Displacements and strain

Consider the displacement of a particle located by \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}(S, y)\) on the parallel curve to the position located by vector \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R^{*}}}}\left(S^{*}, y\right)\) with respect to origin O on the parallel curve in the deformed bar. Let S* denote the arc-length along the parallel curve in the deformed bar, where the \(y\)-coordinate remains unchanged in the deformation. That is, it is assumed the cross section remains rigid in its own plane. We write the position of the particle in the deformed bar with respect to its position in the undeformed bar as \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R^{*}}}} =\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}+\bar{u}\), where the displacement vector is defined by \[\begin{aligned} \label{eq7.5} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{u}}}(S, y)=u_{s}(S, y) \hat{t}(s)+u_{y}(S, y) \hat{j}(s).\end{aligned}\] The differential of \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R^{*}}}}\left(S^{*}, y\right)\) along the parallel curve in the deformed bar is given by \[\begin{aligned} \label{eq7.6} d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}^{*}=d S \hat{t}+\frac{d}{d S}\left(u_{s} \hat{t}+u_{y} j\right) d S=d S \hat{t}+\left[\frac{d}{d s}\left(u_{s} \hat{t}+u_{y} j\right)\right]\left(1+\frac{y}{R(s)}\right)^{-1} d S,\end{aligned}\] in which we use the chain rule of differentiation (i.e., \(\frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{u}}}}{d S}=\frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{u}}}}{d s} \frac{d s}{d S}\)) and eq. ([eq7.4]) to transform the derivative with respect S to the derivative with respect s. Performing the differentiations in eq. ([eq7.6]) and using the relations in eq. ([eq7.3]), we get \[\begin{aligned} \label{eq7.7} d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R}}}^{*}=\left[\hat{t}+\frac{1}{\left(1+\frac{y}{R}\right)} \frac{d}{d s}\left(u_{s} \hat{t}+u_{y} j\right)\right] d S=\left[\hat{t}+\frac{1}{\left(1+\frac{y}{R}\right)}\left[\left(\frac{d u_{s}}{d s}+\frac{u_{y}}{R}\right) \hat{t}+\left(\frac{d u_{y}}{d s}-\frac{u_{s}}{R}\right) \hat{j}\right]\right] d S.\end{aligned}\] The last result is written as \[\begin{aligned} \label{eq7.8} d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R^{*}}}}=[(1+\varepsilon) \hat{t}-\omega \hat{j}] d S,\end{aligned}\] where \[\begin{aligned} \label{eq7.9} \varepsilon=\frac{R}{R+y}\left(\frac{d u_{s}}{d s}+\frac{u_{y}}{R}\right) \quad \omega=-\left(\frac{R}{R+y}\right)\left(\frac{d u_{y}}{d s}-\frac{u_{s}}{R}\right).\end{aligned}\]

The differential arc-length along the parallel curve in the deformed bar is determined by \(\left(d S^{*}\right)^{2}=d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R^{*}}}} \bullet d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R^{*}}}}\), and the unit tangent vector to the parallel curve in the deformed bar is \(\hat{t}^{*}=\left(d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{R^{*}}}}\right) /\left(d S^{*}\right)\). The stretch ratio \(\lambda\) is defined by \(d S^{*}=\lambda d S\), and with regard to eq. ([eq7.8]) \(\lambda=\sqrt{(1+\varepsilon)^{2}+\omega^{2}}\). Consider the binomial series \(\sqrt{1+\xi}=1+\xi / 2-\xi^{2} / 8 \pm \ldots.\) For \(\xi=2 \varepsilon+\varepsilon^{2}+\omega^{2}\), the binomial series for the stretch ratio is \(\lambda=1+\varepsilon+\omega^{2} / 2+\ldots.\) Since the engineering strain is defined by \(\lambda-1=\varepsilon+\omega^{2} / 2+\ldots,\) we see that to the lowest degree in the series expansion of the engineering strain that \(\varepsilon\) is interpreted as the normal strain of a parallel curve for infinitesimal deformation. To interpret the physical meaning of \(\omega\) a graph of eq. ([eq7.8]) is shown in figure [fig7.2](a). Let \(\Omega\) denote the angle between the unit tangent vector \(\hat{t}\) of the parallel curve in the undeformed bar and the unit tangent vector \(\hat{t^{*}}\) of the parallel curve in the deformed bar. From the geometry of the triangle in figure [fig7.2](a) we see that \(\tan \Omega=\omega /(1+\varepsilon)\). Using the small angle approximation for the tangent function, and the assumption of infinitesimal strain, we find \(\Omega \cong \omega\). That is, \(\omega\) represents the rotation of the tangent line element to the parallel curve for infinitesimal deformations of the bar. Let \(\hat{j}^{*}\) denote unit normal vector to \(\hat{t^{*}}\). From figure [fig7.2](a) the unit normal vector \(\hat{j}^{*}=(\cos \Omega) \hat{j}+(\sin \Omega) \hat{t}\). For small angles \(\Omega\), the unit normal vector approximates as \[\begin{aligned} \label{eq7.10} \hat{j}{ }^{*} \cong \hat{j}+\omega \hat{t}.\end{aligned}\] The relationship between the displacement vector of the parallel curve to the displacement vector of the reference axis is based on the assumptions that plane cross sections normal to the reference axis remain normal to the reference axis in the deformed bar and that the cross section remains rigid in its own plane. These assumptions, which are the basis of classical beam theory, are depicted in figure [fig7.2](b). Thus, \[\begin{aligned} \label{eq7.11} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{u}}}(S, y)=\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{u}}}_{C}(s)+y(\hat{j} * \hat{j}),\end{aligned}\] where the displacement vector of the reference axis is defined by \[\begin{aligned} \label{eq7.12} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{u}}}_{C}(s)=w(s) \hat{t}(s)+v(s) \hat{j}(s).\end{aligned}\] Substitute eq. ([eq7.5]) for \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{u}}}(S, y)\) in eq. ([eq7.11]), followed by substitutions of eq. ([eq7.10]) and ([eq7.12]) into eq. ([eq7.11]) to find \[\begin{aligned} \label{eq7.13} u_{s}=w(s)+y \omega(s) \quad u_{y}=v(s).\end{aligned}\] Substitute the displacements of eq. ([eq7.13]) into the expression in eq. ([eq7.9]) for \(\omega\) to get \[\begin{aligned} \label{eq7.14} \omega=-\left(\frac{R}{R+y}\right)\left(\frac{d v}{d s}-\frac{w(s)+y \omega}{R}\right)=-\left(\frac{R}{R+y}\right)\left(\frac{d v}{d s}-\frac{w(s)}{R}\right)+\frac{y \omega}{R+y}.\end{aligned}\] Solve eq. ([eq7.14]) for \(\omega\) to find \[\begin{aligned} \label{eq7.15} \omega=-\left(\frac{d v}{d s}-\frac{w}{R}\right).\end{aligned}\] Finally, substitute the displacements of eq. ([eq7.13]) into the expression for \(\varepsilon\) in eq. ([eq7.9]), followed by the substitution of \(\omega\) from eq. ([eq7.15]), to get the strain-displacement relation for the curved bar as \[\begin{aligned} \label{eq7.16} \varepsilon=\frac{d w}{d s}+\frac{v}{R+y}-\frac{y}{R+y}\left[R\left(\frac{d^{2} v}{d s^{2}}\right)+\frac{w}{R}\left(\frac{d R}{d s}\right)\right].\end{aligned}\] Although the displacements ([eq7.13]) are linear in the thickness coordinate \(y\), note that the strain ([eq7.16]) is nonlinear in \(y\).

