Dissemination of IT for the Promotion of Materials Science (DoITPoMS)

# Symmetrical 3-point bending

The bending moment is given by

$M = \frac{{ - Fx}}{2}$

It follows that

$EI\frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} = \frac{{ - Fx}}{2}$

and the integration procedures lead to

$\begin{array}{l} EI\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = - \frac{{F{x^2}}}{4} + {C_1}\\ {\rm{at}}\;x = \frac{L}{2},\;\;\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 0\quad {\rm{thus}}\;{C_1} = \frac{{F{L^2}}}{{16}} \end{array}$

$\begin{array}{l} EIy = \frac{{ - F{x^3}}}{{12}} - \frac{{F{L^2}x}}{{16}} + {C_2}\\ {\rm{at}}\;x = 0,\;y = 0\quad {\rm{thus}}\;{C_2} = 0 \end{array}$

so the equation for the deflection is

$y = \frac{{Fx}}{{48EI}}(3{L^2} - 4{x^2})$

and deflection of the centre of the beam is given by

$\delta = \frac{{F{L^3}}}{{48EI}}$