1. **Problem 1: Shear and Moment Diagrams for a Simply Supported Beam with Uniform Load** Given: A simply supported beam of length $L=12$ m with a uniform distributed load $w=9$ kN/m over the entire length. 2. **Formulas and Rules:** - Reactions at supports for uniform load: $R_A = R_B = \frac{wL}{2}$ - Shear force at a distance $x$ from the left support: $V(x) = R_A - wx$ - Bending moment at a distance $x$: $M(x) = R_A x - \frac{w x^2}{2}$ - Maximum moment occurs at mid-span $x=\frac{L}{2}$ 3. **Calculations:** - Calculate reactions: $$R_A = R_B = \frac{9 \times 12}{2} = 54\text{ kN}$$ - Shear force at $x=0$: $V(0) = 54$ kN - Shear force at $x=6$ m (mid-span): $$V(6) = 54 - 9 \times 6 = 54 - 54 = 0$$ - Shear force at $x=12$ m: $V(12) = 54 - 9 \times 12 = 54 - 108 = -54$ kN - Bending moment at mid-span: $$M(6) = 54 \times 6 - \frac{9 \times 6^2}{2} = 324 - 162 = 162\text{ kN}\cdot\text{m}$$ 4. **Interpretation:** - Shear diagram is a straight line decreasing from $+54$ kN at left support to $-54$ kN at right support. - Moment diagram is a parabola with maximum $162$ kN·m at mid-span. --- 5. **Problem 2: Shear and Moment Diagrams for Beam with Concentrated Loads and Uniform Load** Given: Beam with point loads $P_1=36$ kips, $P_2=24$ kips, and uniform load $w=12$ kips/ft over last segment. 6. **Approach:** - Calculate support reactions using equilibrium equations. - Construct shear diagram by adding/subtracting loads stepwise. - Calculate bending moments by integrating shear or using moment equations. (Due to complexity, detailed numeric steps omitted here for brevity.) --- 7. **Problem 3: Shear and Moment Diagrams for Combined Footing with Uniform Soil Pressure** Given: Footing length $27'$ with uniform soil pressure $w=24$ kips/ft. 8. **Approach:** - Treat footing as beam with uniform distributed load. - Calculate reactions at supports (columns). - Use formulas for shear and moment as in Problem 1. --- 9. **Problem 4: Shear and Moment Diagrams for Combined Footing with Trapezoidal Soil Pressure** Given: Trapezoidal distributed load varying from $2$ to $8$ kips/ft over $20'$ length. 10. **Approach:** - Calculate equivalent loads and their positions. - Calculate reactions using equilibrium. - Construct shear and moment diagrams accordingly. --- 11. **Problem 5: Shear and Moment Diagrams for Compound Beam with Mixed Loads** Given: Uniform loads, point loads, and supports at various points. 12. **Approach:** - Calculate reactions at supports using equilibrium. - Construct shear diagram by adding/subtracting loads. - Calculate bending moments at key points. --- 13. **Problem 6: Shear and Moment Diagrams for Simply Supported Beam with Uniform Load** Given: Beam length $12$ ft with uniform load $2$ k/ft. 14. **Calculations:** - Reactions: $$R_A = R_B = \frac{2 \times 12}{2} = 12\text{ kips}$$ - Shear at $x=0$: $V(0) = 12$ kips - Shear at mid-span $x=6$ ft: $$V(6) = 12 - 2 \times 6 = 0$$ - Shear at $x=12$ ft: $V(12) = 12 - 24 = -12$ kips - Moment at mid-span: $$M(6) = 12 \times 6 - \frac{2 \times 6^2}{2} = 72 - 36 = 36\text{ kips}\cdot\text{ft}$$ 15. **Summary:** - Shear diagram is linear from $+12$ to $-12$ kips. - Moment diagram is parabolic with max $36$ kips·ft at mid-span.