User:Eas4200c.f08.aeris.guan/hw4

Derivation of the Twist Angle in the Case of a Uniform, Non-Circular Cross Sectional Bar with Varying Shear Flow
Regarding the previous lecture, once $$\theta$$ is found in terms of T, use $$T=GJ\theta$$ or $$J=\frac{T}{G\theta}$$ to find the torsional constant J.  For this homework assignment, the NACA 2415 airfoil, referring to lecture 16.2, is partitioned into three cells. The cells are separated by two walls of assigned thickness. To perform the Matlab analysis similar to the homework 3, more information must be provided in order to properly calculate the "average" areas $$\bar{A}_1, \bar{A}_2, and \bar{A}_3$$ corresponding to the three cells of the airfoil and along with the torque (not indicated but will be treated as a constant), thickness of the walls and shell, contour length and shear modulus will allow us to calculate the shear flow on the airfoil. First, one wall is .25 times the chord length away from the leading edge, and a second wall is .75 times the chord length away from the leading edge. The shear modulus is not needed because the equations used will cancel out these terms. There will be three unknowns $$q_1,q_2, and q_3$$ and using the three following equations, these unknowns will be solved for. $$T=2\sum_{i=1}^3{q_i\bar{A}_i}$$ $$\displaystyle \theta_1=\theta_2$$ $$ \displaystyle \theta_2=\theta_3$$ Then, the torsional constant J will be found by the equation mentioned above. As we have already finished deriving the follow equations, $$T=GJ\theta, T=2q\bar{A}$$ Now we will derive the twist angle equation for a uniform, non-circular cross sectional with an engineering (ad hoc) method. $$\theta=\frac{1}{2G\bar{A}}\oint\frac{q}{t}ds$$ Let's consider this uniform, non-circular cross sectional bar subject to a torque. The bottom two pictures describe this type of bar with infinitesimally small axial length dx and the cross section is shown in the (y,z) plane.  Displacement PP' is due to alpha, and approximating alpha as very small, we have the following expression for PP' and OP in a simple geometric triangular relation. $$\frac{PP'}{OP}=tan(\alpha)$$ Now, project the displacement PP' in the direction orthogonal to OP'. In other words, $$\displaystyle PP' = PP'' cos( \alpha)$$ $$\displaystyle PP=(OPtan(\alpha))cos(\alpha)=(OPcos(\alpha))tan(\alpha)=OPtan(\alpha)$$ Recall, that OP=R, the radial coordinate of the point P and thus $$\rho$$ must be the perpendicular distance from the origin to P which equals, in this case, OP. Knowing that, the following expression can be written. $$\displaystyle PP''=(Rcos(\alpha))tan(\alpha)=\rho \alpha$$ Here, PP" is the displacement of P in the direction "tangent" to the lateral surface of the bar. Strain is $$\gamma=\frac{PP}{dx}$$ where dx is the distance between very close planes on the bar. $$\gamma=\frac{PP}{dx}=\frac{\rho \alpha}{dx}=\rho \theta$$ Here theta is the rate of twist angle.