Centroid C.

The centroid decouples the extension and bending responses of the bar in the material law. Refer to eq. ([eq3.80]) on page . The procedure to locate the centroid is presented in example [ex3.1] on page for an open cross-sectional contour, and in part (a) of example [ex3.4] on page for a closed cross-sectional contour.

Shear center S.C.

The shear center is a point in the cross section through which the plane of the loading must pass for the bar to bend and not twist in torsion. That is, the resultant of the shear forces in the cross section must act through the shear center to prevent torsion. Using energy methods in article [sec5.5.3] it is shown that the shear center decouples the transverse shear and torsion responses of the bar in the material law. Refer to eq. ([eq5.76]) on page . The procedure to locate the shear center is presented in example [ex3.3] on page for an open cross-sectional contour, and in part (c) of example [ex3.4] on page for a closed cross-sectional contour.

Differential equilibrium equations

Let the internal forces acting on the cross section at $z$ be denoted by the vector $\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}(z)$, and let the internal moments acting on the cross section resolved at the centroid be denoted by the vector $\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}(z)$. These vectors of internal actions are $$\begin{aligned} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}(z)=V_{x} \hat{i}+V_{y} \hat{j}+N \hat{k}\mbox{, and}\label{eq3.46}\\ \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}(z)=M_{x} \hat{i}-M_{y} \hat{j}+M_{z C} \hat{k}.\label{eq3.47}\end{aligned}$$ Consider the forces and moments acting on a bar element defined by $z$ and $z+ \Delta z$, $\Delta z>0$, as shown in figure [fig3.9].

The vector equations of equilibrium are $$\begin{aligned} \label{eq3.48} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}(z+\Delta z)-\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}(z)+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{f}}}\left(z^{*}\right) \Delta z=0 \quad \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}(z+\Delta z)-\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}(z)+\Delta z \hat{k} \times \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}(z+\Delta z)+\left(z^{*}-z\right) \hat{k} \times \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{f}}}\left(z^{*}\right) \Delta z+ \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{m}}}_{C}\left(z^{*}\right)=0,\end{aligned}$$ where $z<z^{*}<z+\Delta z$. For a continuous force vector and moment vector with respect to coordinate $z$, eq. ([eq3.48]) can be written as $$\begin{aligned} \label{eq3.49} \left.\frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}}{d z}\right|_{z} \Delta z+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{f}}}\left(z^{*}\right) \Delta z=\left.0 \quad \frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}}{d z}\right|_{z} \Delta z+\Delta z \hat{k} \times \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}(z+\Delta z)+\left(z^{*}-z\right) \hat{k} \times \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{f}}}\left(z^{*}\right) \Delta z+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{m}}}_{C}\left(z^{*}\right) \Delta z=0.\end{aligned}$$ Divide the latter equations by $\Delta z$ and take the limit as $\Delta z \rightarrow 0$ and note that $z^{*} \rightarrow z$ in the limit. The differential equations of equilibrium obtained from the limiting procedure are $$\begin{aligned} \label{eq3.50} \frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}}{d z}+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{f}}}(z)=0 \quad \frac{d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}}{d z}+\hat{k} \times \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}(z)+\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{m}}}_{C}(z)=0.\end{aligned}$$ Expand the differential equation ([eq3.50]) in terms of components to get $$\begin{aligned} \left(\frac{d V_{x}}{d z}+f_{x}\right) \hat{i}+\left(\frac{d V_{y}}{d z}+f_{y}\right) \hat{j}+\left(\frac{d N}{d z}+f_{z}\right) \hat{k}=0 \hat{i}+0 \hat{j}+0 \hat{k}\mbox{, and}\label{eq3.51}\\ \left(\frac{d M_{x}}{d z}-V_{y}+m_{x}\right) \hat{i}+\left(-\frac{d M_{y}}{d z}+V_{x}-m_{y}\right) \hat{j}+\left(\frac{d M_{z C}}{d z}+m_{z}+x_{sc} f_{y}-y_{sc} f_{x}\right) \hat{k}=0 \hat{i}+0 \hat{j}+0 \hat{k}.\label{eq3.52}\end{aligned}$$

Effect of thermal expansion

Consider structures subject not only to external forces, but also subject to heating. Aerospace examples include high-speed flight vehicles and orbiting space structures. Aerothermal loads consisting of pressure, skin friction or shearing stresses, and aerodynamic heating, are exerted on the external surfaces of high-speed flight vehicles. Conduction and radiant heat transfer result in significant thermally induced forces acting on orbiting space structures. These aerospace examples are discussed in detail by Thornton (1996), who provides an historical account, and methods of analysis, of thermal structures for aerospace applications.

It is assumed that a change in temperature (thermal state) causes a change in deformation and stress (mechanical state) in the structure, but a change in deformation does not cause a change in temperature. For example, under adiabatic conditions strain can cause a change in temperature. However, in many structural applications the change in temperature under adiabatic straining is negligible and can be ignored (Fung, 1965, p. 390). Thornton (1996, p.51) defines the change in thermal energy state causing a change in mechanical state, but not the reverse, as one-way thermal-mechanical coupling. Thus, heat conduction and thermoelasticty separate into two separate problems. In this text it is assumed that the heat conduction problem has been solved so that the temperature distribution in the structure is known. The thermoelastic problem is to determine the mechanical state in an elastic structure for the specified temperature distribution and the specified external loads.

For a uniaxial stress state the generalized Hooke’s law including temperature is $$\begin{aligned} \label{eq3.65} \varepsilon_{z z}=\left(\sigma_{z z}+\beta \Delta T\right) / E \quad~\text{or } \quad \sigma_{z z}=E \varepsilon_{z z}-\beta \Delta T,\end{aligned}$$ where $\beta=E \alpha$, and $\alpha$ is the coefficient of thermal expansion. The change in temperature is denoted by $\Delta T=T-T_{0}$, and $T_{0}$ is the spatially uniform temperature in the reference state. The reference state is stress free when the external loads acting on the bar are removed and the spatially uniform temperature $T=T_{0}$. Assume a linear distribution of the change in temperature in the thickness, which we write as $$\begin{aligned} \label{eq3.66} \Delta T(s, z, \zeta)=\Delta T(s, z)+\zeta D T(s, z),\end{aligned}$$ where $$\begin{aligned} \label{eq3.67} \Delta T(s, z)=\Delta T(s, z, 0) \quad~\text{and } \quad D T(s, z)=\left.\frac{\partial \Delta T}{\partial \zeta}\right|_{\zeta=0}.\end{aligned}$$

Material law for extension and bending

Substitute the expression ([eq3.30]) for the axial strain, and substitute eq. ([eq3.66]) for the change in temperature, into Hooke’s law ([eq3.65]) to get the following expression for normal stress. $$\begin{aligned} \label{eq3.68} \sigma_{z z}=E\left[\frac{d w}{d z}+y(s) \frac{d \phi_{x}}{d z}+x(s) \frac{d \phi_{y}}{d z}-\alpha \Delta T(s, z)\right]+\zeta E\left[\frac{d \phi_{x}}{d z} \sin \theta(s)+\frac{d \phi_{y}}{d z} \cos \theta(s)-\alpha D T(s, z)\right].\end{aligned}$$ In the thin-wall bar theory we neglect the distribution of the normal stress and normal strain across the thickness of the wall. Therefore, the normal strain and stress is assumed uniform in the thickness coordinate, and are given by $$\begin{gathered} \varepsilon_{z z}=\frac{d w}{d z}+y(s) \frac{d \phi_{x}}{d z}+x(s) \frac{d \phi_{y}}{d z}\mbox{, and}\label{eq3.69}\\ \sigma_{z z}=E\left[\frac{d w}{d z}+y(s) \frac{d \phi_{x}}{d z}+x(s) \frac{d \phi_{y}}{d z}-\alpha \Delta T(s, z)\right].\label{eq3.70}\end{gathered}$$ In other words, the local bending of the wall represented by the bending moment resultant $m_{zz}$ in ([eq3.36]) is neglected with respect to the membrane stiffness of the wall represented by the normal stress resultant $n_z$. In addition, 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. ([eq3.36])². The definition of the normal stress resultant reduces to $$\begin{aligned} \label{eq3.71} n_{z}=\int_{-t / 2}^{t / 2} \sigma_{z z} d \zeta.\end{aligned}$$ Substitute eq. ([eq3.70]) for the normal stress into the normal stress resultant ([eq3.71]) to get $$\begin{aligned} \label{eq3.72} n_{z}=E t\left[\frac{d w}{d z}+y(s) \frac{d \phi_{x}}{d z}+x(s) \frac{d \phi_{y}}{d z}-\alpha \Delta T(s, z)\right].\end{aligned}$$

