Deflection Coefficients
This section covers deflection coefficients, including deflection due to shear and the use of graphical methods.
Contents
The Method Of Deflection Coefficients.
It can be seen that any beam of length and flexural rigidity which carries a load (no mattter how it is distributed), will have a maximum deflection of ; where is a constant which depends upon the type of loading and supports. The value of has been found for the standard cases of a cantilever and a simply supported beam (See Deflection of Beams Part 1 Example 4 and Part 3 Example 1), and the deflection in other cases may frequently be built up by superposition.Deflection is a term that is used to describe the degree to which a structural element is displaced under a load.
The Principle of Superposition:
This states that where a number of loads act together on an elastic material, the resulting strain is the sum of the individual strains caused by each load acting separately.
Two types of problems will be solved by this method:
Example:
[imperial]
Example  Example 1
Problem
A beam of uniform section and length is simply supported at its ends and by an elastic prop at the centre. If the prop deflects an amount times the load it carries and the beam carries a total uniformly distributed load show that the the load carried by the prop
If and
Find the position and value of the maximum Bending Moment.
If is the load on the prop, then its deflection is carries a uniformly distributed load of 5 tons over its length of 10 ft. The beam is supported by three vertical steel tie rods each 6 ft. long, one at each end and one in the middle, the end rods having diameters of 1 in. and the centre rod 1.25 in.
Calculate the deflection at the centre of the beam below the end points and the stress in each tie rod.
Workings
The downwards deflection due to the load only is :
The upwards deflection due to the prop only is:
By superposition, the net downwards deflection is given by:
Thus,
Substituting the numerical values given: The reaction at the end supportsis given by: And for , For a maximum And
\therefore\;\;\;\;\;\;P\left(\alpha + \frac{l^3}{6\;E\.I} \right) = \left(\frac{5}{48} \right)\left(\frac{W\;l^3}{E\,I} \right)
Substituting the numerical values given: The reaction at the end supportsis given by: And for , For a maximum And
Solution
 The deflection is
 The value of the maximum Bending Moment is
Deflection Due To Shear
It can be shown that the shear stress set up in the transverse section of a beam and the accompanying shear strain will cause a distortion of the crosssection. Since the shear stress varies from zero at the extreme fibres to a maximum at the neutral axis, cross sections can no longer remain plane after bending.
A cantilever is a beam anchored at only one end. The beam carries the load to the support where it is resisted by moment and shear stress. Cantilever construction allows for overhanging structures without external bracing.
In fact the "warping" will be of the form shown in the diagram. The lefthand view being for positive shear and the righthand for negative shear. These strains are incompatible with the theory of pure bending, but nevertheless a good approximation in deflection can be obtained by strain energy methods. It should also be noted that the shear distribution near to the application of a concentrated load must differ considerably from that given by the theory, since there can be no sudden change of shear strain from one type to the other, as would be implied for a simply supported beam with a central load.
Strain Energy due to Shear
And for the whole beam:
where is an element of crosssection and an element of the length.
The integration can only be performed for a particular crosssection over which the variation of is known and rectangular and sections will be calculated.
For A Cantilever With A Load Of W At The Free End W = F
 Thus from equation (3), If is the deflection due to shear, then
For A Cantilever With A Uniformly Distributed Load.
 A load acting on a length (situated at a distance from the fixed end) will produce a deflection due to shear at this point of . For this load alone the distortion produced is indicated in the diagram, and is uniform for the shear force over the length and zero over the rest of the beam . Hence the total deflection due to shear for all the distributed load is given by:
For A Simply Supported Beam With Central Load W
A beam is a horizontal structural element that is capable of withstanding load primarily by resisting bending. The bending force induced into the material of the beam as a result of the external loads, own weight, span and external reactions to these loads is called a bending moment.
The simplified deflection is as shown in the upper diagram and since the shearing force is constant over each half, this case is equivalent to a cantilever of length carrying an end load of .
If the load is not centrally applied but divides the length into and , then we can treat either end as a cantilever with an end load equal to the reaction on that side.
Hence,
A Simply Supported Beam With A Uniformly Distributed Load

Considering a load only at a distance from one end the deflection at the load will be:
Note this has already been proved in equation (5)
By proportion the deflection at the centre of the beam : Then the total central deflection due to shear is:
Isection
 The shear force is treated as being uniformly distributed over the web area.
Thus and and using equation (2) By similar methods to those used from a rectangular section the deflections due to shear may be obtained as follows:
 Cantilever with end load
 Cantilever with distributed load
 Simply supported beam with central load
 Simply supported beam with distributed load
The Strain Energy method known as {Castigliano's Theorem} (See Bending of Curved Bars) may be used where a number of loads exist concurrently, or to find the distributed load by imposing a concentrated load at a deflection point; the latter giving it a value of zero. i.e.
Example:
[imperial]
Example  Example 1
Problem
For a given cantilever of rectangular crosssection, length and depth show that if and are the deflections due to shear and Bending due to a concentrated load at the free end, and find the value for for steel. .
Hence find the least value of if the deflection due to shear is not to exceed 1% of the total.
Workings
It has been shown that:
For a rectangular Section:
If
Then
The least value of
Solution
 is
 The least value of is
Deflection By Graphical Method
It was shown in the pages on "Shearing force and Bending Moment" that a Funicular Polygon could be used to perform a double integration of the load curve and this would produce the Bending Moment diagram.Since , it follows that a double integration of the Bending Moment curve will
produce the Deflection Curve.
If is constant, draw the B.M. diagram and divide it into a number of strips of width. Now draw a vertical line to represent the areas and join this to a pole on the right of the line.
Proceed in the normal way to draw the funicular polygon, which will be a series of straight lines to be smoothed out into a curve.The vertical ordinates on this diagram represent deflection and it will usually be necessary to slew the diagram through an angle in order to produce a horizontal base (e.g. from a simply supported beam)
produce the Deflection Curve.
If the scales are and , then the distance is given by
If then the Deflection scale required is