1. **Stating the problem:** Determine the reaction forces at the supports of the first beam structure and then draw the diagrams of Normal force (N), Shear force (T), and Bending moment (Mf). 2. **Given data and setup:** - Beam length: $1 + 3 + 1 = 5$ m - Uniform distributed load (UDL): 100 N/m over the first 1 m from the left end - Concentrated forces: 300 N at 1 m from left end, 200 N at 1 m from right end - Beam is fixed at the right end 3. **Step 1: Calculate total load from UDL:** $$\text{Load}_{UDL} = 100 \times 1 = 100 \text{ N}$$ 4. **Step 2: Determine the position of the resultant of UDL:** The resultant acts at the midpoint of the 1 m segment, i.e., at 0.5 m from the left end. 5. **Step 3: Define reaction forces:** Let $R_A$ be the vertical reaction at the left end (if any), and $R_B$ be the reaction at the fixed right end (3 + 1 = 4 m from left end). 6. **Step 4: Apply equilibrium equations:** - Sum of vertical forces: $$R_A + R_B = 100 + 300 + 200 = 600$$ - Sum of moments about left end (taking counterclockwise positive): $$-100 \times 0.5 - 300 \times 1 - 200 \times 4 + R_B \times 5 = 0$$ Calculate: $$-50 - 300 - 800 + 5 R_B = 0$$ $$5 R_B = 1150$$ $$R_B = 230$$ 7. **Step 5: Calculate $R_A$:** $$R_A = 600 - 230 = 370$$ 8. **Step 6: Draw diagrams:** - Normal force (N) diagram: Since no axial loads are given, $N=0$ everywhere. - Shear force (T) diagram: Starts at $R_A=370$ N, decreases by 100 N over first 1 m (UDL), drops by 300 N at 1 m, remains constant until 4 m, then drops by 200 N at 4 m, ends at $-R_B = -230$ N. - Bending moment (Mf) diagram: Calculate bending moments at key points using shear force and loads. **Final answer:** $$R_A = 370 \text{ N}, \quad R_B = 230 \text{ N}$$ Normal force $N=0$ everywhere. Shear and bending moment diagrams can be drawn based on these reactions and loads.