Let's break down the solution step by step for the **fixed beam** under the given loading: --- ### **1. Find the Reaction Force and Moment at the Fixed End** #### **Given Data:** - Uniformly distributed load (UDL): \( w = 2 \) kip/ft over \( 6 \) ft - Point load: \( P_1 = 10 \) kip at \( 6 \) ft from left - Point load: \( P_2 = 8 \) kip at \( 10 \) ft from left - Applied moment: \( M_0 = 40 \) kip·ft at the right end - Beam length: \( L = 10 \) ft #### **Step 1: Replace UDL with Equivalent Point Load** - Total UDL force: \( F_{UDL} = w \times 6 = 2 \times 6 = 12 \) kip - Acts at \( 3 \) ft from the left end (centroid of UDL region) #### **Step 2: Equilibrium Equations** Let the reactions at the fixed end (left) be \( A_y \) (vertical) and \( M_A \) (moment). ##### **Sum of Vertical Forces:** \[ A_y - F_{UDL} - P_1 - P_2 = 0 \] \[ A_y - 12 - 10 - 8 = 0 \] \[ A_y = 30 \text{ kip (upward)} \] ##### **Sum of Moments about the Fixed End (A):** Take counterclockwise moments as positive. \[ M_A - (12 \times 3) - (10 \times 6) - (8 \times 10) - 40 = 0 \] \[ M_A - 36 - 60 - 80 - 40 = 0 \] \[ M_A = 36 + 60 + 80 + 40 = 216 \text{ kip·ft (counterclockwise)} \] --- ### **2. Shear Force Profile** #### **From Left to Right:** - **From A to 6 ft (UDL only):** - Shear starts at \( +30 \) kip. - Drops linearly by \( 2 \) kip/ft over 6 ft: \( 2 \times 6 = 12 \) kip. - Shear just left of 6 ft = \( 30 - 12 = 18 \) kip. - **At 6 ft:** - Shear drops by 10 kip at the point load: \( 18 - 10 = 8 \) kip. - **At 10 ft:** - Shear drops by 8 kip at the point load: \( 8 - 8 = 0 \) kip. - **Moment at the end:** - No additional vertical loads after 10 ft. #### **Shear Values:** - At \( x = 0 \): \( +30 \) kip - At \( x = 6^- \): \( +18 \) kip - At \( x = 6^+ \): \( +8 \) kip - At \( x = 10^- \): \( 0 \) kip --- ### **3. Bending Moment Profile** - At \( x = 0 \): \( M_A = +216 \) kip·ft - Bending moment decreases under loads. #### **General equation for bending moment at \( x \) (0 ≤ x ≤ 6):** \[ M(x) = M_A - 2x \cdot \frac{x}{2} \] \[ M(x) = M_A - x^2 \] (where \( M_A = 216 \) and UDL contributes \(-2x \cdot \frac{x}{2}\)) #### **At \( x = 6 \):** \[ M(6) = 216 - (2 \cdot 6 \cdot 3) = 216 - 36 = 180 \text{ kip·ft} \] But also consider the point load at 6 ft for \( x > 6 \): #### **For \( 6 < x < 10 \):** \[ M(x) = M_A - 2 \cdot 6 \cdot (x - 3) - 10(x - 6) \] Alternatively, sum moments due to all left-side forces at any section \( x \). #### **At \( x = 10 \):** \[ M(10) = 216 - (2 \times 10 \times 5) - 10 \times 4 - 8 \times 0 - 40 \] \[ M(10) = 216 - 100 - 40 - 0 - 40 = 36 \text{ kip·ft} \] But also subtract the applied moment: \[ M_{final}(10) = 36 - 40 = -4 \text{ kip·ft} \] --- ### **Summary Table** | Location | Shear (kip) | Moment (kip·ft) | |-------------|-------------|-----------------| | \( x = 0 \) | 30 | 216 | | \( x = 6^- \) | 18 | (calculate using above) | | \( x = 6^+ \) | 8 | (calculate) | | \( x = 10^- \) | 0 | -4 | --- ## **Shear Force and Bending Moment Diagrams** - **Shear Force Diagram:** Starts at +30 kip, drops linearly to +18 at 6 ft, then drops by 10 at point load, then by 8 at 10 ft to 0. - **Bending Moment Diagram:** Starts at +216 kip·ft, parabolic decrease due to UDL, changes slope at point loads, ends at -4 kip·ft at 10 ft. --- **If you need to see the diagrams, let me know and I can sketch them!**