Indhold Bilag 1 2 Bilag 2 4 Bilag 3 9 Bilag 4 22 Bilag 5 27 Bilag 6 38 Bilag 7 42 Bilag 8 46 Bilag 9 49 Bilag 10 52 Bilag 11 63 Bilag 12 66 Bilag 13 73 Bilag 14 75 Bilag 15 88 Bilag 16 101 Bilag 17 117 Bilag 18 121 ii
INDHOLD 1 Bilag 19 123
Bilag 1 Projektbeskrivelse 2
9. februar 2010 PK Afgangsprojektbeskrivelse for Michael Oftedal, s053834, Helgi Gudbjartsson, s060267. Udvikling af ventilsystem til 2-takts forbrændingsmotor. Development of a valve system for a 2-stroke combustion engine. Til DTU s øko-bil projekt ønskes udviklet en 2-taktsmotor. Følgende teoretiske problemstillinger ønskes behandlet: 1. Litteraturstudium om opbygningsprincipper for ventilmekanismer. 2. Teoretiske modeller. En generel matematisk model til analyse ventilmekanismens bevægelse opbygges. En generel model der beskriver kraftforløbet i ventilmekanismen opstilles. 3. Et forslag til motorens opbygning udarbejdes. 4. Ventilmekanismen detailkonstrueres og tilpasses den udformning af motoren som er givet under pkt. 3. Konstruktionsarbejdet udføres i Pro/Engineer, og fremstillingsgrundlaget for alle relevante delkomponenter udarbejdes i form af bearbejdningstegninger. 5. Ventilmekanismens kritiske komponenter dimensioneres mht. relevante kriterier om f.eks. levetid, maximalt tilladelig spænding, deformation etc. 6. Centrale elementers styrkeforhold analyseres vha. FEM. Projektet afleveres den 12. februar 2010. Med venlig hilsen Peder Klit MEK DTU Kim Rene Hansen MEK DTU Institut for Mekanik, Energi og Konstruktion Administrationen Studentertorvet DTU - Bygning 101E 2800 Kgs. Lyngby Tlf 45 25 19 60 Tlf dir. 45 25 62 67 Fax 45 88 43 25 VAT reg. Nr.: 63 39 30 10 Klit@mek.dtu.dk www.mek.dtu.dk
Bilag 2 Løftekurve for 3-4-5 Polynomium knast 4
restart : with plots : Forskydning: sdθ/l$ 10$ Hastighed: D s : vd % Acceleration: D v : a d % Jerk: D a : jd % θ β 3-4-5 Polynomium knastkurve 3 K15$ 4 5 θ θ C6$ β β 10 θ 3 θ/l K 15 θ θ/l θ/l θ/l β 3 30 θ 2 β 3 60 θ β 3 β 4 4 K 60 θ 3 β 4 K 180 θ 2 β 4 60 β 3 K 360 θ β 4 C 6 θ 5 β 5 C 30 θ 4 β 5 C 120 θ 3 β 5 C 360 θ 2 β 5 Knastens løftehøjde m : L d 0.002658114558 0.002658114558 Motorens max omdrejningshastighed RPM : n d 8000 8000 Knastens arbejdsvinkel (fra løftekurvens start til slut) grader φd 120 120 Det ønskede dwell periode er: Dwd20 20 Motorens omdrejningshastighed ωd 2 $π$n 60 : evalf % rad s : 837.7580412 Omdrejningshastighed for knastens samlede løftekurve ω$ 360 φ : evalf % rad s : (1) (2) (3) (4) (5) (6) (7) (8) (9)
2513.274123 Vinkelen for halvdelen af løftekurven rad : φ 2 K Dw $π 2 βd 180 5 18 π plot s, 0..β (10) (11) (12) 0.0025 0.0020 0.0015 0.0010 0.0005 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 plot v, 0..β 0.005 0.004 0.003 0.002 0.001 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 plot a, 0..β
0.02 0.01 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 K0.01 K0.02 plot j, 0..β 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 K0.1 Maximum accelerationen findes: solve j θ = 0,θ xd % 2 zd %% 1 Max accelerationen udtrykt i 0.6882488914, 0.1844157346 0.1844157346 0.6882488914 (13) (14) (15)
