Solution.

The elongation of each bar as determined from eq. ([eq6.6]) is $$\Delta_{m}=\cos \left(\theta_{m}\right) q_{1}+\sin \left(\theta_{m}\right) q_{2} \quad m=1,2,3. \tag{a}$$ Castigliano’s theorem ([eq6.8]) applied to this example yields $$\begin{gathered} Q_{1}=\sum_{m\,=\,1}^3\left(\frac{E A}{L}\right)_{m}\left[\cos \left(\theta_{m}\right) q_{1}+\sin \left(\theta_{m}\right) q_{2}\right]\left[\cos \left(\theta_{m}\right)\right] \tag{b}\\ Q_{2}=\sum_{m\,=\,1}^3\left(\frac{E A}{L}\right)_{m}\left[\cos \left(\theta_{m}\right) q_{1}+\sin \left(\theta_{m}\right) q_{2}\right]\left[\sin \left(\theta_{m}\right)\right]\tag{c}\end{gathered}$$ These results are written in the matrix form $$\left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2}\end{array}\right]=\left[\begin{array}{@{}ll@{}}k_{11} & k_{12} \\k_{21} & k_{22}\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2}\end{array}\right],\tag{d}$$ where the elements of the stiffness matrix are $$k_{11}=\sum_{m\,=\,1}^{3}\left(\frac{E A}{L}\right)_{m}\cos^{2}\left(\theta_{m}\right) \quad k_{22}=\sum_{m\,=\,1}^{3}\left(\frac{E A}{L}\right)_{m} \sin^{2}\left(\theta_{m}\right) \quad k_{12}=k_{21}=\sum_{m\,=\,1}^{3}\left(\frac{E A}{L}\right)_{m} \cos \left(\theta_{m}\right) \sin \left(\theta_{m}\right).\tag{e}$$ Note that this example is statically indeterminate, since there are only two equilibrium equations at the movable joint 1 but three unknown bar forces. For specified nodal forces $Q_1$ and $Q_2$, matrix eq. (d) is solved for the nodal displacements $q_1$ and $q_2$. From eq. (a) the elongation of each bar is then computed, and from these elongations the bar forces are determined from $$\begin{aligned} N_{m}=\left(\frac{E A}{L}\right)_{m} \Delta_{m} \quad m=1,2,3.\tag{f}\end{aligned}$$

[ex6.2]Consider the same three bar-truss of example 6.1, but now assume that bar 1 was too short and had to be stretched an amount $\bar{\Delta}_{1}$ in order to connect it to joint 1. This is a case of lack of fit, and lack of fit is common in the fabrication of structures. That is, before the external loads are applied ($Q_{1}=Q_{2}=0$), the truss bars experience initial forces due to the lack of fit of bar 1. Determine the initial forces in the bars using Castigliano’s first theorem.

Solution.

Lack of fit can be included in the energy analysis by modifying the specified thermal force term in the strain energy ([eq6.7]). For uniform material properties and uniform change in temperature, the thermal force in a truss bar is $N_{T}=E A \alpha \Delta T$. (Refer to eq. ([eq3.75]) on page .) The factor $\alpha \Delta T=\varepsilon_{0}$ is the initial strain due to the temperature change. Note $\varepsilon_{0}$ is dimensionless. Now interpret $\varepsilon_{0}$ as the initial strain specified due to lack of fit. The initial strain due to the specified displacement $\bar{\Delta}$ required to connect a bar to a joint is $\varepsilon_{0}=\bar{\Delta}/ L$. Let $N_{T} \rightarrow N_{\bar{\Delta}}=E A(\bar{\Delta}/ L)$. The strain energy is modified to $$U=\sum_{m\,=\,1}^3\left\{\frac{1}{2}\left(\frac{E A}{L}\right)_{m} \Delta_{m}^{2}-N_{\bar{\Delta} m}\Delta_{m}\right\}.\tag{a}$$ The specified initial strain is only for bar 1, so $$U=\sum_{m\,=\,1}^3 \frac{1}{2}\left(\frac{E A}{L}\right)_{m} \Delta_{m}^{2}-\left(\frac{E A}{L}\right)_{1} \bar{\Delta}_{1} \Delta_{1}.\tag{b}$$ Castigliano’s theorem ([eq6.8]) leads to $$\begin{gathered} Q_{n}=\sum_{m\,=\,1}^3\left[\left(\frac{E A}{L}\right) \Delta_{m}\right] \frac{\partial \Delta_{m}}{\partial q_{n}}-\left(\frac{E A}{L}\right)_{1} \bar{\Delta}_{1}\left(\frac{\partial \Delta_{1}}{\partial q_{n}}\right) \quad n=1,2.\tag{c}\end{gathered}$$ The matrix form of eq. (c) is $$\begin{gathered} \left[\begin{array}{@{}l@{}}Q_{1} \\Q_{2}\end{array}\right]=\left[\begin{array}{@{}ll@{}}k_{11} & k_{12} \\k_{21} & k_{22}\end{array}\right]\left[\begin{array}{@{}l@{}}q_{1} \\q_{2}\end{array}\right]-\left(\frac{E A}{L}\right)_{1}\bar{\Delta}_{1}\left[\begin{array}{@{}l@{}}\cos \left(\theta_{1}\right) \\\sin \left(\theta_{1}\right)\end{array}\right].\tag{d}\end{gathered}$$ Elements of the stiffness matrix are the same as given by eq. (e) of example 6.1. Set $Q_{1}=0$ and $Q_{2}=0$, since no external forces are applied to the joint just after assembly. Then solve the matrix equation (d) for the joint displacements to get $$\left[\begin{array}{@{}l@{}}q_{1} \\q_{2}\end{array}\right]=\left[\begin{array}{@{}ll@{}}k_{11} & k_{12} \\k_{21} & k_{22}\end{array}\right]^{-1}\left[\begin{array}{@{}c@{}}\cos \left(\theta_{1}\right) \\\sin \left(\theta_{1}\right)\end{array}\right]\left(\frac{E A}{L}\right)_{1} \bar{\Delta}_{1}.\tag{e}$$ From this solution for the displacements we can calculate the elongation of each bar after assembly from eq. (a) in example 6.1. The initial bar forces after assembly are computed from $$N_{1}=\left(\frac{E A}{L}\right)_{1}\left(\Delta_{1}-\bar{\Delta}_{1}\right) \quad N_{2}=\left(\frac{E A}{L}\right)_{2} \Delta_{2} \quad N_{3}=\left(\frac{E A}{L}\right)_{3} \Delta_{3}.\tag{f}$$

