User:Eas4200c.f08.carbon.orear/HW2/Lecture Notes

 Comment: This page is identical to the previous one, except that the "Relating Resultant Force to Torque" was added and minor corrections fixed. Also, this comment box was not made correctly before. Here is a comparison between these two versions. Eas4200c.f08.carbon.orear 04:18, 7 November 2008 (UTC)

Motivation
We are motivated to study the box beam with a rectangular cross-section (as in Homework Problem 1.1) as a model for aerospace structures such as the fuselage or wings. As we explore more complex geometries we will decide the helpful and problematic aspects of certain thin-walled cross-sections.

Uniaxial Tension and Mohr's Circle
Taking a closer look at uniaxial tension, we recall the mechanics approach using the Mohr's Circle. This approach begins with reporting the principal stresses. There is no shear in pure uniaxial tension. We plot normal stress on the horizontal axis (remembering that shear is zero at the principal stresses) and enscribe a circle with a diameter of σ.

The image below (also seen under HW Problem 1.1) shows an example of how we get the Mohr's circle for the case of uniaxial tension.



σxx and σyy are the axial and transverse principal stresses, respectively. σxx is denoted on the plot as simply σ and σyy has a value of 0 in this case (which is plotted at the origin. The length between these two points gives us the diameter of our Mohr's circle.

With the vertical axis denoting shear stress, τ, we see that the maximum positive allowable shear stress (τallow) occurs at the top of the circle, and the maximum negative allowable shear stress (-τallow) occurs at the bottom of the circle. In the circle shown these values are denoted 'allowable.'

The relationship σallow = 2 τallow is derived from the Mohr's circle visually once we understand these values. The radius in the vertical direction is multiplied by 2 to equal the diameter in the horizontl direction. Since it is a circle, this calculation makes sense.

Stringers (axial members)
The first question we want to answer is: Should we use rectangular cross-sectional bars?

The answer is No, not always. We should consider the example of a thin-walled open beam. Stringers are structural reinforcement components for the fuselage, and are subjected to a distributed load. It is important to have a lightweight beam that is resistant to bending and buckling. We can create cross-sections both open and closed that have the same properties (or very similar) in resistance of buckling. Why then use open cross-sections? Manufacturing is one important consideration. Stamping flat sheets of metal into beam elements yields a simple, lower cost alternative to extrusion of a closed thin-walled rectangular beam. Also, riveting and hot-dimpling require access that may only be obtained with an open cross-section. Artisans working on the aircraft must be able to easily and quickly work on the stringer component and a closed-wall cross-section becomes a hindrance. We can further examine stringer geometry to make it easy to stack and store many components without making significant changes to the structural properties that are advantageous to our design intentions.

Case 1

Given a circle of radius R, the second area moment of inertia can be derived from the definition:
 * $$I_{y} = \iint_A z^2 \,dA$$

To convert the second moment of inertia equation in to polar coordinates you must substitute r and \theta for y and z.


 * $$ \begin{align}

\\y &= r\ cos(\theta) \\z &= r\ sin(\theta) \\dA &= r\ dr\ d\theta \end{align}$$

Which gives us


 * $$I_{y}^{(1)}= \int_0^R \int_{0}^{2\pi} r^3\ sin^2(\theta)\ d\theta\ dr$$

and integrating
 * $$I_y^{(1)}= \frac{R^4}{4}\ \frac{2\pi}{2}$$


 * $$I_y^{(1)}= \frac{\pi R^4}{4}$$

Case 2

To find the second moment of inertia of the channel shown above we start with the original equation for I.


 * $$I_{y} = \iint_A z^2 \,dA$$


 * $$I_{y}^{(2)}= \int_{-a/2}^{a/2} \int_0^b z^2 \,dy dz - \int_{\frac{-(a-t)}{2}}^{\frac{a-t}{2}} \int_0^{b-t} z^2 \,dy dz $$


 * $$I_y^{(2)}= \frac{1}{3} b(\frac{a^3}{8} + \frac{a^3}{8}) - \frac{1}{3} (b-t)(2(\frac{(a-t)}{8}^3)$$
 * $$I_y^{(2)}= \frac{1}{12} ba^3 - \frac{1}{12} (b-t)(a-t)^3$$


 * $$A^{(2)}= 3at = A^{(1)} = \displaystyle\pi R^2$$
 * $$t=\frac{a}{10} $$
 * $$I_y^{(2)}= \frac{1}{12} ba^3 - \frac{1}{12} (b-\frac{a}{10})(a-\frac{a}{10})^3$$

Solving for R
 * $$R =\sqrt{ \frac{3a^2}{10\displaystyle \pi}}$$
 * $$I_y^{(1)}=\frac{9a^4}{400\displaystyle \pi}$$

When you compare the two cross sections you find that Case 2 has a much higher moment of inertia than Case 1

Hooke's Law for Shear and Simple Shear
Shear Stress is the product of the shear modulus and shear strain.


 * $$\displaystyle \tau = G \gamma$$

The shear modulus (G) is a property of the material in question. The shear strain (γ) is based on the deformation displacement in the transverse (or shear) direction. This direction is parallel to the face of the material acted upon. This means that we can think of shear stress as tangential, or parallel, stress.


