1. **State the problem:** We need to design a compression member 3m long using a pair of angles with long legs back to back, connected by a 10 mm gusset plate at each end. The member carries a load of 450000 N (450 kN) and the stress is limited to 100 MPa. Given: - Modulus of elasticity, $E = 200000$ MPa - Yield strength, $F_y = 230$ MPa - Load, $P = 450000$ N - Allowable stress, $\sigma = 100$ MPa 2. **Calculate the required cross-sectional area:** $$ A = \frac{P}{\sigma} = \frac{450000}{100} = 4500 \text{ mm}^2 $$ This is the minimum required cross-sectional area to carry the load safely. 3. **Proposed section:** Two angles (150 x 100 x 10 mm) back to back with a 10 mm gusset plate. Given properties for one angle: - Area, $A_{angle} = 2329$ mm$^2$ (since two angles total $4658$ mm$^2$) - Radius of gyration, $r = 41$ mm - Distance from neutral axis to outer fiber, $y = 49.3$ mm 4. **Check cross-sectional area:** Total area $A_{total} = 4658$ mm$^2$ which is greater than the required $4500$ mm$^2$, so the proposed section area is sufficient. 5. **Summary:** The chosen section (two 150x100x10 mm angles back to back) connected with a 10 mm gusset plate provides a total cross-sectional area of 4658 mm$^2$, exceeding the required area of 4500 mm$^2$, making it safe to carry the 450 kN load with an allowable stress of 100 MPa. Final answer: Use two 150x100x10 mm angles back to back with a 10 mm gusset plate at each end.