User:Eas4200c.f08.carbon.w

=Aircraft Structures=

Course Information

Website - www.mae.ufl.edu/~vql → "teaching" → "my course website" → "EML 4500 Finite Element Anslysis and Design" EML 4500 Course website Password: fall08.wiki

Course Content

Finite Element Analysis (FEM) makes solving partial differential equations (PDEs) convenient. This is integral to the study of Aerospace Structures, or its structural components.

The emphasis of the course will be to understand the mechanics, formulate problems, judge solution correctness, and avoiding ad-hoc structural analysis methods.

The plan of the course is to maintain confidentiality while using large-scale peer-to-peer domains including Wikiversity and MIT's OpenCourseWare in addition to the textbook. The course will examine the method of work instead of other approaches and such as e-learning and the old approach to EML4500 (10% HW and 30% x 3 for each exam). This format allows students to be introduced to Media Wiki and stimulate collaboration between students with other students, teams, the professor, and the class as a whole.

There are several relevant time zones important to us (UTC, EDT, and EST). HW will be turned in according to the UTC time to maintain consistency with the Wikipedia time stamp format.

HW shortcuts should be created by attaching the URL from the archived version by going to "History" → click on the date to open archived version → copy this address to submit


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!Contribution Metals Stress-Strain Curve, from team Carbon
 * The graph depicts the general graph of the relationship between stresses and strains in metals.
 * The graph depicts the general graph of the relationship between stresses and strains in metals.


 * }

--

Steel alloys have very high stiffness and strength. Titanium alloys have less high stiffness and strength, and Aluminum alloys still less. Modern aluminum lithium alloys are 10% stiffer and 10% lighter than conventional aluminum alloys.

-Aluminum alloys are resistant to fatigue and cyclic tensile stress, and thus are used on the fuselage and the lower wing skins. Aluminum has historically played a major roll as material used in aircraft structures because it is cheap to buy, manufacture and is abundant, as opposed to Titanium which is expensive.

-Titanium is heavier than aluminum and lighter than steel, and double the yield and ultimate stress of aluminum. It is popular for its corrosion resistance and ability to withstand extreme temperatures up to 1000°F.

-Steel Alloys have extremely high stiffness and strength, but is expensive and has poor resistance to corrosion. Thus, despite its unique material property, is general reserved for landing gear and fittings with high loads.

Reference Sun [2006], pg. 14-15

Because of the very low toughness of glass, it has a very limited use on aircraft. Because of the extreme forces acting on the fuselage in-flight, windows are designed to be very small and sturdy. Recent discoveries of new glass-like materials allow for larger windows. Also, the study of metallic glass has become increasingly significant and could prove to have major importance on aircraft in the future. The article below describes some of the issues and discoveries:

Metallic Glass

Fiber-reinforced composites allow for complex loads by introducing fibers embedded in a matrix (polymers, metals, ceramics, etc.). They are stiff, strong, and light, and are ideal to withstand high loads experienced during flight while still minimizing weight to increase the aerodynamic properties. The composite laminates "have excellent fatigue life, damage tolerance, and corrosion resistance". Fiber composites are becoming very popular for their unique material properties and compose 50% of the new Boeing 787. More facts on the 787 can be found at:

Boeing 787 Fact Sheet

Reference Sun [2006], pg. 15-16

=Pb 1.1=

HW PROBLEM 1.1
BELOW IS THE SOLUTION FROM CLASS.

