User:EAS4200C.F08.WIKI.A/My Contribution: 4

=Multicell Airfoil=

In this section we will examin a multicel airfoil. We are interested to find the rate of twist as a function of the torque $$\displaystyle T$$ and the torsional constant $$\displaystyle J$$. The figure depicted below is the object that will be in concideration.



We first have to express the torque of the while object. To do so we will can use the principal of superposition. In other words the torque of the whole object is the summation of the torques from each cell, $$\displaystyle C_1$$ and $$\displaystyle C_2$$.

$$\displaystyle \sum_{i=1}^{N_c}{2q_i\bar{A_i}}= 2q_1\bar{A_1} + 2q_2\bar{A_2}$$

We now must find the rate of twist for each cell. The formulation for cell one is shown below.

$$\displaystyle \theta_1 = \frac{1}{2G\bar{A_1}}\oint_{a}^{a}{\frac{q_1}{t_1} ds} = \frac{1}{2G\bar{A_1}}[\frac{2q_1a}{t_1} + \frac{q_1c}{t_1} + \frac{(q_1-q_2)c}{t_{12}}]$$

When all known variables are put in and noting that $$\displaystyle \bar{A_1} = ac = (30cm)(40cm) = 1200cm^2$$ equals. the expression for the rate of twist for cell one then becomes;

$$\theta_1 = \frac{1}{Gcm^2}[.181q_1 - .042q_2]$$

The same is done for cell two and is performed below.

$$\displaystyle \theta_2 = \frac{1}{2G\bar{A_2}}\oint_{a}^{a}{\frac{q_2}{t_2} ds} = \frac{1}{2G\bar{A_2}}[\frac{2q_2b}{t_2} + \frac{q_2c}{t_2} + \frac{(q_2-q_1)c}{t_{12}}]$$

Substituting values and noting that $$\displaystyle \bar{A_2} = bc = (60cm)(40cm) = 2400cm^2$$, the expression reduces to;

$$\theta_2 = \frac{1}{Gcm^2}[-.021q_1 + .062q_2]$$

Now it is imperative to notice that the rates of twist for each cell are equivalent. Setting them equal to each other and solving for either $$\displaystyle q_1$$ or $$\displaystyle q_2$$ is the next step in solving this problem. the result is as follows. We chose to solve for $$\displaystyle q_1$$.

$$ \frac{1}{Gcm^2}[.181q_1 - .042q_2] = \frac{1}{Gcm^2}[-.021q_1 + .062q_2]$$  $$\displaystyle q_1= .65 q_2$$

Now we pace this expression for $$\displaystyle q_1$$ into the torque equation that was first derived in order to obtain the rate of twist for cell two $$\displaystyle q_2$$ in terms of the torque $$\displaystyle T$$. It is then possible to obtain the rate of twist for cell one $$\displaystyle q_1$$ in terms of the torque.

$$\displaystyle T = 2q_2(.65)(1200cm^2) + 2q_2(2400cm^2) $$  $$\displaystyle q_2 = \frac{1.57x10^{-4}T}{cm^2}$$  $$\displaystyle q_1 = \frac{1.02x10^{-4}T}{cm^2}$$

Now we must put these expressions back into either one of the rates of twist equations. Doing so will put the rate of twist in terms of the torque. This is done below.

$$\theta_1 = \frac{1}{Gcm^2}[.181q_1 - .042q_2] = \frac{1}{Gcm^2}[.181(\frac{1.02x10^{-4}T}{cm^2}) - .042(\frac{1.57x10^{-4}T}{cm^2})]$$  <p style="text-align:center;">$$\theta_1 = \frac{1.18x10^{-5}T}{Gcm^4} = \theta_2$$

Now we must relate the angle of twist to the torsional constant. We know that the angle of twist is equal to the following.

<p style="text-align:center;">$$\displaystyle \theta = \frac{T}{GJ}$$

We can conclude that

<p style="text-align:center;"> $$\displaystyle \frac{1}{J} = \frac{1.18x10^{-6}}{cm^4}$$

And that

<p style="text-align:center;">$$\displaystyle J = \frac{1}{1.18x10^{-6}} = 84745.8cm^4$$