A specific case: $\theta_{1}=30^{\circ}$, $\theta_{2}=45^{\circ}$, $\theta_{3}=60^{\circ}$, and $E A$ is the same for each bar. Take $L_{1}=L$, so that $L_{2}=L / \sqrt{2}$, and $L_{3}=L /(\sqrt{3})$. The solution for the displacements from eq. (b) are $q_{1}=1.458 \bar{\Delta}_{1}$ and $q_{2}=-\bar{\Delta}_{1}$. The elongations are $\Delta_{1}=0.763 \bar{\Delta}_{1}$, $\Delta_{2}=0.324 \bar{\Delta}_{1}$, and $\Delta_{3}=-0.137 \bar{\Delta}_{1}$, and the bar forces from eq. (f) are $$N_{1}=-0.237 E A\left(\bar{\Delta}_{1} / L\right) \quad N_{2}=0.458 E A\left(\bar{\Delta}_{1} / L\right) \quad N_{3}=-0.237 E A\left(\bar{\Delta}_{1} / L\right).\tag{g}$$

A note on static determinacy:

Let $m\,{=}\,$the number of unknown bar forces, $r\,{=}\,$the number of support reactions, and let $j\,{=}\,$the number of joints. There are two independent equilibrium equations per joint. For a statically determinate truss, the number of unknown forces is equal to number of independent equilibrium equations (i.e., $2j = m + r$). For the truss in this example $j = 3$, $m = 3$, and $r = 3$. So it is statically determinate. For the truss in example 6.1, $j = 3$, $m = 3$, and $r = 6$, and $6 < 3 + 6$. So the truss in example 6.1 is statically indeterminate.

Solution.

Free body diagrams of joints 2 and 3 are shown figure 1.2. The diagrams are drawn assuming each bar is in tension, so the reaction of the bar force acting on a joint is an arrow aligned with the bar and pointing away from the joint. The objective is to determine each bar force in terms of external forces $Q_3$ and $Q_4$. Note that $Q_3 = 0$ and $Q_4 = -84{,}000$ N, but we will wait to substitute these numerical values after the derivatives are evaluated in Castigliano’s second theorem.

The only contribution to the complementary strain energy in ([eq5.84]) on page is the axial normal force $N$, which is spatially uniform along the length of the bar. Also, there is no change in temperature from the reference state. Hence, the complementary strain energy for the truss is $$U^{*}=\frac{1}{2}\left[\left(\frac{L}{E A}\right)_{1-2} N_{1-2}^{2}+\left(\frac{L}{E A}\right)_{1-3} N_{1-3}^{2}+\left(\frac{L}{E A}\right)_{2-3} N^{2}_{2-3}\right],\tag{a}$$ Castigliano’s second theorem for the displacement $q_3$ is $$q_{3}=\frac{\partial U^{*}}{\partial Q_{3}}=\left(\frac{L}{E A}\right)_{1-2} N_{1-2} \frac{\partial N_{1-2}}{\partial Q_{3}}+\left(\frac{L}{E A}\right)_{1-3} N_{1-3} \frac{\partial N_{1-3}}{\partial Q_{3}}+\left(\frac{L}{E A}\right)_{2-3} N_{2-3} \frac{\partial N_{2-3}}{\partial Q_{3}}.\tag{b}$$

The terms in eq. (b) are listed in table [tab6.1]. Replace the derivatives of the bar forces with their values listed in the table to get

$$q_{3}=\left(\frac{L}{E A}\right)_{1-2} N_{1-2}(1)+\left(\frac{L}{E A}\right)_{1-3} N_{1-3}(0)+\left(\frac{L}{E A}\right)_{2-3} N_{2-3}(0).\tag{c}$$ Substitute the equation for bar force $N_{1-2}$ from the table in the previous equation and note that $Q_{3}=0$ and $Q_{4}=-84{,}000$ N to get $$q_{3}=\left(\frac{L}{E A}\right)_{1-2}\left(Q_{3}+3 Q_{4}/4\right)=\left(11.90\times 10^{-6}\,\mathrm{mm}/ \mathrm{N}\right)\left(0+\frac{3}{4}(-84{,}000\,\mathrm{N})\right)=-0.750\,\mathrm{mm}.\tag{d}$$ Castigliano’s second theorem for the displacement $q_4$ is $$q_{4}=\frac{\partial U^{*}}{\partial Q_{4}}=\left(\frac{L}{E A}\right)_{1-2} N_{1-2} \frac{\partial N_{1-2}}{\partial Q_{4}}+\left(\frac{L}{E A}\right)_{1-3} N_{1-3} \frac{\partial N_{1-3}}{\partial Q_{4}}+\left(\frac{L}{E A}\right)_{2-3} N_{2-3} \frac{\partial N_{2-3}}{\partial Q_{4}}.\tag{e}$$ Replace the bar forces and their derivatives in eq. (e) with their values listed in table [tab6.1] to get $$q_{4}=\left(\frac{L}{E A}\right)_{1-2}\left(0+3 Q_{4}/4\right)\left(\frac{3}{4}\right)+\left(\frac{L}{E A}\right)_{1-3}\left(Q_{4}\right)(1)+\left(\frac{L}{E A}\right)_{2-3}\left(-5 Q_{4}/4\right)\left(-\frac{5}{4}\right).\tag{f}$$ Substitute numerical values into eq. (f) to get $$q_{4}=\left[(11.90 \times 10^{-6}\,\mathrm{mm} / \mathrm{N}) \frac{9}{16}+(47.62 \times 10^{-6}\,\mathrm{mm} / \mathrm{N})+(14.88 \times 10^{-6}\,\mathrm{mm} / \mathrm{N}) \frac{25}{16}\right](-84{,}000\,\mathrm{N}).\tag{g}$$ The final result from eq. (g) is $$\begin{aligned} q_{3}=-0.750\,\mathrm{mm} \quad q_{4}=-6.516\,\mathrm{mm} \tag{h}\end{aligned}$$