Consider a square tube (seen to the right)




 * wall thickness << height (a)
 * wall thichness << width (b)

ASSUMPTION 1: $$\displaystyle L = 2 * (a + b)$$

Q: Find optimal b/a ratio to maximize load bearing capability of beam assuming:

ASSUMPTION 2: $$\displaystyle M = T$$ ASSUMPTION 3: $$\displaystyle \sigma$$ allowable $$\displaystyle = 2*\tau$$ allowable

In other words, find optimal cross section carrying max bending moment and max torque:

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

Since cross section walls are very thin, we can assume shear stress distribution to be uniform along the wall. See the wikipedia article on shear flow to learn more. Shear Flow



Proof:

$$\displaystyle T = T$$ab $$\displaystyle + T$$bc $$\displaystyle + T$$cd $$\displaystyle + T$$da $$\displaystyle$$

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

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

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

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

Recalling that $$\displaystyle T = T$$ab $$\displaystyle + T$$bc $$\displaystyle + T$$cd $$\displaystyle + T$$da $$\displaystyle$$, this proves that $$\displaystyle T = 2 \tau a b t$$ and  $$\displaystyle  \tau = T/(2 a b t)$$.

$$\displaystyle $$

Case:1
Assume stress (bending normal stress) reaches stress allowable, first. What to verify that $$\displaystyle \tau << \tau$$allowable

Recall: $$\displaystyle Stress = \sigma = (M *z)/I$$

M = bending moment, Z = ordinate of a point on axis perpendicular to neutral bending axis

I = 2nd area moment of inertia

Assume $$\sigma_{max} = \sigma_{allowable}$$, that the maximum bending normal stress increases to the allowable stress first. The goal is to find the maximum allowable bending moment which corresponds to the max bending stress at z=b/2.

Recall: $$\sigma = \frac {Mz} I, ~ Z = \frac b 2$$

=> $$M = \frac {2 I \sigma_{max}} b \frac {2 I \sigma_{allowable}} b$$



To maximize the bending moment, simply maximize the ratio I/b. First, find the moment of inertia with respect to 'b',

Recall from assumption 1:

=> $$a = \frac L 2 - b$$

=> (I/b) fuction of b

Maximize M : Max = max, from assumption 2 (T=M)

According to assumption 3, the allowable shear stress is equal to half of the allowable normal stress. Therefore, if the max shear stress is less than the allowable shear stress, then the optimal ratio is acceptable. If not, then it is not acceptable.

Assume that the maximum bending normal stress reaches the allowable stress first. Then the moment of inertia of the cross section is the sum of the moments fo inertia of the cross section is the sum of the moments of inertia of the four segments of the cross section. If you use the parallel axis theorem, it is as follows:



t^3 << t, hence the approximation



Plot f(b) we get inverted parabola, 0 at 0 and 3L/4. max at 3L/8





We can see that this case can NOT exist! We have to violate a known equation to use this cross-section!

Case 2:
Assume torsional shear stress reaches the allowable shear stress first. Torsional shear stress is constant along the thin wall because the cross section is negligibly thin.

$$\tau^{(1)}_{max} < \tau_{allowable}$$

$$\tau_{max} = \tau_{allowable} = \tau = \frac T {2~a~b~t}$$

Maximize T by maximizing the product of a and b. For a constant perimeter L, the cross section of the box beam is square. The product of a and b is the area of the cross section and a=b, therefore L/4 = a = b. We need to check to make sure that the maximum bending stress for the cross section is below the allowable bending stress. If this is true, then the optimal ratio b/a = 1 is acceptable, otherwise it is not acceptable.

$$\frac {a^{(2)}} {b^{(2)}} = 1$$

$$T^{(2)}_{max} = (2 + \tau_{all}) * (\frac L 4) = \frac 1 8 + L_{tall} = M^{(2)}_{max}$$



$$M_{max} = 2 \sigma_{all} (\frac I b) = \frac {3 + L^2} {32} \sigma_{all}$$

$$\tau_{max} = \frac {M_{max}} {2abt} = \frac {M_{max}} {2 (\frac L 8) (3 \frac L 8) t} = \frac {32M_{max}} {3tL^2} = \sigma_{all}$$

Shear stress due to maximum Torsion(1)=Tmax(1)/(2a(1)b(1)t) = (by assumption 2) Mmax(1)/[2(L/8)(3L/8)t] = σallowable

τmax(1) = σallowable = (by assumption 3) 2τallowable > τallowable

Because τmax(1) > τallowable, therefore this case is not acceptable.

