Let's solve **Case 5 – Uniformly Decreasing Distributed Load** using the **Double Integration Method**. --- ### **Given Diagram** Here is a reconstructed version of your provided diagram for clarity: ``` w2 ↓↓↓↓↓ A-----|///////|-----------------B |<--a4-->|<--b2-->|<----L----->| ``` - **A**: Fixed Support - **B**: Roller Support - **w₂**: Maximum load intensity at the left end of the distributed load - **a₄**: Length over which load decreases from \( w_2 \) to 0 - **b₂**: Length of the remaining span (no load) - **L**: Total span (\( L = a_4 + b_2 \)) --- ### **Step 1: Load Equation** For \( 0 \leq x \leq a_4 \): - Load decreases uniformly from \( w_2 \) to 0: \[ w(x) = w_2 \left( 1 - \frac{x}{a_4} \right) \] For \( a_4 < x \leq L \): - \( w(x) = 0 \) --- ### **Step 2: Differential Equation** The differential equation for deflection \( v(x) \) is: \[ EI \frac{d^4v}{dx^4} = -w(x) \] #### **For \( 0 \leq x \leq a_4 \):** \[ EI \frac{d^4v_1}{dx^4} = -w_2 \left( 1 - \frac{x}{a_4} \right) \] #### **For \( a_4 < x \leq L \):** \[ EI \frac{d^4v_2}{dx^4} = 0 \] --- ### **Step 3: Integrate (Double Integration Method)** #### **For \( 0 \leq x \leq a_4 \):** Integrate 4 times: \[ EI \frac{d^4v_1}{dx^4} = -w_2 \left( 1 - \frac{x}{a_4} \right) \] Integrate once: \[ EI \frac{d^3v_1}{dx^3} = -w_2 \left( x - \frac{x^2}{2a_4} \right) + C_1 \] Integrate again: \[ EI \frac{d^2v_1}{dx^2} = -w_2 \left( \frac{x^2}{2} - \frac{x^3}{6a_4} \right) + C_1 x + C_2 \] Integrate again (slope): \[ EI \frac{dv_1}{dx} = -w_2 \left( \frac{x^3}{6} - \frac{x^4}{24a_4} \right) + \frac{C_1 x^2}{2} + C_2 x + C_3 \] Integrate again (deflection): \[ EI v_1(x) = -w_2 \left( \frac{x^4}{24} - \frac{x^5}{120a_4} \right) + \frac{C_1 x^3}{6} + \frac{C_2 x^2}{2} + C_3 x + C_4 \] #### **For \( a_4 < x \leq L \):** Here, \( w(x) = 0 \), so: \[ EI \frac{d^4v_2}{dx^4} = 0 \] Integrate 4 times: \[ EI v_2(x) = \frac{C_5 x^3}{6} + \frac{C_6 x^2}{2} + C_7 x + C_8 \] --- ### **Step 4: Boundary and Continuity Conditions** #### **At \( x = 0 \) (Fixed end, A):** - \( v_1(0) = 0 \) (deflection is zero) - \( v_1'(0) = 0 \) (slope is zero) #### **At \( x = L \) (Roller end, B):** - \( v_2(L) = 0 \) (deflection is zero) - \( M_B = EI v_2''(L) = 0 \) (moment is zero) #### **At \( x = a_4 \) (Continuity):** - \( v_1(a_4) = v_2(a_4) \) (deflection is continuous) - \( v_1'(a_4) = v_2'(a_4) \) (slope is continuous) - \( v_1''(a_4) = v_2''(a_4) \) (moment is continuous) - \( v_1'''(a_4) = v_2'''(a_4) \) (shear is continuous) --- ### **Step 5: Solve for Constants** You have 8 constants (\( C_1, C_2, C_3, C_4 \) for segment 1, and \( C_5, C_6, C_7, C_8 \) for segment 2). Apply the 8 boundary and continuity conditions above to set up 8 equations and solve for all constants. --- ### **Step 6: Find Values at x1, x2** - For \( x_1 \) in \( 0 \leq x_1 \leq a_4 \): Use \( v_1(x_1) \) - For \( x_2 \) in \( a_4 < x_2 \leq L \): Use \( v_2(x_2) \) --- ### **Summary Table of Key Quantities** | Symbol | Meaning | |----------|-------------------| | \( C_1..C_4 \) | Constants for segment 1 | | \( C_5..C_8 \) | Constants for segment 2 | | \( v_1(x) \), \( v_2(x) \) | Deflection equations in each region | | \( \theta_1 \), \( \theta_2 \) | Slope at \( x_1 \), \( x_2 \) | | \( v_1 \), \( v_2 \) | Deflection at \( x_1 \), \( x_2 \) | --- ### **Final Formulas** #### **Deflection:** \[ v_1(x_1) = \frac{1}{EI} \left[ -w_2 \left( \frac{x_1^4}{24} - \frac{x_1^5}{120a_4} \right) + \frac{C_1 x_1^3}{6} + \frac{C_2 x_1^2}{2} + C_3 x_1 + C_4 \right] \] \[ v_2(x_2) = \frac{1}{EI} \left[ \frac{C_5 x_2^3}{6} + \frac{C_6 x_2^2}{2} + C_7 x_2 + C_8 \right] \] #### **Slope:** \[ \theta_1(x_1) = \frac{dv_1}{dx_1} \] \[ \theta_2(x_2) = \frac{dv_2}{dx_2} \] --- ### **Diagram (Recreated)** ``` w2 ↓↓↓↓↓ A---|///////|---------------B |<--a4-->|<--b2-->|<----L----->| ``` - \( 0 \leq x_1 \leq a_4 \) (Loaded region) - \( a_4 < x_2 \leq L \) (Unloaded region) --- ## **Summary Steps** 1. **Write load equation** \( w(x) \) 2. **Write and integrate differential equation** for each region 3. **Apply boundary and continuity conditions** to solve for constants 4. **Plug in x1, x2** to get deflection and slope --- If you need the explicit solution for all 8 constants, let me know with the values of \( w_2, a_4, b_2, L, EI \) and I can walk you through the algebra step by step!

Certainly! Let's systematically solve **Case 5 — Uniformly Decreasing Distributed Load** using the **Double Integration Method**. We'll include detailed steps, calculations, boundary conditions, continuity conditions, and diagrams. --- ## **Problem Setup** **Given:** - A simply supported beam of length \( L \). - The load \( w(x) \) decreases uniformly from \( w_2 \) at \( x=0 \) to 0 at \( x = a_4 \), then zero from \( x = a_4 \) to \( x = L \). - The load distribution: \[ w(x) = \begin{cases} w_2 \left( 1 - \frac{x}{a_4} \right), & 0 \leq x \leq a_4 \\ 0, & a_4 < x \leq L \end{cases} \] --- ## **Step 1: Load Function** \[ w(x) = \begin{cases} w_2 \left( 1 - \frac{x}{a_4} \right), & 0 \leq x \leq a_4 \\ 0, & a_4 < x \leq L \end{cases} \] --- ## **Step 2: Differential