User:Eas4200c.f08.radsam.z

Stability

December 8th, 2008
pg 40.3 continued...

$$0=q^{(1)}+q^{(2)}+q^{(4)}=-q^{(3)}\ne o,$$

....not true or possible! since....

$$0=q^{(1)}+q^{(2)}+q^{(3)}+q^{(4)}\,$$

$$0=\sum_{e=1}^{4} q^{e}\,$$

pg 38-2:

$$q^{e}=n_{z}Q_{z}^{e}+n_{y}Q_{y}^{e}\,$$

$$n_{z}:= -(k_{y}V_{y}-k_{yz}V_{z})\,$$

$$n_{y}:=-(k_{z}V_{z}-k_{zy}V_{y})\,$$

$$\sum_{e=1}^{4} q^{e}=n_{z} \sum_{e=1}^{4} Q_{z}^{e}+n_{y} \sum_{e=1}^{4} Q_{y}^{e}\,$$

where... the sums above are equal to zero!



$$Qy=(\hat {z})=\int_{A(\hat {z})}zdA\,$$

$$Qy=\int_{A}zdA=0\,$$


 * Recalling the Mean Value Theorem, the above would simply be...

$$Qy=Zc\int_{A}dA\,$$

where...

$$Zc=0\,$$

HW #7: NACA Airfoil - "Back of envelope" verification




 * Looking at the figure above, we can see that...

$$\bar B \bar E=\bar F \bar H=\frac{1}{2}(BE+FH)\,$$

Verify magnitude of: Iy, Iz, Iyz --> (Section pg. 34-4) Angle B of N.A. $$\sigma_{xx}\,$$ @ each stringer

Read and Report


When considering beams with arbitrarily shaped cross-sections, as shown in the figure above, we place the external loads passing through the shear center/center of twist if we are avoiding torsion.

Studying the bidirectional bending in this cross-section, we obtain the longitudinal displacement as a function of x,y,z...

$$u=u_{0}(x)+z\psi_{y}(x)+y\psi_{z}(x)\,$$  (4.22a)

$$v=v_{0}(x)\,$$    (4.22b)

$$w=w_{0}(x)$$       (4.22c)

...where

$$\psi_{y}, \psi_{z}\,$$...are rotations of the cross-section about the y and z axes

The strains can be defined as...

$$\epsilon_{xx}=\frac{\partial u}{\partial x} = \frac{du_{0}}{dx}+z\frac{d\psi_{y}}{dx}+y\frac{d\psi_{z}}{dx}\,$$   (4.23a)

$$\gamma_{xy}=\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}=\frac{dv_{0}}{dx}+\psi_{z}\,$$  (4.23b)

$$\gamma_{xz}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}=\frac{dw_{0}}{dx}+\psi_{y}\,$$  (4.23c)


 * However recalling the assumptions... $$\gamma_{xy}=\gamma_{xz}=0\,$$...

$$\psi_{z}=-\frac{dv_{0}}{dx}\,$$

$$\psi_{y}=-\frac{dw_{0}}{dx}\,$$

...which we would substitute into equation 4.23c, and this yields...

$$\epsilon_{xx}=\frac{du_{0}}{dx}-y\frac{d^{2}v_{0}}{dx^{2}}-z\frac{d^{2}w_{0}}{dx^{2}}\,$$   (4.24)

...recalling     $$\frac{du_{0}}{dx}=0\,$$  ...if...  $$N_{x}=0\,$$...

$$\epsilon_{xx}=-y\frac{d^{2}v_{0}}{dx^{2}}-z\frac{d^{2}w_{0}}{dx^{2}}\,$$          (4.25)


 * The Bending moments about the y and z axes can be written as...

$$M_{y}=\iint_{A}z\sigma_{xx}dA=-E\iint_{A}(yz\frac{d^{2}v_{0}}{dx^{2}}+z^{2}\frac{d^{2}w_{0}}{dx^{2}})dA=-EI_{yz}\frac{d^{2}v_{0}}{dx^{2}}-EI_{y}\frac{d^{2}w_{0}}{dx^{2}}\,$$  (4.26)

$$M_{z}=\iint_{A}y\sigma_{xx}dA=-EI_{z}\frac{d^{2}v_{0}}{dx^{2}}-EI_{yz}\frac{d^{2}w_{0}}{dx^{2}}\,$$ (4.27)

where...

