1. **Problem Statement:** Draw the moment diagrams for the beam ABC using the method of superposition, considering the beam is cantilevered at support B. 2. **Given Data:** - Beam length divided into three segments of 3 m each: A-B, B-center, center-C. - Loads: 12 kN downward at 3 m from A (point between A and B), 4 kN downward at C. - Moment: 12 kN·m clockwise at B. - Support B is the cantilever point. 3. **Method of Superposition:** We analyze the effect of each load and moment separately on the cantilevered beam at B, then sum the moments. 4. **Step 1: Moment due to 12 kN load at 3 m from A (left of B):** - Distance from B to load = 3 m (since A-B = 3 m). - Moment at B due to this load: $$M_{B1} = -12 \times 3 = -36\ \text{kN}\cdot\text{m}$$ (negative sign for clockwise moment on cantilever). 5. **Step 2: Moment due to 4 kN load at C (right end):** - Distance from B to C = 6 m (B to center 3 m + center to C 3 m). - Moment at B due to this load: $$M_{B2} = -4 \times 6 = -24\ \text{kN}\cdot\text{m}$$ (clockwise moment). 6. **Step 3: Moment due to applied 12 kN·m moment at B:** - Given as clockwise moment: $$M_{B3} = -12\ \text{kN}\cdot\text{m}$$ 7. **Step 4: Total moment at B:** $$M_B = M_{B1} + M_{B2} + M_{B3} = -36 - 24 - 12 = -72\ \text{kN}\cdot\text{m}$$ 8. **Step 5: Moment diagram construction:** - At B (cantilever support), moment is maximum negative (clockwise) of -72 kN·m. - Moment at free end A and C is zero. - Moment diagram is linear between loads and supports. 9. **Summary:** The moment diagram is the superposition of three effects: - 12 kN load at 3 m left of B creates a negative moment of -36 kN·m at B. - 4 kN load at C creates a negative moment of -24 kN·m at B. - 12 kN·m moment at B adds -12 kN·m. - Total moment at B is -72 kN·m. This completes the moment diagram using superposition for the cantilevered beam at B.