Normal stress, stress resultants, and strain energy

The material of the bar is assumed to linear elastic, isotropic, and homogeneous. Then, the normal stress \(\sigma\) acting on the cross section is related to the normal strain \(\varepsilon\) via Hooke’s law. As was discussed in article [sec3.7] we take \(\sigma=E \varepsilon\), where E is the modulus of elasticity of the material. The result for the normal stress is written as \[\begin{aligned} \label{eq7.17} \sigma=\frac{E}{R+y}\left\{\left(v+R \frac{d w}{d s}\right)+y\left[\frac{d w}{d s}-\left[R\left(\frac{d^{2} v}{d s^{2}}\right)+\frac{w}{R}\left(\frac{d R}{d s}\right)\right]\right]\right\}.\end{aligned}\] The normal force N and bending moment M about the centroidal \(x\)-axis of the cross section are related to the normal stress by \[\begin{aligned} \label{eq7.18} (N, M)=\int_{A}(1, y) \sigma d A,\end{aligned}\] where A denotes the cross-sectional area of the bar. Substitute Hooke’s law for the normal stress ([eq7.17]) into eq. ([eq7.18]) to get the relation between the resultants and the displacements. In this substitution process the following integrals over the cross section occur: \[\begin{aligned} \label{eq7.19} \int_{A} \frac{d A}{R+y}=\frac{A}{R}(1+Y) \quad \int_{A} \frac{y d A}{R+y}=-Y A \quad \int_{A} \frac{y^{2} d A}{R+y}=Y A R,\end{aligned}\]

in which Y is the dimensionless parameter of Winkler’s curved bar theory. Equivalence of the three integrals([eq7.19]) with respect to parameter Y can be demonstrated by dividing the denominator into the numerator of the integrand in the first two integrals and noting \(\int_A y d A=0\), since \(y\) is measured from the centroid. Langhaar (1962) states that the third expression in eq. ([eq7.19]) is the most convenient for the evaluation of the parameter Y if numerical integration is required. If \(R \gg y\), then the third expression in eq. ([eq7.19]) approximates Y as \[\begin{aligned} \label{eq7.20} Y \cong Y_{\mathrm{appx}}=I /\left(A R^{2}\right) \quad \text { where } \quad I=\int_{A} y^{2} d A.\end{aligned}\]

The equations for the resultants are \[\begin{gathered} \label{eq7.21} N=\frac{E A}{R}\left\{\left(v+R \frac{d w}{d s}\right)+Y\left[v+R\left[R\left(\frac{d^{2} v}{d s^{2}}\right)+\frac{w}{R}\left(\frac{d R}{d s}\right)\right]\right]\right\}, \textrm{ and} \\ M=-E A Y\left\{v+R\left[R\left(\frac{d^{2} v}{d s^{2}}\right)+\frac{w}{R}\left(\frac{d R}{d s}\right)\right]\right\}. \label{eq7.22}\end{gathered}\] From eqs. ([eq7.21]) and ([eq7.22]), the following results are obtained: \[\begin{aligned} \label{eq7.23} v+R \frac{d w}{d s}=\frac{R N+M}{E A} \quad \frac{d w}{d s}-\left[R\left(\frac{d^{2} v}{d s^{2}}\right)+\frac{w}{R}\left(\frac{d R}{d s}\right)\right]=\frac{1}{R E A}\left[R N+\left(1+\frac{1}{Y}\right) M\right].\end{aligned}\] Substitution of eq. ([eq7.23]) into the expression for the normal stress ([eq7.17]), yields the stress in terms of resultants as \[\begin{aligned} \label{eq7.24} \sigma=\frac{N}{A}+\left(1+\frac{y}{(R+y) Y}\right) \frac{M}{R A} \quad \text { or } \quad \sigma=\frac{1}{R A}\left(R N+M+\frac{y}{R+y} \frac{M}{Y}\right).\end{aligned}\] Note the normal stress at the centroid (\(y= 0\)) in pure bending (\(N=0\)) is \(\sigma=M /(R A)\). That is, the centroidal axis does not coincide with the neutral axis of the cross section for the curved bar. The neutral axis of the curved bar is located radially inward from the centroidal axis, i.e., \(y=-R Y /(1+Y)\). For pure bending of a straight bar the \(\sigma=0\) at the centroid, so the centroidal axis and neutral axis coincide.