a x : evalf %..og derfor er max accelerationen i m rad 2 : 0.02015200780 m s 2 : % $ω 2 : evalf % 14143.45564 Forholdet mellem accelerationen på flank og næse: a z : evalf % a x 0.9999999950 Knastens geometriske udforming findes ved at gange beta værdi med 0.05 til 1.0 og indsætte i s: 5 s 18 π $0.05 : evalf % $1000 0.003078428922 (16) (17) (18) (19)
Bilag 3 Kraftligevægt ˆ Frit legeme diagram: - Kraftligevægt - Nedadgående stødstang & opadgående ventil ˆ Frit legeme diagram: - Kraftligevægt - Opadgående stødstang & nedadgående ventil 9
restart : Kraftligevægt Nedadgående stødstang - Opadgående ventil Vippearmsforhold bestemmes for en vippearm med 68.9 mm mellem kontaktpunkterne: Le 1 d 0.027 Le 2 d 0.0689KLe 1 0.027 0.0419 (2) Masseinertimoment omkring vippearmen fås fra Pro/E og er afhængigt af vippearmsforholdet: if Le 1 = 0.03445 then inertid 1.9376318e-05 elif Le 1 = 0.033 then inertid 1.9198068e-05 elif Le 1 = 0.032 then inertid 1.9154988e-05 elif Le 1 = 0.031 then inertid 1.9177092e-05 elif Le 1 = 0.030 then inertid 1.9264376e-05 elif Le 1 = 0.029 then inertid 1.9428855e-05 elif Le 1 = 0.028 then inertid 1.9652808e-05 elif Le 1 = 0.027 then inertid 1.9935866e-05 elif Le 1 = 0.026 then inerti d 2.0284123e-05 else 0 end if 0.000019935866 Knastens maksimum løftehøjde er: Le 1 Hd 0.004125$ Le 2 3-4-5 Polynomium: sdθ/h$ 10$ D s : vd % D v : a d % D a : jd % θ β 0.002658114558 3 4 5 θ θ K15$ C6$ β β 10 θ 3 θ/h K 15 θ θ/h θ/h θ/h β 3 30 θ 2 β 3 60 θ β 3 β 4 4 K 60 θ 3 β 4 K 180 θ 2 β 4 60 β 3 K 360 θ β 4 C 6 θ 5 β 5 C 30 θ 4 β 5 C 120 θ 3 β 5 C 360 θ 2 β 5 (1) (3) (4) (5) (6) (7) (8) Motorens max omdrejningshastighed RPM : n d 8000 8000 Knastens arbejdsvinkel (fra løftekurvens start til slut) grader (9)
φd 120 Det ønskede dwell periode er: Dwd0 120 0 (10) (11) Motorens omdrejningshastighed ωd 2 $π$n 60 : evalf % rad s : 837.7580412 Omdrejningshastighed for knastens samlede løftekurve ω$ 360 φ : evalf % Vinkelen for halvdelen af løftekurven rad : φ 2 K Dw $π 2 βd 180 2513.274123 1 3 π rad s : (12) (13) (14) (15) solve j θ = 0,θ yd % 2 Max accelerationen a y : evalf % m rad 2 :..og derfor er max accelerationen a knast d % $ω 2 : evalf % 0.8258986698, 0.2212988816 0.2212988816 0.01399444988 m s 2 : 9821.844208 (16) (17) (18) (19) (20) Den krævede løfte for ventilen: x ventil d 0.004125 0.004125 Systemets ækvivalent masse (set fra ventil-siden): (21)
m eq d m ventilside C J 1 L 2 $L 2 Cm knastside $ m stang d 0.004589 m ventil d 0.0053053 m vippe d 0.039523 m kf d 0.028409 m fjeder d 0.012852 m fjederskive d 0.00301391 m laasering d 0.000056007 J 1 d 1.9935866e-05 J 2 d 3.4883207e-06 Rd0.02 L 1 d Le 1 L 2 d Le 2 F v0 d 0.05 L 1 L 2 2 m ventilside C J 1 2 L C m knastside L 2 1 2 2 L 2 0.004589 0.0053053 0.039523 0.028409 0.012852 0.00301391 0.000056007 0.000019935866 0.0000034883207 0.02 0.027 0.0419 0.05 (22) (23) (24) (25) (26) (27) (28) (29) (30) (31) (32) (33) (34) (35) m ventilside d m ventil C m fjeder 3 m knastside d J 2 Cm fjederskive C2$m laasering 0.01271522400 (36) R 2 Cm stang 0.01330980175 (37) α 1 d a knast L 1
