Hooke’s law and its corollaries

Consider a body, or structure, supported so that rigid body motion is impossible. If it subject to a force, say by hanging a weight on it, then by Newton’s law of action-reaction the body must resist the force by producing an equal and opposite force. The manner by which the body produces this reactive force is by deforming. That is, the body changes shape under the action of a mechanical force and it is the change in shape that enables it to supply the reactive force. If the force is removed and the body returns to its original shape, then the body iselastic.

R113pt

Consider the action of force \(Q_{1}\) at point 1 on the body, and the action of force \(Q_{2}\) at point 2 on the body shown in figure [fig5.1]. Let the forces be fixed in direction and in point of application. Let the displacements at points 1 and 2 be denoted by \(q_{1}\) and \(q_{2}\), respectively, being measured with respect to a rectangular Cartesian reference frame. Define the displacements \(q_{1}\) and \(q_{2}\) at the points of application to be in the direction of the forces \(Q_{1}\) and \(Q_{2}\), respectively. Displacements \(q_{i}\) and forces \(Q_{i}\), \(i=1,2\), are said to correspond; they are defined at the same point and in the same direction.

For a linear elastic body Hooke’s law governs the response (Robert Hooke, 1635–1703). If only force \(Q_1\) is applied, Hooke’s law is \[\begin{gathered} \label{eq5.1} \begin{array}{@{}ll@{}} q_{1}=c_{11} Q_{1} & \text { for } Q_{2}=0 \\[4pt] q_{2}=c_{21} Q_{1} & \text { for } Q_{2}=0 \end{array}.\end{gathered}\] If force only force \(Q_2\) is applied, Hooke’s law is \[\begin{gathered} \label{eq5.2} \begin{array}{@{}ll@{}} q_{1}=c_{12} Q_{2} & \text { for } Q_{1}=0 \\[4pt] q_{2}=c_{22} Q_{2} & \text { for } Q_{1}=0 \end{array}.\end{gathered}\] The coefficients \(c_{11}\), \(c_{12}\), \(c_{21}\), and \(c_{22}\) are called flexibility influence coefficients, and they depend on the points of application and the direction of the corresponding forces and displacements, and the size, shape, and the material of the body.

Note that the application of the force \(Q_1\) results in a displacement at point 2, and force \(Q_2\) results in a displacement at point 1. Under mechanical load Hooke recognized that the material from which the body is made deforms internally throughout its extent. We now know the scale of deformation is to the level of the distortion of interatomic bonds constituting the material. At the atomic scale the material is not continuous. However, for length scales greater than that of interatomic distances the atomic structure of the body is ignored and the body is idealized as a continuum. Points within the body are identified with the material particles, and continuity is defined in the mathematical sense. Neighboring points remain neighbors under any loading condition.

If both forces \(Q_1\) and \(Q_2\) act on the body, a questions that arises: Is Hooke’s law given by eq. ([eq5.3]) below? \[\begin{gathered} \label{eq5.3} \begin{array}{@{}l@{}} q_{1}=c_{11} Q_{1}+c_{12} Q_{2} \\[4pt] q_{2}=c_{21} Q_{1}+c_{22} Q_{2} \end{array}.\end{gathered}\] From the hypothesis that the body returns to its original shape after the forces are removed it is proved that eq. ([eq5.3]) for two loads is the correct form of Hooke’s law. The proof is given by Fung (1965, p. 3). Also, coefficients \(c_{11}\) and \(c_{21}\) are independent of force \(Q_2\), and coefficients \(c_{12}\) and \(c_{22}\) are independent of force \(Q_1\). This proof leads to the principle of superposition.

Principle of superposition

For a linear elastic body the effects caused by two or more loads are the sum of the loads applied separately.

• The deformations are small, and

• the order of loading is unimportant.

Callstack:
    at (Under_Construction/Aerospace_Structures_(Johnson)/05:_Work_and_energy_methods), /content/body/div/div[1]/div[2]/p[2]/span[14]/span, line 1, column 1

Extensions of Hooke’s law to include a couple and rotation

Hooke’s law, eq. ([eq5.3]), can be extended to include the moment of a couple acting on the body and the rotation of the arm connecting the couple.

R120pt

As shown in figure [fig5.5], the forces \(Q_{3}^{\prime}\) and \(Q_{4}^{\prime}\) form a couple with an arm of length \(a\) if \(Q_{3}^{\prime}=P\) and \(Q_{4}^{\prime}=P\). That is, forces \(Q_{3}^{\prime}\) and \(Q_{4}^{\prime}\) are functions of the force \(P\). Take the partial derivative of the work done by the forces with respect to force P and use the chain rule to get \[\begin{aligned} \frac{\partial W}{\partial P}=\frac{\partial W}{\partial Q_{3}^{\prime}} \frac{\partial Q_{3}^{\prime}}{\partial P}+\frac{\partial W}{\partial Q_{4}^{\prime}} \frac{\partial Q_{4}^{\prime}}{\partial P}.\end{aligned}\] Let \(q_{3}^{\prime}\) denote the displacement corresponding to force \(Q_{3}^{\prime}\), and let \(q_{4}^{\prime}\) denote the displacement corresponding to force \(Q_{4}^{\prime}\). Then, with reference to eq. ([eq5.8]) \(q_{3}^{\prime}=\frac{\partial W}{\partial Q_{3}^{\prime}}\), \(q_{4}^{\prime}=\frac{\partial W}{\partial Q_{4}^{\prime}}\), and note that \(\frac{\partial Q_{3}^{\prime}}{\partial P}=\frac{\partial Q_{4}^{\prime}}{\partial P}=1\). So \[\begin{aligned} \frac{\partial W}{\partial P}=q_{3}^{\prime}+q_{4}^{\prime}.\end{aligned}\]

