Extension and bending

Hooke’s law for extension and bending of the bar is defined relative to the centroid. From eq. ([eq3.80]) on page  the compliance form of Hooke’s law is \[\begin{aligned} \label{eq4.1} \left[\begin{array}{@{}l@{}}d w/d z \\d \phi_{x}/d z \\d \phi_{y}/d z\end{array}\right]=\frac{1}{E}\left[\begin{array}{@{}ccc@{}}1/A & 0 & 0 \\0 & k/I_{x x} & \left(-k n_{x}\right)/I_{y y} \\0 & \left(-k n_{y}\right)/I_{x x} & k/I_{y y}\end{array}\right]\left[\begin{array}{@{}c@{}}N+N_{T} \\M_{x}+M_{x T} \\M_{y}+M_{y T}\end{array}\right].\end{aligned}\] Geometric properties of the cross section are its area A, its first area moments \(Q_{x}\) and \(Q_{y}\), and its second area moments \(I_{xx}, I_{yy}\), and \(I_{xy}\). In eq. ([eq4.1]) the modulus of elasticity of the material is denoted by E. The locus of points on the contour is expressed parametrically by the equations \(x(s)\) and \(y(s)\), where the arc-length of the contrary is denoted by \(s\). Let \(t(s)\) denote the thickness of the contour. See Fig. [fig4.1]. The area and first area moments are given by \[\begin{aligned} \label{eq4.2} A=\int_{c} t(s) d s \quad Q_{x}=\int_{c} y(s) t(s) d s=0 \quad Q_{y}=\int_{c} x(s) t(s) d s=0.\end{aligned}\] First area moments \(Q_{x}\) and \(Q_{y}\) vanish since origin of the x-y-axes is located at the centroid of the cross section. Hence, the definition of the centroid allows decoupling of the extension and bending responses of the bar. That is, the axial strain \(d w/d z\) is independent of the bending moments \(M_{x}\) and \(M_{y}\), and bending rotation gradients \(d \phi_{x}/d z\), and \(d \phi_{y}/d z\) are independent of axial force \(N\). The second area moments are given by, \[\begin{aligned} \label{eq4.3} I_{x x}=\int_{c} y^{2}(s) t(s) d s \quad I_{y y}=\int_{c} x^{2}(s) t(s) d s \quad I_{x y}=\int_{c} x(s) y(s) t(s) ds.\end{aligned}\] The dimensionless parameters are defined by \[\begin{aligned} \label{eq4.4} n_{x}=I_{x y}/I_{x x} \quad n_{y}=I_{x y}/I_{y y} \quad k=\frac{1}{1-n_{x} n_{y}}.\end{aligned}\] The thermal loads \(N_{T}\), \(M_{x T}\), and \(M_{y T}\) appearing in eq. ([eq4.1]) are from the prescribed change in temperature from the reference state. Refer to eqs. ([eq3.75]), and ([eq3.78]) on page .

The axial normal stress \(\sigma_{z z}\) and the shear stress tangent to the contour \(\sigma_{z s}\) are shown in figure [fig4.2]. The axial normal strain \(\varepsilon_{z z}\) and the axial normal stress \(\sigma_{z z}\) are given by eqs. ([eq3.82]) and ([eq3.83]) on page , respectively. These results are repeated below as eqs. ([eq4.5]) and ([eq4.6]), respectively. \[\begin{aligned} \label{eq4.5} \varepsilon_{z z}=\frac{N+N_{T}}{E A}+k \frac{\left(M_{x}+M_{x T}\right)}{E I_{x x}} \bar{y}(s)+k \frac{\left(M_{y}+M_{y T}\right)}{E I_{y y}} \bar{x}(s),\ \text{and}\\ \label{eq4.6} \sigma_{z z}=\frac{N+N_{T}}{A}+k \frac{\left(M_{x}+M_{x T}\right)}{I_{x x}} \bar{y}(s)+k \frac{\left(M_{y}+M_{y T}\right)}{I_{y y}} \bar{x}(s)-\beta \Delta T(s, z).\end{aligned}\] In eq. ([eq4.6]) \(\beta=E \alpha\), where \(\alpha\) is the linear coefficient of thermal expansion of the material. The cross-sectional coordinates of the contour \(\bar{x}(s)\) and \(\bar{y}(s)\) appearing in eq. ([eq4.6]) are defined by \[\begin{aligned} \label{eq4.7} \bar{x}(s)=x(s)-n_{x} y(s) \quad \bar{y}(s)=y(s)-n_{y} x(s).\end{aligned}\]

Shear stresses in open and closed sections

The location of the shear center and the equation for the shear stress depend on whether the cross-sectional contour is open or closed.

Hooke’s law for transverse shear and torsion

Hooke’s law for transverse shear and torsion is defined relative to the shear center. From eq. ([eq5.76]) on page the compliance form of Hooke’s law is \[\begin{aligned} \label{eq4.27} \left[\begin{array}{@{}c@{}}\psi_{x} \\[3pt]\psi_{y} \\[6pt]\frac{d \phi_{z}}{d_{z}}\end{array}\right]=\left[\begin{array}{@{}ccc@{}}c_{x x} & c_{x y} & 0 \\c_{y x} & c_{y y} & 0 \\[3pt] 0 & 0 & 1 /(G J)\end{array}\right]\left[\begin{array}{@{}c@{}}V_{x} \\[3pt] V_{y} \\[3pt] M_{z}\end{array}\right],\end{aligned}\] where \(\psi_{x}(z)\) and \(\psi_{y}(z)\) denote the averaged shear strains. These shear strains are depicted in figure [fig3.6] on page  and are given by \[\begin{aligned} \label{eq4.28} \psi_{x}(z)=\frac{d u}{d z}+\phi_{y}(z) \quad \psi_{y}(z)=\frac{d v}{d z}+\phi_{x}(z).\end{aligned}\] In eq. ([eq4.27]) the compliance coefficients are denoted by \(\left(c_{x x}, c_{y y}, c_{x y}\right)\). From eq. ([eq5.62]) on page  the compliance coefficients for an open cross section are \[\begin{aligned} \label{eq4.29} c_{x x}=\left(\frac{k}{I_{y y}}\right)^{2} \int_{c} {\frac{\left[\bar{Q}_{y}(s)\right]^{2}}{G t} d s} \quad c_{x y}=c_{y x}=\frac{k^{2}}{I_{x x} I_{y y}} \int_{c} {\frac{\left[\bar{Q}_{x}(s) \bar{Q}_{y}(s)\right]}{G t} d s} \quad c_{y y}=\left(\frac{k}{I_{x x}}\right)^{2} \frac{\left[\bar{Q}_{x}(s)\right]^{2}}{G t} d s.\end{aligned}\] For a closed cross-sectional contour the compliance coefficients are given by eq. ([eq5.66]) on page , which are \[\begin{aligned} \label{eq4.30} c_{x x}=\oint\! \frac{F_{x}^{2}(s)}{G t} d s \quad c_{y y}=\oint\! \frac{F_{y}^{2}(s)}{G t} d s \quad c_{x y}=c_{y x}=\oint\! \frac{F_{x}(s) F_{y}(s)}{G t} d s.\end{aligned}\] The shear flow functions defined relative to the shear center result in a decoupling of the transverse shear and torsional responses of the bar as shown in eq. ([eq4.27]). That is, the shear strains are independent of the torque, and the twist per unit length is independent of the shear forces. For an open section the torsion constant is given by eq. ([eq4.14]) and for a single-cell, closed section \(G J\) is given by eq. ([eq4.24]).

Bending moment diagrams

To determine the axial normal stress distribution in bars subject to lateral loading, we have to first find the distribution of the bending moment. Analyses are presented for a cantilever wing and barge in still water. Airplanes and ships can be regarded as vehicles moving in different mediums, the air or water. In this regard, the study of buoyancy distribution acting on ship structures is instructive in determining the distribution of the bending moment in the hull.

[ex4.1] Consider the cantilever wing with tip tank as shown in figure [fig4.4]. Given the weight of the tip tank and its contents W, the distance \(e\) of the weight W from the wing tip, the wing span L, and the value of the distributed load intensity \(f_{r}\) at the wing root, determine the shear force and bending moment along the span.

Solution.The differential equilibrium for the shear force is given by eq. ([eq3.54]) on page . Integration of this differential equation from \(z =0\) to \(z\) results in \[\begin{aligned} \label{ex4.1a} V_{y}(z)=V_{y}(0)-\int_{0}^{z} f_{r}(u/L) d u,\ \text{which evaluates as}\ V_{y}(z)=V_{y}(0)-f_{r}\left(\frac{z^{2}}{2 L}\right).\end{aligned}\] The differential equation for the bending moment is given by eq. ([eq3.55]), where in this example the distributed moment per unit length \(m_{x}=0\). Integrating this differential equation from \(z =0\) to \(z\) results in \[\begin{aligned} \label{ex4.1b} M_{x}(z)=M_{x}(0)+\int_{0}^{z} V_{y}(u) d u,\ \text{which evaluates as}\ M_{x}(z)=M_{x}(0)+V_{y}(0) z-f_{r}\left(\frac{z^{3}}{6 L}\right).\end{aligned}\]

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At the wing tip, equilibrium of the tip tank as shown in figure [fig4.5] leads to \[\begin{aligned} \label{ex4.1c} V_{y}(0)=W \quad M_{x}(0)=e W.\end{aligned}\] Hence, the shear force and bending moment are \[\begin{aligned} \label{ex4.1d} V_{y}(z)=W-f_{r}\left(\frac{z^{2}}{2 L}\right) \quad M_{x}(z)=e W+W z+f_{r}\left(\frac{z^{3}}{6 L}\right) \quad 0 \leq z \leq L.\end{aligned}\]

Take \(L=144\) in. , \(e=6\,\textrm{in}.\) , \(W=500\,\textrm{lb.}\) , and \(f_{r}=70\,\textrm{lb}./\mathrm{in.}\) . Numerical evaluation gives \[\begin{aligned} \label{ex4.1e} V_{y}(z)=500-(35 z^{2})/144 \quad M_{x}(z)=3,000+500 z-(35 z^{3})/432.\end{aligned}\] The shear force and bending moment distributions with respect to the normalized coordinate z/L are plotted in figure [fig4.6]

The shear force equals zero at \(z/L=0.315\), which corresponds to \(z=45.3557\,\textrm{in}.\). At \(z=45.3557\,\textrm{in}.\) the bending moment exhibits a horizontal slope. Thus, the bending moment is a relative maximum at \(z=45.3557\,\textrm{in}.\) with a value of 18,118.6 lb.-in. The largest magnitudes of the shear force and bending moment occur at the wing root where, \(V_{y}(L)=-4{,}540\,\textrm{lb.}\) and \(M_{x}(L)= -166{,}920\). lb.-in.

[ex4.2] Consider a barge at rest in still water with a uniform immersed cross section, and subjected to the symmetrical loads shown in figure [fig4.7]. There is a distributed load acting on the barge due to buoyancy forces produced by displacing the water. Let \(f_{b}\) represent the distributed load intensity due to buoyancy, and \(f_{b}\) is a constant along the barge because the immersed cross section is uniform and the water is still. This is an example of a structure with no boundary supports, and is typical of aerospace and ocean vehicle structures.

1. Plot the shear force and bending moment diagrams for the barge,

2. Determine the maximum axial normal stress.

Solution to part (a).Vertical equilibrium of the entire barge requires that the buoyant upthrust equals 40 kN, so that \(f_{b}=2\,\mathrm{KN}/\mathrm{m}\). The total distributed load intensity is the difference between \(f_{b}\) and the magnitude of the downward acting applied loading intensity. The point force of 10 kN acting at \(z = 10\) m is shown schematically in the \(f_{y}(z)\)-diagram as a downward pointing arrow. Actually, \(f_{y} \rightarrow-\infty\) as \(z \rightarrow 10~\mathrm{m}\), because a point force is a finite load acting over zero length. Point forces are idealizations to actual loads and introduce discontinuities in the mathematical descriptions of some of the dependent variables.

A semigraphical method is used to sketch the shear force and bending moment diagrams. In this approach we first sketch the distributive load \(f_{y}(z)\) (F/L), then the shear force \(V_{y}(z)\), and finally the bending moment \(M_{x}(z)\). The differential equilibrium equations governing the shear force and the bending moment are eqs. ([eq3.54]) and ([eq3.55]) on page . These equations are repeated in eq. ([ex4.2a]) below. \[\begin{aligned} \label{ex4.2a} \frac{d V_{y}}{d z}+f_{y}(z)=0 \quad \frac{d M_{x}}{d z}-V_{y}+m_{x}(z)=0.\end{aligned}\] The prescribed external moment intensity \(m_{x}=0\) (F-L/L) in this example. From eq. ([ex4.2a]) we note that the slope of shear diagram at \(z\) is the negative of the distributed load intensity at \(z\), and that the slope of the moment diagram at \(z\) is the shear force at \(z\). Integrate eq ([ex4.2a]) with respect to \(z\) from \(z = z_1\) to \(z = z_2\) to get \[\begin{aligned} \label{ex4.2b} V_{y}\left(z_{2}\right)-V_{y}\left(z_{1}\right)=-\int_{z_{1}}^{z_{2}} f_{y}(z) d z,\ \text{and}\ M_{x}\left(z_{2}\right)-M_{x}\left(z_{1}\right)=\int_{z_{1}}^{z_{2}} V_{y}(z) d z.\end{aligned}\] Equation ([ex4.2b]) is interpreted in a graphical sense to mean that the difference in the shear force between \(z_2\) and \(z_1\) is the negative of the area under the distributed loading diagram from \(z_1\) to \(z_2\), and the difference in the bending moment between \(z_2\) and \(z_1\) is the area under the shear force diagram from \(z_1\) to \(z_2\). These are not geometrical areas. The area between the \(f_{y}(z)\) curve and the \(z\)-axis has units of force, and may be positive, zero, or negative.