 * $$\displaystyle \gamma = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}$$

Note that v = uy and u = ux.

Simple shear is shown below.

We also want to define engineering shear strain as the total shear strain in the plane (here, x-y).

This value is exactly twice the tensorial shear strain, which only measures strain of x vs. y or y vs. x.

Shear Flow
Shear flow, (q) is force over contoured length.


 * $$\displaystyle q = \tau t$$

This means that shear flow is constant along the wall cross section despite wall thickness. This allows shear stress (τ) along the wall to be represented by shear flow (q) along the wall centerline. Shear flow forms a closed loop; this means that force resultants in the x and y directions due to shear flow are zero although a resultant torque is created as defined below.


 * $$\displaystyle dT = \rho q ds$$

The torque is dependent on the distance (ρ) from the origin to the wall segment (ds) and the shear flow (q) at that segment. Integrated, this becomes the following general equation for relating torque and shear flow.


 * $$\displaystyle T = \oint \rho q ds = \iint_\bar{A} 2 q dA = 2 q \bar{A}$$

Knowing that $$\bar{A} = ab$$ in the ad-hoc method used in problem 1.1 that shear flow is defined as $$q = \tau t$$, we find for problem 1.1, torque is defined as:


 * $$\displaystyle T = 2 (\tau t) (a b)$$

which can be rearranged as:


 * $$\displaystyle \tau = \frac{T}{2 a b t}$$

This proves that the ad-hoc method of elasticity seen in Problem 1.1 fits into our general case of elasticity as derived from shear flow.

Curved Panels
The concepts in this section, while discussed and continued in week three, are applicable and thus included in week two. The will be repeated for emphasis and completion in week three.

A shear panel (figure on right) is a thin sheet of material designed to take on an in-plane shear load. If the panel is of uniform thickness “t”, then the total shear forces in the x-direction and y-direction provided by the panel is given by

$$\displaystyle Vx = \tau*t*a = G*\gamma*t*a$$

$$\displaystyle Vy = \tau*t*b = G*\gamma*t*b$$

G is the shear modulus and $$\gamma$$ is the shear strain. Therefore the shear forces are proportional to the panel’s thickness and lateral dimensions.

The shear forces on a thin-walled curved panel (figure on lower right) under constant shear stress $$\tau$$ can be found in a similar way. Breaking down the shear force in horizontal and vertical components yields

$$\displaystyle Vx = \tau*t*a$$

$$\displaystyle Vy = \tau*t*a $$

Therefore the resulting components of the shear force can be related by

$$\displaystyle \frac{Vx}{Vy} = \frac{a}{b} $$

This proves that to handle a constant shear stress, a flat panel is most efficient because it saves unnecessary material from being used. This reduces weight which is one of the main concerns in the construction of airplanes and other aircraft structures.

The shear force on a thin curved panel can be found using shear flow. Referring to the diagram, shear flow, q, is constant with respect to the y and z axis and is defined by the following

$$\displaystyle q = \tau*t$$

With L being the length of the panel,

$$\displaystyle F = q*L$$

Therefore the force can be calculated by looking at a infinitesimal length of the panel and integrating over the entire length of the panel.

$$\displaystyle dF = q*dL = q*(dLy j + dLz k)$$

$$\displaystyle dF = q*(dL*cos(\Theta) j + dL*sin(\Theta) k)$$

The resultant shear force vector can then be calculated as

$$\displaystyle F = \int_{a}^{b}{dF} = q*[(\int_{a}^{b}{dy}) j + (\int_{a}^{b}{dz}) k]$$

$$\displaystyle F = q*(a j + b k)$$

Breaking it down into components

$$\displaystyle F = Fy j +Fz k$$

$$\displaystyle Fy= q*a$$

$$\displaystyle Fx = q*b$$

$$\displaystyle \frac{Fy}{Fz} = \frac{a}{b}$$

The magnitude of the resultant shear force is

$$\displaystyle \parallel F \parallel = [(Fy)$$2$$\displaystyle + (Fz)$$2$$\displaystyle ]$$1/2$$\displaystyle  = q*[a$$2$$\displaystyle  +b$$2$$\displaystyle ]$$1/2$$$$

Relating Resultant Force to Torque
The resultant force $$R = \displaystyle \parallel F \parallel$$ can be related to the torque by looking at the figure on the left with $$\displaystyle \vec{r}$$ being the position vector and $$\displaystyle \rho$$ being the perpendicular distance from the origin to dF. The torque can be derived by

$$\displaystyle d\vec{T}=\vec{r}Xd\vec{F} \Rightarrow dT  = p*dF $$

$$\displaystyle T = \oint_{}^{}{}dT = q*\oint_{}^{}{}p*dL$$ with p*dL = 2*dA

$$\displaystyle T = 2*q*\int dA$$

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

And therefore the relation can be visualized in the following equation:

$$\displaystyle T = \oint_{}^{}{} \rho *q*ds =\int \int_{\bar{A}}^{}{} 2*q*dA = 2*q*\bar{A}$$

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