Equation** The deflection \( v(x) \) satisfies: \[ EI \frac{d^4 v}{dx^4} = -w(x) \] ### **For \( 0 \leq x \leq a_4 \):** \[ EI \frac{d^4 v_1}{dx^4} = -w_2 \left( 1 - \frac{x}{a_4} \right) \] ### **For \( a_4 < x \leq L \):** \[ EI \frac{d^4 v_2}{dx^4} = 0 \] --- ## **Step 3: Integrate to find \( v(x) \)** ### **For \( 0 \leq x \leq a_4 \):** Integrate four times: \[ EI \frac{d^4 v_1}{dx^4} = -w_2 \left( 1 - \frac{x}{a_4} \right) \] --- ### **First integration:** \[ EI \frac{d^3 v_1}{dx^3} = -w_2 \left( x - \frac{x^2}{2a_4} \right) + C_1 \] *(constant of integration \( C_1 \))* --- ### **Second integration:** \[ EI \frac{d^2 v_1}{dx^2} = -w_2 \left( \frac{x^2}{2} - \frac{x^3}{6a_4} \right) + C_1 x + C_2 \] --- ### **Third integration:** \[ EI \frac{dv_1}{dx} = -w_2 \left( \frac{x^3}{6} - \frac{x^4}{24a_4} \right) + \frac{C_1 x^2}{2} + C_2 x + C_3 \] --- ### **Fourth integration:** \[ EI v_1(x) = -w_2 \left( \frac{x^4}{24} - \frac{x^5}{120 a_4} \right) + \frac{C_1 x^3}{6} + \frac{C_2 x^2}{2} + C_3 x + C_4 \] --- ### **For \( a_4 < x \leq L \):** \[ EI \frac{d^4 v_2}{dx^4} = 0 \] Integrating four times: \[ EI v_2(x) = \frac{C_5 x^3}{6} + \frac{C_6 x^2}{2} + C_7 x + C_8 \] *(constants \( C_5, C_6, C_7, C_8 \))* --- ## **Step 4: Boundary Conditions** ### **At supports:** - **At \( x=0 \)** (fixed support): \[ v_1(0) = 0 \quad \Rightarrow \quad C_4 = 0 \] \[ v_1'(0) = 0 \quad \Rightarrow \quad \left.\frac{dv_1}{dx}\right|_{x=0} = 0 \] \[ \Rightarrow -w_2 \left( 0 - 0 \right) + C_1 = 0 \quad \Rightarrow \quad C_1 = 0 \] ### **At \( x=L \)** (roller support): \[ v_2(L) = 0 \] \[ \Rightarrow \frac{C_5 L^3}{6} + \frac{C_6 L^2}{2} + C_7 L + C_8 = 0 \] --- ## **Step 5: Continuity Conditions at \( x=a_4 \)** - **Deflection continuity:** \[ v_1(a_4) = v_2(a_4) \] - **Slope continuity:** \[ v_1'(a_4) = v_2'(a_4) \] - **Moment continuity:** \[ v_1''(a_4) = v_2''(a_4) \] - **Shear continuity:** \[ v_1'''(a_4) = v_2'''(a_4) \] --- ## **Step 6: Calculations of derivatives** ### **Derivatives of \( v_1(x) \):** \[ \frac{dv_1}{dx} = \frac{1}{EI} \left( -w_2 \left( \frac{x^3}{6} - \frac{x^4}{24a_4} \right) + \frac{C_1 x^2}{2} + C_2 x + C_3 \right) \] \[ \frac{d^2 v_1}{dx^2} = \frac{1}{EI} \left( -w_2 \left( \frac{x^2}{2} - \frac{x^3}{6a_4} \right) + C_1 x + C_2 \right) \] \[ \frac{d^3 v_1}{dx^3} = \frac{1}{EI} \left( -w_2 \left( x - \frac{x^2}{2a_4} \right) + C_1 \right) \] --- ### **Derivatives of \( v_2(x) \):** \[ \frac{dv_2}{dx} = \frac{1}{EI} \left( \frac{C_5 x^2}{2} + C_6 x + C_7 \right) \] \[ \frac{d^2 v_2}{dx^2} = \frac{1}{EI} \left( C_5 