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Homework Help: Change in length of axially loaded member

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]b = 165 \text{ mm}[/itex]
    [itex]k = 950 \text{ N/m}[/itex]


    2. Relevant equations
    [itex]\delta = \frac{PL}{EA}[/itex]
    [itex]\delta = \frac{P}{k}[/itex]

    3. The attempt at a solution

    construct FBD


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

    \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}

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

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

    The answer given in the back of the book is [itex]\theta = 1.325^{\circ}[/itex], 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
  2. jcsd
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