To solve the differential equation for the beam deflection, we will consider the specific case where the beam is simply supported at both ends and subjected to a uniform distributed load $q(x) = q_0$. The general equation for the bending of a beam is given by:

$\frac{d^2}{dx^2} \left( EI \frac{d^2 w}{dx^2} \right) = q(x)$

For a beam with constant flexural rigidity $EI$ and a uniform load $q(x) = q_0$, the equation simplifies to:

$EI \frac{d^4 w}{dx^4} = q_0$

We need to solve this differential equation with the boundary conditions for a simply supported beam:

Boundary Conditions:

• At $x = 0$: $w(0) = 0$ (no deflection at the left support)
• At $x = L$: $w(L) = 0$ (no deflection at the right support)
• At $x = 0$: $M(0) = EI \frac{d^2 w}{dx^2}(0) = 0$ (no moment at the left support)
• At $x = L$: $M(L) = EI \frac{d^2 w}{dx^2}(L) = 0$ (no moment at the right support)

Step 1: Integrate the Differential Equation

Start by integrating the differential equation:

$EI \frac{d^4 w}{dx^4} = q_0$

Integrate once:

$EI \frac{d^3 w}{dx^3} = q_0 x + C_1$

Integrate again:

$EI \frac{d^2 w}{dx^2} = \frac{q_0 x^2}{2} + C_1 x + C_2$

This gives the bending moment equation. Integrate again:

$EI \frac{d w}{dx} = \frac{q_0 x^3}{6} + \frac{C_1 x^2}{2} + C_2 x + C_3$

Finally, integrate once more to obtain the deflection:

$EI w(x) = \frac{q_0 x^4}{24} + \frac{C_1 x^3}{6} + \frac{C_2 x^2}{2} + C_3 x + C_4$

Step 2: Apply Boundary Conditions

Now, apply the boundary conditions to determine the constants $C_1$, $C_2$, $C_3$, and $C_4$.

1. At $x = 0$, $w(0) = 0$: $EI \cdot 0 = C_4 \quad \Rightarrow \quad C_4 = 0$

2. At $x = L$, $w(L) = 0$: $EI \cdot 0 = \frac{q_0 L^4}{24} + \frac{C_1 L^3}{6} + \frac{C_2 L^2}{2} + C_3 L \quad \Rightarrow \quad \frac{q_0 L^4}{24} + \frac{C_1 L^3}{6} + \frac{C_2 L^2}{2} + C_3 L = 0 \quad \text{(Eq. 1)}$

3. At $x = 0$, $M(0) = 0$: $EI \cdot 0 = C_2 \quad \Rightarrow \quad C_2 = 0$

4. At $x = L$, $M(L) = 0$: $EI \cdot 0 = \frac{q_0 L^2}{2} + C_1 L \quad \Rightarrow \quad \frac{q_0 L^2}{2} + C_1 L = 0 \quad \Rightarrow \quad C_1 = -\frac{q_0 L}{2}$

Step 3: Substitute the Constants Back

Using $C_1 = -\frac{q_0 L}{2}$, $C_2 = 0$, $C_3 = 0$, and $C_4 = 0$, we substitute back into the expression for $w(x)$:

$EI w(x) = \frac{q_0 x^4}{24} - \frac{q_0 L x^3}{12}$

Simplifying, the deflection of the beam is:

$w(x) = \frac{q_0}{24 EI} \left( x^4 - 2L x^3 + L^3 x \right)$

Final Solution for the Deflection $w(x)$:

$w(x) = \frac{q_0}{24 EI} \left( x^3 (x - 2L) + L^3 x \right)$

This is the equation for the deflection of a simply supported beam subjected to a uniform load $q_0$.

Would you like any further details or a breakdown of specific steps?

Follow-up Questions:

1. Would you like to see how the shear force and bending moment vary along the beam using the above solution?
2. Are you interested in visualizing the deflection curve for different values of $q_0$, $EI$, and $L$?
3. Should we explore how this solution changes if the beam has different boundary conditions, like a cantilever?
4. Do you want to calculate the maximum deflection and where it occurs along the beam?
5. Are you interested in the physical interpretation of the constants $EI$ and $q_0$ in this context?

Tip:

When solving beam problems, always start by clearly stating and applying the boundary conditions; this step is crucial to obtaining the correct particular solution.