1. **Problem Statement:** Determine the reactions at the supports for the beam ABCD with point loads 150 kN at B, 60 kN at D, and a uniformly distributed load (UDL) of 12 kN/m over the entire length from A to D. 2. **Given Data:** - Loads: 150 kN at B, 60 kN at D, UDL = 12 kN/m over 17 m (6+6+5) - Beam spans: AB = 6 m, BC = 6 m, CD = 5 m - Supports: A (pinned), C (roller) 3. **Goal:** Find vertical reactions at supports A and C, denoted as $R_A$ and $R_C$. 4. **Formulas and Rules:** - Sum of vertical forces must be zero: $$\sum F_y = 0$$ - Sum of moments about any point must be zero: $$\sum M = 0$$ - UDL total load = $12 \times 17 = 204$ kN acting at midpoint of the beam (8.5 m from A) 5. **Step 1: Calculate total vertical loads:** $$\text{Total load} = 150 + 60 + 204 = 414 \text{ kN}$$ 6. **Step 2: Write equilibrium equations:** - Vertical forces: $$R_A + R_C = 414$$ - Moments about A (taking counterclockwise as positive): $$-150 \times 6 - 204 \times 8.5 - 60 \times 17 + R_C \times 12 = 0$$ 7. **Step 3: Calculate moments:** $$-150 \times 6 = -900$$ $$-204 \times 8.5 = -1734$$ $$-60 \times 17 = -1020$$ Sum: $$-900 - 1734 - 1020 + 12 R_C = 0$$ $$12 R_C = 900 + 1734 + 1020 = 3654$$ $$R_C = \frac{3654}{12} = 304.5 \text{ kN}$$ 8. **Step 4: Find $R_A$ using vertical force equilibrium:** $$R_A = 414 - 304.5 = 109.5 \text{ kN}$$ 9. **Answer:** - Reaction at support A: $R_A = 109.5$ kN upward - Reaction at support C: $R_C = 304.5$ kN upward These reactions balance the applied loads and satisfy static equilibrium conditions.