Solution.

From eqs. ([eq5.83]) to ([eq5.85]) on page the complementary strain energy in this example is $$U^{*}=\frac{1}{2} \int_{0}^{L}\left[\frac{\left(N+N_{T}\right)^{2}}{E A}+\frac{\left(M_{x}+M_{x T}\right)^{2}}{E I_{x x}}+\frac{\left(M_{y}+M_{y T}\right)^{2}}{E I_{y y}}+c_{x x} V_{x}^{2}+c_{y y} V_{y}^{2}+\frac{M_{z}^{2}}{G J}\right] d z.\tag{b}$$ The cross-sectional properties were determined in example 6.4, and they are $$\begin{gathered} A=2 \pi a t=2 \pi(0.03812\,\mathrm{m})(7.14 \times 10^{-4}\,\mathrm{m})=171.014 \times 10^{-6}\,\mathrm{m}^{2}, \tag{c}\\ I_{x x}=I_{y y}=\pi a^{3} t=\pi(0.03812\,\mathrm{m})^{3}(7.14 \times 10^{-4}\,\mathrm{m})=124.25 \times 10^{-9}\,\mathrm{m}^{4},\ \text{and}\ J=2 \pi a^{3} t=248.50 \times 10^{-9}\,\mathrm{m}^{4}.\tag{d}\end{gathered}$$ The shear modulus is given by the isotropic formula $G=E /(2(1+v))=25.6~\mathrm{GPa}$, so the transverse shear compliances are $$c_{x x}=c_{y y}=\frac{1}{\pi a t G}=\frac{1}{\pi(0.03812\,\mathrm{m})\left(7.14 \times 10^{-4}\,\mathrm{m}\right)\left(25.6 \times 10^{9}\,\mathrm{N} / \mathrm{m}^{2}\right)}=\frac{456.61 \times 10^{-9}}{\mathrm{~N}}.\tag{e}$$ The thermal axial force is given by eq. ([eq3.75]) on page , and thermal moments are given in eq. ([eq3.78]). Material properties are uniform along the contour and $x=a \cos \theta$, and $y=a \sin \theta$ in the thermal action formulas. The results for these thermal actions are $$\begin{aligned} &N_{T}=E \alpha \int_{0}^{2 \pi} \Delta T t a d \theta=2 \pi a t E \alpha \bar{T} \quad M_{x T}=E \alpha \int_{0}^{2 \pi} \Delta T y t a d \theta=0 \quad M_{y T}=E \alpha \int_{0}^{2 \pi} \Delta T x t a d \theta=\pi a^{2} t E \alpha T_{m},&\tag{f} \\ &N_{T}=2 \pi(0.03812\,\mathrm{m})(7.14 \times 10^{-4}\,\mathrm{m})(68.3 \times 10^{9}\,\mathrm{N} / \mathrm{m}^{2})(23 \times 10^{-6}{}^{\circ} ~K)(462^{\circ}~K)=124.11 \times 10^{3}\,\mathrm{N},\ \text{and}& \tag{g} \\ &M_{y T}=\pi(0.03812\,\mathrm{m})^{2}(7.14 \times 10^{-4}\,\mathrm{m})(68.3 \times 10^{9}\,\mathrm{N} / \mathrm{m}^{2})(23 \times 10^{-6}{ }^{\circ} K)(34^{\circ} ~K)=174.10\,\mathrm{N}-\mathrm{m}.&\tag{h}\end{aligned}$$ The free body diagram of the tube in the x-z plane is shown in figure [fig6.11]. Generalized forces $Q_1$, $Q_3$, and $Q_5$ are introduced at the free end to facilitate computing the corresponding displacements via Castigliano’s theorem, and they are set equal to zero at the end of the procedure (i.e., they are fictitious actions).

Equilibrium of the free body diagram in the x-z plane yields $$N=Q_{3} \quad V_{x}=Q_{1} \quad M_{y}=Q_{5}-Q_{1}(L-z) \quad M_{z}=0 \quad z \in[0, L].\tag{i}$$ Since generalized forces $Q_{2}=Q_{4}=0$, equilibrium in the y-z plane yields $V_{y}=M_{x}=0$ for $z \in[0, L]$. The complementary strain energy reduces to $$U^{*}=\frac{1}{2} \int_{0}^{L}\left[\frac{\left(Q_{3}+N_{T}\right)^{2}}{E A}+\frac{\left[Q_{5}-Q_{1}(L-z)+M_{y T}\right]^{2}}{E I_{y y}}+\frac{Q_{1}^{2}}{\pi a t G}\right] d z.\tag{j}$$

