## Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

# Chapter 11 - Virtual Work - Section 11.3 - Principle of Virtual Work for a System of Connected Rigid Bodies - Problems - Page 595: 19

#### Answer

$k=9.88KN/m$

#### Work Step by Step

We can determine the stiffness of the spring as follows: $F_s=k[(600sin\theta-600sin15)-(200sin\theta-200sin15)]$ $\implies F_s=400k(sin\theta-sin15)$ The virtual displacements are given as $\delta_{yA}=\frac{d(200sin\theta)}{d\theta}=200cos\theta$ $\delta_{yB}=\frac{d(600sin\theta)}{d\theta}=600cos\theta$ and $\delta_{yC}=\frac{d(1600sin\theta)}{d\theta}=1600cos\theta$ Now the virtual-work equation is $\delta U=0$ $\implies P\delta_{yC}+F_s(\delta_{yB}-\delta_{yA})=0$ We plug in the known values to obtian: $600(1600cos\theta)-400k(sin\theta-sin15)(600cos\theta-200cos\theta)=0$ This simplifies to: $k=9.88KN/m$

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