Explanation Given information: The modulus of elasticity of the beam ( 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). E I ( 0 ) = [ 0 − 0 − 0 − 0 + 0 + C 2 ] C 2 = 0 Substitute 0 for C 2 in Equation (5). E I υ = [ 30 x 3 − 25 ( x − 0.25 ) 3 − 10 ( x − 0.5 ) 3 − 25 ( x − 0

11th Edition

ISBN: 9780137605460

Author: Russell C. Hibbeler

Publisher: Pearson Education (US)

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Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and af…

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Chapter 12.3, Problem 45P

To determine

The deflection of the pulley C(υC).

The deflection of the pulley D(υD).

The deflection of the pulley E(υE).

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Chapter 12.3, Problem 44P

Chapter 12.3, Problem 46P

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The deflection of the pulley C ( υ C ) . The deflection of the pulley D ( υ D ) . The deflection of the pulley E ( υ E ) .

Solution Summary: The author calculates the deflection of the pulleys C, D, and E.

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The deflection of the pulley C(υC).

The deflection of the pulley D(υD).

The deflection of the pulley E(υE).

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Chapter 12 Solutions

The deflection of the pulley C ( υ C ) . The deflection of the pulley D ( υ D ) . The deflection of the pulley E ( υ E ) .

Solution Summary: The author calculates the deflection of the pulleys C, D, and E.

Concept explainers

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The deflection of the pulley C(υC). The deflection of the pulley D(υD). The deflection of the pulley E(υE).

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Chapter 12 Solutions

The deflection of the pulley C(υC).

The deflection of the pulley D(υD).

The deflection of the pulley E(υE).