Displacement $q_1$ is determined from $$\begin{aligned} q_{1}=\frac{\partial U^{*}}{\partial Q_{1}}=\int_{0}^{L} &\left[\frac{\left(Q_{3}+N_{T}\right)}{E A} \frac{\partial\left(Q_{3}+N_{T}\right)}{\partial Q_{1}}+\frac{\left[Q_{5}-Q_{1}(L-z)+M_{y T}\right]}{E I_{y y}} \frac{\partial\left[Q_{5}-Q_{1}(L-z)+M_{y T}\right]}{\partial Q_{1}} \right.\nonumber \\ &\quad \left. +\frac{Q_{1}}{\pi a t G}\left(\frac{\partial Q_{1}}{\partial Q_{1}}\right)\right] d z,\tag{k}\end{aligned}$$ where we interchanged the derivative and integral since our functions are continuous. Performing the derivative inside the integral we get $$q_{1}=\frac{1}{E I_{y y}} \int_{0}^{L}\left[Q_{5}-Q_{1}(L-z)+M_{y T}\right][-(L-z)] d z+\frac{1}{\pi a t G} \int_{0}^{L} Q_{1}(1) d z.\tag{l}$$ Now set $Q_1$ and $Q_5$ equal to zero and find $$q_{1}=\left(\frac{1}{E I_{y y}}\right) \int_{0}^{L} M_{y T}[-(L-z)] d z=\frac{-L^{2} M_{y T}}{2 E I_{y y}}=\frac{-(0.8\,\mathrm{m})^{2}(174.10\,\mathrm{N}\mbox{-}\mathrm{m})}{2\left(68.3 \times 10^{9}\,\mathrm{N} / \mathrm{m}^{2}\right)\left(124.25 \times 10^{-9}\,\mathrm{m}^{4}\right)}=-6.565 \times 10^{-3}\,\mathrm{m}.\tag{m}$$ Axial displacement $q_3$ is given by $$\begin{aligned} q_{3}=\frac{\partial U^{*}}{\partial Q_{3}}=\int_{0}^{L}&\left[\frac{\left(Q_{3}+N_{T}\right)}{E A} \frac{\partial\left(Q_{3}+N_{T}\right)}{\partial Q_{3}}+\frac{\left[Q_{5}-Q_{1}(L-z)+M_{y T}\right]}{E I_{y y}} \frac{\partial \left[Q_{5}-Q_{1}(L-z)+M_{y T}\right]}{\partial Q_{3}} \right.\nonumber\\ &\quad\left.+\frac{Q_{1}}{\pi a t G}\left(\frac{\partial Q_{1}}{\partial Q_{3}}\right)\right] d z.\tag{n}\end{aligned}$$ Since $Q_3 = 0$, the latter equation reduces to $$q_{3}=\left.\left(\frac{1}{E A}\right) \int_{0}^{L}\left(Q_{3}+N_{T}\right)(1) d z\right|_{Q_{3}=0}=\frac{L}{E A} N_{T}=\frac{(0.8\,\mathrm{m})\left(124.11 \times 10^{3}\,\mathrm{N}\right)}{\left(68.3 \times 10^{9}\,\mathrm{N} / \mathrm{m}^{2}\right)\left(171.014 \times 10^{-6}\,\mathrm{m}^{2}\right)}=8.5 \times 10^{-3}\,\mathrm{m}.\tag{o}$$ Finally, the rotation in radians about the $y$-axis is given by $$\begin{aligned} q_{5}&=\left.\left(\frac{1}{E I_{y y}}\right)\int_{0}^{L}\left[Q_{5}-Q_{1}(L-z)+M_{y T}\right][1] d z\right|_{Q_{1}\,=\,Q_{5}\,=\,0},\nonumber\\ q_{5}&=\frac{L M_{y T}}{E I_{y y}}=\frac{(0.8\,\mathrm{m})(174.10\,\mathrm{N}\mbox{-}\mathrm{m})}{\left(68.3 \times 10^{9}\,\mathrm{N} / \mathrm{m}^{2}\right)\left(124.25 \times 10^{-9}\,\mathrm{m}^{4}\right)}=16.41 \times 10^{-3}~\mbox{rad.} \tag{p}\end{aligned}$$

[ex6.6]A light airplane experiences a total lift L = 12,000 lb. in a certain symmetric maneuver. Thus, the lift acting on each wing is L/2. Assume the airload is distributed elliptically over the wing, so that the airload intensity $f_{L}$ per unit span is given as $$f_{L}=\frac{2 L}{\pi z_{m a x}} \sqrt{1-\left(\frac{z}{z_{m a x}}\right)^{2}} \quad 0 \leq z \leq z_{\textit{max}},\tag{a}$$ where $z$ is the spanwise coordinate, $z$ = 0 at the root, and $z=z_{\max }$ at the tip of the wing. See figure 1.3(a). The spar of the wing is a uniform, longitudinal, thin-walled beam with a closed section stiffened by four longitudinal stringers as shown in figure 1.3(b). This cross-section is the same one shown in figure [fig3.24] and analyzed in example [ex3.4] on page . Assume the spar is clamped at the root and free at the tip (i.e., a cantilever spar). At the tip of the spar we will use Castigliano’s second theorem to find the vertical displacement of the shear center denoted by $q_2$, and to find the torsional rotation of the cross section denoted by $q_6$. To use the theorem, we introduce a fictitious force $Q_2$ corresponding to displacement $q_2$, and a fictitious torque $Q_6$ corresponding to rotation $q_6$. A typical cross section of the spar with the locations of the centroid ($X_C$), the shear center ($X_{SC}$), and the line of action of the airload ($X_L$) with respect to the vertical web are shown in the left-hand sketch of figure 1.4. The right-hand sketch in figure 1.4 illustrates that the airload is statically equivalent to the external line load intensity $f_{y}$ and line moment intensity $m_{z}$ resolved at the shear center.

