11.4 Castigliano’s Method

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11.4.1 Castigliano’s theorums

• In summary:
  • \(P_i = {\frac{\partial U}{\partial \delta_i}}\)
  • \(\delta_i = {\frac{\partial U^c}{\partial P_i}}\)
  • \(P_i\) and \(\delta_i\) represent generalized loads and displacements.
• The theorems that lead to these equations are useful for conveniently finding the deflection or reaction at a point of interest.

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11.4.2 Example: Castigliano’s method for a cantilever beam with a tip load

Cantilever beam with tip load

\[\begin{equation*} U^c=\int_V \int_{\sigma} {\varepsilon}\; d\sigma \; dV \end{equation*}\] \[\begin{equation*} U^c=\int_V \int_{\sigma} \frac{\sigma}{E} \; d\sigma \; dV \end{equation*}\] \[\begin{equation*} U^c=\int_V \frac{\sigma^2}{2 E} \; dV \end{equation*}\]

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• Recognize the relationship between \(\sigma\) and the moment in the cross section: \[\begin{equation*} \sigma = \frac{M(z) y}{I} \end{equation*}\]
• Therefore: \[\begin{equation*} U^c=\int_V \frac{1}{2 E} \left(\frac{M(z) y}{I}\right)^2 \; dV \end{equation*}\]

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• The moment is constant with respect to the cross section, therefore it can be pulled out such that \(\int_A y^2 dA\) is isolated: \[\begin{equation*} U^c=\int_l \frac{M(z)^2}{2 E I^2} \int_A \left(y^2\right) \; dA \; dz \end{equation*}\] \[\begin{equation*} U^c=\int_l \frac{M(z)^2}{2 E I^2} \left(I\right) \; dz \end{equation*}\] \[\begin{equation*} U^c=\int \frac{1}{2} \frac{M(z)^2}{EI} \; dz \end{equation*}\]

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• The above form of the equation is complementary energy for a beam of uniform cross section (in bending about one axis). You needn’t re-derive this if all the appropriate assumptions apply.
• From Castigliano’s theorem:

\[\begin{equation*} \delta_p=\frac{d}{d P} U^c\left(P\right) \end{equation*}\]

\[\begin{equation*} U^c\left(P\right)=\frac{\int_{0}^{l}{M^{2 }\left(z\right)\;dz}}{2 EI } \end{equation*}\]

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• The solution can be expressed most simply as: \[\begin{equation*} \delta_P = \frac{\partial U^C}{\partial P} = \frac{\int_{0}^{l}{M\left(z\right) \frac{\partial M}{\partial P} \;dz}}{EI} \end{equation*}\]
  • however, in this example the moment will be squared first and then subsequently a derivative will be taken.

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Cantilever beam with tip load

\[\begin{equation*} M\left(z\right)=\left(l-z\right) P \end{equation*}\]

\[\begin{equation*} U^c = \frac{\int_{0}^{l}{\left(l-z\right)^{2} P^{2} \;dz}}{2 EI } \end{equation*}\]

\[\begin{equation*} U^c = \frac{\int_{0}^{l}{\left(l^2-2 l z + z^2\right) P^{2} \;dz}}{2 EI } \end{equation*}\]

\[\begin{equation*} U^c = \frac{{\left.\left(l^2 z- l z^2 + \frac{z^3}{3} \right) P^{2} \right|_0^l}}{2 EI } \end{equation*}\]

\[\begin{equation*} U^c = \frac{l^3 P^2}{6 EI } \end{equation*}\]

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• Therefore, the displacement is: \[\begin{equation*} \delta_P=\frac{d}{d P} \left(\frac{l^3 P^2}{6 EI } \right) \end{equation*}\] \[\begin{equation*} \delta_P= \frac{l^{3} P}{3 EI } \end{equation*}\]
• This is identical to the tip displacement predicted by beam theory.

