# Structural Analysis Problem Breakdown ## Problem Statement **Find the nodal displacements and reaction forces for the case shown in the below figure.** The figure shows a structure with: - A horizontal beam (15 m), loaded with a uniformly distributed load of 250 kN/m. - An inclined member joining the end of the beam to a vertical support 9 m away. - A 50 kN point load at the joint between the beam and the inclined member. --- ## Step-by-Step Solution Outline ### 1. **Identify Support Types and Reactions** - **Left end of beam (A):** Fixed or pinned support (typically provides vertical and horizontal reactions, possibly a moment). - **Right end of inclined member (B):** Vertical support (roller or pinned, providing a vertical reaction). ### 2. **Calculate Equivalent Loads** #### Distributed Load (w = 250 kN/m over 15 m) - **Total load:** \( w \times L = 250 \times 15 = 375 \) kN - **Acts at:** Center of beam, 7.5 m from left #### Point Load - **50 kN** at the joint (inclined member start). --- ### 3. **Free Body Diagram and Equilibrium Equations** Let: - \( A_y \): Vertical reaction at A - \( A_x \): Horizontal reaction at A - \( A_M \): Moment at A (if fixed) - \( B_y \): Vertical reaction at B #### Sum of Forces (Vertical) \[ A_y + B_y = 375 + 50 \] \[ A_y + B_y = 380 \text{ kN} \] #### Sum of Forces (Horizontal) - If there’s no horizontal load, \( A_x = \). #### Sum of Moments (about A) Take moments about point A: \[ \sum M_A = \] \[ -375 \times 7.5 - 50 \times 15 + B_y \times (15 + 9) = \] \[ -28125 - 750 + B_y \times 24 = \] \[ B_y = \frac{28875}{24} = 1203.125 \text{ kN} \] #### Substitute \( B_y \) into vertical force equilibrium: \[ A_y + 1203.125 = 380 \] \[ A_y = 2596.875 \text{ kN} \] --- ### 4. **Nodal Displacements** To find displacements, you need: - Material properties (E, I) - Use methods like slope-deflection, moment distribution, or FEM. Since material properties are not given, you cannot compute numeric displacements. --- ## **Summary of Reaction Forces** - **Vertical Reaction at A:** 2596.88 kN (up) - **Vertical Reaction at B:** 1203.13 kN (up) - **Horizontal Reaction at A:** kN (if no horizontal loads) --- ### **Note** - Actual nodal displacements require more data (material and section properties). - The above is a static equilibrium solution for support reactions. --- **If you need the displacements, please provide Young’s modulus (E) and moment of inertia (I) for the members.**