Numerical data for the cross-sectional dimensions are listed in table [tab6.3]. The material is an aluminium alloy with a Young’s modulus $E=10.5 \times 10^{3}~\mathrm{ksi}$, a shear modulus $G=4.0 \times 10^{3}~\mathrm{ksi}$, and with a yield strength $\sigma_{\text {yield }}=44~\mathrm{ksi}$. Additional cross-sectional properties computed from example [ex3.4] on page  are listed in table [tab6.4].

1. Determine the statically admissible bar resultants in the spar for $0 \leq z \leq z_{\textit{max}}$.

2. Determine the generalized displacements $q_2$ and $q_6$ of the shear center at $z=z_{\textit{max}}$.

3. Tabulate the displacement $q_2$, percentage of the displacement $q_2$ due to transverse shear, and the rotation $q_6$ of part (b) for the following spar lengths: $z_{\textit{max}}=12$, 24, 60, 120, 180, 240, and 300 in.. Also, tabulate the ratio of the maximum von Mises effective stress (eq. ([eq4.31]) on page ) to the yield strength in the semi-circular web, or branch 1, at $z = 0$ for the same set of spar lengths.

Solution to part (a).

The external distributed line load intensities resolved at the shear center are shown figure 1.4. In terms of the specified airload $f_{y}=f_{L}$ and $m_{z}=ef_{L}$, where $e=X_{L}-X_{S C}$. The differential equation for the transverse shear force $V_y$ is given by eq. ([eq3.54]) on page . Substitute the expression for the airload to get the shear force as $$V_{y}=\int\left[-\left(\frac{2 L}{\pi z_{m a x}}\right) \sqrt{1-\left(\frac{z}{z_{m a x}}\right)^{2}}\hspace*{3pt}\right] d z=\left(\frac{L}{\pi}\right)\left[\operatorname{acos}\left(\frac{z}{z_{m a x}}\right)-\left(\frac{z}{z_{m a x}}\right) \sqrt{1-\left(\frac{z}{z_{m a x}}\right)^{2}}\hspace*{3pt}\right]+c_{1}.\tag{a}$$ Note that the integration is facilitated by the substitution $z / z_{\it max }=\cos \theta$, and using trigonometric identities. The constant of integration $c_1$ is determined by the boundary condition $V_{y}(z_{\it max})=Q_{2}$. Hence $c_{1}=Q_{2}$, and the final result for the shear force is $$V_{y}=\left(\frac{L}{\pi}\right)\left[\text{acos}\left(\frac{z}{z_{\it max }}\right)-\left(\frac{z}{z_{\it max }}\right) \sqrt{1-\left(\frac{z}{z_{\it max }}\right)^{2}}\hspace*{3pt}\right]+Q_{2}.\tag{b}$$ The shear force at the root for $Q_2 = 0$ is $V_{y}(0)|_{Q_{2}\,=\,0}=L / 2$.

The bending moment $M_x$ is determined by eq. ([eq3.55]) on page . Substitute the result for the shear force $V_y$ into eq. ([eq3.55]) to find $$M_{x}=\int\left\{\left(\frac{L}{\pi}\right)\left[\operatorname{acos}\left(\frac{z}{z_{\it max }}\right)-\left(\frac{z}{z_{\it max }}\right) \sqrt{1-\left(\frac{z}{z_{\it max }}\right)^{2}}\right]+Q_{2}\right\} d z.$$ Again, the integration for $M_x$ is facilitated by the substitution $z/z_{\it max}=\cos \theta$, and using trigonometric identities to get $$M_{x}=\left(\frac{L z_{\it max}}{\pi}\right)\left[-\sqrt{1-\left(\frac{z}{z_{\it max }}\right)^{2}}+\frac{1}{3}\left[1-\left(\frac{z}{z_{\it max }}\right)^{2}\right]^{3 / 2}+\left(\frac{z}{z_{\it max }}\right) \operatorname{acos}\left(\frac{z}{z_{\it max }}\right)\right]+Q_{2} z+c_{2}.\tag{c}$$ The constant of integration $c_2$ is determined by the boundary condition $M_{x}\left(z_{max}\right)=0$. Hence $c_{2}=-z_{\it max } Q_{2}$, and the final result for the bending moment is $$M_{x}=\left(\frac{L z_{\it max }}{\pi}\right)\left[-\sqrt{1-\left(\frac{z}{z_{\it max }}\right)^{2}}+\frac{1}{3}\left[1-\left(\frac{z}{z_{\it max }}\right)^{2}\right]^{3 / 2}+\left(\frac{z}{z_{\it max }}\right) \operatorname{acos}\left(\frac{z}{z_{\it max }}\right)\right]-Q_{2}\left(z_{\it max }-z\right).\tag{d}$$ The bending moment at the root of the spar for $Q_2 = 0$ is $\left.M_{x}(0)\right|_{Q_{2}=0}=\frac{-2 L z_{m a x}}{3 \pi}$.

From eq. ([eq3.61]) on page , and that ${m}_{z}=e f_{L}$, we can express the torque $M_z$ as $$\frac{d M_{z}}{d z}=-m_{z}=-e f_{L}.\tag{e}$$ Hence, $$M_{z}=-e \int \!\!\frac{2 L}{\pi z_{\it max }} \sqrt{1-\left(\frac{z}{z_{\it max }}\right)^{2}} d z+c_{3}=-e \frac{L}{\pi}\left\{\left(\frac{z}{z_{\it max }}\right) \sqrt{1-\left(\frac{z}{z_{\it max }}\right)^{2}}+\operatorname{asin}\left(\frac{z}{z_{\it max }}\right)\right\}+c_{3}.\tag{f}$$ The constant of integration $c$3 is determined from the boundary condition $M_{z}\left(z_{\it max }\right)=Q_{6}$, which yields $c_{3}=e \frac{L}{2}+Q_{6}$. The final result for the torque is $$M_{z}=\frac{e L}{2}-\frac{e L}{\pi}\left[\left(\frac{z}{z_{m a x}}\right) \sqrt{1-\left(\frac{z}{z_{m a x}}\right)^{2}}+\operatorname{asin}\left(\frac{z}{z_{\it max }}\right)\right]+Q_{6}.\tag{g}$$ The torque at the root of the spar for $Q_6 = 0$ is $\left.M_{z}(0)\right|_{Q_{6}\,=\,0}=e L / 2$. We have determined the statically admissible shear force $V_y$, bending moment $M_x$, and torque $M_z$ in the wing spar in terms of the distributed airload, external force $Q_2$, and external torque $Q_6$.

