Misc. Home project - rotary table calculations

[dislect]

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!

 

[nvn]

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]

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, I am not sure yet about d0 since I am 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]

dislect: Case 1L, p. 463. It also contains the deflection formula.

 

[dislect]

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]

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]

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 can't 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]

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]

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]

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]

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]

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]

(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]

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]

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.

 

[dislect]

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 there's 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]

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]

Thank you, much clearer now.

 

[nvn]

dislect: (1) I checked other sources, and now confirmed that the Roark solution in post 2 is correct.

(1a) I also checked the Roark deflection formula in case 1L, and confirmed that it is correct. The table design in post 2 is indeed governed by deflection, instead of stress.

 

[dislect]

dislect: (1) I checked other sources, and now confirmed that the Roark solution in post 2 is correct.

(1a) I also checked the Roark deflection formula in case 1L, and confirmed that it is correct. The table design in post 2 is indeed governed by deflection, instead of stress.

Thank you so much, don't have words to express my gratitude
I'm still struggling with the stand part, if you'll have any leads on that i would really appreciate it

Sharon

 

[nvn]

dislect: (3) You can compute the force on the leg pipes described in post 14, item 3, as follows. Compute summation of moment, M2, about the x axis, as described in post 14, item 3, and post 15, item 1. After you obtain M2, then compute force P2 = -M2/[x/cos(45 deg)], where x = leg pipe centerline spacing, shown in post 14, item 3. Also compute force P1 = -(pi/theta0)(m1*g)/4, where theta0 = sector angle in radians, which I called theta in post 2. Then, compute P3 = P1 + P2. Force P3 is the maximum axial force on one leg pipe, for your leg configuration shown in post 14, item 3.

The stress on a leg pipe is then sigma3 = P3/A3, where A3 = leg pipe cross-sectional area. Ensure stress level Rty3 = abs(sigma3)/Sta does not exceed 100 %, where Sta = 155 MPa.

 

[dislect]

dislect: (3) You can compute the force on the leg pipes described in post 14, item 3, as follows. Compute summation of moment, M2, about the x axis, as described in post 14, item 3, and post 15, item 1. After you obtain M2, then compute force P2 = -M2/[x/cos(45 deg)], where x = leg pipe centerline spacing, shown in post 14, item 3. Also compute force P1 = -(pi/theta0)(m1*g)/4, where theta0 = sector angle in radians, which I called theta in post 2. Then, compute P3 = P1 + P2. Force P3 is the maximum axial force on one leg pipe, for your leg configuration shown in post 14, item 3.

The stress on a leg pipe is then sigma3 = P3/A3, where A3 = leg pipe cross-sectional area. Ensure Rty3 = abs(sigma3)/Sta does not exceed 100 %, where Sta = 155 MPa.

Thanks,
I couldn't see how post 15 describes how to calc M2 but i guess -
M2: 2m1Xy1 + 2m1Xy2 ?

Could I ask how did you come by the equations:
P1 = -(pi/theta0)(m1*g)/4 P2 = -M2/[x/cos(45 deg)]

 

[nvn]

i guess -
M2: 2m1Xy1 + 2m1Xy2 ?

Yes, close. Multiply each m1 by g.

P1 = axial force on each pipe due to uniform force from masses; pi/theta0 = number of masses on half of table top; m1*g = weight of each mass; you divide by 4 because there are four pipes. P2 = axial force on each pipe due to moment; you divide M2 by the diagonal distance between pipes, x/cos(45 deg), to obtain the axial force on each pipe.

 

[dislect]

a.

ok then, M2=2m1*y1*g + 2m1*y2*g
and, P2=-M2/[x/cos(45 deg)] when x is the distance between pipe centers and 45 is the angle between each weight
P1 = -(pi/theta0)(m1*g)/4 when theta0=0.7854rad
P3=P1+P2= -(pi/theta0)(m1*g)/4 - M2/[x/cos(45 deg)] which is the maximum force on one leg, when taking into account the extreme situation of only half a table with weights instead off a full table (8 weights) where some of the momentum would cancel each other out.

