1. **State the problem:** We have a bar with cross-sectional area $A = 1385$ mm$^2$ under a compressive load $P = 60000$ N (since 60 kN = 60000 N). The bar is cut at an angle $B = 29^\circ$ to the horizontal, and we need to find the direct stress normal to this cut plane. 2. **Formula and explanation:** The direct normal stress $\sigma$ on a plane is defined as the force perpendicular to that plane divided by the area of that plane. Since the load $P$ is applied horizontally, the force component normal to the cut plane is $P \cos B$. The area of the cut plane $A_{cut}$ is larger than the original cross-sectional area because the cut is at an angle. It can be found by: $$ A_{cut} = \frac{A}{\cos B} $$ Therefore, the normal stress on the cut plane is: $$ \sigma = \frac{P \cos B}{A / \cos B} = \frac{P \cos^2 B}{A} $$ 3. **Calculate intermediate values:** - $\cos 29^\circ \approx 0.8746$ - $\cos^2 29^\circ = (0.8746)^2 \approx 0.7650$ 4. **Calculate the normal stress:** $$ \sigma = \frac{60000 \times 0.7650}{1385} = \frac{45900}{1385} \approx 33.14 \text{ N/mm}^2 $$ Since 1 N/mm$^2$ = 1 MPa, the stress is approximately 33.14 MPa. 5. **Final answer:** The direct normal stress on the cut plane is **33.14 MPa**.