The complementary strain energy is given by \[\begin{aligned} \label{eq7.25} U^{*}=\frac{1}{2} \iint \frac{\sigma^{2}}{E}(R+y) d \theta d A,\end{aligned}\] where the volume element is \((R+y) d \theta d A\). Substitute eq. ([eq7.24]) for the normal stress into the complementary strain energy ([eq7.25]), followed by integration over the cross-sectional area, to get \[\begin{aligned} \label{eq7.26} U^{*}=\int \frac{1}{2 E R^{2} A}\left[(R N+M)^{2}+\frac{M^{2}}{Y}\right] d s.\end{aligned}\] The final result for the complementary strain energy ([eq7.26]) was obtained using property of the centroid \(\int_{A} y d A=0\), the third integral in ([eq7.19]), and \(d s=R d \theta\).

[ex7.1]The reference axis is a segment of a circle of radius \(a\). Thus, the radius of curvature is uniform with respect to the arc-length along the reference axis, and \(R=a\). The cross section is a solid rectangle of height \(h\) and width \(b\). The bar is subject to equal and opposite moments M at each end, and no other loads. See figure [fig7.3]. Equilibrium is satisfied by the fact that the bending moment in the bar at each cross section is equal to the applied moment M, and that the circumferential force N vanishes at each cross section. The normal stress distribution through the thickness at a typical cross section is determined from eq. ([eq7.24]).

The formula for the dimensionless Winkler parameter Y is evaluated from the third expression in eq. ([eq7.19]), where the cross-sectional area \(A=bh\). That is, \[Y=\frac{1}{b h a} \int_{-h / 2}^{h / 2} \frac{y^{2}}{a+y} b d y=\frac{1}{h a} \int_{-h / 2}^{h / 2} \frac{y^{2}}{a+y} d y. \label{eq7.1.a}\tag{a}\] For the simple rectangular section the following indefinite integral is obtained from a table of integrals: \[\int \frac{y^{2}}{a+y} d y=(a+y)^{2} / 2-2 a(a+y)+a^{2} \log (a+y). \label{eq7.1.b}\tag{b}\] The natural logarithm is specified in eq. ([eq7.1.b]). After some algebra we get \[Y=\rho \log \left(\frac{2 \rho+1}{2 \rho-1}\right)-1 \quad \text { where } \quad \rho=a / h. \label{eq7.1.c}\tag{c}\] Practical geometries require \(\rho>1 / 2\). That is, \(a>h/2\). If \(a \gg y\), then \(Y_{\mathrm{appx}}=\left(\frac{\left(b h^{3}\right)}{12}\right) /\left(b h a^{2}\right)=1 /\left(12 \rho^{2}\right)\). For selected values of \(\rho\), the values of Y, \(Y_{\text {appx }}\), and percentage error between the exact and approximate values of Y are listed in table [tab7.1]. As can be seen in table [tab7.1], the parameter Y decreases rapidly with increasing radius to thickness ratios \(a / h\), and that the approximate value of Y is less than 0.15 percent of the exact value for \(a / h>10\).

From eq. ([eq7.24]) the normal stress for the curved bar is \(\sigma=\frac{M}{a A}\left[1+\frac{y}{a+y} \frac{1}{Y}\right]\). For a straight bar subject to pure bending the normal stress is given by \(\sigma=(M y) / I\), where the second area moment about the \(x\)-axis is \(I=\left(b h^{3}\right) / 12\). Define the dimensionless thickness coordinate \(\eta=(2 y) / h\), and a dimensionless normal stress by \(\bar{\sigma}=\frac{2 I \sigma}{h M}\). Then \(\bar{\sigma}=\eta\) for the straight bar, and \[\bar{\sigma}=\frac{1}{6 \rho}\left[1+\frac{\eta}{2 \rho+\eta} \frac{1}{Y}\right] \text {, for the curved bar. } \label{eq7.1.d}\tag{d}\] The distributions of the dimensionless stress through the thickness of the curved bar and the straight bar are shown in figure [fig7.4] for radius to thickness ratios of 1 and 5. The normal stress for the curved bar is hyperbolic in the thickness coordinate \(y\), whereas for the straight bar the distribution of the normal stress is linear in coordinate \(y\). For \(\textit{M} > 0\) the maximum tensile stress for the curved bar is less than that of the straight bar, but the magnitude of the maximum compressive stress is larger for the curved bar than for the straight bar. The graph for \(a / h=5\) in figure [fig7.4] shows that the stress distribution in the curved bar is close to the linear distribution of the straight bar. The percentage error of the maximum compressive stress of the straight bar with respect to the curved bar for increasing radius to thickness ratios are listed in table [tab7.2].

The error approximating the maximum stress from straight bar theory with respect to curved bar theory is 3.32 percent for \(a / h=10\), and less than 1.66 percent for \(a / h>20\).

[ex7.2]A semicircular bar of radius \(a\) is connected to a fixed pin support at its left end and a pin support free to move horizontally at its right end. The right end support is subject to a horizontal force Q with a corresponding displacement denoted by \(q\). The cross section of the bar is the thin-walled tube of example [ex6.4] on page  with the mean radius denoted by \(b\), and a wall thickness denoted by \(t\). See figure [fig7.5]. Determine the flexibility influence coefficient \(c\) in the relation \(q=c Q\) using Castigliano’s second theorem.

Solution to part (a).

The arch is symmetric about the vertical axis through the center, so only the left-half section is analyzed. The free body diagram shown in figure [fig7.8] cuts the arch at angle \(\theta\) measured counterclockwise from the vertical axis, and \(0 \leq \theta \leq \beta\). The internal actions on the cut face are the circumferential normal force N, transverse shear force V, and the bending moment M. The cut face is a negative \(\theta\)-face with a positive normal force causing circumferential extension, a positive shear force is defined radially inward, and a positive bending moment causes tension on the radially outboard circumference.