R152pt

For small displacements, \(q_{3}^{\prime}+q_{4}^{\prime}=a q\), where \(q\) is the small rotation of the moment arm in radians, as is shown in figure [fig5.6]. Thus, \(\frac{\partial W}{\partial p}=a q\). Divide this last equation by the length of the moment arm to get\(\frac{1}{a} \frac{\partial W}{\partial P}=q\). Lastly, the moment of a couple is \(Q=a P\), so we get \[\begin{aligned} \label{eq5.9} \frac{\partial W}{\partial Q}=q .\end{aligned}\] Thus, in Hooke’s law if \(q_{i}\) is a rotation, then corresponding “force” \(Q_{i}\) is a moment.

We can consider a concentrated couple as the limiting case of two equal and opposite forces acting in a plane at the surface of the body that approach each other, but maintain a constant moment; i.e., \[\begin{aligned} \underset{a \rightarrow 0}{\operatorname{Limit}}(a P)=Q\end{aligned}\] In this limiting process \(P \rightarrow \infty\). Then, the angle of rotation \(q\) is interpreted as the rotation of an infinitesimal line element in the plane of the couple.

Strain energy density functions

External loads imposed on a body cause it to deform. The energy stored in an elastic body due to deformation is called the strain energy, and the strain energy per unit volume is called the strain energy density. For the thin wall bar theory discussed in article [sec3.4] on page , deformation is quantified by values of the axial normal strain \(\varepsilon_{z z}\), and shear strains \(\gamma_{z s}\) and \(\gamma_{z \zeta}\). The three remaining strains \(\varepsilon_{s s}=\varepsilon_{\zeta \zeta}=\gamma_{s \zeta}=0\). In this article expressions for the strain energy density functions in terms of the non-zero strains are developed.

Strain energy for extension and bending of a thin-walled bar

Assuming that the axial normal strain is uniform through the thickness of the wall, we obtain from eq. ([eq3.30]) on page that the axial normal strain is related to the axial displacement \(w(z)\), and bending rotations \(\phi_{x}(z)\) and \(\phi_{y}(z)\), by \[\begin{aligned} \label{eq5.45} \varepsilon_{z z}=\frac{d w}{d z}+y(s) \frac{d \phi_{x}}{d z}+x(s) \frac{d \phi_{y}}{d z}.\end{aligned}\] Substitute eq. ([eq5.45]) for the normal strain into the strain energy density ([eq5.34]) to get \[\begin{aligned} \label{eq5.46} U_{0}\left(\varepsilon_{z z}\right)=\frac{E}{2}\left(\frac{d w}{d z}+y(s) \frac{d \phi_{x}}{d z}+x(s) \frac{d \phi_{y}}{d z}\right)^{2}-\beta \Delta T\left(\frac{d w}{d z}+y(s) \frac{d \phi_{x}}{d z}+x(s) \frac{d \phi_{y}}{d z}\right).\end{aligned}\] The strain energy per unit axial length is defined by \(\bar{U}=\int_{c} U_{0}\left(\varepsilon_{z z}\right) t(s) d s\). Substitute eq. ([eq5.46]) for the strain energy density into the strain energy per unit axial length, and note the geometric properties listed in eqs. ([eq3.74]) and ([eq3.77]) on page relative to the centroid, to get \[\begin{aligned} \label{eq5.47} \bar{U}=\frac{E}{2}\left[A\left(\frac{d w}{d z}\right)^{2}+I_{x x}\left(\frac{d \phi_{x}}{d z}\right)^{2}+I_{y y}\left(\frac{d \phi_{y}}{d z}\right)^{2}+2 I_{x y}\left(\frac{d \phi_{x}}{d z}\right)\left(\frac{d \phi_{y}}{d z}\right)\right]-N_{T}\left(\frac{d w}{d z}\right)-M_{xT}\left(\frac{d \phi_{x}}{d z}\right)-M_{y T}\left(\frac{d \phi_{y}}{d z}\right).\end{aligned}\] The thermal actions appearing in eq. ([eq5.47]) are given by eqs. ([eq3.75]) and ([eq3.78]) on page .

