1. More Discussions of the Flip Axle (Continued on “Flip Axle” Report #1 by Andy Stefanik) PPD/MD/ME September 08, 2008 Edward Chi

2. 2 Upper Ring, Inner Ring, Flip Axle and their assemblies

3. 3 Partial section view of the inner and upper rings @ Flip axle area (#2150.500E002-J)

4. 4 2D detail drawing #2150.530D201-C from www.noao.edu/ets/Mechanical/

5. 5 The 1 st case model for the force distribution: (See page 11 for the 2nd case model). Hollow shaft supports by two (2) bearings, overhanging concentrated load P, and torsional moment M t applying simultaneously. Where: L = 4.0 in, the center distance between two bearings (KAYDON KG047AR0) a = 7.50 in (See page 12 for the length derivation) P = 20,000 lbs. (See page 14 of report #1) M t = 50% of the weight of the PF Cage x 1.65 x the eccentric distance* = 0.5 x 12,000 lbs x 1.65 x 6.00 in (where1.65, seismic factor) = 59,400 lbs- in M b@R2 = Pa = 150,000 lbs –in (maximum) M r@R2 = (M b ^ 2 + M t ^ 2 )^ 0.5 = 161,333 lbs –in ( Resultant maximum moment ) R1 = 37,500 lbs. ↓ R2 = 57,500 lbs. ↑

6. 6 *: assuming the distance between the gravity ctr. of the PF Cage to the axial axis of the shaft. The geometric properties of the shaft: Between two bearings section: d o = 4.75 in, d i = 3.00 in A cross = 10.65 in^ 2 I r1-r2 = 21 in^ 4, moment of inertia S r1-r2 = I r1-r2 /r o = 8.842 in^ 3, section of modulus J r1-r2 = 42 in^ 4, polar moment of inertia Location where 7.5” away from bearing R 2 (x 2 = 7.50”, cantilever): d o = 5.245 in, d i = 3.50 in A cross = ~10.24 in^ 2 I x2@7.5” = ~21.33 in^ 4, moment of inertia J x2@7.5” = ~50.97 in^ 4, polar moment of inertia Assuming: material of the shaft: AISI C1018 (cold finished steel) E = 29 x 10^ 3 ksi, modulus of elasticity G = 11.5 x 10^ 3 ksi, shear modulus of elasticity F t = 64 ksi (tensile strength), F y = 54 ksi (yielding strength)

7. 7 The calculated deflections: ∆ max@ the cantilever where load P applying (x2=7.50”) = 0.00697 in ∆ max between two bearings (x1 = 2.31”) = 0.00025 in ө twist angle = M t a/JG = 0.000922 radians To find the allowable stresses VS. working stresses by two different approaches: 1 st approach: To find the allowable stresses per ASD of AISC 9th edition: F b = 0.6 F y (1- 0.25) = 24.3 ksi F v = 0.4 F y (1 –0.25)= 16.2 ksi The calculated working stresses: f b = M r@R2 / S r1-r2 = 161,333 lbs-in / 8.842 in^ 3 = 18.247 ksi < F b = 24.3 ksi (max. calculated working bending stress) f v1 = 20,000 lbs / A cross = 10.65 in^2 = 1.878 ksi f v2 = M t c / J r1-r2 = 59,400 lbs-in x 2.375 in / 42 in^ 4 = 3.359 ksi f v = f v1 + f v2 = 5.237 ksi < F v = 16.2 ksi (max. calculated working shear stress)

8. 8 2rd approach: The allowable stresses per ASME code for a shaft design (Maximum shear stress theory): F b1 = F v1 = 0.18 F t (1-.25) = 8.64 ksi → the lesser F b2 = F v2 = 0.30 F y (1-.25) = 12.15 ksi F b = F v = F b1 = F v1 = 8.64 ksi The calculated working stress of the shaft: f b = f v = [16/ π d o ^ 3 (1-k^ 4 )] x [(K b M r )^ 2 + (k t M t )^ 2 ]^ 0.5 = 9.716 ksi > F b = F v = 8.64 ksi The working stress is too large subject to the B.C.! where: k: = d i /d o K b : combined shock and fatigue factor applied to the bending moment, assuming 1.0 K t : combined shock and fatigue factor applied to the torsional moment, assuming 1.0 d o, d i, M r & M t: see page 6 and page 7 for the details.

