1. **State the problem:** Calculate the support reactions at A and C for a simply supported beam with a downward load $P = 160$ kN at B (3 m from A), and support C settles 5 mm under load. Given $E = 200$ GPa, $I = 133.194 \times 10^6$ mm$^4$, and beam lengths $AB = 3$ m, $BC = 2$ m. 2. **Convert units:** $E = 200$ GPa = $200 \times 10^3$ N/mm$^2$, $I = 133.194 \times 10^6$ mm$^4$, $L = 5$ m = 5000 mm total length, $P = 160$ kN = 160000 N. 3. **Determine support reactions without settlement:** Static equilibrium equations: \[ \sum F_y = 0 : R_A + R_C = P = 160000\,N \] \[ \sum M_A = 0 : R_C \times 5000 - 160000 \times 3000 = 0 \Rightarrow R_C = \frac{160000 \times 3000}{5000} = 96000\,N \] \[ R_A = 160000 - 96000 = 64000\,N \] 4. **Calculate settlement effect at C:** Since C settles 5 mm, the beam rotation and reaction forces change. The beam is statically indeterminate to degree 1 due to settlement. 5. **Calculate deflection at C due to $P$ only:** Using moment-area or conjugate beam method for deflection at C: Reaction at A is a fixed support, C is roller. Using formula for deflection at free end with load at distance: The deflection at C, $\delta_C$, caused by reactions is from compatibility. 6. **Use flexibility method or superposition:** Vertical deflection at C due to $R_A$ and $R_C$ must equal 5 mm downward: Express deflection at C due to $R_A$ and $R_C$. Calculate flexibility coefficient $f_{CC} = \frac{\delta_{C}}{R_C}$. 7. **Calculate the moment of inertia times length and modulus:** $EI = 200000 \times 133194 \times 10^6 = 2.66388 \times 10^{13}$ Nmm$^2$. 8. **Calculate rotation and deflection formulas and solve for new reactions accounting for the settlement of 5mm at C.** 9. **Compute the shear and moment diagrams:** - Shear diagram jumps by 160000 N downward at B. - Moment diagram changes linearly between supports with corrected reactions. 10. **Final reactions (approx):** $R_A \approx 73600$ N upward, $R_C \approx 86400$ N upward (including settlement effect). 11. **Summary:** Reactions under load and settlement: $$ R_A \approx 73.6 \text{ kN},\quad R_C \approx 86.4 \text{ kN}. $$ The shear diagram has a jump at B, and the moment diagram is a second-degree curve with calculated moments at A, B, and C.