1. **State the problem:** Calculate the direct axial stress at point A on the top of a rectangular beam with breadth $b=0.03$ m, depth $d=0.21$ m, subjected to an axial compressive load $F=61000$ N and a bending moment $M=4000$ Nm. 2. **Relevant formulas:** The total axial stress $\sigma$ at a point due to axial load and bending moment is given by: $$\sigma = \sigma_{axial} + \sigma_{bending}$$ where $$\sigma_{axial} = \frac{F}{A}$$ $$\sigma_{bending} = \frac{M y}{I}$$ 3. **Calculate cross-sectional area $A$:** $$A = b \times d = 0.03 \times 0.21 = 0.0063\ \text{m}^2$$ 4. **Calculate moment of inertia $I$ about the neutral axis (centroidal axis):** For a rectangle, $$I = \frac{b d^3}{12} = \frac{0.03 \times (0.21)^3}{12}$$ Calculate $d^3$: $$0.21^3 = 0.009261$$ So, $$I = \frac{0.03 \times 0.009261}{12} = \frac{0.00027783}{12} = 2.31525 \times 10^{-5}\ \text{m}^4$$ 5. **Determine $y$ coordinate of point A:** Point A is at the top surface, so distance from neutral axis (center) is half the depth: $$y = \frac{d}{2} = \frac{0.21}{2} = 0.105\ \text{m}$$ 6. **Calculate axial stress:** $$\sigma_{axial} = \frac{F}{A} = \frac{61000}{0.0063} = 9682539.68\ \text{Pa} = 9.6825\ \text{MPa}$$ Since the load is compressive, axial stress is negative: $$\sigma_{axial} = -9.6825\ \text{MPa}$$ 7. **Calculate bending stress:** $$\sigma_{bending} = \frac{M y}{I} = \frac{4000 \times 0.105}{2.31525 \times 10^{-5}} = \frac{420}{2.31525 \times 10^{-5}} = 18136305.5\ \text{Pa} = 18.1363\ \text{MPa}$$ 8. **Determine sign of bending stress:** At the top fiber, bending stress is tensile if moment causes compression at bottom and tension at top, or vice versa. Given the bending moment direction, assume top fiber is in tension, so positive. 9. **Calculate total stress at point A:** $$\sigma = \sigma_{axial} + \sigma_{bending} = -9.6825 + 18.1363 = 8.4538\ \text{MPa}$$ 10. **Final answer rounded to 1 decimal place:** $$\boxed{8.5\ \text{MPa}}$$ This means the top of the beam at point A experiences a net tensile stress of 8.5 MPa.