User:EAS4200C.F08.WIKI.A/My Contribution: 3

=Open Thin Walled Cross-Sections=



The torque of the depicted thin walled structure is

 $$\displaystyle T = 2q\bar{A}$$

This is also equivalent to the location of the resultant force acting parallel to the line connecting the two ends of the object.

=Uniform Circular Cylinder: Non-Warping=

We will now take a look at a uniform circular cylinder cross-section. This cross-section behaves as a rigid disk.



We will look at the infinitesimal area $$\displaystyle dA $$ at a distance $$\displaystyle r $$ away from the origin. The torque here can be described by the following equation;

$$\displaystyle T = \int\int_{A} r\tau dA $$

It must be noted that;

$$\displaystyle \tau = G\gamma $$, $$\displaystyle \tau dA = F $$

where $$\displaystyle G $$ is the strain modulus and $$\displaystyle \gamma $$ is the shear strain. Together they comprise Hooks Law which is used to describe the deformation of the disk. The variable $$\displaystyle F $$ is the force and when multiplied by the displacement produces the moment around the origin.The shear strain is described by the following equation;

 $$\displaystyle \gamma = r \frac{d\alpha}{dx}$$

where $$\displaystyle \frac{d\alpha}{dx} = \theta $$ which is defined as the rate of twist. Since $$\displaystyle \tau $$ is proportional to $$\displaystyle \gamma $$ which is also proportional to $$\displaystyle r $$, the shear stress $$\displaystyle \tau $$ gets smaller in the material closer to the core because the distance away from the core $$\displaystyle r $$ gets smaller. Therefore it is more efficient and cost effective to remove material to make the cylinder lighter. The limits of the material will be tested but if calculations are done correctly then there shouldn't be anything to worry about.

The torque can be represented by the following;



$$\displaystyle T = \int _A r G (r\theta) dA = G\theta\int _Ar^2dA $$

where $$\displaystyle dA $$ is a function of $$\displaystyle dy $$ and $$\displaystyle dz $$.



$$\displaystyle dA = dy * dz = r(dr)(dw)$$

The strain modulus and rate of twist is not a function of $$\displaystyle z $$ or $$\displaystyle y $$ and is why they come out of the integral. The resulting integral is defined as the second polar moment of inertia.

$$\displaystyle J = \int _Ar^2dA $$

=Solid Circular Cross-Section vs Hollow Thin Walled Cross-Section=

In order to compare these two cross-sections, we will be using the figure from the text Mechanics of Aircraft Structures by C.T. Sun. The figure is depicted below with all dimensions labeled on the figure.



We will first compute the area for each cross-section.

$$\displaystyle  A_a = \pi r^2 = \pi (1cm)^2 = 3.1416 cm^2 $$

$$\displaystyle A_b = \pi r_o^2 - \pi r_i^2 = \pi (5.1cm)^2 - \pi (5cm)^2= 3.173cm^2$$

As the computtion shows, the areas are similar despite the size difference. We will next compute the polar moment of inertia for each cross section.



$$\displaystyle J_a = \frac{\pi r^4}{2} = \frac{\pi (1cm)^2}{2} = 1.5707cm^4 $$

$$\displaystyle J_b = \frac{\pi (r_0^4 - r_i^4)}{2} = \frac{\pi ((5.1cm)^4-(5cm)^4)}{2} = 80.927cm^4$$

Comparing the two Polar moment of inertia shows that the thin walled cross-section is a better torsional member than the solid cross-section. Since its polar moment of inertia (measure of its resistance to torsion) is much larger, it is a more desirable cross-section to use. The polar moment of inertia for the solid cross-section is only $$\displaystyle 2\% $$ of the polar moment of inertia for the hollow thin walled cross-section.

$$\displaystyle \frac{J_a}{J_b} = \frac{1.5707cm^4}{80.927cm^4} = .0194 $$

If the thickness to equal $$\displaystyle .02 r_i^{(c)}$$, where $$\displaystyle c $$ represents a new hollow thin walled cross-section, what would this inner radious $$\displaystyle r_i^{(c)}$$ equal such that the polar moment of inertia for figure (a) equals that of the new figure (c)?