R210.9pt

Equilibrium in the direction of the normal force leads to the equation \[N=-(P / 2) \sin \theta-Q \cos \theta. \label{eq7.3.a}\tag{a}\] Equilibrium in the direction of the shear force leads to the equation \[V=(P / 2) \cos \theta-Q \sin \theta. \label{eq7.3.b}\tag{b}\] Moment equilibrium at the \(\theta\)-face leads to the equation \[M=-a(\sin \beta-\sin \theta)(P / 2)+a(\cos \theta-\cos \beta) Q. \label{eq7.3.c}\tag{c}\] The complementary strain energy from eq. ([eq7.29]) reduces to \[U^{*}=2\left(\frac{1}{2}\right) \int_{0}^{\beta}\left[\frac{N^{2}}{2 E A}+\frac{M^{2}}{2 E I}\right] a d \theta. \label{eq7.3.d}\tag{d}\] In the previous integral, integration is performed over the left-hand portion of the arch, and the total complementary energy stored in the arch is accounted for by the factor of two multiplying the integral. Since the displacement in the direction of the horizontal redundant force Q is zero, Castigliano’s second theorem gives \(0=\left(\partial U^{*}\right) /(\partial Q)\). Thus, \[0=\int_{0}^{\mathrm{\beta}}\left[\frac{N}{E A}\left(\frac{\partial N}{\partial Q}\right)+\frac{M}{E I}\left(\frac{\partial M}{\partial Q}\right)\right] a d \theta. \label{eq7.3.e}\tag{e}\] Substitute eq. ([eq7.3.a]) for N and eq. ([eq7.3.c]) for M in eq. ([eq7.3.e]) followed by integration to get \[\begin{aligned} 0=\left(\frac{a}{2 E A}\right) &\left[\sin ^{2} \beta P+(2 \beta+\sin 2 \beta) Q\right]+ \nonumber\\ &\left(\frac{a^{3}}{4 E I}\right)[(1-4 \cos \beta+3 \cos 2 \beta+2 \beta \sin 2 \beta) P+(8 \beta+4 \beta \cos 2 \beta-6 \sin 2 \beta) Q]. \label{eq7.3.f}\tag{f}\end{aligned}\] Solve eq. ([eq7.3.f]) for the redundant force Q to get \[\frac{Q}{P}=\frac{a^{2}(-1+4 \cos \beta-3 \cos 2 \beta-2 \beta \sin 2 \beta) E A-\left(2 \sin ^{2} \beta\right) E I}{a^{2}(8 \beta+4 \beta \cos 2 \beta-6 \sin 2 \beta) E A+(4 \beta+2 \sin 2 \beta) E I}. \label{eq7.3.g}\tag{g}\]

Solution to part (b).

The rise-to-span ratio in terms of the semi-opening angle \(\beta\) is \[\frac{H}{L}=\frac{a(1-\cos \beta)}{2 a \sin \beta}=\frac{(1-\cos \beta)}{2 \sin \beta}. \label{eq7.3.h}\tag{h}\] Note that this ratio is independent of the radius \(a\). The distance L between the supports is fixed at one meter, and \(L=2 a \sin \beta\). Hence, the radius of the arch is a function of the semi-opening angle \(\beta\) given by \[a=(1\,\mathrm{m}) /(2 \sin \beta). \label{eq7.3.i}\tag{i}\] Specifying \(\beta\), we compute \(a\) from eq. (i), H/L from eq. ([eq7.3.h]), and the ratio of Q/P from eq. ([eq7.3.g]). A graph of Q/P versus H/L is shown in figure [fig7.9].

The maximum value of the redundant is 2P for \(\textit{H/L} = 0.0485\) corresponding to \(\beta=11.08^{\circ}\). This maximum occurs for a shallow arch. A shallow arch is characterized by a small value of the angle \(\beta\).

[ex7.4]A uniform, thin ring of radius \(a\) is subject to equal and opposite tension forces labeled P as shown in the left sketch in figure 1.1. This configuration is symmetric about the horizontal diameter AB and the vertical diameter. Determine the bending moment distribution in the upper right-hand quarter segment of the ring where the angle \(\theta\) has the range \(0 \leq \theta \leq \pi / 2\). Include only bending deformations in Castigliano’s second theorem to determine the redundant actions.

Solution.

The right-hand sketch in figure 1.1 shows the free body diagrams obtained by sectioning the ring across diameter AB. Action-reaction pairs of the internal circumferential normal forces, internal transverse shear forces, and internal bending moments are labeled in the free body diagrams. The upper and lower free body diagrams are mirror images of one another, which implies that the transverse shear forces \(V_{A}\) in the upper and lower free body diagrams should be drawn in the same direction at the cut at A. But this mirror image argument is in contradiction with the action-reaction pairing. So to resolve this contradiction means that the transverse shear force \(V_{A}=0\). A similar argument implies that the transverse shear force at cut B must vanish (i.e., \(V_{B}=0\)).

From symmetry about the vertical diameter the circumferential normal forces \(N_{A}\) and \(N_{B}\) are equal and the bending moments \(M_{A}\) and \(M_{B}\) are equal. Vertical force equilibrium yields \(N_{A}=N_{B}=P / 2\). Thus, the only redundant is the bending moment \(M_{A}\). A free body diagram of the section from point A to the cut at \(\theta\) is shown in figure [fig7.11].

Moment equilibrium about the cut at \(\theta\) determines the bending moment as \[M=M_{A}+a(1-\cos \theta)(P / 2). \label{eq7.4.a}\tag{a}\]

The complementary strain energy including bending only is \[U^{*}=\int_{0}^{\frac{\pi}{2}}\left[\frac{M^{2}}{2 E I}\right] a d \theta. \label{eq7.4.b}\tag{b}\] By symmetry, the rotation corresponding to the moment \(M_{A}\) must vanish. Hence, Castigliano’s second theorem gives \[0=\frac{\partial U^{*}}{\partial M_{A}}=\frac{1}{E I} \int_{0}^{\frac{\pi}{2}}[M]\left[\frac{\partial M}{\partial M_{A}}\right] a d \theta=\frac{1}{E I} \int_{0}^{\frac{\pi}{2}}\left[M_{A}+a(1-\cos \theta)(P / 2)\right][1] a d \theta. \label{eq7.4.c}\tag{c}\] Perform the integration in eq. ([eq7.4.c]) to get \[0=\frac{1}{E I}\left[\frac{M_{A} \pi a}{2}-\frac{P a^{2}}{2}+\frac{1}{4} P \pi a^{2}\right]. \label{eq7.4.d}\tag{d}\] Solve eq. ([eq7.4.d]) for the redundant to find \[M_{A}=\frac{P a(2-\pi)}{2 \pi}. \label{eq7.4.e}\tag{e}\] Substitute eq. ([eq7.4.e]) into eq. ([eq7.4.a]) to get the bending moment as \[M=\frac{P a}{\pi}-\frac{1}{2} P a \cos \theta \quad 0 \leq \theta \leq \frac{\pi}{2}. \label{eq7.4.f}\tag{f}\] The bending moment is plotted in figure [fig7.12] as a radial coordinate with respect to the ring, where a positive moment is plotted outside the ring and a negative moment is plotted inside the ring.