Solution to part (b).

From eqs. ([eq5.84]) and ([eq5.85]) on page , the total complementary strain energy for the bar in this example is $$U^{*}=\frac{1}{2} \int_{0}^{z_{\it max}}\left(\frac{M_{x}^{2}}{E I_{x x}}+c_{y y} V_{y}^{2}+c_{z z} M_{z}^{2}\right) d z.\tag{h}$$ Castigliano’s second theorem for the vertical displacement of the shear center is $$q_{2}=\left.\frac{\partial U^{*}}{\partial Q_{2}}\right|_{Q_{2}\,=\,Q_{6}\,=\,0}=\left.\int_{0}^{z_{\it max }}\left[\frac{M_{x}}{E I_{x x}} \frac{\partial M_{x}}{\partial Q_{2}}+c_{y y} V_{y} \frac{\partial V_{y}}{\partial Q_{2}}+c_{z z} M_{z} \frac{\partial M_{z}}{\partial Q_{2}}\right] d z\right|_{Q_{2}\,=\,Q_{6}\,=\,0}.\tag{i}$$

Note that the torque is independent of force $Q_2$, so that $\partial M_{z}/\partial Q_{2}=0$. Let $q_{2}=q_{2 m}+q_{2 v}$, where $q_{2 m}$ is the portion of the displacement due to bending moment $M_x$ and $q_{2 v}$ is the portion due to transverse shear force $V_y$. $$q_{2 m}=\left.\int_{0}^{z_{\it max }}\left[\frac{M_{x}}{E I_{x x}} \frac{\partial M_{x}}{\partial Q_{2}}\right] d z\right|_{Q_{2}\,=\,Q_{6}\,=\,0}=\left.\frac{1}{E I_{x x}} \int_{0}^{z_{\it max }} M_{x}\left[-\left(z_{\it max }-z\right)\right] d z\right|_{Q_{2}\,=\,Q_{6}\,=\,0}.\tag{j}$$ Substitute eq. (d) for $M_x$ with $Q_2 = 0$ into eq. (j), and perform the integration to get $$q_{2 m}=\frac{L(-32+45 \pi) z_{\it max }^{3}}{720 E I_{x x} \pi}.\tag{k}$$ The portion of the displacement due to transverse shear force $V_y$ is $$\begin{aligned} q_{2 v}&=\left.\int_{0}^{z_{\it max }}\left[c_{y y} V_{y} \frac{\partial V_{y}}{\partial Q_{2}}\right] d z\right|_{Q_{2}\,=\,Q_{6}\,=\,0}=c_{y y} \int_{0}^{z_{\it max }}\left\{\left(\frac{L}{\pi}\right)\left[\operatorname{acos}\left(\frac{z}{z_{\it max }}\right)-\left(\frac{z}{z_{\it max }}\right) \sqrt{1-\left(\frac{z}{z_{\it max }}\right)^{2}}\right]\right\}\{1\} d z\nonumber\\ &=\frac{2 c_{y y} L z_{\it max }}{3 \pi}.\tag{l}\end{aligned}$$ Add eqs. (k) and (l) to get the total vertical displacement at the shear center as $$q_{2}=\frac{L(-32+45 \pi) z_{\it max }^{3}}{720 E I_{x x} \pi}+\frac{2 c_{y y} L z_{m a x}}{3 \pi}.\tag{m}$$

Castigliano’s second theorem for the rotation of the cross-section at $z=z_{\it max}$ about the shear center is $$q_{6}=\left.\frac{\partial U^{*}}{\partial Q_{6}}\right|_{Q_{2}\,=\,Q_{6}\,=\,0}=\left.\int_{0}^{z_{\it max }}\left[\frac{1}{E I_{x x}} M_{x} \frac{\partial M_{x}}{\partial Q_{6}}+c_{y y} V_{y} \frac{\partial V_{y}}{\partial Q_{6}}+c_{z z} M_{z} \frac{\partial M_{z}}{\partial Q_{6}}\right] d z\right|_{Q_{2}\,=\,Q_{6}\,=\,0}.\tag{n}$$ The bending moment $M_x$ and transverse shear force $V_y$ are independent of the external torque $Q_6$. Hence, $$q_{6}=\left.\frac{\partial U^{*}}{\partial Q_{6}}\right|_{Q_{2}\,=\,Q_{6}\,=\,0}=c_{z z} \int_{0}^{z_{\it max }}\left\{\frac{e L}{2}-\frac{e L}{\pi}\left[\left(\frac{z}{z_{\it max }}\right) \sqrt{1-\left(\frac{z}{z_{\it max }}\right)^{2}}+\operatorname{asin}\left(\frac{z}{z_{\it max }}\right)\right]\right\}[1] d z.\tag{o}$$ The integration of eq. (o) yields $$q_{6}=\frac{2 c_{z z} e L z_{\it max }}{3 \pi}.\tag{p}$$

Solution to part (c).

