University of Florida/Eas4200c.f08.radsam.d

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1.2 - Basic Structural Elements in Aircraft Structure
Aircraft structures are composed of several structural elements, which are designed to withstand various types of loads. It is the combination of these elements that make the entire structure of an aircraft capable of resisting applied loads.

For better comprehension of structural mechanics, we must introduce a few important definitions:


 * Stiffness: It is an extensive material property that determines the resistance of an elastic body to deflection. In Engineering, the modulus on elasticity or Young's Modulus (E) is an extremely important characteristic of materials, which shows how much a given material will be able deflect before a permanent deformation occurs. The Young's Modulus or is the radio of applied stress to strain: E = $$\displaystyle \sigma$$/$$\displaystyle \epsilon$$


 * Strength: The strength of a material is the ability to resist an applied force. In engineering, the Yield strength ($$\displaystyle\sigma$$y) is defined as "the stress at which a material begins to deform plastically." Elastic or elastoplastic materials are able to withstand certain loads and go back to their original shape when the given load is removed. This effect will occur as long as the applied stress is less than the Yield stress (strength). However, if the applied load exceeds the yield stress, the material will deform permanently even when the load is removed. If the strain or deformation keeps increasing, the material may reach its ultimate rupture stress ($$\displaystyle\sigma$$u) point where it will break.


 * Toughness: The difference between toughness and fracture toughness is that the former is the ability of a material to resist fracture, while the latter is the ability of the material to resist fracture when a crack is already present. If a material has a high value of fracture toughness, it will most likely undergo ductile fracture. This means that it will have the ability to bend (passed the yield strength) before it fractures. On the other hand, if a material has a low value of fracture toughness, such as ceramics, it will undergo brittle fracture.

1.2.1 - Axial Member

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! Axial Member

The length of axial members is significantly larger than their cross-sections and are meant to hold compressive and extensional loads applied along its length (axial direction). Deriving from the slope of the stress vs. strain curve, we may obtain a relation for the applied stress: $$\displaystyle\sigma$$ = E$$\displaystyle\epsilon$$

The applied stress $$\displaystyle\sigma$$ is equal to the applied force per the cross-sectional area of the axial member, so substituting in the above equation we obtain: F = EA$$\displaystyle\epsilon$$

EA is the axial stiffness, which is only a function of the Young's Modulus and the cross-sectional area. This tells us that the stiffness of a member does not vary by changing the shape of a cross-section unless the actual total area increases or decreases. For instance, if a member with a circular cross-section has the same area as a member with a square cross-section (or any other shape), the axial stiffness will be the same for both members. Axial members are good for tensile stresses but may cause buckling failure when compressed. Two ways of improving the buckling strength are to increase the bending stiffness or to shorten the buckle mode by adding supporting members at the joints.

In aircraft structures, aside of the longitudinal axial members, supporting members must be added at the wings and the fuselage to be able to withstand stresses. The two most important parts in a wing structure are the spars (2, refer to figure) and ribs (3). Spars extend lengthwise to the wing covering its entire length (from the fuselage to the wing tip). The main purpose of these members is to provide support to the wing and carry all loads back to the fuselage. The ribs on the other hand, give the aerodynamic shape to the wing while providing support to the spars.

Contribution by: radsam.d
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HW5
For simplification purposes, we look at a a 1-dimensional model first. Equating all forces on the x-direction, we obtain:

 $$\sum\ F_x = 0 = -\sigma_{(x)}\ A + \sigma_{(x + dx)}\ A + f_{(x)} dx\,$$ $$ 0 = A [ \sigma_{(x + dx)}\ - \sigma_{(x)}\ ] + f_{(x)} dx\,$$

Applying Taylor series to the term inside the brackets, it becomes: $$\frac{\partial\sigma_{x}}{\partial\ x} dx + higher~order~temrs\,$$

Recall: $$F_{(x + dx)} = F_{(x)} + \frac{\partial\ F_{(x)}}{\partial\ dx} dx + \frac{1}{2} \frac{\partial ^2 F_{(x)}}{\partial x^2} (dx)^2 + .~.~.\,$$

Neglecting higher order terms: $$\frac{\partial\sigma}{\partial\ x} + \frac{f_{(x)}}{A} = 0Applied~load\,$$

Now, we may look at a non-uniform 3-D field without applied loads, and focusing on the x-direction.

