STUDY NOTE: CASTIGILANO’S METHOD (DISPLACEMENTS IN BEAMS)
Please note that this study note is a combination of Prof Tabakov’s notes and sections
taken from the prescribed book: Mechanics of Engineering Materials by PP Benham &
RJ Crawford.
Remember that the test questions will be very similar to those in this Study Note so
go through them carefully!
Method of strain energy
Consider a structural member subjected to a known tensile load P.
In determining the amount of work done by the load P, it must be understood that the
load is considered to be applied gradually and that the deflection commences as soon as
the load starts to come on the member. For a linear elastic structure, the load-deflection
diagram would be the straight line shown in the figure - at the maximum load P the
deflection is ∆.
Suppose at an intermediate stage when the load on the structure has reached a value of p,
the deflection has reached δ.
Suppose also that a further small increment of load dp causes the displacement to
increase by dδ. The work W done by the load during this further increment of load is, to a
first approximation, given by p x dδ, which will be seen to equal the shaded area in the
figure above. It follows that the load increases from zero to P, the work done by the load
is equal to the area of the triangle under the complete line, thus work done by load P
We know that
The strain energy U stored in a member subjected to an axial force P is
and it is clearly irrelevant whether P is a positive or negative value.
Displacements using the method of strain energy
If a moment M moves through a small angle dθ, the moment remaining constant during
this rotation, the work done by the moment is Md0. If a moment increases gradually from
zero to M and during this application a total rotation of θ occurs, the rotation at any stage
being proportional to the applied moment.
The work done on this section by the bending moment (internal forces) is
The following ratio is valid here
Thus, we can write
Substituting it into expression for the work done, we arrive at
The work done for the entire beam can be obtained by integration of this expression.
Obviously, the work done equals to the accumulated strain energy. Thus, the strain
energy stored in the element:
Often, both E and I are constant along the length of the beam and the expression would
become:
By expressing the strain energy U in terms of the independent displacements ∆i it can be
shown that that a partial derivative with respect to any of the displacements is equal
to the force P acting in the direction of this displacement:
The deflection ∆ is thus
This may also be expressed as
1 𝐿𝐿
𝛿𝛿𝛿𝛿
∆=
� (𝑀𝑀)( )𝑑𝑑𝑑𝑑
𝐸𝐸𝐸𝐸 0
𝛿𝛿𝛿𝛿
for deflection under the load. This is known as the modified Castigliano’s theorem.
The first example below is solved using the strain energy formula but all the rest are
solved using the modified theorem (because its slightly easier because calculating M2 is
𝛿𝛿𝛿𝛿
usually more difficult than calculating (𝑀𝑀)( 𝛿𝛿𝛿𝛿 ).
Example 1:
Find the deflection at the point of the application of the concentrated load P for the
cantilever below.
Solution:
The bending moment at a distance x from the end of the beam is M = Px
Thus
This is a standard solution.
Example 2:
Determine the displacement at the mid-span of a simply supported beam with a uniformly
distributed load (UDL) of q.
Solution:
At any distance x from the left-hand side the bending moment is M = qx/2(L-x), or
To calculate the displacement at midspan, we apply an imaginary concentrated load P at
this point. Then
and
Thus using the modified theorem, the deflection ∆ is (see red box above)
We must put P = 0 to obtain the deflection at midspan:
If we require the deflection due to the point load only, we put q = 0, then
Example 3:
A simply-supported beam carries a concentrated load P at a distance a from the
left-hand support. Determine the deflection of the beam directly underneath the load.
Solution:
We have
The support reactions are
The bending moment for the section of the beam to the LHS of P is:
𝑃𝑃𝑃𝑃
ML = 𝐿𝐿 x
The bending moment for the section of the beam to the RHS of P is:
𝑃𝑃𝑃𝑃
ML = 𝐿𝐿 x
Thus,
On LHS:
1
𝑎𝑎 𝑃𝑃𝑃𝑃
𝑏𝑏
1
𝑎𝑎 𝑃𝑃𝑃𝑃2
𝑎𝑎
1
𝑏𝑏 𝑃𝑃𝑃𝑃2
∆ = 𝐸𝐸𝐸𝐸 ∫0 ( 𝐿𝐿 𝑥𝑥)(𝐿𝐿 𝑥𝑥)𝑑𝑑𝑑𝑑 = 𝐸𝐸𝐸𝐸 ∫0
𝐿𝐿2
𝑥𝑥 2 𝑑𝑑𝑑𝑑 =
𝑃𝑃𝑏𝑏2 𝑎𝑎3
3𝐸𝐸𝐸𝐸𝐿𝐿2
On RHS:
1
𝑏𝑏 𝑃𝑃𝑃𝑃
∆ = 𝐸𝐸𝐸𝐸 ∫0 ( 𝐿𝐿 𝑥𝑥)(𝐿𝐿 𝑥𝑥)𝑑𝑑𝑑𝑑 = 𝐸𝐸𝐸𝐸 ∫0
𝐿𝐿2
𝑥𝑥 2 𝑑𝑑𝑑𝑑 =
𝑃𝑃𝑎𝑎2 𝑏𝑏3
3𝐸𝐸𝐸𝐸𝐿𝐿2
Add these deflections together and rationalise:
∆=
𝑃𝑃𝑏𝑏2 𝑎𝑎3
3𝐸𝐸𝐸𝐸𝐿𝐿2
+
𝑃𝑃𝑎𝑎2 𝑏𝑏3
3𝐸𝐸𝐸𝐸𝐿𝐿2
=
𝑃𝑃𝑏𝑏2 𝑎𝑎2
3𝐸𝐸𝐸𝐸𝐿𝐿2
𝑃𝑃𝑎𝑎2
(𝑎𝑎 + 𝑏𝑏) = 3𝐸𝐸𝐸𝐸𝐸𝐸 (𝐿𝐿 − 𝑎𝑎)2 =
𝑃𝑃𝑎𝑎2 𝑏𝑏2
3𝐸𝐸𝐸𝐸𝐸𝐸
Example 4:
A simple beam AB supports a uniform load of intensity q and a concentrated load P, as
shown below. The load P acts at the midpoint C of the beam. Determine the downward
deflection at the midpoint of the beam.
