More Discussions of the Flip Axle

(Continued on “Flip Axle” Report #1 by Andy Stefanik)

PPD/MD/ME

September 08, 2008

Edward Chi

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Upper Ring, Inner Ring, Flip Axle and

their assemblies

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Partial section view of the inner and upper rings

@ Flip axle area (#2150.500E002-J)

42D detail drawing #2150.530D201-C from www.noao.edu/ets/Mechanical/

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The 1st 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 Mt 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)

Mt = 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

Mb@R2 = Pa = 150,000 lbs –in (maximum)

Mr@R2 = (Mb^2 + Mt^2)^0.5 = 161,333 lbs –in (Resultant maximum moment)

R1 = 37,500 lbs. ↓

R2 = 57,500 lbs. ↑

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*: 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:

do = 4.75 in, di = 3.00 in

Across = 10.65 in^2

Ir1-r2 = 21 in^4, moment of inertia

Sr1-r2 = Ir1-r2 /ro = 8.842 in^3, section of modulus

Jr1-r2 = 42 in^4, polar moment of inertia

Location where 7.5” away from bearing R2 (x2 = 7.50”, cantilever):

do = 5.245 in, di = 3.50 in Across = ~10.24 in^2

Ix2@7.5” = ~21.33 in^4, moment of inertia Jx2@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

Ft = 64 ksi (tensile strength), Fy = 54 ksi (yielding strength)

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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

өtwist angle = Mta/JG = 0.000922 radians

To find the allowable stresses VS. working stresses by two different approaches:

1st approach:

To find the allowable stresses per ASD of AISC 9th edition:

Fb = 0.6 Fy (1- 0.25) = 24.3 ksi Fv = 0.4 Fy (1 –0.25)= 16.2 ksi The calculated working stresses:

fb = Mr@R2 / Sr1-r2 = 161,333 lbs-in / 8.842 in^3

= 18.247 ksi < Fb = 24.3 ksi (max. calculated working bending stress)

fv1 = 20,000 lbs / Across = 10.65 in^2 = 1.878 ksi

fv2 = Mt c / Jr1-r2 = 59,400 lbs-in x 2.375 in / 42 in^4 = 3.359 ksi

fv = fv1 + fv2 = 5.237 ksi < Fv = 16.2 ksi (max. calculated working shear stress)

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2rd approach:

The allowable stresses per ASME code for a shaft design (Maximum shear stress theory):

Fb1 = Fv1 = 0.18 Ft (1-.25) = 8.64 ksi → the lesser

Fb2 = Fv2 = 0.30 Fy (1-.25) = 12.15 ksi

Fb = Fv = Fb1 = Fv1 = 8.64 ksi

The calculated working stress of the shaft:

fb = fv = [16/ π do^3 (1-k^4)] x [(KbMr)^2 + (ktMt)^2]^0.5

= 9.716 ksi > Fb = Fv = 8.64 ksi The working stress is too large subject to the B.C.!

where:

k: = di/do

Kb: combined shock and fatigue factor applied to the bending moment, assuming 1.0

Kt: combined shock and fatigue factor applied to the torsional moment, assuming 1.0

do, di, Mr & Mt: see page 6 and page 7 for the details.

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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

2nd 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.

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References:

• “Flip Axle”, Report #1, by Andy Stefanik, July 23, 2008

• “Allowable Stress Design”, AISC 9th 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, 6th 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.

• See page 13 for the detail information of the KAYDON bearing #KG047AR0 & its link.

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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 2nd model case becomes the same model case as on page 5: (1st case model for the force dist.) Mt = 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)

Mb@R2 = Pa = 150,000 lbs –in (maximum)

The resultant moment Mr@R2 = (Mb^2 + Mt^2)^0.5 = 161,333 lbs –in

So, we only use the model on page 5 to carry out the rest detail calculations.

The 2nd case model for the force dist.:

Hollow shaft supports by 2 bearings, overhanging uniform load m distributes

over length a, torsional moment Mt apply simultaneously.

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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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