User:Eas4200c.f08.radsam/HW3

=Homework #3=

Class Notes
 September $$19^{\mbox{th}}$$ - Engineering Shear Strain  Engineering shear strain, $$\gamma\,$$, is measurement of the total deformation of a material under a shear stress. It is the counterpart of normal strain, $$\epsilon\,$$.

Engineering shear strain can be defined as:

$$\gamma = \frac{du}{dy} + \frac{dv}{dx} = \frac{du_{x}}{dy} + \frac{du_{y}}{dx}\,$$

Which is shown in Figure 1.

Where:

$$u = u_{x}\,$$ = the displacement along the x-direction

$$v = u_{y}\,$$ = the displacement along the y-direction

Contributed by radsam.a and .m.

 September $$22^{\mbox{nd}}$$ - Curved Panels  Figure 1 shows a thin curved panel being exposed to a shear stress. With the help of this diagram, we may derive the following two equations of the book:

$$V_y = \tau\,t a\,$$ (1.4)

$$V_z = \tau\,t b\,$$ (1.5)

where:

$$V_y\,$$ is the shear stress in the y direction.

$$V_z\,$$ is the shear stress in the z direction.

A side view of the curved panel is shown in Figure 2. In this diagram we may see how the force $$d\vec{F}\,$$ acts on a small section $$dl\,$$ at and angle $$\theta\,$$. This force may be decomposed in two components along $$Y\,$$ and $$Z\,$$. Also, knowing that the force is equal to the shear flow $$q\,$$ times the length, we obtain the following relation:

$$q = \tau\, t\,$$

So,

$$d\vec{F} = q dl = q (dl_y + dl_z)\,$$

Following the geometry on Figure 2, we know:

$$dl_y = dl cos \theta\,$$

$$dl_z = dl sin \theta\,$$

Therefore:

$$d\vec{F} = q (dl cos\theta\, + dl sin\theta\,)\,$$

$$d\vec{F} = q (dy + dz)\,$$



A detailed drawing of the shear force being applied on the small element $$dl\,$$ is shown in Figure 3. We may clearly see how $$d\vec{F}\,$$ is out an an angle and how it may be decomposed into components. To obtain the total value of $$\vec{F}\,$$, we integrate $$d\vec{F}\,$$ from point $$A\,$$ to $$B\,$$. The resultant force is computed as follows: $$\vec{F} = \iint_{A}^{B} d\vec{F}\, = q ((\iint_{A}^{B} dy)\hat{j}\, + (\iint_{A}^{B} dz)\hat{k}\,)\,$$

If we look at Figure 2 again, one may see that the horizontal distance from $$A\,$$ to $$B\,$$ is equal to $$a\,$$, and that the vertical distance between $$A\,$$ and $$B\,$$ is equal to $$b\,$$. Therefore whe may solve the two integrals above and obtain an expression for $$\vec{F}\,$$. $$\vec{F} = q (a\hat{j} + b\hat{k})\,$$

This last equation corresponds to equations 1.4 and 1.5, because we know that:  $$q = \tau\, t$$ The expression for F shown above is still decomposed for 'y' and 'z' coordinates, but one may compute the total resultant force in the actual direction of stress:

$$|| \vec{F} || = (F_y^2 + F_z^2)^{(1/2)}\,$$

$$|| \vec{F} || = (q (a^2 + b^2)^{(1/2)})\,$$ Where:

$$d = (a^2 + b^2)^{(1/2)}\,$$ -- Length of a straight line.

So we conclude:

$$\vec{F} = q d\,$$ -- Which is essentially Eq. 3.49a

Content by radsam.d Drawings by radsam.a

 September $$22^{\mbox{nd}}$$ - Proof of equation 3.48  Figure 1 shows a constant flow on a closed thin-walled section. Equation 3.48 implies that the torque 'T' is equal to 2 times the shear flow and the enclosed area. It is written as follows:  $$T = 2 q \bar A\,$$ Where $$\bar A\,$$ is the enclosed area and $$q\,$$ is the shear flow. We know that the torque at a point on the thin wall is the radius times the force being applied. Also, the force is equal to the shear flow times the length of the element.  $$dT = \vec r\ x d \vec F\,$$ $$dT = \rho\, dF\, = \rho\, (q dl)\,$$ Figure 2 shows a closer view of the element $$dl\,$$ and the force $$dF$$. Due to the irregular shape of the cross-section, the radius may not always be perpendicular to the wall, so we draw a vector $$\rho\,$$ perpendicular to $$dF\,$$. This forms the triangle shown in Figure 2, so the area of the triangle $$dA$$ may be computed:  $$dA = \frac{1}{2} \rho\, dl\,$$ The total torque $$T\,$$ applied to the thin-walled is the integral of $$dT\,$$

