# Castigliano's theorem for deflection

Consider a prismatic bar fixed at one end and acted upon by a force P at the other. According to Castigliano's theorem we can determine the deflection of point A by partially differentiating the total strain energy of the bar with respect to the load P.

If I apply the theorem for the point B, then I would obtain the same result as obtained for free end at A, however the fixed end doesn't undergo any deflection. Why this contradiction? What am I missing?

It is because of the boundary condition - point B is fixed against deformation, so if you apply the load at point B, it results in zero strain, thus zero energy.

To my understanding for the Castigliano theorem you are calculating the strain energy for each point based on a reference. There tables that can help out to make life easier which explain how each different load affects the strain rate (See below )

Figure: Energy and deflection equation (source: msu.edu)

So for a simple axial case with a single material E, and cross-section A, you would use

$$U = \frac{P^2L}{2 E A}$$

I.e. The strain energy

• for point A, considers the entire length of the bar (from B to A).
• for point B, (from B to B) a zero length (L=0) is considered which means zero strain energy.

For an arbitrary point C (somewhere between B and A), you will also calculate a different energy $$U_C = \frac{P^2}{2 E A}L_C$$

Therefore each point is depended on the position and therefore it will have a different displacement.

• Pardon me but doesn't the statement of castigliano's theorem is that the total strain energy of the member when differentiated partially with respect to the applied load gives the deflection of point of application of load in the direction of load? As written in one of the books that I'm referring, we can make use of a fictitious load to calculate the deflection of C. I would calculate the total strain energy of bar by considering a 'dummy' load acting on C and then differentiate wrt that dummy load, and then put that dummy load = 0, to get the deflection of C. Commented Dec 22, 2021 at 20:15
• I think you are not considering the strain energy properly. The idea is that you create a section at C (Between B and A) , and if you want to find out the displacement of C, then you consider the strain energy in the part BC. (Keep in mind that in this simple case, the axial force at C is still P.) Otherwise (if you consider the strain energy in the entire beam) then all points would have the same displacement.
– NMech
Commented Dec 23, 2021 at 11:36

U in your equation, $$\ x_{b}=\frac{dU}{dP}$$ is the deflection at B which is zero..

Because B is fixed the deflection zero and dU is zero. So $$x_{b}=\frac{0}{dP}=0$$