9. 9 Conclusions (under the current defined boundary condition): The existing bearings of the shaft (KAYDON Bearing #KG047AR0) will not meet the requirements for applying the shaft under the new design configurations (see page 5 for the calculated results). The existing shaft will not meet the new design criteria as it defined (the working stress is higher than the allowable stress, see page 8 from the 2 nd approach calculations). However, to select a medium carbon, hot rolled, more ductile steel with higher strength should fulfill such design requirements, also the shaft design should minimize the stress concentration. The deflection values (see page 7) will not change under the same defined boundary condition (with the various steel materials) in general.

10. 10 References: “Flip Axle”, Report #1, by Andy Stefanik, July 23, 2008 “Allowable Stress Design”, AISC 9 th edition. “Theory and Problems of Machine Design”, by Hall, Holowenko, Laughlin, 1961. Section 2.10 and Table 37 of “Roark’s Formulas for Stress & Strain”, by Warren C. Young, 6 th edition. “Ryerson Stock List”, by J.T. Ryerson & Son, Inc. 1991. Link http://www.noao.edu/ets/Mechanical/ for the details of the related 2D fab. drawings.http://www.noao.edu/ets/Mechanical/ See page 13 for the detail information of the KAYDON bearing #KG047AR0 & its link.

11. 11 Where: L = 4.0 in, the center distance between two bearings (KAYDON KG047AR0) a = 8.75 in, the contact length of the shaft where taking the uniform load m. b = 3.125”, the dist. between the ctr. of bearing R2 and the edge of the inner ring. m = 20,000 lbs./8.75 in = 2,286 lbs/in (See page 12 for the value calculations of a & b) By transferring the uniform load m to the equivalent concentrate load P, it finds that: ½ a + b = 4.375” + 3.125” = 7.50” (The overhanging dist. from R2 to the equivalent concentrated load P) The above 2 nd model case becomes the same model case as on page 5: (1 st case model for the force dist.) M t = 50% of the weight of the PF Cage x 1.65 x the eccentric distance* = 0.5 x 12,000 lbs x 1.65 x 6.00 in = 59,400 lbs- in (see page 5) M b@R2 = Pa = 150,000 lbs –in (maximum) The resultant moment M r@R2 = (M b ^ 2 + M t ^ 2 )^ 0.5 = 161,333 lbs –in So, we only use the model on page 5 to carry out the rest detail calculations. The 2 nd case model for the force dist.: Hollow shaft supports by 2 bearings, overhanging uniform load m distributes over length a, torsional moment M t apply simultaneously.

12. 12 The dimension derivation for force distribution model case 1 ( ref. page 5): 1. L = 5.937” – 0.937” – 2 x0.50” (between two ctr. Line of the bearings) = 4.00” (See page 4 and page13 drawing #2150.500E002) 2. a = [8.50”-6.875” (per dwg. #2150.500E036) +1.0” (gap between the inner & outer rings) + 0.50” (half width of the bearing) ] + [97.0” –(85.75” + 6.50”) (per #2150.500E020 ) – 0.375” (per #2150.500C027) ] = 3.125” +4.375” = 7.50” The dimension derivation for force distribution model case 2 ( ref. page 11): 1. L = 5.937” – 0.937” – 2 x0.50” (between two crt. Line of the bearings)) = 4.00” (See page 4 and page13 drawing #2150.500E002) 2. b = 8.50” – 6.875” (per dwg. #2150.500E036) + 1.0” (gap between inner & outer rings) + 0.50” (half width of the bearing) = 3.125” 3. a = 11.625” (per page 4/#2150.530D203) – 3.125” (b) + 0.50” (1/2 width of the bearing) -0.25” (chamfer size of the shaft) = 8.75”

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