Let's break down the problem **step-by-step**: --- ### **Step 1: Identify the Support Reactions** - **A**: Pinned support (forces in both \(x\) and \(y\) directions: \(A_x\), \(A_y\)) - **B**: Roller support (force in \(y\) direction: \(B_y\)) --- ### **Step 2: Break Down the Loads** - **12 k** at 20 ft from A, at \(60^\circ\) from horizontal: - \(12 \cos(60^\circ) = 6 \text{ k (horizontal, right)}\) - \(12 \sin(60^\circ) = 10.392 \text{ k (vertical, down)}\) - **6 k** vertical point load at 25 ft from A. --- ### **Step 3: Draw the Free Body Diagram and Write Equilibrium Equations** #### **Sum of Forces in X:** \[ \sum F_x = 0: \quad A_x - (12 \cos 60^\circ) = 0 \implies A_x - 6 = 0 \implies A_x = 6 \text{ k (left)} \] #### **Sum of Forces in Y:** \[ \sum F_y = 0: \quad A_y + B_y - (12 \sin 60^\circ) - 6 = 0 \] \[ A_y + B_y - 10.392 - 6 = 0 \] \[ A_y + B_y = 16.392 \quad ...(1) \] #### **Sum of Moments about A:** (Moments caused by forces to the right of A are positive (counterclockwise)) - **12 k at 20 ft from A:** - Vertical component (\(10.392\) k) at 10 ft from A: Moment = \(10.392 \times 10 = 103.92\) k-ft (clockwise, so negative) - Horizontal component (\(6\) k) passes through A (no moment arm), so **no moment**. - **6 k at 25 ft from A:** Moment = \(6 \times 25 = 150\) k-ft (clockwise, so negative) - **B_y** at 20 ft from A: Moment = \(B_y \times 20\) (counterclockwise, so positive) \[ \sum M_A = 0: \quad -10.392 \times 10 - 6 \times 25 + B_y \times 20 = 0 \] \[ -103.92 - 150 + 20B_y = 0 \] \[ 20B_y = 253.92 \] \[ B_y = \frac{253.92}{20} = 12.696 \text{ k} \] Now substitute \(B_y\) into equation (1): \[ A_y + 12.696 = 16.392 \] \[ A_y = 16.392 - 12.696 = 3.696 \text{ k} \] --- ### **Step 4: Support Reactions (Final Answers)** \[ \boxed{ \begin{aligned} A_x &= 6 \text{ k (left)} \\ A_y &= 3.70 \text{ k (up)} \\ B_y &= 12.70 \text{ k (up)} \\ \end{aligned} } \] --- ### **Step 5: Shear, Moment, and Normal Force Diagrams** #### **Shear Force Diagram (V):** - **From A to 10 ft:** Only reactions at A act. \(V = +3.70\) k - **At 10 ft:** Subtract \(12\sin(60^\circ) = 10.392\) k (downward load) - \(V = 3.70 - 10.392 = -6.692\) k - **From 10 ft to 20 ft:** \(V = -6.692\) k - **At 20 ft:** Add \(B_y = 12.70\) k (upward reaction) - \(V = -6.692 + 12.70 = 6.008\) k - **From 20 ft to 25 ft:** \(V = 6.008\) k - **At 25 ft:** Subtract \(6\) k (downward load) - \(V = 6.008 - 6 = 0.008 \approx 0\) k #### **Moment Diagram (M):** Calculate at key points (relative to A): - **At A:** \(M = 0\) - **At 10 ft:** \(M = 3.70 \times 10 = 37\) k-ft (area under V from 0 to 10 ft) - **Just right of 10 ft:** Drop in slope due to vertical load (\(-6.692\) k shear) - **At 20 ft:** \(M = 37 + [(-6.692) \times 10] = 37 - 66.92 = -29.92\) k-ft - **Just right of 20 ft:** Add moment due to \(B_y\) (shear jumps to \(+6.008\) k) - **At 25 ft:** \(M = -29.92 + [6.008 \times 5] = -29.92 + 30.04 = 0.12\) k-ft #### **Normal Force Diagram (N):** - Only horizontal load is at 10 ft (\(12 \cos 60^\circ = 6\) k rightward). - \(A_x = 6\) k leftward balances it, so normal force in beam is 6 k in compression from A to 10 ft, then zero thereafter. --- ### **Step 6: Diagrams (Sketch)** #### **Shear Diagram** ``` V (k) | | _______ | | | | | |____________ |_____| | |_________ 0 10 20 25 x (ft) 3.70 -6.69 6.01 0 ``` #### **Moment Diagram** ``` M (k-ft) | | /\ | / \ | / \ |________/ \_________ 0 10 20 25 x (ft) 0 37 -29.92 0.12 ``` #### **Normal Force Diagram** ``` N (k) | |------- | | |_______|______________ 0 10 25 x (ft) 6 0 ``` --- ## **Final Answers** ### **Support Reactions** \[ \boxed{ A_x = 6 \text{ k (left)},\quad A_y = 3.70 \text{ k (up)},\quad B_y = 12.70 \text{ k (up)} } \] ### **Diagrams** - **Normal Force:** 6 k (compression) from A to 10 ft, then 0. - **Shear Force:** 3.7 k (A to 10 ft), -6.69 k (10 ft to 20 ft), 6.01 k (20 ft to 25 ft), drops to 0 at end. - **Moment:** Peaks at 37 k-ft at 10 ft, drops to -29.92 k-ft at 20 ft, returns near zero at the end. --- **Let me know if you need the diagrams hand-drawn or in a specific format!**