1. **Problem Statement:** We have a simply supported beam with a uniform distributed load (UDL) of 1.5 kN/m over its entire length of 3 m. Supports: pinned at left end (A), roller at 2 m from left end (B). We need to draw shear and bending moment diagrams and find the distance $b$ from the left end where the bending moment is zero between the supports. 2. **Step 1: Calculate reactions at supports** Total load $W = 1.5 \times 3 = 4.5$ kN. Let reaction at A be $R_A$ and at B be $R_B$. Taking moments about A: $$R_B \times 2 = 1.5 \times 3 \times \frac{3}{2} = 4.5 \times 1.5 = 6.75$$ $$R_B = \frac{6.75}{2} = 3.375 \text{ kN}$$ Sum of vertical forces: $$R_A + R_B = 4.5 \Rightarrow R_A = 4.5 - 3.375 = 1.125 \text{ kN}$$ 3. **Step 2: Shear force diagram (V)** - At A (x=0): $V = R_A = 1.125$ kN - Between A and B (0 < x < 2): $$V = R_A - 1.5x$$ - At B (x=2): $$V = 1.125 - 1.5 \times 2 = 1.125 - 3 = -1.875 \text{ kN}$$ - Between B and end (2 < x < 3): $$V = 1.125 - 1.5x + R_B = 1.125 - 1.5x + 3.375 = 4.5 - 1.5x$$ - At end (x=3): $$V = 4.5 - 1.5 \times 3 = 0$$ 4. **Step 3: Bending moment diagram (M)** Moment at A (x=0): $M=0$ Between A and B (0 < x < 2): $$M = R_A x - \frac{1.5 x^2}{2} = 1.125 x - 0.75 x^2$$ Between B and end (2 < x < 3): $$M = R_A x - \frac{1.5 x^2}{2} - R_B (x-2) = 1.125 x - 0.75 x^2 - 3.375 (x-2)$$ Simplify: $$M = 1.125 x - 0.75 x^2 - 3.375 x + 6.75 = -2.25 x - 0.75 x^2 + 6.75$$ 5. **Step 4: Find $b$ where bending moment is zero between supports (0 < b < 2)** Set moment formula between A and B to zero: $$1.125 b - 0.75 b^2 = 0$$ Factor: $$b (1.125 - 0.75 b) = 0$$ Non-zero solution: $$1.125 - 0.75 b = 0 \Rightarrow b = \frac{1.125}{0.75} = 1.5 \text{ m}$$ **Final answer:** The bending moment is zero at $b = 1.5$ m from the left end. --- **Desmos function for moment between supports:** $$y = 1.125 x - 0.75 x^2$$