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Strain Energy

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

For the whole beam

The product EI is called the flexural Rigidity of the beam

Example 1

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.

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.

Similarly by taking a variable X measured from C Total

But if \displaystyle \delta is the deflection under the load, the strain energy must be equal to the work done by the load if it is gradually applied.

For a Central Load \displaystyle a\;=\;b\;=\;\frac{l}{2} Hence

It should be noted that this method of finding deflection is limited to cases where only one concentrated load is applied ( i.e. doing work)and then only gives the deflection under the load A. For a more general application of strain energy to deflection see Castigliano's Theorem. This can be found under Engineering Materials Curved Beams.

Example 2

Compare the strain energy of a beam, simply supported at its ends and loaded with a uniformly distributed load, with that of the same beam centrally loaded and having the same value of maximum bending moment. (U.L.)

If l is the span and EI the Flexural Rigidity, then for a uniformly distributed load w, the end reactions are \displaystyle \frac{wl}{2} and at a distance x from one end:-

Using Equation (6)

Using Equation (13) fro a central load W

The Maximum Bending Stress \displaystyle =\;\frac{\hat{M}}{Z} and for a given beam depends upon the maximum Bending Moment. ( See Engineering materials Shear Force and Bending Moments)

Equating maximum Bending Moments:-

Using Equations (22) and (23), the ratio

Using Equation (25)

Application To Impact

Example 3

A concentrated load W is gradually applied to a horizontal beam simply supported at its ends, produces a deflection y at the load point. If this falls through a distance h onto the beam find an expression for the maximum deflection produced.

In a given beam, for a load W, y = 0.2 in. and the maximum stress is 4 tons/sq.in.. Find the greatest height from which a load of 0.1 W can be dropped without exceeding the elastic limit of 18 tons/sq.in. (U.L.)

The loss of Potential Energy by the load = The gain in Strain Energy by the beam

i.e. where \displaystyle \delta is the maximum deflection caused by dropping the load W onto the beam and P is the equivalent gradually applied load which would produce the same deflection.

But a gradually applied load of W would produce a deflection of y and hence by proportion :- or

Substituting in equation(29) or

The Energy equation for dropping 0.1W is:-

But the equivalent gradually applied load and hence the deflection \displaystyle \delta 'is proportional to the maximum stress i.e.

Substituting in equation (34) Thus

Deflection By Calculus

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

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


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 or

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:-

(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 4

Obtain expressions for the maximum slope and deflection of a cantilever of length l carrying (a) a concentrate4d load W at its free end and (b) a uniformly distributed load w along its whole length.

(a) If the origin is taken through the free end and the X axis through the fixed end then at a distance x from the origin:-

And using equation (43)

Integrating:- but

Integrating again:-

At x = l y = 0

The slope and deflection at the free end ( Where they are at a maximum) are given by the values of dy/dx and y when x = 0

And the deflection \displaystyle \mathbf{=-\frac{Wl^3}{3E\,I}} (Note the negative sign indicating downwards)



When x = l dy/dx = 0

Integrating again:-

When x = l y = 0

Putting x = 0

The Maximum Deflection\displaystyle =\mathbf{\frac{w\;l^3}{6\,E\,I}}

The Maximum Deflection\displaystyle =\mathbf{\frac{w\;l^4}{8\,E\,I}}