x + C_6 \right) \] \[ \frac{d^3 v_2}{dx^3} = \frac{C_5}{EI} \] --- ## **Step 7: Apply Conditions at \( x = a_4 \)** Set \( v_1(a_4) = v_2(a_4) \): \[ \text{From above:} \] \[ v_1(a_4) = -w_2 \left( \frac{a_4^4}{24} - \frac{a_4^5}{120 a_4} \right) + \frac{C_1 a_4^3}{6} + \frac{C_2 a_4^2}{2} + 0 \] \[ = -w_2 \left( \frac{a_4^4}{24} - \frac{a_4^4}{120} \right) + \frac{C_1 a_4^3}{6} + \frac{C_2 a_4^2}{2} \] \[ = -w_2 \left( \frac{5 a_4^4}{120} - \frac{a_4^4}{120} \right) + \frac{C_1 a_4^3}{6} + \frac{C_2 a_4^2}{2} \] \[ = -w_2 \left( \frac{4 a_4^4}{120} \right) + \frac{C_1 a_4^3}{6} + \frac{C_2 a_4^2}{2} \] \[ = -w_2 \left( \frac{a_4^4}{30} \right) + \frac{C_1 a_4^3}{6} + \frac{C_2 a_4^2}{2} \] Similarly, \[ v_2(a_4) = \frac{C_5 a_4^3}{6} + \frac{C_6 a_4^2}{2} + C_7 a_4 + C_8 \] --- ## **Continuity of slope:** \[ v_1'(a_4) = v_2'(a_4) \] \[ \Rightarrow \frac{1}{EI} \left( -w_2 \left( \frac{a_4^3}{6} - \frac{a_4^4}{24 a_4} \right) + \frac{C_1 a_4^2}{2} + C_2 a_4 \right) = \frac{1}{EI} \left( \frac{C_5 a_4^2}{2} + C_6 a_4 + C_7 \right) \] Simplify the left: \[ -w_2 \left( \frac{a_4^3}{6} - \frac{a_4^3}{24} \right) = -w_2 \left( \frac{4 a_4^3}{24} - \frac{a_4^3}{24} \right) = -w_2 \frac{3 a_4^3}{24} = -w_2 \frac{a_4^3}{8} \] Thus, \[ \frac{1}{EI} \left( -w_2 \frac{a_4^3}{8} + \frac{C_1 a_4^2}{2} + C_2 a_4 \right) = \frac{1}{EI} \left( \frac{C_5 a_4^2}{2} + C_6 a_4 + C_7 \right) \] --- ## **Continuity of moment:** \[ v_1''(a_4) = v_2''(a_4) \] \[ \Rightarrow \frac{1}{EI} \left( -w_2 \left( \frac{a_4^2}{2} - \frac{a_4^3}{6 a_4} \right) + C_1 a_4 + C_2 \right) = \frac{1}{EI} \left( C_5 a_4 + C_6 \right) \] Simplify: \[ -w_2 \left( \frac{a_4^2}{2} - \frac{a_4^2}{6} \right) = -w_2 \left( \frac{3 a_4^2}{6} - \frac{a_4^2}{6} \right) = -w_2 \frac{2 a_4^2}{6} = -w_2 \frac{a_4^2}{3} \] So, \[ \frac{1}{EI} \left( -w_2 \frac{a_4^2}{3} + C_1 a_4 + C_2 \right) = \frac{1}{EI} \left( C_5 a_4 + C_6 \right) \] --- ## **Continuity of shear:** \[ v_1'''(a_4) = v_2'''(a_4) \] \[ \Rightarrow \frac{1}{EI} \left( -w_2 \left( a_4 - \frac{a_4^2}{2 a_4} \right) + C_1 \right) = \frac{C_5}{EI} \] \[ -w_2 \left( a_4 - \frac{a_4}{2} \right) + C_1 = C_5 \] Simplify: \[ -w_2 \left( \frac{a_4}{2} \right) + C_1 = C_5 \] Recall \( C_1 = 0 \), so \[ C_5 = - \frac{w_2 a_4}{2} \] --- ## **Summary of constants:** - \( C_1 = 0 \) - \( C_5 = - \frac{w_2 a_4}{2} \) - \( C_4 = 0 \) Remaining constants \( C_2, C_3, C_6, C_7, C_8 \) are found from boundary conditions at \( x=L \) and the continuity equations. --- ## **Step 8: Boundary conditions at \( x=L \):** -