Explanation The figure below shows the free body diagram of the member A C . Figure-(1) The figure below shows the free body diagram of member C B . Figure-(2) Write the expression for moment at C for member A C . ∑ M c = 0 A y × l A C − 1 2 × w × l A C × l A C 3 = 0 ......... (I). Here, reaction at A is A y , the length of member A C is l A C and load intensity is w . Write the expression for equilibrium forces along vertical direction for member A C . ∑ F y = 0 C y + A y − 1 2 × w × l A C = 0 ......... (II) Here, vertical reaction at C is C y . Write the expression for horizontal equilibrium condition for member A C . ∑ F x = 0 C x = 0 ......... (III) Here, the horizontal reaction at C is C x . Write the expression for vertical equilibrium for member C B . ∑ F y = 0 B y − C y − 1 2 × w × l C B = 0 ......... (IV) Here, the vertical reaction at B is B y and length of member C B is l C B . Write the expression for horizontal equilibrium condition for member C B . ∑ F x = 0 B x − C x = 0 ......... (V) Here, the horizontal reaction at B is B x . Write the expression for net reaction at B . B = B x 2 + B y 2 ......... (VI) Write the expression for moment at C for member C B . ∑ M c = 0 M B − ( B y × l C B ) + 1 2 × ( w l C B ) ( l C B 3 ) = 0 .......... (VII) Here, moment about point B is M B . Calculation: Substitute 6 m for l A C , 4 kN / m for w in Equation (I)

10th Edition

ISBN: 9780134610672

Author: Russell C. Hibbeler

Publisher: PEARSON

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Question

Chapter 2, Problem 2.21P

To determine

The reaction at the supports A and B.

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Chapter 2 Solutions

Structural Analysis (10th Edition)

Ch. 2 - Prob. 2.1P Ch. 2 - Prob. 2.2P Ch. 2 - Prob. 2.3P Ch. 2 - Prob. 2.4P Ch. 2 - Prob. 2.5P Ch. 2 - Prob. 2.6P Ch. 2 - Prob. 2.7P Ch. 2 - Prob. 2.8P Ch. 2 - Prob. 2.9P Ch. 2 - Prob. 2.10P

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