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This material is for your information only and not for the exam.
1. Solving Statically Indeterminate Structures by Castigliano's Second Theorem
For simplicity, the method will be illustrated for structures with indeterminacy of
order 1. The derivation here is readily extended to structures with indeterminacy of
order . For simplicity, let us confine ourselves to a two-dimensional structure, say a
two-dimensional Bernoulli-Euler beam structure subjected to systems of concentrated
and distributed loads denoted symbolically by . Among all the reactions acting on
the structure, let us arbitrarily choose one reaction to be the redundant one, and call it
. The remaining three reactions could be expressed in terms of and
the external loads by means of the three equilibrium conditions
(1.1)
where denotes the forces in the direction, denotes the forces in the
direction, and denotes the moments about an arbitrary point .
The determination of by Castigliano's second theorem is carried out as follows.
We "release" the supports affiliated with , and regard the reaction together with
the external load as an independent set of external loads acting on the statically
determinate structure which has resulted by releasing the support affiliated with .
The deflection (or rotation) corresponding to can be computed by Castigliano’s
second theorem. Let the internal energy be denoted by ( ). The
deflection (or rotation) affiliated to is computed and then set equal to zero:
(1.2)
Equations (1.1) and (1.2) allow the determination of the four unknowns
and .
It is important to mention here that before carrying out the derivative in (1.2), the
reactions are to be determined in terms of and the other external loads
through the equilibrium equations. Thus, in evaluating (1.2), the derivatives of
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with respect to need to be taken into consideration. In other words, if
we represent the internal energy by:
with
(1.3)
we can write:
(1.4)
from which can be determined in terms of the external loads .
Example 1.1 :
Consider a beam of length L, simple supported on the its left end A and clamped
on its right end B. Let the beam be subjected to a uniform load (see Figure 1).
Find the reaction at the left hand support.
Choose the left hand reaction to be the redundant one. We will call it . The right
hand reactions at the clamped support will be called and . The force will be
chosen to be upward, and the moment is taken in the clockwise direction. From
equilibrium conditions we get:
2
(1.5)
We will find the reaction in two ways: a) working from the left support, b)
working from the right support. Let us first determine by working from the left
support. Choose to be a coordinate pointing from the left support to right. The
moment is:
(1.6)
Let us write the internal energy :
(1.7)
and implement now equation (1.4). Note that since we are working from the left, the
expression contains only and , and does not involve and . Thus
implementation of (1.4) provides:
(1.8)
This gives:
(1.9)
Now let us recover the same expression by working from right support. Let be a
coordinate running left from the clamped support. The moment is given by:
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(1.10)
where and are given in (1.5)
Thus,
(1.11)
(1.12)
with
, , , . (1.13)
Thus, one gets
(1.14)
Finally, the values of the other reactions are obtained from (1.5):
) 1.15( ,
In order to clarify further the implementation of (1.4), suppose we wish to
determine the reaction in (1.15) without having determined the reaction at the left
end. Now the vertical force at the clamped will be chosen to be the redundant one and
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will be called (the notation prime has been used to distinguish between the
reactions in the present alternative from the reactions defined before). The remaining
reactions are the moment at the right end which will be now called and the force at
the left end which will be called (see Figure 2). From equilibrium, there is:
(1.16)
Proceeding from right end we have:
(1.17)
Now we can implement equation (1.4):
(1.18)
with
(1.19)
Which leads to:
(1.20)
Carrying out the integration provides
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(1.21)
Which the same answer (for given in (1.15).
2. Finding the Deflection in a Statistically Indeterminate Beam by Using
Castigliano's Second Theorem
Let us consider the indeterminate beam (of degree 1) described above. Suppose we
want to find the deflection at a point A where there is no concentrated load acting in
the given structure. We therefore put a load at point A , which will eventually be
set to zero. This results in a new system of reactions and . All these
reactions depend on the external load on the beam and on the additionally added
load . There are two ways to proceed in order to find the deflection at point A. We
will call the first way "the long way" and the second one "the short way".
The long way necessitates to solve again the reactions in the statistically
indeterminate structure which is now under the presence of the external loads and
the additional load . Suppose the solution is indicated by :
with (2.1)
Then the internal energy will be a function of and denoted by
The deflection at point A can then be found by carrying out the derivative with
respect , and setting afterwards
(2.2)
Now it will be shown that there exists a "short way" in which solving again the
statically indeterminate structure can be avoided. Suppose that the explicit solutions
for the reactions in the form of (2.1) have not been found a priori. Even though we
will not determine now in the course of this short way, it is important to describe
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how it could be determined, in principle, in a simultaneous manner with the
deflection . To achieve that, we release the support affiliated to , and consider
the resulting statically determinate structure to be loaded now by and , and the
unknown redundant reaction . The reactions can be solved in terms of , ,
and through the use of the equilibrium equations. Let us denote the solutions for
symbolically by . Then the internal energy can be written in the
form:
with
(2.3)
The reaction can, in principle, be found by demanding that the deflection
affiliated to it be zero:
(2.4)
Equation (2.4) allows to determine , but we will not do so. We only indicate that,
in view of (2.4), will turn out to be a function of and in the form of
Substituting this form into (2.3) provides:
(2.5)
The deflection at point A is obtained by taking the derivative of the internal
energy with respect to and then setting This gives:
(2.6)
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Note that in view of (2.4) the first term in the right hand-side of (2.6) vanishes and
one is left with:
(2.7)
The derivatives on the right hand side of (2.7) can be identified as follows: the
derivative consists in taking the derivative of with respect to the specific
denoted in bold in , and then
everything is evaluated at . This means that one can define an internal energy
built on: a) the values of , b) The added load c) the new
quantities which are obtained from the equilibrium equations by using , , and
the external load . Performing the derivative with respect to gives:
(2.8)
In summary, the following strategy can be applied:
1) Arbitrarily choose the redundant reaction.
2) Apply the load at point A.
3) Using the equilibrium equations, find the values of arising from the
presence of together with the other external loads , and the value which
was present before the load was applied.
4) Finally take the derivative of the internal energy with respect to .
This means that if one knows already the reaction before the load was
applied, one does not need to resolve the statically indeterminate structure. All what
one needs to do is to find the reactions simply for the equilibrium
conditions. This certainly is a much shorter way , specially if one has an indeterminate
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structure of degree to which the present analysis can readily be extended. In the
case of a structure with indeterminacy , one needs to use the values of arbitrarily
chosen redundant reactions which present in the structure before the load was
added.
Exercise 2.1
Find the rotation at the left hand support in the indeterminate structure of
Example 1.1
Apply a clockwise moment at support A. Chose the left hand reaction to be the
redundant one, use its value , find the other reactions by loading the beam
with and the external load . The reactions are
(2.9)
Now, proceed from the left support:
(2.10)
This leads to:
(2.11)
Let us recover the same answer by proceeding from the right support. Now we have:
(2.12)
Thus
(2.13)
Which is the same answer.
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Exercise 2.2
Consider a beam of length supported at its left end A and its right end E by simple supports, and at its middle C by a roller. Let the beam be loaded uniformly by a load per unit length. Determine the deflection at point B located at distance from the left end. Suppose this indeterminate structure
has been solved and the reactions from left to right are
Let us choose the rightmost reaction at point D to be the redundant one. Thus, the reactions in the presence of will be named as follows:
At point A:
At point C:
At point E:
Use the value of , load the beam at point B, with a load , and w . The
reactions will be:
.
Define a coordinate system running towards the left from point E.
The moment is:
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These moments lead to the following integrals and deflection of the point B:
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