Free body diagrams of the barge at each end are shown in figure [fig4.8]. The water pressure varies linearly with the depth of the immersed cross section and acts in \(z\)-direction. We assume the moment about the \(x\)-axis caused by the water pressure is small and can be neglected. As the infinitesimal distance \(\varepsilon \rightarrow 0\), the distributed loading acting at each end vanishes. In the limit we get the equilibrium conditions \[\begin{aligned} \label{ex4.2c} V_{y}(0)=0 \quad M_{x}(0)=0,\ and\ V_{y}(20)=0 \quad M_{x}(20)=0.\end{aligned}\]

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From eq. ([ex4.2c]) the shear force diagram begins at zero, and the slope \(d V_{y}/d z\) at \(z = 0\) is equal to 1 kN/m. The slope is constant between \(0<z<5 m\), thus \({V}_y(z)\) is a straight line in this range of \(z\). The difference in the shear force between \(z = 5\) m and \(z = 0\) is equal to the negative of the area under the \(f_{y}(z)\) curve, which is 5 kN. Thus \({V}_y(5) = 5\) kN since \({V}_y(0) = 0\). At \(z = 5^+\) m the loading intensity jumps to \(+2\) kN/m. The slope of the shear force jumps from 1 kN/m to \(-2\) kN/m at \(z = 5\) m, but the shear force is itself continuous. The difference \(V_y(10) - V_y(5)\) is equal to the negative of the area between the \(f_{y}(z)\)-curve and the \(z\)-axis between \(z = 5\) m and \(z = 10\) m. Thus \({V}_y(10) - {V}_y(5) = -10\) kN, so \({V}_y(10) = -5\) kN. Note the shear force is zero at \(z = 7.5\) m. At \(z = 10\) m the point force of 10 kN acts. As shown in figure [fig4.9], vertical equilibrium at \(z = 10\) m yields at jump in the shear force \({V}_y(10^+) - {V}_y(10^-) = 10\) kN, so that \({V}_y(10^+) = 5\) kN. The slope of the shear at \({z} = 10\) m is \(+2\) kN/m, and remains constant until \(z = 15\) m. The difference \({V}_y(15) - {V}_y(10^+) = -10\) kN, so that \({V}_y(15) = -5\) kN. Finally, the slope changes to \(+1\) kN/m at \(z = 15^+\) m and remains constant in the range \(15 < z < 20\). The difference \({V}_y(20) - {V}_y(15) = 5\) kN, so that \({V}_y(20) = 0\). The shear force equal to zero at \(z = 20\) m is expected from the result in eq. ([ex4.2c]). The shear force diagram is drawn below the loading intensity diagram in figure [fig4.10].

From eq. (c) the bending moment at \(z = 0\) equals zero, and its slope \(z = 0\) is equal to zero since the shear force is zero at \(\textit{z }= 0\). The slope \(d M_{x}/d z\) of the moment diagram increases linearly from zero at \(z = 0\) to the 5 kN, which is the value of the shear force at \(z = 5\) m. The difference \(\textit{M}_x(5) - \textit{M}_x(0)\) is equal to the area under shear diagram from \(z = 0\) to \(z = 5\) m. Hence, \(M_{x}(5) - M_{x}(0) = 12.5\) kNm, and \(M_{x}(5) = 12.5\) kNm since \(M_{x}(0) = 0\). From \(z = 5\) to \(z = 7.5\) the slope of the moment decreases from 5 kN to zero. At \(z = 7.5, M_{x}\) is a local maximum with a magnitude of 18.75 kNm. The slope of \(M_{x}(z)\) for \(7.5 < z < 10\) is negative, decreasing linearly from zero to \(-5\) kN. The difference \(M_{x}(10) - M_{x}(7.5) = -6.25\) kNm, so that \(M_{x}(10) = 12.50\) kNm. The slope of \(M_{x}(z)\) at \(z = 10\) m jumps from a \(-5\) kN to \(+5\) kN as shown in figure [fig4.10], but the moment itself is continuous. That is, the bending moment exhibits a cusp at \(z = 10\) m. The bending moment diagram in the range \(10 < z < 20\) is completed in a manner similar to the description of its construction in the range \(0 < z < 10\). In this example the shear force diagram is anti- symmetric about \(z = 10\) m and the bending moment is symmetric about \(z = 10\) m. This follows from the symmetrical loading on the barge.

Solution part (b).Let us assume an open cross section of the barge is as shown in figure [fig4.11]. The thickness of the three branches is 5 mm, and the section is symmetric about the \(y\)-axis so the product area moment \(I_{x y}=0\). From ([eq4.4]) the cross-sectional coefficients \(n_{x}=n_{y}=0\), \(k=1\), and \(\bar{y}=y\). At \(z=7.5~\mathrm{m}\) the shear force is equal to zero and the bending moment has a maximum value of 18.75 kNm. Hence, the shear stress \(\sigma_{z s}=0\) and the axial normal stress ([eq4.6]) is given by \[\begin{aligned} \label{ex4.2d} \sigma_{z z}=\frac{M_{x}}{I_{x x}} y \quad-\frac{5}{12}~\mathrm{m} \leq y \leq \frac{25}{12}~\mathrm{m},\end{aligned}\] where the second area moment about the \(x\)-axis \(I_{x x}=(5/128)\,\mathrm{m}^{4}\). The maximum magnitude of the normal stress occurs at \(y=(25/12)\,\mathrm{m}\), and it is a tensile stress with the value of \[\begin{aligned} \label{ex4.2e} \left.\sigma_{z z}\right|_{\max }=\frac{18.75\,\textrm{kNm}(25/12~\mathrm{m})}{(5/128)\,\mathrm{m}^{4}}=1{,}000 \times 10^{3}(\mathrm{N}/\mathrm{m}^{2})=1.0\,\textrm{MPa}.\end{aligned}\]

Buoyancy force distribution on ships

The simple uniform buoyancy distribution acting on the barge in example 1.2 is an exception to the buoyancy distributions found in practice. It is true that equilibrium requires the total buoyant upthrust to equal the weight of the ship and its contents. However, the distribution of the buoyancy and weight along the length of the ship is not necessarily the same. The difference in the magnitudes of the buoyancy and weight distribution intensities is the applied load intensity \(f_{y}(z)\). In ship design three conditions are recognized to compute \(f_{y}(z)\) for the same ship. These conditions are called

• the still water condition,

• the sagging condition, and

• the hogging condition.

A more detailed account of these conditions on the longitudinal bending of the ship is given by Muckle (1967) and Zubaly (1996), and here we only summarize the basic ideas.

A ship in still water is shown in figure [fig4.12], and a section A-A between \(z\) and \(z + \textit{dz}\) is also shown. Archimedes’s principle asserts that the buoyant upthrust is equal to the weight of the fluid displaced. Let A(z) denote the submerged cross section at \(z\), and let \(\gamma\) denote the specific weight (force per volume) of the fluid. The differential buoyancy force \(dF_{b}\) acting on the ship over a differential length dz is \[\begin{aligned} \label{eq4.32} dF_{b}=\gamma A(z) d z.\end{aligned}\] Consequently, the buoyant upthrust per unit ship length, which we designate \(f_{b}\), is equal to \(\gamma\textit{A(z)}\); i.e., \[\begin{aligned} \label{eq4.33} f_{b}=\frac{d F_{b}}{d z}=\gamma A(z).\end{aligned}\] A curve of \(f_{b}\) for a ship as well as the weight per unit length is shown in figure [fig4.13]. Overall equilibrium requires the area under these curves to have the same magnitude. If the submerged cross section is uniform in \(z\), as is the case for the barge in example 1.2, the distribution of the buoyancy per unit length \(f_{b}\) is a constant.

At sea a ship is subject to waves, and this alters the buoyancy distribution. For longitudinal bending of the ship two extreme static conditions are assumed: sagging and hogging. In each condition, the length of the wave is assumed to be the length of the ship. This is an “accepted” assumption for the worst buoyancy distribution causing the most severe bending of the ship.

The sagging condition is shown in figure [fig4.14](a). The wave crests are at the bow and stern, and the wave trough is amidships. A schematic of the buoyancy per unit length is shown below the ship in figure [fig4.14](a). The immersed cross section is the largest at or near the wave crests, and is least near the trough. The intensity of the buoyancy distribution reflects this. In this condition the deck sags and is in compression while the bottom is in tension. The worst location to concentrate the cargo in the ship is amidships, as this will result in the largest bendingmoment.

The hogging condition is depicted in figure [fig4.14](b). Here the wave troughs are at bow and stern, and the crest is amidships. The immersed cross section is greatest near amidships and is least near bow and stern. The distribution of the buoyancy per unit length \(f_{b}\), shown in figure [fig4.14](b), reflects this situation. In hogging the deck is in tension and the bottom is in compression. The worst possible locations to concentrate cargo is fore and aft, as this will produce the greatest bending moment in the ship.

Properties of plane areas

First and second area moments of the cross-sectional area need to be determined before evaluating eq. ([eq4.6]) for the normal stress. Analytical procedures were used in Example [ex3.1] on page to compute first and second area moments for a thin-walled bar. Frequently, the composite area technique in conjunction with the parallel axis theorem are used to determine these geometric properties.

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Neutral axis of the cross section

For the case of a vanishing axial normal force N, and neglecting temperature effects, the axial normal stress ([eq4.6]) reduces to \[\begin{aligned} \label{eq4.48} \sigma_{z z}=\frac{k M_{x}}{I_{x x}} \bar{y}+\frac{k M_{y}}{I_{y y}} \bar{x}=\frac{k M_{x}}{I_{x x}}\left(y-n_{y} x\right)+\frac{k M_{y}}{I_{y y}}\left(x-n_{x} y\right)=k\left(\frac{-n_{y} M_{x}}{I_{x x}}+\frac{M_{y}}{I_{y y}}\right) x+k\left(\frac{M_{x}}{I_{x x}}-\frac{n_{x} M_{y}}{I_{y y}}\right) y.\end{aligned}\]

At the centroid where \(x=y=0\) the axial normal stress vanishes. Set the axial normal stress equal to zero to get \[\begin{aligned} \label{eq4.49} \left(\frac{-n_{y} M_{x}}{I_{x x}}+\frac{M_{y}}{I_{y y}}\right) x+\left(\frac{M_{x}}{I_{x x}}-\frac{n_{x} M_{y}}{I_{y y}}\right) y=0.\end{aligned}\] Equation ([eq4.49]) defines a straight line in the cross section that is called the neutral axis. Substitute the definitions of \(n_{x}\) and \(n_{y}\) from eq. ([eq4.4]) into eq. ([eq4.49]), and then solve for \(y\) in terms of \(x\). Let these coordinates be denoted as \(\left(x_{\mathrm{NA}}, y_{\mathrm{NA}}\right)\). After some algebra we find \[\begin{aligned} \label{eq4.50} y_{\mathrm{NA}}=-\left(\frac{-I_{x y} M_{x}+I_{x x} M_{y}}{I_{y y} M_{x}-I_{x y} M_{y}}\right) x_{\mathrm{NA}}=-(\tan \beta) x_{\mathrm{NA}},\end{aligned}\] where \[\begin{aligned} \label{eq4.51} \tan \beta=\frac{-I_{x y} M_{x}+I_{x x} M_{y}}{I_{y y} M_{x}-I_{x y} M_{y}}.\end{aligned}\]

[ex4.4] The cantilever beam shown in figure [fig4.19] is subject to a bending moment M at its tip. The cross section is the thin-walled zee shown in figure [fig4.18]. The second area moments about the centroidal axes are given by eqs. ([ex4.4c]) to ([ex4.4e]) in example [fig4.3]. Determine the neutral axis in the cross section and the distribution of the bending normal stress.

Solution.First, note that the components of the prescribed bending moment are \(M_{x}=-M\) and \(M_{y}=0\). Thus, the cantilever beam is subject to loading in the y-z plane. The equation of the neutral axis is \(y_{\mathrm{NA}}=-(\tan \beta) x_{\mathrm{NA}}\), where eq. ([eq4.51]) is \[\begin{aligned} \label{ex4.4a} \tan \beta=\frac{(-I_{x y})(-M)}{I_{y y}(-M)}=\frac{-(-b^{3} t)}{\displaystyle\frac{2}{3} b^{3} t}=\displaystyle\frac{3}{2}.\end{aligned}\] The angle \(\beta=56.31^{\circ}\).

To compute the distribution of the normal stress ([eq4.6]) in the cross section, we begin by evaluating section coefficients ([eq4.4]): \[\begin{aligned} \label{ex4.4b} n_{x}=I_{x y}/I_{x x}=-3/8 \quad n_{y}=I_{x y}/I_{y y}=-3/2 \quad k=\frac{1}{1-n_{x} n_{y}}=16/7.\end{aligned}\] The coordinates in each branch of the cross section defined by eq. ([eq4.7]) are \[\begin{aligned} \label{ex4.4c} \bar{x}_{i}(s)=x_{i}(s)-n_{x} y_{i}(s) \quad \bar{y}_{i}(s)=y_{i}(s)-n_{y} x_{i}(s) \quad i=1,2,3.\end{aligned}\] The normal stress ([eq4.6]) in i-th branch is given by \[\begin{aligned} \label{ex4.4d} \left(\sigma_{z z}\right)_{i}=\left[\frac{k(-M)}{I_{x x}}\right] \bar{y}_{i}(s)=\left[\frac{k(-M)}{I_{x x}}\right]\left(y_{i}-n_{y} x_{i}\right)=\frac{6}{7 b^{3} t}(-M)\left(y_{i}+\frac{3}{2} x_{i}\right).\end{aligned}\] Coordinates in branch 1 are \(\left(x_{1}, y_{1}\right)=\left(x_{1},-b\right)\), in branch 2 \(\left(x_{2}, y_{2}\right)=\left(0, y_{2}\right)\), and in branch 3 \(\left(x_{3}, y_{3}\right)=\left(x_{3}, b\right)\). The normal stress in each branch is a linear function of the contour coordinate. The results are \[\begin{gathered} \left(\sigma_{z z}\right)_{1}=\frac{6}{7 b^{3} t}(-M)\left(-b+\frac{3}{2} x_{1}\right) \quad 0 \leq x_{1} \leq b,\label{ex4.4e}\\[6pt] \left(\sigma_{z z}\right)_{2}=\frac{6}{7 b^{3} t}(-M)\left(y_{2}-n_{y} x_{2}\right)=\frac{6}{7 b^{3} t}(-M) y_{2} \quad-b \leq y_{2} \leq b,\ \text{and}\label{ex4.4f}\\[6pt] \left(\sigma_{z z}\right)_{3}=\frac{6}{7 b^{3} t}(-M)\left(y_{3}-n_{y} x_{3}\right)=\frac{6}{7 b^{3} t}(-M)\left(b+\frac{3}{2} x_{3}\right) \quad-b \leq x_{3} \leq 0.\label{ex4.4g}\end{gathered}\] The neutral axis and the bending normal stress distribution are shown in figure [fig4.20].