=HW3=

uy = -(PP')sinβ. Because PP'=(OP)α

→ uy = -α(OP)sinβ, where (OP)sinβ=zp

=Warping=

Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.

θ=α/x=rate of twist.

The following are mathematical definitions for the warping along the x,y, and z-axes.

1: y-displacement of a point on the cross section, horizontal

→ uy=-rαsinβ=-αy=-θxz

2: z-displacement of a point on the cross section, vertical

→ uz=v=rαcosβ=+(PP')cosβ=+αyp=+θxy, where PP'=OPα, α=θx, and yp=y because the variable is general for all coordinates

3: x-displacement, along centerline

→ ux=θψ(y,z), where this value is proportional to the rate of twist (θ)

These three equations compose the kinematic assumptions of warping.


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!Contribution Physical Representation of Torsion, from team Carbon
 * Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.
 * Torsion relates the rate of twist of an object and the displacement of any point along a non-centerline axis to an applied torque.



Contribution by carbon.w
 * }

Roadmap for Torsional Analysis of Aircraft Wing
Below is a picture of a cross section of a wing with a two-cell stringer-skin-web section with a thin, variable wall.

A) Kinematic Assumptions described above with the three equations.

B) Strain-displacement relationship:

εxx = εyy = εzz = γxy = 0

σxx = σyy = σzz = τxy = 0

$$\gamma _{xz} = \frac{\partial w}{\partial x} + \frac{\partial u}{\partial z}$$

$$\gamma _{yz} = \frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}$$

C) Equilibrium Equation for Stresses:

$$\frac{\partial \tau _{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} = 0$$

$$\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \tau _{yx}}{\partial y} + \frac{\partial \tau _{zx}}{\partial z} = 0$$

$$\frac{\partial \tau _{xy}}{\partial x} + \frac{\partial \sigma _{yy}}{\partial y} + \frac{\partial \tau _{zy}}{\partial z} = 0 $$

$$\frac{\partial \tau _{xz}}{\partial x} + \frac{\partial \tau _{yz}}{\partial y} + \frac{\partial \sigma _{zz}}{\partial z} = 0 $$

$$\tau _{xy} = \tau _{yx}$$

$$\tau _{yz} = \tau _{zy}$$

$$\tau _{xz} = \tau _{zx}$$

D) Prandtl Stress Funtion, $$\phi $$

$$\tau _{xz} = \frac{\partial \phi }{\partial y}$$

$$\tau _{yz} = \frac{\partial \phi }{\partial x}$$

E) Strain Compatibility

$$\frac{\partial \gamma _{yz}}{\partial x} - \frac{\partial \gamma _{xz}}{\partial y} = 2\theta $$

F) Equation for $$\phi $$

$$\frac{\partial ^{2}\phi }{\partial x^{2}} = -2G\theta$$

G) Boundary Conditions for $$\phi$$

$$\frac{\partial \phi }{\partial s} = 0$$ or $$\phi $$ = constant (on the lateral surface)

If cross section is solid and thus has only a single contour boundary, then the constant can be chosen arbitrarily as zero: $$\phi$$ = 0 on lateral surface of the bar

H) Torque, $$T = 2\int \int_{A}^{}{\phi dA}$$

→ $$T = GJ\theta $$, where using the above equation and the equation for $$\phi $$ we obtain $$J = -\frac{4}{\bigtriangledown^{2}\phi }\int \int_{A}^{}{\phi dxdy}$$ and $$(\phi dxdy = dA)$$