The constitutive equation for the axial normal force $N$ is obtained by substituting eq. ([eq3.72]) for the normal stress resultant in the expression ([eq3.39]). The result is $$\begin{aligned} \label{eq3.73} N=E A \frac{d w}{d z}+E Q_{x} \frac{d \phi_{x}}{d z}+E Q_{y} \frac{d \phi_{y}}{d z}-N_{T},\end{aligned}$$ where $A$ denotes the cross-sectional area, $Q_{x}$ the first area moment about the $x$-axis, $Q_{y}$ the first area moment about the $y$-axis, and $N_{T}(z)$ the thermal axial force. These geometrical measures of the cross section are given by the formulas $$\begin{aligned} \label{eq3.74} A=\int_{c} t d s \quad Q_{x}=\int_{c} y t d s=0 \quad Q_{y}=\int_{c} x t d s=0.\end{aligned}$$ The first area moments $Q_x$ and $Q_{y}$ are equal to zero because the origin of the x-y coordinate system is located at the centroid. The thermal axial force is given by the expression $$\begin{aligned} \label{eq3.75} N_{T}(z)=\int_{c} \beta \Delta T(s, z) t(s) d s.\end{aligned}$$

The constitutive equations for the bending moments $M_{x}$ and $M_{y}$ are obtained by substituting eq. ([eq3.72]) for $n_z$ into the definitions of the bending moments in eq. ([eq3.39]), with the contribution of $m_{zz}$ neglected. The result is $$\begin{aligned} \label{eq3.76} M_{x}=E Q_{x} \frac{d w}{d z}+E I_{x x} \frac{d \phi_{x}}{d z}+E I_{x y} \frac{d \phi_{y}}{d z}-M_{x T} \quad M_{y}=E Q_{y} \frac{d w}{d z}+E I_{x y} \frac{d \phi_{x}}{d z}+E I_{y y} \frac{d \phi_{y}}{d z}-M_{y T},\end{aligned}$$ where $I_{xx}$, $I_{yy}$, and $I_{xy}$ denote the second area moments of the cross section with respect to the centroidal x-y coordinate system. The second area moments are given by the formulas $$\begin{aligned} \label{eq3.77} I_{x x}=\int_{c} y^{2} t d s \quad I_{y y}=\int_{c} x^{2} t d s \quad I_{x y}=\int_{c} x y t d s.\end{aligned}$$ The thermal bending moments in eq. ([eq3.76]) are given by the expressions $$\begin{aligned} \label{eq3.78} M_{x T}(z)=\int_{c} \beta \Delta T(s, z) y(s) t(s) d s \quad M_{y T}(z)=\int_{c} \beta \Delta T(s, z) x(s) t(s) d s.\end{aligned}$$ Since the origin of the $x$-$y$ system is taken at the centroid of the cross section, the first area moments are zero by the definition of the centroid. For $Q_{x}=Q_{y}=0$, eqs. ([eq3.73]) and ([eq3.76]) reduce to $$\begin{aligned} \label{eq3.79} \left[\begin{array}{@{}c@{}}N+N_{T} \\M_{x}+M_{x T} \\M_{y}+M_{y T}\end{array}\right]=E\left[\begin{array}{@{}ccc@{}}A & 0 & 0 \\0 & I_{x x} & I_{x y} \\0 & I_{x y} & I_{y y}\end{array}\right]\left[\begin{array}{@{}l@{}}d w / d z \\d \phi_{x} / d z \\d \phi_{y} / d z\end{array}\right].\end{aligned}$$

Open cross-sectional contour

From eq. ([eq3.97]) the shear flow at the contour origin is $q(0, z)=q_{0}(z)$. For most open cross sections, there is a longitudinal edge of the bar that is free of external tractions. If the contour origin is taken at the location of the free longitudinal edge, then $q(0, z)=q_{0}(z)=0$.The shear flow for an open cross section with the contour origin located at the longitudinal free edge and $q_{T}=0$ is given by $$\begin{aligned} \label{eq3.98} q(s, z)=-\frac{k}{I_{y y}} V_{x} \bar{Q}_{y}(s)-\frac{k}{I_{x x}} V_{y} \bar{Q}_{x}(s).\end{aligned}$$

[ex3.2]Take the change in temperature $\Delta T(s, z)=0$, $0 \leq s \leq S$, and $0 \leq z \leq L$. Hence, $q_{T}=0$ for the shear flow expression given by eq. ([eq3.98]). Second area moments were computed in example [ex3.1] page with the results listed in eqs. (l) to (n). Cross-sectional properties that depend on the second area moments, eq. ([eq3.81]), are $$\begin{aligned} \label{ex3.2a} n_{x}=\frac{I_{x y}}{I_{x x}}=\frac{0.724359 a^{3} t}{3.36086 a^{3} t}=0.215528 \quad n_{y}=\frac{I_{x y}}{I_{y y}}=\frac{0.724359 a^{3} t}{0.604984 a^{3} t}=1.19732 \quad k=1.34781.\end{aligned}$$

The first area moments of the segment of branch 1 from $s_{1}=0$ to $s_{1} \in[0, a]$ with respect the centroidal coordinates are $$\begin{aligned} \label{ex3.2b} Q_{x 1}=\int_{0}^{s_{1}} y_{1}\left(s_{1}\right) t d s_{1} \quad Q_{y 1}=\int_{0}^{s_{1}} x_{1}\left(s_{1}\right) t d s_{1}.\end{aligned}$$ From eq. (j) of example [ex3.1] $x_{1}\left(s_{1}\right)=-0.482906 a$ and $y_{1}\left(s_{1}\right)=-1.63782 a+s_{1}$. Performing the integrals in the first area moments we get $$\begin{aligned} \label{ex3.2c} Q_{x 1}=\frac{t\left(-a-4 a \pi+s_{1}+\pi s_{1}\right) s_{1}}{2(1+\pi)}=-1.63782 a t s_{1}+0.50 t s_{1}^{2} \quad Q_{y 1}=\frac{-2 a t}{1+\pi} s_{1}=-0.482906 a t s_{1}.\end{aligned}$$ The distribution functions of the segment with respect to the $\bar{x}$-$\bar{y}$ system are given by eq. ([eq3.93]), which for branch 1 results in $$\begin{gathered} \bar{Q}_{x 1}=-1.63782 a t s_{1}+0.50 t s_{1}^{2}-(1.19732)\left(-0.482906 a t s_{1}\right)=-1.05963 a t s_{1}+0.5 t s_{1}^{2},\label{ex3.2d} \tag{d}\\ \bar{Q}_{y 1}=(-0.482906 a t s_{1})-(0.215528)(-1.63782 a t s_{1}+0.50 t s_{1}^{2})=-0.12991 a t s_{1}-0.107764 t s_{1}^{2}.\label{ex3.2e} \tag{e}\end{gathered}$$ At $s_1 = 0$, the longitudinal free edge condition requires $q_0 = 0$. The shear flow in branch 1 can now be computed from eq. ([eq3.98]). The result is $$\begin{aligned} \label{ex3.2f} q_{1}\left(s_{1}\right)=\left[0.289419\left(\frac{s_{1}}{a}\right)+0.240081\left(\frac{s_{1}}{a}\right)^{2}\right] \frac{V_{x}}{a}+\left[0.424944\left(\frac{s_{1}}{a}\right)-0.200516\left(\frac{s_{1}}{a}\right)^{2}\right] \frac{V_{y}}{a} \quad 0 \leq s_{1} \leq a. \tag{f}\end{aligned}$$

The first area moments of the cross-sectional area consisting of branch 1 and a segment of branch 2 are given by $$\begin{aligned} \label{ex3.2g} Q_{x 2}(\theta)=Q_{x 1}(a)+\int_{-\pi / 2}^{\theta} y_{2}(\theta) \textit{tad} \theta \quad Q_{y 2}(\theta)=Q_{y 1}(a)+\int_{-\pi / 2}^{\theta} x_{2}(\theta) \textit{tad} \theta. \tag{g}\end{aligned}$$ where from eq. ([ex3.2c]) we find $$\begin{aligned} \label{ex3.2h} Q_{x 1}(a)=(-1.13782) a^{2} t \quad Q_{y 1}(a)=(-0.482906) a^{2} t. \tag{h}\end{aligned}$$ From eq. (k) of example [ex3.1] $x_{2}(\theta)=-0.482906 a+a \cos \theta$ and $y_{2}(\theta)=0.36218 a+a \sin \theta$. Evaluating the integrals in the first area moments for branch 2 we get $$\begin{aligned} \int_{-\pi / 2}^{\theta} y_{2}(\theta) \textit{tad} \theta=(0.56891+0.36218 \theta-\cos \theta) a^{2} t \quad \int_{-\pi / 2}^{\theta} x_{2}(\theta) t a d \theta=(0.241453-0.482906 \theta+\sin \theta) a^{2} t.\end{aligned}$$ Thus, the first area moments of the cross-sectional area consisting of branch 1 and a segment of branch 2 are $$\begin{gathered} \label{ex3.2i} \begin{split} Q_{x 2}=((-1.13782)+0.56891+0.36218 \theta-\cos \theta) a^{2} t=(-0.56891+0.36218 \theta-\cos \theta) a^{2} t \\ Q_{y 2}=((-0.482906)+0.241453-0.482906 \theta+\sin \theta) a^{2} t=(-0.241453-0.482906 \theta+\sin \theta) a^{2} t. \end{split}\tag{i}\end{gathered}$$ Note that at $\theta=\pi / 2$ both $Q_{x 2}=Q_{y 2}=0$, since the origin of the x-y system is at centroid of the cross section (i.e., the first area moments of the entire cross-sectional area about the centroidal coordinate system vanish).