Assuming that the axial normal stress is uniform through the thickness of the wall, then the axial normal stress is given by eq. ([eq3.83]) on page Substitute eq. ([eq3.83]) for the normal stress into the expression ([eq5.41]) for the complementary strain energy density to get \[\begin{aligned} U_{0}^{*}\left(\sigma_{z z}\right)=&\left(\frac{1}{2 E}\right)\left\{\left[\frac{N+N_{T}}{A}+k \frac{\left(M_{x}+M_{x T}\right)}{I_{x x}} \bar{y}(s)+k \frac{\left(M_{y}+M_{y T}\right)}{I_{y y}} \bar{x}(s)-\beta \Delta T\right]^{2}\right.\\ &+\left.2 \beta \Delta T\left[\frac{N+N_{T}}{A}+k \frac{\left(M_{x}+M_{x T}\right)}{I_{x x}} \bar{y}(s)+k \frac{\left(M_{y}+M_{y T}\right)}{I_{y y}} \bar{x}(s)-\beta \Delta T\right]\right\},\end{aligned}\] in which the quadratic term in the temperature change of eq. ([eq5.41]) is neglected. Expand and simplify the latter expression to find \[\begin{aligned} \label{eq5.48} U_{0}^{*}\left(\sigma_{z z}\right)=&\frac{1}{2 E}\left\{\left(\frac{N+N_{T}}{A}\right)^{2}+\left[k \frac{\left(M_{x}+M_{x T}\right)}{I_{x x}} \bar{y}(s)\right]^{2}+\left[k \frac{\left(M_{y}+M_{y T}\right)}{I_{y y}} \bar{x}(s)\right]^{2}+2\left(\frac{N+N_{T}}{A}\right)\left(k \frac{M_{x}+M_{x T}}{I_{x x}} \bar{y}(s)\right)\right. \nonumber\\ &+\left.2\left(\frac{N+N_{T}}{A}\right)\left(k \frac{\left(M_{y}+M_{y T}\right)}{I_{y y}} \bar{x}(s)\right)+2\left(k \frac{M_{x}+M_{x T}}{I_{x x}} \bar{y}(s)\right)\left(k \frac{\left(M_{y}+M_{y T}\right)}{I_{y y}} \bar{x}(s)\right)\right\}.\end{aligned}\] Again, all terms in the simplification of eq. ([eq5.48]) that contain only the temperature are neglected. The complementary strain energy per unit axial length is defined by \(\bar{U}^{*}=\int_{c} U_{0}^{*}\left(\sigma_{z z}\right) t(s) d s\). Substitute eq. ([eq5.48]) for the complementary strain energy density into the complementary strain energy per unit length. In the evaluation of \(\bar{U}^{*}\) we use the definitions given by eqs. ([eq3.74]), ([eq3.81]), and ([eq3.84]) in article [sec3.7.2] on page to determine the following integrals: \[\begin{gathered} \int_{c} \bar{x}(s) t d s=Q_{y}-n_{x} Q_{x}=0 \quad \int_{c} \bar{y}(s) t d s=Q_{x}-n_{y} Q_{y}=0\mbox{, and}\label{eq5.49}\\ \int_{c} \bar{y}^{2} t d s=\frac{I_{x x}}{k} \quad \int_{c} \bar{x}^{2} t d s=\frac{I_{y y}}{k} \quad \int_{c} \bar{x} \bar{y} t d s=\frac{-I_{x y}}{k}.\label{eq5.50}\end{gathered}\] The final result for the complementary strain energy per unit length is \[\begin{aligned} \label{eq5.51} \bar{U}^{*}=\frac{1}{2 E}\left[\frac{\left(N+N_{T}\right)^{2}}{A}+k \frac{\left(M_{x}+M_{x T}\right)^{2}}{I_{x x}}+k \frac{\left(M_{y}+M_{y T}\right)^{2}}{I_{y y}}-2 k I_{x y} \frac{\left(M_{x}+M_{x T}\right)}{I_{x x}} \frac{\left(M_{y}+M_{y T}\right)}{I_{y y}}\right].\end{aligned}\]

Strain energy for shear and torsion of a thin-walled bar

Consider the strain energy densities due to shear ([eq5.44]). Integrate these strain energy densities over the cross-sectional area to get the strain energies per unit axial length.That is, \[\begin{aligned} \label{eq5.52} \bar{U}=\frac{1}{2} \int_{c}\left[\int_{-t / 2}^{t / 2} G(\gamma_{z s}^{2}+\gamma_{z \zeta}^{2}) d \zeta\right] d s \quad \bar{U}^{*}=\frac{1}{2} \int_{c}\left[\int_{-t / 2}^{t / 2} \frac{1}{G}(\sigma_{z s}^{2}+\sigma_{z \zeta}^{2}) d \zeta\right] d s.\end{aligned}\] If we substituted the shear strains \(\gamma_{z s}\) and \(\gamma_{z \zeta}\) from eq. ([eq3.31]) on page into the strain energy per unit length and performed the integration over the cross section we would get the strain energy function per unit length in the form \[\begin{aligned} \label{eq5.53} \bar{U}=\bar{U}\left[\psi_{x}, \psi_{y}, \frac{d \phi_{z}}{d z}\right].\end{aligned}\] Partial derivatives of \(\bar{U}\) with respect to the transverse shears and twist per unit length determine the material law for the transverse shear forces and torque. That is, \[\begin{aligned} \label{eq5.54} V_{x}=\frac{\partial \bar{U}}{\partial \psi_{x}} \quad V_{y}=\frac{\partial \bar{U}}{\partial \psi_{y}} \quad M_{z}=\frac{\partial \bar{U}}{\partial\left(\frac{d \phi_{z}}{d_{z}}\right)}.\end{aligned}\]

