the work due to external load is stored as energy in the bar
\(U=W\)
Therefore:
\(U=\int_0^y P dy\)
\(P = {\frac{\partial U}{\partial y}}\)
This equation is commonly known as “Castigliano’s 1st
theorem.” (1879)
Complementary energy
We also define the term complementary energy as:
\[\begin{equation*}
U^c_0 = \int_0^\sigma {\varepsilon}d \sigma
\end{equation*}\]\[\begin{equation*}
U^c = \int_V U^c_0 \; dV
\end{equation*}\]
It can be shown that (for our uniaxial bar)
\[\begin{equation*}
y = {\frac{\partial U^c}{\partial P}}
\end{equation*}\]
This equation is commonly known as “Castigliano’s 2nd theorem.”
\(P_i\) 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.
Example: Castigliano’s method for a cantilever beam with a tip load
\[\begin{equation*}
U^c=\int_V \int_{\sigma} {\varepsilon}\; d\sigma \; dV
\end{equation*}\]\[\begin{equation*}
U^c=\int_V \int_{\sigma} \frac{\sigma}{E} \; d\sigma \; dV
\end{equation*}\]\[\begin{equation*}
U^c=\int_V \frac{\sigma^2}{2 E} \; dV
\end{equation*}\]
Recognize the relationship between \(\sigma\) and the moment in the cross
section:
\[\begin{equation*}
\sigma = \frac{M(z) y}{I}
\end{equation*}\]
Therefore:
\[\begin{equation*}
U^c=\int_V \frac{1}{2 E} \left(\frac{M(z) y}{I}\right)^2 \; dV
\end{equation*}\]
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:
\[\begin{equation*}
U^c=\int_l \frac{M(z)^2}{2 E I^2} \int_A \left(y^2\right) \; dA \; dz
\end{equation*}\]\[\begin{equation*}
U^c=\int_l \frac{M(z)^2}{2 E I^2} \left(I\right) \; dz
\end{equation*}\]\[\begin{equation*}
U^c=\int \frac{1}{2} \frac{M(z)^2}{EI} \; dz
\end{equation*}\]
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.
\[\begin{equation*}
U^c\left(P\right)=\frac{\int_{0}^{l}{M^{2 }\left(z\right)\;dz}}{2 EI }
\end{equation*}\]
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.
\[\begin{equation*}
M\left(z\right)=\left(l-z\right) P
\end{equation*}\]
\[\begin{equation*}
U^c = \frac{\int_{0}^{l}{\left(l-z\right)^{2} P^{2} \;dz}}{2 EI }
\end{equation*}\]
\[\begin{equation*}
U^c = \frac{\int_{0}^{l}{\left(l^2-2 l z + z^2\right) P^{2} \;dz}}{2 EI }
\end{equation*}\]
\[\begin{equation*}
U^c = \frac{{\left.\left(l^2 z- l z^2 + \frac{z^3}{3} \right) P^{2} \right|_0^l}}{2 EI }
\end{equation*}\]
\[\begin{equation*}
U^c = \frac{l^3 P^2}{6 EI }
\end{equation*}\]
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.
Castigliano’s method for displacement at the mid-point load due to a tip load
The total complementary strain energy has two parts:
\[\begin{equation*}
U^c = U_1^c + U_2^c
\end{equation*}\]\[\begin{equation*}
U_1^c=\int_{0}^{\frac{l}{2}} \frac{{M_1^{2 }\left(z\right)}}{2 EI } \; dz
\end{equation*}\]\[\begin{equation*}
U_2^c=\int_{\frac{l}{2}}^{l} \frac{{M_2^{2 }\left(z\right)}}{2 EI } \; dz
\end{equation*}\]
\[\begin{equation*}
\small
U^c=\int_{0}^{\frac{l}{2}} \frac{{M_1^{2 }\left(z\right)}}{2 EI } \; dz+
\int_{\frac{l}{2}}^{l} \frac{{M_2^{2 }\left(z\right)}}{2 EI } \; dz
\end{equation*}\]
\[\begin{equation*}
\small
M_2\left(z\right)=\left(l-z\right) P
\end{equation*}\]
\[\begin{equation*}
\small
M_1\left(z\right)=\left(\frac{l}{2}-z\right) Q+\left(l-z\right) P
\end{equation*}\]
\[\begin{equation*}
\small
U^c= \int_{0}^{\frac{l}{2}} \frac{{\left(\left(\frac{l}{2}-z\right) Q+\left(l-z\right) P\right)^{2}}}{2 EI } \; dz
+\int_{\frac{l}{2}}^{l} \frac{{\left(l-z\right)^{2}\;dz} P^{2}}{2 EI }
\end{equation*}\]
\[\begin{equation*}
\small
\delta_Q=
\frac{d}{d Q} \left(\frac{\int_{0}^{\frac{l}{2}}{\left(\left(\frac{l}{2}-z\right) Q+\left(l-z\right) P\right)^{2}}}{2 EI } \; dz+\frac{\int_{\frac{l}{2}}^{l}{\left(l-z\right)^{2}\;dz} P^{2}}{2 EI }\right)
\end{equation*}\]
\[\begin{equation*}
\small
\delta_Q=\frac{d}{d Q} \left(\frac{2 l^3 Q^2}{96 EI }+\frac{5 l^3 P Q}{48 EI }\right) = \frac{2 l^{3} Q+5 l^{3} P}{48 EI }
\end{equation*}\]
Castigliano for a statically indeterminate system
(Castigliano for more complex systems)
Castigliano for statically indeterminate system
The total complementary energy is:
\[\begin{equation*}
U^c=\frac{\int_{0}^{l}{F_b^2\left(x\right)\;dx}}{2\,\mathrm{EA}}+\frac{F_s^2}{2\,k}
\end{equation*}\]
The force through the bar:
\[\begin{equation*}
\label{eq:barplusspring}
F_b\left(x\right)=P+F_s
\end{equation*}\]
Since \(P\) is the applied force, we will take derivatives with
respect to it. We must also involve the spring stiffness, thus, we
substitute \(F_s = - \delta_P k\):
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.