The first area moment $\bar{Q}_{x}$ given in eq. ([eq3.93]) for the cross-sectional area consisting of branch 1 plus a segment of branch 2 is computed as $$%\begin{align} \bar{Q}_{x 2}=(-0.56891+0.36218 \theta-\cos \theta) a^{2} t-(1.19732)((-0.241453-0.482906 \theta+\sin \theta) a^{2} t). %\end{align}$$

Combining terms we get $$\begin{aligned} \label{ex3.2j} \bar{Q}_{x 2}=(-0.279814+0.940372 \theta-\cos \theta-1.19732 \sin \theta) a^{2} t.\tag{j}\end{aligned}$$ The first area moment $\bar{Q}_{y}$ given in eq. ([eq3.93]) for the cross-sectional area consisting of branch 1 plus a segment of branch 2 is computed as $$\begin{aligned} \label{ex3.2k} \bar{Q}_{y 2}=(-0.241453-0.482906 \theta+\sin \theta) a^{2} t-(0.215528)((-0.56891+0.36218 \theta-\cos \theta) a^{2} t).\tag{k}\end{aligned}$$ Combining terms we get $$\begin{aligned} \label{ex3.2l} \bar{Q}_{y_{2}}=(-0.118837-0.560966 \theta+0.215528 \cos \theta+\sin \theta) a^{2} t. \tag{l}\end{aligned}$$ The shear flow in branch 2 can now be computed from eq. ([eq3.98]), which yields $$\begin{aligned} \label{ex3.2m} q_{2}(\theta)=& [0.26475+1.24974 \theta-0.480162 \cos \theta-2.22784 \sin \theta] \frac{V_{x}}{a}\nonumber\\[-3pt] & +[0.112214-0.377118 \theta+0.401031 \cos \theta+0.480162 \sin \theta] \frac{V_{y}}{a}. \tag{m}\end{aligned}$$ Note that $q_{2}(\pi / 2)=0$, which is consistent with the vanishing of the shear flow at the top free edge. Shear flow distributions are plotted normal to the contour in figure [fig3.11].

Location of the shear center for an open cross section

The shear flow given by eq. ([eq3.98]) is determined by transverse shear forces $V_x$ and $V_y$, and is independent of the torque $M_z$. For transverse bending the shear flow $q$ is the dominate term in the expression ([eq3.40]) for the torque. Hence, the contribution of the twisting moment resultant $m_{zs}$ and the transverse stress resultant $q_z$ are neglected in eq. ([eq3.40]) with respect to the shear flow.⁴ The torque with respect to the shear center resulting from the shear flow is then $$\begin{aligned} \label{eq3.99} M_{z}=\int_{c} r_{n} q d s=\int_{c} r_{n}(s)\left[-\frac{k}{I_{y y}} V_{x} \bar{Q}_{y}(s)-\frac{k}{I_{x x}} V_{y} \bar{Q}_{x}(s)\right] d s.\end{aligned}$$ Expanding eq. ([eq3.99]), we get $$\begin{aligned} \label{eq3.100} M_{z}=-\left[\left(\frac{k}{I_{y y}}\right) \int_{c} r_{n}(s) \bar{Q}_{y}(s) d s\right] V_{x}-\left[\left(\frac{k}{I_{x x}}\right) \int_{c} r_{n}(s) \bar{Q}_{x}(s) d s\right] V_{y}.\end{aligned}$$ The contribution of the shear forces acting at the shear center to the torque in eq. ([eq3.100]) must vanish by the definition of the shear center. Thus, $$\begin{aligned} \label{eq3.101} -\left(\frac{k}{I_{y y}}\right)\left[\int_{c} r_{n}(s) \bar{Q}_{y}(s) d s \right] V_{x}-\left(\frac{k}{I_{x x}}\right)\left[\int_{c} r_{n}(s) \bar{Q}_{x}(s) d s \right] V_{y}=0 \quad \forall V_{x} \& V_{y}.\end{aligned}$$ Equation ([eq3.101]) can only be satisfied if $$\begin{aligned} \label{eq3.102} \int_{c} r_{n}(s) \bar{Q}_{y}(s) d s=0 \quad \int_{c} r_{n}(s) \bar{Q}_{x}(s) d s=0.\end{aligned}$$

To locate the shear center relative to the centroid, substitute the expression for the normal coordinate $r_{n}(s)$ from eq. ([eq3.10]) into the preceding geometric properties of the shear center to get $$\begin{gathered} \int_{c} r_{n}(s) \bar{Q}_{y}(s) d s=\int_{c} r_{n c}(s) \bar{Q}_{y}(s) d s-x_{s c} \int_{c} \bar{Q}_{y}(s) \frac{d y}{d s} d s+y_{s c} \int_{c} \bar{Q}_{y}(s) \frac{d x}{d s} d s=0\mbox{, and}\label{eq3.103}\\ \int_{c} r_{n}(s) \bar{Q}_{x}(s) d s=\int_{c} r_{n c}(s) \bar{Q}_{x}(s) d s-x_{s c} \int_{c} \bar{Q}_{x}(s) \frac{d y}{d s} d s+y_{s c} \int_{c} \bar{Q}_{x}(s) \frac{d x}{d s} d s=0.\label{eq3.104}\end{gathered}$$ With the aid of eqs. ([eq3.93]), ([eq3.84]), ([eq3.81]), and ([eq3.77]), integrate by parts the following terms in eqs. ([eq3.103]) and ([eq3.104]) to find $$\begin{aligned} \label{eq3.105} \int_{c} \bar{Q}_{y}(s)\left(\frac{d x}{d s}\right) d s=\frac{-I_{y y}}{k} \quad \int_{c} \bar{Q}_{x}(s)\left(\frac{d x}{d s}\right) d s=0 \quad \int_{c} \bar{Q}_{y}(s)\left(\frac{d y}{d s}\right) d s=0 \quad \int_{c} \bar{Q}_{x}(s)\left(\frac{d y}{d s}\right) d s=\frac{-I_{x x}}{k}.\end{aligned}$$ Substitute the results from eq. ([eq3.105]) into eqs. ([eq3.103]) and ([eq3.104]), and then solve for the coordinates of the shear center relative to the centroid as $$\begin{aligned} \label{eq3.106} \fbox{$\displaystyle x_{s c}=-\left(\frac{k}{I_{x x}}\right)\left[\int_{c} r_{n c}(s) \bar{Q}_{x}(s) d s\right] \quad y_{s c}=\frac{k}{I_{y y}}\left[\int_{c} r_{n c}(s) \bar{Q}_{y(s)} d s\right]$}.\end{aligned}$$ Note that normal coordinate $r_{n c}(s)$ is computed from the second of eq. ([eq3.11]) once the contour coordinates with respect to the centroid are established.

