Análise de Eixo de Laminador Industrial: Diagramas de Momento e Forças

388 Elementos de máquinas de 74 Um laminador industrial de engrenagens mostrado na figura é movido a 300 revmin por uma força F atuando em um círculo de diâmetro primitivo de 75 mm O rolo exerce uma força normal de 5200 N na face do componente do rolo sobre o material seco puxado O material passa sob o rolo O coeficiente de fricção é de 040 Desenvolva os diagramas de momento de flexão e momento torçante no eixo considerando a força do rolo como a uma força concentrada no centro do rolo e b uma força uniformemente distribuída ao longo do rolo Essas diagramas apareceram em dois planos ortogonais Problema 74 O material se move sob o rolo Dimensões em polegadas 75 Desenhe um eixo para a situação do laminador industrial do Problema 74 com um fator de projeto de 2 uma medida de confiabilidade de 0999 com uma falha por milha por rolete e para um manual de esfera à esquerda e um manual de rolos cilíndricos à direita Para a deformação use um fator de segurança 2 76 A figura mostra um projeto proposto para o eixo do rolo industrial do Problema 74 Mancais de filme hidro 74 Seja C o vão central do rolo 𝐹𝐶 𝑦 5200 𝑁 𝑚 200 1000 𝑚 1040𝑁 𝐹𝐶 𝑧 𝐹𝐶 𝑦 𝜇 1040 04 416𝑁 Torque 𝑎 𝐶𝑇 𝑇𝐶 𝐹𝐶 𝑧 𝑑𝑟𝑜𝑙𝑜 2 416𝑁 100 1000𝑚 2 208𝑁 𝑚 Torque de 𝐵1 𝑇𝐵 𝐹𝐵 𝑍 𝑑4 2 208𝑁 𝑚 𝑇𝐶 𝑇𝐵 𝐹𝐵 𝑍 75 1000𝑚 2 208 𝐹𝐵 𝑧 55467𝑁 𝐹𝐵 𝑦 𝐹𝐵 𝑧𝑡𝑎𝑛20 55467 𝑡𝑎𝑛20 20188𝑁 a Plano XY 𝑅𝑂 𝑌 56873𝑁 𝑅𝐴 𝑌 26939𝑁 Plano XZ 𝑅𝑂 𝑍 7411𝑁 𝑅𝐴 𝑍 89656𝑁 b Plano XY 𝑅𝑂 𝑌 56873𝑁 𝑅𝐴 𝑌 26939𝑁 Maximo local no ponto x 15437mm Mmax15437 5687315437 52015437 4522 5669412 Nmm Plano XZ 𝑊1 𝜇 52𝑁 𝑚𝑚 208𝑁𝑚𝑚 𝑅𝑂 𝑍 7411𝑁 𝑅𝐴 𝑍 89656𝑁 Maximo local do ponto x 8063 mm Mmax8063 74118063 2088063 4522 465555 Nmm X Z RO W1 208 RA B Z B 55467 45 mm 200 mm 45 mm 70 mm 1 sec X1 2 sec X2 3 sec X3 4 sec X4 55467 Qy 7411 34189 Mx 333514 465555 0 2344204 388269 Draw the forces on the roller as in Figure 1 y FBy FBz z FCz FCy Calculate the forces acting on the roller Fc F l Here the force acting on centre of the span of roller is Fc normal force is F and length of the roller is l Substitute 30 lbfin for F and 8 in for l Fc 30 lbfin 8 in Fc 240 lbf Calculate the force acting at C along z axis Fc Fzc μ Fc Here coefficient of friction is μ Substitute 04 for μ and 240 lbf for Fc Fzc 04 240 lbf Fzc 96 lbf Calculate the torque T T Fzc d 2 Here diameter of roller is d Substitute 96 lbf for Fzc and 4 in for d T 96 lbf 4 in 2 T 192 lbf in Calculate the force at B along z axis FzB Fzb T dc 2 Here diameter of pitch circle is dc Substitute 192 lbf in for T and 3 in for dc Fzb 192 lbf in 3 in 2 Fzb 128 lbf Calculate the force at B along y axis FyB FyB Fzb tan 20 Substitute 128 lbf for Fzb FyB 128 lbf tan 20 FyB 466 lbf Draw the free body diagram and bending moment diagram of the beam in xy plane as in Figure 2 Comment Step 4 of 18 Calculate moments about the point O Mo M0 0 Fc575 in FA115 in FyB1425 in 0 Substitute 240 lbf for Fc and 466 lbf for FyB 240 lbf 575 in FA 115 in 