Moment Shear Diagram 4C03E8

1. **Problem Statement:** We have a structural frame/beam with a $10$ kN point load, distributed loads of $5$ kN/m and $8$ kN/m, and spans of $5$ m, $8$ m, $2$ m, $3$ m, and $2$ m. We need to find the shear force and bending moment diagrams using Hibbler's structural analysis method. 2. **Step 1: Calculate Reactions at Supports** - Use equilibrium equations: $$\sum F_x=0, \sum F_y=0, \sum M=0$$ - Calculate equivalent point loads for distributed loads: - For $5$ kN/m over $5$ m: $$5 \times 5 = 25\text{ kN}$$ acting at midpoint $2.5$ m from left. - For $8$ kN/m over $8$ m: $$8 \times 8 = 64\text{ kN}$$ acting at midpoint $4$ m from left of that span. 3. **Step 2: Apply Equilibrium Equations** - Sum vertical forces: $$R_A + R_B = 10 + 25 + 64$$ $$R_A + R_B = 99\text{ kN}$$ - Sum moments about left support $A$: $$\sum M_A = 0 = -10 \times 0 + 25 \times 2.5 + 64 \times (5 + 4) - R_B \times (5 + 8)$$ $$0 = 0 + 62.5 + 576 - R_B \times 13$$ $$R_B = \frac{638.5}{13} = 49.12\text{ kN}$$ - Then, $$R_A = 99 - 49.12 = 49.88\text{ kN}$$ 4. **Step 3: Construct Shear Force Diagram (V)** - Start from left support $A$ with $+49.88$ kN. - Subtract point load $10$ kN at $0$ m: $$49.88 - 10 = 39.88\text{ kN}$$ - Subtract distributed load $5$ kN/m over $5$ m: $$V = 39.88 - 5x, \quad 0 \leq x \leq 5$$ At $x=5$: $$39.88 - 5 \times 5 = 14.88\text{ kN}$$ - Jump down by $25$ kN equivalent load at midpoint. - Subtract distributed load $8$ kN/m over $8$ m: $$V = 14.88 - 8y, \quad 0 \leq y \leq 8$$ At $y=8$: $$14.88 - 8 \times 8 = -49.12\text{ kN}$$ - At right support $B$, shear returns to zero. 5. **Step 4: Construct Bending Moment Diagram (M)** - Integrate shear diagram: - From $0$ to $5$ m: $$M(x) = 49.88x - \frac{5x^2}{2} - 10(x - 0)H(x - 0)$$ - From $5$ to $13$ m: $$M(x) = M(5) + \int_5^x V(x) dx$$ - Calculate moments at key points: - At $x=0$: $$M=0$$ - At $x=5$: $$M = 49.88 \times 5 - \frac{5 \times 5^2}{2} - 10 \times 5 = 249.4 - 62.5 - 50 = 136.9\text{ kNm}$$ - At $x=13$: $$M = 0$$ (support) 6. **Step 5: Summary** - Reactions: $$R_A = 49.88\text{ kN}, R_B = 49.12\text{ kN}$$ - Shear force varies linearly under distributed loads and jumps at point loads. - Bending moment diagram is parabolic under distributed loads and linear between loads. This completes the shear and moment diagram calculation using Hibbler's structural analysis approach.