Let's solve the problem **step-by-step** using Laplace transforms. --- ### **Given:** - Length of beam, \( L = 5 \) m - Simply supported at both ends - Load \( P = 2EI \) N applied at \( x = 2 \) m - To find: **Deflection \( y(x) \) as a function of \( x \) using Laplace transforms** --- ## **Step 1: Write the Governing Equation** For a beam with a point load at \( x = a \): \[ EI \frac{d^4 y}{dx^4} = P \delta(x - a) \] Where: - \( EI \) = flexural rigidity - \( P = 2EI \) - \( a = 2 \) - \( \delta(x-a) \) = Dirac delta function So, \[ EI \frac{d^4 y}{dx^4} = 2EI \delta(x-2) \] \[ \frac{d^4 y}{dx^4} = 2 \delta(x-2) \] --- ## **Step 2: Take Laplace Transform** Let \( \mathcal{L}\{y(x)\} = Y(s) \). Recall: - \( \mathcal{L}\left\{ \frac{d^4 y}{dx^4} \right\} = s^4 Y(s) - s^3 y(0) - s^2 y'(0) - s y''(0) - y'''(0) \) - \( \mathcal{L}\{\delta(x-a)\} = e^{-as} \) So, \[ s^4 Y(s) - s^3 y(0) - s^2 y'(0) - s y''(0) - y'''(0) = 2 e^{-2s} \] --- ## **Step 3: Apply Boundary Conditions** For a simply supported beam: - \( y(0) = 0 \) (deflection at \( x=0 \)) - \( y(L) = 0 \) (deflection at \( x=L \)) - \( y''(0) = 0 \) (moment at \( x=0 \)) - \( y''(L) = 0 \) (moment at \( x=L \)) From Laplace, we need conditions at \( x=0 \): \[ y(0) = 0 \] \[ y''(0) = 0 \] Also, \( y'(0) \) and \( y'''(0) \) are unknowns. So, equation becomes: \[ s^4 Y(s) - s^2 y'(0) - y'''(0) = 2 e^{-2s} \] \[ Y(s) = \frac{2 e^{-2s} + s^2 y'(0) + y'''(0)}{s^4} \] \[ Y(s) = 2 e^{-2s} s^{-4} + y'(0) s^{-2} + y'''(0) s^{-4} \] --- ## **Step 4: Inverse Laplace Transform** Recall: - \( \mathcal{L}^{-1}\{ e^{-as} s^{-n} \} = \frac{(x-a)^{n-1}}{(n-1)!} u(x-a) \) - \( \mathcal{L}^{-1}\{ s^{-2} \} = x \) - \( \mathcal{L}^{-1}\{ s^{-4} \} = \frac{x^3}{6} \) So: \[ \mathcal{L}^{-1}\left\{ 2 e^{-2s} s^{-4} \right\} = 2 \cdot \frac{(x-2)^3}{6} u(x-2) = \frac{(x-2)^3}{3} u(x-2) \] \[ \mathcal{L}^{-1}\left\{ y'(0) s^{-2} \right\} = y'(0) x \] \[ \mathcal{L}^{-1}\left\{ y'''(0) s^{-4} \right\} = y'''(0) \frac{x^3}{6} \] So, \[ y(x) = \frac{(x-2)^3}{3} u(x-2) + y'(0)x + y'''(0)\frac{x^3}{6} \] Where \( u(x-2) \) is the unit step function. --- ## **Step 5: Apply Remaining Boundary Conditions** Recall: - \( y(0) = 0 \) - \( y(L) = y(5) = 0 \) - \( y''(0) = 0 \) - \( y''(5) = 0 \) ### **Find \( y'(0) \) and \( y'''(0) \):** First, compute derivatives: \[ y(x) = \begin{cases} y'(0)x + y'''(0)\frac{x^3}{6}, & x < 2 \\ \frac{(x-2)^3}{3} + y'(0)x + y'''(0)\frac{x^3}{6}, & x \geq 2 \end{cases} \] For \( x < 2 \), \( u(x-2)=0 \). Let’s compute \( y''(x) \): \[ y''(x) = \begin{cases} y'''(0)x, & x < 2 \\ 2(x-2) + y'''(0)x, & x \geq 2 \end{cases} \] Wait, let's write the derivatives more precisely. ### For \( x < 2 \): \[ y(x) = y'(0)x + y'''(0)\frac{x^3}{6} \] \[ y'(x) = y'(0) + y'''(0)\frac{x^2}{2} \] \[ y''(x) = y'''(0)x \] \[ y'''(x) = y'''(0) \] ### For \( x \geq 2 \): \[ y(x) = \frac{(x-2)^3}{3} + y'(0)x + y'''(0)\frac{x^3}{6} \] \[ y'(x) = (x-2)^2 + y'(0) + y'''(0)\frac{x^2}{2} \] \[ y''(x) = 2(x-2) + y'''(0)x \] \[ y'''(x) = 2 + y'''(0) \] --- ### Now Apply Boundary Conditions #### 1. \( y(0) = 0 \): \[ y(0) = 0 + 0 = 0 \] (Satisfied automatically.) #### 2. \( y(5) = 0 \): For \( x = 5 > 2 \): \[ y(5) = \frac{(5-2)^3}{3} + y'(0) \cdot 5 + y'''(0)\frac{125}{6} = \frac{27}{3} + 5y'(0) + \frac{125}{6}y'''(0) = 9 + 5y'(0) + \frac{125}{6}y'''(0) \] Set to zero: \[ 9 + 5y'(0) + \frac{125}{6}y'''(0) = 0 \tag{1} \] #### 3. \( y''(0) = 0 \): For \( x = 0 < 2 \): \[ y''(0) = y'''(0) \cdot 0 = 0 \] (Satisfied automatically.) #### 4. \( y''(5) = 0 \): For \( x = 5 > 2 \): \[ y''(5) = 2(5-2) + y'''(0) \cdot 5 = 6 + 5y'''(0) \] Set to zero: \[ 6 + 5y'''(0) = 0 \implies y'''(0) = -\frac{6}{5} \] --- ### **Step 6: Find \( y'(0) \) Using Eq (1)** Plug \( y'''(0) = -\frac{6}{5} \) into Eq (1): \[ 9 + 5y'(0) + \frac{125}{6} \left(-\frac{6}{5}\right) = 0 \] \[ 9 + 5y'(0) - 25 = 0 \] \[ 5y'(0) = 25 - 9 = 16 \implies y'(0) = \frac{16}{5} \] --- ## **Step 7: Write the Final Deflection Function** Plug values into \( y(x) \): \[ y(x) = \frac{(x-2)^3}{3} u(x-2) + \frac{16}{5}x - \frac{6}{5} \cdot \frac{x^3}{6} \] \[ y(x) = \frac{(x-2)^3}{3} u(x-2) + \frac{16}{5}x - \frac{x^3}{5} \] --- ### **Final Answer** \[ \boxed{ y(x) = \frac{(x-2)^3}{3} \cdot u(x-2) + \frac{16}{5}x - \frac{x^3}{5} } \] Where \( u(x-2) \) is the **unit step function**: - For \( x < 2 \): \( y(x) = \frac{16}{5}x - \frac{x^3}{5} \) - For \( x \geq 2 \): \( y(x) = \frac{(x-2)^3}{3} + \frac{16}{5}x - \frac{x^3}{5} \) --- ### **Explanation Summary:** 1. **Set up** the differential equation for the beam. 2. **Apply Laplace transforms**. 3. **Use boundary conditions** to solve for constants. 4. **Inverse Laplace** to get the deflection formula. Let me know if you need a plot or further explanation!