I) Thin Walled Cross Section

$$\tau = \frac{T}{2abt}$$ Ad-hoc assumption of shear flow

$$T = \oint_{}^{}{\rho qds} = \int \int_{\bar{A}}^{}{2qdA} = 2q\bar{A}$$

J) Twist Angle, θ: Method 1

\theta = \frac{1}{2G\bar{A}}\oint_{}^{}{\tau ds} = \frac{1}{2G\bar{A}}\oint_{}^{}{\frac{q}{t}ds}

=Multicell Airfoil=

To consider an example of an airfoil separated by spars (creating multicells), one must first consider the simplified case using right angles. In the figure below:

t1 = .3cm, t2 = .5cm, t3 = .4cm



Looking back to the Roadmap Section K, and using the equation $$\theta_i = \frac{1}{2G_i\bar{A_i}} \oint \frac{q_i}{t_i} ds$$ we can find θ as a function of T and J (torsional constant).

(1) $$T = T_{1} + T_{2} = 2q_{1}\bar{A_{1}} + 2q_{2}\bar{A_{2}}$$, where $$\bar{A_{1}} = ac$$ and $$\bar{A_{2}} = bc$$

(3) $$\theta_1 = \frac{1}{2G\bar{A_1}}[\frac{2q_{1}a}{t_{1}} + \frac{q_{1}c}{t_{1}} + \frac{(q_{1} - q_{2})c}{t_{12}}]$$

(4) $$\theta_2 = \frac{1}{2G\bar{A_2}}[\frac{2q_{2}a}{t_{2}} + \frac{q_{2}c}{t_{2}} + \frac{(q_{2} - q_{1})c}{t_{12}}]$$

Because the two cells are connected the angle of each cell must rotate together. In other words, C1 and C2 have the same rate of twist angle, where q1 and q2 are two unknowns.

(2) $$\theta_1 = \theta_2$$ Equations (1) and (2) can be used to solve for an expression of q1 and q2 in terms of the resultant of T.


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!HW Torsion of Multicell Airfoil, from team Carbon
 * $$q_{1} = \frac{T_{1}}{2\bar{A_{1}}}$$, where $$\beta _{1} = \frac{1}{2\bar{A_{1}}} = \frac{1}{2*a*c} \approx 4.167m$$
 * $$q_{1} = \frac{T_{1}}{2\bar{A_{1}}}$$, where $$\beta _{1} = \frac{1}{2\bar{A_{1}}} = \frac{1}{2*a*c} \approx 4.167m$$

→ $$q_{1} = 4.167T_{1}$$

$$q_{2} = \frac{T_{2}}{2\bar{A_{2}}}$$, where $$\beta _{2} = \frac{1}{2\bar{A_{2}}} = \frac{1}{2*b*c} \approx 2.083m$$

→ $$q_{2} = 2.083T_{2}$$

From (3): $$\theta = \frac{1}{2G\bar{A_1}}[\frac{2q_{1}a}{t_{1}} + \frac{q_{1}c}{t_{1}} + \frac{(q_{1} - q_{2})c}{t_{12}}]$$, where $$\theta _1 = \theta _2 = \theta$$

→ $$\theta = \frac{1}{2Gac}[\frac{24.167T_{1}a}{t_{1}}  \frac{4.167T_{1}c}{t_{1}}   \frac{(4.167T_{1} - 2.083T_{2})c}{t_{12}}]$$

→ $$\theta = \frac{1}{.24G}[T_{1}(833.4 + 333.36) + 416.7T_{1} - 208.3T_{2}]$$

→ $$\theta = \frac{1}{.24G}(1583.46T_{1} - 208.3T_{2})$$

Using $$T = GJ \theta $$ and $$\theta _1 = \theta _2 = \theta$$, plug into the equation for $$\theta$$

→ $$\theta = \frac{1}{.24G}(1583.46GJ \theta - 208.3GJ \theta)$$

→ $$\theta = 5729.83J \theta$$

→ $$J = 1.745e-4$$

Contribution by carbon.w
 * }

Use $$T = T_{1} + T_{2}$$ to find the first equation:

$$\bar{A}_{1} = a*c = (.3m)(.4m) = .12m^{2}$$ and $$\bar{A}_{2} = b*c = (.6m)(.4m) = .24m^{2}$$