The shear stresses enter the definitions of the shear flow \(q\), twisting moment resultant \(m_{zs}\), and the transverse stress resultant \(q_z\), given by eq. ([eq3.37]) on page . For a thin, curved wall we neglect the term \(\zeta / R_{s}\) in the factor \(\left(1+\zeta / R_{s}\right)\) appearing in the integrand of eq. ([eq3.37]). It follows that the shear stresses consistent with stress resultant definitions are \[\begin{aligned} \label{eq5.55} \sigma_{z s}(s, z, \zeta)=\frac{q(s, z)}{t(s)}+\frac{12}{t^{3}(s)} m_{z s}(s, z) \zeta \quad \sigma_{z \zeta}(s, z)=q_{z}(s, z) / t(s).\end{aligned}\] Substitute eq. ([eq5.55]) for the stresses in complementary energy per unit length ([eq5.52]), followed by integration through the thickness to get \[\begin{aligned} \label{eq5.56} \bar{U}^{*}=\frac{1}{2} \int_{c} \frac{q^{2}}{G t} d s+\frac{1}{2} \int_{c} \frac{1}{G}\left[\frac{12}{t^{3}} m_{z s}^{2}+\frac{q_{z}^{2}}{t}\right] d s.\end{aligned}\]

Total strain energy expressions for a thin-walled bar

The total strain energy is obtained from strain energy per unit axial length by integration with respect to axial coordinate \(z\), \(0 \leq z \leq L\), where \(L\) is the length of the bar. The total strain energy is written as \[\begin{aligned} \label{eq5.80} U=U_{\varepsilon}+U_{\gamma},\end{aligned}\] where the strain energy obtained from axial normal strain \(\varepsilon_{z z}\) is denoted by \(U_{\varepsilon}\), and the strain energy obtained from shear strains \(\gamma_{z s}\) and \(\gamma_{z \xi}\) is denoted by \(U_{\gamma}\). From eqs. ([eq5.47]) and ([eq5.79]) these strain energies are \[\begin{aligned} \label{eq5.81} U_{\varepsilon}=\int_{0}^{L}&\left\{\frac{E}{2}\left[A\left(\frac{d w}{d z}\right)^{2}+I_{x x}\left(\frac{d \phi_{x}}{d z}\right)^{2}+I_{y y}\left(\frac{d \phi_{y}}{d z}\right)^{2}+2 I_{x y}\left(\frac{d \phi_{x}}{d z}\right)\left(\frac{d \phi_{y}}{d z}\right)\right]\right.\nonumber\\ &\qquad\left.-N_{T}\left(\frac{d w}{d z}\right)-M_{x T}\left(\frac{d \phi_{x}}{d z}\right)-M_{y T}\left(\frac{d \phi_{y}}{d z}\right)\right\} d z,\end{aligned}\] and \[\begin{aligned} \label{eq5.82} U_{\gamma}=\frac{1}{2} \int^L_0\left[s_{x x} \psi_{x}^{2}+2 s_{x y} \psi_{x} \psi_{y}+s_{y y} \psi_{y}^{2}+G J\left(\frac{d \phi_{z}}{d z}\right)^{2}\right] d z.\end{aligned}\] The stiffness coefficients in eq. ([eq5.82]) are computed from the compliance coefficients as shown in eq. ([eq5.78]).

The total complementary strain energy is written as \[\begin{aligned} \label{eq5.83} U^{*}=U_{\sigma}^{*}+U_{\tau}^{*},\end{aligned}\] where the complementary strain energy obtained from axial normal stress \(\sigma_{z z}\) is denoted by \(U_{\sigma}^{*}\), and the complementary strain energy obtained from shear stresses \(\sigma_{z s}\) and \(\sigma_{z \zeta}\) is denoted by \(U_{\tau}^{*}\). From eqs. ([eq5.51]) and ([eq5.61]) these complementary strain energies are \[\begin{aligned} \label{eq5.84} U_{\sigma}^{*}=\frac{1}{2 E} \int_{0}^{L}\left[\frac{\left(N+N_{T}\right)^{2}}{A}+k \frac{\left(M_{x}+M_{x T}\right)^{2}}{I_{x x}}+k \frac{\left(M_{y}+M_{y T}\right)^{2}}{I_{y y}}-2 k I_{x y} \frac{\left(M_{x}+M_{x T}\right)}{I_{x x}} \frac{\left(M_{y}+M_{y T}\right)}{I_{y y}}\right] d z,\end{aligned}\] and \[\begin{aligned} \label{eq5.85} U_{\tau}^{*}=\frac{1}{2} \int_{0}^{L}\left[c_{x x} V_{x}^{2}+2 c_{x y} V_{x} V_{y}+c_{y y} V_{y}^{2}+\frac{M_{z}^{2}}{G J}\right] d z.\end{aligned}\] The compliance coefficients in eq. ([eq5.85]) are determined by eq. ([eq5.62]) for the open contour and by eqs. ([eq5.65]) and ([eq5.66]) for the closed contour.