[ex4.5] Determine the lateral displacement functions \(u(z)\) and \(v(z)\), \(0 \leq z \leq L\), for the zee section beam of example [ex4.4].

Solution.First note that this a statically determinate beam subject to pure bending. The equilibrium equations are satisfied for \(M_{x}=-M\) and \(M_{y}=0\) for all \(z \in(0, L)\), where M denotes the specified end moment. To find the lateral displacements \(u(z)\) and \(v(z)\), \(0 \leq z \leq L\), we begin with matrix eq. ([eq4.1]), which for this examplereduces to \[\begin{aligned} \label{ex4.5a} \left[\begin{array}{@{}l@{}}d w/d z \\[3pt]d \phi_{x}/d z \\[3pt]d \phi_{y}/d z\end{array}\right]=\frac{1}{E}\left[\begin{array}{@{}ccc@{}}1/A & 0 & 0 \\[3pt]0 & k/I_{x x} & \left(-k n_{x}\right)/I_{y y} \\[3pt]0 & \left(-k n_{y}\right)/I_{x x} & k/I_{y y}\end{array}\right]\left[\begin{array}{@{}c@{}}0 \\[3pt]-M \\[3pt]0\end{array}\right].\end{aligned}\] For this pure bending case the derivatives of the rotations are given by \[\begin{gathered} \frac{d \phi_{x}}{d z}=\frac{k(-M)}{E I_{x x}}=\frac{16/7}{(8/3) E b^{3} t}(-M)=-\frac{6}{7}\left(\frac{M}{E b^{3} t}\right),\ \text{and}\label{ex4.5b}\\[6pt] \frac{d \phi_{y}}{d z}=-\frac{k n_{y}}{E I_{x x}}(-M)=-\frac{(16/7)(-3/2)}{(8/3) E b^{3} t}(-M)=-\frac{9}{14}\left(\frac{M}{E b^{3} t}\right).\label{ex4.5c}\end{gathered}\] Integrate eqs. ([ex4.5b]) and ([ex4.5c]) from \(z = 0\) to \(z\) to get \[\begin{gathered} \phi_{x}(z)-\phi_{x}(0)=\int_{0}^{z}\left[-\frac{6}{7}\left(\frac{M}{E b^{3} t}\right)\right] d z=\left[-\frac{6}{7}\left(\frac{M}{E b^{3} t}\right)\right] z,\ \text{and}\label{ex4.5d}\\[6pt] \phi_{y}(z)-\phi_{y}(0)=\int_{0}^{z}\left[-\frac{9}{14}\left(\frac{M}{E b^{3} t}\right)\right] d z=\left[-\frac{9}{14}\left(\frac{M}{E b^{3} t}\right)\right] z.\label{ex4.5e}\end{gathered}\] At \(z =\)0 the beam is clamped to the rigid wall so that the rotations of the cross section at the wall are \(\phi_{x}(0)=\phi_{y}(0)=0\). The results for the rotations are \[\begin{aligned} \label{ex4.5f} \phi_{x}(z)=-\frac{6}{7}\left(\frac{M}{E b^{3} t}\right) z \quad \phi_{y}(z)=-\frac{9}{14}\left(\frac{M}{E b^{3} t}\right) z.\end{aligned}\] The cantilever beam is subject to pure bending and by equilibrium the transverse shear forces \(V_{x}=V_{y}=0\), for \(0 \leq z \leq L\). Hooke’s law ([eq4.27]) then yields that the transverse shear strains \(\psi_{x}=\psi_{y}=0\), for \(0 \leq z \leq L\). Vanishing of the transverse shear strains in eq. ([eq4.28]) leads to \[\begin{aligned} \label{ex4.5g} \psi_{x}=\frac{d u}{d z}+\phi_{y}=0 \quad \psi_{y}=\frac{d v}{d z}+\phi_{x}=0.\end{aligned}\] From eqs. (f) and (g) the derivatives of the displacements are given by \[\begin{aligned} \label{ex4.5h} \frac{d u}{d z}=-\phi_{y}=\frac{9}{14}\left(\frac{M}{E b^{3} t}\right) z \quad \frac{d v}{d z}=-\phi_{x}=\frac{6}{7}\left(\frac{M}{E b^{3} t}\right) z.\end{aligned}\] Integrate eq. (h) from \(z=0\) to \(z\) to get \[\begin{aligned} \label{ex4.5i} u(z)-u(0)=\frac{9}{14}\left(\frac{M}{E b^{3} t}\right) \frac{z^{2}}{2} \quad v(z)-v(0)=\frac{6}{7}\left(\frac{M}{E b^{3} t}\right) \frac{z^{2}}{2}.\end{aligned}\] At the clamped end the beam displacements equal zero. Thus, the displacements are \[\begin{aligned} \label{ex4.5j} \left.u(z)=\frac{9}{14}\left(\frac{M}{E b^{3} t}\right) \frac{z^{2}}{2} \quad v(z)=\frac{6}{7} \frac{M}{E b^{3} t}\right) \frac{z^{2}}{2}.\end{aligned}\] The view of the lateral displacements of the beam at \(z=L\) are shown in figure [fig4.21]. As a result of \(I_{x y} \neq 0\) the beam displaces both vertically and horizontally for a load that is applied in y-z plane.

[ex4.6] Consider a uniform cantilever beam from example [ex4.4] subject to a vertical force F acting through the shear center at its free end as shown in figure [fig4.22]. There is no change in temperature from the stress-free state. The cross section is the thin-walled zee shown in figure [fig4.18]. Determine the displacements of the cantilever.

Solution.The cantilever is statically determinate, and from equilibrium the shear force and bending moment are \[\begin{aligned} \label{ex4.6a} V_{y}=F \quad M_{x}=-(L-z) F \quad 0 \leq z \leq L.\end{aligned}\] From Hooke’s law ([eq4.1]) we find the derivatives of the rotations are \[\begin{aligned} \label{ex4.6b} \frac{d \phi_{x}}{d z}=\frac{k M_{x}}{E I_{x x}}=\frac{-k}{E I_{x x}}(L-z) F \quad \frac{d \phi_{y}}{d z}=-\frac{k n_{y}}{E I_{x x}} M_{x}=\frac{k n_{y}}{E I_{x x}}(L-z) F.\end{aligned}\] Integrate the rotations in eq. (b) with respect to coordinate \(z\) and impose \(\phi_{x}(0)=\phi_{y}(0)=0\) at the clamped end of the cantilever to get \[\begin{aligned} \label{ex4.6c} \phi_{x}=\frac{-3 z(2 L-z)}{7 E b^{3} t} F \quad \phi_{y}=\frac{-9 z(2 L-z)}{14 b^{3} t E} F.\end{aligned}\] From eq. ([eq4.27]) the shear strains of the cantilever beam are \[\begin{aligned} \label{ex4.6d} \psi_{x}=c_{x y} F \quad \psi_{y}=c_{y y} F.\end{aligned}\] The compliance coefficients ([eq4.29]) for this example are \[\begin{aligned} c_{x y}=\frac{k^{2}}{I_{x x} I_{y y}}\left[\int\limits_{0}^{b} \bar{Q}_{x 1} \bar{Q}_{y 1} d s_{1}+\int\limits_{0}^{2b} \bar{Q}_{x 2} \bar{Q}_{y 2} d s_{2}+\int\limits_{0}^{b} \bar{Q}_{x 3} \bar{Q}_{y 3} d s_{3}\right] \frac{1}{G t},\ \text{and}\label{ex4.6e}\\ c_{y y}=\left(\frac{k}{I_{x x}}\right)^{2}\left[\int\limits_{0}^{b} \bar{Q}_{x 1}^{2} d s_{1}+\int\limits_{0}^{2b} \bar{Q}_{x 2}^{2} d s_{2}+\int\limits_{0}^{b} \bar{Q}_{x 3}^{2} d s_{3}\right] \frac{1}{G t}.\label{ex4.6f}\end{aligned}\] To evaluate the compliance coefficients in eq. ([ex4.6d]) we need the parametric equations of the contour of the cross section with respect to centroidal coordinates. The equations are listed in table [tab4.3].

We need to determine the distribution functions \(\bar{Q}_{x i}\left(s_{i}\right)\) and \(\bar{Q}_{y i}\left(s_{i}\right)\), \(i=1,2,3\), beginning with the contour origin from the general expression in eq. ([eq4.9]). The contour origin is in branch 1 where \(s_{1}=0\) at its intersection with the traction-free longitudinal edge. As we move along the contour each branch is cut at a generic value of its contour coordinate. The area of the contour preceding the cut determines the range of integration for the distribution function as shown in figure [fig4.23]. Using the parametric equations in table [tab4.3] the results for branch one are \[\begin{aligned} \bar{Q}_{x 1}\left(s_{1}\right)&=\int_{0}^{s_{1}}\left[y_{1}\left(s_{1}\right)-n_{y} x_{1}\left(s_{1}\right)\right] t d s_{1}=\frac{1}{4}\left(2 b-3 s_{1}\right) s_{1} t,\ \text{and}\nonumber\\ \bar{Q}_{y 1}\left(s_{1}\right)&=\int_{0}^{s_{1}}\left[x_{1}\left(s_{1}\right)-n_{x} y_{1}\left(s_{1}\right)\right] t d s_{1}=\frac{1}{8}\left(5 b-4 s_{1}\right) s_{1} t.\label{ex4.6g}\end{aligned}\] For a cut in branch 2, the integration includes all of branch 1 and the integration of the segment in branch 2. The distribution functions for the cut in branch 2 are \[\begin{gathered} \bar{Q}_{x 2}\left(s_{2}\right)=\bar{Q}_{x 1}(b)+\int_{0}^{s_{2}}\left[y_{2}\left(s_{2}\right)-n_{y} x_{2}(s_{2})\right] t d s_{2}=-\frac{1}{4}(b^{2}+4 b s_{2}-2 s_{2}^{2}) t,\ \text{and}\label{ex4.6h}\\ \bar{Q}_{y 2}\left(s_{2}\right)=\bar{Q}_{y 1}(b)+\int_{0}^{s_{2}}\left[x_{2}\left(s_{2}\right)-n_{x} y_{2}\left(s_{2}\right)\right] t d s_{2}=\frac{1}{16}(2 b^{2}-6 b s_{2}+3 s_{2}^{2}) t,\,\textrm{where } 0 \leq s_{2} \leq 2 b.\label{ex4.6i}\end{gathered}\] For a cut in branch 3, the integration includes all of branch 1, all of branch 2, and the integration of the segment in branch 3. The distribution functions for the cut in branch 3 are \[\begin{gathered} \bar{Q}_{x 3}\left(s_{3}\right)=\bar{Q}_{x 2}(2 b)+\int_{0}^{s_{3}}\left[y_{3}\left(s_{3}\right)-n_{y} x_{3}\left(s_{3}\right)\right] t d s_{3}=-\frac{1}{4}(b^{2}-4 b s_{3}+3 s_{3}^{2}) t,\ \text{and}\label{ex4.6j}\\ \bar{Q}_{y 3}\left(s_{3}\right)=\bar{Q}_{y 2}(2 b)+\int_{0}^{s_{3}}\left[x_{3}\left(s_{3}\right)-n_{x} y_{3}\left(s_{3}\right)\right] t d s_{3}=\frac{1}{8}(b^{2}+3 b s_{3}-4 s_{3}^{2}) t,\ \text{where}\ 0 \leq s_{3} \leq b.\label{ex4.6k}\end{gathered}\] Substitute the distribution functions in eqs. ([ex4.6g]) to ([ex4.6k]) into the compliance coefficients of eqs. ([ex4.6e]) and ([ex4.6f]), followed by integration, to find \[\begin{aligned} c_{x y}=\frac{9}{196 b G t} \quad c_{y y}=\frac{267}{490 b G t}.\label{ex4.6l}\end{aligned}\] From the definition of the shear strains ([eq4.28]), we get the following integrals for the displacements that satisfy the boundary conditions \(u(0)=v(0)=0\): \[\begin{aligned} \label{ex4.6m} u(z)=\int_{0}^{z}\left(\psi_{x}-\phi_{y}\right) d z \quad v(z)=\int_{0}^{z}\left(\psi_{y}-\phi_{x}\right) d z.\end{aligned}\] Substitute the compliance coefficients from eq. ([ex4.6i]) into Hooke’s law ([ex4.6d]), followed by substituting the result for shear strains into eq. ([ex4.6m]). Then substitute eq. ([ex4.6c]) for the rotations into eq. ([ex4.6m]). Perform the integration with respect to \(z\) to get: \[\begin{aligned} \label{ex4.6n} u(z)=\frac{9 F z}{196 b G t}+\frac{9 L F z^{2}}{14 b^{3} t E}-\frac{3 F z^{3}}{14 b^{3} t E} \quad v(z)=\frac{267 F z}{480 b t G}+\frac{3 L F z^{2}}{7 b^{3} t E}-\frac{F z^{3}}{7 b^{3} t E}.\end{aligned}\] The vertical displacement at point of application of the force F is \[\begin{aligned} \label{ex4.6o} v(L)=\left(\frac{2 L^{3}}{7 b^{3} t E}+\frac{267 L}{490 b t G}\right) F=\left(\frac{3 L^{3} F}{7 b^{3} t E}\right)\left[1+\frac{267 b^{2}}{140 L^{2}} \frac{E}{G}\right],\end{aligned}\] where the last result was obtained by factoring out the first term on the right-hand side. For aluminum alloys \(E/G \approx 2.5\), eq. ([ex4.6o]) can be manipulated to the form \[\begin{aligned} \label{ex4.6p} v(L)=\frac{16 L^{3} F}{21 E I_{x x}} \underbrace{\left[1+\frac{4.77}{(L/b)^{2}}\right]}_{=\,g(L/b)}.\end{aligned}\] The function \(g(L/b)\) is evaluated for several ratios of \(L/b\), and the results are listed in table [tab4.4]. The function \(g(L/b) \rightarrow 1\) for values of \(L/b>10\), which implies the displacement \(v(L) \rightarrow \frac{16 L^{3} F}{21 E I_{x x}}\) for \(L/b>10\).