[ex3.3]Method 1.For the open section consisting of two branches, the coordinates of the shear center relative to the centroid from eq. ([eq3.106]) are given by $$\begin{aligned} x_{s c}=\left(-k / I_{x x}\right)\left(\int_{0}^{a} r_{n c 1} \bar{Q}_{x 1} d s_{1}+\int_{-\pi / 2}^{\pi / 2} r_{n c 2} \bar{Q}_{x 2} a d \theta\right) \quad y_{s c}=\left(k / I_{y y}\right)\left(\int_{0}^{a} r_{n c 1} \bar{Q}_{y 1} d s_{1}+\int_{-\pi / 2}^{\pi / 2} r_{n c 2} \bar{Q}_{y 2} a d \theta\right).\end{aligned}$$

Some of the terms in these formulas are listed as eqs. ([ex3.1l]) and ([ex3.1m]) in example [ex3.1] page , and eqs. ([ex3.2d]), ([ex3.2c]), (j)), and (l) in example [ex3.2] on page . We list these terms for convenience as follows: $$\begin{gathered} k=1.34781 \quad I_{x x}=3.36086 a^{3} t \quad I_{y y}=0.604984 a^{3} t,\\ \bar{Q}_{x 1}=-1.05963 a t s_{1}+0.5 t s_{1}^{2} \quad \bar{Q}_{y 1}=-0.12991 a t s_{1}-0.107764 t s_{1}^{2}\mbox{, and}\\ \bar{Q}_{x 2}=(-0.279814+0.940372 \theta-\cos \theta-1.19732 \sin \theta) a^{2} t \\ \bar{Q}_{y 2}=(-0.118837-0.560966 \theta+0.215528 \cos \theta+\sin \theta) a^{2} t.\end{gathered}$$ From eq. ([eq3.11]) the coordinates normal to the contour relative to the centroid are given by $$\begin{aligned} r_{n c 1}=x_{1}\left(\frac{d y_{1}}{d s_{1}}\right)-y_{1}\left(\frac{d x_{1}}{d s_{1}}\right) \quad r_{n c 2}=\frac{x_{2}}{a}\left(\frac{d y_{2}}{d \theta}\right)-\frac{y_{2}}{a}\left(\frac{d x_{2}}{d \theta}\right).\end{aligned}$$ The Cartesian coordinates of the contour for each branch are given by eqs. ([ex3.1j]) and ([ex3.1k]) in example [ex3.1], which are repeated below: $$\begin{gathered} \begin{array}{lll} x_{1}\left(s_{1}\right)=-0.482906 a & y_{1}\left(s_{1}\right) =-1.63782 a+s_{1} & 0 \leq s_{1} \leq a \\ x_{2}(\theta)=-0.482906 a+a \cos \theta & y_{2}(\theta) =0.36218 a+a \sin \theta & -\pi / 2 \leq \theta \leq \pi / 2\end{array}.\end{gathered}$$ The results for the coordinates normal to the contour are $$\begin{aligned} r_{n c 1}=-0.482906 a \quad r_{n c 2}=a(1-0.482906 \cos \theta+0.36218 \sin \theta).\end{aligned}$$ The shear center coordinates are determined from the following integrals $$\begin{aligned} x_{s c}=&\left(\frac{-1.34781}{3.36086 a^{3} t}\right)\left[\int_{0}^{a}(-0.482906 a)(-1.05963 a t s_{1}+0.5 t s_{1}^{2}) d s_{1}\right. \\ &+\!\!\left.\int_{-\pi / 2}^{\pi / 2}[a(1-0.482906 \cos \theta+0.36218 \sin \theta)]\left[(-0.279814+0.940372 \theta-\cos \theta-1.19732 \sin \theta) a^{2} t\right] a d \theta\right]\\ y_{s c}=&\left(\frac{1.34781}{0.604984 a^{3} t}\right)\left[\int_{0}^{a}(-0.482906 a)(-0.12991 a t s_{1}-0.107764 t s_{1}^{2}) d s_{1}\right.\\ &+\!\!\left.\int_{-\pi / 2}^{\pi / 2}\![a(1-0.482906 \cos \theta+0.36218 \sin \theta)]\! \left[(-0.118837-0.560966 \theta+0.215528 \cos \theta+\sin \theta) a^{2} t\right] a d \theta\right]\!.\end{aligned}$$ The preceding integrals were evaluated in Mathematica to get $$\begin{aligned} \label{ex3.3a} x_{s c}=0.67169 a \quad y_{s c}=0.490767 a. \tag{a}\end{aligned}$$

Method 2.The shear flow distributions were determined in example [ex3.2] on page with the results given by eq. (f) for branch 1 and by eq. (m) for branch 2. These shear flows are illustrated in the left-hand sketch in figure [fig3.12]. It is convenient to determine the resultant of the shear flow distribution at point $O$ first. As shown in figure [fig3.12] the components of the resultant force are $F_{X}$ and $F_{Y}$, and the torque is $M_{z 0}$. Under transverse bending the contributions of the transverse shear resultant $q_{z}$ and twisting moment resultant $m_{z s}$ are negligible with respect to the shear flow q in the expressions for the shear forces and the torque⁵ in eq. ([eq3.40]). Hence, the force components and the torque are given by the following integrals of the shear flow over the contour. $$\begin{aligned} \label{ex3.3b} F_{X}=\int_{-\pi / 2}^{\pi / 2} q_{2}(\theta)(-\sin \theta) a d \theta \quad F_{Y}=\int_{0}^{a} q_{1}\left(s_{1}\right) d s_{1}+\int_{-\pi / 2}^{\pi / 2} q_{2}(\theta)(\cos \theta) a d \theta \quad M_{z_{0}}=\int_{-\pi / 2}^{\pi / 2} a\left[q_{2}(\theta) a d \theta\right]. \tag{b}\end{aligned}$$ The line of action of the shear flow in branch 1 is parallel to the $Y$-direction and so does not contribute to the force component $F_{X}$ in eq. ([ex3.3b]). From example [ex3.2] substitute eq. ([ex3.2d]) for $q_{1}$ and eq. ([ex3.2h]) for $q_{2}$ into eq. ([ex3.3b]) above, then perform the integrations, to find $F_{X}=V_{x}$ and $F_{Y}=V_{y}$. It is expected that the force components would be equal to their respective transverse shear components, since the shear flows were determined from equilibrium conditions with respect to the transverse shear forces in article 1.8. Only the shear flow in branch 2 contributes to the torque about point $O$, since the line of action of the shear flow in branch 1 passes through point $O$. The moment arm to the differential force $q_{2} a d \theta$ in branch 2 about point $O$ is simply the radius $a$. From example [ex3.2] substitute eq. ([ex3.2h]) for $q_{2}$ in the expression for the torque in eq. ([ex3.3b]) above, and perform the integration to get $$\begin{aligned} \label{ex3.3c} M_{z 0}=-0.128587 a V_{x}+1.15459 a V_{y}. \tag{c}\end{aligned}$$ Now add and subtract the shear forces $V_x$ and $V_y$ at the shear center (S.C.) in order to preserve force equivalence as is shown in figure [fig3.12]. The upward force $V_y$ at point $O$ and the downward force $V_y$ at S.C. form a clockwise couple $X_{s c} V_{y}$ and no net vertical force. Similarly, force $V_x$ at point $O$ and the equal and opposite force $V_x$ at the S.C. form a counterclockwise couple $Y_{s c} V_{x}$ and no net horizontal force. The total counterclockwise torque in the cross section must vanish by the definition of the S.C.; i.e., $M_{z 0}-X_{s c} V_{y}+Y_{s c} V_{x}=0$. Substitute eq. ([ex3.3c]) for $M_{z 0}$ in the total torque to get $$\begin{aligned} \label{ex3.3d} \left(-0.128587 a+Y_{s c}\right) V_{x}+\left(1.15459 a-X_{s c}\right) V_{y}=0 \quad \forall\left(V_{x}, V_{y}\right). \tag{d}\end{aligned}$$ Therefore, the location of the shear center relative to point $O$ is given by $$\begin{aligned} \label{ex3.3e} X_{s c}=1.15459 a \quad Y_{s c}=0.128587 a. \tag{e}\end{aligned}$$ The coordinates of the shear center relative to the centroid are given by $x_{s c}=X_{s c}-X_{c}$ and $y_{s c}=Y_{s c}-Y_{c}$, where the coordinates of the centroid relative to point $O$ is given by eq. ([ex3.1i]) in example [ex3.1] on page . Thus, $$\begin{aligned} \label{ex3.3f} x_{s c}=1.15459 a-0.48290 a=0.67169 a \quad y_{s c}=0.128587 a-(-0.36218 a)=0.490767 a, \tag{f}\end{aligned}$$ which is the same result obtained in eq. ([ex3.3a]) by method 1.

Notes concerning the shear center

• The resultant of the shear flow distribution over the contour is a force with components $V_x$ and $V_y$ acting through the shear center such that there is no torque acting at the shear center. If the cross section is subject to a torque, this torque cannot be balanced by the shear flow which, according to eq. ([eq3.98]), is uniquely determined by the shear forces $V_x$ and $V_y$.

