Let's go step by step: ## **Given Data** - **Diameter of wire, \( d \) = 1/8 in = 0.125 in** - **Radius of arc, \( R = 5.2 \) in** - **Length of straight segment, \( l = 4 \) in** - **Force, \( P = 1.04 \) lbf** - **Young’s modulus for steel (assuming), \( E = 30 \times 10^6 \) psi** - **We are to use Castigliano’s Theorem to find the horizontal deflection at \( A \).** --- ### **Step 1: Moment of Inertia** The moment of inertia for a circular section is: \[ I = \frac{\pi}{64} d^4 = \frac{\pi}{64} (0.125)^4 = 0.00024 \text{ in}^4 \] --- ### **Step 2: Strain Energy (Castigliano’s Theorem)** For a structure subjected to a force, **the deflection in the direction of the force is:** \[ \delta = \frac{\partial U}{\partial P} \] where \( U \) is the strain energy. #### **Strain energy for bending:** \[ U = \int \frac{M^2}{2EI} \, ds \] We **neglect axial and shear strain energy** (as they contribute much less compared to bending for slender wires). --- ### **Step 3: Find Bending Moment Expressions** #### **For Arc (from C to B):** At a point at angle \( \theta \) from the fixed end, the moment due to force \( P \) at \( A \): \[ M(\theta) = P \cdot R \sin(\theta) \] where \( \theta \) goes from 0 to \( \frac{\pi}{2} \). #### **For Straight Segment (from B to A):** At a distance \( x \) from B, \[ M(x) = P \cdot x \] where \( x \) goes from 0 to \( l \). --- ### **Step 4: Total Strain Energy** \[ U = \int_{\text{arc}} \frac{M_{arc}^2}{2EI} \, R d\theta + \int_{\text{straight}} \frac{M_{straight}^2}{2EI} dx \] #### **Arc:** \[ U_{arc} = \int_0^{\pi/2} \frac{(P R \sin\theta)^2}{2 E I} R d\theta \] \[ = \frac{P^2 R^3}{2 E I} \int_0^{\pi/2} \sin^2\theta \, d\theta \] \[ \int_0^{\pi/2} \sin^2\theta \, d\theta = \frac{\pi}{4} \] \[ U_{arc} = \frac{P^2 R^3}{2 E I} \cdot \frac{\pi}{4} = \frac{P^2 R^3 \pi}{8 E I} \] #### **Straight Segment:** \[ U_{straight} = \int_0^l \frac{(P x)^2}{2 E I} dx = \frac{P^2}{2 E I} \int_0^l x^2 dx = \frac{P^2}{2 E I} \cdot \frac{l^3}{3} \] \[ = \frac{P^2 l^3}{6 E I} \] --- ### **Step 5: Total Strain Energy** \[ U = U_{arc} + U_{straight} = \frac{P^2 R^3 \pi}{8 E I} + \frac{P^2 l^3}{6 E I} \] --- ### **Step 6: Deflection at \( A \) (Horizontal, direction of \( P \))** By Castigliano’s theorem: \[ \delta_A = \frac{\partial U}{\partial P} \] \[ = \frac{\partial}{\partial P} \left[ \frac{P^2 R^3 \pi}{8 E I} + \frac{P^2 l^3}{6 E I} \right] \] \[ = \frac{2P R^3 \pi}{8 E I} + \frac{2P l^3}{6 E I} \] \[ = \frac{P R^3 \pi}{4 E I} + \frac{P l^3}{3 E I} \] --- ### **Step 7: Substitute Values** - \( P = 1.04 \) lbf - \( R = 5.2 \) in - \( l = 4 \) in - \( E = 30 \times 10^6 \) psi - \( I = 0.00024 \) in\(^4\) - \( \pi = 3.1416 \) #### First term: \[ \frac{P R^3 \pi}{4 E I} = \frac{1.04 \times (5.2)^3 \times 3.1416}{4 \times 30 \times 10^6 \times 0.00024} \] \[ (5.2)^3 = 140.608 \] \[ 1.04 \times 140.608 \times 3.1416 = 459.312 \] \[ 4 \times 30 \times 10^6 \times 0.00024 = 28800 \] \[ \frac{459.312}{28800} = 0.01596 \] #### Second term: \[ \frac{P l^3}{3 E I} = \frac{1.04 \times (4)^3}{3 \times 30 \times 10^6 \times 0.00024} \] \[ (4)^3 = 64 \] \[ 1.04 \times 64 = 66.56 \] \[ 3 \times 30 \times 10^6 \times 0.00024 = 21600 \] \[ \frac{66.56}{21600} = 0.00308 \] --- ### **Step 8: Add Both Terms** \[ \delta_A = 0.01596 + 0.00308 = \boxed{0.019 \text{ in}} \] --- ## **Final Answer** \[ \boxed{0.019 \text{ in}} \] --- ### **Justification for Neglected Strain Energy** - **Axial and shear energies are much smaller than bending energy for slender wire with loading as shown, so are neglected here (common assumption).**