Solution.

The circular skin-stringer fuselage is symmetric with respect to the \(x\)- and \(y\)-axes through its center, so the centroid and shear center are both located at the center of the circle. The resultant downward force acting on the frame is \(P_{y}=12 \times 86.6=1,039.2 \textrm{ lb}\). When the frame is removed from the fuselage the action of the frame on the fuselage is represented by a jump in the shear force in the fuselage. From eq. ([eq7.41]) \(\Delta V_{y}=P_{y}\). A free body diagram of the skin-stringer structure is shown in figure [fig7.20].

The expression ([eq7.45]) for the interface shear flow in this example reduces to \(\bar{q}=-F_{y}(s) P_{y}\). The contour coordinate \(s\) is related to the polar angle \(\theta\) by \(s=a \theta\), where the radius of the circular contour \(a=50 \text { in. }\), and \(0 \leq \theta<2 \pi\) radians. The contour distribution function \(F_{y}(s)\), or \(F_{y}(\theta)\), is given by eq. ([eq7.46]), which is a evaluated by computing the first area moment with respect to the \(x\)-axis of a portion of the skin-stringer fuselage from \(\theta=-\pi / 2\) to a value in the range \(-\pi / 2<\theta<3 \pi / 2\). The expression for the first area moment \(Q_{x}(\theta)\) is facilitated by accumulating the first area moments in each of the six segments of the contour.

Beginning at \(\theta=-\pi / 2\), and proceeding counterclockwise around the contour, the first area moments of the skin-stringer structure with respect to the \(x\)-axis are denoted by \(Q_{x i}(\theta)\), \(i=1,2, \ldots, 6\). The first area moments are calculated as follows: \[\begin{gathered} Q_{x 1}(\theta)=-a\left(2 A_{f}\right)+\int_{-\pi / 2}^{\theta}(a \sin \theta) t a d \theta=-a\left(2 A_{f}\right)-a^{2} t \cos \theta \quad-\pi / 2 \leq \theta<-\pi / 6, \label{eq7.6.a}\tag{a} \\ Q_{x 2}(\theta)=Q_{x 1}\left(-\frac{\pi}{6}\right)+\left(-\frac{a}{2}\right) A_{f}+\int_{-\pi / 6}^{\theta}(a \sin \theta) \operatorname{tad} \theta=-\frac{a}{2}\left(5 A_{f}+2 a t \cos \theta\right) \quad-\pi / 6 \leq \theta<\pi / 6, \label{eq7.6.b}\tag{b} \\ Q_{x 3}(\theta)=Q_{x 2}\left(\frac{\pi}{6}\right)+\left(\frac{a}{2}\right) A_{f}+\int_{\pi / 6}^{\theta}(a \sin \theta) \operatorname{ta} d \theta=-a\left(2 A_{f}+a t \cos \theta\right) \quad \pi / 6 \leq \theta<\pi / 2, \label{eq7.6.c}\tag{c} \\ Q_{x 4}(\theta)=Q_{x 3}\left(\frac{\pi}{2}\right)+a\left(2 A_{f}\right)+\int_{\pi / 2}^{\theta}(a \sin \theta) \operatorname{tad} \theta=-a^{2} t \cos \theta \quad \pi / 2 \leq \theta<5 \pi / 6, \label{eq7.6.d}\tag{d} \\ Q_{x 5}(\theta)=Q_{x 4}\left(\frac{5 \pi}{6}\right)+\left(\frac{a}{2}\right) A_{f}+\int_{5 \pi / 6}^{\theta}(a \sin \theta) \operatorname{tad} \theta=\frac{a}{2}\left(A_{f}-2 a t \cos \theta\right) \quad 5 \pi / 6 \leq \theta<7 \pi / 6, \textrm{ and}\label{eq7.6.e}\tag{e}\\ Q_{x 6}(\theta)=Q_{x 5}\left(\frac{7 \pi}{6}\right)+\left(-\frac{a}{2}\right) A_{f}+\int_{7 \pi / 6}^{\theta}(a \sin \theta) \operatorname{ta} d \theta=-a^{2} t \cos \theta \quad 7 \pi / 6 \leq \theta<3 \pi / 2. \label{eq7.6.f}\tag{f}\end{gathered}\] Note that the first area moment of the entire cross section about the \(x\)-axis vanishes, or \(Q_{x 6}(3 \pi / 2)=0\), since the \(x\)-axis passes through the centroid.

