User:Eas4200c.f08.spars.lee/Homework 3

22 Sep notes
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)

comparing J
The area of the two cylinders are determined.

case a


 * $$\pi \,\! r^2 = \pi \,\! (0.01\mathbf{m})^2 = (0.0001)\pi \,\!$$ $$ \mathbf{m^2}$$

case b


 * $$\pi \,\! r_o^2 - r_i^2 =\pi \,\! [(0.051\mathbf{m})^2 - (0.05\mathbf{m})^2] =(0.000101)\pi \,\!$$ $$ \mathbf{m^2}$$

As the difference in area is negligible the two areas can be approximated as being equal which means that in both cases the same amount of material is used such that the shape resulting in the more efficient torsion member can be determined.

The torsional constant for the two cylinders are determined as follows,

case a


 * $$J=\frac{1}{2}\pi \,\!r^4 = \frac{1}{2}\pi \,\!(0.01\mathbf{m})^4 = (0.000000005)\pi \,\!$$ $$\mathbf{m^4}$$

case b

$$J = \frac{1}{2}\pi \,\!(r_o^4 - r_i^4) = \frac{1}{2}\pi \,\!(r_o - r_i)(r_o - r_i)(r_o^2 + r_i^2) = \frac{1}{2}\pi \,\!(0.001\mathbf{m})(0.101\mathbf{m})(0.05101\mathbf{m}) = (0.0000002576)\pi \,\!$$ $$\mathbf{m^4}$$

Comparing the two,


 * $$\frac{J^{(a)}}{J^{(b)}} = 0.0194099002$$

In other words the torsional constant of the shape in case a is less than two percent of the torsional constant of the shape in case b despite both shapes having the same area. To illustrate this consider a hollow cylinder of interior radius $$r_i$$ and thickness $$ t = 0.02 r_i$$ where the torsional constant is equal to the torsional constant of case a.


 * $$J = \frac{1}{2}\pi \,\![r_i(1+0.02)-r_i][r_i(1+0.02)+r_i][[r_i(1+0.02)]^2+r_i^2] = (0.000000005)\pi \,\! = \frac{1}{2}\pi \,\!(0.02r_i)(2.02r_i)(2.0404r_i^2) = 0.04121608r_i^4\pi \,\!$$

Which leads to


 * $$\!r_i = 0.0186627582$$ $$\mathbf{m}$$


 * $$\!t =  0.000373255$$ $$\mathbf{m}$$

Where the area is determined by


 * $$A=\pi \,\!(0.0190360134^2-0.0186627582^2)=(0.00001407126)\pi \,\!$$ $$\mathbf{m^2}$$
 * $$A=\pi \,\!(0.0190360134^2-0.0186627582^2)=(0.00001407126)\pi \,\!$$ $$\mathbf{m^2}$$

Comparing this area to the area of case a


 * $$\frac{A^{(a)}}{A^{(c)}} = 7.10668352$$

Which indicates that although both cases have the same torsional constant the area of this third case is over seven times less than the area of case a.

Big Picture Road Map
In order to build up to the torsional analysis of multicell structures Dr. Vu-Quoc has suggested the following set of steps.


 * Kinematic assumptions
 * Strain displacement relationship
 * Equilbrium equation for stresses
 * Prondtl stress function $$\phi \,\! (x,y)$$
 * Strain compatability equation
 * Equation fo $$\phi \,\!$$
 * Boundary conditions for $$\phi \,\!$$
 * Relationship between $$t$$ and $$\phi \,\!$$
 * Thin-walled cross section
 * Twist angle $$\theta \,\!$$: method one
 * '''Section 3.6 multicell thin-walled cross section