1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change in length of axially loaded member

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Given:
    [itex]b = 165 \text{ mm}[/itex]
    [itex]k = 950 \text{ N/m}[/itex]
    2r5eo9v.png

    dpcys8.png


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

    3. The attempt at a solution

    construct FBD

    35mnp7t.png

    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]

    [itex]\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*}[/itex]

    Then
    [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]
    and
    [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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Change in length of axially loaded member
  1. Axial Deformation (Replies: 0)

  2. Transverse loading (Replies: 0)

Loading...