Home project - rotary table calculations


by dislect
Tags: calculations, home, project, rotary, table
dislect
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#1
Feb24-13, 02:44 AM
P: 150
Hi guys,

I'm a mechanical engineering student and my dad want's a simplified motorized rotating table. I want to take it as a chance to implement my study in it, and i'm doing it all the way modeled on solidworks and animated.

The calculation part is holding me back, there's a couple of things I need your help with :-)

In the attached picture (sorry, haven't started modeling the thing yet):

http://s13.postimage.org/43lpw52rb/image.png

1. on each sector of the table, which you see divided to eight, 45 degree parts, is a weight with mass M1 with a C.G distanced R from the center of the table. How do I calculate the minimum thickness of the table (in general, material will probably be aluminium) before the load passes the yield strength of the table?
2. regarding the last question, I would also like to know how to calculate the strain of the table in a given material and thickness (t), and see if it passes 1% ?
3. It's kind of simplified but the table is sitting on a bearing on top of a stationary axis and legs who support the weight. Given that the axis is a pipe with diameter D0 and thickness t2, how do I find what should be the pipe's dimension in order to support the weight of the table and loads on top of it without yielding?


Thank you guys so much!
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nvn
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#2
Feb24-13, 12:27 PM
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dislect: (1) The stress on your table top is in Roark. The table top at D0 is assumed to be essentially fixed (let us know if this is not the case, such that your table top is instead essentially simply supported at D0). For a table top essentially fixed at D0, the table top stress is sigma1, listed below, where t1 = table top thickness. Your applied mass on each sector is m1.
nu = Poisson's ratio = 0.33.
theta = table top sector angle = 0.7854 rad.
w = m1*g/(theta*R).
C8 = 0.665 + 0.335(D0/D)^2.
C9 = (0.1675*D0/D)[1 + 3.970*ln(D/D0) - (D0/D)^2].
L9 = (0.3350*R/D)[1 + 3.970*ln(0.5*D/R) - 4(R/D)^2].
sigma1 = (3*w*D/C8)[(2*C9*R/D0) - L9]/(t1^2).
The aluminum and mild steel have about the same tensile yield strength. You would not want to exceed an allowable stress of Sta = 155 MPa, for the aluminum and for the mild steel. Therefore, ensure Rty = sigma1/Sta does not exceed 100 %.

Do you have some example numerical values? E.g., just as a completely arbitrary example, if D = 1050 mm, D0 = 75 mm, R = 420 mm, t1 = 1.7 mm, and m1 = 2.1 kg, then I currently get w = 2.1*9.81/(0.7854*420) = 0.06245 N/mm, C8 = 0.6667, C9 = 0.1373, L9 = 0.1669, and sigma1 = (3*0.06245*1050/0.6667)[(2*0.1373*420/75) - 0.1669]/(1.7^2) = 139.96 MPa. Rty = sigma1/Sta = 139.96/155 = 90.29 %, which does not exceed 100 %, and therefore currently indicates the table top is not overstressed. (I have not independently checked to see if the above solution in Roark is correct.)

(1a) The table top might be governed by deflection (?), instead of stress. I currently do not know, because I have not checked deflection. Plus, I do not know if you want a deflection limit, and what table top outer edge vertical deflection limit you want.

(2) You do not need to check strain, because if you do not exceed Sta = 155 MPa, your strain will not exceed (155 MPa)/(68 950 MPa) = 0.225 %.

(3) I currently have not investigated item 3.

Could you decrease the width of your picture in post 1? Or, just post the text link, in post 1, to the picture.
dislect
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#3
Feb24-13, 02:09 PM
P: 150
Hi, thank you very much for the help.
Could you direct me to the original formulas you have used? There's a booklet on it here, you could direct to me to the page or write it down :

<< Link to copyrighted book deleted by Moderators >>

since i don't understand what c8 c9 l9 are

the diameter of the table is about 2m, im not sure yet about d0 since im suppose to place a bearing there and so the stand might be made out of a rectangular shape welded from a couple of rods since i'll leave that aside for the moment along with (3)

About the deflection part i'm not sure i want to limit that but i sure do want to know how to calculate it :)
How do i do that for a loaded disk instead of a beam, not using cad software?

Thanks again,
Sharon

nvn
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#4
Feb24-13, 02:21 PM
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Home project - rotary table calculations


dislect: Case 1L, p. 463. It also contains the deflection formula.
dislect
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#5
Feb24-13, 02:36 PM
P: 150
Quote Quote by nvn View Post
dislect: Case 1L, p. 463. It also contains the deflection formula.
Does it take into account the 8 weights around the plate? because it is meant for edge free plates.
I understand that y_a is the deflection, what are Q_b, Theta_a, M_rb and what is the final formula for total deflection and total sigma ? I couldn't figure out where
sigma1 = (3*w*D/C8)[(2*C9*R/D0) - L9]/(t1^2).
came from

Thank you for your patience

Sharon
nvn
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#6
Feb24-13, 02:56 PM
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dislect: Yes, it takes into account the eight weights around your table top plate; those weights are the applied load, w, listed in post 2.

The parameters are described on p. 457. The final formula for total deflection is y_a in column 2 of p. 463. The final formula for total sigma is sigma1 in post 2. sigma1 comes from the Mrb formula. See p. 457.
dislect
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#7
Feb24-13, 03:05 PM
P: 150
Quote Quote by nvn View Post
dislect: Yes, it takes into account the eight weights around your table top plate; those weights are the applied load, w, listed in post 2.

