User:Eas4200c.f08.blue.a/Lecture 12

Assigned Reading from Mechanics of Aircraft Structures (C. T. Sun): pg. 451 - Shear Panel pg 85-86 - Shear Flow (Eq. 3.49, 3.50) pg. 20 - Fig. 2.2 pg. 64 - Fig, 3.2 pg 115 - Fig. 4.1

The lecture began with a discussion of shear in non-uniform objects such as the one in Figure 1 below.



Figure 1

Then, curved panels were discussed and reference was made to equations 1.4 and 1.5 (Sun, 5). To illustrate the concept of shear in a curved panel, problem 1.5 from the text book (Sun, 17) was discussed.

Problem 1.5
Problem statement: Derive the relations given by equations 1.4 $$\left( V_{x}=\tau ta\right)$$ and 1.5 $$\left( V_{y}=\tau tb\right)$$ in a curved panel as in Figure 2 below.



Figure 2

A Closer look is taken at the z,y plane and the shear flow on the edge of the curve as seen below:



Figure 3

Now let q be equal to the shear flow traveling from point A to point B as shown by the direction of the arrows above in Figure 3. $$q= \left( \tau \right) \left( t \right)$$

The infinitesimal change of shear force $$\left( d\vec{F}\right)$$ is shown below as

$$d\vec{F}=qd\vec{l}=q(dl_{y}\hat{j}+dl_{z}\hat{k})=q(dlcos\theta \hat{j}+dlsin\theta \hat{k})=qdydz$$

A closer look at $$d\vec{F}$$is shown in Figure 4 below.



Figure 4

The Resultant shear force vector is found as

$$\vec{F}=\int_{B}^{A}{d\vec{F}}=q\left(\left(\int_{B}^{A}{dy}\right)\hat{j}+\left(\int_{A}^{B}{dz} \right) \hat{k}\right)$$

Substituting equations 1.4 and 1.5,

$$\vec{F}=q\left(a\hat{j} +b\hat{k}\right)=F_{y}\hat{j}+F_{z}\hat{k}$$

where

$$F_{y}=\left( q\right)\left(a\right)$$ as in equation 3.49b

and

$$F_{z}=\left( q\right)\left(b\right)$$ as in equation 3.49c

yielding

$$\frac{F_{y}}{F_{z}}=\frac{a}{b}$$

The Resultant magnitude is

$$\left|\left|\vec{F} \right| \right|=\sqrt{\left[\left(F_{y} \right)^{2} +\left(F_{z} \right)^{2}\right]}=q\left[a^{2} +b^{2}\right]^{1/2}$$

$$R=\left|\left|\vec{F} \right| \right|$$

$$T=2\left( q\right)\left( A\right)$$

Now, a thin walled cross section can be analyzed in the same way as the curved panel. The thin walled cross section is seen in the z,y plane below.



Figure 5

a close up of the section in question:



Figure 6

It is assumed that the Torque (T) is in the x direction ie. $$\vec{T}=T\hat{i}$$

$$\vec{T}=\vec{r}\bowtie d\vec{F}$$

It follows that

$$dT = \rho dF=\rho \left(qdl \right)$$

and therefore

$$T=\oint_{}^{}{dT}=q\oint_{}^{}{\rho dl}$$

From Figure 6 above, it can be seen that $$ dA=\frac{1}{2}\rho dl$$

Therefore

$$T=\oint_{}^{}{dT}=q\oint_{}^{}{2dA}=2q\oint_{}^{}{dA}=2q\bar{A}$$

Where $$\bar{A}$$ is equal to the average area as measured from the middle of the thickness dl.