36857.81039 π 2 α 2 d a knast R 49758.04402 π 2 Den ækvivalent masse bliver således: m eq Ventilens acceleration findes: a ventil dα 1 $L 2 Kraften som fjederen skal kunne yde: a ventil $m eq simplify units 0.02959751077 1544.342255 π 2 45.70868652 π 2 (38) (39) (40) (41) (42) Minimum fjederkonstant findes: k fjeder d % x ventil : evalf % Momentligevægt om vippearm f+: vadf v1 $L 2 KJ 1 $α 1 Kg$L 1 = 0 1.093640373 10 5 (43) solve va, g : g d % 0.0419 F v1 K0.7347923690 π 2 K0.027 g = 0 1.551851852 F v1 K268.5966666 Den maksimum kraft der virker på vippearmens L1 arm under ventillukning er: simplify units Momentligevægt om knastfølger f+: kfdf kn2 $RKF kn1 $RKJ 2 $α 2 = 0 solve kf, F kn2 : F kn2 d % Kraftligevægt for stødstang +: std F kn2 KF kn3 Cm stang $a knast = 0 solve st, F kn3 : F kn3 d % 1.551851852 F v1 K268.5966666 0.02 F kn2 K0.02 F kn1 K0.1735720149 π 2 = 0 F kn1 C85.65435611 F kn1 C85.65435611KF kn3 C4.566793280 π 2 = 0 130.7267992CF kn1 (44) (45) (46) (47) (48) (49) (50)
Kraftligevægt for ventil +: vedkm ventil $a ventil KF v1 Ck fjeder $x ventil CF v0 = 0 solve ve, F v1 : F v1 d % Momentligevægt om vippearm f+: vadf v1 $L 2 CJ 1 $α 1 KF kn3 $L 1 = 0 Alle kræfterne i systemet kendes nu: solve va, F kn1 : F kn1 d % F kn2 F kn3 F v1 37.51548754 π 2 KF v1 C0.05 = 0 370.3130209 11.98649200C0.7347923690 π 2 K0.027 F kn1 = 0 712.5408148 798.1951709 843.2676140 370.3130209 Den oprindelige momentligevægt om vippearmen burde nu give 0 (eller tæt på): F v1 $L 2 CJ 1 $α 1 KF kn3 $L 1 : evalf % 1. 10-9 (51) (52) (53) (54) (55) (56) (57) (58)
restart : Kraftligevægt Opadgående stødstang - Nedadgående ventil Vippearmsforhold bestemmes for en vippearm med 68.9 mm mellem kontaktpunkterne: Le 1 d 0.027 Le 2 d 0.0689KLe 1 0.027 0.0419 Fjederkonstanten fås fra den første kraftligevægt: k fjeder d 1.093640373 10 5 1.093640373 10 5 (3) Masseinertimoment omkring vippearmen fås fra Pro/E og er afhængigt af vippearmsforholdet: if Le 1 = 0.03445 then inertid 1.9376318e-05 elif Le 1 = 0.033 then inertid 1.9198068e-05 elif Le 1 = 0.032 then inertid 1.9154988e-05 elif Le 1 = 0.031 then inertid 1.9177092e-05 elif Le 1 = 0.030 then inertid 1.9264376e-05 elif Le 1 = 0.029 then inertid 1.9428855e-05 elif Le 1 = 0.028 then inertid 1.9652808e-05 elif Le 1 = 0.027 then inertid 1.9935866e-05 elif Le 1 = 0.026 then inerti d 2.0284123e-05 else 0 end if 0.000019935866 Knastens maksimum løftehøjde er: Le 1 Hd 0.004125$ Le 2 sdθ/h$ 10$ D s : vd % D v : a d % D a : jd % θ β 3 K15$ 0.002658114558 4 5 θ θ C6$ β β 10 θ 3 θ/h K 15 θ 4 θ/h θ/h θ/h β 3 30 θ 2 β 3 60 θ β 3 β 4 K 60 θ 3 β 4 K 180 θ 2 β 4 60 β 3 K 360 θ β 4 Motorens max omdrejningshastighed RPM : C 6 θ 5 β 5 C 30 θ 4 β 5 C 120 θ 3 β 5 C 360 θ 2 β 5 (1) (2) (4) (5) (6) (7) (8) (9)