Stress on one leg is sigma3 = P3/A3, when A3=(1/4)*pi*(D_outer_diameter - D_inner_diameter)^2 and i need to make sure that Rty3 = abs(sigma3)/155 < 100%

Is there a place where I can read to full description of what Rt criteria is and how it was developed?

b.

I tried calculating the table thickness using your guidelines and i came up with a weird, negative result. Could you please take a look?
At the end t1 is the sqrt of -19 ...

Thanks a lot

 

[nvn]

dislect: (a) No, sorry, I do not have a place with a good description. Perhaps look for "factor of safety" (FS) in the index of mechanics of materials books, etc. Perhaps try http://school.mech.uwa.edu.au/~dwright/DANotes/SSS/safety/safety.html#factors, item 2.

Stress level R_ty = sigma/S_ta, where S_ta = allowable tensile stress = S_ty/FS_y, where S_ty = tensile yield strength, and FS_y = yield factor of safety. For your components currently under analysis, you currently could use S_ty = 250 MPa, and FS_y = 1.613. Hence, your allowable tensile stress is currently, S_ta = (250 MPa)/1.613 = 155 MPa.

By the way, always leave a space between a numeric value and its following unit symbol. E.g., 250 N, not 250N. See the international standard for writing units[/color] (ISO 31-0[/color]).

(b) In your L9 equation, D0 should be D. The units of w should be N/mm, not N; i.e., w = 333 N/m = 0.333 N/mm. Why did you put 100 in the denominator of your last equation? 100 % is just the number 1, not 100. First try fixing these, and then let's see if it gets you any closer.

 

[dislect]

dislect: (a) No, sorry, I do not have a place with a good description. Perhaps look for "factor of safety" (FS) in the index of mechanics of materials books, etc. Perhaps try http://school.mech.uwa.edu.au/~dwright/DANotes/SSS/safety/safety.html#factors, item 2.

Stress level R_ty = sigma/S_ta, where S_ta = allowable tensile stress = S_ty/FS_y, where S_ty = tensile yield strength, and FS_y = yield factor of safety. For your components currently under analysis, you currently could use S_ty = 250 MPa, and FS_y = 1.613. Hence, your allowable tensile stress is currently, S_ta = (250 MPa)/1.613 = 155 MPa.

By the way, always leave a space between a numeric value and its following unit symbol. E.g., 250 N, not 250N. See the international standard for writing units[/color] (ISO 31-0[/color]).

(b) In your L9 equation, D0 should be D. The units of w should be N/mm, not N; i.e., w = 333 N/m = 0.333 N/mm. Why did you put 100 in the denominator of your last equation? 100 % is just the number 1, not 100. First try fixing these, and then let's see if it gets you any closer.

(a) Thanks I understand where it comes from now

(b) Now I received L9=0.198, t=4.72 [mm], which I guess makes sense now.
I'm just wondering, if the calculations take into account the weight of the disk itself since its around 40 Kg?
Thanks again!

 

[nvn]

dislect: Excellent work. Also check deflection. No, the calculations currently do not include the disk self weight. Perhaps just include the self mass of each sector in m1. By the way, the unit symbol for kilogram is spelled kg, not Kg. Capital K means kelvin.

 

[dislect]

Making progress by the minute now :-)

About the deflection, once I calculate y_a what do I need to do in order to check if it is reasonable? is there some sort of a failure criteria for that?

 

[nvn]

dislect: There is no standard deflection criterion for this application, to my current knowledge. You might need to make one up. I.e., you could set your own limit, to design it however you prefer. E.g., you might be able to put shims under two table legs, to determine the maximum table top slope (deflection) you consider acceptable.

Do you have a current deflection limit value in mind that you consider acceptable?

 

[dislect]

I tried calculating deflection but the result is crazy, over 3 million. I must be doing something wrong, the a^3 marked part takes the number way too high:

And regarding the calculations of the support legs, it doesn't make sense that the length of the legs is not taken into account, right?