1. **Problem Statement:** We have a simply supported beam of length $L$ with supports at $A$ (pin) and $C$ (roller). The beam is loaded with a triangular distributed load starting at zero at $A$, increasing linearly to $w_0$ at midspan $B$ ($L/2$), then symmetrically decreasing back to zero at $C$. 2. **Goal:** Draw the shear force $V(x)$ and bending moment $M(x)$ diagrams and find the expressions for $V(x)$ and $M(x)$ along the beam. 3. **Load Description:** The load intensity $w(x)$ varies linearly: - From $A$ to $B$ ($0 \leq x \leq L/2$): $w(x) = \frac{2w_0}{L}x$ - From $B$ to $C$ ($L/2 < x \leq L$): $w(x) = 2w_0 - \frac{2w_0}{L}x$ 4. **Reactions at Supports:** By symmetry and equilibrium: - Total load $W = \text{area under load} = \frac{1}{2} \times \frac{L}{2} \times w_0 + \frac{1}{2} \times \frac{L}{2} \times w_0 = \frac{w_0 L}{2}$ - Since load is symmetric, reactions at $A$ and $C$ are equal: $$ R_A = R_C = \frac{W}{2} = \frac{w_0 L}{4} $$ 5. **Shear Force $V(x)$:** Shear is the reaction minus the load up to point $x$. - For $0 \leq x \leq L/2$: $$ V(x) = R_A - \int_0^x w(t) dt = \frac{w_0 L}{4} - \int_0^x \frac{2w_0}{L} t dt = \frac{w_0 L}{4} - \frac{w_0}{L} x^2 $$ - For $L/2 < x \leq L$: Calculate load from 0 to $L/2$ plus from $L/2$ to $x$: $$ \int_0^{L/2} w(t) dt = \frac{w_0 L}{4} $$ $$ \int_{L/2}^x w(t) dt = \int_{L/2}^x \left(2w_0 - \frac{2w_0}{L} t\right) dt = 2w_0(x - \frac{L}{2}) - \frac{w_0}{L}(x^2 - \frac{L^2}{4}) $$ So, $$ V(x) = R_A - \frac{w_0 L}{4} - \left[2w_0(x - \frac{L}{2}) - \frac{w_0}{L}(x^2 - \frac{L^2}{4})\right] = \frac{w_0 L}{4} - \frac{w_0 L}{4} - 2w_0(x - \frac{L}{2}) + \frac{w_0}{L}(x^2 - \frac{L^2}{4}) $$ Simplify: $$ V(x) = -2w_0 x + w_0 L + \frac{w_0}{L} x^2 - \frac{w_0 L}{4} = w_0 L - 2w_0 x + \frac{w_0}{L} x^2 - \frac{w_0 L}{4} $$ 6. **Bending Moment $M(x)$:** Moment is integral of shear: - For $0 \leq x \leq L/2$: $$ M(x) = \int_0^x V(t) dt = \int_0^x \left(\frac{w_0 L}{4} - \frac{w_0}{L} t^2\right) dt = \frac{w_0 L}{4} x - \frac{w_0}{L} \frac{x^3}{3} = \frac{w_0 L}{4} x - \frac{w_0 x^3}{3L} $$ - For $L/2 < x \leq L$: $$ M(x) = M(\frac{L}{2}) + \int_{L/2}^x V(t) dt $$ Calculate $M(L/2)$: $$ M(\frac{L}{2}) = \frac{w_0 L}{4} \cdot \frac{L}{2} - \frac{w_0}{L} \frac{(L/2)^3}{3} = \frac{w_0 L^2}{8} - \frac{w_0 L^2}{24} = \frac{w_0 L^2}{12} $$ Integral of $V(t)$ from $L/2$ to $x$: $$ \int_{L/2}^x \left(w_0 L - 2w_0 t + \frac{w_0}{L} t^2 - \frac{w_0 L}{4}\right) dt = \int_{L/2}^x \left(\frac{3w_0 L}{4} - 2w_0 t + \frac{w_0}{L} t^2\right) dt $$ Calculate: $$ = \frac{3w_0 L}{4} (x - \frac{L}{2}) - w_0 (x^2 - \frac{L^2}{4}) + \frac{w_0}{3L} (x^3 - \frac{L^3}{8}) $$ So, $$ M(x) = \frac{w_0 L^2}{12} + \frac{3w_0 L}{4} (x - \frac{L}{2}) - w_0 (x^2 - \frac{L^2}{4}) + \frac{w_0}{3L} (x^3 - \frac{L^3}{8}) $$ 7. **Summary:** $$ V(x) = \begin{cases} \frac{w_0 L}{4} - \frac{w_0}{L} x^2 & 0 \leq x \leq \frac{L}{2} \\ w_0 L - 2w_0 x + \frac{w_0}{L} x^2 - \frac{w_0 L}{4} & \frac{L}{2} < x \leq L \end{cases} $$ $$ M(x) = \begin{cases} \frac{w_0 L}{4} x - \frac{w_0 x^3}{3L} & 0 \leq x \leq \frac{L}{2} \\ \frac{w_0 L^2}{12} + \frac{3w_0 L}{4} (x - \frac{L}{2}) - w_0 (x^2 - \frac{L^2}{4}) + \frac{w_0}{3L} (x^3 - \frac{L^3}{8}) & \frac{L}{2} < x \leq L \end{cases} $$ These expressions define the shear and moment diagrams for the beam under the given triangular load.