n d 8000 8000 Knastens arbejdsvinkel (fra løftekurvens start til slut) grader φd 120 120 Det ønskede dwell periode er: Dwd0 0 (10) (11) (12) Motorens omdrejningshastighed ωd 2 $π$n 60 : evalf % rad s : 837.7580412 Omdrejningshastighed for knastens samlede løftekurve ω$ 360 φ : evalf % Vinkelen for halvdelen af løftekurven rad : φ 2 K Dw $π 2 βd 180 2513.274123 1 3 π rad s : (13) (14) (15) (16) solve j θ = 0,θ yd % 2 Max accelerationen a y : evalf % m rad 2 :..og derfor er max accelerationen a knast d % $ω 2 : evalf % 0.8258986698, 0.2212988816 0.2212988816 0.01399444988 m s 2 : 9821.844208 (17) (18) (19) (20) (21)
m eq d m knastside C J 1 L 2 2 Cm ventilside $ L 2 L 1 m stang d 0.004589 m ventil d 0.0053053 m vippe d 0.039523 m kf d 0.028409 m fjeder d 0.012852 m fjederskive d 0.00301391 m laasering d 0.000056007 2 m knastside C J 1 2 L C m ventilside L 2 2 2 2 L 1 0.004589 0.0053053 0.039523 0.028409 0.012852 0.00301391 0.000056007 (22) (23) (24) (25) (26) (27) (28) (29) J 1 d 1.9935866e-05 J 2 d 3.4883207e-06 Rd0.02 L 1 d Le 1 L 2 d Le 2 F v0 d 0.05 m ventilside d m ventil C m fjeder 3 m knastside d J 2 0.000019935866 0.0000034883207 0.02 0.027 0.0419 0.05 Cm fjederskive C2$m laasering 0.01271522400 (30) (31) (32) (33) (34) (35) (36) R 2 Cm stang 0.01330980175 (37)
α 1 d a knast L 1 36857.81039 π 2 (38) α 2 d a knast R 49758.04402 π 2 a ventil dα 1 $L 2 x ventil d 0.004125 Den ækvivalent masse bliver således: m eq 1544.342255 π 2 0.004125 0.05528668669 (39) (40) (41) (42) Momentligevægt om knastfølger f+: kfdf kn2 $RKF kn1 $RCJ 2 $α 2 = 0 solve kf, F kn2 : F kn2 d % Kraftligevægt for stødstang +: std F kn2 KF kn3 Km stang $a knast = 0 solve st, F kn3 : F kn3 d % 0.02 F kn2 K0.02 F kn1 C0.1735720149 π 2 = 0 F kn1 K85.65435611 F kn1 K85.65435611KF kn3 K4.566793280 π 2 = 0 K130.7267992CF kn1 Kraftligevægt for ventil +: vedm ventil $a ventil KF v1 Ck fjeder $x ventil CF v0 = 0 solve ve, F v1 : F v1 d % Momentligevægt om vippearm f+: vadf v1 $L 2 CJ 1 $α 1 KF kn3 $L 1 = 0 Alle kræfterne i systemet kendes nu. solve va, F kn1 : F kn1 d % F kn2 8.193198965 π 2 KF v1 C451.1766539 = 0 532.0402865 25.82211158C0.7347923690 π 2 K0.027 F kn1 = 0 1224.971170 1139.316814 (43) (44) (45) (46) (47) (48) (49) (50) (51)
F kn3 F v1 1094.244371 532.0402865 Den oprindelige momentligevægt om vippearmen burde nu give 0 (eller tæt på): F v1 $L 2 CJ 1 $α 1 KF kn3 $L 1 : evalf % K1.9 10-8 (52) (53) (54)
Bilag 4 Analyse af vippearms- og dwellforhold ˆ Mulige design variationer ved 8.000 RPM - Diagram for kræfter ved 0 dwell ˆ Mulige design variationer ved 6.000 RPM 22
