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The question is as follows:

The curved surface of a circular cone of radius 2 sitting on the plane $z=0$ is defined vectorially as $$\underline{r}(R,\phi) = R\cos\phi\underline{\hat{\imath}} + R\sin\phi\underline{\hat{\jmath}} + (8-4R)\underline{\hat{k}}$$ Find the total force on this curved surface under the stress distribution $$\bf{T} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \end{bmatrix} = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & x^2 + y^2 \end{bmatrix}$$

So far I have that the $z$ component is given as $z=8-4R$ so when $R=2$, $z=0$ as it is given and that when $R=0$, $z=8$ so that is where the peak is.

Now, because the coordinate system is cylindrical, I can see that $$\bf{T} = \begin{bmatrix} R\cos\phi & 0 & 0 \\ 0 & R\sin\phi & 0 \\ 0 & 0 & R^2 \end{bmatrix}$$ The equation we have for $z$ I think will be helpful for the integration but what I need some help with is how to find the integrand $\underline{\tau}$ because I don't know how to find the unit normal $\underline{\hat{n}}$ so that I can compute $$\underline{\tau}^{n}_{i} = \sigma_{ij}n_j$$

MRT
  • 603

2 Answers2

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Hint:

In Cartesian coordinates, the surface is given by $\{(x,y,z)\}$ where $z = f(x,y) = 8 - 4\sqrt{x^2 + y^2}$ and $0 \leqslant R = \sqrt{x^2 + y^2} \leqslant 2$, with outward unit normal vector

$$\mathbf{n} = \frac{-\frac{\partial f}{\partial x}}{\sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2+ \left(\frac{\partial f}{\partial y}\right)^2}}\, \mathbf{i} + \frac{-\frac{\partial f}{\partial y}}{\sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2+ \left(\frac{\partial f}{\partial y}\right)^2}}\, \mathbf{j} + \frac{1}{\sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2+ \left(\frac{\partial f}{\partial y}\right)^2}}\, \mathbf{k}$$

In general, when a surface is defined implicitly as $F(x,y,z) = 0$, the gradient vector $\nabla F$ is normal to the surface. Hence, if the surface is specified as $F(x,y,z) = z - f(x,y)=0$ we get the formula above after rescaling to be a unit vector.

RRL
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  • Wait I'm confused now. So the unit normal is you're answer for $\mathbf{n}$. I don't see how you found this? Could you please explain more? @RRL – MRT Feb 09 '18 at 15:21
  • Well trying to use what you have for $\mathbf{n}$ I get that: $$\large\frac{\frac{4x}{\sqrt{x^2+y^2}}}{\sqrt{1 + \frac{16(x^2+y^2)}{(x^2+y^2)}}} \underline{\hat{\imath}} + \frac{\frac{4y}{\sqrt{x^2+y^2}}}{\sqrt{1 + \frac{16(x^2+y^2)}{(x^2+y^2)}}} \underline{\hat{\jmath}} + \frac{1}{\sqrt{1 + \frac{16(x^2+y^2)}{(x^2+y^2)}}} \underline{\hat{k}} = \frac{4}{\sqrt{17}}\left(\frac{4R\cos\phi}{R}, \frac{4R\sin\phi}{R}, \frac{1}{4}\right) = \frac{4}{\sqrt{17}}\left(\cos\phi, \sin\phi, \frac{1}{4}\right)$$ @RRL – MRT Feb 09 '18 at 15:46
  • That looks correct. By unit normal vector, I mean a vector that is normal to the surface of length $1$. We also want it to point outward. You can check those conditions hold with your answer. Where did I find this? Basic concept from multivariable calculus: the gradient $\nabla F$ is normal to the surface given by $F(x,y,z) = 0$. – RRL Feb 09 '18 at 16:19
  • Oh right okay I'm with you now. Okay so I have worked through this now, If I post what I have would you be able to compare with your own answer please? thanks @RRL – MRT Feb 09 '18 at 16:26
  • Sure. See if you can finish and I'll help if necessary. It seems you only want to compute the net force on the curved surface and not the base of the cone. – RRL Feb 09 '18 at 16:29
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Okay so after working I have that the Total force will be the surface force $\underline{F}_S$ which I have found using the stress vector $\underline{\tau}$. Now I have that

$$\underline{\tau} = \left(\frac{4R}{\sqrt{17}}\cos^{2}\phi, \frac{4R}{\sqrt{17}}\sin^{2}\phi, \frac{R^2}{\sqrt{17}}\right)$$

which I then integrate to find $\underline{F}_S$. So doing that gives me

$$\underline{F}_S = \int^{\phi=2\pi}_{\phi=0}\int^{R=2}_{R=0}\left(\frac{4R}{\sqrt{17}}\cos^{2}\phi, \frac{4R}{\sqrt{17}}\sin^{2}\phi, \frac{R^2}{\sqrt{17}}\right)\sqrt{17}RdRd\phi$$

$$ = \int^{\phi=2\pi}_{\phi=0}\left[\frac{4}{3}R^3\cos^{2}\phi, \frac{4}{3}R^3\sin^{2}\phi, \frac{R^4}{4}\right]^{2}_{0}d\phi$$

$$ = \frac{1}{3}\Big[16\cos\phi\sin\phi + 16\phi, 16\phi - 16\cos\phi\sin\phi, 12\phi\Big]^{2\pi}_{0}$$

$$ = \frac{1}{3}\Big[0 + 16\times2\pi, 16\times2\pi - 0, 24\pi\Big] = \frac{32\pi}{3}\Big(1,1,\frac{3}{4}\Big)$$

Does this answer for the Force on the surface match yours? @RRL

MRT
  • 603
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    The bigger error is you are integrating over the disk, when you should be integrating over the conical surface. The differential element of surface area is $dS = \sqrt{1 + (\partial_x f)^2 + (\partial_y)^2} R dR d \phi = \sqrt{17} R dR d\phi$. That extra factor will cancel the $\sqrt{17}$ in the denominator of your answer. The projection of the element between $R$ and $R + dR$ and $\phi$ + $\phi + d\phi$ onto the surface has a bigger area than the planar element. – RRL Feb 09 '18 at 22:25
  • Okay I have made some changes. Does this look correct now? I am still a little hazy as why I was only integrating over the disk... I don't understand how you know about this differential element and how it will be the integrand $\underline{\tau}\cdot\sqrt{17}$ @RRL – MRT Feb 09 '18 at 22:39
  • answer is good. Draw a picture. – RRL Feb 09 '18 at 23:16
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    Draw a picture with a rectangular element of area $dA = dx dy$ in the plane $z = 0$ Draw lines perpendicular to the plane from the corners of the element and look at the intercepted element on the surface $z = f(x,y)$. The surface area of the element is $dS = \alpha dA$ where $\alpha$ is a factor related to the cosine of the angle between the normal vector $\mathbf{n}$ at the surface and the vector $\mathbf{k}$ perpendicular to the plane. That's how $\sqrt{1 + (\partial_x f)^2 + (\partial_y f)^2}$ comes up. You will find this in any advanced calculus book that covers surface integrals. – RRL Feb 09 '18 at 23:35
  • Find the formula here: https://en.wikipedia.org/wiki/Surface_integral – RRL Feb 09 '18 at 23:38
  • Oh yes that link helps. It's been so long since I've worked with vector calculus. But if my answer matches yours then that's great! Thanks for all your help @RRL – MRT Feb 10 '18 at 14:21