The parameters are described on p. 457.

sigma1 comes from the Mrb formula. See p. 457.
1. I see Mra is the reacions, but not Mrb explained. is Mrb the radial load?

2. I also see that Sigma=6M/t^2, but what does M include? I guess its some sort of a combination of Mr and Mt? still cant see how Mrb is related to Sigma

3. In the attached picture, what are the red circled parts?
http://i46.tinypic.com/8xprbb.png
nvn
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#8
Feb24-13, 03:29 PM
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dislect: Mrb is Mr on p. 457. Mrb = unit "radial" bending moment at b = 0.5*D0. Roark confusingly calls Mr "radial" bending moment, whereas a good book would call Mr tangential bending moment.

In sigma = 6*M/(t^2), M, in this case, is Mrb.

The red circled parts are a generic diagram of a generic annular plate.
dislect
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#9
Feb24-13, 03:33 PM
P: 150
Quote Quote by nvn View Post
dislect: Mrb is Mr on p. 457. Mrb = unit "radial" bending moment at b = 0.5*D0. Roark confusingly calls Mr "radial" bending moment, whereas a good book would call Mr tangential bending moment.

In sigma = 6*M/(t^2), M, in this case, is Mrb.

The red circled parts are a generic diagram of a generic annular plate.
Thank you so much, much clearer now.
I'm still trying to figure out the schematic above, don't understand if the black lines with the forces acting upon - is actually a part of the disk or some sort of a weird section cut?
nvn
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#10
Feb24-13, 03:48 PM
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dislect: The free-body diagram is a section cut through the center of a generic annular plate. It shows a side view of a generic annular plate.
dislect
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#11
Feb24-13, 03:52 PM
P: 150
Thank you much more clear now.
Last thing before I start my calculations, isn't w suppose to be w = 8*m1*g/(theta*R) instead of w = m1*g/(theta*R) ?
nvn
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#12
Feb24-13, 04:01 PM
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dislect: Only if you change theta in post 2 to 2*pi rad, then you would need to multiply by 8 in the w formula numerator. But as currently written in post 2, no, do not multiply by 8.
nvn
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#13
Feb24-13, 11:06 PM
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(3) To compute the stress on your upright pipe, probably load only half of the table top, plus add some force corresponding to someone pushing downward on the table top edge. Compute the summation of moment about the table top centerline, M2. Then compute the bending stress on the pipe, sigma2 = M2*(0.5*D0)/I2, where I2 = pipe moment of inertia (second moment of area). Ensure stress level Rty2 = sigma2/Sta does not exceed 100 %, where Sta = 155 MPa.

E.g., just as an arbitrary example, if D = 1050 mm, D0 = 75 mm, R = 420 mm, m1 = 2.1 kg, t2 = pipe wall thickness = 2.4 mm, and a person presses downward with a force of 250 N, then M2 = 2*m1*g*R*sin(22.5 deg) + 2*m1*g*R*sin(67.5 deg) + (250 N)(0.5*D) = 153 860 N*mm. I2 = (pi/64)[D0^4 - (D0 - 2*t2)^4] = 361 040 mm^4. Therefore, sigma2 = (153 860 N*mm)(0.5*75 mm)/(361 040 mm^4) = 15.98 MPa. Therefore, stress level is Rty2 = 15.98/155 = 10.31 %, which does not exceed 100 %, and therefore currently indicates the upright pipe is not overstressed.
dislect
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#14
Feb25-13, 12:59 AM
P: 150
Thanks! very helpful!
two questions:
1. why are the angles 22.5 and 67.5?

2. regarding previous calculations of table disk thickness, how would they change if i want to add to consideration 250N at the circumference edge of the table every 45 degress?

3. how would the calculations change if i was to create the "stand" out of 4 pipes connected together to create a shape of an even square? i want to compute the stress on two out of four connected pipes while table is loaded on the half of it, supported by those pipes. picture:



(distance x is between pipe centers)

Thank you so much nvn, you helped a lot more than I expected coming in
nvn
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#15
Feb25-13, 09:28 AM
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dislect: (1) In Figure 1 of the attached file, you want to compute summation of moment about the table top x axis, with only half of the table top loaded. You need moment arms y1 and y2, shown in Figure 1. In Figure 1, you can see, theta1 = 22.5 deg, and theta2 = 67.5 deg, for a 45 deg sector angle. Therefore, e.g., moment arm y1 = R*sin(theta1).

(2) To consider a 250 N applied load at the circumference edge, every 45 deg, simply change m1 in post 2 to m1 = 25.48 kg, and assign to R a value of R = 0.5*D.

(3) I currently have not investigated item 3 in post 14. By the way, you do not necessarily need to switch to four pipes just because (if) your original, single pipe is overstressed. You could simply increase the single pipe D0 and/or t2 value, until your original, single pipe is not overstressed, if you wish.
Attached Thumbnails
tabletop01.png  
dislect
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#16
Feb25-13, 10:54 AM
P: 150
Thanks,

(2) This might be a dumb question but wasn't R half of D in the beginning? :-) , also, how did you come by 25.48kg if theres 250N * 8 + m1 * 8 ?

(3) I'm switching to four because i want to mount my electric motor on two of them, and supply the table with a better basis on the ground

Again, thanks a lot for the effort and help
nvn
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#17
Feb25-13, 11:08 AM
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dislect: (2) No, R is shown in your diagram in post 1. Changing m1 to 25.48 kg in post 2 places only eight m1 masses at radius R, one mass per sector. For 45 deg sector angles, that is a table top total force of 8*m1*g = 2000 N.
dislect
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#18
Feb25-13, 11:49 AM
P: 150
Thank you, much clearer now.


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