This result means the contribution of the displacement due to transverse shear deformation is negligible with respect to the component due to bending for beams that are long with respect to their cross-sectional dimensions. To neglect the influence of transverse shear means we can set shear strains \(\psi_{x}=0\) and \(\psi_{y}=0\). Setting the shear strains equal to zero in Hooke’s law for transverse shear in ([eq4.27]) means the shear forces equal zero. However, the shear forces are necessary for beam equilibrium. Hence, we omit Hooke’s law for transverse shear in a theory where the shear strains are assumed equal to zero. The beam theory neglecting transverse shear strains is called the Euler-Bernoulli beam theory.

Solution to part (a).

The parametric equations of the semicircular contour are \[\begin{aligned} \label{ex4.7a} X(\theta)=a \cos \theta \quad Y(\theta)=a \sin \theta \quad s=a \theta \quad-\pi/2 \leq \theta \leq \pi/2.\end{aligned}\] We compute the area A, the first area moment about the Y-axis \(Q_{Y}\), and the location of the centroid as follows. \[\begin{aligned} \label{ex4.7b} A=\int_{-\pi/2}^{\pi/2} t a d \theta=a \pi t \quad Q_{Y}=\int_{-\pi/2}^{\pi/2} X(\theta) t a d \theta=2 a^{2} t \quad X_{C}=Q_{Y}/A=2 a/\pi.\end{aligned}\] The coordinates relative to the centroid, and the second area moment about the \(x\)-axis through the centroid are \[\begin{aligned} \label{ex4.7c} x(\theta)=X(\theta)-X_{C}=a \cos \theta-2 a/\pi \quad y(\theta)=Y(\theta) \quad I_{x x}=\int_{-\pi/2}^{\pi/2} y^{2}(\theta) t a d \theta=a^{3} \pi t/2.\end{aligned}\]

Solution to part (b).

The distribution function for the shear flow with the contour origin at \(\theta=-\pi/ 2\) is given by \[\begin{aligned} \label{ex4.7d} Q_{x}(\theta)=\int_{-\pi/2}^{\theta} y(\theta) \textit{tad} \theta=-a^{2} t \cos \theta \quad-\pi/2 \leq \theta \leq \pi/2.\end{aligned}\] Since the product area moment \(I_{x y}=0\), eq. ([eq4.4]) yields the parameters \(n_{x}=n_{y}=0\) and \(k=1\). The shear flow in eq. ([eq4.13]) on page  reduces to \[\begin{aligned} \label{ex4.7e} q(s, z)=-\frac{V_{y}}{I_{x x}} Q_{x}(s),\end{aligned}\] and the explicit equation for the shear flow is \[\begin{aligned} \label{ex4.7f} q(\theta)=-V_{y} Q_{x}(\theta)/I_{x x}=\frac{2 V_{y}}{(a \pi)} \cos \theta.\end{aligned}\] The shear flow is plotted normal to the contour in figure [fig4.25], and it apparent in the figure that the shear flow is a maximum at \(\theta = 0\).

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Solution to part (c).

The coordinates of the shear center relative to the centroid is given in eq. ([eq4.8]) on page . In this example the equation for the coordinate \(x_{S c}\) reduces to \[\begin{aligned} \label{ex4.7g} x_{s c}=-\left(\frac{1}{I_{x x}}\right)\left[\int_{c} r_{n c}(\theta) Q_{x}(\theta) a d \theta \mid\right.\!,\end{aligned}\] where the coordinate normal to the contour with respect to the centroid is given in eq. ([eq4.10]). In this example the normal coordinate is \[\begin{aligned} \label{ex4.7h} r_{n c}=x(s) \frac{d y}{d s}-y(s) \frac{d x}{d s}=a-\frac{2 a}{\pi} \cos \theta.\end{aligned}\] Substitute eq. (h) for the normal coordinate and substitute eq. (d) for the distribution function, into eq. (g) to find \[\begin{aligned} x_{s c}=\frac{-1}{I_{x x}} \int r_{n c}(\theta) Q_{x}(\theta) a d \theta=\frac{2 a}{\pi}.\end{aligned}\] Relative to point O the coordinate of the shear center is given by \(X_{S C}=X_{C}+x_{s c}=(4 a)/\pi\). The shear center lies outside of the circular contour. As a check on the shear flow we compute its torque with respect to the shear center by \[\begin{aligned} \label{ex4.7i} M_{z}=\int_{-\pi/2}^{\pi/2} r_{n}(\theta) q(\theta) ad \theta,\end{aligned}\] where \(r_{n}(\theta)\) is the coordinate normal to the line of action of the shear flow as shown in figure [fig4.26]. The normal coordinate is given by eq. ([eq4.11]), where \(r_{n c}(s)\) is the coordinate normal to the line of action of the shear flow with

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respect to the centroid as shown in figure [fig4.26]. In this example the normal coordinate with respect to the centroid is \[\begin{aligned} \label{ex4.7j} r_{n}=a\left(1-\frac{4}{\pi} \cos \theta\right),\end{aligned}\] Substitute eq. ([ex4.7k]) for \(r_{n}(\theta)\) in eq. (i) followed by substitution eq. ([ex4.7f]) for the shear flow to get \[\begin{aligned} \label{ex4.7k} M_{z}&=\int_{-\pi/2}^{\pi/2}\left[a\left(1-\frac{4}{\pi} \cos \theta\right)\right]\left[\frac{2 V_{y}}{(a \pi)} \cos \theta\right] a d \theta\nonumber\\[5pt] &=\frac{2 a}{\pi} V_{y} \int_{-\pi/2}^{\pi/2}\left[\cos \theta-\frac{4}{\pi} \cos ^{2} \theta\right] d \theta \nonumber\\[2pt] &=\left.\frac{2 a}{\pi} V_{y}\left[\sin \theta-\frac{4}{\pi}\left(\frac{\theta}{2}+\frac{1}{4} \sin 2 \theta\right)\right]\right|_{-\pi/2} ^{\pi/2}.\end{aligned}\] Evaluating eq. (k) we find \[\begin{aligned} \label{ex4.7l} M_{z} =\frac{2 a}{\pi} V_{y}\left[1-(-1)-\frac{4}{\pi}\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)+\frac{1}{4}(0-0)\right]\right] =\frac{2 a}{\pi} V_{y}\left[2-\frac{4}{\pi}\left[\frac{\pi}{2}+\frac{1}{4}(0-0)\right]\right]=\frac{2 a}{\pi} V_{y}[2-2]=0.\end{aligned}\] The result of the integration given by eq. (l) verifies that the shear flow results in no torque at the shear center.

Solution to part (d).

The shear stress \(\sigma_{z s}\) is the sum of the shear stresses from the transverse shear force \({V}_y\) and from the torque \(M_{z}\), and is given by eq. ([eq4.12]) on page . For this example we find \[\begin{aligned} \label{ex4.7m} \sigma_{z s}=q/t+2\left(M_{z}/J\right) \zeta=2\left(V_{y}/A\right) \cos \theta+2\left(M_{z}/J\right) \zeta \quad-\pi/2 \leq \theta \leq \pi/2 \quad-t/2 \leq \zeta \leq t/2.\end{aligned}\] From eq. ([eq4.14]) the torsion constant \(J=\left(b t^{3}\right)/3=\left(a \pi t^{3}\right)/3\). The shear stress (m) is a maximum at \(\theta = 0\), where the shear flow is maximum. At \(\theta = 0\) the maximum shear stress is determined from \[\begin{aligned} \label{ex4.7n} \left.\sigma_{z s}\right|_{\theta=0}=2\left(V_{y}/A\right) \pm\left(M_{z}/J\right) t.\end{aligned}\]

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[ex4.8] A closed contour in the shape of the letter D is shown in figure [fig4.27]. The origin of the X-Y system is taken at the center of the semicircular branch. The cross section is symmetric about the X-axis, so the centroid and shear center lie on the X-axis. To locate the shear center we take shear force \(V_{x}=0\), and note that the shear force \(V_{y} \neq 0\) and torque \(M_{z} \neq 0\) act at the shear center.

1. Locate the centroid,

2. determine the shear flow with respect to the centroid,

3. locate the shear center,

4. establish the shear flow with respect to the shear center, and

5. determine the shear stress \(\sigma_{z s}\).

Solution to part (a).

The cross section is composed of two branches. Branch 1 is the semicircular contour, and branch 2 is the vertical web connecting the ends of branch 1. The parametric equations of the contour for each branch are \[\begin{aligned} \label{ex4.8a} \begin{array}{@{}cc@{}}{\left[X_{1}(\theta), Y_{1}(\theta)\right]=a[\cos (\theta), \sin (\theta)]} & -\pi/2 \leq \theta \leq \pi/2 \\{\left[X_{2}(s), Y_{2}(s)\right]=[0, a-s]} & 0 \leq s \leq 2 a\end{array}.\end{aligned}\] Note that the contour coordinate \(s\) in branch 2 is defined in the negative \(y\)-direction. We compute the area A, the first area moment about the Y-axis \(Q_{Y}\), and the location of the centroid as follows: \[\begin{aligned} \label{ex4.8b} A=\int_{-\pi/2}^{\pi/2} t a d \theta+\int_{0}^{2a} t d s=(2+\pi) a t \quad Q_{Y}=\int_{-\pi/2}^{\pi/2} X_{1} t a d \theta+\int_{0}^{2a} X_{2} t d s=2 a^{2} t \quad X_{C}=\frac{Q_{Y}}{A}=\frac{2 a}{2+\pi}.\end{aligned}\]

Solution part to (b).

The shear flow relative to the centroid is given by eq. ([eq4.17]) on page . For this example this shear flow function is \[\begin{aligned} \label{ex4.8c} q_{C}(s, z)=\frac{M_{z c}(z)}{2 A_{c}}-F_{y c}(s) V_{y}(z),\end{aligned}\] The function \(F_{y c}(s)\) is given by eq. ([eq4.19]). For this symmetric section the product area moment \(I_{x y}=0\), and from eq. ([eq4.4]) on page  the cross-sectional coefficients \(n_{x}=n_{y}=0\) and \(k=1\). Also, from eq. ([eq4.7]) \(\bar{y}(s)=y(s)\). The function \(F_{y c}(s)\) in eq. ([eq4.19]) reduces to \[\begin{aligned} \label{ex4.8d} F_{y c}(s)=\frac{1}{I_{x x}}\left[Q_{x}(s)-\frac{1}{\left(2 A_{c}\right)} \oint\! r_{n c}(s) Q_{x}(s) d s\right].\end{aligned}\] The distribution function given by eq. ([eq4.23]) in this example is simplified to \[\begin{aligned} \label{ex4.8e} \bar{Q}_{x}=Q_{x}=\int_{0}^{s} y(s) t(s) d s,\end{aligned}\] and the coordinate normal to the contour \(r_{n c}(s)\) in eq. ([ex4.8d]) is given in eq. ([eq4.10]). The parametric equations of the contour with respect to the centroid are \[\begin{aligned} \label{ex4.8f} \begin{array}{@{}cc@{}}x_{1}(\theta)=X_{1}(\theta)-X_{C}=\displaystyle\frac{-2 a}{2+\pi}+a \cos \theta & y_{1}(\theta)=Y_{1}(\theta) \\[10pt] x_{2}(s)=X_{2}(s)-X_{C}=\displaystyle\frac{-2 a}{2+\pi} \quad y_{2}(s)=Y_{2}(s)\end{array}.\end{aligned}\] The second area moment of the cross section about the \(x\)-axis through the centroid is \[\begin{aligned} \label{ex4.8g} I_{x x}=\int_{-\pi/2}^{\pi/2} y_{1}^{2} t a d \theta+\int_{0}^{2 a} y_{2}^{2} t d s=\left(\frac{4+3 \pi}{6}\right) a^{3} t=2.23746 a^{3} t.\end{aligned}\]

The distribution function about the \(x\)-axis beginning at the contour origin at \(\theta=-\pi/2\) and going counterclockwise around the contour are determined for each branch from eq. ([ex4.8e]) as \[\begin{aligned} \label{ex4.8h} Q_{x 1}(\theta)&=\int_{-\pi/2}^{\theta} y_{1} t a d \theta=-a^{2} t \cos \theta \quad-\pi/2 \leq \theta \leq \pi/2, \text{and}\\ Q_{x 2}(s)&=Q_{x 1}(\pi/2)+\int_{0}^{s} y_{2} t d s=a t s-t\left(\frac{s^{2}}{2}\right) \quad 0 \leq s \leq 2 a.\label{ex4.8i}\end{aligned}\]

