Explanation Given information: The modulus of elasticity of the shaft ( E ) is 200 GPa . The diameter of the shaft ( d ) is 30 mm . The point load act at a first pulley of the shaft ( P 1 ) is 150 N . The point load act at a second pulley of the shaft ( P 2 ) is 60 N . The point load act at a third pulley of the shaft ( P 3 ) is 150 N . The moment of inertia of the shaft is I . The length of the shaft ( L ) is 1000 mm . Calculation: Show the free body diagram of the shaft as in Figure 1. Refer 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 − 150 − 60 − 150 = 0 R A + R B − 360 = 0 R A + R B = 360 (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 × 1000 ) − ( 150 × 250 ) − ( 60 × 500 ) − ( 150 × 750 ) = 0 1 , 000 R B − 180 , 000 = 0 1 , 000 R B = 180 , 000 R B = 180 , 000 1 , 000 R B = 180 N Substitute 180 N for R B in Equation (1). R A + 180 = 360 R A = 360 − 180 R A = 180 N Calculate the moment function with respect of the x distance M ( x ) : M ( x ) = [ − ( − 180 ) ( x − 0 ) − 150 ( x − 0.25 ) − 60 ( x − 0.5 ) − 150 ( x − 0.75 ) ] = [ 180 x − 150 ( x − 0.25 ) − 60 ( x − 0.5 ) − 150 ( x − 0.75 ) ] Elastic Curve and slope is as follows: E I d 2 υ d x 2 = M ( x ) (2) Substitute [ − 180 x − 150 ( x − 0.25 ) − 60 ( x − 0.5 ) − 150 ( x − 0.75 ) ] for M ( x ) in Equation (2). E I d 2 υ d x 2 = [ 180 x − 150 ( x − 0.25 ) − 60 ( x − 0.5 ) − 150 ( x − 0.75 ) ] (3) Integrate both sides of the Equation (3). E I d υ d x = [ 180 2 x 2 − 150 2 ( x − 0.25 ) 2 − 60 2 ( x − 0.5 ) 2 − 150 2 ( x − 0.75 ) 2 + C 1 ] = [ 90 x 2 − 75 ( x − 0.25 ) 2 − 30 ( x − 0.5 ) 2 − 75 ( x − 0.75 ) 2 + C 1 ] (4) Integrate both sides of the Equation (4). E I υ = [ 90 3 x 3 − 75 3 ( x − 0.25 ) 3 − 30 3 ( x − 0.5 ) 3 − 75 3 ( x − 0.75 ) 3 + C 1 x + C 2 ] = [ 30 x 3 − 25 ( x − 0.25 ) 3 − 10 ( x − 0.5 ) 3 − 25 ( x − 0.75 ) 3 + C 1 x + C 2 ] (5) Boundary conditions are as follows: υ = 0 at x = 0 υ = 0 at x = 1 m Apply the first boundary condition in Equation (5)

11th Edition

ISBN: 9780137605460

Author: Russell C. Hibbeler

Publisher: Pearson Education (US)

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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 12.3, Problem 46P

To determine

The slope of the shaft at A (θA) .

The slope of the shaft at B (θB) .

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Chapter 12.3, Problem 45P

Chapter 12.3, Problem 47P

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The slope of the shaft at A ( θ A ) . The slope of the shaft at B ( θ B ) .

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The slope of the shaft at A (θA) . The slope of the shaft at B (θB) .

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Chapter 12 Solutions

The slope of the shaft at A (θA) .

The slope of the shaft at B (θB) .