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

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Symmetrical 3-point bending

Symmetrical 3-point bend loading

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

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