Let's solve the **step-by-step** problem for the beam shown in the image. --- ## **Step 1: Understanding the Problem** - A simply supported beam rests on supports at \( A \) and \( B \). - There are two point loads of \( 250\, \text{lb} \) each, at both ends. - There is a **uniformly distributed load (UDL)** of \( 150\, \text{lb/ft} \) over the central \( 6\, \text{ft} \) of the beam. - The distances are marked as \( 4\, \text{ft} \) from each end to the start of the UDL, and \( 6\, \text{ft} \) is the length of the UDL. - The positions of supports \( A \) and \( B \) are to be determined (denoted by \( x \)), assuming they are under the beam between the loads. --- ## **Step 2: Drawing the Free-Body Diagram** - Let's denote the **reactions at A and B** as \( R_A \) and \( R_B \), respectively. - The total length of the beam: \( 4 + 6 + 4 = 14\, \text{ft} \). --- ## **Step 3: Calculating the Total Loads** - **Point loads**: \( 250\, \text{lb} \) each at both ends. - **UDL total load**: \( 150\, \text{lb/ft} \times 6\, \text{ft} = 900\, \text{lb} \). - UDL acts at the center of the 6 ft section, i.e., at \( 4 + 3 = 7\, \text{ft} \) from the left end. --- ## **Step 4: Applying Equilibrium Equations** ### **Sum of Vertical Forces = ** \[ R_A + R_B = 250 + 900 + 250 = 140\, \text{lb} \] --- ### **Sum of Moments About A = ** Let's take moments about \( A \): - **Moment arm for left 250 lb**: \( x \) (but it acts before A, so negative direction?) - **Moment arm for UDL**: The UDL (900 lb) acts at the centroid of the distributed load, which is \( 4 + 3 = 7\, \text{ft} \) from the left end, so distance from A is \( 7 - x \). - **Moment arm for right 250 lb**: It acts at \( 14\, \text{ft} \) from the left, so distance from A is \( 14 - x \). - **Reaction at B**: Distance from A to B is \( 6\, \text{ft} \). \[ \begin{align*} \sum M_A = : & \\ -250 \cdot x & \quad \text{(counterclockwise)} \\ + 900 \cdot (7-x) & \quad \text{(clockwise)} \\ + 250 \cdot (14-x) & \quad \text{(clockwise)} \\ - R_B \cdot 6 & \quad \text{(counterclockwise)} \\ = \end{align*} \] But \( x \) is not given in the original image as a variable; the supports are at 4 ft from the left end (A) and 10 ft from the left end (B). Let's clarify: - \( A \) is at 4 ft from the left end. - \( B \) is at \( 4 + 6 = 10 \) ft from the left end. So, use these values: - Distance from A to left end = 4 ft. - Distance from B to left end = 10 ft. --- ### **Rewriting the Moment Equation About A** Take moments about \( A \) (which is 4 ft from the left end): - Left end 250 lb: 4 ft to the left of A (acts at -4 ft from A, but moment is clockwise). - UDL: 3 ft from A (since UDL centroid is at 7 ft from the left end, A is at 4 ft, so 7 - 4 = 3 ft from A). - Right end 250 lb: 14 - 4 = 10 ft from A. - B: 10 - 4 = 6 ft from A. Now, sum moments about A (counterclockwise positive): \[ \begin{align*} \sum M_A = : \\ +250 \cdot (-4) \\ + 900 \cdot 3 \\ + 250 \cdot 10 \\ - R_B \cdot 6 \\ = \end{align*} \] \[ -100 + 270 + 250 - 6R_B = \] \[ (-100 + 270 + 250) - 6R_B = \] \[ 420 - 6R_B = \] \[ 6R_B = 420 \] \[ R_B = \frac{420}{6} = 700\, \text{lb} \] --- ### **Sum of Vertical Forces** \[ R_A + R_B = 140 \] \[ R_A + 700 = 140 \] \[ R_A = 700\, \text{lb} \] --- ## **Step 5: Final Answer** ### **Reactions at Supports** \[ \boxed{ R_A = 700\, \text{lb} \quad \text{(upward)} } \] \[ \boxed{ R_B = 700\, \text{lb} \quad \text{(upward)} } \] --- ## **Summary Table** | Support | Reaction (lb) | |---------|--------------| | A | 700 | | B | 700 | --- ### **Explanation** - The reactions at both supports are equal because the loading is symmetric about the center of the beam. - All loads and distances were summed and moments about support A were used to solve for the unknown reactions. --- **Let me know if you need shear/moment diagrams or further details!**