Numerical evaluation of the displacements yields $$q_{2 m}=\left(\frac{5.438 \times 10^{-7}}{\text {in.}^{2}}\right) z_{\it max}^{3} \quad q_{2 v}=(1.0489 \times 10^{-3}) z_{\it max } \quad q_{6}=(4.3905 \times 10^{-5} \text { in.}^{-1}) z_{\it max}.\tag{q}$$

The expression for the shear flow is given by eq. ([eq3.163]) on page . At the root cross section the equation for the shear flow reduces to $$q(s, 0)=M_{z}(0) /\left(2 A_{c}\right)-F_{y}(s) V_{y}(0).\tag{r}$$

The torque results in a spatially uniform component to the shear flow around the contour equal to $$q_{t}=M_{z}(0) /\left(2 A_{c}\right)=76.9188~\mathrm{lb} / \mathrm{in}.\,.\tag{s}$$ (Refer to eq. ([eq3.165]) on p. ). The total shear flow in each branch is $$q_{i}\left(s_{i}, 0\right)=76.9188~\mathrm{lb}/\text{in. }-F_{y i}\left(s_{i}\right)(6000.0 \text { lb.}) \quad i=1,2,3,4,\tag{t}$$ where the contour coordinates $s_i$ are shown in figure [fig3.24](b) on page , and the shear flow distribution functions $F_{y i}\left(s_{i}\right)$ are given by eqs. (ab) to (ae) in part c of example [ex3.4]. The shear stress distribution along the contour in each branch is given by $$\sigma_{z s i}\left(s_{i}, 0\right)=[q_{i}(s_{i}, 0)] / t \quad i=1,2,3,4.\tag{u}$$ In this example $I_{x y}=n_{x}=n_{y}=0$ and $k=1$ in the axial normal stress given by eq. ([eq4.6]) on page . For no change in temperature and $N=M_{y}=0$, the axial normal stress eq. ([eq4.6]) at the root cross section in each branch reduces to $$\sigma_{z i}\left(s_{i}, 0\right)=\frac{M_{x}(0) y_{i}\left(s_{i}\right)}{I_{x x}}=\left[\left(-25.0591 \frac{\mathrm{lb} .}{\mathrm{in.}^{4}}\right) z_{\it max }\right] y_{i}\left(s_{i}\right) \quad i=1,2,3,4.\tag{v}$$ The parametric equations for the $y$-coordinates of the contour in each branch are $$y_{1}\left(s_{1}\right)=-a \cos \left(s_{1} / a\right) \quad y_{2}=a \quad y_{3}\left(s_{3}\right)=a-s_{3} \quad y_{4}=-a. \tag{w}$$ (Refer to figure [fig3.24] on page .) From eq. ([eq4.31]) the von Mises effective stress is $$\left(\sigma_{\text {Mises }}\right)_{i}=\sqrt{\left[\sigma_{z i}\left(s_{i}, 0\right)\right]^{2}+3\left[\sigma_{z s i}\left(s_{i}, 0\right)\right]^{2}} \quad i=1,2,3,4.\tag{x}$$ The von Mises effective stress normalized by the yield strength is plotted with respect to the contour coordinate for $z_{\it max} = 60$ inches in figure [fig6.14]. As shown in the figure the maximum normalized effective stress is 0.383 at $s = 4.696$ in. in branch 1. In terms of an angular measurement in the semi-circular branch 1, we note the location as $s_{1} / a=(4.696 / 6)\left(180^{\circ} / \pi\right)=44.8^{\circ}$. For the other values of $z_{\it max}$, the maximum value of the von Mises stress also occurs in branch 1 but at different angular locations. Discontinuities in the von Mises stress with respect to the contour coordinate are a result of the jumps in the shear flow across the stringers. (Refer to eq. ([eq3.135]) on p. ).

Numerical results are listed in table [tab6.5].

Note that as the length of the spar increases the percentage of the vertical displacement at the tip due to transverse shear decreases and the von Mises effective stress increases. At $z_{\it max} = 300$ in. the von Mises stress exceeds the yield strength of the material indicating failure by material yielding.

Solution.

The structural model of the left-hand wing and bracing shown in figure 1.6(a) consists of five truss bars. The turnbuckle displacements are determined from the horizontal position of the wing. Free body diagrams of joints 1 and 4 are shown in figure 1.6(b). A vertical external force $Q_2$ is introduced at joint 1 so that its corresponding displacement $q_2$ can be determined in the application of Castigliano’s theorem. Displacement $q_2$ is specified from the wing’s required dihedral. That is $q_{2}=L \sin \Gamma$, and after its determination external force $Q_{2}$ is set to zero.

From eq. ([eq5.84]) on page  the complementary energy for a homogenous truss bar subject to a uniform change in temperature is $$\begin{aligned} U^{*}=\frac{L}{2 E A}\left(N+N_{T}\right)^{2}.\end{aligned}$$ To account for the displacements of the turnbuckles in Castigliano’s second theorem we modify the axial temperature term in the complementary strain energy. The thermal axial force in the truss bar is obtained from eq. ([eq3.75]) on page : $$\begin{aligned} N_{T}=\int_{c} \beta \Delta T(s, z) t(s) d s=E A \alpha \Delta T.\end{aligned}$$ Let the thermal strain $\alpha \Delta T$ be replaced by the initial strain induced by the turnbuckle displacement $\Delta_{T}=2 n p$ divided by the length of the bar containing the turnbuckle. That is, $\alpha \Delta T \rightarrow \Delta_{T} / L$. Then the complementary strain energy in eq. (a) that includes the displacement caused by the turnbuckle is $$\begin{aligned} U^{*}=\frac{c}{2}\left(N+\Delta_{T} / c\right)^{2},\end{aligned}$$ where the flexibility influence coefficient $c=L /(E A)$.

