Question

A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in tension and shear are 14,500 psi and 7,100 psi, respectively. Determine the maximum permissible load Pmax.

Answer 1

Explanation:

Value of the cross-sectional area is as follows.

        A = $1.5 \times 2.30$

           = 3.45 $in^{2}$

The given data is as follows.

          Allowable stress = 14,500 psi

          Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

           $P_{max} = \sigma_{a}A$

                       = $14500 \times 3.45$

                       = 50025 lb

Now, maximum load from shear stress is as follows.

           $P_{max} = 2 \times \tau_{a} \times A$

                      = $2 \times 7100 \times 3.45$

                      = 48990 lb

Hence, $P_{max}$ will be calculated as follows.

       $P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})$

                  = 48990 lb

Thus, we can conclude that the maximum permissible load $P_{max}$ is 48990 lb.