Note that \(Q_{x 2}(2 a)=0\), since this represents the first area moment of the entire cross section about the centroidal \(x\)-axis. From eq. ([eq4.10]) the coordinates normal to the contour for each branch are \[\begin{aligned} \label{ex4.8j} r_{n c 1}=x_{1}\left(\frac{d y_{1}}{a d \theta}\right)-y_{1}\left(\frac{d x_{1}}{a d \theta}\right)=a\left(1-\frac{2}{2+\pi} \cos \theta\right) \quad r_{n c 2}=x_{2}\left(\frac{d y_{2}}{d s}\right)-y_{2}\left(\frac{d x_{2}}{d s}\right)=\frac{2 a}{2+\pi}.\end{aligned}\] The area enclosed by the closed contour \(A_{c}\) is given by eq. ([eq4.18]), and we get the expected result as shown below. \[\begin{aligned} \label{ex4.8k} A_{c}=\int_{-\pi/2}^{\pi/2} r_{n c 1} a d \theta+\int_{0}^{2a} r_{n c 2} d s=\pi a^{2}/2.\end{aligned}\] Let the term containing the integral over the entire contour in eq. (d) be denoted by \(F_{y c 0}\). We compute \(F_{y c 0}\) for this cross section as \[\begin{aligned} \label{ex4.8l} F_{y c 0}=\frac{1}{\left(2 A_{c}\right)} \oint\! r_{n c}(s) Q_{x}(s) d s=\frac{1}{2 A_{c}}\left(\int_{-\pi/2}^{\pi/2} r_{n c 1} Q_{x 1} a d \theta+\int_{0}^{2a} r_{n c 2} Q_{x 2} d s\right)=-\frac{8+3 \pi}{3 \pi(2+\pi)}(a^{2} t).\end{aligned}\] We now compute the shear flow distribution functions for each branch by substituting results from eqs. ([ex4.8h]), ([ex4.8i]), and ([ex4.8l]) into eq. ([ex4.8d]). The results are \[\begin{aligned} \label{ex4.8m} F_{y c 1}&=\frac{\left(Q_{x 1}-F_{y c 0}\right)}{I_{x x}}=\frac{1}{a}(0.16071-0.446935 \cos \theta), \text{and} \\ F_{y c 2}&=\frac{(Q_{x 2}-F_{y c 0})}{I_{x x}}=\frac{1}{a}(0.16071+0.446935(s/a)-0.223467(s/ a)^{2}).\label{ex4.8n}\end{aligned}\] The shear flows with respect to the centroid in branch 1 and branch 2 are \[\begin{aligned} \label{ex4.8o} q_{C 1}(s, z)=\frac{M_{z c}(z)}{2 A_{c}}-F_{y c 1}(\theta) V_{y}(z) \quad q_{C 2}(s, z)=\frac{M_{z c}(z)}{2 A_{c}}-F_{y c 2}(\theta) V_{y}(z).\end{aligned}\]

Solution to part (c).

The equation for the location of the shear center relative to the centroid is given in eq. ([eq4.23]). The shear modulus of the material is denoted by G in eq. ([eq4.23]), and we assume it is uniform around the contour in this example. Then the shear center relative to the centroid is \[\begin{aligned} \label{ex4.8p} x_{s c}=\frac{2 A_{c}}{\oint\! d s}\left[\int_{-\pi/2}^{\pi/2} F_{y c 1} a d \theta+\int_{0}^{2a} F_{y c 2} d s\right], \text{where}\ \oint\! d s=\int_{-\pi/2}^{\pi/2} a d \theta+\int_{0}^{2 a} d s=(2+\pi) a,\end{aligned}\] Performing the integrals for \(x_{s c}\) in eq. ([ex4.8p]) we get \[\begin{aligned} \label{ex4.8q} x_{s c}=\frac{\pi a^{2}/2}{(2+\pi) a}\left[\frac{16-2 \pi}{4 \pi+3 \pi^{2}}\right]=\frac{(16-2 \pi)}{(2+\pi)(4+3 \pi)} a=0.140773 a.\end{aligned}\] The location of the shear center relative to point O is \(X_{S C}=X_{C}+x_{s c}=0.529757 a\).

Solution to part (d).

The shear flow relative to the shear center is given by eq. ([eq4.25]), which in this example is \[\begin{aligned} \label{ex4.8r} q(s, z)=\frac{M_{z}(z)}{2 A_{c}}-F_{y}(s) V_{y}(z),\end{aligned}\]

where the shear flow distribution function relative to the shear center is given by eq. ([eq4.26]). The results for the shear flow distribution functions relative to the shear center for each branch are as follows: \[\begin{gathered} \label{ex4.8s} F_{y 1}=-\left(\frac{x_{s c}}{2 A_{c}}\right)+F_{y c 1}=\frac{1}{a}(0.1159-0.445635 \cos \theta), \text{and} \\[6pt] F_{y 2}=-\left(\frac{x_{s c}}{2 A_{c}}\right)+F_{y c 2}=\frac{1}{a}\left[0.1159+0.446935\left(\frac{s}{a}\right)-0.223467\left(\frac{s}{a}\right)^{2}\right].\label{ex4.8t}\end{gathered}\] Finally, from eq. (r) the shear flows in each branch are given by \[\begin{gathered} q_{1}=\frac{M_{z}}{2 A_{c}}-F_{y 1} V_{y}=\frac{M_{z}}{\pi a^{2}}-\frac{1}{a}(0.1159-0.445635 \cos \theta) V_{y}, \text{and} \label{ex4.8u}\\[6pt] q_{2}=\frac{M_{z}}{2 A_{c}}-F_{y 2} V_{y}=\frac{M_{z}}{\pi a^{2}}-\frac{1}{a}\left[0.1159+0.446935\left(\frac{s}{a}\right)-0.223467\left(\frac{s}{a}\right)^{2}\right].\label{ex4.8v}\end{gathered}\]

A check on the shear flows is to compute the resultant for \(F_{y}\) and the resultant moment about the shear center \(C_{z}\). Resolving the shear flows in the positive \(y\)-direction in each branch we compute the integrals \[F_{y}=\int_{-\pi/2}^{\pi/2} q_{1} \cos \theta a d \theta-\int_{0}^{2a} q_{2} d s=V_{y}.\] Thus, the resultant \(F_{y}\) is equal to the shear force. Note that \(q_2\) is positive in the negative \(y\)-direction. To compute the moment \(C_{z}\) we need the normal coordinate to the contour with respect to the shear center. From eq. ([eq4.11]) on page  the normal coordinates for each branch are determined by \[\begin{aligned} \label{ex4.8w} r_{n 1}=r_{n c 1}-x_{s c}\left(\frac{d y_{1}}{a d \theta}\right)=a(1-0.529757 \cos \theta) \quad r_{n 2}=r_{n c 2}-x_{s c}\left(\frac{d y_{2}}{d s}\right)=0.529757 a.\end{aligned}\] The resultant moment is \[\begin{aligned} \label{ex4.8x} C_{z}=\int_{-\pi/2}^{\pi/2} r_{n 1} q_{1} a d \theta+\int_{0}^{2a} r_{n 2} q_{2} d s=M_{z}.\end{aligned}\] The moment of the shear flows about the shear center equal the torque \(M_{z}\). The shear flows in eqs. ([ex4.8u]) and ([ex4.8v]) are statically equivalent to the shear force \(V_{y}\) and the torque \(M_{z}\) resolved at the shear center.

Solution to part (e).

The shear flows are plotted normal to the contour in figure [fig4.28]. The shear flow from the torque is spatially uniform and equal to \(M_{z} /\left(2 A_{c}\right)\). The shear stress is equal to the shear flow divided by the thickness of the branch. From figure [fig4.28] it is apparent that the maximum magnitude of the shear stress occurs either at \(\theta=0\) in the semicircular contour or at \(s=a\) in the vertical contour. These shear stress components are \[\begin{aligned} \sigma_{z s 1}(0)=\frac{M_{z}}{\pi a^{2} t}+\frac{0.3297 V_{y}}{a t} \quad \sigma_{z s 2}(a)=\frac{M_{z}}{\pi a^{2} t}-\frac{0.3394 V_{y}}{a t}.\end{aligned}\]

[ex4.9] A thin-walled circular tube with contour radius \(a\) and wall thickness \(t\) is subject to a torque \(M_{z}\). The wall of a second identical tube is cut parallel to its longitudinal axis along its entire length to make the cross section of this second tube an open circular arc. See figure [fig4.29]. Assume the saw kerf is very small. Compare the unit twist and maximum shear stress in the closed section to the open section.

Solution. For the closed section the shear flow \(q=\frac{M_{z}}{2\left(\pi a^{2}\right)}\) as shown in figure [fig4.28], and the torsion constant \(J=2 \pi a^{3} t\) from eq. ([eq3.128]) on page . Hence the maximum shear stress and unit twist are \[\begin{aligned} \label{ex4.9a} \left(\sigma_{z s}\right)_{\textit{closed }}=\frac{M_{z}}{2 \pi a^{2} t} \quad\left(\frac{d \theta_{z}}{d z}\right)_{\textit{closed }}=\frac{M_{z}}{G J}=\frac{M_{z}}{G\left(2 \pi a^{3} t\right)}.\end{aligned}\]

For the open section the developed length \(b\) of the contour is essentially \(2 \pi a\), since the saw kerf is assumed to be very small. By the membrane analogy discussed in article [sec3.9.1] on page , the torsional response is the same as the thin-walled rectangular section of length \(b\) and thickness \(t\). The maximum shear stress is given by eq. ([eq3.129]) on page , and the torsion constant is given in eq. ([eq3.128]). For \(b=2 \pi a\), we have \[\begin{aligned} \label{ex4.9b} \left(\sigma_{z s}\right)_{\textit{open}}=\frac{3 M_{z}}{2 \pi a t^{2}} \quad\left(\frac{d \theta_{z}}{d z}\right)_{\textit{open }}=\frac{M_{z}}{G\left(\frac{1}{3} 2 \pi a t^{3}\right)}.\end{aligned}\]

Forming the ratio of the maximum shear stress of the open section to the closed section we find \[\begin{aligned} \label{ex4.9c} \frac{(\sigma_{z s})_{\textit{open }}}{(\sigma_{z s})_{\textit{closed }}}=\frac{3 M_{z}}{2 \pi a t^{2}} \cdot \frac{2 \pi a^{2} t}{M_{z}}=3 \frac{a}{t} \gg 1.\end{aligned}\] Likewise, the ratio of the unit twists are \[\begin{aligned} \label{ex4.9d} \frac{(d \theta_{z}/d z)_{\textit{open }}}{(d \theta_{z}/d z)_{\textit{closed }}}=\frac{3 M_{z}}{G 2 \pi a t^{3}} \cdot \frac{G 2 \pi a^{3} t}{M_{z}}=3\left(\frac{a}{t}\right)^{2} \gg 1.\end{aligned}\] Since the ratio of the radius to thickness is greater than ten for a thin-walled section, the above results show that the shear stress ([ex4.9c]) and unit twist ([ex4.9d]) of the open circular section are much larger than for the closed section if both sections are subject to the same torque.

Hence, if a bar is to resist torsional loading, a closed section is preferable to an equivalent open section bar. That is, the unit twist is smaller for the closed section bar (it is stiffer), and the maximum shear stress is smaller, than for the equivalent open section bar subject to the same torque.

Resultant of uniform shear flow

In torsion problems it is often necessary to find the resultant of a constant shear flow along the contour of a curved branch. This situation is depicted in figure [fig4.30], where the curved branch begins at point A with coordinates \(\left(X_{A}, Y_{A}\right)\) and ends at point B with coordinates \(\left(X_{B}, Y_{B}\right)\). The resultant of the shear flow is resolved at point A in this figure.

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The sense of the arc-length \(s\) and the shear flow \(q\) are assumed positive from A to B along the contour of the branch. The shear flow acts tangent to the contour, and the unit tangent vector to the contour is denoted by \(\hat{t}(s)\). The unit tangent vector is given by eq. ([eq3.4]) on page , where we note that \(d X/d s=d x/d s\) and \(d Y/d s=d y/d s\). The differential force obtained from the shear flow is \[\begin{aligned} \label{eq4.52} d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}=q d s \hat{t}=q d s\left[\left(\frac{d X}{d s}\right) \hat{i}+\left(\frac{d Y}{d s}\right) \hat{j}\right]=q[(d X) \hat{i}+(d Y) \hat{j}].\end{aligned}\] Integrate eq. ([eq4.52]) from point A to point B on the contour to get \[\begin{aligned} \label{eq4.53} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}=q\left[\int_{A}^{B}\left(d X \hat{i}+d Y_{\hat{j}}\right)\right]=q\left[\left(X_{B}-X_{A}\right) \hat{i}+\left(Y_{B}-Y_{A}\right) \hat{j}\right]=\left(q L_{B/A}\right) \hat{u}_{B/A},\end{aligned}\] where the length of the chord connecting the ends of contour is denoted by \(L_{B/A}\). The length of the chord and the unit vector \(\hat{u}_{B/A}\) are given by \[\begin{aligned} \label{eq4.54} L_{B/A}=\sqrt{\left(X_{B}-X_{A}\right)^{2}+\left(Y_{B}-Y_{A}\right)^{2}} \quad \hat{u}_{B/ A}=\frac{\left(X_{B}-X_{A}\right)}{L_{B/A}} \hat{i}+\frac{\left(Y_{B}-Y_{A}\right)}{L_{B/A}} \hat{j}.\end{aligned}\] The differential torque about point A is \[\begin{aligned} \label{eq4.55} d \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}_{A}=r(s) \hat{n}(s) \times q d s \hat{t}=q r(s) d s \hat{k},\end{aligned}\]

where the position vector from point A to the line of action of the shear flow is \(r(s) \hat{n}(s)\), and \(\hat{n}(s)\) is the unit vector normal to the contour at \(s\). The product of \(r\) times ds is twice the enclosed area of the triangle with base ds and height \(r\). As shown in figure [fig4.30] \(r(s) d s=2 d A_{B/A}\). Integrate eq. ([eq4.55]) from point A to point B to get \[\begin{aligned} \label{eq4.56} \smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{M}}}_{A}=q \int_{A}^{B} r(s) d s \hat{k}=q \int_{A}^{B} 2 d A_{B/A} \hat{k}=2 q A_{B/A} \hat{k}.\end{aligned}\] The area between the contour and the chord is denoted by \(A_{B/A}\). The force and torque resolved at point A is shown in figure [fig4.31](a). The force and torque at point A are statically equivalent to the force acting along a line of action that is parallel to the chord at a perpendicular distance \(e\) from the chord. The distance is determined from \(M_{A}=e F=e q L_{A/B}\). Solve for \(e\) and substitute \(2 q A_{B/A}\) for the torque to get

\[\begin{aligned} \label{eq4.57} e=\frac{M_{A}}{q L_{B/A}}=\frac{2 q A_{B/A}}{q L_{B/A}},\ \text{or}\ e=\frac{2 A_{B/A}}{L_{B/A}}.\end{aligned}\]

The resultant of a constant shear flow is shown in figure [fig4.31](b).

