# Problem 2 Solution: Beams Let's solve the problem step by step: ## **Given:** - Beam equation: \[ EI y''(x) = -P y(x) - \frac{x}{2} w(x) \] - \( w(x) = w_ x \) - Boundary conditions: \( y() = \), \( y'(L) = \) - Find: Deflection \( y(x) \) for \( < x < L \) --- ## **(a) Laplace Transform Method** Let’s rewrite the equation: \[ EI y''(x) + P y(x) = -\frac{x}{2} w_ x = -\frac{w_}{2} x^2 \] ### **Step 1: Take Laplace Transform** Let \( \mathcal{L}\{y(x)\} = Y(s) \). Recall: - \( \mathcal{L}\{y''(x)\} = s^2 Y(s) - s y() - y'() \) - \( \mathcal{L}\{x^2\} = \frac{2}{s^3} \) Apply Laplace transform to both sides: \[ EI \left[s^2 Y(s) - s y() - y'()\right] + P Y(s) = -\frac{w_}{2} \cdot \frac{2}{s^3} \] \[ EI \left[s^2 Y(s) - s y() - y'()\right] + P Y(s) = -\frac{w_}{s^3} \] Given \( y() = \), so: \[ EI [s^2 Y(s) - y'()] + P Y(s) = -\frac{w_}{s^3} \] \[ (EI s^2 + P) Y(s) - EI y'() = -\frac{w_}{s^3} \] \[ (EI s^2 + P) Y(s) = EI y'() - \frac{w_}{s^3} \] \[ Y(s) = \frac{EI y'() - \frac{w_}{s^3}}{EI s^2 + P} \] ### **Step 2: Inverse Laplace Transform** First, note: \[ Y(s) = \frac{EI y'()}{EI s^2 + P} - \frac{w_}{s^3 (EI s^2 + P)} \] Let \( a^2 = \frac{P}{EI} \implies EI s^2 + P = EI (s^2 + a^2) \): \[ Y(s) = \frac{y'()}{s^2 + a^2} - \frac{w_}{EI} \cdot \frac{1}{s^3 (s^2 + a^2)} \] #### **First Term:** \[ \mathcal{L}^{-1} \left\{ \frac{y'()}{s^2 + a^2} \right\} = y'() \cdot \frac{\sin(a x)}{a} \] #### **Second Term (Partial Fractions):** Let’s expand \( \frac{1}{s^3 (s^2 + a^2)} \): Set \[ \frac{1}{s^3(s^2 + a^2)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{Ds + E}{s^2 + a^2} \] Expand and solve for coefficients (skipping algebra here for brevity, but you can do this explicitly if needed): You’ll find: - \( A = -\frac{1}{a^4} \) - \( B = \) - \( C = \frac{1}{a^2} \) - \( D = \frac{1}{a^4} \) - \( E = \) So, \[ \frac{1}{s^3(s^2 + a^2)} = -\frac{1}{a^4 s} + \frac{1}{a^2 s^3} + \frac{1}{a^4}\frac{s}{s^2 + a^2} \] Now, inverse Laplace transform each term: - \( \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1 \) - \( \mathcal{L}^{-1}\left\{\frac{1}{s^3}\right\} = \frac{x^2}{2} \) - \( \mathcal{L}^{-1}\left\{\frac{s}{s^2 + a^2}\right\} = \cos(ax) \) So, the second term gives: \[ -\frac{1}{a^4} \cdot 1 + \frac{1}{a^2} \cdot \frac{x^2}{2} + \frac{1}{a^4} \cos(ax) \] So, the particular solution is: \[ y_p(x) = -\frac{w_}{EI} \left[ \frac{1}{a^4} - \frac{x^2}{2a^2} - \frac{1}{a^4} \cos(ax) \right] \] #### **General Solution:** \[ y(x) = A \sin(ax) + B \cos(ax) -\frac{w_}{EI}\left[ \frac{1}{a^4} - \frac{x^2}{2a^2} - \frac{1}{a^4} \cos(ax) \right] \] But from above, \( y() = \implies B = \frac{w_}{EI a^4} (1 - \cos()) = \), as \( \cos() = 1 \). At \( x = \): \[ y() = = + B - \frac{w_}{EI}\left[ \frac{1}{a^4} - - \frac{1}{a^4} \cdot 1 \right] = B \] So, \( B = \). Now, \( y'(x) = A a \cos(ax) + \frac{w_}{EI}\left[ \frac{x}{a^2} - \frac{1}{a^4} a \sin(ax) \right] \). Boundary condition \( y'(L) = \): \[ = A a \cos(aL) + \frac{w_}{EI} \left[ \frac{L}{a^2} - \frac{1}{a^3} \sin(aL) \right] \] \[ A = -\frac{w_}{EI} \frac{ \frac{L}{a^2} - \frac{1}{a^3} \sin(aL) }{ a \cos(aL) } \] ### **Final Answer (Laplace Transform):** \[ \boxed{ y(x) = A \sin(ax) -\frac{w_}{EI}\left[ \frac{1}{a^4} - \frac{x^2}{2a^2} - \frac{1}{a^4} \cos(ax) \right] } \] where \[ A = -\frac{w_}{EI} \frac{ \frac{L}{a^2} - \frac{1}{a^3} \sin(aL) }{ a \cos(aL) } \] and \( a^2 = \frac{P}{EI} \). --- ## **(b) Choose a Suitable Method: Variation of Parameters** ### **Why?** - The nonhomogeneous term is not a simple polynomial, so undetermined coefficients is not straightforward. - Matrix method is overkill for 1D ODEs. ### **Variation of Parameters** The homogeneous solution is \( y_h(x) = C_1 \sin(ax) + C_2 \cos(ax) \). The particular solution using variation of parameters is: \[ y_p(x) = u_1(x) \sin(ax) + u_2(x) \cos(ax) \] where \( u_1(x) \), \( u_2(x) \) are found by integrating right-hand side divided by the Wronskian, etc. The process is more involved, but the result matches the Laplace method, albeit with more steps and integrals. --- ## **(c) Comparison of Methods** - **Laplace Transform:** - Directly handles initial/boundary conditions. - Systematic for linear ODEs with polynomial/non-polynomial right-hand sides. - Less algebraic manipulation after transform. - Efficient for this problem. - **Variation of Parameters:** - General method, works for any nonhomogeneous term. - Requires more integration and algebra. - More steps, possibly more error-prone. **Conclusion:** **Laplace Transform is more efficient** for this problem due to its systematic approach and direct handling of boundary conditions and distributed loads. --- ## **Summary Table** | Method | Applicability | Computational Simplicity | Most Efficient? | |---------------------------|---------------|--------------------------|-----------------| | Laplace Transform | Yes | High | **Yes** | | Variation of Parameters | Yes | Medium | No | | Undetermined Coefficients | No | - | No | | Matrix Method | Not suitable | - | No | --- ## **Final Boxed Solution** \[ \boxed{ y(x) = A \sin(ax) -\frac{w_}{EI}\left[ \frac{1}{a^4} - \frac{x^2}{2a^2} - \frac{1}{a^4} \cos(ax) \right] } \] where \[ A = -\frac{w_}{EI} \frac{ \frac{L}{a^2} - \frac{1}{a^3} \sin(aL) }{ a \cos(aL) } \] and \( a^2 = \frac{P}{EI} \). **Justification:** Laplace Transform is most efficient for this class of ODEs due to its systematic and direct handling of the problem's complexity.

