# Home project - rotary table calculations

by dislect
Tags: calculations, home, project, rotary, table
 HW Helper Sci Advisor P: 2,009 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.
P: 149
 Quote by 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.

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
 HW Helper Sci Advisor P: 2,009 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.
P: 149
 Quote by 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 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)]
HW Helper
P: 2,009
 Quote by dislect 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.
 P: 149 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
 HW Helper Sci Advisor P: 2,009 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 here, item 2. Stress level Rty = sigma/Sta, where Sta = allowable tensile stress = Sty/FSy, where Sty = tensile yield strength, and FSy = yield factor of safety. For your components currently under analysis, you currently could use Sty = 250 MPa, and FSy = 1.613. Hence, your allowable tensile stress is currently, Sta = (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 (ISO 31-0). (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.
P: 149
 Quote by 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 here, item 2. Stress level Rty = sigma/Sta, where Sta = allowable tensile stress = Sty/FSy, where Sty = tensile yield strength, and FSy = yield factor of safety. For your components currently under analysis, you currently could use Sty = 250 MPa, and FSy = 1.613. Hence, your allowable tensile stress is currently, Sta = (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 (ISO 31-0). (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!
 HW Helper Sci Advisor P: 2,009 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.
 P: 149 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?
 HW Helper Sci Advisor P: 2,009 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?
 P: 149 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?
 HW Helper Sci Advisor P: 2,009 dislect: D in the denominator of y_a is not your D. See p. 457. You will need to give their D a different name, since you are already using the name D. Your answers for C3 and L3 are currently incorrect. You forgot to multiply by the beginning coefficient, before the braces. Try again. Also, I was not able to obtain your w value yet. By the way, generally always maintain four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits. You do not need the pipe length to obtain the axial force on the vertical pipes.
P: 149
 Quote by nvn dislect: D in the denominator of y_a is not your D. See p. 457. You will need to give their D a different name, since you are already using the name D. Your answers for C3 and L3 are currently incorrect. You forgot to multiply by the beginning coefficient, before the braces. Try again. Also, I was not able to obtain your w value yet. By the way, generally always maintain four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits. You do not need the pipe length to obtain the axial force on the vertical pipes.
- My a=1000mm, b=85mm, r0=750mm
I corrected D, using E=69 GPa. thanks for noticing !
Corrected calculation but still result is way too high :

http://www.siz.co.il/my/tdzh4aw2lzyi.png

-- About the support leg pipes,
1. I tried calculating the minimum x distance between the 4 legs and came up with a result that seems a bit big, plus as written in green in the picture its weird that if i reduce the number of legs i actually minimize the distance x ?
Also, it seems like the dimensions of the pipes barely make a difference on the end result even with t2=0 (full pipe)

http://www.siz.co.il/my/tayjnd0jn1em.png

2.what about calculating the bend momentum and the center of the pipe (L/2) to see if it breaks at the weakest spot and how much it bends due to the momentum of the weights + table mass?

I know it's a lot but i'm really feeling like I made progress thanks to your help, getting there really fast now!
 P: 27 You should also check for buckling of the leg(s). If the slenderness ratio is high enough the legs will buckle well below the listed yield strength of the material.
 HW Helper Sci Advisor P: 2,009 dislect: Your value for E is correct in the text of post 32, but wrong in your first attached file. Check your units. And you did not update C3 and L3 in your y_a equation. Let's see if that gets you closer on your first attached file. I will review your second attached file within one day. I could not exactly match your M_table (nor w) answer yet. I do not know if you are using some strange value for pi, or something like that. By the way, list four significant digits for each intermediate value. Leading zeros do not count as significant digits. E.g., list 0.002251 for L3, four significant digits, not 0.002, which is only one significant digit. As another example, list 0.2393 for C2, not 0.239.
P: 149
 Quote by nvn dislect: Your value for E is correct in the text of post 32, but wrong in your first attached file. Check your units. And you did not update C3 and L3 in your y_a equation. Let's see if that gets you closer on your first attached file. I will review your second attached file within one day. I could not exactly match your M_table (nor w) answer yet. I do not know if you are using some strange value for pi, or something like that. By the way, list four significant digits for each intermediate value. Leading zeros do not count as significant digits. E.g., list 0.002251 for L3, four significant digits, not 0.002, which is only one significant digit. As another example, list 0.2393 for C2, not 0.239.

E: im using E=69 GPA which as i checked it 69 * 10^6 [N/m] and for the calculation of D i used meter instead of mm, so i converted 10mm^3 to 0.01m^3 and came up with the result written. When i used it in y_a equation i went back to mm so i multiplied by 1000. Did i do something wrong?
C3 & L3: I updated the values and my y_a is now - 10,701 mm .... still doesn't make sense.
M_table : 83.6 kg total ; so every 45 degree section of the table is m1*=20 kg + 83.6/8 kg = 30.45 kg and then i use it to calculate w*=0.507 [N/mm]
 HW Helper Sci Advisor P: 2,009 dislect: 69 GPa is not 69e6 Pa. Try again. (By the way, do not worry too much yet about analyzing the pipes. First, get the preliminary things computed correctly.)

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