User:Eas4200c.f08.vqcrew.d/hw4

Ad-hoc (Engineering) Derivation of Rate of Twist Formula
To derive the twist angle formula $$\theta = \frac{1}{2G\bar{A}}\oint_{C}^{}{\frac{q}{t}ds}$$ we will consider a uniform bar with non-circular cross section subjected to twisting, as shown in Figure XX

To begin, the displacement $$PP'$$ is found to be

$$PP'=OPtan \mathbf{\alpha}$$ Then, project $$PP'$$ into the direction perpendicular to $$OP'$$ : $$PP''=PP'cos \mathbf{\alpha}$$

Now the displacement of $$P$$ in the direction tangent to the lateral surface of the beam, $$PP$$, can be expressed in terms of $$OP$$ : $$PP=(OPtan \mathbf{\alpha})cos \mathbf{\alpha} = OPcos \mathbf{\alpha}tan \mathbf{\alpha} = OPtan \mathbf{\alpha}$$ As in Figure XX, $$OP''$$ is defined as the perpendicular distance to the surface, $$\mathbf{\rho}$$. If we assume that the twist angle, $$\mathbf{\alpha}$$, is a small angle, $$PP''$$ simplifies to: $$PP''=\mathbf{\rho}\mathbf{\alpha}$$

The shear strain, $$\mathbf{\gamma}$$, can be defined as the ratio of this displacement distance, $$PP''$$, to a small displacement in the axial (x) direction: $$\gamma = \frac{PP''}{dx}$$ We can then substitute for $$PP''$$, and since we assumed $$\mathbf{\alpha}$$ is very small it can be replaced by it's differential element $$d\mathbf{\alpha}$$. We can also define the rate of twist, $$\mathbf{\theta}$$, as the differential twist angle $$d\mathbf{\alpha}$$ per differential change in the axial direction $$dx$$ : $$\gamma = \frac{\rho d\alpha}{dx} = \rho \theta$$

We can use Hooke's Law to relate the shear stress to the shear strain: $$\mathbf{\tau =G \gamma =G \rho \theta}$$ $$\mathbf{\tau(s) =G \rho (s) \theta (x)}$$

If we take the contour integral along C, the equation becomes: $$\oint_{C}^{}{\tau(s)ds} = G \theta(x)\oint_{C}^{}{\rho (s)ds}$$ Next we can substitute in values from previous derivations to arrive at the final form for $$\mathbf{\theta}$$: $$\oint_{C}^{}{\frac{q(s)}{t(s)}ds} = G \theta(x)2\bar{A}$$ $$\Rightarrow \theta = \frac{1}{2G\bar{A}}\oint_{C}^{}{\frac{q}{t}ds}$$

What makes this derivation "ad hoc"?

 * 1) The shear strain $$\mathbf{\gamma}$$ must be obtained using the displacement of $$P$$ in the direction tangent to $$C$$ at $$P$$, but $$PP''$$ is not necessarily tangent to C (but is close).
 * 2) The equation for the shear stress, $$\mathbf{\tau} =\frac{q}{t}$$, was obtained from an ad hoc assumption that $$\mathbf{\tau}$$ is uniform across the wall thickness.
 * 3) There is an inconsistency in the assumption on the size of $$\mathbf{\alpha}$$. To get line $$PP'$$, we assumed $$\mathbf{\alpha}$$ was small, but to get $$PP''$$ we assumed a finite $$\mathbf{\alpha}$$ to use the relation $$\mathbf{\rho}=OPcos \mathbf{\alpha}$$. We then reintroduced the small angle approximation for $$\mathbf{\alpha}$$ in the rest of the derivation.