Let's solve the frame analysis problem step by step as requested: --- ## **Step 1: Support Reactions (Global Equilibrium Equations)** ### **Frame Geometry** - \( AB = 4\, \text{m} \) - \( BC = 2\, \text{m} \) - \( CD = 4\, \text{m} \) - \( EF = 3\, \text{m} \) - \( DE = 1\, \text{m} \) ### **Loads** - Moment at B: \( M_B = 20\, \text{kNm} \) (counterclockwise) - Point load at F: \( P_F = 50\, \text{kN} \) (to the right) ### **Supports** - **A**: Pinned (Resists vertical and horizontal forces) - **D**: Roller (Resists only vertical force) ### **Unknown Reactions** - At **A**: \( A_x, A_y \) - At **D**: \( D_y \) #### **Equilibrium Equations** #### 1. \(\sum F_x = 0\): \[ A_x - 50 = 0 \implies A_x = 50\,\text{kN} \] #### 2. \(\sum F_y = 0\): \[ A_y + D_y = 0 \] #### 3. \(\sum M_A = 0\) (Counterclockwise +): - The 20 kNm moment at B acts directly. - The 50 kN force at F acts at a horizontal distance of \(2 + 1 = 3\,\text{m}\) from A, and at vertical height of 1 m from base. - Moment arm from A to F is 4 m up, then 2 m right, then 3 m down, then 1 m right to F. But let's clarify: F is located 3 m below B, so its vertical position is 1 m above the base (since total height is 4 m). The horizontal distance from A is 2+1 = 3 m. So, moment arm from A to F: \(3\,\text{m}\) horizontally, \(3\,\text{m}\) vertically up from A. The moment about A from the 50 kN force: - The **vertical** distance from A to F is 1 m above the base (F at (3, 1)), so the force acts horizontally (to the right). - The moment arm is \( y = 1\,\text{m} \) above A. - Moment = \( 50 \times 1 = 50\,\text{kNm} \) (clockwise, so negative in our sign convention). The vertical reaction at D is \( D_y \) at a distance of \(2 + 1 = 3\,\text{m}\) from A. Sum moments about A: \[ \sum M_A = 0: \quad 20 - 50 \times 1 - D_y \times 3 = 0 \] \[ 20 - 50 - 3D_y = 0 \] \[ -30 - 3D_y = 0 \] \[ D_y = -10\,\text{kN} \] A negative value means D_y acts **downwards**. Plug into vertical force equilibrium: \[ A_y + D_y = 0 \implies A_y = -D_y = 10\,\text{kN} \] --- ### **Summary of Support Reactions** - \( A_x = 50\,\text{kN} \) (→) - \( A_y = 10\,\text{kN} \) (↑) - \( D_y = -10\,\text{kN} \) (↓) --- ## **Step 2: Sketch Deflected Shape** - **Moment at B** will cause BC to want to rotate: Top beam (BC) bends. - **Horizontal force at F** pushes middle column (EF) to the right, causing sway. - **D** will move horizontally (since it's a roller in y), the whole frame sways right. --- ## **Step 3: BMD, SFD, AFD** ### **A. Shear Force Diagram (SFD)** - **AB**: Vertical member, only vertical reactions. - **BC**: Horizontal, moment applied, no vertical loads. - **CD, DE, EF**: Carry vertical and horizontal forces as per frame action. ### **B. Bending Moment Diagram (BMD)** - **Moment at B**: 20 kNm (CCW). - **Moment at F**: Induced by horizontal load. - **Moment at A**: Calculated from equilibrium. You would draw: - Maximum moment at B (20 kNm). - Moments induced by the 50 kN force at F in EF and DE. ### **C. Axial Force Diagram (AFD)** - **AB**: Axial force due to horizontal load (A_x). - **BC, CD, DE, EF**: As per frame action. --- ## **Step 4: Unit-Load Method for Horizontal Deflection at D** ### **Horizontal Deflection at D (\(\Delta_D\)), Unit-Load Method** Use **Virtual Work Principle**: \[ \Delta_D = \int \frac{M \cdot m}{EI} dx \] Where: - \( M \): Moment due to actual loading. - \( m \): Moment due to unit load at D (to the right). #### **Step 4.1: Find \( M \) (Real Loading)** - Use moment diagrams from the actual loading. #### **Step 4.2: Find \( m \) (Unit Load at D)** - Apply a unit horizontal load at D (to the right). - Find resulting moments in each member. #### **Step 4.3: Integrate** - For each member, calculate the product \( M \cdot m \), integrate over the length, and sum for all members. **In practice:** 1. For vertical members, the moment due to the unit load is linear along the height. 2. For horizontal members, calculate as per statics. \[ \Delta_D = \frac{1}{EI} \sum \int M(x) m(x) dx \] --- ## **Step 5: Rotation at C (\(\theta_C\)), Unit-Load Method** Similarly, apply a unit moment at C, calculate the virtual moments, and use: \[ \theta_C = \frac{1}{EI} \sum \int M(x) m'(x) dx \] Where \( m'(x) \) is the moment distribution due to a unit moment at C. --- ## **Final Answers (Summary)** ### **Support Reactions** - \( A_x = 50\,\text{kN} \) (→) - \( A_y = 10\,\text{kN} \) (↑) - \( D_y = -10\,\text{kN} \) (↓) ### **Deflected Shape** - Frame sways to the right, top beam rotates CCW at B. ### **BMD, SFD, AFD** - See above for qualitative descriptions (draw diagrams as per calculated values). ### **Horizontal Deflection at D (\(\Delta_D\))** \[ \Delta_D = \frac{1}{EI} \sum \int M(x) m(x) dx \] ### **Rotation at C (\(\theta_C\))** \[ \theta_C = \frac{1}{EI} \sum \int M(x) m'(x) dx \] --- **If you want detailed calculation for \(\Delta_D\) and \(\theta_C\) with all member-by-member integrations, let me know!**