 $$T = \oint dT\, = q \oint \rho\, dl\,$$ But, $$\rho\, dl\, = 2~dA\,$$

Therefore:

$$T = 2q \int_{A} dA =  2q \bar A\,\,$$

Content by radsam.d Drawings by radsam.a

 September $$26^{\mbox{th}}$$ - Thin-Walled Cross Sections 

First, the torque of an open thin-walled cross section (Figure 1) is examined.



From the torque equations applied in the last lecture:

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

The total torque applied on this cross section generates a resultant force, which can be expressed as:

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

Secondly, the cross section of a uniform bar is examined. Ultimately we want to compare a thin-walled cross section to a solid cross section and determine which performs better when placed under a torsional load.

The equation for the torsion in a cross sectional area (as shown in Figure 2) can be found by taking the integral of the area:

$$T=\int \int_{A}^{}{}r\tau dA=\int \int_{A}^{}{}rG\gamma dA$$ (from Hooke's Law)

The angle gamma can be defined as: $$\gamma = \frac{r d\alpha }{dx}$$

Where: $$\frac{d\alpha }{dx} \equiv \theta$$ (rate of twist)

Thus, torsion is: $$T = \int_{A}^{}{rG(r\theta )dA}=G\theta \int_{A}^{}{r^{2}dA}$$

Where: $$\int_{A}^{}{r^{2}dA} = J$$ = 2nd polar area moment of inertia.


 * See homework section below for the examination of J for a solid circular cross-section: Proof of Moment of Inertia for a solid circular cross-section

Now, we need to consider how all this can apply to aircraft structures. Typically, the fuselage of an airplane is NOT a solid cylinder, but rather a hollow thin-walled circular cross section. In order to express the torsion for such a case, our equation for J needs to be revised somewhat.


 * First, we must note that a thin wall can be defined in terms of an inner radius and an outer radius. Observing Figure 6 to the right: rinner = a    router = b  and  b-a=t where t is the thickness.


 * Notice that because the structure is a thin wall, we can assume t<<a and t<<b. Also, the average radius will be defined as:


 * $$\bar{r}=r_{avg}=\frac{a+b}{2}$$


 * Using our expression for a solid cylinder: $$J=\frac{1}{2}\pi a^{4}$$


 * We can re-express the equation for a thin-walled cylinder (as shown in Figure 3): $$J=\frac{1}{2}\pi (b^{4}-a^{4})=\frac{1}{2}\pi ((b^{2})^{2}-(a^{2})^2)=\frac{1}{2}\pi (b-a) (b+a) (b^2 + a^2)$$


 * Notice that $$\left(b-a \right)=t$$


 * $$(b+a)=2\bar{r}$$


 * and $$(b^2 + a^2) = 2\bar{r}^2$$  This last expression is proven rigorously in the homework section below titled "Proof of Moment of Inertia for a hollow circular cross-section"

Therefore:

$$J=2\pi t\bar{r}^{3}$$

It is also interesting to note that this equation can be expressed as $$J=(2\pi ^{-\frac{1}{2}}t)(\pi \bar{r}^{2})^{3/2}$$. Noticing that $$(\pi \bar{r}^{2})$$ is an expression for the average area of the cross-section, it can be deduced that

J is proportional to $$\bar{A}^{3/2}$$ with a $$\left( 2\pi ^{-1/2}t\right)$$ proportionality factor.

 September $$29^{\mbox{th}}$$ - NACA Airfoil 

ns = # of segments (as depicted in Figure 2) to discretize the y-axis.



As shown in Figure 1 and Figure 3:

$$dT = \vec r\ x ~d \vec F\,$$

$$ = \vec r\ x ~\vec {PQ}\,$$

$$ = q\vec r\ x ~\vec {PQ}\,$$ Torsion of uniform, non-circular bars (leads to warping of cross-section) Warping = Axial displacement x-axis (i.e, along the bar length) of a point on the deformed (rotated) cross-section.