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11.4.3 Castigliano’s method for displacement at the mid-point load due to a tip load

Cantilever beam with tip and middle loads

• Recall: \[\begin{equation*} \delta_Q=\frac{d}{d Q} U^c \end{equation*}\]

• The total complementary strain energy has two parts: \[\begin{equation*} U^c = U_1^c + U_2^c \end{equation*}\] \[\begin{equation*} U_1^c=\int_{0}^{\frac{l}{2}} \frac{{M_1^{2 }\left(z\right)}}{2 EI } \; dz \end{equation*}\] \[\begin{equation*} U_2^c=\int_{\frac{l}{2}}^{l} \frac{{M_2^{2 }\left(z\right)}}{2 EI } \; dz \end{equation*}\]

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\[\begin{equation*} \small U^c=\int_{0}^{\frac{l}{2}} \frac{{M_1^{2 }\left(z\right)}}{2 EI } \; dz+ \int_{\frac{l}{2}}^{l} \frac{{M_2^{2 }\left(z\right)}}{2 EI } \; dz \end{equation*}\]

\[\begin{equation*} \small M_2\left(z\right)=\left(l-z\right) P \end{equation*}\]

\[\begin{equation*} \small M_1\left(z\right)=\left(\frac{l}{2}-z\right) Q+\left(l-z\right) P \end{equation*}\]

\[\begin{equation*} \small U^c= \int_{0}^{\frac{l}{2}} \frac{{\left(\left(\frac{l}{2}-z\right) Q+\left(l-z\right) P\right)^{2}}}{2 EI } \; dz +\int_{\frac{l}{2}}^{l} \frac{{\left(l-z\right)^{2}\;dz} P^{2}}{2 EI } \end{equation*}\]

\[\begin{equation*} \small \delta_Q= \frac{d}{d Q} \left(\frac{\int_{0}^{\frac{l}{2}}{\left(\left(\frac{l}{2}-z\right) Q+\left(l-z\right) P\right)^{2}}}{2 EI } \; dz+\frac{\int_{\frac{l}{2}}^{l}{\left(l-z\right)^{2}\;dz} P^{2}}{2 EI }\right) \end{equation*}\]

\[\begin{equation*} \small \delta_Q=\frac{d}{d Q} \left(\frac{2 l^3 Q^2}{96 EI }+\frac{5 l^3 P Q}{48 EI }\right) = \frac{2 l^{3} Q+5 l^{3} P}{48 EI } \end{equation*}\]

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11.4.4 Castigliano for a statically indeterminate system

11.4.5 (Castigliano for more complex systems)

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11.4.5.1 Castigliano for statically indeterminate system

This example comes from Prof John Shaw AE513 notes.

Bar with spring

The total complementary energy is: \[\begin{equation*} U^c=\frac{\int_{0}^{l}{F_b^2\left(x\right)\;dx}}{2\,\mathrm{EA}}+\frac{F_s^2}{2\,k} \end{equation*}\]

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• The force through the bar: \[\begin{equation*} \label{eq:barplusspring} F_b\left(x\right)=P+F_s \end{equation*}\]

• Thus: \[\begin{equation*} U^c = \frac{l\,\left(P+F_s\right)^2}{2\,\mathrm{EA}}+\frac{F_s^2}{2\,k} \end{equation*}\]

• Since \(P\) is the applied force, we will take derivatives with respect to it. We must also involve the spring stiffness, thus, we substitute \(F_s = - \delta_P k\):

\[\begin{equation*} U^c = \frac{l\,\left(P-\delta_P\,k\right)^2}{2\,\mathrm{EA}}+\frac{\delta_P^2\,k}{2} \end{equation*}\]

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• Taking the derivative: \[\begin{equation*} \delta_P=\frac{l\,\left(P-\delta_P\,k\right)}{\mathrm{EA}} \end{equation*}\]

• Solving for \(\delta_P\): \[\begin{equation*} \left[ \delta_P=\frac{l\,P}{\mathrm{EA}+k\,l} \right] \end{equation*}\]

• Note we can also take \[\begin{equation*} \frac{\partial U^c}{\partial F_s} = \frac{l\,\left(P+F_s\right)}{\mathrm{EA}}+\frac{F_s}{k} = 0 \end{equation*}\]

• This equation would recover the compatibility equation, that is, the displacement requirement which would balance the forces.