The interplane strut subject to force $N_{1-4}$ is assumed to be rigid. Its flexibility influence coefficients vanishes and it does not contribute the elastic complementary strain energy. The complementary strain energy is $$\begin{aligned} U^{*}=\frac{1}{2} c_{W}(N_{1-2})^{2}+\frac{1}{2} c_{S T}(N_{1-3}+\Delta_{L} / c_{S T})^{2}+\frac{1}{2} c_{S T}(N_{2-4}+\Delta_{F} / c_{S T})^{2}+\frac{1}{2} c_{W}(N_{3-4})^{2}.\end{aligned}$$ The flexibility influence coefficients for the two wing spars is $$\begin{aligned} c_{W}=\frac{L}{E_{W} A_{W}}=\frac{120}{\left(1.5 \times 10^{6}\right) 1.25}=64 \times 10^{-6}~\mathrm{in} . / \mathrm{lb},\end{aligned}$$ and the flexibility influence coefficients for the two wires is $$\begin{aligned} c_{S T}=\frac{\sqrt{120^{2}+51.6^{2}}}{\left(30 \times 10^{6}\right) \pi(0.125 / 2)^{2}}=\frac{130.624}{\left(30 \times 10^{6}\right)(0.012272)}=354.81 \times 10^{-6}~\mathrm{in} . / \mathrm{lb}.\end{aligned}$$

Equilibrium equations at joint 1 in figure 1.6(b) are $$\begin{aligned} N_{1-3} \cos \theta+N_{1-2}=0 \quad Q_{2}+N_{1-4}+N_{1-3} \sin \theta=0,\end{aligned}$$ and equilibrium equations at joint 4 are $$\begin{aligned} N_{3-4}+N_{2-4} \cos \theta=0 \quad N_{1-4}+N_{2-4} \sin \theta=0.\end{aligned}$$ The trigonometric functions of the angle $\theta$ are $$\begin{aligned} \cos \theta=\lambda=\frac{120}{\sqrt{120^{2}+51.6^{2}}}=0.91866919 \quad \sin \theta=\mu=\frac{51.6}{\sqrt{120^{2}+51.6^{2}}}=0.39502775.\end{aligned}$$ Now eliminate the bar force $N_{1-4}$ between the four equilibrium equations to get the three equations $$\begin{aligned} N_{1-3} \lambda+N_{1-2}=0 \quad N_{3-4}+N_{2-4} \lambda=0 \quad Q_{2}-N_{2-4} \mu+N_{1-3} \mu=0.\end{aligned}$$ The force $N_{2-4}$ in the flying wire is taken as the redundant. Solve the remaining bar forces from eq. (j) in terms of the redundant and force $Q_2$ to get $$\begin{aligned} N_{1-2}=-N_{2-4} \lambda+Q_{2} \lambda / \mu \quad N_{1-3}=N_{2-4}-Q_{2} / \mu \quad N_{3-4}=-N_{2-4} \lambda.\end{aligned}$$

Substitute the results for $N_{1-2}$, $N_{1-3}$, and $N_{3-4}$ from eq. (k) into the complementary strain energy (d) to find the energy in the form $U^{*}\left[N_{2-4}, Q_{2}\right]$ with turnbuckle displacements $\Delta_L$ and $\Delta_F$ appearing in $U_c$ as parameters. $$\begin{gathered} q_{2}=\left.\frac{\partial U^{*}}{\partial Q_{2}}\right|_{Q_{2}\,=\,0}=-\big(c_{S T}+c_{W} \lambda^{2}\big)\left(N_{2-4} / \mu\right)-(\Delta L) / \mu.\\ \frac{\partial U^{*}}{\partial N_{2-4}}=2\big(c_{S T}+c_{w} \lambda^{2}\big) N_{2-4}-\big(c_{S T}+c_{W} \lambda^{2}\big)\left(Q_{2} / \mu\right)+\Delta_{L}+\Delta_{F}=0.\end{gathered}$$ Set $q_{2}=L \sin \Gamma$ in eq. (l) and solve for the landing wire turnbuckle displacement, followed by solving eq. (m) for the to find flying wire turnbuckle displacement. The results are $$\begin{aligned} \Delta L=-\big(c_{S T}+c_{w} \lambda^{2}\big) N_{2-4}-\mu L \sin \Gamma\quad \text{and}\quad \Delta F=-\big(c_{S T}+c_{w} \lambda^{2}\big) N_{2-4}+\mu L \sin \Gamma.\end{aligned}$$ Set $N_{2-4}=400~\mathrm{lb.}$ to obtain the numerical results for the turnbuckle displacements and their number of turns as $$\begin{aligned} \Delta L=-3.41912 \text { in. } \quad n_{L}=-17.0956 \text { turns } \quad \Delta F=3.19425 \text { in. } \quad n_{F}=15.9713 \text { turns }.\end{aligned}$$ The landing wire turnbuckle decreases the length between joints 1 and 3, and the flying wire turnbuckle increases the length between joints 2 and 4. The bar forces are $$\begin{aligned} N_{1-2}=N_{3-4}=-367.468~\mathrm{lb} . \quad N_{1-3}=N_{2-4}=400~\mathrm{lb} . \quad N_{1-4}=-158.011~\mathrm{lb.}. \end{aligned}$$

[sec6.5]

Bruhn, E. F., 1973, Analysis and Design of Flight Vehicle Structures, Jacobs Publishing, Inc., Carmel, Indiana, 46032, p. A8.42. (ISBN# 0-9615234-0-9)

Dowling, Norman E., 1993, Mechanical Behavior of Materials, Prentice Hall, Englewood Cliffs, New Jersey 07632, pp. 245–247.

Thornton, E. A., 1996, Thermal Structures for Aerospace Applications, AIAA Education Series, American Institute of Aeronautics and Astronautics, Inc., Reston, Virginia, pp. 118–121.

Warwick, Graham. “Big Fuel Savings Demand New Configurations,” November 7, 2011. Aviation Week.com.