User:Eas4200c.f08.nine.s/Lecture 6

Homework 1

Homework 2

Homework 3

Homework 4

Homework 5

Homework 6

 Group nine - Homework 6 

Prandtl Stress Function
The Prandtl stress function is given by the following equation:

$$\sigma_{yx} = \frac{\delta \phi}{\delta z}, \, \, \sigma_{zx} = - \frac{\delta \phi}{\delta y}$$

where $$\phi$$ plays the role of a potential function.

This equation automatically satisfies $$\frac{\delta}{\delta y}\left(\frac{\delta\phi}{\delta z}\right) + \frac{\delta}{\delta z}\left(-\frac{\delta\phi}{\delta y}\right)$$

$$\,\, = \frac{\delta^{2}\phi}{\delta y\delta z} - \frac{\delta^{2}\phi}{\delta z\delta y} = 0$$

$$\phi$$ is constant and smooth so the second mixed derivative is interchangeable.

Now, let us take a look at the strain-displacement relations,

$$\gamma_{yx} = \frac{\delta u}{\delta y} + \frac{\delta v}{\delta x}, \,\,\,\,\, \gamma_{zx} = \frac{\delta u}{\delta z} + \frac{\delta w}{\delta x}$$

$$ \Rightarrow \,\,\,\,\begin{matrix}(1a)\\(1b)\end{matrix}\,\,\,\,\,\left\{\begin{matrix}\gamma_{yx} = \frac{\delta u}{\delta y} - \theta_{z}\\ \gamma_{zx} = \frac{\delta u}{\delta z} + \theta_{y} \end{matrix}\right.$$

We can now obtain the compatibility equation for torsion shown below:

$$\Rightarrow \,\,\,\,\frac{\delta\gamma_{zx}}{\delta y} - \frac{\delta\gamma_{yx}}{\delta z} = 2\theta$$

And now by using the stress-strain relations, we obtain the following

$$\gamma_{zx} = \frac{1}{G}\sigma_{zx}, \,\,\,\,\,\gamma_{yx} = \frac{1}{G}\sigma_{yx}$$

And from the compatibility equation for torsion, we get

$$\frac{\delta\sigma_{zx}}{\delta y} - \frac{\delta\sigma_{yx}}{\delta z} = 2G\theta$$

And now taking the Laplacian of $$\phi$$ and putting back in terms of the Prandtl stress function, we obtain the following,

$$(2)\,\,\,\,\,\frac{\delta^{2}\phi}{\delta x^{2}} + \frac{\delta^{2}\phi}{\delta y^{2}} = -2G\theta$$

Now, the torsion problem is reduced to finding the stress function $$\phi$$ while using the given boundary conditions. Also, their is no applied loads on the lateral surface of the bar, and therefore, the stress vector  $$t$$  (traction) must vanish.

$$\left\{t\right\}_{3x1} = \left[\sigma\right]_{3x3}\left\{n\right\}_{3x1}$$

By using the unit normal  $$n$$ , the stress vector can be evaluated on the lateral surface where  $$n_{x} = 0$$ . Therefore,

$$\begin{Bmatrix}t_{x}\\t_{y}\\t_{z}\end{Bmatrix} = \begin{bmatrix}0&\sigma_{xy}&\sigma_{xz}\\\sigma_{xy}&0&0\\\sigma_{zx}&0&0\end{bmatrix} \begin{Bmatrix}0\\n_{y}\\n_{z}\end{Bmatrix}$$

$$\Rightarrow \,\,\,\,t_{y} = 0 \,\,\,\,\,t_{z} = 0$$

$$\Rightarrow \,\,\,\,t_{x} = \sigma_{yx}n_{y} + \sigma_{zx}n_{z} = -\frac{\delta\phi}{\delta z}n_{y} + \frac{\delta\phi}{\delta y}n_{z}$$