466 lbf 1425 in 0 FA 240 lbf 5745 in 466 lbf 1425 in 115 in FA 623 lbf Calculate moments about the point A MA 0 Fo 115 in FyB 275 in Fc 575 in 0 Here moment about point A is MA and force acting at O along y axis Fo Substitute 240 lbf for Fc and 466 lbf for FyB Fo 115 in 466 lbf 275 in 240 lbf 575 in 0 Fo 466 lbf 275 in 240 lbf 575 in 115 in Fo 1311 lbf Step 5 of 18 Calculate moments at point O along z axis Mo Mo 0 lbf in Calculate moments at point C along z axis Mc Mc Fo 575 in Here force acting at point C along y axis is Fo Substitute 1311 lbf for Fo Mc 1311 lbf 575 in Mc 7538 lbf in Calculate moments at point A along z axis MA MA F 275 in Here normal force is F Substitute 466 lbf for F MA 466 lbf 275 in MA 12815 lbf in Calculate moments at point B along z axis MB MB 0 lbf in Comment Step 6 of 18 Draw the free body diagram and bending moment diagram of the beam in xz plane as in Figure 3 Calculate moments about the point O Mo M00 Fc2575 inFz2115 inFb21425 in0 Substitute 96 lbf for Fc2 and 128 lbf for Fb2 96 lbf575 inFz2115 in128 lbf1425 in0 Fz2 96 lbf575 in128 lbf1425 in 115 in Fz22066 lbf Calculate moments about the point A Ma MA0 Fz0115 inFb2275 inFc2575 in0 Substitute 128 lbf for Fb2 and 96 lbf for Fc2 Fz0115 in128 lbf275 in96 lbf575 in0 Fz0 96 lbf575 in128 lbf275 in 115 in Fz0174 lbf Comment Step 8 of 18 Calculate moments at point O along z axis Mo Mo0 lbfin Calculate moments at point C along z axis Mc McFz0575 in Here force acting at point C along y axis is Fy0 Substitute 174 lbf for Fy0 Mc174 lbf575 in Mc10005 lbfin Calculate moments at point A along z axis Ma MaFb2275 in Substitute 128 lbf for Fb2 Ma128 lbf275 in Ma352 lbfin Calculate moments at point B along z axis Mb Mb0 lbfin Calculate the total moment at C McMc2yMc2x Here total moment at point C is Mc bending moment at C in xy plane is Mcy and bending moment at C in xz plane is Mcx Substitute 100 lbfin for Mcy and 754 lbfin for Mcx Mc100 lbfin2754 lbfin2 Mc761 lbfin Calculate the total moment at D MdMDy2MDx2 Here total moment at point D is MD bending moment at D in xy plane is MDy and bending moment at D in xz plane is MDx Substitute 128 lbfin for MDy and 352 lbfin for MDx MD128 lbfin2352 lbfin2 MD375 lbfin The maximum moment Mmax is 761 lbfin at x be 575 in Hence torque rises from 0 to 192 lbfin and then becomes constant to B Comment Step 10 of 18 Write the design analysis The midrange and alternating moments Ma is 761 lbfin and torque Tm is 192 lbfin MmTa0 From table 71 First Iteration Estimates for Stress Concentration Factors Kt and Kts For End Milled Key Seats Concentration factor for bending stress Kt is 214 Concentration factor for torsional stress Kts is 3 As the material specified is of low strength AISI 1020 CD steel From Figure 620 notch sensitivity q is 075 From figure 621 notch sensitivity for material in reversed torsion qs is 08 A typical ratio of rd002 at the bottom of the key seat and a shaft diameter up to 3in Calculate the fatigue