Question

A specimen of aluminum having a rectangular cross section 9.6 mm × 12.4 mm (0.3780 in. × 0.4882 in.) is pulled in tension with 35000 N (7868 lbf) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 × 106 psi). Calculate the resulting strain.

Answer 1

Answer:

0.00426

Explanation:

Young's modulus is given by the following expression,

$Y=\frac{Stress}{Strain} =\frac{F/A}{\Delta l/l}$

where,

$Y$ = Young's modulus

$F$ = force applied on the material

$A$ = cross-sectional area

$l$ = length of the material

$\Delta l$ = elongation due to stress applied

For our problem, we need to find the strain of the aluminum specimen,

$Strain=\frac{Stress}{Y} =\frac{F/A}{Y}=\frac{7868/(0.3780\times0.4882)}{10\times10^6} =0.00426$

Strain is a ratio and hence a dimensionless quantity, so even if calculate using the values in S.I. units, we shall get the same value.