Figure goes here

By using the Figure to the right, we can see that,

$$ \, dy = sin\theta ds$$

$$ \, dz = cos\theta ds$$

$$\,\,\,\,\,\Rightarrow \,\,\,\,n_{y} = cos\theta$$

$$\,\,\,\,\,\Rightarrow \,\,\,\,n_{z} = sin\theta$$

Using the above relations we obtain the following for $$t_{x}$$

$$t_{x} = \frac{\delta\phi}{\delta z}\frac{dz}{ds} + \frac{\delta\phi}{\delta y}\frac{dy}{ds} = \frac{d\phi}{ds}$$

And now the traction free boundary condition $$\left(t_{x} = 0\right)$$ is given by the following,

$$\frac{d\phi}{ds} = 0\,\,\,\,\,\Rightarrow \,\,\,\,\phi = $$ constant on the lateral surface of the bar.

Bars with Circular Cross-Sections
We are now going to take at look a uniform bar with a circular cross-section and show that there is no warping under torsion. The origin is chosen to be at the centroid of the bar, so that the boundary contour is given by the following equation.

$$y^{2} + z^{2} = a^{2}\,\,\,$$ where a is the radius of the circular boundary.

The stress function satisfying the boundary condition of the previous section is given below:

$$\phi(y,z) = C\left(\frac{y^{2}}{a^{2}} + \frac{z^{2}}{a^{2}} - 1\right)$$

Substituting this equation into Eqn.(2), we obtain the following,

$$C = -\frac{1}{2}a^{2}G\theta$$

We can now solve for the torque as shown,

$$ \begin{align} T & = 2C\int\int_A\left(\frac{y^{2}}{a^{2}} + \frac{z^{2}}{a^{2}} - 1\right)dydz\\ & = 2C\int\int_A\left(\frac{r^{2}}{a^{2}} - 1\right)dA\\ & = 2C\left(\frac{I_{p}}{a^{2}} - A\right) \end{align} $$

where $$I_{p} = \int\int_A r^{2}dA = \frac{1}{2}\pi a^{4}$$  and  $$\,\,A = \pi a^{2}$$

Because $$a^{2}A = 2I_{p}$$  we have

$$T = -\left(\frac{2CI_{p}}{a^{2}}\right) = \theta GI_{p} = \theta GJ$$

The shear stresses become

$$\begin{matrix}(3a)\\(3b)\end{matrix}\,\,\,\,\, \left\{\begin{matrix}\sigma_{yz} = \frac{\delta\phi}{\delta z} = 2C\frac{z}{a^{2}} = -G\theta z\\ \,\,\,\,\,\sigma_{xz} = -\frac{\delta\phi}{\delta y} = -2C\frac{y}{a^{2}} = G\theta y \end{matrix}\right.$$

By taking a look stress vector t on the lateral surface of the bar and having $$n_{y} = \frac{y}{r}$$ and $$n_{z} = \frac{z}{r}$$ and by also using Eqns(3a, 3b), we obtain the following,

$$t_{x} = -G\theta\frac{yz}{r} + G\theta\frac{yz}{r} = 0$$

Figures go Here

Now let us take a wedge out of the bar as shown in the Figure. So the unit vector on the wedged surface is given by

$$(4)\,\,\,\,\, n_x = 0 \,\,\,\,\,\,\,\,\,\, n_y = -cos\beta = -\frac{z}{r} \,\,\,\,\,\,\,\,\,\, n_z = sin\beta = \frac{y}{r}$$

Substituting Eqns(3a, 3b) with Eqn(4) into $$\,t_z = \sigma_{yx}n_y + \sigma_{zx}n_z$$ we obtain the following

$$\,t_z -G\theta r\,\,\,\,\,\Rightarrow\,\,\,\,\,\tau = -t_x = G\theta r$$

Using $$\,T = \theta GJ$$, we can eliminate $$\,\theta$$ to obtain the above expression in terms of the torque.

$$\tau = \frac{Tr}{J}$$

Finally, using Eqn(1) and Eqn(3) and the stress-strain relations, we can see that the warping is zero.

$$\,w = 0$$

Therefore, a uniform bar with a circular cross-section has no warping under torsion.

Contributing Team Members
The following students contributed to this report:

--Eas4200C.f08.nine.d (talk) 02:16, 19 November 2008 (UTC)