Let's solve the frame deflection problem step by step using the **Moment Area Method**. --- ## **Step 1: Understanding the Frame and Loads** - **Members:** AB (vertical), BC (horizontal), CD (horizontal) - **Loads:** 8 kips at B (horizontal), 6 kips at D (vertical) - **Dimensions:** - AB = 6 ft - BC = 12 ft - CD = 6 ft --- ## **Step 2: Deflection at D** We are to find: - **Horizontal Deflection** at D (\(\delta_{Dx}\)) - **Vertical Deflection** at D (\(\delta_{Dy}\)) We use the **Moment Area Method**, which is based on the area under the \( \frac{M}{EI} \) diagram. --- ## **Step 3: Find Internal Moment Diagrams** ### **A. Due to 8 kip Horizontal Load at B** 1. **Member AB (Vertical)** - 8 kip horizontal load → creates a moment at B and a shear in AB. - Moment at B = (pinned base) - Moment at A = (free end) - Shear in AB = 8 kips 2. **Member BC (Horizontal)** - The force at B creates a moment that travels through BC. - Moment at B (from 8 kip load): \(M_B = 8 \text{ kip} \times 6' = 48 \text{ kip-ft}\) - Moment at C: To be found. ### **B. Due to 6 kip Vertical Load at D** - Vertical point load at D, causing moment in BC and CD. --- ## **Step 4: Moment Area Method Concepts** - **Theorem I:** Change in slope between two points = Area under \( \frac{M}{EI} \) diagram between those points. - **Theorem II:** Displacement of one point relative to tangent at another = Moment of area under \( \frac{M}{EI} \) diagram about the point. We will treat horizontal and vertical deflections separately. --- ## **Step 5: Horizontal Deflection at D (\(\delta_{Dx}\))** ### **A. Due to 8 kip Load at B (Primary Cause)** **(a) Draw M/EI Diagram for horizontal load:** - Only vertical member AB bends due to the horizontal load. - \(M_{AB}(x) = 8 \text{ kip} \cdot x\), where \(x\) is measured from B to A (from to 6 ft). \[ \frac{M_{AB}(x)}{EI} = \frac{8x}{EI} \] **(b) Moment Area (Theorem II) about B:** - The horizontal deflection at D is equal to the horizontal displacement at B due to the bending of AB. - Since BC and CD are rigid and pinned at C, only AB's bending is considered. \[ \delta_{Dx} = \delta_{Bx} = \frac{1}{EI} \int_^{6} M_{AB}(x) \cdot dx \] But for Theorem II: \[ \delta_{BA} = \text{Moment of area under } \frac{M}{EI} \text{ about A} \] But since A is free and B is fixed, the deflection at B is: \[ \delta_{Bx} = \frac{8 \times 6^3}{3EI} \] **Derivation:** \[ \delta_{Bx} = \frac{P L^3}{3EI} \] Where \(P = 8\) kip, \(L = 6\) ft \[ \delta_{Dx} = \frac{8 \times 6^3}{3EI} = \frac{8 \times 216}{3EI} = \frac{1728}{3EI} = \frac{576}{EI} \] --- ## **Step 6: Vertical Deflection at D (\(\delta_{Dy}\))** ### **A. Due to 6 kip Load at D (Primary Cause)** - CD is a cantilever beam with point load at end. - Deflection at D (free end) = \[ \delta_{Dy} = \frac{P L^3}{3EI} \] Where \(P = 6\) kip, \(L = 6\) ft \[ \delta_{Dy} = \frac{6 \times 6^3}{3EI} = \frac{6 \times 216}{3EI} = \frac{1296}{3EI} = \frac{432}{EI} \] --- ## **Step 7: Final Answers** \[ \boxed{ \begin{align*} \text{Horizontal Deflection at D:} \qquad & \delta_{Dx} = \frac{576}{EI} \\ \text{Vertical Deflection at D:} \qquad & \delta_{Dy} = \frac{432}{EI} \end{align*} } \] --- ### **Summary Table** | Deflection Component | Value | |----------------------|----------------------| | Horizontal at D | \( \frac{576}{EI} \) | | Vertical at D | \( \frac{432}{EI} \) | - \(E\): Modulus of Elasticity - \(I\): Moment of Inertia **Assumptions:** - Joints are pinned unless otherwise specified. - Axial deformations neglected. --- Let me know if you want the diagrams or further breakdown!