# Derive an expression for the deflection between two forces

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1. Sep 27, 2016

### cementite

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:
$$\delta_P = \frac{Pl}{AE} (1-\frac{2x}{a}+\frac{3x^2}{2a^2})$$

Thanks!

2. Sep 28, 2016

### Staff: Mentor

I get $$\delta_p=\delta_1+(\delta_2-\delta_1)\frac{x}{a}$$

3. Sep 28, 2016

### cementite

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!