Solution:
Once again, as in example 2, because the beam and its loading are symmetrical about
the midpoint, the strain energy for the entire beam is equal to twice the strain energy
for the left-hand half of the beam. Therefore, we need to analyse only the left-hand
half of the beam.
The reaction at the left-hand support is
The bending moment is
Then
𝛿𝛿𝛿𝛿
𝛿𝛿𝛿𝛿
= x/2 and
1 𝐿𝐿/2
𝑥𝑥
∆=
� (𝑀𝑀)( )𝑑𝑑𝑑𝑑
𝐸𝐸𝐸𝐸 0
2
This answer makes perfect sense as it is the same for example 2 above (when P ≠ 0 in
that example).
Example 5.
A simple beam with an overhang supports a uniform load of intensity q on span AB
and a concentrated load P at end C of the overhang. Determine the deflection δ at C.
Solution:
The reaction at support A is
Taking moments for span AB:
where xi is measured from support A.
The bending moment in the overhang is
where x2 is measured from support C.
The partial derivates 𝛿𝛿𝛿𝛿/𝛿𝛿𝛿𝛿 are:
Using the modified theorem
By substitution we get:
After performing the integrations and combining terms we get
Example 6.
An overhanging beam ABC supports a concentrated load P
at the end of the overhang. Span AB has length L and the
overhang has length a. Determine the deflection δC at the
end of the overhang.
Solution:
The reaction at support A is
Taking moments about B, the moment to the left is:
ML = −
𝑃𝑃𝑃𝑃
𝐿𝐿
x
And to the right of B:
MR = -Px
Thus, on LHS
1
𝐿𝐿 𝑃𝑃𝑃𝑃
1
𝑎𝑎
𝑎𝑎
1
𝐿𝐿 𝑃𝑃𝑃𝑃2
∆ = 𝐸𝐸𝐸𝐸 ∫0 ( 𝐿𝐿 𝑥𝑥)(𝐿𝐿 𝑥𝑥)𝑑𝑑𝑑𝑑 = 𝐸𝐸𝐸𝐸 ∫0
On RHS
1
𝐿𝐿2
𝑎𝑎
𝑥𝑥 2 𝑑𝑑𝑑𝑑 =
∆ = 𝐸𝐸𝐸𝐸 ∫0 (𝑃𝑃𝑃𝑃)(𝑥𝑥)𝑑𝑑𝑑𝑑 = 𝐸𝐸𝐸𝐸 ∫0 𝑃𝑃𝑥𝑥 2 𝑑𝑑𝑑𝑑 =
𝑃𝑃𝑎𝑎2 𝐿𝐿
3𝐸𝐸𝐸𝐸
𝑃𝑃𝑎𝑎3
3𝐸𝐸𝐸𝐸
Adding the two deflections together and rationalising
∆=
𝑃𝑃𝑎𝑎3 𝐿𝐿
3𝐸𝐸𝐸𝐸
𝑃𝑃𝑎𝑎3
+ 3𝐸𝐸𝐸𝐸 =
𝑃𝑃𝑎𝑎2
3𝐸𝐸𝐸𝐸
(𝐿𝐿 + 𝑎𝑎)
Example 7.
The cantilever beam shown supports a triangularly distributed load of maximum
intensity q0. Determine the deflection of the beam at the free end.
Solution:
As in Example 2 we must add an imaginary load P at the free end B of the cantilever, as
that is where we need to determine the deflection.
Thus:
Taking moments about A we get
𝑀𝑀 = 𝑃𝑃𝑃𝑃 +
𝑞𝑞𝑥𝑥 3
6𝐿𝐿
𝑞𝑞𝑥𝑥3
(or 𝑀𝑀 = −𝑃𝑃𝑃𝑃 − 6𝐿𝐿 depending on what you regard as the
positive direction)
Thus,
1
𝐿𝐿
∆ = 𝐸𝐸𝐸𝐸 ∫0 (𝑃𝑃𝑃𝑃 +
𝑞𝑞𝑥𝑥 3
1
𝐿𝐿
)(𝑥𝑥)𝑑𝑑𝑑𝑑 = 𝐸𝐸𝐸𝐸 ∫0 (𝑃𝑃𝑥𝑥 2 +
6𝐿𝐿
𝑞𝑞𝑥𝑥 4
Now, setting P = 0, we get the deflection at B as
∆=
𝑞𝑞𝐿𝐿4
30𝐸𝐸𝐸𝐸
𝑃𝑃𝐿𝐿3
𝑞𝑞𝐿𝐿4
)𝑑𝑑𝑑𝑑 = 3𝐸𝐸𝐸𝐸 + 30𝐸𝐸𝐸𝐸
6𝐿𝐿