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Finite Element Analysis of the18 Turn Beam

Finite Element Analysis of the18 Turn Beam. H. F. Fan November 5, 2004. 1000 lb. Nodes couples in Dy. Dy=0 at support. Dz=0. Dx=0. 18 Turn Beam FEA Model. 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

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Finite Element Analysis of the18 Turn Beam

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  1. Finite Element Analysis of the18 Turn Beam H. F. FanNovember 5, 2004

  2. 1000 lb Nodes couples in Dy Dy=0 at support Dz=0 Dx=0 18 Turn Beam FEA Model • 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

  3. Vertical Displacement Uy for 4000 lb Load Undeformed shape Conductor only Whole model

  4. Von Mises Stress and Axial Stress of Conductor

  5. Axial Stress of Turn Insulation and Ground Wrap

  6. 5.1” 1.43” Simple Beam Pure Bending Stress Calculations • Calculate 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*SAc*(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*SAi1*(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

  7. 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 = a*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

  8. 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, 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

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