$$I_{y}=\iint_{A}z^{2}dA\,$$(4.28a) moment of inertia about y-axis

$$I_{z}=\iint_{A}y^{2}dA\,$$ (4.28b) moment of inertia about z-axis

$$I_{yz}=\iint_{A}yzdA\,$$ (4.28c) product of inertia

 Matrix Form: 

$$\begin{Bmatrix} M_{y} \\ M_{z} \end{Bmatrix}= \begin{vmatrix} I_{y} & I_{yz} \\ I_{yz} & I_{z} \end{vmatrix} \begin{Bmatrix} -E\chi_{z} \\ -E\chi_{y} \end{Bmatrix}\,$$

$$\epsilon_{xx}=-y\chi_{y}-z\chi_{z}\,$$

$$\epsilon_{xx}=\begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -\chi_{z} \\ -\chi_{y} \end{Bmatrix}\,$$

$$\sigma_{xx}=\Epsilon \epsilon_{xx}=\epsilon \begin{bmatrix} z & y \end{bmatrix} \begin{Bmatrix} -\chi_{z} \\ -\chi_{y} \end{Bmatrix} \,$$

$$I^{-1}=\frac{1}{D}\begin{bmatrix} I_{z} & -I_{yz} \\ -I_{yz} & I_{y} \end{bmatrix}\,$$

where...

$$D=I_{y}I_{z}-(I_{yz})^{2}\,$$

Shear flow:

$$q=-\int_{A} \frac{d\sigma_{xx}}{dx} dA\,$$

$$\frac{d\sigma_{xx}}{dx}=\begin{bmatrix} z & y \end{bmatrix} I^{-1}\begin{Bmatrix} \frac{dM_{y}}{dx} \\ \frac{dM_{z}}{dx} \end{Bmatrix}\,$$

$$V_{y}=\frac{dM_{y}}{dx}\,$$

$$V_{z}=\frac{dM_{z}}{dx}\,$$

$$Q_{y}=\int_{A}zdA\,$$

$$Q_{z}=\int_{A}ydA\,$$

November 24th, 2008
Solving the problem: Equilibrium of each stringer

 Stringer #3 



$$\sum F_{x}=0=\int_{A_{3}} [\sigma_{xx}(x+dx)-\sigma_{xx}(x)]dA_{3}\,$$

The terms... $$[\sigma_{xx}(x+dx)-\sigma_{xx}(x)]dA_{3}\,$$ is the Taylor series...

$$\frac{\partial \sigma_{xx}}{dx}dx+h.o.t\,$$

$$[-\tilde{q}_{23} -\tilde{q}_{43} +\tilde{q}_{31}]dx\,$$


 * Now, observing the contribution to shear flow by stringer 3

$$q^{3}=-\int_{A_{3}} \frac{\partial \sigma_{xx}}{dx} A_{3}\,$$

...recalling,

$$V_{y}=\frac{\partial d M_{z}}{dx}\,$$

and

$$V_{z}=\frac{\partial d M_{y}}{dx}\,$$

$$q^{3}=-(KyVy-KyzVz)Qz^{3}-(KzVz-KyzVy)Qy^{3}\,$$

$$Qz^{3}=\int_{A_{3}} ydA_{3}\,$$

$$Qy^{3}=\int_{A_{3}} zdA_{3}\,$$


 * Now for stringer #2...

$$\tilde {q}_{23}=\tilde {q}_{12} - \tilde {q}_{24} + \tilde {q}^{4}\,$$

...where the first term and second term on the right equal the shear flow in and shear flow out respectively

$$Qy^{2}=\int_{A_{2}} zdA_{2}\,$$


 * Now for stringer #4...

$$\tilde {q}_{43}=\tilde {q}_{24} - \tilde {q}_{41} + \tilde {q}^{4}\,$$

$$Qy^{4}=\int_{A_{4}} zdA_{4}\,$$

Superposition

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

$$q_{12}=\tilde {q}_{12} + q_{1}\,$$

...where the first term and second term to the right are the known and unknown parameters of the equation


 * Therefore...

$$q_{23}=\tilde {q}_{23} + q_{1}\,$$

$$q_{31}=\tilde {q}_{31} + q_{1}\,$$

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

$$q_{43}=\tilde {q}_{43} + q_{2} - q_{3}\,$$

$$q_{41}=\tilde {q}_{41} + q_{3}\,$$

...where we have 3 unknowns $$q_{1}, q_{2}, q_{3}\,$$


 * Thus we need 3 equations...