The expression for the shear flow distribution function from eq. ([eq7.46]) is written as \[F_{y i}(\theta)=\frac{1}{I_{x x}}\left[Q_{x i}(\theta)-\frac{1}{2 A_{c}} \oint\left(r_{n} Q_{x}\right) a d \theta\right] \quad i=1,2 \ldots, 6. \label{eq7.6.g}\tag{g}\] The integral of \(r_{n} Q_{x}\) around the entire contour, where \(r_{n}=a\), is given by \[\begin{gathered}\oint\left(r_{n} Q_{x}\right) a d \theta=\int_{-\pi / 2}^{-\pi / 6} a Q_{x 1}(\theta) a d \theta+\int_{-\pi / 6}^{\pi / 6} a Q_{x 2}(\theta) a d \theta+\int_{\pi / 6}^{\pi / 2} a Q_{x 3}(\theta) a d \theta+\int_{\pi / 2}^{5\pi / 6} a Q_{x 4}(\theta) a d \theta \\ + \int_{5 \pi / 6}^{7 \pi / 6} a Q_{x 5} (\theta) a d \theta+\int_{7 \pi / 6}^{3\pi / 2} a Q_{x 6}(\theta) a d \theta\end{gathered}, \label{eq7.6.h}\tag{h}\] which after integration yields \[\oint\left(r_{n} Q_{x}\right) a d \theta=-2 \pi a^{3} A_{f}. \label{eq7.6.i}\tag{i}\] The area enclosed by the contour is \(A_{c}=a^{2} \pi\). Hence, the interface shear flows are given by \[\bar{q}_{i}(\theta)=-\left(\frac{1}{I_{x x}}\right)\left[Q_{x i}(\theta)+a A_{f}\right] P_{y} \quad i=1,2 \ldots, 6. \label{eq7.6.j}\tag{j}\] The second area moment about the \(x\)-axis is given by \[I_{x x}=\int_{0}^{2 \pi}(a \sin \theta)^{2} t a d \theta+\sum_{i=1}^{6}\left(y f_{i}\right)^{2} A f_{i}=5 a^{2} A_{f}+\pi a^{3} t=7,765.49 \textrm{ in.}^{4}\label{eq7.6.k}\tag{k}\] Results for the interface shear flows with dimensional units of lb./in. are listed below: \[\begin{gathered} \bar{q}_{1}(\theta)=\left[\frac{a(A_{f}+a t \cos \theta)}{I_{x x}}\right] P_{y}=1.00367+5.01836 \cos \theta \quad-\pi / 2 \leq \theta<-\pi / 6, \label{eq7.6.l}\tag{l} \\[3pt] \bar{q}_{2}(\theta)=\left[\frac{a(3 A_{f}+2 a t \cos \theta)}{2 I_{x x}}\right] P_{y}=1.50551+5.01836 \cos \theta \quad-\pi / 6 \leq \theta<\pi / 6, \label{eq7.6.m}\tag{m} \\[3pt] \bar{q}_{3}(\theta)=\frac{a[A_{f}+a t \cos \theta]}{I_{x x}} P_{y}=1.00367+5.01836 \cos \theta \quad \pi / 6 \leq \theta<\pi / 2, \label{eq7.6.n}\tag{n} \\[3pt] \bar{q}_{4}(\theta)=\frac{-a(A_{f}-a t \cos \theta)}{I_{x x}} P_{y}=-1.00367+5.01836 \cos \theta \quad \pi / 2 \leq \theta<5 \pi / 6, \label{eq7.6.o}\tag{o} \\[3pt] \bar{q}_{5}(\theta)=\frac{-a(3 A_{f}-2 a t \cos \theta)}{2 I_{x x}} P_{y}=-1.50551+5.01836 \cos \theta \quad 5 \pi / 6 \leq \theta<7 \pi / 6, \textrm{ and}\label{eq7.6.p}\tag{p}\\[3pt] \bar{q}_{6}(\theta)=\frac{-a(A_{f}-a t \cos \theta)}{I_{x x}} P_{y}=-1.00367+5.01836 \cos \theta \quad 7 \pi / 6 \leq \theta<3 \pi / 2. \label{eq7.6.q}\tag{q}\end{gathered}\] The shear flows are plotted normal to the contour in the graph shown in figure [fig7.21].

[ex7.7]We use symmetry of the frame about the \(y\)-axis and draw a free body diagram of the right half of the frame as shown in figure [fig7.22]. Consequently, the frame has three vertical cuts: cut \(a\) is at the bottom of the frame, cut \(b\) at the top, and cut \(c\) at the center of the floor beam. The action of the left half of the frame on the right half is represented by three forces \(N_{a}\), \(N_{b}\), and \(N_{c}\), and by three moments \(M_{a}\), \(M_{b}\), and \(M_{c}\) at the location of the three cuts as shown in figure [fig7.22]. Shear forces at cuts \(a\), \(b\), and \(c\) vanish by symmetry conditions similar to what was discussed in example [ex7.4]. These six unknown actions at the cuts are related by the three independent equations of equilibrium for the overall free body diagram in figure [fig7.22]. Hence, this is a statically indeterminate structure. The redundants are determined from conditions of compatibility imposed via Castigliano’s second theorem. The solution process is divided into six steps:

1. Overall frame equilibrium

2. Solutions to the differential equilibrium equations for the frame

3. Bending moments in segments 1, 2, and 3 of the frame

4. Floor beam equilibrium

5. Equilibrium at the junction of the floor beam and frame

6. Compatibility conditions

Solution to part 1.

Overall frame equilibrium

Force equilibrium of the right half frame segment in the negative \(x\)-direction requires \[N_{a}+N_{b}+N_{c}+ \underbrace{\int_{-\pi / 2}^{-\pi / 6}\left(\bar{q}_{1}(\theta) \sin \theta\right) a d \theta+\int_{-\pi/6}^{\pi/6}\left(\bar{q}_{2}(\theta) \sin \theta\right) a d \theta+\int_{\pi/6}^{\pi/2}\left(\bar{q}_{3}(\theta) \sin \theta\right) a d \theta}_{=R_x} =0. \label{eq7.7.a}\tag{a}\] The resultant force \(R_{x}=0\) as determined from the interface shear flows given in example [ex7.6]. Hence, force equilibrium in the \(x\)-direction reduces to \[N_{a}+N_{b}+N_{c}=0. \label{eq7.7.b}\tag{b}\] Force equilibrium of the right half frame segment in the \(y\)-direction requires \[\underbrace{\int_{-\pi / 2}^{-\pi / 6}\left(\bar{q}_{1}(\theta) \cos \theta\right) a d \theta+\int_{-\pi/6}^{\pi/6}\left(\bar{q}_{2}(\theta) \cos \theta\right) a d \theta+\int_{-\pi/6}^{\pi/2}\left(\bar{q}_{3}(\theta) \cos \theta\right) a d \theta}_{=R_y}- P_y /2=0. \label{eq7.7.c}\tag{c}\] Substituting the interface shear flows from example [ex7.6] we find the force \(R_{y}=P_{y} / 2\). So force equilibrium in the \(y\)-direction is identically satisfied. Torsional equilibrium about the origin yields \[M_{z}+M_{b}+M_{c}-M_{a}+a N_{b}-(a / 2) N_{c}-a N_{a}-(L / 4)\left(P_{y} / 2\right)=0, \label{eq7.7.d}\tag{d}\] where the torque from the interface shear flows is given by \[M_{z}=\int_{-\pi / 2}^{-\pi / 6} a \bar{q}_{1}(\theta) a d \theta+\int_{-\pi / 6}^{\pi / 6} a \bar{q}_{2}(\theta) a d \theta+\int_{\pi / 6}^{\pi / 2} a \bar{q}_{3}(\theta) a d \theta=34{,}289.4 \text { lb.-in. }\label{eq7.7.e}\tag{e}\]

Solution to part 2.

