1. **Problem Statement:** Determine the reaction forces at supports A and B of the beam shown in Figure 1, and then find the shear force and bending moment at point B (just before the support). 2. **Given Data:** - Moment at left end (A): 36 kNm (counterclockwise) - Point load at right end: 12 kN downward - Beam spans: 4 segments each 1.5 m, total length 6 m - Supports at A, C, and B (A at 0 m, C at 3 m, B at 4.5 m) 3. **Step 1: Define reactions** Let reaction forces be: - $R_A$ vertical reaction at A - $R_B$ vertical reaction at B - $R_C$ vertical reaction at C 4. **Step 2: Equilibrium equations** Sum of vertical forces: $$R_A + R_C + R_B - 12 = 0$$ Sum of moments about A (taking counterclockwise as positive): $$-36 + R_C \times 3 + R_B \times 4.5 - 12 \times 6 = 0$$ Sum of moments about B: $$36 - R_A \times 4.5 - R_C \times 1.5 + 12 \times 1.5 = 0$$ 5. **Step 3: Solve for reactions** From vertical forces: $$R_A + R_C + R_B = 12$$ From moments about A: $$-36 + 3R_C + 4.5R_B - 72 = 0 \Rightarrow 3R_C + 4.5R_B = 108$$ From moments about B: $$36 - 4.5R_A - 1.5R_C + 18 = 0 \Rightarrow -4.5R_A - 1.5R_C = -54$$ Rewrite: $$4.5R_A + 1.5R_C = 54$$ 6. **Step 4: Express $R_A$ from last equation:** $$4.5R_A = 54 - 1.5R_C \Rightarrow R_A = \frac{54 - 1.5R_C}{4.5} = 12 - \frac{1}{3}R_C$$ 7. **Step 5: Substitute $R_A$ into vertical forces equation:** $$12 = R_A + R_C + R_B = \left(12 - \frac{1}{3}R_C\right) + R_C + R_B = 12 + \frac{2}{3}R_C + R_B$$ Simplify: $$\frac{2}{3}R_C + R_B = 0 \Rightarrow R_B = -\frac{2}{3}R_C$$ 8. **Step 6: Substitute $R_B$ into moment about A equation:** $$3R_C + 4.5R_B = 108$$ $$3R_C + 4.5 \times \left(-\frac{2}{3}R_C\right) = 108$$ $$3R_C - 3R_C = 108$$ $$0 = 108$$ This is a contradiction, indicating a statically indeterminate beam or missing data. **Re-examining the problem:** Since there are three supports (A, C, B) and only two equilibrium equations for vertical forces and moments, the system is statically indeterminate. **Assuming support C is a roller (no vertical reaction) or ignoring $R_C$ for simplicity:** Set $R_C = 0$. Then: From vertical forces: $$R_A + R_B = 12$$ From moments about A: $$-36 + 4.5R_B - 72 = 0 \Rightarrow 4.5R_B = 108 \Rightarrow R_B = 24$$ From vertical forces: $$R_A + 24 = 12 \Rightarrow R_A = -12$$ Negative reaction at A is not physically possible, so this assumption is invalid. **Therefore, the problem requires more information or a different approach (e.g., considering beam deflection or support types).** 9. **Step 7: Shear and moment at B (just before support)** Assuming reactions $R_A$, $R_C$, and $R_B$ are known, shear at B is: $$V_B = R_B - \text{loads to the left of B}$$ Moment at B is: $$M_B = \text{sum of moments to the left of B}$$ Since reactions are indeterminate, exact values cannot be computed here. --- **Final answer:** The beam is statically indeterminate with three supports and given loads. Additional information or methods (e.g., compatibility equations) are needed to find reactions and internal forces. ---