• The location of the shear center in the cross section is determined by the pattern of the shear flow distribution and not on the magnitude of the transverse shear forces.

• Transverse shear forces $V_x$ and $V_y$ act in the plane of loading to equilibrate the externally applied lateral load intensities $f_{x}(z)$ and $f_{y}(z)$. (Refer to equilibrium equations ([eq3.54]) and ([eq3.56]).) Thus, the line of action of the external lateral loads must pass through the shear center to bend the bar without twisting it in torsion.

• The shear center is located on an axis of symmetry of the cross section if there is one. If there are two axes of symmetry in the cross section the shear center and the centroid lie on the intersection of the symmetry axes.

• For an open cross section with straight branches and one junction the shear center is at the junction, since the torque from the shear flows at the junction vanishes. See figure 1.2.

Displacements and strains.

Saint-Venant assumed that as the bar twists the cross section is displaced normal to the $s$-$\zeta$ plane (i.e., it warps) but its projection on the $s$-$\zeta$ plane rotates as a rigid body. To prevent rigid body displacement, the displacement components of the centroid are set equal to zero. Then, the in-plane displacements given by eqs. ([eq3.20]) and ([eq3.22]) reduce to $$\begin{aligned} \label{eq3.107} u_{\zeta}(s, z, \zeta)=-s \phi_{z}(z) \quad u_{s}(s, z, \zeta)=\zeta \phi_{z}(z).\end{aligned}$$ The out-of-plane displacement given by eq. ([eq3.26]) is $$\begin{aligned} \label{eq3.108} u_{z}(s, z, \zeta)=-s \phi_{y}(z)+\zeta\left[\phi_{x}(z)\right].\end{aligned}$$ However, this out-of-plane displacement is changed to account for the warping of the cross section in uniform torsion. It is assumed that the rotation about the $\textbf{\textit{x}}$-axis, or the negative $\textbf{\textit{s}}$-axis, is independent of the $\textbf{\textit{z}}$-coordinate but is an unknown function of the $\textbf{\textit{s}}$-coordinate. Consequently, the out-of-plane displacement in eq. ([eq3.108]) is changed to $$\begin{aligned} \label{eq3.109} u_{z}(s, z, \zeta)=\zeta \phi_{x}(s)-s \phi_{y}(z).\end{aligned}$$

The only non-zero strains determined from the displacements ([eq3.107]) and ([eq3.109]) are shear strains $\gamma_{z s}$ and $\gamma_{z \zeta}$, From the strain-displacement relations given by eq. ([eq3.28]) these shear strain-displacement relations are $$\begin{aligned} \label{eq3.110} \gamma_{z s}=\left(\frac{d \phi_{z}}{d z}+\frac{d \phi_{x}}{d s}\right) \zeta-\phi_{y} \quad \gamma_{z \zeta}=\phi_{x}-s \frac{d \phi_{z}}{d z}.\end{aligned}$$ Hooke’s laws for the shear stresses are $\sigma_{z s}=G \gamma_{z s}$ and $\sigma_{z \zeta}=G \gamma_{z \zeta}$.

Stress resultants and equilibrium.

The stress resultants associated with these non-zero shear strains ([eq3.110]) are determined from Hooke’s law, and the definition of stress resultants $q$, $m_{z s}$, and $q_{z}$ in eq. ([eq3.37]). The expressions for the stress resultants are $$\begin{aligned} \label{eq3.111} &\left(q, m_{z s}\right)=\hspace*{-3pt}\int_{-t / 2}^{t / 2}(1, \zeta) \sigma_{z s} d \zeta=G\hspace*{-3pt} \int_{-t / 2}^{t / 2}(1, \zeta)\left[\left(\frac{d \phi_{z}}{d z}+\frac{d \phi_{x}}{d s}\right) \zeta-\phi_{y}\right] d \zeta \nonumber\\ &q_{z}=\int_{-t / 2}^{t / 2}\hspace*{-4pt} \sigma_{z \zeta} d \zeta=G \int_{-t / 2}^{t / 2}\left(\phi_{x}-s \frac{d \phi_{z}}{d z}\right) d \zeta.\end{aligned}$$ Performing the integrations through the thickness in eq. ([eq3.111]) we find the stress resultants are given by $$\begin{aligned} \label{eq3.112} q=-G t \phi_{y} \quad m_{z s}=\frac{G t^{3}}{12}\left(\frac{d \phi_{z}}{d z}+\frac{d \phi_{x}}{d s}\right) \quad q_{z}=G t\left(\phi_{x}-s \frac{d \phi_{z}}{d z}\right).\end{aligned}$$

Since $V_{x}=V_{y}=q_{T}=0$, the shear flow from eq. ([eq3.97]) reduces to $q(s, z)=q_{0}(z)$. That is, the shear flow is spatially uniform in the $s$-coordinate. Furthermore, the longitudinal edges at $s=\pm b / 2$ are free of tractions, which means the shear flow vanishes. Hence, $q=0$ for $-b / 2 \leq s \leq b / 2$. It follows from the first equation in ([eq3.112]) that rotation $\phi_{y}=0$.

R196pt

The twisting moment resultant $m_{s z}$ and the transverse shear resultant $q_{z}$ are related by moment equilibrium about the $s$-axis for a differential element ds by dz cut from the wall. From the free body diagram for differential element of the wall shown in figure [fig3.15] moment equilibrium gives $$\begin{aligned} &\left.\frac{d z}{2}\left(q_{z} d s\right)\right|_{z+d z}\!\!+\left.\frac{d z}{2}\left(q_{z} d s\right)\right|_{z}\!-[\left.(m_{s z} d z)\right|_{s+d s}-\left.(m_{s z} d z)\right|_{s}] \nonumber\\ &\quad =0. \label{eq3.113}\end{aligned}$$ Division of eq. ([eq3.113]) by area element ds by dz, followed by the limit as $d s \rightarrow 0$ and $d z \rightarrow 0$ yields the moment equilibrium differential equation $$\begin{aligned} \label{eq3.114} q_{z}-\frac{d m_{s z}}{d s}=0.\end{aligned}$$

Governing boundary value problem.

Substitute $m_{z s}$ from eq. ([eq3.112]) into the differential equation ([eq3.114]), followed by substitution of $q_{z}$ from eq. ([eq3.112]) into ([eq3.114]). After these substitutions and re-arrangement, the result is $$\begin{aligned} \label{eq3.115} \frac{d^{2} \phi_{x}}{d s^{2}}-\frac{12}{t^{2}} \phi_{x}=-\left(\frac{12}{t^{2}} \frac{d \phi_{z}}{d z}\right) s \quad \phi_{x}=\phi_{x}(s) \quad-b / 2<s<b / 2.\end{aligned}$$ The longitudinal edges at $s=\pm b / 2$ are free of tractions, which additionally means the twisting moment $m_{s z}$ vanishes at $s=\pm b / 2$. From eq. ([eq3.112]) the vanishing of the twisting moment at the end points leads to the boundary conditions $$\begin{aligned} \label{eq3.116} \frac{d \phi_{z}}{d z}+\frac{d \phi_{x}}{d s}=0 \quad~\text{at } s=\pm b / 2.\end{aligned}$$ The solution to differential equation ([eq3.115]) subject to boundary conditions ([eq3.116]) is $$\begin{aligned} \label{eq3.117} \phi_{x}(s)=\left[s-\frac{2}{k} \frac{\sinh k s}{\cosh \lambda}\right] \frac{d \phi_{z}}{d z},\end{aligned}$$

where $$\begin{aligned} \label{eq3.118} k=\frac{2 \sqrt{3}}{t} \quad \lambda=\frac{k b}{2}=\sqrt{3}\left(\frac{b}{t}\right).\end{aligned}$$

Substitute the solution for $\phi_{x}(s)$ from ([eq3.117]) into the expressions for the twisting moment $m_{z s}$ and transverse shear $q_{z}$ listed in ([eq3.112]) to find $$\begin{aligned} \label{eq3.119} m_{z s}=\frac{G t^{3}}{6}\left(1-\frac{\cosh k s}{\cosh \lambda}\right)\left(\frac{d \phi_{z}}{d z}\right) \quad q_{z}=-2 G t \frac{\sinh k s}{k \cosh \lambda}\left(\frac{d \phi_{z}}{d z}\right).\end{aligned}$$