→ $$T = 2q_{1}\bar{A}_{1} + 2q_{2}\bar{A}_{2} = .24q_{1} + .48q_{2}$$ (1)

Now solve for $$\theta _{1}$$ and $$\theta _{2}$$:


 * $$\theta _{1} = \frac{1}{2G\bar{A_1}}[\frac{2q_{1}a}{t_{1}} + \frac{q_{1}c}{t_{1}} + \frac{(q_{1} - q_{2})c}{t_{12}}]$$, where $$G_{1} = G_{2} = G$$

$$\theta _{1} = \frac{1}{.24G}[200q_{1} + 133.33q_{1} + 100(q_{1} - q_{2})]$$

$$\theta _{1} = \frac{1}{.24G}(433.33q_{1} - 100q_{2})$$


 * $$\theta _{2} = \frac{1}{2G\bar{A_2}}[\frac{2q_{2}a}{t_{2}} + \frac{q_{2}c}{t_{2}} + \frac{(q_{2} - q_{1})c}{t_{12}}]$$

$$\theta _{2} = \frac{1}{.48G}[240q_{2} + 80q_{2} + 100(q_{2} - q_{1})]$$

$$\theta _{2} = \frac{1}{.48G}(420q_{2} - 100q_{1})$$


 * $$\theta _{1} = \theta _{2}$$

$$\frac{1}{.24}(433.33q_{1} - 100q_{2}) = \frac{1}{.48}(420q_{2} - 100q_{1})$$

$$2013.875q_{1} = 1291.67q_{2}$$, where G cancels. (2)


 * Solving the system of equations of (1) and (2):

from (1): $$q_{1} = 4.167T - 2q_{2}$$

substitute into (2): $$(2013.87)(4.167T - 2q_{2}) = 1291.67q_{2}$$

→ $$q_{2} = .000783T$$


 * Substitute into (2):

→ $$q_{1} = .0005024T$$

Since $$\theta _{1} = \theta _{2} = \theta $$, we can plug into either $$\theta _{1}$$ or $$\theta _{2}$$

$$\theta _{2} = \theta = \frac{1}{.48G}(420q_{2} - 100q_{1})$$

$$\theta = \frac{.58}{G}T$$


 * Because $$\theta = \frac{T}{2GJ}$$, therefore $$\frac{.58}{G}T = \frac{T}{2GJ}$$

→ → $$J = .862m^{4}$$

$$\theta = \frac{1}{2G\bar{A_1}}[\frac{2q_{1}a}{t_{1}} + \frac{q_{1}c}{t_{1}} + \frac{(q_{1} - q_{2})c}{t_{12}}]$$

Lecture 22
$$\boldsymbol{\varepsilon } = \begin{bmatrix} \varepsilon _{11} & \varepsilon _{12} & \varepsilon _{13}\\ \varepsilon _{21} & \varepsilon _{22} & \varepsilon _{23}\\ \varepsilon _{31} & \varepsilon _{32} & \varepsilon _{33} \end{bmatrix} = [\varepsilon _{ij}]$$ (tensorial notation), where x → 1, y → 2, z → 3, i = row, j = column, and i,j = 1,2,3

By using the tensorial notation, it allows you to only need to memorize one equation for $$\boldsymbol{\varepsilon }$$ rather than three:

$$\varepsilon _{ij} = \frac{1}{2}(\frac{\partial u_{i}}{\partial x_{j}}  \frac{\partial u_{j}}{\partial x_{i}})$$

where x → x1, y → x2, z → x3

$$\varepsilon _{11} = \varepsilon _{xx} = \frac{1}{2}(\frac{\partial u_{1}}{\partial x_{1}}  \frac{\partial u_{1}}{\partial x_{1}}) = \frac{\partial u_{1}}{\partial x_{1}} = \frac{\partial u_{x}}{\partial x}$$


