Question

A specimen of aluminum having a rectangular cross-section of 10 mm x 12.7 mm is pulled in tension with 35,500 N of force, producing only elastic deformation. Calculate the resulting strain.

Answer 1

Explanation:

The given data is as follows.

            Cross-section area of the rectangular is as follows.

      Area = $10 mm \times 12.7 mm$

              = $127 \times 10^{-6} m^{2}$

              = $1.27 \times 10^{-4} m^{2}$

Tension applied on the specimen is 35,500 N

Now, formula to calculate the modulus of elasticity for a material of aluminium is as follows.

       E = 70 GPa

          = $70 \times 10^{9} Pa$

or,      = $70 \times 10^{9} N/m^{2}$

Now, stress on the specimen is as follows.

           $\sigma = \frac{P}{A}$

                      = $\frac{35000}{1.27 \times 10^{-4}}$

                      = $275.59 \times 10^{6} N/m^{2}$

According to Hook's law, we will calculate the resulting stain as follows.

           $\sigma = E \epsilon$

       $\epsilon = \frac{\sigma}{E}$

                  = $\frac{275.59 \times 10^{6}}{70 \times 10^{9}}$

                  = $3.937 \times 10^{-3} mm/mm$

Thus, we can conclude that the resulting strain is $3.937 \times 10^{-3} mm/mm$.