From the third expression in eq. ([eq3.40]) the torque about the $z$-axis, counterclockwise positive, is given by $$\begin{aligned} \label{eq3.120} M_{z}=\int_{-b / 2}^{b / 2}\left(m_{z s}-s q_{z}\right) d s.\end{aligned}$$ Substituting the results for $m_{z s}$ and $q_{z}$ from eq. ([eq3.119]) into the expression for the torque we write the result as $$\begin{aligned} \label{eq3.121} M_{z}=G J\left(\frac{d \phi_{z}}{d z}\right),\end{aligned}$$ where the torsion constant $J$ is given by the integral $$\begin{aligned} J=\int_{-b / 2}^{b / 2}\left[\frac{t^{3}}{6}\left(1-\frac{\cosh (k s)}{\cosh \lambda}\right)+2 t s\left(\frac{\sinh (k s)}{k \cosh \lambda}\right)\right] d s.\end{aligned}$$ After performing the integration, the result for the torsion constant is $$\begin{aligned} \label{eq3.122} J=\frac{b t^{3}}{3}\left(1-\frac{\tanh \lambda}{\lambda}\right).\end{aligned}$$ For a thin, elongated rectangular cross section the value of the ratio of $b / t \gg 1$, which from the expression for $\lambda$ in eq. ([eq3.118]) implies $\lambda \gg 1$. In the limiting case of $\lambda \rightarrow \infty$ we find $$\begin{aligned} \label{eq3.123} J=\frac{b t^{3}}{3} \quad~\text{as } \quad \lambda \rightarrow \infty.\end{aligned}$$ In the simplified theory of thin-walled open section bars, the torsion constant in each open branch is given by eq. ([eq3.123]).

The distribution of the twisting moment resultant $m_{z s}$ over the length of the contour for $b/t = 20$ is shown in figure [fig3.16]. As shown in the plot the distribution of $m_{z s}$ is symmetric with respect to the contour coordinate, attains a uniform magnitude over the majority of the contour, and decreases rapidly to zero near the boundaries of the contour where $s=\pm b / 2$. The distribution of the transverse shear resultant $q_{z}$ over the contour is shown in figure [fig3.17]. The distribution of $q_{z}$ is antisymmetric with respect to the contour coordinate, it is essentially equal to zero over the majority of the contour, and its maximum magnitude occurs in the narrow boundaries of the contour at $s=\pm b / 2$.

From Hooke’s law and eq. ([eq3.110]), the shear stress component tangent to the contour is $$\begin{aligned} \sigma_{z s}=G \gamma_{z s}=G\left(\frac{d \phi_{z}}{d z}+\frac{d \phi_{x}}{d s}\right) \zeta.\end{aligned}$$ Substitute the solution for $\phi_{x}$ from eq. ([eq3.117]) into the previous equation to get $$\begin{aligned} \label{eq3.124} \sigma_{z s}(s, \zeta)=2 G\left(\frac{d \phi_{z}}{d z}\right)\left(1-\frac{\cosh (k s)}{\cosh \lambda}\right) \zeta.\end{aligned}$$ Note that this shear stress vanishes on the contour where $\zeta=0$ and attains its maximum magnitude along the top and bottom edges where $\zeta=\pm t / 2$. Shear stress $\sigma_{Z s}$ is the dominate shear stress in the rectangular cross section subject to uniform torsion, since through-the-thickness shear stress $\sigma_{z \zeta}=q_{z} / t$ is essentially zero over most of the contour. For large values of $b/t$, we neglect the shear stress $\sigma_{z \zeta}$ with respect to $\sigma_{z s}$, and we use the following approximation $$\begin{aligned} \label{eq3.125} \sigma_{z s}=2 G\left(\frac{d \phi_{z}}{d z}\right) \zeta \quad b / t \gg 1.\end{aligned}$$

Warping of the cross section.

Substitute eq. ([eq3.117]) for $\phi_{x}(s)$ in eq. ([eq3.109]), and recall that $\phi_{y}=0$, to find that the warping displacement $u_{z}(s, \zeta)$ is given by $$\begin{aligned} \label{eq3.126} u_{z}(s, \zeta)=\zeta\left[s-\frac{t}{\sqrt{3}} \frac{\sinh k s}{\cosh \lambda}\right]\left(\frac{d \phi_{z}}{d z}\right).\end{aligned}$$ A contour plot of the warping displacement divided by $u_{z}(b / 2, t / 2)$ for $b / t=4$ and $\frac{d \phi_{z}}{d z}>0$ is shown in figure [fig3.18], where $$\begin{aligned} u_{z}(b / 2, t / 2)=0.7113 t^{2}\left(\frac{d \phi_{z}}{d z}\right) \quad b / t=4.\end{aligned}$$ Along the $s$-axis and the $\zeta$-axis the warping displacement is zero, and it attains maximum magnitude near the corners of the rectangular cross section. For a positive unit twist, $u_{z}>0$ if the product $s \zeta>0$, and $u_{z}<0$ if the product $s \zeta<0$.

For a thin rectangular cross section a good approximation to the warping function is $$\begin{aligned} \label{eq3.127} u_{z} \approx u_{z a}=s \zeta \quad b / t \gg 1.\end{aligned}$$ To show eq. ([eq3.127]) is a good approximation, let $$\begin{aligned} I_{e}=\left(\int_{-t / 2}^{t/2}\int_{-b / 2}^{b/2}\left[u_{z}(s, \zeta)\right]^{2} d s d \zeta\right)^{1 / 2}\mbox{ and } I_{a}=\left(\int_{-t / 2}^{t/2}\int_{-b / 2}^{b/2}\left[u_{z a}(s, \zeta)\right]^{2} d s d \zeta\right)^{1 / 2}.\end{aligned}$$ Define the percentage error between the approximate warping function and the exact one by $\textrm{error} = (I_{a}-I_{e})\break 100 / I_{e}$. For $b / t=20$ the error is 0.482 percent, and for $b / t=40$ the error is 0.123 percent.

Torsion of built-up open sections

For large values of the ratio of b/t, the analysis of thin-walled rectangular section of article 1.9 results in the following formulas given by eqs ([eq3.121]), ([eq3.123]), and ([eq3.125]): $$\begin{aligned} \label{eq3.128} M_{z}=G J\left(\frac{d \phi_{z}}{d z}\right) \quad J=\frac{b t^{3}}{3} \quad \sigma_{z s}=2 G\left(\frac{d \phi_{z}}{d z}\right) \zeta.\end{aligned}$$ The maximum magnitude of the shear stress $\sigma_{z s}$ occurs at $\zeta=\pm t / 2$. Then from the previous equations for large values of the ratio of b/t this maximum shear stress can be expressed as $$\begin{aligned} \label{eq3.129} \left.\sigma_{z s}\right|_{\max }=\frac{3 M_{z}}{b t^{2}}.\end{aligned}$$

Now consider torsion of open section bars of more complex shape as are shown in figure [fig3.19]. Understanding the torsional response of these bars with complex, open cross-sectional shapes is facilitated by an analogy to the response of an initially flat membrane supported on its edges over an opening, where the edges of the opening are in the same shape as the cross section. The membrane is stretched under a uniform tension, and then subject to an internal pressure to cause the membrane to deflect. The deflected shape of this pressurized membrane is analogous to the torsion problem in that level contours on the surface of the deflected membrane coincide with the lines of action of the shear stresses, and that the slope of the membrane normal to the level contour is proportional to the magnitude to the shear stress. Also, the volume between the x-y plane and the deflected surface of the membrane is proportional to the total torque carried by the section. The following text excerpted from Oden and Ripperger (1981, p. 46) summarizes this analogy.

This analogy was first discovered by Ludwig Prandtl in 1903 and is known as Prandtl’s membrane analogy. Prandtl took full advantage of the analogy and devised clever experiments with membranes. By measuring the volumes under membranes formed by a soap film subject to a known pressure, he was able to evaluate torsional constants. By obtaining the contour lines of the membranes he determined stress distributions.

Torsional constants and the maximum shearing stress can be found for complex cross sections by using the results for the thin-walled rectangular section. The membrane analogy shows that the torsional load carrying capacity of the complex open section is nearly the same as the narrow rectangular section, because the volumes under the membranes are nearly the same if we neglect the small error introduced at the corners or junctions. In this way, the membrane analogy implies that the complex open cross section has about the same torsional load carrying capacity as a thin-walled rectangular section with a length equal to the total arc length of the contour of the complex section.