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!HW Strain Symmetry, from team Carbon
 * By symmetry, $$\varepsilon _{ij} = \varepsilon _{ji}$$. If we make this assumption, then:
 * By symmetry, $$\varepsilon _{ij} = \varepsilon _{ji}$$. If we make this assumption, then:

$$\varepsilon _{ij} = \frac{1}{2}(\frac{\partial u_{i}}{\partial x_{j}} + \frac{\partial u_{j}}{\partial x_{i}}) = \varepsilon _{ji} = \frac{1}{2}(\frac{\partial u_{j}}{\partial x_{i}} +  \frac{\partial u_{i}}{\partial x_{j}})$$

Contribution by carbon.w
 * }

Use $$T = T_{1} + T_{2}$$ to find the first equation:

$$\bar{A}_{1} = a*c = (.3m)(.4m) = .12m^{2}$$ and $$\bar{A}_{2} = b*c = (.6m)(.4m) = .24m^{2} $$

→ $$T = 2q_{1}\bar{A}_{1} + 2q_{2}\bar{A}_{2} = .24q_{1} + .48q_{2}$$ (1)

Now solve for $$\theta _{1}$$ and $$\theta _{2}$$:

=Equation of Equilibrium=

Using dimensional analysis, we can find the dimension, which is not the same as the unit. All measures have units, but some are dimensionless.

To find the dimension of $$\begin{bmatrix} \frac{f}{A} \end{bmatrix}$$, let's analyze each individual component:

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

→ $$\begin{bmatrix} \frac{f}{A} \end{bmatrix} = \frac{F}{L^3}$$

This same methodology can be applied to any measurement such as stress and strain. The ad-hoc proof below show that you can approach the same dimension no matter what form of a measure you use:

$$[\sigma ] = \frac{F}{L^{2}}$$ and $$[dx] = L \Rightarrow \begin{bmatrix} \frac{d\sigma }{dx} \end{bmatrix} = \frac{[d\sigma ]}{[dx]} = \frac{F/L^{2}}{L} = \frac{F}{L^{3}} = [f]$$

Below is the dimensional analysis for strain, and shows that strain is non-dimensional (but still has a unit).

Unit: $$\varepsilon = \frac{du}{dx} = \frac{\Delta L}{L}$$

Dimension: $$[\varepsilon ] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$ (non-dimensional)

Also, $$\nu = -\frac{\varepsilon _{yy}}{\varepsilon _{xx}} \Rightarrow [\nu ] = \frac{[\varepsilon _{yy}]}{[\varepsilon _{xx}]} = 1$$


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!Contribution Eigenvalues and Buckling, from team Carbon
 * Eigenvalues are very important in determining material capabilities and making judgments for better designs. A study at Rice University using eigenvalue analysis showed that a column would be strongest if it were widest at the top, middle, and bottom of the column and skinny 1/4 distance from the top and 1/4 from the bottom.
 * Eigenvalues are very important in determining material capabilities and making judgments for better designs. A study at Rice University using eigenvalue analysis showed that a column would be strongest if it were widest at the top, middle, and bottom of the column and skinny 1/4 distance from the top and 1/4 from the bottom.



This study showed that this is the most material-conscious and strength-conscious way to build a column. Eigenvalues showed that naturally a column, regardless of its shape buckles at the top and lower quarter location. To conserve material, the thinning of these sections would not make a contribution to the buckling.

Contribution by carbon.w
 * }


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!HW Particularize symmetrical cross section, from team Carbon
 * For the second moment of inertia about the y-axis then the z-axis, the symmetry would be at the centroid if the cross section is symmetrical. Because there is symmetry about the centroid, integration on either side would cancel out and equal zero.
 * For the second moment of inertia about the y-axis then the z-axis, the symmetry would be at the centroid if the cross section is symmetrical. Because there is symmetry about the centroid, integration on either side would cancel out and equal zero.

