This problem is also straightforward as the previous one. The problem is sketched in Figure 2.16.

The resisting area is (annular portion) =\frac{\pi}{4}\left(50^2-40^2\right)

= 706.85 mm²

The normal stress =\sigma=\frac{55 \times 10^3}{706.85}

= 77.80 N/mm².