stress concentration factor for bending kf kf1qkt1 Substitute 075 for q and 214 for Kt kf10752141 kf19 Calculate the fatigue stress concentration factor for torsion kfs kfs1qskts1 Substitute 08 for qs and 3 for Kts kfs10831 kfs26 Calculate the endurance limit Se Se05Sut Substitute 68 kpsi for Sut Se0568 kpsi Se34 kpsi Comment Step 12 of 18 From table 62 Parameters for Marin Surface Condition Factor For Cold Rolled Surface Finish The factor a is 27 The exponent b is 0265 Calculate the surface factor ka kaaSb ut Here minimum tensile strength is Sut factor is a and exponent is b Substitute 68 kpsi for Sut 27 for a and 0265 for b ka2768 kpsi0265 ka0883 Calculate size factor kb kbd030107 Substitute 2 in for d kb2 in030107 kb0816 From table 65 Reliability factors Kₑ corresponding to 8 percent standard deviation of the endurance limit For a reliability percentage of 999 the reliability factor Kₑ is 0753 Calculate endurance limit at critical location of machine part Sₑ Sₑ kₐkbkcSe Here surface condition modification factor is kₐ size modification factor is kb reliability factor is Kₑ and rotarybeam test specimen endurance limit is Se Substitute 0883 for kₐ 0816 for kb 0753 for Kₑ and 36 kpsi for Se Sₑ 08830816075334 kpsi Sₑ 1844 kpsi Comment Step 14 of 18 Calculate the conservative first design selecting the DEGoodman criteria d 16n π 1Se 4KfMa π212 1Sm 3KfTm π21213 Here diameter is d factor of safety is n endurance limit is Sₑ stress concentration factor for bending is kf alternating moment is Ma tensile strength is Sm stress concentration factor for torsion is kf and torque is Tm Substitute 2 for n 18440 psi for Sₑ 19 for kf 761 lbfin for Ma 68000 psi for Sm 26 for kf and 192 lbfin for Tm d 16 2 π 118440 psi 4 19 761 lbfin212 168000 psi 3 26 192 lbfin21213 d 32 π 57836 18440 psi 14976 68000 psi13 d 15065 in Hence the diameter is 15065 in Since the notch sensitivity and size factor are conservative for the obtained diameter but close enough for a first iteration until deflections are checked Calculate Von Mises maximum stress σmax σmax 32KfMa πd32 3 16KfTm πd3212 Substitute 19 for kf 761 lbfin for Ma 26 for kf 15065 in for d and 192 lbfin for Tm σmax 3219 761 lbfin π15065 in32 31626 192 lbfin π15065 in3212 σmax 4496 kpsi Comment Step 16 of 18 Calculate factor of safety ny ny Sy σmax Here yield strength is Sy Substitute 57 kpsi for Sy and 4496 kpsi for σmax ny 57 kpsi 4496 kpsi ny 127 Hence it is not within range and tends to redesign the shaft Calculate Von Mises stress σmax σmax 32KfMa πd32 3 16KfTm πd3212 Substitute 19 for kf 761 lbfin for Ma 26 for kf and 192 lbfin for Tm σmax 3219 761 lbfin πd3 2 31626 192 lbfin πd3212 σmax 1 d153219 761 lbfin πd32 31626 192 lbfin πd3212 σmax 15372 d15 kpsi Comment Step 18 of 18 Calculate factor of safety ny ny Sy σmax Here yield strength is Sy Substitute 2 for ny 57 kpsi for Sy and 15372 d15 kpsi for σmax 2 57 kpsi 15372 d15 kpsi d 06626 in Hence the nearest 18th in diameter is 075 in