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Now consider a continuous contour that does not intersect itself except that point B coincides with point A as shown in figure [fig4.32]. Since \(L_{B/A}=0\) in eq. ([eq4.53]) the force \(\smash{\mathord{\buildrel{\lower 3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over{F}}}=0\). For a closed contour let \(A_{B/A}=A_{c}\) and \(M_{A}=M_{z}\) in eq. ([eq4.56]). For a single-cell cross section subject to a torque \(M_{z}\) and no shear forces (pure torsion), the shear flow is given by \[\begin{aligned} \label{eq4.58} q=\frac{M_{z}}{2 A_{c}}.\end{aligned}\]

Equation ([eq4.58]) is called Bredt’s formula. (Also see eq. ([eq3.165]) on page ,)

A constant shear flow in a closed contour is statically equivalent to a torque, and this torque is the same for any point in the plane about which moments are computed. The fact that the torque is a “free vector” is depicted in figure [fig4.33], in which it is shown that some of the enclosed area used in Bredt’s formula can add as a negative quantity if the torque produced by the constant shear flow is clockwise over a segment of the branch. About point O in this figure, the torque produced by the shear flow from point B to A in the right half of the contour is counterclockwise, and the torque produced by the shear flow from A to B in the left half is clockwise. Hence, the total torque is the sum of these two torques with due respect to the sign. This summation shows that the total torque is proportional to the area enclosed by the contour.

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Torsion of a hybrid section

Consider a hybrid section composed of a single closed cell and open branches, or fins, as shown in figure [fig4.34]. The total torque carried by the section is the sum of the torques carried by the closed cell and open branches. For \(n\) open branches, we have \[\begin{aligned} \label{eq4.59} M_{z}=\left(M_{z}\right)_{\textit{closed}}+\sum_{i=1}^{n}\left(M_{z}\right)_{i},\end{aligned}\] where \[\begin{aligned} \label{eq4.60} \left(M_{z}\right)_{\textit{closed }}=(G J)_{\textit{closed }}\left(\frac{d \phi_{z}}{d z}\right)\ \text{and}\ \left(M_{z}\right)_{i}=(G J)_{i}\left(\frac{d \phi_{z}}{d z}\right).\end{aligned}\] The torsional stiffness for the closed cell is \[\begin{aligned} \label{eq4.61} (G J)_{\textit{closed }}=\frac{4 A_{c}^{2}}{\oint\! \frac{d s}{G t}},\end{aligned}\] and for each open branch the torsional stiffness is \[\begin{aligned} \label{eq4.62} (G J)_{i}=G_{i}\left(\frac{1}{3} b_{i} t_{i}^{3}\right).\end{aligned}\]

Combining eqs. ([eq4.59]) and ([eq4.60]), we have \[\begin{aligned} \label{eq4.63} M_{z}=\left[(G J)_{\textit{closed }}+\sum_{i=1}^{n}(G J)_{i}\right\rfloor \frac{d \phi_{z}}{d z}.\end{aligned}\] Comparing eq. ([eq4.63]) to the standard torsional formula \(M_{z}=(G J)_{\textit{eff}}\left(\frac{d \phi_{z}}{d z}\right)\), the effective torsional stiffness for the entire section is \[\begin{aligned} \label{eq4.64} (G J)_{\textit{eff}}=(G J)_{\textit{closed}}+\sum_{i=1}^{n}(G J)_{i}.\end{aligned}\] where the closed and open parts of the torsional stiffness are given by eqs. ([eq4.61]) and ([eq4.62]).

The shear stress in the closed cell is \(\left(\sigma_{z s}\right)_{\textit{closed}}=\left(M_{z}\right)_{\textit{closed}} /\left(2 A_{c} t\right)\), where the portion of the torque carried by the closed cell is \(\left(M_{z}\right)_{\textit{closed }}=(G J)_{\textit{closed }}\left(d \phi_{z}/d z\right)\). The unit twist is given by \(\left(d \phi_{z}/d z\right)=M_{z} /(G J)_{\textit{eff}}\). Combining these results, the shear stress in the closed cell is \[\begin{aligned} \label{eq4.65} \left(\sigma_{z s}\right)_{\textit{closed}}=\frac{(G J)_{\textit{closed}}}{(G J)_{\textit{eff}}} \cdot \frac{M_{z}}{2 A_{c} t}\end{aligned}\] The shear stress in open branches is given by \[\begin{aligned} \label{eq4.66} \left(\left.\sigma_{z s}\right|_{\max }\right)_{i}=\frac{3 M_{z i}}{b_{i} t_{i}^{2}},\end{aligned}\] which was derived as eq. ([eq3.132]) on page . Substitute the second equation in ([eq4.60]) for the torque carried by the branch, and then substitute eq. ([eq4.62]) for the torsional stiffness of the branch, to write the shear stress as \[\begin{aligned} \label{eq4.67} \left(\sigma_{z s}\right)_{i}=\frac{3\left(M_{z}\right)_{i}}{b_{i} t_{i}^{2}}=\frac{3}{b_{i} t_{i}^{2}} \cdot\left(G_{i} \frac{1}{3} b_{i} t_{i}^{3}\right) \frac{d \phi_{z}}{d z}.\end{aligned}\] Substitute \(\frac{d \phi_{z}}{d z}=\frac{M_{z}}{(G J)_{\textit{eff}}}\) for the unit twist in eq. ([eq4.67]) to get \[\begin{aligned} \label{eq4.68} \left(\sigma_{z s}\right)_{i}=\frac{G_{i} t_{i}}{(G J)_{\textit{eff}}} M_{z}.\end{aligned}\] If the shear modulus is the same in all branches, then eqs. ([eq4.65]) and ([eq4.68]) reduce to \[\begin{aligned} \label{eq4.69} \left(\sigma_{z s}\right)_{\textit{closed}}=\frac{J_{\textit{closed}}}{J_{\textit{eff}}} \cdot \frac{M_{z}}{2 A_{c} t} \quad\left(\sigma_{z s}\right)_{i}=\frac{t_{i} M_{z}}{J_{\textit{eff}}},\end{aligned}\] where \(J_{\textit{eff}}=J_{\textit{closed }}+\displaystyle\sum\limits_{i=1}^{n} J_{i}\).

[ex4.10]In multicell cross sections subject to pure torsion the shear flow is constant in each branch. In general the shear flows are different from branch to branch. Consider a cross section consisting of two cells with a horizontal axis of symmetry shown in figure [fig4.35]. It is subject to a torque \({M}_z\). Cell 1 is enclosed by a semicircular exterior web of radius \(a\), and a vertical web of length 2a, which is common with cell 2. Cell 2 is enclosed by an isosceles triangle with equal exterior webs of length \(c\) and the common web of length \(2a\). Take \(a = 5\) in., \(b = 12\) in., and \(c = 13\) in. The thickness of the exterior webs \(t = 0.040\) in., and the thickness of the common web is \(t/2\). The contour is composed of four branches. Branch 1 is the semicircular web with the contour coordinate denoted by \(s_1\), branch 2 is the upper exterior straight web of cell 2 with contour coordinate \(s_2\), branch 3 is the lower exterior straight web of cell 2 with contour coordinate \(s_3\), and branch 4 is the vertical common web between cells with the contour coordinate denoted by \(s_{\textrm{1-2}}\).The Cartesian coordinate system X-Y has its origin at point O, the center of the semicircular web. The X-axis is the axis of symmetry.

A free body diagram of the junction between branches 1, 2, and 1-2 is shown in figure [fig4.36]. The sum of forces in the axial direction is \(q_{\textrm{1-2}} \Delta z+q_{2} \Delta z-q_{1} \Delta z=0\), Hence, axial equilibrium per unit \(z\)-length determines the shear flow in the common web as \[\begin{aligned} \label{ex4.10a} q_{\textrm{1-2}}=q_{1}-q_{2}.\tag{a}\end{aligned}\]

A simple rule to determine the axial equilibrium of the junction is to observe that the shear flow into the junction must equal the shear flow out of the junction. At the junction of branches 2 and 3 this rule results in \(q_{3}=q_{2}\). At the junction of branches 3, 1-2, and 1 the rule gives \(q_{\textrm{1-2}}+q_{3}-q_{1}=0\), but \(q_{3}=q_{2}\), which leads us to the same relation given by eq. ([ex4.10a]). We started with four unknown shear flows \(q_1\), \(q_2\), \(q_3\), and \(q_{\textrm{1-2}}\). Axial equilibrium at the three junctions results in two relations between the four shear flows. Two shear flows remain unknown, say \(q_1\) and \(q_2\), at this point in the analysis.

The remaining equation of static equivalence is to equate the torque due to the shear flows to \({M}_z\). The constant shear flows in each branch are shown in figure [fig4.37](a), and the resultants of these shear flows are determined by the analysis presented in article 1.4.1. As shown in figure [fig4.37](b) the resultant of the shear flow \(q_{1}\) in branch 1 is a vertical upward force of magnitude \(2 a q_{1}\), the resultant of the shear flow \(q_{2}\) in branches 2 and 3 is a vertical downward force of magnitude \(2 a q_{2}\), and the resultant of the shear flow in the common branch is a downward force of magnitude \(2 a\left(q_{1}-q_{2}\right)\). The locations of the lines of action of the force resultants with respect to the common branch are also shown in figure [fig4.37](b). The vertical force F shown in figure [fig4.37](c) is determined from the branch forces by equilibrium. The result is \[\begin{aligned} \label{ex4.10b} F=2 a q_{1}-2 a q_{2}-2 a\left(q_{1}-q_{2}\right)=0.\tag{b}\end{aligned}\] Take the moment of the branch forces in figure [fig4.37](b) about the common web to determine the torque \({M}_z\) shown in figure [fig4.37](c). The result for static equivalence of the torque is \[\begin{aligned} \label{eq4.70} \left(\frac{2 A_{c 1}}{2 a}\right) 2 a q_{1}+\left(\frac{2 A_{c 2}}{2 a}\right) 2 a q_{2}=M_{z},\ \text{or}\nonumber\\ 2 A_{c 1} q_{1}+2 A_{c 2} q_{2}=M_{z}.\end{aligned}\] Equation ([eq4.70]) is the extension of Bredt’s formula in eq. ([eq4.58]) to two cells.

