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Strain energy due to bending and deflection calculated using Calculus

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Strain Energy due to Bending.

Consider a short length of beam $\displaystyle \delta x$ under the action of a Bending Moment M. If f is the Bending Stress on an element of the cross section of area $\displaystyle \delta A$ at a distance y from the Neutral Axis, then the Strain energy of the length $\displaystyle \delta x$ is given by:-

$\delta U = \int \left(\frac{f^2}{2E} \right)\times volume$

(1)

$= \delta x\int f^2\times\frac{dA}{2E}$

(2)

$=\left(\frac{f^2}{2E} \right)\int M^2\,y^2\;\frac{dA}{I^2}$

(3)

$But\;\;\;\;\;\;\;\int y^2\times dA = I$

(4)

$\therefore\;\;\;\;\;\delta U = \left(\frac{M^2}{2EI} \right)dx$

(5)

For the whole beam

$\mathbf{U = \int\frac{ M^2}{2EI}\;dx}$

(6)

The product EI is called the flexural Rigidity of the beam

Deflection by Calculus

In "Bending Stress" equation (3) it the general equation on bending was written. From this it can be seen that:-

$\frac{M}{E\;I}=\frac{1}{R}$

(7)

And that in terms of the co-ordinates x and y

$\frac{1}{R}=\frac{\pm D^2y/dx^2}{\left[1+(dy/dx)^2 \right]}$

(8)

The sign depends upon the convention for axes. For beams met with in normal engineering practice the slope $\displaystyle \frac{dy}{dx}$ is everywhere very small and may be neglected in comparison to 1 in the denominator.

Taking y as positive upwards, under the action of a positive Bending Moment, the curvature of the beam is shown in the diagram. It can be seen that dy/dx is increasing as x increases. $\displaystyle i.e.\;\;\;\;\frac{d^2y}{dx^2}\;is\;positive\;and\;\therefore\frac{1}{R} = \frac{d^2y}{dx^2}$

Hence

$\frac{M}{E\;I} = \frac{d^2y}{dx^2}$

(9)

$M = \mathbf{{E\;I}\;\frac{d^2y}{dx^2}}$

(10)

Thus provided that M can be expressed as a function of x equation(43) can be integrated to give the slope dy/dx and the deflection y can be found for any value of x. Two constants of integration will be involved and these can be found by substituting known values of slope or deflection at particular points. A mathematical expression is thus obtained for the form of the deflected beam. ( Also known as The Elastic Line

Notes on Application

Take the X axis through the level of the supports.
Take the origin at one end of the beam or at a point of zero slope.
For built in or fixed end beams or when the deflection is a maximum. the slope dy/dx=0
For points on the X axis( Usually the supports) the deflection y = 0

For those working in Imperial Units.

It is convenient to use the following units

E in lb./sq.in. ( or tons/sq.in.
I in $\displaystyle in.^4$
y in in.
M in lb.ft (or tons-ft.)
x in ft.

After the integration, one side of the Equation $\displaystyle E\;I\;\frac{dy}{dx}$ has units $\displaystyle lb.in^2$ and the other side has units $\displaystyle lb.ft.^2$ . Hence in numerical questions, the right hand side has to be multiplied by 144. After the second integration ELy has units $\displaystyle lb.ft.^3$ and the corresponding right-hand side must be multiplied by 1728

It is possible to differentiate equation (43) and in which case:-

$E\;I\;\frac{d"y}{dx^2}=\frac{dM}{dx}=F$

(11)

$E\;I\;\frac{d^3y}{dx^3}=\frac{dF}{dx}=- w$

(12)

(See the paragraph on the relationship between F M and w in Engineering Materials Shearing Force and Bending Moment.)

These forms are of use in some cases although generally the Bending Moment relationship is the most convenient.

Example 1

Problem

A simply supported beam of length l carries a concentrated load W at distances of a and b from the two ends. Find expressions for the total strain energy of the beam and the deflection under load.

Workings

The integration for strain energy can only be applied over a length of beam for which a continuous expression for M can be obtained. This usually implies a separate integration for each section between two concentrated loads or reactions.

For the section AB.

\f[M\;=\;\left(\frac{Wb}{l} \right)x\f] \f[U_a\;=\;\int_{0}^{a}{\frac{W^2\;b^2\;x^2}{2\;l^2\;EI}\;dx}\f] \f[\therefore\;\;\;\;\;U_a\;=\;\frac{W^2\;b^2}{2\;l^2\;EI}\left[\frac{x^3}{3} \right]_0^a\f] \f[\therefore\;\;\;\;\;U_a\;=\;\frac{W^2\;a^3\;b^2}{6\,E\,I\,l^2}\f]Similarly by taking a variable X measured from C \f[\therefore\;\;\;\;\;U_b\;=\;\int_{0}^{b}{\frac{W^2\;a^2\;X^2}{2\;l^2\,E\,I}dX}\;=\;\frac{W^2\;a^2\;b^3}{6\,E\,I\,l^2}\f] Total \f[U=U_a+U_b=\left( \frac{W^2\;a^2\;b^2}{6\,E\,I\,l^2} \right)(a\;+\;b)\f] \f[\therefore\;\;\;\;\;\;U\;=\;\frac{W^2\;a^2\;b^2}{6\;E\;I\;l}\f]

But if \f $\displaystyle \delta \f$ is the deflection under the load, the strain energy must be equal to the work done by the load if it is gradually applied. \f[\frac{1}{2}W\,δ =\frac{W^2\;a^2\;b^2}{6\,E\,I\,l} \f] \f[\therefore\;\;\;\;\;δ =\frac{W\;a^2\;b^2}{3\,E\,I\,l} \f]

For a Central Load \f $\displaystyle a\;=\;b\;=\;\frac{l}{2}\f$ \f[ \therefore\;\;\;\;\;δ \;=\;\left(\frac{W}{3\,E\,I\,l} \right)\left(\frac{l^2}{4} \right)\left(\frac{l^2}{4} \right)\f] Hence \f[\mathbf{δ =\frac{W\;l^3}{48\;E\;I}}\f]