User:Eas4200c.f08.wiki.b/HW6

=Mtg 33 Wednesday, November 12=



Torque can be expressed as

 $$\displaystyle T = 2 \int_A \phi dA $$

From the theory of elasticity, the torque becomes

 $$\displaystyle T = 2C ( {J \over a^2} - A) $$

Where the area, $$\displaystyle A $$, since the radius is $$\displaystyle a $$, is simply

 $$\displaystyle A = \pi a^2 $$

And where $$\displaystyle J $$ is

 $$\displaystyle J = \int_A r^2 dA = {1 \over 2} \pi a^4 $$

Using the ad-hoc method

 $$\displaystyle T = GJ \theta $$

Recall that the Prandtl Stress function of $$\displaystyle \phi $$ states that the stress in the x-direction normal to the y-axis is

 $$\displaystyle \sigma_{yx} = {\partial \phi \over \partial z} $$

Taking this partial derivative, you obtain

 $$\displaystyle \sigma_{yx} = 2C {z \over a^2} $$

Where substitution for $$\displaystyle C $$ yields

 $$\displaystyle \sigma_{yx} = - G \theta z $$  (1.1)

Similarly, the stress in the x-direction normal to the z-axis is

 $$\displaystyle \sigma_{zx} = - {\partial \phi \over \partial y} $$

Again, taking this partial derivative, you obtain

 $$\displaystyle \sigma_{zx} = -2C {y \over a^2} $$

Where, again, substitution for $$\displaystyle C $$ yields

 $$\displaystyle \sigma_{zx} = G \theta y $$  (1.2)

Also note that shear stress, $$\displaystyle \tau $$, is

 $$\displaystyle \tau = {Tr \over J} $$

Proof of No Warping
Now, using the above equations in addition to the stress strain relation, we can prove that no warping takes place.

First, recall that

 $$\displaystyle \gamma = {\sigma \over G} $$

Or more specifically in the context of this problem

 $$\displaystyle \gamma_{yx} = {\sigma_{yx} \over G} $$

 $$\displaystyle \gamma_{zx} = {\sigma_{zx} \over G} $$

Also recall that

 $$\displaystyle \gamma_{yx} = {\partial u_x \over \partial y} - \theta z $$

<p style="text-align:center;"> $$\displaystyle \gamma_{zx} = {\partial u_x \over \partial z} + \theta y $$

Thus,

<p style="text-align:center;"> $$\displaystyle {\sigma_{yx} \over G} = {\partial u_x \over \partial y} - \theta z $$  (2.1)

<p style="text-align:center;"> $$\displaystyle {\sigma_{zx} \over G} = {\partial u_x \over \partial z} + \theta y $$  (2.2)

Now, substituting equation (1.1) into equation (2.1), and similarly substituting equation (1.2) into equation (2.2) the result is

<p style="text-align:center;"> $$\displaystyle {-G \theta z \over G} = {\partial u_x \over \partial y} - \theta z $$

<p style="text-align:center;"> $$\displaystyle {G \theta y \over G} = {\partial u_x \over \partial z} + \theta y $$

In both equations, each $$\displaystyle G $$ cancels out. Also, $$\displaystyle \theta x $$ and $$\displaystyle \theta y $$ cancel out when subtracted from each side of the respective equation, thus leaving

<p style="text-align:center;"> $$\displaystyle {\partial u_x \over \partial y} = 0 $$

<p style="text-align:center;"> $$\displaystyle {\partial u_x \over \partial z} = 0 $$

Thus, we can conclude that there is no warping since

<p style="text-align:center;"> $$\displaystyle u_x (y,z) = 0 $$