solutions

The slop at A ( θ A ) . The slop at B ( θ B ) .

Solution Summary: The author calculates the slop at A and B by applying the equation of equilibrium: Sum of moments about point A is equal to 0.

Concept explainers

The slop at A(θA). The slop at B(θB).

Want to see the full answer?

Chapter 12 Solutions

The slop at A(θA).

The slop at B(θB).