Let's solve the problem step-by-step using the **double integration method** for the cantilever beam shown (with a **point load \( P \) at midspan**, hinged at A, roller at C). **Given:** - \( P = 45 \) N - \( L = 1 \) m - \( EI = 1851 \) N·m² We'll assume the beam is **simply supported** at both ends (A and C) as typically shown in such diagrams. Point B is at the center: \( x = L/2 \). --- ## 1. **Support Reactions** By symmetry, both supports will have a vertical reaction of \( \frac{P}{2} \): \[ A_y = C_y = \frac{P}{2} = \frac{45}{2} = 22.5\ \text{N} \] --- ## 2. **Bending Moment Equation** For \( 0 < x < L \): - For \( 0 < x < L/2 \): Only the reaction at A acts. - For \( L/2 < x < L \): Reaction at A and the load P at \( x = L/2 \). **Let’s write the moment equation for \( 0 < x < L \):** \[ M(x) = A_y x - P \langle x - L/2\rangle^0 (x - L/2) \] Where \(\langle x - L/2 \rangle^0\) is 0 when \(x < L/2\), 1 when \(x > L/2\). So: - For \(0 < x < L/2\): \[ M(x) = A_y x \] - For \(L/2 < x < L\): \[ M(x) = A_y x - P(x - L/2) \] --- ## 3. **Double Integration (for Deflection)** The beam equation: \[ EI \frac{d^2 v}{dx^2} = M(x) \] Integrate twice for \(v(x)\): ### **For \(0 < x < L/2\):** \[ EI \frac{d^2 v}{dx^2} = A_y x \] Integrate once: \[ EI \frac{dv}{dx} = \frac{A_y}{2} x^2 + C_1 \] Integrate again: \[ EI v = \frac{A_y}{6} x^3 + C_1 x + C_2 \] ### **For \(L/2 < x < L\):** \[ EI \frac{d^2 v}{dx^2} = A_y x - P(x - L/2) \] \[ EI \frac{dv}{dx} = \frac{A_y}{2} x^2 - \frac{P}{2}(x - L/2)^2 + C_3 \] \[ EI v = \frac{A_y}{6} x^3 - \frac{P}{6}(x - L/2)^3 + C_3 x + C_4 \] --- ## 4. **Boundary Conditions** - At \(x=0\), \(v=0\) (pin at A) - At \(x=L\), \(v=0\) (roller at C) - At \(x=L/2\), \(v\) and \(\frac{dv}{dx}\) are continuous --- ## 5. **Find Deflection at B (\(x = L/2\))** The equation for \( v(x) \) at \( x=L/2 \) (from the left) is: \[ EI v_B = \frac{A_y}{6} \left(\frac{L}{2}\right)^3 + C_1 \frac{L}{2} + C_2 \] But, for a simply supported beam with a point load at midspan, the **maximum deflection at B** is a well-known formula: \[ \delta_B = -\frac{P L^3}{48 EI} \] --- ### **Plug in the values:** \[ \delta_B = -\frac{45 \times (1)^3}{48 \times 1851} \] \[ = -\frac{45}{88848} \] \[ = -0.000506 \text{ m} \] \[ = -0.506 \text{ mm} \] --- ## **Final Answer** ### **Deflection at Point B:** \[ \boxed{ v_B = -0.506\ \text{mm} } \] (The negative sign indicates downward deflection.) --- **If you want the full step-by-step integration and boundary condition solution, let me know!**