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:

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(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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