Castigliano's Second Theorem for Beam Deflection

1. Castigliano's Second Theorem

In simple terms, Castigliano's Second Theorem states that the displacement of a structure at the point of application of a force, in the direction of that force, is equal to the partial derivative of the total strain energy with respect to that force.

Let's watch a short video that introduces and formally states the theorem.

As the video explained, the theorem gives us two key relationships for a linearly elastic structure:

1. For Deflection ($\delta$): If a force P is applied at a point, the deflection at that point in the direction of P is:
   $\delta =\frac{\partial U}{\partial P}$
2. For Rotation ($\theta$): If a moment M₀ is applied at a point, the rotation at that point is:
   $\theta =\frac{\partial U}{\partial M_0}$

Here, U is the total strain energy stored in the structure, which we know for bending is $U=\int \frac{M^2}{2EI}dx$.

2. The Practical Formula for Beam Deflection

Now, let's derive the practical formula you'll use for your exam. We start by combining Castigliano's theorem with the strain energy formula.

1. We start with the theorem:
   $\delta =\frac{\partial U}{\partial P}$

2. We substitute the formula for bending strain energy:
   $\delta =\frac{\partial }{\partial P}\left(\int \frac{M^2}{2EI}dx\right)$

3. For our purposes, we can bring the partial derivative inside the integral:
   $\delta =\int \frac{\partial }{\partial P}\left(\frac{M^2}{2EI}\right)dx$

4. Since E and I are constants with respect to P, we only need to differentiate the M² term. Using the chain rule from calculus, $\frac{\partial }{\partial P}(M^2)=2M\frac{\partial M}{\partial P}$. Substituting this in gives:
   $\delta =\int \frac{1}{2EI}\left(2M\frac{\partial M}{\partial P}\right)dx$

5. Finally, canceling the 2s gives us the practical, or "modified," version of Castigliano's theorem for beam deflection:
   $\delta =\int \frac{1}{EI}M\frac{\partial M}{\partial P}dx$

This is the integral you will be solving. To use it, you need to find two things: the moment expression M(x) and its partial derivative ∂M/∂P.

3. Worked Example: Deflection of a Cantilever Beam

Let's apply this to a classic problem: finding the deflection at the tip of a cantilever beam of length L with a point load P at the free end.

The following video segment walks through this exact problem.

Let's break down the procedure for calculating the deflection due to bending:

1. Goal: Find the vertical deflection, ${\delta }_A$, at the free end (point A). Since the force P is already applied at this point, we can use it directly.

2. Moment Expression, M(x): We make a cut at a distance x from the free end. The moment at the cut is caused by the force P acting at a lever arm of x. By standard sign convention (causing tension at the top), the moment is negative.
   $M(x)=-Px$

3. Partial Derivative, ∂M/∂P: We differentiate the moment expression with respect to P. We treat x as a constant during this differentiation.
   $\frac{\partial M}{\partial P}=\frac{\partial }{\partial P}(-Px)=-x$

4. Set up and Solve the Integral: Now we plug M and ∂M/∂P into the practical formula and integrate over the length of the beam, from x=0 to x=L.
   ${\delta }_A={\int }_0^L\frac{1}{EI}(M)\left(\frac{\partial M}{\partial P}\right)dx$
   ${\delta }_A={\int }_0^L\frac{1}{EI}(-Px)(-x)dx=\frac{P}{EI}{\int }_0^L{x}^{2}dx$
   ${\delta }_A=\frac{P}{EI}{\left[\frac{x^3}{3}\right]}_0^L=\frac{P}{EI}\left(\frac{L^3}{3}-0\right)=\frac{P{L}^3}{3EI}$
   This result, ${\delta }_A=\frac{P{L}^3}{3EI}$, is the standard formula for the deflection of a cantilever beam, which confirms our method works perfectly.

4. The "Dummy" Load Technique

What if you need to find the deflection at a point where there is no external force applied? Castigliano's theorem requires a force P to differentiate with respect to.

The solution is to use a fictitious or "dummy" force.

Here is the procedure:

1. Apply a "dummy" force, let's call it Q, at the point and in the direction where you want to find the deflection.
2. Write your moment expression(s) M(x) including this dummy force Q.
3. Find the partial derivative of M(x) with respect to Q, which is ∂M/∂Q.
4. Set up the Castigliano integral: $\delta =\int \frac{1}{EI}M\left(\frac{\partial M}{\partial Q}\right)dx$.
5. Crucial Step: In the M term inside the integral, you can now set Q = 0, since the dummy force isn't actually there. This greatly simplifies the expression before you integrate. The ∂M/∂Q term will typically be a non-zero constant or function of x.

Let's see this technique in action.

This dummy load technique is essential for your exam, as it allows you to find the deflection at any point on a beam, regardless of the loading.

Conclusion

In this lesson, you have learned the principle and practical application of Castigliano's Second Theorem. This is the theoretical engine that will drive our solutions for indeterminate structures.

Key Takeaways:

• Castigliano's Second Theorem states that the deflection at a point is the partial derivative of the strain energy with respect to the force at that point: $\delta =\partial U/\partial P$.
• For beams, this simplifies to a practical integral formula: $\delta =\int \frac{1}{EI}M\frac{\partial M}{\partial P}dx$.
• The key to solving problems is to correctly determine the bending moment expression M(x) and its partial derivative ∂M/∂P.
• If you need to find the deflection at a point with no load, you must use a "dummy" force Q and set its value to zero after setting up the integral.

We have been using the theorem to find an unknown deflection. In the next lesson, we will pivot to the main goal: using this same theorem to find an unknown force—specifically, a redundant reaction in an indeterminate beam. We will see that by setting a known deflection (like the zero deflection at a support) equal to the Castigliano integral, we can create an equation to solve for the unknown reaction.