Let’s walk through each part step by step for the continuous beam shown above. #### **Given:** - \( E = 29,000 \) ksi - \( I = 600 \) in\(^4\) - The beam is supported at **A** (fixed), **B** (roller), and **C** (roller). - Two point loads \( P \) are applied at \( L/2 \) from A and C, respectively. - Span AB = BC = \( L/2 \), total length = \( L \). - Treat reaction at C (\( C_y \)) as **redundant** (use the **force method**). --- ## (a) **Determine the reactions at A, B, and C (using \( C_y \) as redundant)** ### **Step 1: Primary Structure (remove \( C_y \))** - Remove support at C (make it a simple support or leave as a roller). - The beam is now simply supported at A (fixed) and B (roller), with load \( P \) at midspan of AB and at B. - Calculate reactions (**let's call them \( R_A', R_B' \)**) ignoring \( C_y \). ### **Step 2: Compatibility Equation** - The vertical deflection at C due to all loads + \( C_y \) must be zero (since C is a support). ### **Step 3: Set up equations** #### **1. Reactions for Primary Structure (without Cy):** - Let's denote the reaction at B as \( R_B' \), at A as \( R_A' \), and at C as 0. Sum of vertical forces: \[ R_A' + R_B' = 2P \] Take moment about A: \[ P \cdot \frac{L}{2} + P \cdot L - R_B' \cdot L = 0 \] \[ P \left( \frac{L}{2} + L \right) = R_B' L \] \[ P \left( \frac{3L}{2} \right) = R_B' L \] \[ R_B' = \frac{3P}{2} \] So, \[ R_A' = 2P - \frac{3P}{2} = \frac{P}{2} \] #### **2. Deflection at C using unit load method:** - The **deflection at C** due to: - The real loads (\( \delta_{C, \text{loads}} \)) - A redundant reaction \( C_y \) at C (\( \delta_{C, Cy} \)) The sum must be zero: \[ \delta_{C, \text{loads}} + C_y \delta_{C, 1} = 0 \] Where \( \delta_{C, 1} \) is the deflection at C due to a unit upward force at C. ##### **Calculate \( \delta_{C, \text{loads}} \):** For a simply supported beam of length \( L \) with point load \( P \) at \( a \) from the left, deflection at distance \( x \) from left is: \[ \delta = \frac{P b x (L^2 - b^2 - x^2)}{6 E I L} \] where \( b = L - a \). But since both loads are symmetrical, let's sum up the deflections at C (at \( x = L \)) due to both loads. - Load at \( L/2 \): \( a = L/2 \), \( b = L/2 \), deflection at \( x = L \): \[ \delta_{C,1} = \frac{P \cdot (L/2) \cdot L \cdot (L^2 - (L/2)^2 - L^2)}{6 E I L} = \frac{P \cdot (L/2) \cdot L \cdot (- (L/2)^2)}{6 E I L} = -\frac{P L^2}{24 E I} \] Similarly, load at \( L \) (which is at C itself) causes zero deflection at C. So, \[ \delta_{C, \text{loads}} = -\frac{P L^2}{24 E I} \] ##### **Calculate \( \delta_{C, 1} \):** For a unit load at C on a simply supported beam, the deflection at C is: \[ \delta_{C,1} = \frac{L^3}{3 E I} \] ##### **Set up the compatibility equation:** \[ -\frac{P L^2}{24 E I} + C_y \frac{L^3}{3 E I} = 0 \] \[ C_y = \frac{P L^2}{24 E I} \cdot \frac{3 E I}{L^3} = \frac{P}{8 L} \] #### **3. Find all Reactions** Now add the effect of \( C_y \): - The reaction at C is \( C_y = \frac{P}{8 L} \). - The additional upward force at A and B due to \( C_y \) can be found by superposition (using statics). --- ## (b) **Shear Force at A, B, and C** Shear at A: \( V_A \) Shear at B: \( V_B \) Shear at C: \( V_C \) Calculate these by summing forces to the left of each point. --- ## (c) **Bending Moment at Midspan of AB and BC** - At \( x = L/4 \) (midspan of AB) - At \( x = 3L/4 \) (midspan of BC) Use \( M = R_A \cdot x - P \cdot (x - L/2) \) (if \( x > L/2 \)), etc. --- ## (d) **Vertical Displacement Below Load P in AB** Use **deflection formula** for point load on simply supported span. --- # **Summary Table of Formulas** | Quantity | Formula | |---|---| | Reaction at A | \( R_A = \frac{P}{2} + \text{contribution from } C_y \) | | Reaction at B | \( R_B = \frac{3P}{2} + \text{contribution from } C_y \) | | Reaction at C | \( C_y = \frac{P}{8 L} \) | | Shear at A | \( V_A = R_A \) | | Shear at B | \( V_B = R_B - P \) | | Shear at C | \( V_C = R_C \) | | Moment at mid-AB | \( M = R_A \cdot \frac{L}{4} \) | | Moment at mid-BC | Similar formula | | Deflection under P | \( \delta = \frac{P a b (L^2 - a^2 - b^2)}{6 E I L} \) at location of P | --- If you provide values for \( P \) and \( L \), I can compute the numerical answers! If you need the full symbolic solution, let me know, and I’ll continue step-by-step for each part!