Piece-Wise Deflection of a Crank Arm Deflection of the loaded end point is the sum of three deflections: 1. Bending of the 6” Rod due to the 300 lb load. yFlEI133= 2. Twisting of the 6” Rod due to the 1200 in.lb torque, with rotation of the 4” Rod. θ θ= =TlJGy r,2 3. Bending of the 4” Rod due to the 300 lb load. yFlEI333= 4 in 6 in 300 LB 3/4 in 4 in Rod 1200 in.lb Moment 300 lb 300 lb 6 in Rod 1200 in.lb Torque 300 lb y3 y1 y2 θ r IrinJrI inE psi G psi= = == = == × = ×0 37540 0155322 0 0310630 10 115 1044446 6. " .., .ππy iny r iny iny y y y intotal136263361 2 3300 63 30 10 0 015530 04644 1200 60 03106 115 100 0806300 43 30 10 0 015530 013701407=×== =×==×== + + =( )( )( )( . ).( )( )( )( . )( . ).( )( )( )( . )..θCrank Deflection Analysis by Castigliano Strain EnergyTOTAL = Bending (in 4”) + Bending (in 6”) + Torsion (in 6”) dxGJTdxEIMdxEIMUBaseElbowBaseElbowElbowTip∫∫∫++=22222624 ∫∫∫++=6026024022)4()(21)(21dxGJPdxPxEIdxPxEIU ∫∫∫++=6026022402221622dxGJPdxxEIPdxxEIPU602603240328312312xGJPxEIPxEIPU ++=GJPEIPEIPGJPEIPEIPU22223232486216664686646++=++= ++=++=GJEIEIPGJEIEIPU483667.1048621666422 ++=∂∂=GJEIEIPPUP483667.102δ ×+×+=)03106.0)(105.11(48)01553.0)(1030(3667.1060066Pδ )0001344.000007727.000002289.0(600 ++=Pδ .1407.0)08064.004636.001374.0( inP=++=δ Torsion in 6” Rod 0 0 0 4 6 6 1200 Lb in, T = 4P Bending in 6” Rod Bending in 4” Rod 1800 Lb in, M6 = (P)*(x) 1200 Lb in, M4 = (P)*(x)Finite Element Analysis of the Crank Arm with 300LB Tip Load Total Deflection at Tip = 0.141
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