Explanation Given information: The moment act at end C of the shaft is M 0 . The flexural rigidity of the shaft is EI . The length of segment AB is L . The length of segment BC is L 2 . Calculation: Show the free body diagram of the shaft as in Figure 1. Refer to Figure 1. Calculate the support reactions A and B ( A y and B y ) : Calculate the vertical forces by applying the Equation of equilibrium: Sum of vertical forces is equal to zero. ∑ F y = 0 A y + B y = 0 (1) Calculate the moment about point A by applying the Equation of equilibrium: Sum of moments about point A is equal to zero. ∑ M A = 0 ( B y × L ) − M 0 = 0 B y L = M 0 B y = M 0 L Substitute M 0 L for B y in Equation (1). A y + M 0 L = 0 A y = − M 0 L A y = M 0 L ↓ Show the free body diagram of the moment function of the shaft’s cut segment 1 as in Figure 2. Refer Figure 2. Calculate the moment function with respect of the x 1 distance M ( x 1 ) : M ( x 1 ) = − M 0 x 1 L Show the free body diagram of the moment function of the shaft’s cut segment 2 as in Figure 3. Refer Figure 3. Calculate the moment function with respect to the x 2 distance M ( x 2 ) : M ( x 2 ) = − M 0 Write the bending equation to determine the bending moment as follows: E I d 2 υ d x 2 = M ( x ) (2) For the coordinate x 1 . Substitute ( − M 0 x 1 L ) for M ( x ) in equation (2). E I d 2 υ 1 d x 1 2 = − M 0 x 1 L (3) Integrate both sides of the Equation (3). E I d υ 1 d x 1 = − M 0 x 1 2 2 L + C 1 (4) Integrate both sides of the Equation (4). E I υ 1 = − M 0 x 1 3 6 L + C 1 x 1 + C 2 (5) For the coordinate x 2 . Substitute ( − M 0 ) for M ( x ) in equation (2). E I d 2 υ 2 d x 2 2 = − M 0 (6) Integrate both sides of the Equation (6). E I d υ 2 d x 2 = − M 0 x 2 + C 3 (7) Integrate both sides of the Equation (7). E I υ 2 = − M 0 x 2 2 2 + C 3 x 2 + C 4 (8) Boundary conditions are, Condition 1: υ 1 = 0 at x 1 = 0 Condition 2: υ 1 = 0 at x 1 = L Condition 3: υ 2 = 0 at x 2 = L 2 Apply the first boundary condition in Equation (5). Substitute 0 for υ 1 and 0 for x 1 in Equation (5). E I ( 0 ) = − M 0 ( 0 ) 3 6 L + C 1 ( 0 ) + C 2 C 2 = 0 Substitute 0 for C 2 in Equation (5). E I υ 1 = − M 0 x 1 3 6 L + C 1 x 1 + 0 E I υ 1 = − M 0 x 1 3 6 L + C 1 x 1 (9) Apply the second boundary condition in Equation (9). Substitute 0 for υ 1 and L for x 1 in Equation (9). E I ( 0 ) = − M 0 L 3 6 L + C 1 L C 1 L = M 0 L 2 6 C 1 = M 0 L 2 6 L = M 0 L 6 Apply the third boundary condition in Equation (8). Substitute 0 for υ 2 and L 2 for x 2 in Equation (8). E I ( 0 ) = − M 0 ( L 2 ) 2 2 + C 3 ( L 2 ) + C 4 0 = − M 0 L 2 8 + L 2 C 3 + C 4 (10) Consider x 1 = L and x 2 = L 2

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 12.2, Problem 12.17P

To determine

The elastic curve in terms of the x1 and x2 coordinates (υ1 and υ2) .

The deflection of end C of the shaft (υC) .

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Chapter 12.2, Problem 12.16P

Chapter 12.2, Problem 12.18P

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The elastic curve in terms of the x 1 and x 2 coordinates ( υ 1 and υ 2 ) . The deflection of end C of the shaft ( υ C ) .

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The elastic curve in terms of the x1 and x2 coordinates (υ1 and υ2) . The deflection of end C of the shaft (υC) .

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Chapter 12 Solutions

The elastic curve in terms of the x1 and x2 coordinates (υ1 and υ2) .

The deflection of end C of the shaft (υC) .