Using SCHNORR Part# 015900 De = 60. mm Di = 30.5 mm t = 3.5 mm lo = 5.004 mm ho = 1.5 mm Max permitted s =.96 mm, limit for sigma ll = 1,250. N/mm^2.

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Presentation transcript:

Using SCHNORR Part# De = 60. mm Di = 30.5 mm t = 3.5 mm lo = mm ho = 1.5 mm Max permitted s =.96 mm, limit for sigma ll = 1,250. N/mm^2 BAUER Item # is exactly the same

Calculations in (N and mm) units Calcs Based on the SCHNORR data formulas. Note: These calculations are for SCHNORR #15900 De - Ouside diameter60mm Di - Inside diameter30.5mm pi3.14 1/pi /pi /pi del=De/Di Schnorr Formula # lndel=ln(De/Di) del del (del-1/del)^ (del+1)/(del-1) (2/lndel) K1 =1/pi(((del-1/del)^2))/((del+1)/((del-1)-(2/lndel))) Formula # K2=6/pi(((del-1)/(lndel))-1)))/(lndel) Formula # K3=3/pi(del-1/lndel) Formula # K4=1 since thikness is less than 6 mm Formula #6 (see chapter 1.3)1 FORCE and STRESS Calculations lo= total height5.004mm ho= lo-t1.499mm t=average spring thickness3.505mm

s= deflection of the single spring (from 75% of ho))0.918mm E= modulus for steel206000N/mm^2 mu= Poisson's ratio0.3 1-mu^ E/(1-mu^2) t^4/(K1*de^2) s/t ho/t s/2t Formula #8a N Formula #9 Stresses at the center of rotation (point OM) N/mm^2 ( should be less then -1600N/mm^2) t^2/(K1*de^2) formula #10 Stress at point (l) N/mm^2 formula #11 Stress at point (ll) N/mm^2 formula #12 Stress at point (lll) N/mm^2 1/del (K2-2K3) (ho/t-s/2t) Formula #13 Stress at point (lV) N/mm^2 t^3/(K1*de^2) Formula #14 Spring Rate (dF/ds) N/mm Formula #15 Spring Work (Integral from 0 to s, of F*ds) N-mm 2E/(1-mu^2) t^5/(K1*de^2) Basic requirements for a good Disk spring design For the above basic equations to work This Spring is linear because ho/t=.428 (since for ho/t<0.4 are linear)slightly non-LINEAR del=De/Di= and it should be between 1.75 and 2.5O.K. Outside diameter De, Inside diameter Di

ho/t=(lo-t)/t=.428 and it should be between 0.4 to 1.3O.K. ho=lo-t cone height, t=disk thickness De/t=17.12 and it should be between 16. and 40.O.K. Conclusion: This spring under the constant load of N is O.K. Basic machine operation requirements: 1) TF coil inner leg maximum thermal expansion was calculated at 8.4mm 2) OH coil requires a preload of 20,000.0 lbs total or 6,428.5 N (14 St) per stack. Must calculate the required deflection (s) for the preload force? 3) OH coil thermal expansion was calculated at 6.0mm So, the total each spring stack travel is = calculated from minimum preload ? For Fatigue predictions, using fatigue life diagrams for disk springs, one requires the Pre and Maximum loads stresses at points ll or lll (depending on del and ho/t).

For SCHNORR Part # and 12 Stacks with 26 disksForce=20,000.0lbs disp. mmtot disp.Force Nsigma ll N/mm^2 26* , Preload , ,100.01,300.0FAIL Prmt. Limt.26* ,790.01,250.0 Total Space required26*lo is 26*.197 = in8.0 inch available For SCHNORR Part # and 14 Stacks with 26 disks disp. mmtot disp.Force Nsigma ll N/mm^2 26* , Preload , ,164.61,183.4O.K. Prmt. Limt.26* ,790.01,250.0 Total Space required26*lo is 26*.197 = in "no change"

14 Stacks 12 Stacks