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Finite Element Analysis of the18 Turn Beam H. F. Fan November 5, 2004

Finite Element Analysis of the18 Turn Beam H. F. Fan November 5, 2004

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Page 1: Finite Element Analysis of the18 Turn Beam H. F. Fan November 5, 2004

Finite Element Analysis of the18 Turn Beam

H. F. FanNovember 5, 2004

Page 2: Finite Element Analysis of the18 Turn Beam H. F. Fan November 5, 2004

18 Turn Beam FEA Model Nodes couples in Dy

1000 lb

Dx=0

Dz=0

Dy=0 at support

• Beam is 42” long and 40” in span• ¼ of beam was modeled (half width & half length)• Conductor is 0.500 x 0.625 inches• Turn insulation is 0.03 inches• Ground wrap is 0.03 inches• E of conductor is 9.5E6 psi• E of turn insulation is 1.5E6 psi• E of ground wrap is 1.5E6 psi• Poissons’ ratio is 0.31

Boundary conditions and loading:• Dz = 0 on mid-span surface• Dx = 0 on mid-width surface• Dy = 0 at support• Nodes on top surface within 0.5” of mid-span are coupled in Dy for a vertical force of -1000 lb

Page 3: Finite Element Analysis of the18 Turn Beam H. F. Fan November 5, 2004

Vertical Displacement Uy for 4000 lb Load

Undeformed shape

Whole model Conductor only

Page 4: Finite Element Analysis of the18 Turn Beam H. F. Fan November 5, 2004

Von Mises Stress and Axial Stress of Conductor

Page 5: Finite Element Analysis of the18 Turn Beam H. F. Fan November 5, 2004

Axial Stress of Turn Insulation and Ground Wrap

Page 6: Finite Element Analysis of the18 Turn Beam H. F. Fan November 5, 2004

Simple Beam Pure Bending Stress CalculationsCalculate equivalent Ie:

• Ratio of Ei/Ec = 1.5E6/9.5E6 = 0.158• Width of turn insulation and ground wrap is

multiplied by the factor of 0.158• Ac of a conductor = 0.3125 in^2• Ic of a conductor = 0.00651 in^4• I1 = 18xIc + 4*Ac*(i*0.56)^2 where i=1,2,3,4 = 11.88 in^4• Ai1 = 0.625*0.06 = 0.0375 in^2• Ii1 = 0.625*0.06^3/12 = 1.125E-5 in^4• Ii2 = 0.06*5.1^3/12 = 0.6633 in^4• I2 = 0.158*[3*Ii2 + 20xIi1 + 4*Ai1*(i*0.28)^2] where i=1,3,5,7,9 = 0.62 in^4 • Ie = I1 + I2 = 12.50 in^4

• Bending Moment M = 4000*40/4 = 40000 in-lb• Distance from neutral axis y = 2.49 in• Max. bending stress = My/Ie = 7969 psi

5.1”

1.43”

Page 7: Finite Element Analysis of the18 Turn Beam H. F. Fan November 5, 2004

Vertical Displacement Calculations

Max. displacement produced by the bending moment:

dmax = 4000*40^3/(48*E*Ie) = 0.04492 in

Max. displacement produced by the shear force:

dmax = *4000*L/(4*A*G)Assuming a = 3/2

A = 5.625 (neglecting insulation) G = 3.0E6

then dmax = (3/2)*4000*40/(4*5.625*3.0E6) = 0.00356 in

Total maximum displacement = 0.04848 in

Page 8: Finite Element Analysis of the18 Turn Beam H. F. Fan November 5, 2004

Discussions:• The calculated maximum bending stress is 7969 psi

• If we consider the applied load is uniformly distributed in 1”, then the maximum bending moment become:

Max. bending moment = 4000*40/4 -2000*0.5/2 = 39500 lb-in, and

Maximum bending stress = 7870 psi

• The FEA maximum stress occurs at the corner of the upper conductor

• The peak FEA axial stress on the conductor are -10060 psi and 7843 psi

• Higher stress in the compression side is due to the Poisson’s effect from the applied loading

• The calculated bending stress assumed the neutral axis is at the mid-height

•The calculated maximum vertical displacement is 0.04848 inches, which is the superposition of the bending and shear effects.

• The vertical displacement produced by the shear force depend on the assumed values of parameters , A, and G. The values are approximate numbers. However, the contribution of the shear displacement is much less then the bending displacement in this case.

• The maximum displacement from FEA is 0.04864 inches