Castigliano’s first theorem

The work function can be written in terms of the generalized displacements and stiffness influence coefficients by substituting \(n=2\) in eq. ([eq5.10]), then substituting the result for the generalized forces into the work eq. ([eq5.5]). The result is \[\begin{aligned} \label{eq5.86} W=\left(\frac{1}{2}\right)\left(k_{11} q_{1}^{2}+2 k_{12} q_{1} q_{2}+k_{22} q_{2}^{2}\right).\end{aligned}\] Take the partial derivative of the work function with respect to \(q_1\) and \(q_2\) to get \[\begin{aligned} \frac{\partial W}{\partial q_{1}}=k_{11} q_{1}+k_{12} q_{2} \quad \frac{\partial W}{\partial q_{2}}=k_{12} q_{1}+k_{22} q_{2}.\end{aligned}\] Comparing these results for the partial derivatives of \(W\) to the eq. ([eq5.10]) for \(n=2\), we find \[\begin{aligned} \label{eq5.87} Q_{1}=\frac{\partial W}{\partial q_{1}} \quad Q_{2}=\frac{\partial W}{\partial q_{2}}.\end{aligned}\] That is, the partial derivative of the work function with respect to a displacement equals the corresponding force. Equation ([eq5.87]) yields the stiffness law, and it is the conjugate to eq. ([eq5.8]) that yields the compliance law.

For adiabatic deformation of a Hookean material the first law of thermodynamics shows that the work done per unit volume is equal to the strain energy density. Refer to eq. ([eq5.36]) on page . For the body composed of a Hookean material, we take the work done by the external loads equal to the strain energy of the entire body. Then \(W=U\) in eq. ([eq5.87]), which leads to Castigliano’s first theorem in terms of generalized displacements and forces as follows:

Castigliano’s first theorem

If the strain energy of an elastic structure is expressed in terms of the independent generalized displacement components \(q_{i}\), \(i=1,2, \ldots, n\), in the direction of the prescribed generalized point forces \(Q_{1}, Q_{2}, \ldots Q_{n}\), then the first partial derivative of the strain energy with respect to the displacement \(q_{i}\) is equal to the corresponding force \(Q_{i}\) or \[\begin{aligned} Q_{i}=\frac{\partial U}{\partial q_{i}} \quad i=1,2, \ldots, n\end{aligned}\]

For the thin-walled bar the strain energy \(U\) is given by eqs. ([eq5.80]) to ([eq5.82]), and it has the form \[\begin{aligned} U=\int_{0}^{L} \bar{U}(w^{\prime}, \phi_{x}^{\prime}, \phi_{y}^{\prime}, \phi_{z}^{\prime}, \psi_{x}, \psi_{y}) d z.\end{aligned}\] The prime indicates ordinary derivative with respect to coordinate \(z\) (e.g. \(w^{\prime}=\frac{d w}{d z}\)), and the averaged shears from eq. ([eq3.32]) on page are repeated below. \[\begin{aligned} \psi_{x}(z)=\frac{d u}{d z}+\phi_{y}(z) \quad \psi_{y}(z)=\frac{d v}{d z}+\phi_{x}(z).\end{aligned}\] The procedure in the application of the first theorem to structural analysis is to assume functions for the displacement components \(u(z), v(z)\), and \(w(z)\) and rotations \(\phi_{x}(z), \phi_{y}(z)\), and \(\phi_{z}(z)\) that satisfy the three conditions below.

1. The displacement and rotations must be continuous functions of the coordinate \(z\), so that their derivatives with respect to the coordinate, or strains, exist and are integrable over the domain of the bar.

2. The displacement and rotations must satisfy any prescribed conditions at the boundaries \(z = 0\) and \(z = L\).

3. The displacement and rotation functions are selected such that they equate to the generalized displacements \(q_{i}\), \(i=1,2, \ldots, n\), at their defined points of application.

Displacement and rotation functions that satisfy continuity conditions and prescribed displacement boundary conditions are said to be kinematically admissible. Kinematically admissible displacements lead to a compatible deformation. Compatibility of a deformable body means the displacements are continuous and single-valued (i.e., no gaps or overlaps of material result in the deformed state). Castigliano’s first theorem is a condition of equilibrium consistent with the assumed kinematically admissible displacements.

[ex5.1]The cantilever beam shown in figure [fig5.11] is subject to a vertical displacement \(q_1\) and a clockwise ration \(q_2\) at its tip, and a uniform thermal moment \(M_{xT}\) along its length. The cross section is symmetric with respect to the y-z plane.