Since each branch of the open section is equivalent to a narrow rectangular section with the same developed length and thickness, we can sum the torques carried by each branch in the following way \[\begin{aligned} \label{eq3.130} M_{z}=\sum_

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Effect of stringers on the shear flow distribution

The shear flow exiting the stringer location is denoted by $q^{(+)}$, the shear flow entering the stringer location by $q^{(-)}$, and the increase in the axial force in the stringer by $\Delta N_{f}$. See figure [fig3.20]. Sum the forces in the $z$-direction of the free body diagram shown in figure [fig3.20] to get $$\begin{aligned} q\big(s_{f}^{(+)}\big) \Delta z-q\big(s_{f}^{(-)}\big) \Delta z+\Delta N_{f}=0.\end{aligned}$$ Divide this equilibrium equation by the incremental length $\Delta z$, then let $\Delta z \rightarrow 0$ to get in the limit $$\begin{aligned} \label{eq3.135} q\left(s_{f}^{(+)}\right)-q\left(s_{f}^{(-)}\right)+\frac{d N_{f}}{d z}=0.\end{aligned}$$ Combine eqs. ([eq3.95]) and ([eq3.97]) to write the shear flow as $$\begin{aligned} \label{eq3.136} q(s)=q_{0}-\frac{k}{I_{y y}}\left(V_{x}+V_{x T}\right) \bar{Q}_{y}(s)-\frac{k}{I_{x x}}\left(V_{y}+V_{y T}\right) \bar{Q}_{x}(s)-\frac{A(s)}{A} \frac{d N_{T}}{d z}+\int_{0}^{s} \beta \frac{\partial \Delta T}{\partial z} t(s) d s.\end{aligned}$$ Equation ([eq3.89]) was used to identify the derivative of the thermal force in the previous result. Then the jump in the shear flow across the stringer is $$\begin{aligned} q\left(s_{f}^{(+)}\right)-q\left(s_{f}^{(-)}\right)=&-\left\{\frac{k}{I_{y y}}\left(V_{x}+V_{x T}\right)\left[\bar{Q}_{y}\left(s_{f}^{(+)}\right)-\bar{Q}_{y}\left(s_{f}^{(-)}\right)\right]\right\}-\left\{\frac{k}{I_{x x}}\left(V_{y}+V_{y T}\right)\left[\bar{Q}_{x}\left(s_{f}^{(+)}\right)-\bar{Q}_{x}\left(s_{f}^{(-)}\right)\right]\right\} \nonumber\\[6pt] &-\frac{\left[A\left(s_{f}^{(+)}\right)-A\left(s_{f}^{(-)}\right)\right]}{A} \frac{d N_{T}}{d z}+\left(\int_{s_{f}^{(-)}}^{s_{f}^{(+)}} \beta \frac{\partial \Delta T}{\partial z} t(s) d s\right) .\label{eq3.137}\end{aligned}$$ Note that $$\begin{aligned} \label{eq3.138} A\left(s_{f}^{(+)}\right)-A\left(s_{f}^{(-)}\right)=A_{f} \quad~\text{and } \quad \int_{s_{f}^{(-)}}^{s_{f}^{(+)}} \beta \frac{\partial \Delta T}{\partial z} t(s) d s=\left.A_{f} \beta \frac{\partial \Delta T}{\partial z}\right|_{s_{f}}.\end{aligned}$$ Under the assumption made to model the stringer $N_{f}=\sigma_{z z} A_{f}$. The axial normal stress in the stringer is given in eq. ([eq3.83]) on page . Thus, $\frac{d N_{f}}{d z}=A_{f} \frac{d \sigma_{z z}}{d z}$. Derivatives of the axial force and bending moments with respect to z appearing in $d \sigma_{z z} / d z$ were replaced by equilibrium differential equations ([eq3.53]), ([eq3.55]), and ([eq3.57]), respectively. The result for the derivative of the normal force in the stringer is $$\begin{aligned} \label{eq3.139} \frac{d N_{f}}{d z}=\frac{k}{I_{y y}}\left(V_{x}+V_{x T}\right) A_{f} \bar{x}_{f}+\frac{k}{I_{x x}}\left(V_{y}+V_{y T}\right) A_{f_{f}} \bar{y}_{f}+\frac{A_{f}}{A} \frac{d N_{T}}{d z}-\left.A_{f} \beta \frac{\partial \Delta T}{\partial z}\right|_{s_{f}}.\end{aligned}$$ Substitute eqs. ([eq3.137]) and ([eq3.139]) into ([eq3.135]) to find $$\begin{aligned} \label{eq3.140} -\left\{\frac{k}{I_{y y}}\left(V_{x}+V_{x T}\right)\left[\bar{Q}_{y}\left(s_{f}^{(+)}\right)-\bar{Q}_{y}\left(s_{f}^{(-)}\right)-A_{f} \bar{x}_{f}\right]\right\}&-\left\{\frac{k}{I_{x x}}\left(V_{y}+V_{y T}\right)\left[\bar{Q}_{x}\left(s_{f}^{(+)}\right)-\bar{Q}_{x}\left(s_{f}^{(-)}\right)-A_{f} \bar{y}_{f}\right]\right\} \nonumber\\[6pt] &-\frac{A_{f}}{A} \frac{d N_{T}}{d z}+\frac{A_{f}}{A} \frac{d N_{T}}{d z}+\left.A_{f} \beta \frac{\partial \Delta T}{\partial z}\right|_{s_{f}}-\left.A_{f} \beta \frac{\partial \Delta T}{\partial z}\right|_{s_{f}}=0.\end{aligned}$$ Equation ([eq3.140]) simplifies to $$\begin{aligned} \label{eq3.141} -\left\{\frac{k}{I_{y y}}\left(V_{x}+V_{x T}\right)\left[\bar{Q}_{y}\left(s_{f}^{(+)}\right)-\bar{Q}_{y}\left(s_{f}^{(-)}\right)-A_{f} \bar{x}_{f}\right]\right\}-\left\{\frac{k}{I_{x x}}\left(V_{y}+V_{y T}\right)\left[\bar{Q}_{x}\left(s_{f}^{(+)}\right)-\bar{Q}_{x}\left(s_{f}^{(-)}\right)-A_{f} \bar{y}_{f}\right]\right\}=0.\end{aligned}$$

Equation ([eq3.141]) is valid for every choice of the shear actions $\left(V_{x}+V_{x T}\right)$ and $\left(V_{y}+V_{y T}\right)$. Then to satisfy eq. ([eq3.141]), the coefficients of the shear actions must vanish, which leads to $$\begin{aligned} \label{eq3.142} \bar{Q}_{y}\left(s_{f}^{(+)}\right)-\bar{Q}_{y}\left(s_{f}^{(-)}\right)-A_{f} \bar{x}_{f}=0 \quad \bar{Q}_{x}\left(s_{f}^{(+)}\right)-\bar{Q}_{x}\left(s_{f}^{(-)}\right)-A_{f} \bar{y}_{f}=0.\end{aligned}$$ Relations ([eq3.84]) and ([eq3.93]) evaluated at $s=s_{f}$ are repeated in the following relations: $$\begin{aligned} \label{eq3.143} \bar{x}_{f}=x_{f}-n_{x} y_{f} \quad \bar{y}_{f}=y_{f} - n_{y} x_{f} \quad \bar{Q}_{x}\left(s_{f}^{-}\right)=Q_{x}\left(s_{f}^{-}\right)-n_{y} Q_{y}\left(s_{f}^{-}\right) \quad \bar{Q}_{y}\left(s_{f}^{-}\right)=Q_{y}\left(s_{f}^{-}\right)-n_{x} Q_{x}\left(s_{f}^{-}\right).\end{aligned}$$ After substituting the relations ([eq3.143]) into eq. ([eq3.142]) we get $$\begin{aligned} \label{eq3.144} Q_{y}\left(s_{f}^{(+)}\right)-Q_{y}\left(s_{f}^{(-)}\right)-A_{f} x_{f}=0, \quad Q_{x}\left(s_{f}^{(+)}\right)-Q_{x}\left(s_{f}^{(-)}\right)-A_{f} y_{f}=0.\end{aligned}$$ Equation ([eq3.144]) shows that the jump in the shear flows exiting and entering the stringer ([eq3.135]) is equivalent to a jump in value of the first area moments across the stringer area.