Solutions to the differential equilibrium equations for the frame.

The differential equilibrium equations ([eq7.35]), ([eq7.36]), and ([eq7.40]) are solved for \(\textit{R} = a\), \(\textit{ds} = ad\theta\), \(p=0\), and \(q=\bar{q}(\theta)\). Differentiate eq. ([eq7.35]) with respect to \(\theta\), and then substitute eq. ([eq7.36]) for the derivative of the shear force to get \[\frac{d^{2} N}{d \theta^{2}}+N+a \frac{d \bar{q}}{d \theta}=0. \label{eq7.7.f}\tag{f}\] Each interface shear flow has the functional form \[\bar{q}_{i}(\theta)=A_{i}+B \cos \theta \quad i=1,2,3, \label{eq7.7.g}\tag{g}\] where \[A_{1}=A_{3}=\frac{a\left(a_{f}\right) P_{y}}{I_{x x}}=1.0037 \textrm{ lb} . / \mathrm{in} . \quad A_{2}=\frac{3}{2} A_{1}=1.50555 \textrm{ lb} / \mathrm{in} . \quad B=\frac{a^{2} t P_{y}}{I_{x x}}=5.01836 \textrm{ lb} . / \mathrm{in.}\label{eq7.7.h}\tag{h}\] Substitute eq. ([eq7.7.g]) for the shear flow in eq. ([eq7.7.f]), then the resulting differential equation is solved for \(N(\theta)\) to get \[N(\theta)=c_{1} \cos \theta+c_{2} \sin \theta-(a B / 2) \theta \cos \theta. \label{eq7.7.i}\tag{i}\] Equilibrium equations ([eq7.35]) and ([eq7.40]) determine the shear force and bending moment as \[V(\theta)=-a \bar{q}_{i}(\theta)-\frac{d N}{d \theta} \quad M(\theta)=a \int V(\theta) d \theta+c_{3}. \label{eq7.7.j}\tag{j}\] It is convenient to determine the three constants \(c_1\), \(c_2\), and \(c_3\) in terms of the forces and moment at the point \(\theta=\theta_{\alpha}\) by the following conditions \[N\left(\theta_{\alpha}\right)=N_{\alpha} \quad V\left(\theta_{\alpha}\right)=V_{\alpha} \quad M\left(\theta_{\alpha}\right)=M_{\alpha}. \label{eq7.7.k}\tag{k}\] The point \(\theta_{\alpha}\) on the contour is either the initial point or final point of the contour segment. The statically admissible solution for the normal force, shear force, and bending is \[\left[\begin{array}{@{}c@{}}N\left(\theta ; \theta_{\alpha}\right) \\V\left(\theta ; \theta_{\alpha}\right) \\M\left(\theta ; \theta_{\alpha}\right)\end{array}\right]=N_{\alpha}\left[\begin{array}{@{}c@{}}\cos \left(\theta-\theta_{\alpha}\right) \\\sin \left(\theta-\theta_{\alpha}\right) \\a\left[1-\cos \left(\theta-\theta_{\alpha}\right)\right]\end{array}\right]+V_{\alpha}\left[\begin{array}{@{}c@{}}-\sin \left(\theta-\theta_{\alpha}\right) \\\cos \left(\theta-\theta_{\alpha}\right) \\a \sin \left(\theta-\theta_{\alpha}\right)\end{array}\right]+M_{\alpha}\left[\begin{array}{@{}l@{}}0 \\0 \\1\end{array}\right]+\left[\begin{array}{@{}c@{}}N_{q}\left(\theta ; \theta_{\alpha}\right) \\V_{q}\left(\theta ; \theta_{\alpha}\right) \\M_{q}\left(\theta ; \theta_{\alpha}\right)\end{array}\right], \label{eq7.7.l}\tag{l}\]

where \[\left[\!\begin{array}{@{}c@{}}N_{q}(\theta ; \theta_{\alpha}) \\V_{q}(\theta ; \theta_{\alpha}) \\M_{q}(\theta ; \theta_{\alpha})\end{array}\!\right]=A_{i}\!\left[\begin{array}{@{}c@{}}-a \sin (\theta-\theta_{\alpha}) \\-a\left[1-\cos (\theta-\theta_{\alpha})\right] \\-a^{2}\left[\theta-\theta_{\alpha}-\sin (\theta-\theta_{\alpha})\right]\end{array}\right]-\left(\!\frac{a B}{4}\!\right)\left[\begin{array}{@{}c@{}}2(\theta-\theta_{\alpha}) \cos \theta+2 \cos \theta_{\alpha} \sin (\theta-\theta_{\alpha}) \\\cos \theta-\cos (\theta-2 \theta_{\alpha})+2(\theta-\theta_{\alpha}) \sin \theta \\a\left[-2(\theta-\theta_{\alpha}) \cos \theta+3 \sin \theta-\sin (\theta-2 \theta_{\alpha})-4 \sin \theta_{\alpha}\right]\end{array}\right]. \label{eq7.7.m}\tag{m}\]

Solution to part 3.

