Explanation Given information: The point load acting at right end of the beam is P . The flexural rigidity of the beam is EI . The length of the segment AB is L . Calculation: Show the free body diagram of the beam 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 − P = 0 A y + B y = P (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 ) − P ( L + L 2 ) = 0 ( B y × L ) = P ( 3 L 2 ) B y = P ( 3 L 2 ) L B y = 3 P 2 Substitute 3 P 2 for B y in Equation (1). A y + 3 P 2 = P A y = P − 3 P 2 A y = − P 2 A y = P 2 ( ↓ ) Show the free body diagram of the moment function of the beam’s cut segment 1 as in Figure 2. Refer to Figure 2. Calculate the moment function with respect of the x 1 distance M ( x 1 ) : M ( x 1 ) = − P 2 x 1 Show the free body diagram of the moment function of the beam’s cut segment 2 as in Figure 3. Refer Figure 3. Calculate the moment function with respect of the x 3 distance M ( x 3 ) : M ( x 3 ) = P x 3 − 3 P L 2 Elastic Curve and slope is as follows: E I d 2 υ d x 2 = M ( x ) (2) For the coordinate x 1 , Substitute ( − P 2 x 1 ) for M ( x ) in Equation (2), E I d 2 υ 1 d x 1 2 = − P x 1 2 (3) Integrate both the sides of the Equation (3). E I d υ 1 d x 1 = − P x 1 2 4 + C 1 (4) Integrate both the sides of the Equation (4). E I υ 1 = − P x 1 3 12 + C 1 x 1 + C 2 (5) For the coordinate x 2 , Substitute ( P x 3 − 3 P L 2 ) for M ( x ) in Equation (2), E I d 2 υ 3 d x 2 2 = P x 3 − 3 P L 2 (6) Integrate both the sides of the Equation (6). E I d υ 3 d x 3 = P x 3 2 2 − 3 P L x 3 2 + C 3 (7) Integrate both the sides of the Equation (7). E I υ 3 = P x 3 3 6 − 3 P L x 3 2 4 + C 3 x 2 + C 4 (8) Apply the Boundary conditions. Condition 1: υ 1 = 0 at x 1 = 0 Condition 2: υ 1 = 0 at x 1 = L Condition 3: υ 3 = 0 at x 3 = L Apply the first boundary condition in Equation (5). Substitute 0 for υ 1 and 0 for x 1 in Equation (5). E I ( 0 ) = − P ( 0 ) 3 12 + C 1 ( 0 ) + C 2 C 2 = 0 Substitute 0 for C 2 in Equation (5). E I υ 1 = − P x 1 3 12 + C 1 x 1 + 0 E I υ 1 = − P x 1 3 12 + 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 ) = − P ( L ) 3 12 + C 1 ( L ) 0 = − P L 3 12 + L C 1 L C 1 = P L 3 12 C 1 = P L 2 12 Apply the third boundary condition in Equation (8). Substitute 0 for υ 3 and L for x 3 in Equation (8). E I ( 0 ) = P L 3 6 − 3 P L ( L ) 2 4 + C 3 ( L ) + C 4 0 = P L 3 6 − 3 P L 3 4 + C 3 L + C 4 0 = − 7 P L 3 12 + C 3 L + C 4 (10) Consider x 1 = x 3 = L

9th Edition

ISBN: 2810014920922

Author: HIBBELER

Publisher: PEARSON

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Chapter 12.2, Problem 12.6P

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The equations of the elastic curve for the beam (υ1 and υ2) .

The maximum deflection for the beam (υmax) .

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Chapter 12.2, Problem 12.5P

Chapter 12.2, Problem 12.7P

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The equations of the elastic curve for the beam ( υ 1 and υ 2 ) . The maximum deflection for the beam ( υ max ) .

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The equations of the elastic curve for the beam (υ1 and υ2) . The maximum deflection for the beam (υmax) .

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Chapter 12 Solutions

The equations of the elastic curve for the beam (υ1 and υ2) .

The maximum deflection for the beam (υmax) .