Explanation Given information: The point load act at right end of the rod is P . The moment of inertia of the rods ( AB and BC ) are I A B and I A B . The Young’s modulus of the of the rod is E . The length of the rod is L . The length of the segment BC is l . Calculation: Show the free body diagram of the moment function of the rod’s cut segment 1 as in Figure 1. Refer Figure 1. Calculate the moment function with respect to the x 1 distance M ( x 1 ) : M ( x 1 ) = − P x 1 Show the free body diagram of the moment function of the rod’s cut segment 2 as in Figure 2. Refer Figure 2. Calculate the moment function with respect to the x 2 distance M ( x 2 ) : M ( x 2 ) = − P x 2 Write the bending equation to determine the moment as follows: E I d 2 υ d x 2 = M ( x ) (1) For the coordinate x 1 , Substitute ( − P x 1 ) for M ( x ) and I A B for I in equation (1), E I B C d 2 υ 1 d x 1 2 = − P x 1 (2) Integrate both sides of the Equation (2). E I B C d υ 1 d x 1 = − P x 1 2 2 + C 1 (3) Integrate both sides of the Equation (3). E I B C υ 1 = − P x 1 3 6 + C 1 x 1 + C 2 (4) For the coordinate x 2 , Substitute ( − P x 2 ) for M ( x ) and I B C for I in equation (1), E I A B d 2 υ 2 d x 2 2 = − P x 2 (5) Integrate both sides of the Equation (5). E I A B d υ 2 d x 2 = − P x 2 2 2 + C 3 (6) Integrate both sides of the Equation (6). E I A B υ 2 = − P x 2 3 6 + C 3 x 2 + C 4 (7) Apply the Boundary conditions, d υ 2 d x 2 = 0 at x 2 = L υ 2 = 0 at x 2 = L Apply the first boundary condition in Equation (6). Substitute 0 for d υ 2 d x 2 and L for x 2 in Equation (6). E I A B ( 0 ) = − P L 2 2 + C 3 C 3 = P L 2 2 Apply the second boundary condition in Equation (7). Substitute 0 for υ 2 and L for x 2 in Equation (7). E I A B ( 0 ) = − P ( L ) 3 6 + C 3 ( L ) + C 4 C 4 = P L 3 6 − C 3 L (8) Substitute P L 2 2 for C 3 in Equation (8). C 4 = P L 3 6 − ( P L 2 2 ) L = P L 3 6 − P L 3 2 = − P L 3 3 Consider x 1 = x 2 = l . Continuity conditions are as follows: d υ 1 d x 1 = d υ 2 d x 2 (9) Substitute 1 E I B C ( − P x 1 2 2 + C 1 ) for d υ 1 d x 1 and 1 I A B ( − P x 2 2 2 + C 3 ) for d υ 2 d x 2 in Equation (9). 1 I B C ( − P x 1 2 2 + C 1 ) = 1 I A B ( − P x 2 2 2 + C 3 ) − P x 1 2 2 + C 1 = I B C I A B ( − P x 2 2 2 + C 3 ) C 1 = I B C I A B ( − P x 2 2 2 + C 3 ) + P x 1 2 2 (10) Substitute l for x 1 , l for x 2 , and P L 2 2 for C 3 in Equation (10). C 1 = I B C I A B ( − P l 2 2 + P L 2 2 ) + P l 2 2 Consider x 1 = x 2 = l Continuity conditions are as follows: υ 1 = υ 2 (11) Substitute 1 I B C ( − P x 1 3 6 + C 1 x 1 + C 2 ) for υ 1 and 1 I A B ( − P x 2 3 6 + C 3 x 2 + C 4 ) for υ 2 in Equation (11). 1 I B C ( − P x 1 3 6 + C 1 x 1 + C 2 ) = 1 I A B ( − P x 2 3 6 + C 3 x 2 + C 4 ) − P x 1 3 6 + C 1 x 1 + C 2 = I B C I A B ( − P x 2 3 6 + C 3 x 2 + C 4 ) C 2 = I B C I A B ( − P x 2 3 6 + C 3 x 2 + C 4 ) + P x 1 3 6 − C 1 x 1 (12) Substitute l for x 1 , l for x 2 , [ I B C I A B ( − P l 2 2 + P L 2 2 ) + P l 2 2 ] for C 1 , P L 2 2 for C 3 , and ( − P L 3 3 ) for C 4 in Equation (12)

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ISBN: 2810014920922

Author: HIBBELER

Publisher: PEARSON

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The maximum deflection of the beam ( υ max ) .

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The maximum deflection of the beam (υmax) .

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