1. **Problem Statement:** A simply supported beam AB of length 5 m has a uniformly distributed load (UDL) of 10 kN/m over a 2 m length starting 1 m from support A. 2. **Find support reactions:** - Total load $W = 10 \times 2 = 20$ kN acting at the midpoint of the loaded segment, which is 1 m + 1 m = 2 m from A. - Let reactions at supports A and B be $R_A$ and $R_B$. - Using equilibrium equations: - Sum of vertical forces: $R_A + R_B = 20$ - Taking moments about A: $R_B \times 5 = 20 \times 2$ so $R_B = \frac{40}{5} = 8$ kN - Then $R_A = 20 - 8 = 12$ kN 3. **Find Shear Force (SF) and Bending Moment (BM):** - Shear force just left of C (start of load, 1 m from A): $SF = R_A = 12$ kN - Shear force just right of C: $SF = 12 - 10 \times 0 = 12$ kN - Shear force at end of load (3 m from A): $SF = 12 - 10 \times 2 = 12 - 20 = -8$ kN - Shear force at B: $SF = -8 + R_B = -8 + 8 = 0$ kN - Bending moment at A: $BM_A = 0$ - Bending moment at C (1 m from A): $BM_C = R_A \times 1 = 12 \times 1 = 12$ kNm - Bending moment at 3 m (end of load): $$BM = R_A \times 3 - 10 \times 2 \times (3 - 1) / 2 = 12 \times 3 - 20 \times 1 = 36 - 20 = 16$$ kNm - Bending moment at B: $BM_B = 0$ 4. **Draw Shear Force Diagram (SFD) and Bending Moment Diagram (BMD):** - SFD starts at 12 kN at A, remains constant until C, decreases linearly to -8 kN at 3 m, then jumps to 0 at B. - BMD starts at 0 at A, rises to 12 kNm at C, peaks at 16 kNm at 3 m, then 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