Twist per unit longitudinal length

The shear strain $\gamma_{z s}$ evaluated at the contour from eq. ([eq3.31]) on page is $$\begin{aligned} \label{eq3.152} \gamma_{z s}(s, z)=\psi_{x}(z) \frac{d x}{d s}+\psi_{y}(z) \frac{d y}{d s}+r_{n}(s) \frac{d \phi_{z}}{d z} \quad~\text{at } \quad \zeta=0,\end{aligned}$$

where eq. ([eq3.3]) was used to write the trigonometric functions in terms of the derivatives of the contour coordinate functions. Integrate the shear strain ([eq3.152]) around the closed contour to get $$\begin{aligned} \label{eq3.153} \oint\!\gamma_{z s} d s=\psi_{x}(z) \underbrace{\oint{\frac{d x}{d s}} d s}_{=\;0}+\psi_{y}(z) \underbrace{\oint{\frac{d y}{d s} d s}}_{=\;0} +\underbrace{\oint r_{n} d s}_{=\;2 A_{c}} \frac{d \phi_{z}}{d z}.\end{aligned}$$

Continuity of the contour and the eq. ([eq3.148]) for the enclosed area results in $$\begin{aligned} \frac{d \phi_{z}}{d z}=\frac{1}{2 A_{c}} \oint \gamma_{z s} d s.\end{aligned}$$

Hooke’s law relates the shear strain to the shear stress by $\gamma_{z s}=\sigma_{z s} / G$, where $G$ is the shear modulus of the material. In torsion of a closed cross-sectional contour the shear stress is assumed uniform through the thickness of the wall. Hence, the shear stress is determined by the shear flow divided by the thickness of the wall, or $\sigma_{z s}=q / t$. Substitute $\gamma_{z s}=q /(G t)$ into the equation for the twist per unit length to get. $$\begin{aligned} \label{eq3.154} \fbox{$\displaystyle\frac{d \phi_{z}}{d z}=\frac{1}{2 A_{c}} \oint\left(\frac{q}{G t}\right) d s$}.\end{aligned}$$

Location of the shear center and the final expression for the shear flow

Substitute eq. ([eq3.150]) for the shear flow in eq. ([eq3.154]) to find $$\begin{aligned} \label{eq3.155} \frac{d \phi_{z}}{d z}=\frac{M_{z C}}{4 A_{c}^{2}} \oint \frac{d s}{G t}-\frac{V_{x}}{2 A_{c}} \oint \frac{F_{x c}}{G t} d s-\frac{V_{y}}{2 A_{c}} \oint \frac{F_{y c}}{G t} d s.\end{aligned}$$ The torque $M_{zc}$ and shear forces $V_x$ and $V_y$ are resolved at the centroid. We can find a statically equivalent torque and force system resolved at the shear center: Simply add and subtract the shear forces at the shear center which does not change static equivalence as shown in figure [fig3.23](a). The upward force $V_y$ at point $C$ and the downward force $V_y$ at S.C. form a clockwise couple $x_{sc} V_{y}$ and no net vertical force. Similarly, rightward force $V_x$ at point $C$ and the leftward force $V_x$ at the S.C. form a counterclockwise couple $y_{sc} V_{x}$ and no net horizontal force. The total counterclockwise torque in the cross section is $M_{z c}-x_{s c} V_{y}+y_{s c} V_{x}$. Formulating these couples leave forces $V_x$ and $V_y$ at the shear center as shown in part (b) of figure [fig3.23] and a counterclockwise torque. Thus, the torque at the shear center must be $$\begin{aligned} \label{eq3.156} M_{z}=M_{z C}-x_{s c} V_{y}+y_{s c} V_{x}.\end{aligned}$$

Solve eq. ([eq3.156]) for the torque at the centroid and substitute the result for $M_{zC}$ into eq. ([eq3.155]) to get $$\begin{aligned} \label{eq3.157} \frac{d \phi_{z}}{d z}=\frac{M_{z}}{4 A_{c}^{2}} \oint \frac{d s}{G t}-\left[\frac{y_{s c}}{4 A_{c}^{2}} \oint \frac{d s}{G t}+\frac{1}{2 A_{c}} \oint \frac{F_{x c}}{G t} d s\right] V_{x}+\left[\frac{x_{s c}}{4 A_{c}^{2}} \oint \frac{d s}{G t}-\frac{1}{2 A_{c}} \oint \frac{F_{y c}}{G t} d s\right] V_{y}.\end{aligned}$$ At the shear center the twist per unit length depends on the torque resolved at the shear center and not on the shear forces. In other words, the shear forces acting at the shear center do not contribute to torsion. This requirement means the coefficients of the shear forces in eq. ([eq3.157]) must vanish. Equating these coefficients to zero determines the coordinates of the shear center relative to the centroid as $$\begin{aligned} \label{eq3.158} x_{s c}=\left[\frac{2 A_{c}}{\oint \frac{d s}{G t}} \oint\left(\frac{F_{y c}(s)}{G t}\right) d s\right] \quad y_{s c}=-\left[\frac{2 A_{c}}{\oint \frac{d s}{G t}} \oint\left(\frac{F_{x c}(s)}{G t}\right) d s\right].\end{aligned}$$ Equation ([eq3.157]) reduces to the form $$\begin{aligned} \label{eq3.159} \frac{d \phi_{z}}{d z}=\frac{M_{z}}{(G J)_{\mathrm{eff}}},\end{aligned}$$ where the effective torsional stiffness is $$\begin{aligned} \label{eq3.160} (G J)_{\mathrm{eff}}=\frac{4 A_{c}^{2}}{\oint \frac{d s}{G t}}.\end{aligned}$$ If the shear modulus is uniform around the contour, then $$\begin{aligned} \label{eq3.161} \frac{d \phi_{z}}{d z}=\frac{M_{z}}{G J}\mbox{ and the torsion constant is }J=\frac{4 A_{c}^{2}}{\oint \frac{d s}{t}}.\end{aligned}$$

Substitute the solution for the torque at the centroid from eq. ([eq3.156]) into the expression for the shear flow in eq. ([eq3.150]). In the process, we drop the “C” subscript on $q_C$ to indicate that we are formulating the shear flow relative to the shear center. The result is $$\begin{aligned} \label{eq3.162} q=\frac{M_{z}+x_{s c} V_{y}-y_{s c} V_{x}}{2 A_{c}}-F_{x c}(s) V_{x}(z)-F_{y c}(s) V_{y}(z).\end{aligned}$$ Equation ([eq3.162]) is written in the form $$\begin{aligned} \label{eq3.163} \fbox{$\displaystyle q(s, z)=\frac{M_{z}(z)}{2 A_{c}}-F_{x}(s) V_{x}(z)-F_{y}(s) V_{y}(z)$},\end{aligned}$$ where the shear flow distribution functions relative to the shear center are defined by $$\begin{aligned} \label{eq3.164} F_{x}(s)=\frac{y_{s c}}{2 A_{c}}+F_{x c}(s) \quad F_{y}(s)=-\left(\frac{x_{s c}}{2 A_{c}}\right)+F_{y c}(s).\end{aligned}$$

In pure torsion only torque $M_z$ acts on the section, and the shear terms in eq. ([eq3.163]) vanish. Then in pure torsion the shear flow is spatially uniform around the contour and leads to $$\begin{aligned} \label{eq3.165} q=\frac{M_{z}}{2 A_{c}} \quad~\text{or } \quad M_{z}=2 A_{c} q.\end{aligned}$$ Equation ([eq3.165]) is called Bredt’s formula, or the Bredt-Batho formula, and it relates the torque to the uniform shear flow in a single-cell section subject to torsion only.

[ex3.4]A uniform bar of length $L$ with a closed, cross-sectional contour is stiffened by four axial stringers. The configuration and the associated nomenclature is shown in figure [fig3.24](a), where the $X$-axis is an axis of symmetry. The areas of the stringer flanges are denoted by $A_{f1}$ and $A_{f2}$, and the wall thickness $t$ is uniform along the entire contour. As shown in figure [fig3.24](b), the contour is divided into four branches. Branch 1 is a semicircle segment of radius a between the lower stringer $A_{f1}$ and the upper stringer $A_{f1}$ of length $a\pi$, branch 2 is the horizontalsegment between upper stringer $A_{f1}$ and upper stringer $A_{f2}$ of length $b$, branch 3 is the vertical segmentbetween the upper stringer $A_{f2}$ and the lower stringer $A_{f2}$ of length 2$a$, and branch 4 is the horizontal segment between lower stringer $A_{f2}$ and lower stringer $A_{f1}$ of length $b$. Dimensional data are $$\begin{aligned} a=6~\text{in. } \quad b=7~\text{in. } \quad t=0.03~\text{in.} \quad A_{f 1}=0.30~\text{in.}^{2} \quad~\text{and } \quad A_{f 2}=0.70~\text{in.}^{2}\end{aligned}$$ The numerical results presented in the solution of this example were performed in Mathematica.

1. Determine the location of the centroid (C) and the second area moments $I_{x x}$ and $I_{y y}$.

2. Determine the shear flow distribution functions $F_{x c}(s)$ and $F_{y c}(s)$ with respect to the centroid.

3. Determine the location of the shear center (S.C.) relative to the centroid.