Virtual Work
Recall our definition of a potential function:
Potential (\(V\))–a function for relating a force-like quantity to a
gradient.
\[\begin{equation*}
F_i = -{\frac{\partial V}{\partial i}}
\end{equation*}\]
Therefore: define \(V\) as a potential function of the external forces
(acting through external displacements).
\[\begin{equation*}
\delta V = - \delta W_e
\end{equation*}\](Think of \(\delta\) as an arbitrary increment.)
Call \(U\) a potential function of the internal forces (acting through
internal displacements):
\[\begin{equation*}
\delta U = - \delta W_i
\end{equation*}\]
We can now examine a “work” equation such that:
\[\begin{equation*}
\delta W = + \delta W_e +\delta W_i =0
\end{equation*}\]
We can refer to this equation as “Virtual Work”.
Note that this means:
\[\begin{equation*}
\delta U + \delta V=0
\end{equation*}\]\[\begin{equation*}
\delta U = \delta W_e
\end{equation*}\]
Definitions
\(\delta W ( = \Delta W)\) “Virtual work” - The “work” done by a force
(moment) due to a “virtual displacement”.
\(\delta u ( = \Delta u)\) “Virtual displacement” - A kinematically
admissible displacement
A displacement that does not violate the kinematic boundary
conditions
The displacement is arbitrary
It need not be related to the actual displacement that results
from the applied loads.
\(\delta F ( = \Delta F)\) “Virtual force”
Any force field that is zero on the boundary
It must not do work on the boundary
The force is arbitrary
It need not be related to the actual force that results from the
displacements.
Could also be a virtual stress
“The principle of virtual work”
A system is in static equilibrium if (and only if) the virtual
work is zero for all independent virtual displacements.
\(\delta W=0\)
Rigid body example
Note: since we are considering this as a rigid body and don’t want to
track strain energy (spring energy), we can assume that the spring
applies an external force based on the spring force formula.
Step 1: Describe the current position of the forces:
\[\begin{equation*}
\begin{split}
y_P\left(\theta\right) &= \, +\frac{l\,\sin \theta}{2} \\
y_k\left(\theta\right) &= \, -\frac{l\,\sin \theta}{2} \\
y_Q\left(\theta\right) &= \, +\frac{l\,\sin \theta}{2} \\
\end{split}
\end{equation*}\]
Step 2: Take the variation. (You are assuming that a small virtual
displacement can be added to the equilibrium displacement. How does
the position change assuming a small change in theta? This is very
similar to taking a derivative.)
\[\begin{equation*}
\begin{split}
\delta y_P &= \, +\frac{l\,\cos \theta\,\delta \theta}{2} \\
\delta y_k &= \, -\frac{l\,\cos \theta\,\delta \theta}{2} \\
\delta y_Q &= \, +\frac{l\,\cos \theta\,\delta \theta}{2} \\
\end{split}
\end{equation*}\]
Step 3: Write the virtual work, ie the forces times the virtual
displacements.
\[\begin{equation*}
\delta W_e=\delta y_Q\,F_Q+\delta y_P\,F_P+\delta y_k\,F_k
\end{equation*}\]\[\begin{equation*}
\delta W_e=\frac{l\,F_Q\,\cos \theta\,\delta
\theta}{2}+\frac{l\,F_P\,\cos \theta\,\delta
\theta}{2}-\frac{l\,F_k\,\cos \theta\,\delta \theta}{2}
\end{equation*}\]
Plug these forces into the virtual work equation:
\[\begin{equation*}
\frac{l\,F_Q\,\cos \theta\,\delta \theta}{2}+\frac{l\,F_P\,\cos
\theta\,\delta \theta}{2}-\frac{l\,F_k\,\cos \theta\,\delta \theta}{2}=0
\end{equation*}\]
Since we know that \(\delta \theta\) is arbitrary:
\[\begin{equation*}
\frac{l\,\cos \theta\,\left(Q-P-\frac{l\,k\,\sin
\theta}{2}\right)}{2}=0
\end{equation*}\]
Now linearize the problem with the small angle assumptions:
\[\begin{align}
\sin \theta =& \; \theta \\
\cos \theta =& \; 1\\
\end{align}\]
Since \(\delta q_i\) are virtual displacements, or arbitrary
kinematically admissible displacements:
\[\begin{equation*}
{\frac{\partial \Pi}{\partial q_i}} \delta q_i = 0
\end{equation*}\]
Of all the kinematically admissible displacements, those corresponding
to a stable equilibrium position make \(\Pi\) a minimum. (Each and every
\(q_i\) must minimize the energy.)
