# Dimensions of a bar when it is subjected to axial pull

Determine the change in dimensions of a rectangular bar of length 2 m, width 200 mm, and thickness 100 mm, when it is subjected to an axial pull of 20 kN in the direction of its length.

Take $$E = 2\times10^5\text{ N/mm}^2$$ and $$\mu = 0.3$$.

I have gotten up to $$\delta L = 0.01$$. Then using $$\mu = \dfrac{δb/b}{δL/L}$$, I got $$\delta b = \dfrac{\mu bL}{δL}$$, which is 0.3×200×2000/0.01 = 12million!

I don't know where I have mistaken but it is frustrating me.

• Please show your effort in working towards a solution, then people may help. Commented Jul 24, 2019 at 9:16
• I have gotten up to δL = 0.01
– 2hn
Commented Jul 24, 2019 at 9:21
• then using μ = (δb/b)/(δL/L) ; I got δb = μbL/δL ; which is = 0.3×200×2000/0.01 = 12million!!!! I don't know where I have mistaken but it is frustrating me. I don't have a tutor. please help.
– 2hn
Commented Jul 24, 2019 at 9:25

Let's do it over. First we calculate the tension stress. $$\sigma=P/A=\frac{20kN}{100*200}=1N/mm^2$$
$$\epsilon=\frac{\sigma}{E}=\frac{1N}{2×10^5}=0.000005$$ and $$lateral\ shrinkage=\mu*0.000005= 0.00000166$$ And then $$200*0.00000166= 0.000332mm$$