A steel bolt is insertede into a copper tube as shown in the 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.

Explanation

When a nut is turned, the tube is compressed and so its length decreases. The bolt is subjected to tensile force, so is length increase. The sum of these two lengths is the distance through which the nut moves.
Now `Y_(S)=(F//A_(S))/(/_\l//l)` or `/_\l=(Fxxl)/(Y_(S)A_(S))`
Similarly for the copper tube
`/_\l=(Fl)/(Y_(C)A_(C))`
`:. h=(Fl)/(Y_(S)A_(S))+(Fl)/(Y_(C)A_(C))`
`impliesF=h/l (Y_(S)Y_(C)A_(S)A_(C))/((Y_(S)A_(S)+Y_(C)A_(C)))`