1) Moment Equations: Take moments of $$V_{y} V_{z}\,$$ and $$[q_{12}, ... q_{41}]\,$$ about any convenient point. (This point usually lies where lines of action of $$V_{y} V_{z}\,$$ intersect)

2)$$\theta_{1}=\theta_{2}\,$$ 3)$$\theta_{2}=\theta_{3}\,$$

NACA Airfoil



November 12th, 2008


If we were to look at a cross section of a homogeneous circular bar, and place the origin of a coordinate system in its center...

$$x^2+y^2=a^2\,$$
 * where a is the radius of the circular boundary

Now, if we were to study the lateral surface of the bar, we would observe that the stress vector (traction) would be equal to zero. This means that the traction free boundary condition states that that the change of the stress function over the lateral surface is zero...

$$\frac{d\phi}{ds}=0\,$$

or

$$\phi=constant\,$$

...This would mean that for single contour boundary-solid sections, the constant is arbitrary and thus can be zero. Hence the boundary condition can be defined on the lateral surface of the bar as...

$$\phi=0\,$$

Now that we have defined the boundary condition (stated above), we write the stress function in a form that satisfies this condition...

$$\phi = C(\frac{x^2}{a^2}+\frac{y^2}{a^2}-1)\,$$

Next we take the Prandtl stress function...

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

Having obtained the compatibility equation above, we plug into it the stress function, which yields in...

$$C=-\frac{1}{2}a^2G\theta\,$$

The stress function and the equation for C above, indicate the fact that the center of the circular cross section is the center of twist.


 * Recalling the definition of torque...

$$T= 2\iint_A \phi dx dy\,$$

...we have the torque as...

$$T=2C\iint_A (\frac{x^2}{a^2} + \frac{y^2}{a^2}-1)dxdy\,$$

$$T=2C\iint_A (\frac{r^2}{a^2}-1)dA\,$$

$$T=2C(\frac{I_p}{a^2}-A)\,$$

...where

$$I_p=\iint_A r^2dA= \frac{1}{2}\pi a^4\,$$

...''where the above eq. corresponds to the polar moment of inertia and also...''

$$A=\pi a^2\,$$ since that is the cross sectional are of the circular bar.

Since...

$$a^2A=2I_p\,$$

Therefore...

$$T=-\frac{2CI_p}{a^2} = \theta GI_p\,$$

and...

$$J = I_p\,$$


 * The shear stresses are...

$$\tau_{xz}=\frac{\partial \phi}{\partial y} = 2C\frac{y}{a^2}=-G\theta y\,$$

$$\tau_{yz}=-\frac{\partial \phi}{\partial x} = -2C\frac{x}{a^2} = G\theta x\,$$

It is also worth noting that the shear stress can be expressed as...

$$\tau={Tr}{J}\,$$


 * We can also show that for bars with circular cross-sections under torsion, no warping occurs...

$$\gamma_{xz}=\frac{\sigma_{xz}}{G}=\frac{\partial w_{x}}{\partial z} - \theta_{y}\,$$

$$\gamma_{yz}=\frac{\sigma_{yz}}{G}=\frac{\partial w_x}{\partial y} + \theta_{x}\,$$



...now recalling the shear stresses derived above...

$$\sigma_{xz}=-G\theta y\,$$

$$\sigma{yz}=G\theta x\,$$

...we can see that the warping is zero...

$$w=0\,$$

Flexural Shear Flow in thin-walled sections

(Chapter 5) --> 5.1, 5.1.1, 5.1.2, 5.1.3, 5.3, 5.3.1, 5.3.2, 5.4

HW #5: Effects of M(x) on NACA airfoil

HW #6: Plate buckling, compressive in-plane load NACA airfoil

HW #7: Effects of V(x) on NACA airfoil, Plate Buckling, Shear Load.