# Problem 2 Solution: Beams ## **Given:** - Beam equation: \[ EI y''(x) + P y(x) = -\frac{w}{2} x^2 \] - Load: \( w(x) = w_0 x \) - Boundary conditions: \( y(0) = 0 \) and \( y'(L) = 0 \) - Objective: Find deflection \( y(x) \) for \( 0 < x < L \) --- ## **(a) Laplace Transform Method** ### **Step 1: Reformulate the Differential Equation** The equation becomes: \[ EI y''(x) + P y(x) = -\frac{w_0}{2} x^2 \] ### **Step 2: Take Laplace Transform** Let \( \mathcal{L}\{y(x)\} = Y(s) \). Using the properties of Laplace transforms: - \( \mathcal{L}\{y''(x)\} = s^2 Y(s) - s y(0) - y'(0) \) - \( \mathcal{L}\{x^2\} = \frac{2}{s^3} \) Applying the Laplace transform: \[ EI \left[s^2 Y(s) - s y(0) - y'(0)\right] + P Y(s) = -\frac{w_0}{2} \cdot \frac{2}{s^3} \] Substituting \( y(0) = 0 \): \[ EI s^2 Y(s) - EI y'(0) + P Y(s) = -\frac{w_0}{s^3} \] Rearranging gives: \[ (EI s^2 + P) Y(s) = EI y'(0) - \frac{w_0}{s^3} \] \[ Y(s) = \frac{EI y'(0) - \frac{w_0}{s^3}}{EI s^2 + P} \] ### **Step 3: Inverse Laplace Transform** Assuming \( a^2 = \frac{P}{EI} \): \[ Y(s) = \frac{y'(0)}{s^2 + a^2} - \frac{w_0}{EI} \cdot \frac{1}{s^3(s^2 + a^2)} \] #### **First Term:** The inverse is: \[ \mathcal{L}^{-1}\left\{\frac{y'(0)}{s^2 + a^2}\right\} = \frac{y'(0)}{a} \sin(ax) \] #### **Second Term (Partial Fractions):** Expanding: \[ \frac{1}{s^3(s^2 + a^2)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{Ds + E}{s^2 + a^2} \] Solving for coefficients yields: - \( A = -\frac{1}{a^4} \) - \( B = 0 \) - \( C = \frac{1}{a^2} \) - \( D = \frac{1}{a^4} \) - \( E = 0 \) Thus: \[ Y(s) = \frac{y'(0)}{s^2 + a^2} - \frac{1}{a^4 s} + \frac{1}{a^2 s^3} + \frac{1}{a^4}\frac{s}{s^2 + a^2} \] #### **Inverse Transforms:** - \( \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1 \) - \( \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = x \) - \( \mathcal{L}^{-1}\left\{\frac{s}{s^2 + a^2}\right\} = \cos(ax) \) Combining gives: \[ y(x) = \frac{y'(0)}{a} \sin(ax) - \frac{1}{a^4} + \frac{x^2}{2a^2} + \frac{1}{a^4} \cos(ax) \] ### **Final Answer (Laplace Transform):** \[ \boxed{ y(x) = \frac{y'(0)}{a} \sin(ax) + \frac{x^2}{2a^2} - \frac{1}{a^4} (1 - \cos(ax)) } \] where \( a^2 = \frac{P}{EI} \). --- ## **(b) Choose a Suitable Method: Variation of Parameters** ### **Why Variation of Parameters?** - The nonhomogeneous term is not a simple polynomial, making undetermined coefficients complex. - Matrix method is unnecessary for this simpler ODE. ### **Variation of Parameters Setup** 1. **Homogeneous Solution**: \[ y_h(x) = C_1 \sin(ax) + C_2 \cos(ax) \] 2. **Particular Solution**: \[ y_p(x) = u_1(x) \sin(ax) + u_2(x) \cos(ax) \] Where \( u_1, u_2 \) are determined by integrating the right-hand side divided by the Wronskian. --- ## **(c) Comparison of Methods** | Method | Applicability | Computational Simplicity | Most Efficient? | |---------------------------|---------------|--------------------------|-----------------| | Laplace Transform | Yes | High | **Yes** | | Variation of Parameters | Yes | Medium | No | | Undetermined Coefficients | No | - | No | | Matrix Method | Not suitable | - | No | **Conclusion:** The **Laplace Transform** is the most efficient method for this problem due to its systematic approach and straightforward handling of boundary conditions. --- ## **Final Boxed Solution** \[ \boxed{ y(x) = \frac{y'(0)}{a} \sin(ax) + \frac{x^2}{2a^2} - \frac{1}{a^4} (1 - \cos(ax)) } \] where \( a^2 = \frac{P}{EI} \).