Considering if $$M_{z} = 0$$ and that $$I_{yz} = 0$$. From previous lectures, it was shown that by definition

$$D:= I_{y}I_{z} - (I_{yz})^{2}$$. Therefore, $$D = I_{y}I_{z}$$

$$\Rightarrow k_{y} = \frac{I_{y}}{D} = \frac{I_{y}}{I_{y}I_{z}} = \frac{1}{I_{z}}$$

$$k_{yz} = 0$$

$$k_{z} = \frac{I_{z}}{I_{y}I_{z}} = \frac{1}{I_{y}}$$

$$\sigma _{xx} = \begin{bmatrix} z & 0 \end{bmatrix}\begin{bmatrix} \frac{1}{I_{y}} & 0\\ 0 & \frac{1}{I_{z}} \end{bmatrix}\begin{Bmatrix} M_{y}\\ 0 \end{Bmatrix}$$

$$\sigma _{xx} = \begin{bmatrix} z & 0 \end{bmatrix} \begin{Bmatrix} \frac{M_{y}}{I_{y}}\\ 0 \end{Bmatrix}$$

$$\Rightarrow \sigma _{xx} = \frac{zM_{y}}{I_{y}}$$

Contribution by carbon.w
 * }


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!HW Matrix Form of q(s), from team Carbon
 * The following equation can can be written in matrix form $$q(s) = -(k_{y}V_{y} - k_{yz}V_{z})Q_{z} - (k_{z}V_{z} - k_{yz}V_{y})Q_{y}$$
 * The following equation can can be written in matrix form $$q(s) = -(k_{y}V_{y} - k_{yz}V_{z})Q_{z} - (k_{z}V_{z} - k_{yz}V_{y})Q_{y}$$

$$q(s) = \begin{bmatrix} -Q_{z} & -Q_{y} \end{bmatrix}\begin{bmatrix} k_{y} & -k_{yz}\\ -k_{yz} & k_{z} \end{bmatrix}\begin{Bmatrix} V_{y}\\ V_{z} \end{Bmatrix}$$

To arrive at the particular case where $$q(s) = -\frac{V_{z}Q_{y}}{I_{y}}$$, make the assumption that $$Q_{z} = 0$$ and $$I_{yz} = 0$$

Therefore $$D = I_{y}I_{z}$$ and this can be plugged into the k-values:

$$\Rightarrow k_{y} = \frac{I_{y}}{D} = \frac{I_{y}}{I_{y}I_{z}} = \frac{1}{I_{z}}$$

$$k_{yz} = 0$$

$$k_{z} = \frac{I_{z}}{I_{y}I_{z}} = \frac{1}{I_{y}}$$

The matrix then becomes

$$q(s) = \begin{bmatrix} 0 & -Q_{y} \end{bmatrix}\begin{bmatrix} \frac{1}{I_{z}} & 0\\ 0 & \frac{1}{I_{y}} \end{bmatrix}\begin{Bmatrix} V_{y}\\ V_{z} \end{Bmatrix} = q(s) = -\frac{V_{z}Q_{y}}{I_{y}}$$

Contribution by carbon.w
 * }

Lecture 38
$$Q_{y}^{(3)} = \int_{A_{3}}^{}{zdA_{3}}$$

$$Q_{y}^{(2)} = z_{2}A_{2}$$

Resuperposition of Problem 2
The basic formula for shear flow along the skin:

$$q_{ij} = \tilde{q}_{ij} + q_{k}$$

Following this formula, several equations can be found:


 * $$q_{12} = \tilde{q}_{12} + q_{1}$$, where $$\tilde{q}_{12}$$ is known as zero and $$q_{1}$$ is unknown


 * $$q_{23} = \tilde{q}_{23} + q_{1} - q_{2}$$


 * $$q_{31} = \tilde{q}_{31} + q_{1} - q_{3}$$


 * $$q_{24} = \tilde{q}_{24} + q_{2}$$


 * $$q_{43} = \tilde{q}_{43} + q_{3} - q_{4}$$


 * $$q_{41} = \tilde{q}_{41} + q_{4}$$

Looking at all these equations, there are 3 unknowns, and thus we need 3 equations.

