# Change in length of axially loaded member

1. Apr 14, 2012

### pirateboy

1. The problem statement, all variables and given/known data
Given:
$b = 165 \text{ mm}$
$k = 950 \text{ N/m}$

2. Relevant equations
$\delta = \frac{PL}{EA}$
$\delta = \frac{P}{k}$

3. The attempt at a solution

construct FBD

First find an expression for the force of the spring acting on the bar at B
$\displaystyle\delta = \frac{F_s - W_s}{k}\quad\Rightarrow\quad F_s = \delta k + W_s$

\begin{align*} \underline{^{+}\curvearrowleft\Sigma M_A = 0:}&& -W_p \cos\theta (3b/4) + F_s\cos\theta(b) &= 0\\ && -W_p \cos\theta (\frac{3b}{4}) + (\delta k +W_s)\cos\theta(b) &= 0\\ && (\delta k + W_s)\cos\theta(b)&=W_p \cos\theta (\frac{3b}{4})\\ &&(\delta k + W_s) &= W_p(\frac{3}{4})\\ && \delta &= \frac{1}{k}\left( \frac{3}{4}W_p - W_s \right)\\ &&&= \frac{1}{950 \text{ N/m}}\left(0.75\cdot3 \text{ N} - 2.75\text{ N}\right)\\ &&\delta&=-\frac{1}{1900}\text{ m} \end{align*}

Then
$\displaystyle\theta = \arctan\left(\frac{\delta}{b}\right) = \arctan\left(\frac{-1/1900\text{ m}}{0.165\text{m}}\right)\approx-0.1828^{\circ}$

Now assuming I made a sign error somewhere, and sould've added the weight of the spring, I'd get
$\delta = \frac{1}{190}\text{ m}$
and
$\theta\approx 1.828^{\circ}$

The answer given in the back of the book is $\theta = 1.325^{\circ}$, so I'm pretty sure I'm doing something wrong here. I just can't figure out what. Any help would be greatly appreciated.

Last edited: Apr 14, 2012