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.

Show Answer

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.