Explanation Given information: The momentat left end of the beam ( M ) is M 0 . The flexural rigidity of the beam is E I . The length between the supports of the beam ( L A B and L B C ) is L . The length of the beam ( L A C ) is 2 L . Calculation: The compatibility condition for the beam is as follows: ( υ B ) 1 + ( υ B ) 2 = 0 (1) Apply the method of superposition as shown below. ( υ B ) 1 = M L B C 6 E I L A C ( L A C 2 − L B C 2 ) (2) ( υ B ) 2 = P L A C 2 48 E I (3) Here, P is the applied load. Substitute M 0 for M , L for L B C , and 2 L for L A C in Equation (2). ( υ B ) 1 = − M 0 L 6 E I × 2 L [ ( 2 L ) 2 − L 2 ] = − M 0 12 E I × ( 4 L 2 − L 2 ) = − M 0 12 E I × 3 L 2 = − M 0 L 2 4 E I Negative sign indicates the downward deflection

Statics and Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (5th Edition)

5th Edition

ISBN: 9780134301006

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 16.5, Problem 10FP

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Find the reaction at the rollersupport B.

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Statics and Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (5th Edition)

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Find the reaction at the rollersupport B .

Solution Summary: The author analyzes the reaction at the rollersupport B. The momentat left end of the beam is M_0.

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Find the reaction at the rollersupport B.

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