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

Step-by-step explanation:

Value of the cross-sectional area is as follows.

A =
`1.5 * 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 * 3.45`

= 50025 lb

Now, maximum load from shear stress is as follows.

`P_(max) = 2 * \tau_(a) * A`

=
`2 * 7100 * 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.