Assume a kinematically admissible displacement and bending rotation as \[\begin{aligned} \label{ex5.1a} v(z)=q_{1}(z / L) \quad \phi_{x}(z)=q_{2}(z / L) \quad 0 \leq z \leq L.\end{aligned}\] Continuous and differentiable functions for the generalized displacements are ensured by employing polynomial functions in coordinate \(z\). Also, assumptions ([ex5.1a]) satisfy the prescribed end conditions \(v(0)=0\) and \(\phi_{x}(0)=0\). Therefore, assumptions ([ex5.1a]) satisfy the conditions of kinematic admissibility. The strain energy reduces to a function generalized displacements \(q_1\) and \(q_2\). For this example the expression for the strain energy from ([eq5.81]) and([eq5.82]) is \[\begin{aligned} \label{ex5.1b} U=\int_{0}^{L} \bar{U}\left[\phi_{x}^{\prime}, \psi_{y}\right] d z\mbox{ where }\bar{U}\left[\phi_{x}^{\prime}, \psi_{y}\right]=\frac{1}{2} E I_{x x}\left(\phi_{x}^{\prime}\right)^{2}-M_{x T}\left(\phi_{x}^{\prime}\right)+\frac{1}{2} s_{y y} \psi_{y}^{2}.\end{aligned}\] The derivatives of the functions in ([ex5.1a]) with respect to \(z\), and the evaluation of the transverse shear strain are \[\begin{aligned} \label{ex5.1c} v^{\prime}(z)=q_{1} / L \quad \phi_{x}^{\prime}(z)=q_{2} / L \quad \psi_{y}(z)=q_{1} / L+q_{2}(z / L).\end{aligned}\] Substitute eq. ([ex5.1c]) into eq. ([ex5.1b]) to get \[\begin{aligned} \label{ex5.1d} \bar{U}=\frac{E I_{x x}}{2 L^{2}} q_{2}^{2}+\frac{s_{y y}}{2}\left(\frac{q_{1}}{L}+\frac{q_{2}}{L} z\right)^{2}-\frac{M_{x T}}{L} q_{2}.\end{aligned}\] The definite integral of eq. ([ex5.1d]) over the length of beam yields the discrete form of the strain energy ([ex5.1b]). The result is \[\begin{aligned} \label{ex5.1e} U\left(q_{1}, q_{2}\right)=\frac{E I_{x x}}{2 L} q_{2}^{2}+\frac{s_{y y}}{2 L} q_{1}^{2}+\frac{s_{y y}}{2} q_{1} q_{2}+\frac{s_{y y} L}{6} q_{2}^{2}-M_{x T}\hspace*{1pt}q_{2}.\end{aligned}\] The strain energy expression in eq. ([ex5.1b]) is called a functional because its value is determined by the functions \(\phi_{x}(z)\) and \(\psi_{y}(z)\). Assumption ([ex5.1a]) results in a strain energy function given by eq. ([ex5.1e]) where the generalized displacements \(q_{1}\) and \(q_{2}\) are the independent variables. The generalized forces \(Q_{1}\) and \(Q_{2}\) corresponding to \(q_{1}\) and \(q_{2}\), respectively, are determined by Castigliano’s first theorem. That is, \[\begin{aligned} \label{ex5.1f} Q_{1}=\frac{\partial U}{\partial q_{1}}\mbox{ and }Q_{2}=\frac{\partial U}{\partial q_{2}}.\end{aligned}\] Take the partial derivatives of eq. ([ex5.1e]) with respect to \(q_{1}\) and \(q_{2}\) to find \[\begin{aligned} \label{ex5.1g} Q_{1}=\frac{s_{y y}}{L} q_{1}+\frac{s_{y y}}{2} q_{2}\mbox{ and }Q_{2}=\frac{s_{y y}}{2} q_{1}+\left(\frac{E I_{x x}}{L}+\frac{L s_{y y}}{3}\right) q_{2}-M_{x T}.\end{aligned}\]

The expressions in eq. ([ex5.1g]) are written in the matrix form \[\begin{aligned} \label{ex5.1h} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2}\end{array}\right]=\left[\begin{array}{@{}cc@{}}\frac{s_{y y}}{L} & \frac{s_{y y}}{2} \\\frac{s_{y y}}{2} & \frac{E I_{x x}}{L}+\frac{L s_{y y}}{3}\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2}\end{array}\right]-\left[\begin{array}{@{}l@{}}0 \\1\end{array}\right] M_{x T}.\end{aligned}\] The elements of the 2X2 stiffness matrix in eq. ([ex5.1h]) are the stiffness influence coefficients. Also note than the stiffness matrix is symmetric.