"Picture is worth a thousand words" - Vu-Quoc

The figure shows an infinitesimal element in which the stress is not uniform. However, the element must remain in equilibrium, therefore the six equations of equilibrium must be satisfied. For example, forces along the x-direction are:

 $$\sum\ F_x = 0 = dy dz [ -\sigma_{xx}\ (x,y,z) + \sigma_{xx}\ (x+dx,y,z)]\,$$ Facets with normal X

$$ + dz dx [ -\sigma_{yx}\ (x,y,z) + \sigma_{yx}\ (x,y+dy,z)]\,$$ Facets with normal Y

$$ + dx dy [ -\sigma_{zx}\ (x,y,z) + \sigma_{zx}\ (x,y,z+dz)]\,$$ Facets with normal Z

$$0 = (dx dy dz) \left [ \frac{\partial\sigma_{xx}}{\partial\ x} + \frac{\partial\sigma_{yx}}{\partial\ y} + \frac{\partial\sigma_{zx}}{\partial\ z} \right ]\,$$

HW 6
Consider plate of dimensions:

a = dimension along the x-axis b = dimension along the y-axis

First, we look at a 1-D case:

 $$[f] = \frac{F}{L}[A] = L^2\,\,$$

Therefore:

 $$\left [ \frac{f}{A} \right ] = \frac{F}{L^3}$$ $$[ \sigma ] = \frac{F}{L^2} \Rightarrow \left [ \frac{\partial\sigma}{\partial x} \right ] = \frac{\frac{F}{L^2}}{L}\,$$ $$\epsilon = \frac{du}{dx}and\epsilon = \frac{\triangle L}{L}\Rightarrow [\epsilon ] = \frac{[du]}{[dx]} = \frac{L}{L} = 1\,$$

$$\nu = -\frac{\epsilon_{yy}}{\epsilon_{xx}}\Rightarrow [\nu] = 1\,$$

From book...pg 71

The vector $$t$$ vanishes because no loads are applied on the lateral surface: $${t} = [\sigma] {n}$$

this way, the stress vector may be evaluated on the lateral surface, knowing that $$n_z = 0$$, thus: $$\begin{Bmatrix} t_x\\ t_y\\ t_z \end{Bmatrix} = \begin{bmatrix} 0 &0 & \tau _{xz}\\ 0 &0 &\tau _{yz} \\ \tau _{xz} &\tau_{yz} &0 \end{bmatrix} \begin{Bmatrix} n_x\\ n_y\\ 0 \end{Bmatrix}\, $$

Therefore, we have: $$t_x = 0t_y = 0$$ $$t_z = \tau_{xz} n_x + \tau_{yz} n_y = \frac{\partial \phi }{\partial y}n_x - \frac{\partial \phi}{\partial y}n_y$$

If we look at Figure 2, we may easily derive: $$n_x = \sin \eta = \frac{dy}{ds}\,$$ $$n_y = \cos \eta = \frac{dx}{ds}$$

So we may express $$t_z\,$$ as:

$$t_z = \frac{\partial \phi}{\partial y}\frac{dy}{ds} + \frac{\partial \phi}{\partial x}\frac{dx}{ds} = \frac{d\phi}{ds}\,$$ So, the free boundary condition $$t_z = 0\,$$ is given by: $$\frac{d\phi}{ds} = 0 or  \phi = constant$$ on the lateral surface.

Note: For solid sections with a single contour boundary, this constant may be approximated to zero.

We are interested in the shear stresses $$\tau_{xz}~and~\tau_{yz}\,$$ and the resultant torque. Considering a small area $$dA = dx dy\,$$, the torque is:

$$dT = x \tau_{yz} dA - y \tau_{xz} dA\,$$ $$dT = \begin{pmatrix} -x\frac{\partial \phi}{\partial x} -y\frac{\partial \phi}{\partial y} \end{pmatrix} dA\,$$