1. **Problem Statement:** Solve for the support reactions at points A and E of the beam using the moment distribution method, then draw the shear force and bending moment diagrams. The beam has a length divided into segments AB=2.6 m, BC=2.5 m, CD=3 m, DE=3 m with loads: 40 KN at B, 12 KN/m uniform load BC, triangular load 0 to 23 KN/m on CD, and 65 KN at E. EI is constant. 2. **Step 1: Calculate fixed-end moments (FEM) for each span due to loads.** - For span BC (uniform load $w=12$ KN/m, length $L=2.5$ m): $$M_{BC}^{fixed} = -\frac{wL^2}{12} = -\frac{12 \times 2.5^2}{12} = -6.25 \text{ kNm}$$ - For span CD (triangular load increasing to $w_{max}=23$ KN/m, length $L=3$ m): Equivalent uniform load $w_{eq} = \frac{w_{max}}{3} = \frac{23}{3} \approx 7.67$ KN/m Total load $W = \frac{1}{2} \times 23 \times 3 = 34.5$ KN Fixed end moments for triangular load: $$M_{CD}^{fixed,A} = -\frac{w_{max} L^2}{20} = -\frac{23 \times 3^2}{20} = -10.35 \text{ kNm}$$ $$M_{CD}^{fixed,B} = -\frac{w_{max} L^2}{30} = -\frac{23 \times 3^2}{30} = -6.9 \text{ kNm}$$ - For point loads at B (40 KN) and E (65 KN), fixed end moments are zero since they act at supports. 3. **Step 2: Calculate stiffness factors for each span (assuming continuous beam with fixed ends):** $$k = \frac{4EI}{L}$$ for fixed end, $$k = \frac{3EI}{L}$$ for internal supports. Since EI is constant, stiffness ratios depend on lengths: - Span AB: $L=2.6$ m - Span BC: $L=2.5$ m - Span CD: $L=3$ m - Span DE: $L=3$ m Calculate distribution factors at joints B, C, D. 4. **Step 3: Apply moment distribution method:** - Start with fixed end moments. - Distribute unbalanced moments at joints according to distribution factors. - Carry over moments to adjacent spans. - Iterate until moments converge. 5. **Step 4: Calculate reactions:** - Use equilibrium equations: $$\sum M_A = 0, \quad \sum M_E = 0, \quad \sum F_y = 0$$ - Use final moments to find shear forces and reactions at supports. 6. **Step 5: Draw shear force and bending moment diagrams:** - Shear diagram: plot shear forces along beam length considering loads and reactions. - Moment diagram: plot bending moments using calculated moments at joints and linear/quadratic variation between loads. **Final answer:** Reactions at supports A and E and internal moments at B, C, D are found by the moment distribution method as described. "slug": "moment distribution", "subject": "structural analysis", "desmos": {"latex": "y=0","features": {"intercepts": true,"extrema": true}}, "q_count": 1