From eq. ([eq3.79]) Hooke’s law for the bending moment is \[\begin{aligned} \label{ex5.1i} M_{x}=E I_{x x} \phi_{x}^{\prime}(z)-M_{x T},\end{aligned}\] and from eq. ([eq5.77]) the material law for the transverse shear force \(i\) \[\begin{aligned} \label{ex5.1j} V_{y}=s_{y y} \psi_{y}.\end{aligned}\] Substitute \(\phi_{x}^{\prime}\) and \(\psi_{y}\) from eq. ([ex5.1c]) into eqs. ([ex5.1i]) and ([ex5.1j]) to get \[\begin{aligned} \label{ex5.1k} M_{x}=E I_{x x}\left(q_{2} / L\right)-M_{x T}\mbox{ and }V_{y}=s_{y y}\left(q_{1} / L+q_{2}(z / L)\right).\end{aligned}\] Equilibrium differential equations ([eq3.54]) and ([eq3.55]) are \[\begin{aligned} \label{ex5.1l} \frac{d V_{y}}{d z}=0\mbox{ and }\frac{d M_{x}}{d z}-V_{y}=0\mbox{ for }0<z<L.\end{aligned}\] Substitute the bending moment and shear force from eq. ([ex5.1k]) into eq. ([ex5.1l]) to get \[\begin{aligned} \label{ex5.1m} \frac{d V_{y}}{d z}=\frac{s_{y y}}{L} q_{2} \neq 0\mbox{ and }\frac{d M_{x}}{d z}-V_{y}=0-\left(\frac{s_{y y}}{L} q_{2} z+c\right) \neq 0,\end{aligned}\] where \(c\) denotes a constant of integration for the shear force. The differential equations are not satisfied for the assumption in eq. ([ex5.1a]). Of all the possible kinematically admissible displacement functions those given by ([ex5.1a]) lead to an approximate equilibrium solution but not the exact equilibrium solution.

Castigliano’s second theorem

The generalized displacements from Hooke’s law in eq. ([eq5.3]) are substituted into the work eq. ([eq5.5]) to express the work function as \[\begin{aligned} W=\left(\frac{1}{2}\right)\big(c_{11} Q_{1}^{2}+c_{12} Q_{1} Q_{2}+c_{21} Q_{2} Q_{1}+c_{22} Q_{2}^{2}\big).\end{aligned}\] Take the partial derivative of the work function with respect to \(Q_1\) and \(Q_2\) to get \[\begin{aligned} \frac{\partial W}{\partial Q_{1}}=c_{11} Q_{1}+c_{12} Q_{2} \quad \frac{\partial W}{\partial Q_{2}}=c_{12} Q_{1}+c_{22} Q_{2}.\end{aligned}\] Comparing these results for the partial derivatives of W to the two equations ([eq5.3]), we find \[\begin{aligned} \label{eq5.88} q_{1}=\frac{\partial W}{\partial Q_{1}}\mbox{ and }q_{2}=\frac{\partial W}{\partial Q_{2}}.\end{aligned}\] If the work function is written in terms of the forces and flexibility influence coefficients, then partial derivatives of the work with respect to a force equals the corresponding displacement. The work done on a Hookean body is equal to the change in energy stored due to elastic deformation. We use the complementary strain energy \(U^{*}\) in this case to represent the change in energy since \(U^{*}=0\) in the reference state. Hence, set \(W=U^{*}\), and we arrive at Castigliano’s second theorem in terms of generalized displacements and forces, which is as follows:

Castigliano’s second theorem

If an elastic structure is mounted such that rigid body displacements are impossible and certain generalized point forces \(Q_{1}, Q_{2}, \ldots, Q_{n}\) act on the structure, in addition to distributed loads and thermal strains, the displacement component \(q_{i}\), \(i=1,2, \ldots, n\) of the point of application of force \(Q_{i}\) in the direction of \(Q_{i}\) is determined by the equation \[\begin{aligned} q_{i}=\frac{\partial U^{*}}{\partial Q_{i}}.\end{aligned}\]

For a thin-walled bar the complementary strain energy given by eqs. ([eq5.83]) to ([eq5.85]) has the form \[\begin{aligned} \label{eq5.89} U^{*}=\int_{0}^{L} \bar{U}^{*}\left(V_{x}, V_{y}, N, M_{x}, M_{y}, M_{z}\right) d z.\end{aligned}\] The procedure in the application of the second theorem to the analysis of thin-walled bars is to assume functions in the axial coordinate \(z\) for the bar resultants \(V_{x}, V_{y}, N, M_{x}, M_{y}\), and \(M_{z}\), that satisfy the three conditions below.

1. The bar resultants must be represented by functions of the Cartesian coordinates that satisfy the differential equations of equilibrium of article [sec3.6.1] on page for \(0<z<L\).

2. The functions for the bar resultants must satisfy their prescribed values on the boundaries at \(z = 0\) and \(z = L\).

3. The functions for the bar resultants must contain the generalized forces \(Q_{i}\), \(i=1,2, \ldots, n\), as parameters.

Functions for the bar resultants that satisfy the differential equations of equilibrium and prescribed force boundary conditions are said to be statically admissible. Castigliano’s second theorem is the condition of compatibility for the statically admissible bar resultants. Compatibility of a deformable body means the displacements are continuous and single-valued (i.e., no gaps or overlaps of material result in the deformed state). Castigliano’s second theorem is useful in determining displacements of a structure.

