1. **Problem Statement:** A simply supported beam AB of length 5 m has a uniformly distributed load (UDL) of 10 kN/m applied from 1 m to 3 m from point A (i.e., over 2 m length). 2. **Find support reactions:** - Total load $W = 10 \times 2 = 20$ kN. - Load acts at the midpoint of the loaded length, which is at $1 + \frac{2}{2} = 2$ m from A. - Let reactions at supports A and B be $R_A$ and $R_B$. Using equilibrium equations: $$\sum F_y = 0: R_A + R_B = 20$$ $$\sum M_A = 0: R_B \times 5 = 20 \times 2$$ $$R_B = \frac{40}{5} = 8 \text{ kN}$$ $$R_A = 20 - 8 = 12 \text{ kN}$$ 3. **Find Shear Force (SF) and Bending Moment (BM):** - Shear force just left of 1 m (start of load): $V = R_A = 12$ kN. - Shear force just right of 1 m: $V = 12 - 10 \times 0 = 12$ kN. - Shear force just right of 3 m (end of load): $V = 12 - 10 \times 2 = 12 - 20 = -8$ kN. - Shear force just left of B: $V = -8$ kN. - Shear force just right of B: $V = -8 + R_B = -8 + 8 = 0$ kN. Bending moment at key points: - At A (0 m): $M_A = 0$ - At 1 m (start of load): $M = R_A \times 1 = 12 \times 1 = 12$ kNm - At 3 m (end of load): $$M = R_A \times 3 - 10 \times 2 \times \frac{2}{2} = 12 \times 3 - 20 \times 1 = 36 - 20 = 16 \text{ kNm}$$ - At B (5 m): $M_B = 0$ 4. **Draw Shear Force Diagram (SFD) and Bending Moment Diagram (BMD):** - SFD: Starts at 12 kN at A, constant till 1 m, linearly decreases to -8 kN at 3 m, constant till B, jumps to 0 at B. - BMD: Starts at 0, rises to 12 kNm at 1 m, peaks at 16 kNm at 3 m, returns to 0 at B. 5. **Find maximum Bending Moment:** - Maximum bending moment is $16$ kNm at 3 m from A. Final answers: - Support reactions: $R_A = 12$ kN, $R_B = 8$ kN. - Maximum bending moment: $16$ kNm.