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I have to solve the differential equation for radial disks, and therefore I need the shear force distribution. The disk is rotationally symmetric.

I want to use the superpositions principle and set 3 areas.

  1. $0 \le r \le D_i / 2$

  2. $D_i / 2 < r \le k / 2$

  3. $k/2 < r \le D / 2$

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The shear forces are for area 1

$\sum F_z = 0$

$F_{rz1} \pi r - \frac{\pi r^2}{2}p = 0$

$F_{rz1} = \frac{p r}{2}p$

Area 2

$0 = -F_{rz2} \pi r + F_s \pi k $

$F_{rz2} = F_s \frac{k}{r} $

Area 3

$0 = F_{rz3} $

The solution confirms $F_{rz1}$ and $F_{rz3}$ for me. However it states that only after $F_{rz3}$ is known one can solve for $F_{rz2}$ for the area $D_1/2 < r \le D / 2$ The solution should be $F_{rz2} = F_s \frac{k}{2r}$

Did I cut correctly?
What am I missing here?

I know realized I had to insert $k/2$ instead of $k$ and then I get the correct solution for

$0 = -F_{rz2} \pi r + F_s \pi k/2 $

$F_{rz2} = F_s \frac{k}{2r} $

However the question that remains is, why the hint to extend the area? Why is it not possible to calculate $F_{rz2}$ before $F_{rz3}$ when it apparantly is possible?

  • $\begingroup$ I partly found the solution myself. I think I can now calculate $F_{rz2}$ myself, however some questions remain. I edited my question accordingly. $\endgroup$
    – idkfa
    Jun 19 '15 at 14:56
  • $\begingroup$ A bit confused on what you're asking. If you're asking why you can't run your Dif EQ in the second region until you solve the outer region, that's because the outer region will affect what's inside. If you're asking why you can't solve the inside region distribution diagram without the first, you did - you solved for it immediately and saw it was 0 and used that to check that the sum of the forces in the middle would have to reach 0 on the outside. If the forces didn't balance, or something else was outside of that, then you would need to solve for the outside force. $\endgroup$
    – Mark
    Jun 19 '15 at 18:28
  • $\begingroup$ I'm not sure if I get you right. My problem is, if I start with area 1, I can calculate the forces. I don't see why I can't do that with area 2. I'm suspecting it has something to do with what side of the cut I look at. If I would use the other side, then I indeed would need two equations since I then have Frz2 and 3 which are unknown. To specify this: is that necessary for some reason or can I just go my way, solving Frz2 before looking at Frz3? It seems easier than going the other way, especially if Frz3 wouldn't be 0. $\endgroup$
    – idkfa
    Jun 19 '15 at 18:52
  • $\begingroup$ I see. I typed it wrong. What I meant was how you worded it. You can go whichever way you want - just need to remember to carry over the remaining shear force so the total body sums to 0. (in other words: Frz1 = F1(r), the integral of Frz1 from 0 to r + F2(r) = Frz2, Frz3 = F3(r) + integral of Frz2 from r2 to r, etc.) For three regions, it's easier to solve for the inside, then the outside, then the middle, but that's just preference for the special case of 3 regions. $\endgroup$
    – Mark
    Jun 19 '15 at 19:31
  • $\begingroup$ Well thank you! Maybe that's the reason why this approach was chosen. If you want you can type up an answer for me to select. I think I got the concept. $\endgroup$
    – idkfa
    Jun 20 '15 at 0:12

When working with superposition method, any way of approaching the disc works - so long as you account for the reaction forces properly. In this context, it is important to ensure that the sum of the forces is always 0, so unknown reactions (like your simple support in the middle) can be factored in appropriately.


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