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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.

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  • $\begingroup$ Please show your effort in working towards a solution, then people may help. $\endgroup$
    – Solar Mike
    Commented Jul 24, 2019 at 9:16
  • $\begingroup$ I have gotten up to δL = 0.01 $\endgroup$
    – 2hn
    Commented Jul 24, 2019 at 9:21
  • $\begingroup$ 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. $\endgroup$
    – 2hn
    Commented Jul 24, 2019 at 9:25

1 Answer 1

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Let's do it over. First we calculate the tension stress. $$\sigma=P/A=\frac{20kN}{100*200}=1N/mm^2 $$

then as per definition of Young modulus,

$$ \epsilon=\frac{\sigma}{E}=\frac{1N}{2×10^5}=0.000005 $$ and $$ lateral\ shrinkage=\mu*0.000005= 0.00000166$$ And then $$200*0.00000166= 0.000332mm $$

the amount the 200mm side narrows is small, 33/100000mm.

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