1. **Problem Statement:** We have a beam supported at points A, C, and D with distributed loads varying linearly and a moment at point E. The beam is divided into sections of lengths 4.0m, 4.0m, 4.0m, and 2.0m. We need to draw the shear force and bending moment diagrams. 2. **Given Data:** - Distributed loads (T/m): - A to B: from 1.0 to 2.0 - B to C: from 2.0 to 1.0 - C to D: from 1.0 to 2.0 - D to E: uniform 2.0 - Lengths (m): 4.0, 4.0, 4.0, 2.0 - Moment at E: 5.0 T-m clockwise 3. **Formulas and Concepts:** - Shear force $V(x)$ is the integral of the distributed load $w(x)$ along the beam. - Bending moment $M(x)$ is the integral of the shear force $V(x)$. - For linearly varying distributed load $w(x) = w_1 + \frac{w_2 - w_1}{L}x$ over length $L$. - Calculate reactions at supports by equilibrium equations: $\sum F_y=0$, $\sum M=0$. 4. **Stepwise Solution:** **Step 1: Calculate equivalent point loads for each distributed load segment** - Segment AB (4m): average load $= \frac{1.0 + 2.0}{2} = 1.5$ T/m - Load magnitude $= 1.5 \times 4 = 6$ T - Acts at $\frac{4}{3}$ m from A (centroid of triangle + rectangle) - Segment BC (4m): average load $= \frac{2.0 + 1.0}{2} = 1.5$ T/m - Load magnitude $= 6$ T - Acts at $\frac{4}{3}$ m from B - Segment CD (4m): average load $= \frac{1.0 + 2.0}{2} = 1.5$ T/m - Load magnitude $= 6$ T - Acts at $\frac{4}{3}$ m from C - Segment DE (2m): uniform load $= 2.0$ T/m - Load magnitude $= 2.0 \times 2 = 4$ T - Acts at 1 m from D **Step 2: Calculate support reactions at A, C, and D** - Use equilibrium equations: - $\sum F_y = 0$ - $\sum M_A = 0$ - $\sum M_C = 0$ **Step 3: Construct shear force diagram** - Start from left end A, add/subtract loads and reactions moving right. - Shear changes linearly under linearly varying loads. **Step 4: Construct bending moment diagram** - Integrate shear force diagram. - Apply moment at E (5.0 T-m clockwise reduces moment by 5 at E). 5. **Summary:** - The shear force diagram will be piecewise linear with slopes equal to the distributed load. - The bending moment diagram will be piecewise quadratic. - The moment at E causes a jump of 5 T-m in the moment diagram. **Final Note:** Due to complexity, plotting is best done with software using the above calculations.