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Derive an expression for the deflection between two forces

  1. Sep 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Two rods of the same material, of equal length l, and of cross section A and 2A respectively, are mounted between two rigid (i.e., nondeformable) cross frames. The frames are pulled by a pair of forces P, located at distance x from the thin bar. Derive an expression for the deflection between the two forces P, that is, for the elastic increase in distance between them.
    http://imgur.com/a/SVZgS

    2. Relevant equations
    Hooke's Law

    $$ s = E \frac{\delta}{l} $$

    3. The attempt at a solution
    Well first of all I obtained the reactions in each rod:
    $$\sum M = 0$$
    $$-Px +R_2 a = 0$$
    $$R_2 = \frac{Px}{a}$$
    $$\sum F_y=0$$
    $$-R_1 + P - R_2 =0$$
    $$R_1 = P (1-\frac{x}{a})$$
    According to Hooke's Law, δ will be for each force:
    $$\frac{R_2}{2A}= \delta_2 \frac{E}{l}$$
    $$\delta_2 = \frac{P l}{A E} \frac{x}{2a}$$
    And similar ##\delta_1## will be
    $$\delta_1 = \frac{Pl}{AE} (1- \frac{x}{a})$$
    Then to obtain the elongation between the two forces, I propose the following relation (where I believe it's the error)
    http://imgur.com/a/GTWT2
    So according to my diagram:
    $$\frac{\delta_1 + \delta_2}{a} = \frac{\delta_P + \delta_2}{a-x} $$
    doing the algebra I end up with a wrong answer:
    $$\delta_P = \frac{Pl}{AE} (1-\frac{2x}{a}+\frac{x^2}{2a^2})$$

    The correct solution is something similar:
    $$\delta_P = \frac{Pl}{AE} (1-\frac{2x}{a}+\frac{3x^2}{2a^2})$$
    :oldgrumpy:
    Thanks!
     
  2. jcsd
  3. Sep 28, 2016 #2
    I get $$\delta_p=\delta_1+(\delta_2-\delta_1)\frac{x}{a}$$
     
  4. Sep 28, 2016 #3
    Hi Chestermiller thanks for your reply! It seems that I was confused about how it would be the deformation for this case...
    By the way this is problem 3 from the book Strength of materials by Den Hartog.
    Thanks again!
     
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