Figure 4 and Figure 5 depict: $$Vy ~= ~y~-~component ~of ~displacement~vector~\vec {PP'}\,$$ $$Vz ~= ~z~-~component ~of ~displacement ~vector~\vec {PP'}\,$$ $$~\vec {OP}\ = ~\vec {PP'}\ = R,$$ $$Vz = Rsin \theta\, = R \alpha\, (small~\alpha\,),$$ $$Vy = R(1-cos \alpha\,) =~ 0,$$

 October $$1^{\mbox{st}}$$ - Torsion and Angle of Twist 

Torque is a type of load typically found in aircraft structures, most typically in beams and bars that serve as supports in aircraft. Torque occurs in the longitudinal axis of a shaft or beam.

To further analyze the torsion and rotation of a bar undergoing torsion (Figure 1), we shall use Saint-Venant's displacement theorem, which is based on assumptions of the displacement field.

If we observe at the figure, we can see that the rate of twist is defined as...

$$\theta=\frac{\alpha}{z}\,$$ Where $$\alpha\,$$ is equal to the total angle of rotation at $$z\,$$ in relation to the end where $$z\,$$ is equal to 0.

"Saint-Venant assumed that during torsional deformation, plane sections warp, but their projections on the x-y plane rotate as a rigid body" From this assumption we can conclude that u and v, the in-plane displacement components, follow a rigid body rotation as well. For example, if we refer back to our figure above, point P is displaced through the angle alpha to P'. Then we can see, if the rotation angle, alpha is small, that the displacement components are...

$$u=-r\alpha sin\beta = -\alpha y = -\theta zy\,$$

$$v=r\alpha cos\beta = \alpha x = \theta zx\,$$

Where $$r\,$$ = distance from origin to point $$P\,$$

The displacement in the z-direction is assumed to be independent of z:

$$w(x,y) = \theta \psi (x,y)\,$$

Where $$\psi (x,y)\,$$ is the warping function.
 * Note: The rate of twist is independent of z

Road Map for Torsional Analysis of Aircraft Wing:


 * (A) Kinematic Assumption
 * (B) Strain Displacement
 * (C) Equilibrium Equation (Stresses)
 * (D) Prandtl Stress Function $$\phi\,$$
 * (E) Strain Compatibility Equation
 * (F) Equilibrium
 * (G) Boundary Condition for $$\phi\,$$
 * (H) $$T=2\iint_{A} \phi \,dA\,$$

$$T=GJ\theta\,$$

$$J=\frac{-4}{\nabla^2 \phi}\,$$


 * (I) Thin-Walled cross section (Problem 1.1 Ad Hoc assumption on shear flow)

$$T=2q\bar B\,$$


 * (J)Twist angle $$\theta\,$$ : Method 1

$$\theta=\frac{1}{2G\bar A} \oint\frac{q}{t} ds\,$$

 October $$3^{\mbox{rd}}$$ - Multicell Section 
 * (K) Multicell Section

$$all~i= 1,2,3,....,n_c$$


 * (K1)   For a discrete sum:

$$T = 2 \sum_{i=1}^{n_c} q_i \bar A_i$$

$$\displaystyle q_i$$ = shear flow in all i

$$\displaystyle A_i$$ = "average" area in all i

Define: $$\displaystyle T_i =2 q_i A_i$$ torque generated by one cell

$$T = \sum_{i=1}^{n_c} T_i $$


 * (K2) Shape of airfoil is "rigid" in the plane (y,z) but can warp out of plane.

$$\theta = \theta_1 = \theta_2 =~...~= \theta_{n_c}$$

$$\theta_i=\frac{1}{2G_i\bar A_i} \oint\frac{q_i}{t_i} ds\,$$

$$\displaystyle G_i$$ = shear modulus in all i

$$\displaystyle t_i$$ = thickness in all i

$$\displaystyle T =2 q A$$ So shear flow is: $$\displaystyle q =\frac{T}{2\bar A}$$

where $$\displaystyle T $$ is the variable.

Twist Angle $$\displaystyle \theta $$:

$$\theta = \frac{1}{2 G \bar A} \sum_{i=1}^{3} \frac{q_i l_i}{t_i}$$

HW 3 Contributors
Eas4200c.f08.radsam.a. 22:06, 7 October 2008 (UTC) Eas4200c.f08.radsam.z 04:16, 8 October 2008 (UTC) Eas4200c.f08.radsam.m 16:55, 8 October 2008 (UTC) Eas4200c.f08.radsam.s 17:50, 8 October 2008 (UTC) Eas4200c.f08.radsam.d 18:50, 7 October 2008 (UTC) Eas4200c.f08.radsam.r 18:40, 8 October 2008 (UTC)