1) The first equation would be one equation; finding the moments of Vy and Vz, and $${q_{12}, q_{23}, q_{31}, q_{24}, q_{43}, q_{41}}$$ about any convenient point.  The point about which the moment is taken should be where the lines of action of $$V_{y}$$ and $$V_{z}$$ intersect.

Because each of the three sections are combined, if one section twists by a certain angle, then the other two must also twist by the same angle. Therefore,

2) $$\theta _{1} = \theta _{2}$$ 3) $$\theta _{2} = \theta _{3}$$

Lecture 39
To solve problem 2, follow each step for each cell:


 * Follow the path "$$s_{i}$$"
 * Write the equilibrium of each stringer on the path "$$s_{i}$$"

There are two ways to write the equilibrium:

1) Method shown in previous section, using FBD 2) The consequence of the first method is shown below:



The equilibrium equation is: $$\tilde{q}_{j6} = \tilde{q}_{3j} - \tilde{q}_{j5} - \tilde{q}_{j7}  q^{j} $$



The equilibrium equation is: $$-\tilde{q}_{6j} = \tilde{q}_{3j} - \tilde{q}_{j5} - \tilde{q}_{j7}  q^{j} $$

Note:

Consider what would happen if the cell walls were cut such that one stringer was isolated. Theoretically, it cannot be solved.



$$\tilde{q}_{23} = \tilde{q}_{31} = \tilde{q}_{34} = 0$$



$$\tilde{q}_{31} = \tilde{q}_{23} - \tilde{q}_{34} + q^{(3)}$$

This means that $$q^{(3)} = 0$$, but this is not true!

A second example is shown below:



$$\tilde{q}_{24} = \tilde{q}_{23} = \tilde{q}_{12} = 0$$

$$\tilde{q}_{23} = \tilde{q}_{12} - \tilde{q}_{24} + q^{(2)}$$

This shows that $$q^{(2)} = 0$$ which also is not true!

Using the peace sign, which was popular in the '60s:




 * $$\tilde{q}_{15} = \tilde{q}_{54} = \tilde{q}_{42} = \tilde{q}_{21} = 0$$

$$\tilde{q}_{15} = \tilde{q}_{31} + \tilde{q}_{21} + q^{1}$$


 * $$\Rightarrow \tilde{q}_{31} = -q^{(1)}$$


 * $$\tilde{q}_{53} = q^{(5)}$$


 * $$\tilde{q}_{43} = q^{(4)}$$


 * $$\tilde{q}_{23} = q^{(2)}$$


 * $$\tilde{q}_{15} = \tilde{q}_{31} + \tilde{q}_{21} + q^{1}$$

$$\tilde{q}_{31} = \tilde{q}_{53} + \tilde{q}_{43} + \tilde{q}_{23} + q^{(3)}$$

$$\Rightarrow \tilde{q}_{31} = q^{(5)} + q^{(4)} + q^{(2)} - q^{(1)}$$

There are four unknowns, so there will need to be four equations described in the previous section:

1) Moment equation

2) $$\theta _{1} = \theta _{2}$$

3) $$\theta _{2} = \theta _{3}$$

4) $$\theta _{3} = \theta _{4}$$



$$\tilde{q}_{53} = \tilde{q}_{43} = \tilde{q}_{23} = \tilde{q}_{31} = 0$$

$$\tilde{q}_{31} = \tilde{q}_{53} + \tilde{q}_{43} + \tilde{q}_{23} + q^{(3)}$$

This leads to $$q^{(3)} = 0$$ which is not true.