Let's walk through the process step by step: ### 1. **Identify the Loads and Reactions** - **Point Load:** 80 lb at 3 ft from A. - **Distributed Load:** Triangular, starting from 0 lb/ft at 3 ft from A, up to 100 lb/ft at B (9 ft from A). ### 2. **Calculate Support Reactions** #### **a. Equivalent Load of the Triangular Distributed Load** - **Base:** 6 ft (from 3 ft to 9 ft) - **Height:** 100 lb/ft - **Area (Total Load):** \[ W = \frac{1}{2} \times 6\, \text{ft} \times 100\, \frac{\text{lb}}{\text{ft}} = 300\, \text{lb} \] - **Location:** From the left of the triangle (at 3 ft from A), the centroid is at \( \frac{1}{3} \) the base from the larger end (B), so: - Distance from B: \( \frac{1}{3} \times 6 = 2\, \text{ft} \) - Distance from A: \( 9 - 2 = 7\, \text{ft} \) #### **b. Take Moments About A** Let \( R_A \) and \( R_B \) be the reactions at A and B. Sum of moments about A: \[ R_B \times 9 - 80 \times 3 - 300 \times 7 = 0 \] \[ R_B \times 9 = 240 + 2100 = 2340 \] \[ R_B = \frac{2340}{9} = 260\, \text{lb} \] Sum of vertical forces: \[ R_A + R_B = 80 + 300 \implies R_A = 380 - 260 = 120\, \text{lb} \] --- ### 3. **Shear Force Diagram (SFD)** - **From A to 3 ft:** Shear is \( +120 \) lb (no load). - **At 3 ft:** Drop by 80 lb (point load): \( 120 - 80 = 40 \) lb. - **From 3 ft to 9 ft:** The triangular load increases linearly: - Shear at 9 ft: \[ V_{9} = 40 - \text{Area under triangular load} = 40 - 300 = -260\, \text{lb} \] - **At B:** Shear jumps up by reaction at B (\(+260\) lb), ending at 0. --- ### 4. **Bending Moment Diagram (BMD)** - **At A (x=0):** \( M_A = 0 \) - **From A to 3 ft:** Linear increase (area under shear diagram): \[ M_{3} = 120 \times 3 = 360\, \text{lb}\cdot\text{ft} \] - **At 3 ft:** Jump in shear (not moment), then from 3 to 9 ft, area under the shear curve (which reduces parabolically due to the triangular load). #### **Between 3 ft and 9 ft:** Let \( x \) be distance from A (\( 3 \leq x \leq 9 \)): - Shear at any \( x \): \[ V(x) = 40 - \frac{1}{2} \times (x-3) \times \frac{100}{6} \times (x-3) \] But for clarity, integrate the triangular load: The intensity at a distance \( x \) from A is: \[ w(x) = \frac{100}{6}(x-3)\ \text{lb/ft} \] So, the total load from 3 to \( x \) is: \[ W(x) = \frac{1}{2}(x-3)w(x) = \frac{1}{2}(x-3)\frac{100}{6}(x-3) = \frac{50}{6}(x-3)^2 \] Shear at \( x \) (for \( x > 3 \)): \[ V(x) = 40 - \frac{50}{6}(x-3)^2 \] Bending moment at \( x \): \[ M(x) = M_3 + \int_{3}^{x} V(x) dx \] \[ M(x) = 360 + \int_{3}^{x} \left[40 - \frac{50}{6}(x-3)^2 \right] dx \] \[ = 360 + 40(x-3) - \frac{50}{18}(x-3)^3 \] - **At B (x = 9):** \[ M_{B} = 360 + 40(6) - \frac{50}{18}(6)^3 = 360 + 240 - \frac{50}{18}(216) \] \[ = 600 - 600 = 0 \] (As expected at the simple support.) --- ### 5. **Diagrams** #### **Shear Force Diagram (SFD):** - Horizontal line at \( +120 \) lb from A to 3 ft. - Drop to \( +40 \) lb at 3 ft. - Parabolic decrease to \( -260 \) lb at B (9 ft). - Jump to 0 at B. #### **Bending Moment Diagram (BMD):** - Linear increase from 0 to 360 lb·ft from A to 3 ft. - Then a cubic curve, peaking before B and dropping back to 0 at B. --- ### 6. **Sketch** #### **SFD:** ``` | ________ | | | |____________ | / \ |_____/ \________ A 3' 9' B ``` #### **BMD:** ``` | | /\ | / \ | / \ |_____/ \_______ A 3' 9' B ``` *(The actual curve from 3' to 9' is cubic, peaking after 3' and returning to 0 at 9')* --- If you need the diagrams plotted, let me know!