Bidirectional Bending (Sec 4.2)
Skin and spar web do not participate in bending; only stringers

Need areas of stringers: $$A_{B} A_{E} A_{H} A_{F}\,$$

Bidirectional Bending Recipe



$$M_{y}=\int_{A} z\sigma_{xx} dA\,$$

Similarly for $$M_{z}\,$$

Moment of Inertia Tensor:

$$I_{y}=I_{yy}=I_{22}\,$$

$$I_{z}=I_{zz}=I_{33}\,$$

$$I_{yz}=I_{23}\,$$

$$I_{y}=\int_{A} z^{2} dA\,$$

$$I_{z}=\int_{A} y^{2} dA\,$$

$$I_{yz}=\int_{A} yz dA\,$$

$$\sigma_{xx}=\Epsilon \varepsilon_{xx} = \frac{I_{y}M_{z} - I_{yz}M_{y}}{I_{y}I_{z}-(I_{yz})^{2}}y + \frac{I_{z}M_{y}-I_{yz}M_{z}}{I_{y}I_{z}-I_{yz}^{2}}z\,$$


 * Recall...


 * $$\mathbf{I} = \begin{bmatrix}

I_{11} & I_{12} & I_{13} \\ I_{21} & I_{22} & I_{23} \\ I_{31} & I_{32} & I_{33} \\\end{bmatrix}$$

Note: $$D:= I_{y}-I_{z}-(I_{yz})^{2} = I_{22}-I_{33}-(I_{23})^2\,$$

...is the determinant of


 * $$\begin{bmatrix}

I_{22} & I_{23} \\ I_{32} & I_{33} \\\end{bmatrix}\,$$

$$I_{32}=I_{zy}:= \int_{A} zy dA = I_{yz}=I_{23}\,$$

Neutral Axis: where  $$\sigma_{xx}=0\,$$

$$\sigma_{xx}=m_{yy}+m_{zz}=0\,$$

--> $$z=(\frac{-m_{y}}{m_{z}}y = (\tan{\beta})y\,$$

where...$$\frac{-m_{y}}{m_{z}}y \approx \tan{\beta}=-\tan{\alpha}\,$$

Example 4.1, Pg. 124
Problem Statement

The cross section of a single-cell box beam with four stringers is shown in Figure 4.7. The Contribution of the thin sheets to bending is assumed to be negligible. Thus, only the areas of the stringers are considered in the bending analysis. The areas of the stringers are considered in the bending analysis. The areas of the stringers are...

$$A_{1}= 6x10^{-4} m^{2}\,$$ $$A_{2}= 5x10^{-4} m^{2}\,$$ $$A_{3}=A_{4}= 4x10^{-4} m^{2}\,$$

First, the centroid of the effective cross-sectional areas (i.e., those of the stringers) must be determined. Denoting the coordinates of the stringers by $$(\bar y_{j}, \bar z_{i})\,$$ with respect to the $$\bar y-\bar z$$ system, we have the coordinates of the centroid:

$$\bar y_{c}=\frac{\sum \bar y_{i}A_{i}}{\sum A_{i}} = \frac{0A_{1}+0A_{2}+0.5A_{3}+0.5A_{4}}{A_{1}+A_{2}+A_{3}+A_{4}}= \frac{4x10^{-4}}{19x10^{-4}}=0.21m\,$$



$$\bar z_{c}= \frac{\sum \bar z_{i}A_{i}}{\sum A_{i}}=\frac{0.4A_{1}+0A_{2}+0A_{3}+0.3A_{4}}{A_{1}+A_{2}+A_{3}+A_{4}}=\frac{3.6}{1.9}=0.19m$$

Thus, the location of the centroid is (0.21, 0.19) in the $$\bar y - \bar z\,$$ system.

