# Vertical Displacement at Point B Using Castigliano's Theorem Given: - The beam is continuous, fixed at A and simply supported at the midpoint. - \( AB = 2L \), with support at the midpoint C. - For \( < x < L \): \( EI = 2EI \) - For \( L < x < 2L \): \( EI = EI \) - Load \( F \) acts downwards at \( B \). We are to find the **vertical displacement at \( B \)** using the **energy method (Castigliano's theorem)**. --- ## 1. **Castigliano's Theorem Overview** Castigliano’s theorem states: \[ \delta_B = \frac{\partial U}{\partial F} \] Where: - \( \delta_B \) = vertical displacement at B - \( U \) = strain energy in the beam --- ## 2. **Step 1: Calculate the Moment at Any Section** Let’s define: - \( A \) is at \( x = \) - \( C \) is at \( x = L \) - \( B \) is at \( x = 2L \) Let’s denote the reaction at support \( C \) as \( R_C \). Let’s also denote the moment at A as \( M_A \). ### **Equilibrium Equations** Let’s do moment equilibrium about A: \[ \sum M_A = \implies R_C \cdot L + - F \cdot 2L = \] \[ R_C = 2F \] Vertical equilibrium: \[ R_A + R_C = F \implies R_A + 2F = F \implies R_A = -F \] So, the fixed support at A provides an upward reaction of \( -F \). ### **Moment at a Section** #### For \( < x < L \) (Segment AC, \( EI = 2EI \)): Take section at distance \( x \) from A: \[ M(x) = -R_A \cdot x = F x \] #### For \( L < x < 2L \) (Segment CB, \( EI = EI \)): \[ M(x) = -R_A \cdot x - R_C \cdot (x - L) = F x - 2F (x - L) = F x - 2F x + 2F L = -F x + 2F L \] --- ## 3. **Step 2: Expression for Strain Energy** Strain energy for a beam: \[ U = \int_^{2L} \frac{M(x)^2}{2EI} dx \] But note \( EI \) is different for the two spans. \[ U = \int_^{L} \frac{[F x]^2}{2 (2EI)} dx + \int_L^{2L} \frac{[-F x + 2F L]^2}{2(EI)} dx \] Simplify: - For \( < x < L \): \( M(x) = F x \), \( EI = 2EI \) - For \( L < x < 2L \): \( M(x) = -F x + 2F L \), \( EI = EI \) \[ U = \int_^L \frac{F^2 x^2}{4EI} dx + \int_L^{2L} \frac{( -F x + 2F L )^2}{2EI} dx \] --- ## 4. **Step 3: Differentiate with Respect to \( F \) (Castigliano)** Because \( U = U(F) \): \[ \delta_B = \frac{\partial U}{\partial F} \] Calculate each term: ### **First Integral:** \[ U_1 = \int_^L \frac{F^2 x^2}{4EI} dx = \frac{F^2}{4EI} \int_^L x^2 dx = \frac{F^2}{4EI} \cdot \frac{L^3}{3} = \frac{F^2 L^3}{12 EI} \] ### **Second Integral:** Expand \( ( -F x + 2F L )^2 = F^2 (x^2 - 4L x + 4L^2) \): \[ U_2 = \int_L^{2L} \frac{F^2 (x^2 - 4L x + 4L^2)}{2EI} dx = \frac{F^2}{2EI} \int_L^{2L} (x^2 - 4L x + 4L^2) dx \] Compute each term: - \( \int_L^{2L} x^2 dx = \left[ \frac{x^3}{3} \right]_L^{2L} = \frac{(2L)^3}{3} - \frac{L^3}{3} = \frac{8L^3 - L^3}{3} = \frac{7L^3}{3} \) - \( \int_L^{2L} x dx = \left[ \frac{x^2}{2} \right]_L^{2L} = \frac{(2L)^2}{2} - \frac{L^2}{2} = \frac{4L^2 - L^2}{2} = \frac{3L^2}{2} \) - \( \int_L^{2L} dx = (2L - L) = L \) So, \[ \int_L^{2L} (x^2 - 4L x + 4L^2) dx = \frac{7L^3}{3} - 4L \cdot \frac{3L^2}{2} + 4L^2 \cdot L = \frac{7L^3}{3} - 6L^3 + 4L^3 = \frac{7L^3}{3} - 6L^3 + 4L^3 \] Calculate: \[ -6L^3 + 4L^3 = -2L^3 \] \[ \frac{7L^3}{3} - 2L^3 = \frac{7L^3 - 6L^3}{3} = \frac{L^3}{3} \] So, \[ U_2 = \frac{F^2}{2EI} \cdot \frac{L^3}{3} = \frac{F^2 L^3}{6 EI} \] --- ## 5. **Step 4: Total Strain Energy** \[ U = U_1 + U_2 = \frac{F^2 L^3}{12 EI} + \frac{F^2 L^3}{6 EI} = \frac{F^2 L^3}{12 EI} + \frac{2 F^2 L^3}{12 EI} = \frac{3 F^2 L^3}{12 EI} = \frac{F^2 L^3}{4 EI} \] --- ## 6. **Step 5: Displacement at Point B** \[ \delta_B = \frac{\partial U}{\partial F} = \frac{\partial}{\partial F} \left( \frac{F^2 L^3}{4 EI} \right ) = \frac{2 F L^3}{4 EI} = \frac{F L^3}{2 EI} \] --- ## 7. **Final Answer** \[ \boxed{ \text{The vertical displacement at point B is } \delta_B = \frac{F L^3}{2 E I} } \] --- ### **Summary Table** | Step | Description | Expression | |------|-------------|------------| | 1 | Strain energy in left span \(( \to L)\) | \(U_1 = \frac{F^2 L^3}{12 E I}\) | | 2 | Strain energy in right span \((L \to 2L)\) | \(U_2 = \frac{F^2 L^3}{6 E I}\) | | 3 | Total strain energy | \(U = \frac{F^2 L^3}{4 E I}\) | | 4 | Displacement at B | \( \delta_B = \frac{F L^3}{2 E I} \) | --- ### **Method Used** - **Energy method (Castigliano's Theorem)** - Accounted for different \( E, I \) values in each segment. --- **If you need the deflection at any other point or using another method, let me know!**