Question

A circular bar of steel is pulled in tension with a force (F) of 88 kN. If the original diameter the bar is 8 mm, calculate the change in diameter (in mm) of the bar.

(Assume E =207 GPa and υ = 0.33 for the bar material)

Answer 1

\[1)\sigma=E*\epsilon\]\[2)\epsilon _{z}=-\upsilon \epsilon_{x}\]\[3)\sigma=E(-\upsilon \epsilon_{x})\]\[4)(F/A_{i}) =-\upsilon E (\Delta d/d_{i})\]\[5)\Delta d = -(F/A_{i})(d_{i}/(\upsilon E))\]

Here are the mathematical steps that you would go through to get to the answer. Now for the explanation:
1) We start out with the basic Hooke's Law for stress strain relations. The force applied is tensile along the longitudinal axis of the bar (I will name this the z-axis...see drawing for clarification of axis).
2) We know that the not only will the length increase but the diameter will decrease. The change in diameter is not in the same direction as the force applied and so we need to relate the strain in the x and y ( they are the same since we are dealing with a circular cross section) to the strain inline with the force.
3) Plug in equation 2) into equation 1).

4) Start breaking down terms. We know that \[\sigma = (F/A)\] and that \[\epsilon =\Delta d/d \] Insert them into the equation.

5) Last step is to now solve for \[\Delta d\] and bing bada boom...you have your answer.