This cross section consisting of four stringers is not symmetric with respect to either the y- or the z-axis. Hence, the general bending equations must be used. The moments of inertia of the effective cross sectional area of the box beam with respect to the coordinate system y-z are calculated according to (4.28). Denoting the coordinates of each stringer by $$(y_{i},z_{i})$$ with respect to the y-z system, we have

$$I_{y} = \sum A_{i}Z_{i}^{2} = A_{1}(0.4-0.19)^2 + (A_{2}+A_{3})(0.19)^{2}+A_{4}(0.3-0.19)^{2}=0.63x10^{-4}m^{4}\,$$

$$I_{z}=\sum A_{i}y_{i}^{2}=(A_{1}+A_{2})(0.21)^{2}+(A_{3}+A_{4})(0.29)^{2}=1.16x10^{-4}m^{4}\,$$

$$I_{yz}=\sum A_{i}y_{i}z_{i}=A_{1}(-0.21)(0.21)+A_{2}(-0.21)(-0.19)+A_{3}(0.29)(-0.19)+A_{4}(0.29)(0.11)=-0.15x10^{-4}m^{4}\,$$

Consider the loading $$M_{z}=0\,$$ and $$M_{y}\ne 0\,$$ The neutral plane is given by...

$$\tan{\alpha} = -\frac{I_{yz}M_{y}}{I_{z}M_{y}} = \frac{0.15x10^{-4}}{1.16x10^{-4}}=0.13\,$$

which yields...

$$\alpha = 7^\circ$$

measured clockwise from the y-axis.

The Structural Beam
!Bending Member (Beam) -Contribution from Chapter 1.2.3 by Radsam.Z

The beam is an elemental structure used in all walks of life, from civil engineering to aerospace engineering. Beams provide structural support by providing resistance to bending moments. "A beam can also be used as an axial member carrying longitudinal tension and compression."



The bending moment is mathematically described to beam deflection as...

$$M=-EI\frac{d^2w}{dx^2}$$

where... EI= Bending Stifness of the beam I= Moment of Inertia

The Beam in Aerospace Structures
An example of a beam used in the aerospace industry is that in some airplane wings. This type of beam is called a cantilever beam, which is only supported in one end, carries both a bending moment and a transverse shear stress. However for a beam of large span/depth ratio, as in the case in aircraft wings, the bending moment is much greater than the shear stress.

The maximum bending moment at the root... $$\sigma=\frac{M_{max}(\frac{h}{2})}{I}$$

The maximum transverse shear stress distribution at the neutral plane... $$\tau_{max}=\frac{3}{2}\frac{V}{bh}$$

If we take the ratio of both the maximum bending moment and shear stress... $$\frac{\sigma_{max}}{\tau_{max}}=\frac{4L}{h}$$

From this it is clear that the bending moment is the ruling factor in aircraft wings. So in order to address this issue aircraft engineers focus in the improvement of airfoils depending on the type of aircraft. For instance high performance military jet-aircraft, that undergo high-loading forces would definitely need stronger wings (cantilever beams) than say a cargo aircraft. In order to make an aircraft more "resistant" to these high bending forces, engineers design these type of aircraft with a low aspect ratio ...

$$AR = {b^2 \over S}$$                    where... b = wingspan S = planform area

Problem Explanation and Proof of Shear Stress

 * {| class="toccolours collapsible collapsed" width="100%" style="text-align:left"

!Bending Member (Beam) -Contribution from Chapter 1.2.3 by Radsam.Z The closed-walled beam under scrutiny is one of negligible thickness, thus one can assume that the shear stress distribution is of a uniform nature...

Therefore adding respectively the torque on each side of the box...

$$T= T_{AB}+T_{BC}+T{CD}+T{DA}$$

...where torque equals the lever arm times the shear stress by the wall thickness... $$T_{AB}=\frac{b}{2}(va) = \frac{1}{2}\tau(abt)$$

$$T_{BC}=\frac{a}{2}(vb) = \frac{1}{2}\tau(abt)$$

$$T_{CD}=\frac{b}{2}(va) = \frac{1}{2}\tau(abt)$$

$$T_{DA}=\frac{a}{2}(vb) = \frac{1}{2}\tau(abt)$$

and recalling...

$$T= T_{AB}+T_{BC}+T{CD}+T{DA}$$

$$T=2\tau(abt)$$

or

$$\tau=\frac{T}{2abt}$$


 * }

Problem 1.1 - Case#1

 * {| class="toccolours collapsible collapsed" width="100%" style="text-align:left"