We have used all the conditions of static equivalence, but the two shear flows \(q_1\) and \(q_2\) remain unknown. An additional equation relating the shear flows is based on the assumed rigidity of the cross section in its own plane. In torsion, this rigid cross section condition implies the unit twist of each cell is the same. The unit twist for a single cell is given by eq. ([eq4.20]) on page . Since the shear modulus is the same for all branches in the cross section eq. ([eq4.20]) reduces to \[\begin{aligned} \label{eq4.71} \frac{d \phi_{z}}{d z}=\frac{1}{2 A_{c} G} \oint\! \frac{q}{t} ds.\end{aligned}\] This unit twist formula was derived on the basis that a counterclockwise shear flow tends to produce a counter-clockwise unit twist. Apply this equation to cell 1 to get \[\begin{aligned} \label{ex4.10c} \left(\frac{d \phi_{z}}{d z}\right)_{1}=\frac{1}{2 A_{c 1} G}\left[\left(\frac{\pi a}{t}\right) q_{1}+\left(\frac{2 a}{t/ 2}\right)\left(q_{1}-q_{2}\right)\right].\tag{c}\end{aligned}\] For cell 2, the unit twist is \[\begin{aligned} \label{ex4.10d} \left(\frac{d \phi_{z}}{d z}\right)_{2}=\frac{1}{2 A_{c 2} G}\left[\left(\frac{2 c}{t}\right) q_{2}-\left(\frac{2 a}{t/ 2}\right)\left(q_{1}-q_{2}\right)\right].\tag{d}\end{aligned}\] Note that the contribution of the common branch shear flow is negative in the unit twist formula for cell 2. Relative to an observer in cell 2, a positive value for the shear flow \(q_{\textrm{1-2}}\) tends to produce a clockwise unit twist, and hence is negative by the convention that counterclockwise is positive. Since the unit twist of each cell is the same, we equate eqs. ([ex4.10c]) and ([ex4.10d]) to get \[\begin{aligned} \label{ex4.10e} \frac{1}{A_{c 1}}\left[\left(\frac{2 a}{t}\right) q_{1}+\left(\frac{2 a}{t/ 2}\right)\left(q_{1}-q_{2}\right)\right]=\frac{1}{A_{c 2}}\left[\left(\frac{2 c}{t}\right) q_{2}-\left(\frac{2 a}{t/ 2}\right)\left(q_{1}-q_{2}\right)\right].\tag{e}\end{aligned}\] After some algebraic manipulations eq. ([ex4.10e]) is written in the form \[\begin{aligned} \label{ex4.10f} \left[\frac{4 a+a \pi}{A_{c 1} t}+\frac{4 a}{A_{c 2} t}\right] q_{1}-\left[\frac{4 a}{A c_{1} t}+\frac{4 a+2 c}{A_{c 2} t}\right] q_{2}=0.\tag{f}\end{aligned}\] The enclosed areas of each cell are \(A_{c 1}=\pi a^{2}\). and \(A_{c 2}=\frac{1}{2} b 2 a=b a\). Numerical evaluations of eqs. ([eq4.70]) and ([ex4.10f]) are \[\begin{aligned} \label{ex4.10g} \begin{gathered}(50 \pi\,\textrm{in}.^{2}) q_{1}+(120\,\textrm{in}.^{2}) q_{2}=M_{z} \\ (19.6995\,\textrm{in}.^{-2}) q_{1}-(25.5329\,\textrm{in}.^{-2}) q_{2}=0\tag{g}\end{gathered}.\end{aligned}\] The solutions of the two equations in ([ex4.10g]) for the shear flows are \[\begin{aligned} \label{ex4.10h} q_{1}=(4.00538 \times 10^{-3}\,\textrm{in}.^{-2}) M_{z} \quad q_{2}=(3.00903 \times 10^{-3}\,\textrm{in}.^{-2}) M_{z} \quad q_{\textrm{1-2}}=(0.915085 \times 10^{-3}\,\textrm{in}.^{-2}) M_{z}.\tag{h}\end{aligned}\] The shear stresses in each branch are obtained from \(\left(\sigma_{z s}\right)_{1}=q_{1}/t\), \(\left(\sigma_{z s}\right)_{2}=q_{2}/t\), and \(\left(\sigma_{z s}\right)_{\textrm{1-2}}=q_{\textrm{1-2}} /(t/2)\). The results are \[\begin{aligned} \label{ex4.10i} (\sigma_{z s})_{1}=(0.100135\,\textrm{in}.^{-3}) M_{z} \quad(\sigma_{z s})_{2}=(0.0772575\,\textrm{in}.^{-3}) M_{z} \quad(\sigma_{z s})_{\textrm{1-2}}=(0.0457542\,\textrm{in}.^{-3}) M_{z}.\tag{i}\end{aligned}\] The twist per unit length for the entire cross section is obtained from either eq. ([ex4.10c]) or eq. ([ex4.10d]). For the shear flows given in eq. ([ex4.10h]) the evaluation of unit twist is \[\begin{aligned} \label{ex4.10j} \frac{d \phi_{z}}{d z}=\left(\frac{d \phi_{z}}{d z}\right)_{1}=\left(\frac{d \phi_{z}}{d z}\right)_{2}=(0.0129263\,\textrm{in}.^{-4})(M_{z}/G).\tag{j}\end{aligned}\] Finally, we compute the torsion constant J by the following relation \[\begin{aligned} \label{ex4.10k} J=\frac{M_{z}}{G\left(\frac{d \phi_{z}}{d z}\right)}=77.3619\,\textrm{in}.^{4}\tag{k}\\[-30pt] \ \nonumber\end{aligned}\]

[ex4.11]Consider the cross section composed of three cells shown in figure [fig4.38](a). All branches have the same thickness \(t\) and shear modulus G. It is convenient to define circuit shear flows in each cell. Circuit shear flows are assumed to be positive counterclockwise in each cell and are equal to the actual shear flows in the exterior branches of the cell, if there are any exterior branches. However, the shear flow in a common branch between cells is the difference between the circuit shear flows sharing the common branch. At the junction of the three branches shown in figure [fig4.38](b) the shear flow into the junction is \(q_{1}-q_{2}\), and this equals the shear flow out of the junction, which is \(\left(q_{1}-q_{3}\right)+\left(q_{3}-q_{2}\right)\). The method of defining circuit shear flows automatically satisfies axial equilibrium at the junction.

For an applied torque \(M_{z}\), determine

• shear flows \(q_1\), \(q_2\), and \(q_3\),

• the torsion constant J, and

• magnitude of the torque at the initiation of yielding.

Solution.Bredt’s formula for the section composed of two cells in eq. ([eq4.70]) is extended to three cells in this example to get \[\begin{aligned} \label{ex4.11a} 2 A_{c 1} q_{1}+2 A_{c 2} q_{2}+2 A_{3} q_{3}=M_{z}.\end{aligned}\] The areas enclosed by cells are \[\begin{aligned} \label{ex4.11b} A_{c 1}=2 a^{2} \quad A_{c 2}=3 a^{2}/2 \quad A_{c 3}=a^{2}/2.\end{aligned}\] From eq. ([eq4.71]) the twist per unit length for this example reduces to \[\begin{aligned} \label{ex4.11c} \frac{d \phi_{z}}{d z}=\frac{1}{2 A_{c} G t} \oint\! q d s=\frac{1}{2 A_{c} G t} \sum \Delta s q.\end{aligned}\] We apply eq. ([ex4.11c]) to each cell and note that a shear flow is positive counterclockwise consistent with a positive counterclockwise unit twist. The results for each cell are as follows. \[\begin{aligned} \left(\frac{d \phi_{z}}{d z}\right)_{1} &=\frac{1}{2 A_{c 1} G t}\left[4 a q_{1}+\frac{3 a}{2}\left(q_{1}-q_{2}\right)+\frac{a}{2}\left(q_{1}-q_{3}\right)\right] \label{ex4.11d}\\ \left(\frac{d \phi_{z}}{d z}\right)_{2} &=\frac{1}{2 A_{c 2} G t}\left[\frac{5 a}{2} q_{2}+a\left(q_{2}-q_{3}\right)+\frac{3a}{2}\left(q_{2}-q_{1}\right)\right] \label{ex4.11e}\\ \left(\frac{d \phi_{z}}{d z}\right)_{3}&=\frac{1}{2 A_{c 3} G t}\left[\frac{3 a}{2} q_{3}+\frac{a}{2}\left(q_{3}-q_{1}\right)+a\left(q_{3}-q_{2}\right)\right].\label{ex4.11f}\end{aligned}\] Since the cross section is assumed to be rigid in its own plane, the unit twist of each cell must be the same. This kinematic condition can be written between cells 1 and 2 as \[\begin{aligned} \label{ex4.11g} \left(\frac{d \phi_{z}}{d z}\right)_{1}-\left(\frac{d \phi_{z}}{d z}\right)_{2}=0.\end{aligned}\] Evaluation of eq. ([ex4.11g]) leads to \[\begin{aligned} \label{ex4.11h} \frac{48 q_{1}-49 q_{2}+5 q_{3}}{24 G a t}=0.\end{aligned}\] Compatibility of the unit twist between cells 2 and 3 is \[\begin{aligned} \label{ex4.11i} \left(\frac{d \phi_{z}}{d z}\right)_{2}-\left(\frac{d \phi_{z}}{d z}\right)_{3}=0.\end{aligned}\] Evaluation of eq. ([ex4.11i]) leads to \[\begin{aligned} \label{ex4.11j} \frac{2\left(4 q_{2}-5 q_{3}\right)}{3 G a t}=0.\end{aligned}\] Solve eqs. ([ex4.11a]), ([ex4.11h]), and ([ex4.11j]) for the shear flows to find \[\begin{aligned} \label{ex4.11k} q_{1}=\frac{75}{604} \frac{M_{z}}{a^{2}} \quad q_{2}=\frac{20}{151} \frac{M_{z}}{a^{2}} \quad q_{3}=\frac{16}{151} \frac{M_{z}}{a^{2}}.\end{aligned}\] The unit twist of the section can be determined by substituting the shear flows ([ex4.11k]) into any one of the eqs. ([ex4.11d]), ([ex4.11e]), or ([ex4.11f]). The unit twist is \[\begin{aligned} \label{ex4.11l} \frac{d \phi_{z}}{d z}=\frac{149}{1208 G a^{3} t} M_{z}.\end{aligned}\] Compare this to the standard formula \(\frac{d \phi_{z}}{d z}=\frac{M_{z}}{G J}\) to find to find the torsion constant J. \[\begin{aligned} \label{ex4.11m} J=\frac{1208 a^{3} t}{149}.\end{aligned}\] The shear flows in the common branches are \[\begin{aligned} \label{ex4.11n} \left\{\left(q_{1}-q_{2}\right),\left(q_{1}-q_{3}\right),\left(q_{3}-q_{2}\right)\right\}=\{-0.0083,0.0182,-0.0265\}(M_{z}/a^{2}).\end{aligned}\] The magnitude of the largest shear flow is in the exterior branches of cell 2, which is also the location of the maximum shear stress. The maximum shear stress is \[\begin{aligned} \label{ex4.11o} \left(\sigma_{z s}\right)_{\max }=q_{2}/t=0.13245 M_{z} /(a^{2} t).\end{aligned}\] According to von Mises criterion ([eq4.31]) yield initiates at \(\sqrt{3}\left(\sigma_{z s}\right)_{\max }=\sigma_{\textrm{yield }}\), so the magnitude of the torque at the initiation of yielding is \[\begin{aligned} \label{ex4.11p} \left|M_{z}\right|_{\max }=4.359(a^{2} t \sigma_{\textrm{yield }}).\end{aligned}\]

If all the common branches were removed to make the section shown in figure [fig4.38](a) a single-cell, square section 2a by 2a, then from eq. ([eq3.161]) on page the torsion constant is \[\begin{aligned} \label{ex4.11q} \left.J\right|_{\textrm{single cell }}=\frac{4 A_{c}^{2}}{\oint\hspace*{-3pt}\frac{d s}{t}}=\frac{4(4 a^{2})^{2}}{\frac{8 a}{t}}=8 a^{3} t.\end{aligned}\] For this example, subdividing the single-cell section into three cells shown in figure [fig4.38](a) increases the torsional stiffness ([ex4.11m]) by only 1.32 percent with respect to the single-cell section, while the weight of the three-cell section increases by 37.5 percent with respect to the weight of the single-cell section. However, a multicell section may be required for improved damage tolerance; i.e., if we modeled damage as a longitudinal fracture, or cut, of an exterior branch, then the loss of torsional stiffness of the single-cell would be substantial since it becomes an open section. Damage to an exterior branch of a multicell section on the other hand results in less of a reduction in torsional load carrying capability since some closed cells remain intact to carry the torsional load.

[ex4.12]Consider the cross section of example [ex4.10] on page subject to a shear force \({V}_y\) with \(V_{x}=M_{z}=0\). There are two cells with a horizontal axis of symmetry as shown in figure [fig4.39]. I. Cell 1 is enclosed by a semicircular exterior web of radius \(a\), and a vertical web of length 2a, which is common with cell 2. Cell 2 is enclosed by an isosceles triangle with equal exterior webs of length \(c\) and the common web of length 2\(a\). Take \(a = 5\) in., \(b = 12\) in., and \(c = 13\) in. The thickness of the exterior webs \(t = 0.040\) in., and the thickness of the common web is \(t/2\). The contour is composed of four branches. Branch 1 is the semicircular web with the contour coordinate denoted by \(s_1\), branch 2 is the upper exterior straight web of cell 2 with contour coordinate \(s_2\), branch 3 is the lower exterior straight web of cell 2 with contour coordinate \(s_3\), and branch 4 is the vertical common web between cells with the contour coordinate denoted by \(s_{\textrm{1-2}}\). The Cartesian coordinate system X-Y has its origin at point O, the center of the semicircular web. The X-axis is the axis of symmetry. The Cartesian coordinates of each branch as a function of the contour coordinate are listed in table [tab4.5].

