Let's break down the problem and **find the value of the integration constant \( C_3 \)** by applying the boundary conditions. --- ### **Given:** - **Distributed load:** \( w = 1.5\, \text{kN/m} \) - **Moment at free end:** \( M_0 = 9\, \text{kN}\cdot\text{m} \) - **Length:** \( AB = 5\, \text{m},\ BC = 3\, \text{m} \) - **Elastic curves for:** - **Section AB (\(0 \leq x_1 \leq 5\)):** \[ v' = \frac{1}{EI}\left(\frac{3.15}{2}x_1^2 + 13.13 x_1\right) \] \[ v = \frac{1}{EI}\left(\frac{3.15}{6}x_1^3 + \frac{13.13}{2}x_1^2\right) \] - **Section BC (\(0 \leq x_2 \leq 3\)):** \[ v' = \frac{1}{EI}\left(\frac{1.5}{2}x_2^2 + 9x_2 + C_3\right) \] \[ v = \frac{1}{EI}\left(\frac{1.5}{6}x_2^3 + \frac{9}{2}x_2^2 + C_3x_2 + C_4\right) \] --- ## **Boundary Condition to Find \(C_3\):** The slope (\( v' \)) must be **continuous at point \( C \)**. At the point \( C \) (junction between AB and BC): - For **AB**, \( x_1 = 5 \) - For **BC**, \( x_2 = 0 \) So, \[ v'_{AB}(x_1 = 5) = v'_{BC}(x_2 = 0) \] ### **Calculating \(v'_{AB}(x_1 = 5)\):** \[ v'_{AB}(5) = \frac{1}{EI} \left( \frac{3.15}{2} (5)^2 + 13.13 \times 5 \right) \] \[ = \frac{1}{EI} \left( \frac{3.15}{2} \times 25 + 65.65 \right) \] \[ = \frac{1}{EI} \left( 39.375 + 65.65 \right) \] \[ = \frac{1}{EI} (105.025) \] ### **Calculating \(v'_{BC}(x_2 = 0)\):** \[ v'_{BC}(0) = \frac{1}{EI} (0 + 0 + C_3) = \frac{C_3}{EI} \] ### **Setting them equal:** \[ v'_{AB}(5) = v'_{BC}(0) \] \[ \frac{1}{EI}(105.025) = \frac{C_3}{EI} \] So, \[ C_3 = 105.025 \] --- ## **Final Answer:** \[ \boxed{C_3 = 105.03} \] (Rounded to two decimal places, **sign is positive**.) --- **Let me know if you need help with other constants or next steps!**