User:Eas4200C.f08.spars.mcgilvray/homework3

=Homework #3=

Big Picture Road Map
In order to build up to the torsional analysis of multi-cell structures Dr. Vu-Quoc has suggested the following set of steps. The points in bold represent areas of particular importance.


 * Kinematic assumptions
 * Strain displacement relationship
 * Equilibrium equation for stresses
 * Prondtl stress function $$\phi \,\! (x,y)$$
 * Strain compatibility equation
 * Equation for $$\phi \,\!$$
 * Boundary conditions for $$\phi \,\!$$
 * Relationship between $$t$$ and $$\phi \,\!$$
 * Thin-walled cross section
 * Twist angle $$\theta \,\!$$: method one
 * Section 3.6 multi-cell thin-walled cross section

Meeting 1
A shear panel (fig 15) is a thin sheet of material that is intended to withstand in-plane shear forces. The shear force in the panel is distributed across the depth of the panel, but because the panel is assumed to be very thin this force may be viewed as being constant across the depth. With this assumption the internal forces can be seen as a shear flow (q) represented by the equation
 * $$ q=\tau \!t$$

Where $$\tau \!$$ is the uniform shear stress and $$\!t$$ is the thickness of the panel.

With this in mind the panel may be viewed as a line in the (Z,Y) plane (fig 16). The shear force acting in the panel can therefore be broken up into an infintessimal force ($$\overrightarrow{dF}$$) acting on an infintessimal length ($$\overrightarrow{dF}$$) represented by the following equation.


 * $$ \overrightarrow{dF} = q\overrightarrow{dL} = q(dL_y\overrightarrow{j}+dL_z\overrightarrow{k})$$

Zooming in on an infintessimal length (fig. 18) allows this equation to be taken further.


 * $$\overrightarrow{dF} = q(dL cos \theta \overrightarrow{j} + dL sin\theta\overrightarrow{k}) = q(dy\overrightarrow{j} + dz\overrightarrow{k})$$

The resultant shear force vector can then be determined by integrating this infintessimal force along the line representing the panel as follows.


 * $$\overrightarrow{F}=\int_A^B\overrightarrow{dF}=q[(\int_A^Bdy)\overrightarrow{j} + (\int_A^Bdz)\overrightarrow{k}]=q(a\overrightarrow{j} + b\overrightarrow{k})$$

Breaking this force into components


 * $$\overrightarrow{F}=F_y\overrightarrow{j} + F_z\overrightarrow{k}$$ → $$F_y=qa\!$$ $$F_z=qb\!$$

Recalling that $$ q=\tau \!t$$ this equation is the same as equations (1.4) and (1.5) (Sun, pg.5). Calculating the resultant shear force magnitude


 * ||$$\overrightarrow{F}$$|| $$=[(F_y)^2 + (F_z)^2]^{1/2}= q[a^2+b^2]^{1/2} = qd \!$$

Where d is the length of a straight line from point A to point B and does not depend of the curve. This equation is the same as equation (3.49a) (sun, pg.85)

Relating the force (||$$\overrightarrow{F}$$||) to torque (T)


 * $$T=2q\overline{A}$$

Which is proven by considering a thin walled cross-section with a shear flow along that wall.


 * $$T=\oint \rho q ds = \int \int_{\overline{A}} 2q dA = 2q\overline{A}$$ (equation 3.48 Sun, pg 85)

Meeting 2


The equations previously derived for a closed thin walled cross-section are also valid for an open thin walled cross-section (fig 19).


 * $$T_0=2q\overline{A}$$ → $$Re=T=2q\overline{A}$$

In considering a uniform bar of circular cross section assume the cross-section behaves like a rigid disc without warping(as it turns out this is the first point on Dr. Vu-Quoc's "road map", the kinematic assumptions). In other words after a torque is applied the planal section of the shaft both remains in the plane and circular, also the diameter in the plane remains straight. From this assumption it can be derived that shear strain is a linear function of the radial distance between a point and the center of the circular section (figs. 20 and 21). Note that these assumptions are only valid for circular cross-sections.


 * $$T=\int\int_A rG\gamma dA = \int\int_A r\tau dA $$ (Hooke's Law)
 * $$\gamma=r\frac{d\alpha}{dx}$$, $$\frac{d\alpha}{dx}=: \theta$$ rate of twist
 * $$T=\int_A rG(r\theta)dA=G\theta\int_Ar^2dA\!$$
 * $$\int_Ar^2dA= J$$

Where $$ J $$ is the second polar area moment of inertia.

