1. **Problem Statement:** We have a simply supported beam of length 3 m with a uniform distributed load (UDL) of 1.5 kN/m acting downward over the entire length. We need to draw the shear force and bending moment diagrams and find the distance $b$ from the left end where the bending moment is zero between the supports. 2. **Given Data:** - Length of beam, $L = 3$ m - Uniform load, $w = 1.5$ kN/m - Supports at $x=0$ m (left end) and $x=3$ m (right end) 3. **Step 1: Calculate reactions at supports** The total load on the beam is: $$ W = w \times L = 1.5 \times 3 = 4.5 \text{ kN} $$ Since the beam is simply supported and the load is uniform, the reactions at both supports are equal: $$ R_A = R_B = \frac{W}{2} = \frac{4.5}{2} = 2.25 \text{ kN} $$ 4. **Step 2: Shear force diagram (V)** - At $x=0$, shear force $V = +2.25$ kN (upward reaction) - Moving right, shear decreases linearly due to UDL: $$ V(x) = R_A - w x = 2.25 - 1.5 x $$ - At $x=3$ m, shear force: $$ V(3) = 2.25 - 1.5 \times 3 = 2.25 - 4.5 = -2.25 \text{ kN} $$ 5. **Step 3: Bending moment diagram (M)** - Moment at left support $M(0) = 0$ (simply supported) - Moment at any point $x$ is: $$ M(x) = R_A x - \frac{w x^2}{2} = 2.25 x - 0.75 x^2 $$ - Moment at right support $M(3) = 2.25 \times 3 - 0.75 \times 9 = 6.75 - 6.75 = 0$ 6. **Step 4: Find $b$ where bending moment is zero between supports** Set $M(b) = 0$: $$ 2.25 b - 0.75 b^2 = 0 $$ Factor out $b$: $$ b (2.25 - 0.75 b) = 0 $$ Solutions: $$ b = 0 \quad \text{or} \quad 2.25 - 0.75 b = 0 $$ Solve second: $$ 0.75 b = 2.25 \Rightarrow b = \frac{2.25}{0.75} = 3 $$ The bending moment is zero at the supports $b=0$ and $b=3$ m only. 7. **Interpretation:** The bending moment is zero only at the supports, so there is no point between the supports where the bending moment is zero. **Final answer:** - Reactions: $R_A = R_B = 2.25$ kN - Shear force: $V(x) = 2.25 - 1.5 x$ - Bending moment: $M(x) = 2.25 x - 0.75 x^2$ - Distance $b$ where bending moment is zero between supports: None (only at supports $b=0$ and $b=3$ m)