The cross-sectional area A, first area moment \({Q}_Y\), and location of the centroid \({X}_c\) are computed as follows: \[\begin{aligned} A &=\int_{0}^{a \pi} t d s_{1}+\int_{0}^{c} t d s_{2}+\int_{0}^{c} t d s_{3}+\int_{0}^{2a}(t/2) d s_{\textrm{1-2}}=(a \pi+2 c+a) t=1.86832\,\mathrm{in}.^{2}\label{ex4.12a}\\ Q_{Y} &=\int_{0}^{a \pi} X_{1} t d s_{1}+\int_{0}^{c} X_{2} t d s_{2}+\int_{0}^{c} X_{3} t d s_{3}+\int_{0}^{2a} X_{12}(t/ 2) d s_{1-s}=(2 a^{2}-b c) t=-4.24\,\textrm{in}.^{3}\label{ex4.12b}\\ X_{c} &=Q_{Y}/A=-2.26942\,\textrm{in}.\label{ex4.12c}\end{aligned}\] Symmetry about X-axis results in \(Q_{X}=0\), so that \(Y_{c}=0\). The Cartesian coordinates of the branches with respect to the centroid are \(x_{i}\left(s_{i}\right)=X_{i}\left(s_{i}\right)-X_{c}\), and \(y_{i}\left(s_{i}\right)=Y_{i}\left(s_{i}\right)\), \(i=1,2,3,1\)-\(2\). The second area moment about the \(x\)-axis is \[\begin{aligned} \label{ex4.12d} I_{x x}=\int_{0}^{a \pi} y_{1}^{2} t d s_{1}+\int_{0}^{c} y_{2}^{2} t d s_{2}+\int_{0}^{c} y_{3}^{2} t d s_{3}+\int_{0}^{2a} y_{12}^{2}(t/2) d s_{\textrm{1-2}}=\frac{\pi a^{3} t}{2}+\frac{2 a^{2} c t}{3}+\frac{a^{3} t}{3}=18.1873\,\textrm{in}.^{4}.\end{aligned}\]

The shear flow is given by eq. ([eq4.15]) on page , and it is repeated as eq. ([ex4.12e]) below. \[\begin{aligned} \label{ex4.12e} q(s, z)=q_{0}(z)-\frac{k}{I_{y y}} V_{x} \bar{Q}_{y}(s)-\frac{k}{I_{x x}} V_{y} \bar{Q}_{x}(s).\end{aligned}\] In this example the product area moment \(I_{x y}=0\). From eq. ([eq4.4]) cross-sectional coefficients \(n_{x}=n_{y}=0\) and \(k=1\). The distribution function defined in eq. ([eq4.9]) simplifies to \(\bar{Q}_{x}(s)=Q_{x}(s)\) since \(\bar{y}(s)=y(s)\). Hence, the shear flow equation in the i-th branch reduces to the form \[\begin{aligned} \label{ex4.12f} q_{i}\left(s_{i}\right)=q_{0 i}-\left(V_{y}/I_{x x}\right) Q_{x i}\left(s_{i}\right).\end{aligned}\] At the contour origin of the \(i\)-th branch where \(s_{i}=0\) the shear flow in eq. ([ex4.12f]) is denoted by \(q_{0 i}\), and the distribution function is given by \[\begin{aligned} \label{ex4.12g} Q_{x i}\left(s_{i}\right)=\int_{0}^{s_{i}} y_{i}\left(s_{i}\right) t_{i} d s_{i} \quad i=1,2,3,\textrm{1-2}.\end{aligned}\]

L122pt

Axial equilibrium per unit \(z\)-length at the three junctions connecting the branches leads to \[\begin{aligned} \label{ex4.12h} q_{1}(a \pi)=q_{02}+q_{012} \quad q_{2}(c)=q_{03} \quad q_{\textrm{1-2}}(2 a)+q_{3}(c)=q_{01}.\end{aligned}\] The shear flows acting at the three junctions are shown in figure [fig4.40]. We use the first two expressions in eq. ([ex4.12h]) to eliminate \(q_{012}\) and \(q_{03}\). After computing the first area moment functions, the shear flows are as follows: \[\begin{aligned} q_{1}\left(s_{1}\right)&=q_{01}+\left(V_{y}/I_{x x}\right) a^{2} t \sin \left(s_{1}/a\right)\label{ex4.12i}\\ q_{2}\left(s_{2}\right)&=q_{02}-\left(V_{y}/I_{x x}\right)\left[a t s_{2}-\frac{a t s_{2}^{2}}{2 c}\right]\label{ex4.12j} \\ q_{3}\left(s_{3}\right)&=q_{02}-\left(V_{y}/I_{x x}\right)\left[\frac{a c t}{2}-\frac{a t s_{3}^{2}}{2 c}\right]\label{ex4.12k}\\ q_{\textrm{1-2}}\left(s_{\textrm{1-2}}\right)&=q_{01}-q_{02}-\left(V_{y}/I_{x x}\right)\left[\frac{a t s_{\textrm{1-2}}}{2}-\frac{t s_{\textrm{1-2}}^{2}}{4}\right].\label{ex4.12l}\end{aligned}\] Note: If eqs. ([ex4.12k]) and ([ex4.12l]) are substituted into the third junction condition of eq. ([ex4.12h]), then we obtain the identity \(q_{01}=q_{01}\). Hence, the shear flows from the axial equilibrium conditions contain two unknowns \(q_{01}\) and \(q_{02}\).

The resultant force acting on the section from the shear flows is given by the general relation \[\begin{aligned} \label{ex4.12m} F_{X} \hat{i}+F_{Y} \hat{j}=\int_{c} q \hat{t} d s=\left[\int_{c} q \frac{d x}{d s} d s\right] \hat{i} +\left[\int_{c} q \frac{d y}{d s} d s\right]\hat{j}.\end{aligned}\]

Evaluation of the resultant forces gives \[\begin{aligned} F_{X} &=\int_{0}^{a \pi} q_{1}\left(\frac{d x_{1}}{d s_{1}}\right) d s_{1}+\int_{0}^{c} q_{2}\left(\frac{d x_{2}}{d s_{2}}\right) d s_{2}+\int_{0}^{c} q_{3}\left(\frac{d x_{3}}{d s_{3}}\right) d s_{3}+\int_{0}^{2a} q_{\textrm{1-2}}\left(\frac{d x_{12}}{d s_{12}}\right) d s_{\textrm{1-2}}=0, \text{and}\label{ex4.12n} \\ F_{Y} &=\int_{0}^{a \pi} q_{1}\left(\frac{d y_{1}}{d s_{1}}\right) d s_{1}+\int_{0}^{c} q_{2}\left(\frac{d y_{2}}{d s_{2}}\right) d s_{2}+\int_{0}^{c} q_{3}\left(\frac{d y_{3}}{d s_{3}}\right) d s_{3}+\int_{0}^{2a} q_{\textrm{1-2}}\left(\frac{d y_{12}}{d s_{12}}\right) d s_{\textrm{1-2}}=V_{y}.\label{ex4.12o}\end{aligned}\] The shear flow is statically equivalent to the shear force, as expected. No new information to determine \(q_{01}\) and \(q_{02}\) is obtained. The shear force acting at the shear center implies the twist per unit axial length of the cross section vanishes. This condition leads to two equations governing the shear flows in each cell. For a uniform shear modulus the twist per unit length is \[\begin{aligned} \label{ex4.12p} \frac{d \phi_{z}}{d z}=\frac{1}{2 A_{c} G} \oint\! \frac{q}{t} d s.\end{aligned}\] Evaluate the twist per unit length for each cell and equate them to zero: \[\begin{gathered} \label{ex4.12q} \left(\left.\frac{d \phi_{z}}{d z}\right|_{\textrm{cell } 1}=0\right) \rightarrow \frac{1}{t} \int_{0}^{a \pi} q_{1} d s_{1}+\frac{2}{t} \int_{0}^{2a} q_{\textrm{1-2}} d s_{\textrm{1-2}}=0 \\ \left(\left.\frac

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With the shear flows known we compute the torque about the centroid due to the shear flows by \[\begin{aligned} \label{ex4.12aa} M_{z c}=\int_{0}^{a \pi} r_{n c 1} q_{1} d s_{1}+\int_{0}^{c} r_{n c 2} q_{2} d s_{2}+\int_{0}^{c} r_{n c 3} q_{3} d s_{3}+\int_{0}^{2a} r_{n c 12} q_{\textrm{1-2}} d s_{\textrm{1-2}},\tag{aa}\end{aligned}\] where \(r_{n c i}\left(s_{i}\right)\) is the normal coordinate to the contour with respect to the centroid in the \(i\)-th branch. From eq. ([eq4.10]) the normal coordinates are determined from \[\begin{aligned} \label{ex4.12ab} r_{n c i}=x_{i}\left(s_{i}\right) \frac{d y_{i}}{d s_{i}}-y_{i}\left(s_{i}\right) \frac{d x_{i}}{d s_{i}}.\tag{ab}\end{aligned}\] The results for the normal coordinates are \[\begin{aligned} \label{ex4.12ac} r_{n c 1}=a-X_{c} \sin \left(s_{1}/a\right) \quad r_{n c 2}=a\left(b+X_{c}\right)/c \quad r_{n c 3}=a\left(b+X_{c}\right)/c \quad r_{n c 1\mbox{-}2}=X_{c}.\tag{ac}\end{aligned}\] Substitute eqs. ([ex4.12w]) to ([ex4.12z]) for the shear flows into eq. ([ex4.12y]), followed by substitution of eq. ([ex4.12ac]) for the normal coordinates. Numerical evaluation of the integrals after the substitutions leads to the expression for the torque in the form \[\begin{aligned} \label{ex4.12ad} M_{z c}=(2.50157\,\textrm{in}.) V_{y}.\tag{ad}\end{aligned}\] The resultant force and torque at the centroid are shown in figure [fig4.41](a). We also added and subtracted the shear force at the shear center in figure [fig4.41](a), which does not change the static state. The upward shear force at the centroid and the downward shear force at the shear center form a clockwise couple whose moment is \(x_{s c} V_{y}\). In figure [fig4.41](b) we resolved the torque \(M_{z}=(2.50157\,\textrm{in}.) V_{y}-x_{s c} V_{y}\) and shear force at the shear center. Since the torque at the shear center is equal to zero in this case, we can solve for the shear center location relative to the centroid to get \[\begin{aligned} \label{ex4.12ae} x_{s c}=2.50157\,\textrm{in}.\tag{ae}\end{aligned}\]

We perform one last check on the solution by computing the torque at the shear due to the shear flows from the equation \[\begin{aligned} \label{ex4.12af} M_{z}=\int_{0}^{a \pi} r_{n 1} q_{1} d s_{1}+\int_{0}^{c} r_{n 2} q_{2} d s_{2}+\int_{0}^{c} r_{n 3} q_{3} d s_{3}+\int_{0}^{2a} r_{n 12} q_{\textrm{1-2}} d s_{\textrm{1-2}},\tag{af}\end{aligned}\]

where the coordinates normal to contour with respect to the shear center are denoted by \(r_{n i}\left(s_{i}\right)\). From eq. ([eq3.10]) on page the normal coordinates are related by \[\begin{aligned} \label{ex4.12ag} r_{n i}=r_{n c i}-x_{s c} \frac{d y_{i}}{d s_{i}}+y_{s c} \frac{d x_{i}}{d s_{i}}.\tag{ag}\end{aligned}\] In this example \(y_{s c}=0\), and the results for the coordinates \(r_{n i}\left(s_{i}\right)\) are given in eqs. ([ex4.12ah]) and ([ex4.12ai]) below. \[\begin{gathered} r_{n 1}=r_{n c 1}-x_{s c}\left(\frac{\partial y_{1}}{\partial s_{1}}\right)=5-0.232149 \sin \left(s_{1}/5\right) \quad r_{n 2}=r_{n c 2}-x_{s c}\left(\frac{\partial y_{2}}{\partial s_{2}}\right)=4.70467\,\textrm{in}. \label{ex4.12ah}\tag{ah}\\ r_{n 3}=r_{n c 3}-x_{s c}\left(\frac{\partial y_{3}}{\partial s_{3}}\right)=4.70467\,\textrm{in. } \quad r_{n 1\mbox{-}2}=r_{n c 1\mbox{-}2}-x_{s c}\left(\frac{\partial y_{\textrm{1-2}}}{\partial s_{\textrm{1-2}}}\right)=0.232149\,\textrm{in}.\label{ex4.12ai}\tag{ai}\end{gathered}\] Substitute the shear flows from eqs. ([ex4.12w]) to ([ex4.12z]) into eq. ([ex4.12af]), followed by the substitution of the coordinates normal to the contour in eqs. ([ex4.12ah]) and ([ex4.12ai]). After these substitutions we perform the integrations indicated in eq. ([ex4.12af]) to find the result for torque at the shear center as \[\begin{aligned} \label{ex4.12aj}\tag{aj} M_{z}=\left(-1.421 \times 10^{-14}\right) \frac{V_{y}}{I_{x x}} \approx 0.\end{aligned}\] Hence, the numerical result for the torque at the shear center with respect to finite precision arithmetic is equal to zero¹.

r116pt

[ex14.13] Now consider that the cross section of example [ex4.10] and example [ex4.12] is subject to a torque \(M_{z}\) and a shear force \(V_{y}\) at the shear center as shown in figure [fig4.42]. We simply add the results for the shear flows due to the torque from example [ex4.10] to the shear flows due to the transverse shear force \(V_{y}\) from example [ex4.12].

The results are \[\begin{aligned} q_{1}(s_{1})&=(0.0040053\,\textrm{in}.^{-2}) M_{z}+\left[0.00342199\,\textrm{in}.^{-1}+(0.0549834\,\textrm{in}.^{-1})\right.\\[4pt] &\quad \left.\sin (s_{1}/5)\right] V_{y}, 0 \leq s_{1} \leq(5 \pi)\,\textrm{in}.,\\[4pt] q_{2}(s_{2})&=(0.00300903\,\textrm{in}.^{-2}) M_{z}+\left[0.0244374\,\textrm{in}.^{-1}-(0.0109967\,\textrm{in}.^{-2}) s_{2}\right.\\[4pt] &\quad +\left.(0.000422949\,\textrm{in}.^{-3}) s_{2}^{2}\right] V_{y},\quad 0 \leq s_{2} \leq 13\,\textrm{in}.\\[4pt] q_{3}(s_{3})&=(0.00300903\,\textrm{in}.^{-2})\; M_{z}+\left[-0.047041\,\textrm{in}.^{-1}+(0.000422949\,\textrm{in}.^{-3}) s_{3}^{2}\right] V_{y} \quad 0 \leq s_{3} \leq 13\,\textrm{in}., \text{ and}\\[4pt] q_{\textrm{1-2}}(s_{\textrm{1-2}})&=(0.000915085\,\textrm{in}.^{-2}) M_{z}+\left[-0.0210154\,\textrm{in}.^{-1}-(0.00549834\,\textrm{in}.^{-2}) s_{\textrm{1-2}}\right.\\[4pt] &\quad +\left.(0.00549834\,\textrm{in}.^{-3}) s_{\textrm{1-2}}^{2}\right] V_{y}\quad 0 \leq s_{\textrm{1-2}} \leq 10\,\textrm{in}.\end{aligned}\]

[sec4.5] Dowling, N. E. Mechanical Behavior of Materials: Engineering Methods for Deformation, Fracture, and Fatigue. Englewood Cliffs, NJ: Prentice Hall, 1993, Chapter 7.[Dowling]

Muckle, W. Strength of Ships’ Structures. London: Edward Arnold, Inc., 1967, pp. 5–12, 27–69.

Zubaly, R.M. Applied Naval Architecture. The Society of Naval Architects and Marine Engineers and Cornell Maritime Press, Inc., 1996, pp. 195–237.