Question

A steel bolt is inserted into a copper tube as shown in figure. Find the forces induced in the bolt and in the tube when the nut is turned through one revolution. Assume that the length of the tube is $l$, the pitch of the bolt thread is $h$ and the cross-sectional areas of the steel bolt and the copper tube are $A_s$ and $A_c$, respectively

Answer 1

When a nut is turned, the tube is compressed and so its length decreases. The bolt is subjected to tensile force, so its length increases. The sum of these two lengths is the distance through which the nut moves.
Now $Y_s=\frac{\left(F/A_s\right)}{\frac{\Delta l}{l}}or\Delta l=\frac{F\times l}{Y_s{A}_s}$
Similarly, for the copper tube
$\Delta l=\frac{Fl}{Y_c{A}_c}$
$\therefore h=\frac{Fl}{Y_s{A}_s}+\frac{Fl}{Y_c{A}_c}$
$\Rightarrow F=\frac{h}{l}\frac{Y_s{Y}_c{A}_s{A}_c}{(Y_s{A}_s+Y_c{A}_c)}$