It can then be shown that $$J$$ is proportional to $$\overline{A}^{\frac{3}{2}}$$ with a proportionality factor of $$2\pi^{-\frac{1}{2}}t$$.

Meeting 3
In discussing the plotting of an airfoil Dr. Vu-Quoc recommended taking the quadrature (dividing an area into a number of rectangles). In this manner the area of a complex shape can be reduced down to a sum of a number of rectangles that fit into that shape(fig. 22). Of course the more rectangles the area is divided into the more accurate the approximation becomes. From this approximation
 * $$\overrightarrow{dT}=\overrightarrow{r}\times \overrightarrow{dF}=\overrightarrow{r} \times(q\overrightarrow{PQ})$$
 * $$\overrightarrow{dT}=q(\overrightarrow{r}\times \overrightarrow{PQ})=q(2dA\overrightarrow{i})$$

As opposed to a uniform, circular bar under torsion a uniform, non-circular bar experiences what is known as warping. Warping is the axial displacement of a point on the deformed cross section along the length of the bar.

Before continuing with warping it is helpful to make a few notes on small angle approximations.

Assuming the angle $$\alpha\!$$ is very small (fig. 24)


 * $$\overline{PP'}\perp\overline{OP}$$
 * $$\overline{OP'}>\overline{OP}$$
 * $$v=-(\overline{PP'})sin(\beta)=-\alpha(\overline{OP})sin(\beta)$$

Recalling that


 * $$Rsin(\alpha)\!$$≈$$R\alpha\!$$
 * $$Rcos(\alpha)\!$$≈$$R\!$$

Meeting 4
Continuing with the torsion of a uniform bar with a non-circular cross-section (fig.25), The rate of twist ($$\theta\!$$) can then be given by the total angle of rotation over the length in the direction of the axis normal to the cross-section.


 * $$\theta\!=\frac{\alpha}{x}$$

For small displacement angles the components of the displacement can be expressed as follows


 * $$v=u_y=-\overline{PP'}sin(\beta)=-\overline{OP}\alpha sin(\beta)=-\alpha x_p=-\theta xz$$
 * $$w=u_z= \overline{PP'}cos(\beta)= \overline{OP}\alpha cos(\beta)= \alpha y_p= \theta xy$$

The displacement in the x-direction is independent of x, therefore


 * $$u=u_x=\theta\psi(y,z)\!$$

Note that the displacement in the (y,z) plane is the same as if it were a rigid disc, while the displacement in the x-direction varies with the warping function $$\psi(y,z)$$. These equations illustrate the kinematic assumptions in the first point of the previously mentioned "road map".

Meeting 5
Two equations form the basis of the torsional analysis of a multi-cell thin walled cross-section.


 * $$T=2q\overline{A}\!$$
 * $$\theta\!=\frac{1}{2G\overline{A}}\oint\frac{q}{t}ds$$

Noting that $$T=\sum_{i=1}^n T_i$$ and that $$\theta=\sum_{i=1}^{n}\theta_i$$ these equations become


 * $$T=2\sum_{i=1}^nq_i\overline{A_i}$$
 * $$\theta_i=\frac{1}{2G_i\overline{A_i}}\oint\frac{q_i}{t_i}ds$$

Where $$q_i$$ is the shear flow in cell i, $$\overline{A_i}$$ is the average area in the cell,$$G_i$$ is the shear modulus of cell i, and $$t_i$$ is the thickness of cell i.

Note that $$t_i\!$$ can be viewed as a function of the curvilinear coordinates along the cell wall.

Chapter 3: Torsion
Chapter 3 begins the in-depth coverage of torsion, and it's effect on aircraft structures. The mechanisms that are covered in the chapter are as follows

Matlab Project
=Contributing Team Members= --Eduardo Rondon Eas4200c.f08.spars.rondon 11:52, 8 October 2008 (UTC)

--Eas4200c.f08.spars.prey 16:48, 8 October 2008 (UTC)

-- Thomas McGilvray --Eas4200C.f08.spars.mcgilvray 18:39, 8 October 2008 (UTC)

--Michael Lee Eas4200c.f08.spars.lee 19:15, 8 October 2008 (UTC)

=Placeholder for Figures=

=References=

2. Sun, C.T. (2006). Mechanics of Aircraft Structures John Wiley & Sons, New York. ISBN 0471699667.

3. Vu-Quoc, Loc (2008) Lecture Notes University of Florida