[ex5.2]

R107pt

Consider the cantilever beam in example [ex5.1] on page again. Take the end loads \(Q_1\) and \(Q_2\) to be specified, and use Castigliano’s second theorem to determine the corresponding displacements \(q_1\) and \(q_2\). We note that the beam in this example is statically determinate. From equilibrium of the free body diagram shown in figure [fig5.12] the statically admissible internal actions are. \[\begin{aligned} \label{ex5.2a} V_{y}=Q_{1} \quad M_{x}=Q_{2}-Q_{1}(L-z) \quad 0 \leq z \leq L.\end{aligned}\] The complementary strain energy from ([eq5.84]) and ([eq5.85]) is \[\begin{aligned} \label{ex5.2b} U^{*}=\int_{0}^{L} \bar{U}^{*}\left[V_{y}, M_{x}\right] d z\mbox{, where }\bar{U}^{*}\left[V_{y}, M_{x}\right]=\frac{1}{2} c_{y y} V_{y}^{2}+\frac{1}{2 E I_{x x}}\left(M_{x}+M_{x T}\right)^{2}.\end{aligned}\]

Substitute the shear force and bending moment from eq. ([ex5.2a]) into the complementary strain energy per unit axial length given in eq. ([ex5.2b]) to get \[\begin{aligned} \label{ex5.2c} \bar{U}^{*}=\frac{c_{y y}}{2} Q_{1}^{2}+\frac{\left[Q_{2}-(L-z) Q_{1}+M_{x T}\right]^{2}}{2 E I_{x x}}.\end{aligned}\] Integrate eq. (c) over the length of the beam to find the complementary strain energy as \[\begin{aligned} U^{*}=\left(\frac{c_{y y} L}{2}+\frac{L^{3}}{6 E I_{x x}}\right) Q_{1}^{2}-\frac{L^{2}}{2 E I_{x x}} Q_{1} Q_{2}+\frac{L}{2 E I_{x x}} Q_{2}^{2}-\frac{L^{2}}{2 E I_{x x}} M_{x T} Q_{1} +\frac{L}{E I_{x x}} M_{x T} Q_{2}+\frac{L M_{x T}^{2}}{2 E I_{x x}}. \label{ex5.2d}\end{aligned}\] The statically admissible functions assumed in eq. ([ex5.2a]) transform the complementary strain energy functional given by eq. ([ex5.2b]) to a function of the independent variables \(Q_{1}\) and \(Q_{2}\) in eq. ([ex5.2d]). Castiglinao’s second theorem determines the generalized displacements \(q_{1}\) and \(q_{2}\) corresponding to the generalized forces \(Q_{1}\) and \(Q_{2}\). That is, \[\begin{aligned} \label{ex5.2e} q_{1}=\frac{\partial {U}^{*}}{\partial Q_{1}}\mbox{ and }q_{2}=\frac{\partial {U}^{*}}{\partial Q_{2}}.\end{aligned}\] Substitute the complementary strain energy ([ex5.2d]) into Castigliano’s second theorem ([ex5.2e]) to get \[\begin{gathered} q_{1}=\left(c_{y y} L+\frac{L^{3}}{3 E I_{x x}}\right) Q_{1}-\frac{L^{2}}{3 E I_{x x}} Q_{2}-\frac{L^{2}}{2 E I_{x x}} M_{x T}\mbox{, and}\label{ex5.2f}\\ q_{2}=-\frac{L^{2}}{E I_{x x}} Q_{1}+\frac{L}{E I_{x x}} Q_{2}+\frac{L}{E I_{x x}} M_{x T}.\label{ex5.2g}\end{gathered}\] Equations ([ex5.2f]) and ([ex5.2g]) are written in the matrix form as \[\begin{aligned} \label{ex5.2h} \left[\begin{array}{@{}l@{}}q_{1} \\q_{2}\end{array}\right]=\left[\begin{array}{@{}cc@{}}c_{y y} L+\displaystyle\frac{L^{3}}{3 E I_{x x}} & \displaystyle-\frac{L^{2}}{2 E I_{x x}} \\[8pt]\displaystyle-\frac{L^{2}}{2 E I_{x x}} & \displaystyle\frac{L}{E I_{x x}}\end{array}\right]\left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2}\end{array}\right]+\left[\begin{array}{@{}c@{}}\displaystyle-\frac{L^{2}}{2 E I_{x x}} \\[10pt] \displaystyle\frac{L}{E I_{x x}}\end{array}\right] M_{x T}.\end{aligned}\] The elements of the 2X2 compliance matrix in eq. ([ex5.2h]) are the flexibility influence coefficients. Also note that the compliance matrix is symmetric.

Allen, D. H., and Haisler, W. E. Introduction to Aerospace Structural Analysis, 2d ed. London:, John Wiley & Sons, 1985, pp. 101, 102, & 287.

Donaldson, B. K. Analysis of Aircraft Structures: An Introduction. New York: McGraw-Hill, Inc., 1993, p. 510.

Fung, Y. C. Foundations of Solid Mechanics. Englewood Cliffs, NJ: Prentice-Hall, Inc., 1965, pp. 1–7, 270–278, 348.

Lanczos, C. The Variational Principles of Mechanics. 4th ed. New York: Dover Publications, Inc., 1970, pp. 54–60, 68–73. (Originally published by the University of Toronto Press).

Langhaar, H. L. Energy Methods in Applied Mechanics. New York: John Wiley & Sons, Inc., 1962, pp. 75–89, & 120.

Malvern, L. E. Introduction to the Mechanics of a Continuous Medium. Englewood Cliffs, NJ: Prentice-Hall, Inc., 1969, p.229.