# 7.5: Beam deflections from applied bending moments

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

he second derivative (curvature) of the neutral axis line (dotted line in diagram)

$\kappa=\frac{d^{2} y}{d x^{2}}$ The approximation involved in equating beam curvature to the curvature of the neutral axis of a beam.

It follows that

$M=\kappa E I=E I \frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}$

Since the moment at the section concerned can also be written, for a cantilever beam, as M = F (L - x)

it follows that

$E I \frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=F(L-x)$

This second order differential equation can be integrated (twice), with appropriate boundary conditions, to find the deflection of the beam at different points along its length. For a cantilever beam, this operation is shown below.

$E I \frac{\mathrm{d} y}{\mathrm{d} x}=F L x-\frac{F x^{2}}{2}+C_{1}$

$$\text { at } x=0, \frac{\mathrm{d} y}{\mathrm{d} x}=0, \text { thus } C_{1}=0$$

$E I y=\frac{F L x^{2}}{2}-\frac{F x^{3}}{6}+C_{2}$

$$\text { at } x=0, y=0, \text { thus } C_{2}=0$$

which can be rearranged to give

$y=\frac{F x^{2}}{6 E I}(3 L-x)$

For example, at the loaded end ( x = L ), this gives

$\delta=\frac{F L^{3}}{3 E I}$

Symmetrical 3-point bending The bending moment is given by

$M=\frac{-F x}{2}$

It follows that

$E I \frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=\frac{-F x}{2}$

and the integration procedures lead to

$E I \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{F x^{2}}{4}+C_{1}$

$$\text { at } x=\frac{L}{2}, \quad \frac{d y}{d x}=0 \text { thus } C_{1}=\frac{F L^{2}}{16}$$

$E I y=\frac{-F x^{3}}{12}-\frac{F L^{2} x}{16}+C_{2}$

$$\text { at } x=0, y=0 \text { thus } C_{2}=0$$

so the equation for the deflection is

$y=\frac{F x}{48 E I}\left(3 L^{2}-4 x^{2}\right)$

and deflection of the centre of the beam is given by

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