Bending moments in segments 1, 2, and 3 of the frame

The range for the solution interval in frame segment 1 is \(-\pi / 2 \leq \theta \leq-\pi / 6\), and we let \(\theta_{\alpha} \rightarrow-\pi / 2\), \(N_{\alpha} \rightarrow N_{a}\), \(V_{\alpha} \rightarrow 0\), and \(M_{\alpha} \rightarrow M_{a}\) in eq. ([eq7.7.i]). The numerical result for the bending moment in segment 1 is \[M_{1}(\theta)=50 N_{a}(1+\sin \theta)+M_{a}-16{,}487.8-2509.25 \theta+12,363.1 \cos \theta+6{,}273.1 \theta \cos \theta-12{,}546.3 \sin \theta. \label{eq7.7.n}\tag{n}\]

The range for the solution interval in frame segment 3 is \(\pi/6 \leq \theta \leq \pi/2\), and we let \(\theta_{\alpha} \rightarrow \pi / 2\), \(N_{\alpha} \rightarrow N_{b}\), \(V_{\alpha} \rightarrow 0\), and \(M_{\alpha} \rightarrow M_{b}\) in eq. (i). The numerical result for the bending moment in segment 3 is \[M_{3}(\theta)=50 N_{b}(1-\sin \theta)+M_{b}+16{,}487.8-2{,}509.25 \theta-12{,}363.1 \cos \theta+6{,}273.13 \theta \cos \theta-12{,}546.3 \sin \theta. \label{eq7.7.o}\tag{o}\]

The range for the solution interval in frame segment 2 is \(-\pi / 6 \leq \theta \leq \pi / 6\), and we let \(\theta_{\alpha} \rightarrow \pi / 6\), \(N_{\alpha} \rightarrow N_{2 a}\), \(V_{\alpha} \rightarrow V_{2 a}\), and \(M_{\alpha} \rightarrow M_{2 a}\) in eq. ([eq7.7.i]). At the junction of segment 2 and 3 where \(\theta=\pi / 6\), we impose continuity conditions \[N_{3}(\pi / 6)=N_{2 a} \quad V_{3}(\pi / 6)=V_{2 a} \quad M_{3}(\pi / 6)=M_{2 a}. \label{eq7.7.p}\tag{p}\] The result of imposing the previous continuity conditions is to express \(N_{2a}, V_{2a}, M_{2a}\) in terms of \(N_

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Solution to part 4.

Floor beam equilibrium

The internal shear force in the floor beam is denoted by \(V_4\) and the bending moment by \(M_4\). Equilibrium of the floor beam yields \[V_4 (x)=P_y(x/L)\quad M_4(x)=M_c+\frac{P_y}{2L}x^{2}\quad 0\leq x \leq L/2. \label{eq7.7.r}\tag{r}\] The axial force \(\textit{N}_

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Solution to part 5.

Equilibrium at the junction of the floor beam and frame

A free body diagram of the junction of the frame and the floor beam is shown in figure [fig7.23]. Force equilibrium in the tangential direction to the frame yields \[N_{2}(-\pi / 6)-N_{1}(-\pi / 6)-N_{c} \sin 30^{\circ}-\left(P_{y} / 2\right) \cos 30^{\circ}=0. \label{eq7.7.s}\tag{s}\] Equation ([eq7.7.s]) leads to \(N_{c}=-N_{a}-N_{b}\), which also satisfies eq. ([eq7.7.b]). Force equilibrium in the normal direction to the frame yields \[V_{2}(-\pi / 6)-V_{1}(-\pi / 6)-N_{c} \cos 30^{\circ}+\left(P_{y} / 2\right) \sin 30^{\circ}=0. \label{eq7.7.t}\tag{t}\]

Equation ([eq7.7.t]) leads to \(N_{a}+N_{b}+N_{c}=0\), which is satisfied by the solution for \(\textit{N}_

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Solution to part 6.

Compatibility conditions

To determine the redundants we impose compatibility by using Castigliano’s second theorem. Neglect the complementary strain energies due to the tangential force N and the shear force V, which implies that complementary strain energy in bending is the main contributor to the deflections of the frame. Assume for simplicity that the flexural stiffness EI of the frame and the floor beam are the same. Symmetry about the vertical axis of the frame requires the generalized displacements correspond to the redundants to vanish. Then, setting the displacement corresponding to redundant \(N_{\textrm{a}}\) equal to zero is imposed by \[0=\int_{-\pi / 2}^{-\pi / 6} M_{1}\left(\frac{\partial M_{1}}{\partial N_{a}}\right) a d \theta+\int_{-\pi / 6}^{\pi / 6} M_{2}\left(\frac{\partial M_{2}}{\partial N_{a}}\right) a d \theta+\int_{\pi / 6}^{\pi / 2} M_{3}\left(\frac{\partial M_{3}}{\partial N_{a}}\right) a d \theta+\int_{0}^{L /2} M_{4}\left(\frac{\partial M_{4}}{\partial N_{a}}\right) d x. \label{eq7.7.ab}\tag

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After substituting eqs. ([eq7.7.x]) to ([eq7.7.aa]) into eq. ([eq7.7.ab]) we find \[0=-2.10978 \times 10^{7}+1{,}535.46 M_{a}-1082.53 M_{b}+33{,}969.8 N_{a}-81{,}189.9 N_{b}. \label{eq7.7.ac}\tag

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[sec7.5] Bruhn, E. F. Analysis and Design of Flight Vehicle Structures. Carmel IN: Jacobs Publishing, Inc., 1973, p. A13.17., and Chapter A21

Chidamparam, P, and A. W. Leissa. “Vibrations of Planar Curved Beams, Rings, and Arches.” American Society of Mechanical Engineers, Applied Mechanics Reviews 46, no. 9 (September 1993).

Curtis, Howard. D. Fundamentals of Aircraft Structural Analysis. Jefferson City, MO: Richard D. Irwin, a Times Mirror Higher Education Group, Inc. Company, 1997, pp. 269, 270, 384–389.

Langhaar, H. L. Energy Methods in Applied Mechanics. New York: John Wiley & Sons, 1962, pp. 48–50.

Megson, T. H. G. Aircraft Structures for Engineering Students. 3d ed. London and New York: Arnold and John Wiley & Sons, Inc., 1999 Section 10.4.