!Solution to Case 1 by Radsam.Z If we assume that the maximum bending stress is the allowable bending stress then we have...

$$\sigma_{max}=\sigma_{allowable}$$

By recalling the bending stress formula...

$$\sigma=\frac{Md}{I}$$


 * sigma=bending stress
 * M= moment at the neutral axis
 * d=distance measured perpendicular to the neutral axis
 * I=the moment of inertia

Now by observing the rectangular beam we can determine that the bending stress would equal...

$$\sigma=\frac{Mz}{I}$$

or by our assumption of maximum bending stress being equal to the allowable stress

$$M=\frac{\sigma I}{z}$$ where $$z=\frac{b}{2}$$

After some arithmetic manipulation and solving for bending moment...

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

Therefore as we can see from the above equation if we wish to maximize M, the only term of the equation that can be optimized is the ratio of the inertia over b

$$M_{max}=(2\sigma_{allowable})(\frac{I}{b})_{max}$$

recall : the ratio is a function of a and b...since the perimeter length $$L=2(a+b)=constant$$

Following the parallel axis theorem we find... "The parallel axes rule applies to the second moment of area (area moment of inertia);
 * $$I_z = I_x + Ad^2.\,$$

where:

Iz is the area moment of inertia through the parallel axis,

Ix is the area moment of inertia through the centre of mass of the area,

A is the surface of the area, and

d is the distance from the new axis z to the centre of gravity of the area."

...Now if we apply this to our rectangular beam as follows...

$$I=\sum_{1}^4 [\frac{b_{i}(h_{i})^3}{12} + A_{i}(d_{i})^2]$$



$$=2\frac{(tb)^3}{12}+2 [\frac{at^3}{12}+(at)(\frac{b}{2})^2]$$ $$=\frac{a}{12}[t^3+t(\frac{12}{4})b^2]$$ $$=\frac{at}{12}[t^2+3b^2] =3\frac{ab^2t}{12}$$

considering t&lt;&lt;b   and   t^2&lt;&lt;b^2    and   t^2&lt;&lt;3b^2

$$I= 2\frac{tb^3}{12}+\frac{ab^2t}{2} = \frac{b^2t}{2}(\frac{b}{3}+a) = \frac{tb^2}{6}(3a+b)$$

$$\frac{I}{b}=\frac{tb}{6}(3a+b)$$

where $$a=\frac{L}{2}-b$$

$$=\frac{tb(3L-4b)}{12} = \beta_{0}+\beta_{1}b'+\beta_{2}b^2$$ $$\beta_{0}=0$$ $$\beta_{1}=3\frac{Lt}{12}$$ $$\beta_{2}=\frac{-4}{12}t=-\frac{t}{3}$$

and

$$\frac{d^2f}{db^2}(b)=2\beta_{2}=-\frac{2t}{3}<0$$ (this tells us that the second derivative is negative and if we recall from Calculus this means that our graph will be concave down and our point of interest will be a maximum!)

...thus finding the maximum by evaluating the graph



$$f(b)=0$$ and $$b=0$$ and $$b=\frac{3L}{4}$$ (the endpoints of the circle)

and as we see in the in the graph the maximum point is $$b=3\frac{L}{8}$$

$$\frac{df}{db}(b)=\beta_{1}+2\beta_{2}b = aprox = 0$$

therefore the solution for case #1 is...

solving for b in the above equation

$$b=-\frac{\beta_{1}}{2\beta_{2}} = \frac{3L}{8}$$

and if we recall the definition

$$L=2(a+b)$$

thus, we obtain...

$$a^(1) = \frac{L}{2}-b^(1)=\frac{L}{8}$$

...and we can obtain a ratio of a to b...

$$\frac{b^1}{a^1} = 3$$

and recalling that in order to maximize the Torque we needed to maximize the ratio...

$$(\frac{I}{b})_{max}=\frac{I^1}{b^1}$$

$$=\frac{tb}{6} (3a+b) = \frac{tb}{6} (3a + 3a) = \frac{tb}{6} (6a)$$

which reduces to...

