Let's break down the solution into two steps as requested: --- ### **Step 1: Calculate the support reactions** #### **1.1: Find total uniformly distributed load (UDL)** - The beam is loaded with **5 kN/m** over a **4 m** span. - Total load, \( W = 5 \times 4 = 20\, \text{kN} \) - The load acts at the center of the beam, i.e., at 2 m from A. #### **1.2: Draw and set up equilibrium equations** Let: - \(A\) is a pin support (vertical and horizontal reaction: \(A_y, A_x\)) - \(C\) is a roller (vertical reaction: \(C_y\)) - The cable at \(B\) and \(E\) provides tension but will be resolved after vertical reactions. Take moments about point A: \[ \sum M_A = 0: -W \cdot 2 + C_y \cdot 3 + T_{BC} \cdot 0.5 = 0 \] But, let's first just find the vertical reactions, assuming the system is in equilibrium. \[ \sum F_y = 0: \quad A_y + C_y = 20 \] Take moments about A to solve for \(C_y\): \[ \sum M_A = 0: \quad -20 \times 2 + C_y \times 3 = 0 \] \[ -40 + 3C_y = 0 \] \[ 3C_y = 40 \implies C_y = \frac{40}{3} \approx 13.33\, \text{kN} \] Now, substitute back: \[ A_y + 13.33 = 20 \implies A_y = 6.67\, \text{kN} \] **Explanation:** We determined the total vertical reactions at A and C by using equilibrium equations, considering the distributed load. --- ### **Step 2: Analyze Section at Point D (2 m from A)** #### **2.1: Cut the beam at D and analyze the left section** - Distance from A to D = 2 m. - UDL to the left of D: \(5\, \text{kN/m} \times 2\, \text{m} = 10\, \text{kN}\) **Shear force (\(V_D\)):** \[ V_D = A_y - (\text{UDL left of D}) = 6.67\, \text{kN} - 10\, \text{kN} = -3.33\, \text{kN} \] **Normal force (\(N_D\)):** - There is **no horizontal load** acting to the left of D (since the cable force is not acting horizontally on this part). \[ N_D = 0 \] **Bending moment (\(M_D\)):** \[ M_D = A_y \cdot 2 - (5 \cdot 2) \cdot (2/2) \] \[ M_D = 6.67 \cdot 2 - 10 \cdot 1 \] \[ M_D = 13.34 - 10 = 3.34\, \text{kN}\cdot\text{m} \] **Explanation:** We isolated the left segment at D and calculated the internal forces by considering equilibrium of this segment. Shear is negative (downward), normal is zero (no axial force), and the bending moment is positive (sagging). --- ## **Final Answers:** \[ \boxed{ V_D = -3.33\, \text{kN} \\ N_D = 0 \\ M_D = 3.34\, \text{kN}\cdot\text{m} } \] Let me know if you want to analyze the right section or include the effect of the cable tension!