1. **Problem Statement:** We need to draw the shear and moment diagrams for a compound beam ABC with a pin at B. The beam has a uniformly distributed load (UDL) of 400 lb/ft over segment AB (4 ft), a moment of 900 lb·ft at C, and reaction forces at supports A and B. 2. **Key Formulas and Rules:** - Shear force $V(x)$ is found by summing vertical forces. - Bending moment $M(x)$ is found by summing moments about a section. - For UDL, shear changes linearly and moment is quadratic. - At supports, sum of vertical forces and moments must be zero (static equilibrium). 3. **Calculate Reactions at Supports:** - Sum moments at B: $\sum M_B=0 \Rightarrow -900 + C_y \times 3=0 \Rightarrow C_y=300$ lb - Sum vertical forces: $\sum F_y=0 \Rightarrow B_y=300$ lb - Sum moments at A: $M_A = (1600)(2) + (300)(3) = 7400$ lb·ft - Sum vertical forces at A: $A_y - 1600 + 300=0 \Rightarrow A_y=1300$ lb 4. **Shear and Moment in Segment AB (0 < x_1 < 4 ft):** - Shear: $V_1 = 1300 - 400 x_1$ - Moment: $M_1 = -200 x_1^2 + 1300 x_1 - 1400$ - Shear zero at $x_1 = \frac{1300}{400} = 3.25$ ft - Maximum moment at $x_1=3.25$ ft: $$M_{max} = -200 (3.25)^2 + 1300 (3.25) - 1400 = 712.5 \text{ lb-ft}$$ 5. **Shear and Moment in Segment BC (0 < x_2 < 2 ft):** - Shear: $V_2 = -300$ lb (constant) - Moment: $M_2 = 300 x_2$ 6. **Shear and Moment in Segment CC (0 < x_3 < 3 ft):** - Shear: $V_3 = -300$ lb - Moment: $M_3 = -300 x_3$ 7. **Summary:** - Shear diagram starts at +1300 lb, decreases linearly to -300 lb at 3.25 ft, remains constant at -300 lb, then returns to zero. - Moment diagram is parabolic in AB, peaks at 712.5 lb-ft at 3.25 ft, then linear in BC and CC segments. This step-by-step approach uses static equilibrium and section cuts to find shear and moment values along the beam, enabling the drawing of accurate diagrams.