$$= abt = \frac{btL}{8} = \frac{3tL^2}{64}$$

...and finally replacing this ratio into the maximum torque equation...

$$M^1_{max}=(2\sigma_{allowable})\frac{I^1}{b^1}$$

...yields the final result...

$$M^1_{max}=3\frac{tL^2}{32} \sigma_{allowable}$$


 * Which also means --->

$$T_{max}= 3\frac{tL^2}{32} \sigma_{allowable}$$


 * $$\tau_{max}=\frac{T_{max}}{2\frac{L}{8}\frac{3L}{8}t}$$


 * $$\sigma_{allowable} = \frac{M_{max}}{2\frac{L}{8}\frac{3L}{8}t}$$

$$\tau^1_{max} = \sigma_{allowable} = 2\tau_{allowable} > \tau_{allowable}$$

$$\tau^1_{max} > \tau_{allowable}$$ is NOT acceptable

This is the conclusion of Case #1

October 10th, 2008 Lecture
{| class="toccolours collapsible collapsed" style="width:100%" ! Lecture from Friday - October 10th, 2008

Recalling...

Torsion : $$T = GJ\theta\,$$

[ http://en.wikipedia.org/wiki/Torsion_(mechanics)]
 * T is the torque (N·m or ft·lbf)
 * G is the shear modulus or more commonly the modulus of rigidity and is usually given in gigapascals (GPa), lbf/in2 (psi), or lbf/ft2
 * J is the torsion constant for the section . It is identical to the polar moment of inertia for a round shaft or concentric tube only. For other shapes J must be determined by other means. For solid shafts the membrane analogy is useful, and for thin walled tubes of arbitrary shape the shear flow approximation is fairly good, if the section is not re-entrant. For thick walled tubes of arbitrary shape there is no simple solution, and FEA may be the best method.
 * $$\theta\,$$ is the angle of twist in radians.

Once $$\theta\,$$ is found as a function of T, use...

$$\theta = \frac{T}{GJ}\,$$

or

$$J=\frac{T}{G\theta\,}$$

...to find J

HW #4:


 * 3-cell NACA 2415 airfoil
 * 2 partition walls
 * 1st wall at quarter chord from leading edge
 * 2nd wall at three quarter chord from leading edge

3 unknowns... $$q_{1}, q_{2}, q_{3}\,$$:


 * 1) $$T=2\sum_{i=1}^3 q_{i} \bar{A}\,$$
 * 2) $$\theta_{1}=\theta_{2}\,$$
 * 3) $$\theta_{2}=\theta_{3}\,$$

...to find J

Finding J...

Note: need thickness t, but G cancels out...


 * Now Theory derivation

$$T=GJ\theta\,$$ (done)

$$T=2q\bar{A}$$  (done)

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

Contribution by "Eas4200c.f08.radsam.z 23:45, 19 October 2008 (UTC)"

Uniform bar w/non-circular cross section subject twist




Displacement $$\underline{PP'}$$ due to $$\alpha\,$$

$$\frac{PP'}{O\underline{P}} = \alpha\,$$ (for alpha small)

Projection Displacement $$\underline{PP'}$$ on the direction perpendicular to $$O\underline{P'}:$$

$$\underline{PP''}=\underline{PP'} cos{\alpha}\,$$

$$\underline{PP''}= (O \underline{P} tan{\alpha}) cos{\alpha}\,$$

$$=(O\underline{P} cos{\alpha}) tan{\alpha}$$

''where... $$O\underline{P} cos{\alpha} = O\underline{P}\,$$

Recall:

$$O\underline{P} = r \,$$ (radial coord)

$$O\underline{P''} = \rho \,$$

$$PP''= (r cos{d}) tan{\alpha}\,$$


 * where: PP is displacement of $$\underline{P}$$ in direction tangent to lateral surface of bar


 * $$r( cos{d})\,$$ is approximately $$\rho\,$$


 * $$tan{\alpha}\,$$ is approximately $$\alpha\,$$

Strain:

$$\gamma=\frac{\underline{PP''}}{dx} = \frac{\rho \alpha}{dx} = \rho \theta\,$$

with... $$\theta = \frac{\alpha}{dx}$$ rate of twist angle (alpha very small!)