# Problem 904 | Combined Axial and Bending

**Problem 904**

To avoid interference, a link in a machine is designed so that its cross-sectional area is reduced one half at section A-B as shown in Fig. P-904. If the thickness of the link is 50 mm, compute the maximum force P that can be applied if the maximum normal stress on section A-B is limited to 80 MPa.

**Solution 904**

$M = 20P$

$\sigma = \sigma_a + \sigma_f$

From the top of section A-B

$\sigma = \dfrac{P}{A} + \dfrac{6M}{bd^2}$

$80 = \dfrac{P}{50(40)} + \dfrac{6(20P)}{50(40^2)}$

$80 = \dfrac{P}{2\,000} + \dfrac{3P}{2\,000}$

$80 = \dfrac{P}{500}$

$P = 40\,000 ~ \text{N} = 40 ~ \text{kN}$ *answer*