How to Solve Deflections Using the Partial Derivative Method?

Deriving Castigliano's Second Theorem

To show how the second theorem was derived, we consider a beam loaded with three different forces: \(F_1\), \(F_2\), and \(F_3\) applied at \(A\), \(B\), and \(C\) respectively. Also shown in the figure are the corresponding vertical deflections \(\Delta_{A_v}\), \(\Delta_{B_v}\) and  \(\Delta_{C_v}\). Say we're interested in finding the vertical deflection at point \(A\).

To adequately explain this method, we represent the deflection at \(A\) with a placeholder force. It is something we will use temporarily for the unknown.

Then, we need to observe the behavior of the beam under two loading sequences: (1) we apply the actual structural loads first, then the said placeholder load, and (2) we apply the placeholder load first, then apply the actual structural loads.

We assume the application of loads was gradual. In that way, the beam still behaves in the elastic range.

First Loading Sequence

Let's start with the first loading sequence. We're interested in observing the beam's total strain energy \(U\) in this sequence. 

When the actual loads act on it, there is external work \(W\). From energy conservation, the structure will store a strain energy \(U\) equal to \(W\). We can graph these using a \(P-\Delta\) diagram.

At this point, the strain energy \(U_T\) of the whole system is:

\(U_T=U\)

Let's move on and apply a small force \(dF_1\) at point \(A\). This additional load will cause an increment in external work and also strain energy \(U_{add}\). Again, its purpose is to act as a placeholder for the deflection component we want to investigate.

\(U_{add}=\Delta_{add}\times{dF_1}=\frac{\partial{U}}{\partial{F_1}}dF_1\)

In the previous expression, we expressed the additional deflection \(\Delta_{add}\) in terms of its partial derivative. Remember, deflection is dependent on both force and strain energy: \(U=F\times\Delta\) or \(\Delta=\frac{U}{F}\). In this case, we express it as dependent on the placeholder force \(dF_1\).

As a result, the total \(U_T\) in this system is now:

\(U_T=U+U_{add}\)

Equation 1: \(U_T=U+\frac{\partial{U}}{\partial{F_1}}dF_1\)

Second Loading Sequence

Let's reverse the sequence of loads seen in the following figure. In this sequence, we will investigate the external work on the beam system.

We apply the placeholder load \(dF_1\) first before \(F_1\), \(F_2\), and \(F_3\).

At the first stage, the small force will create translation \(\delta_{A_v}\). At this point, there is work \(W\) due to \(dF_1\); The total work \(W_T\) at this stage is:

\(W_T=dF_1{\delta_{A_v}}\)

Let's move on to the next stage. We place the structural forces \(F_1\), \(F_2\), and \(F_3\) on the beam. As a result, the beam's deflection will be more prevalent than when the diminutive placeholder force is only applied. It results in additional work \(W_{add}\) due to \(F_1\), \(F_2\), and \(F_3\).

The entire work of the system is now:

Equation 2: \(W_T=dF_1\delta_v+W_{add}\)

Key Idea: General Partial Derivative Equation

We're ready to move on after analyzing both sequences. One important thing to note is that we should still arrive at the same beam behavior no matter what loading sequence we use. It implies that the first loading sequence is the same as the second one. As a result, we can equate Equations 1 and 2:

\(dF_1\delta_v+W_{add}=U+\frac{\partial{U}}{\partial{F_1}}dF_1\)

Because we've applied the same actual loading conditions for both sequences, both \(W_{add}\) and \(U\) are equal; hence, we have the following:

\(dF_1\delta_v+\cancel{W_{add}}=\cancel{U}+\frac{\partial{U}}{\partial{F_1}}dF_1\)

\(\delta_v=\frac{\partial{U}}{\partial{F_1}}\)

• \(\delta\) or \(\theta\) is the deflection we want to investigate.
• \(U\) is the strain energy of the structure.
• \(F\) or \(M\) is the placeholder force or moment.

This equation is equal to the second theorem, as discussed earlier.

We will need to do two things to solve the deflection. First, introduce a placeholder force or moment representing the deflection we want to investigate. Second, we have to get the partial derivative of the strain energy to said placeholder force or moment.

Strain Energy

Like the general work methods, solving for deflection will depend on the strain energy \(U\). This value would entirely depend on the primary stress the structure is experiencing, whether flexural, axial, or torsional.

It Expands the Real Work Method

The second theorem is another approach to finding deflections using real work. What makes this different from the latter is that we can introduce a placeholder force (or moment) at the point where we would like to find the deflection. Meaning we can see the deflection at any position of the structure without relying on imaginary forces.

There is a parallelism between this method and virtual work because we have to introduce placeholders to find the deflection. The unit-load process requires you to create a virtual structure for the placeholder unit load, while this method directly places the placeholder load at the actual structure.

Summary

Let's summarize:

Castigliano's second theorem makes it possible to use real work to solve for deflections.

It states that "the partial derivative of the total strain energy U to a force (or moment) at a point is equal to the translation (or rotation) at said point in the line of direction of the force (or moment)."

This theorem is only valid if the structure is observing elastic behavior and we neglect environmental effects.

Symbolically, it is equal to \(\delta_v=\frac{\partial{U}}{\partial{F_1}}\).

We will need to do two things to solve the deflection. First, introduce a placeholder force or moment representing the deflection we want to investigate. Second, we have to get the partial derivative of the strain energy to said placeholder force or moment.

Like the general work methods, solving for deflection will depend on the strain energy \(U\).