A beam under loading will deflect from the axis os symmetry \( y=0 \) according to the DE (for small deflections):
\( EI \cfrac{d^4y}{dx^4} = w(x) \)
Where \(E\) is the modulas of elasticity, \(I\) is the moment of inertia and \(w\) is the load.
Here, we have boundary conditions which provide information on how to solve for the constants in the general solution to the DE. See the table below for conditions at \( x = a \).
| Boundary | Conditions |
| Embedded | \( y(a) = 0, \cfrac{dy(a)}{dx} = 0\) |
| Free | \( \cfrac{d^2y(a)}{dx^2} = 0, \cfrac{d^3y(a)}{dx^3} = 0 \) |
| Hinged (Simply Supported) | \( y(a) = 0, \cfrac{d^2y(a)}{dx^2} = 0 \) |
Determine the deflection equation for a beam under constant load that is embedded on the left end and free on the right end.
A thin vertical column under constant load, \( P \), will buckle according to the DE:
\( EI \cfrac{d^2y)}{dx^2} + Py = 0\)
This is a linear equation that can be solved using a characteristic equation:
\( m^2 + \cfrac{EI}{P} = 0 \)
\( m = \sqrt{\cfrac{EI}{P}} i \)
\( y(x) = c_1 \cos {(\cfrac{EI}{P} x )} + c_2 \sin {(\cfrac{EI}{P} x)} \)
The boundary conditions are \( y(0) = 0, y(L) = 0\) since there is no deflection at the fixed ends.