Kraftligevægt: 8.000 RPM Stødstang nedadgående Ventil opadgående Stødstang opadgående Ventil nedadgående [grader] [m/s 2 ] [mm] [mm] [N/mm] [N] [N] [N] [N] [N] [N] [N] [N] Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 0 15242 34,45 34,45 1 158 625 758 828 572 1193 1060 990 734 0 14011 33 35,9 1,087879 146 636 758 822 520 1185 1063 998 682 0 13218 32 36,9 1,153125 138 645 760 821 489 1183 1068 1007 650 0 12467 31 37,9 1,222581 131 655 764 821 460 1185 1076 1019 622 0 11755 30 38,9 1,296667 125 667 770 824 434 1190 1087 1033 596 0 11078 29 39,9 1,375862 119 681 777 828 411 1198 1101 1051 573 0 10435 28 40,9 1,460714 114 696 787 835 390 1210 1119 1071 551 0 9822 27 41,9 1,551852 109 713 798 843 370 1225 1139 1094 532 0 9238 26 42,9 1,65 105 731 812 854 353 1244 1163 1121 514 Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 10 18139 34,45 34,45 1 188 744 902 985 681 1420 1261 1178 873 10 16674 33 35,9 1,087879 173 757 902 979 619 1410 1265 1188 812 10 15731 32 36,9 1,153125 164 768 905 977 582 1408 1271 1199 774 10 14837 31 37,9 1,222581 156 780 909 977 548 1410 1281 1213 740 10 13989 30 38,9 1,296667 149 794 916 980 517 1416 1294 1230 710 10 13184 29 39,9 1,375862 142 810 925 986 489 1426 1311 1250 682 10 12418 28 40,9 1,460714 136 828 936 993 464 1440 1331 1274 656 10 11689 27 41,9 1,551852 130 848 950 1004 441 1458 1356 1302 633 10 10994 26 42,9 1,65 125 870 966 1017 420 1480 1385 1334 612 Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 20 21949 34,45 34,45 1 228 900 1091 1192 824 1717 1526 1425 1056 20 20176 33 35,9 1,087879 210 916 1092 1184 749 1706 1530 1438 982 20 19033 32 36,9 1,153125 199 929 1095 1182 704 1704 1538 1451 937 20 17953 31 37,9 1,222581 189 944 1100 1183 663 1706 1550 1467 896 20 16927 30 38,9 1,296667 180 961 1108 1186 626 1713 1566 1488 858 20 15953 29 39,9 1,375862 172 980 1119 1192 592 1725 1586 1513 825 20 15026 28 40,9 1,460714 164 1002 1133 1202 561 1742 1611 1542 794 20 14143 27 41,9 1,551852 157 1026 1149 1214 533 1764 1641 1576 766 20 13302 26 42,9 1,65 151 1053 1169 1230 508 1791 1675 1614 741 Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 30 27097 34,45 34,45 1 281 111 1347 1472 1017 2120 1884 1759 1304 30 24908 33 35,9 1,087879 259 1130 1348 1462 925 2106 1889 1775 1212 30 23499 32 36,9 1,153125 245 1147 1351 1459 869 2104 1899 1791 1156 30 22164 31 37,9 1,222581 233 1165 1358 1460 818 2107 1913 1812 1106 30 20897 30 38,9 1,296667 222 1186 1368 1464 772 2115 1933 1837 1060 30 19695 29 39,9 1,375862 212 1210 1382 1472 731 2130 1958 1868 1018 30 18550 28 40,9 1,460714 203 1237 1399 1484 693 2151 1989 1904 980 30 17461 27 41,9 1,551852 194 1267 1419 1499 658 2178 2025 1945 946 30 16422 26 42,9 1,65 187 1300 1443 1519 627 2212 2068 1993 914 Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 40 34295 34,45 34,45 1 356 1406 1705 1863 1287 2683 2384 2227 1651 40 31524 33 35,9 1,087879 328 1431 1706 1850 1170 2666 2391 2246 1534 40 29741 32 36,9 1,153125 311 1451 1710 1847 1100 2662 2403 2267 1463 40 28051 31 37,9 1,222581 295 1475 1719 1848 1036 2666 2422 2293 1399 40 26448 30 38,9 1,296667 281 1501 1732 1853 977 2677 2447 2325 1341 40 24926 29 39,9 1,375862 268 1531 1749 1863 925 2696 2478 2364 1289 40 23478 28 40,9 1,460714 257 1565 1770 1878 877 2722 2517 2409 1241 40 22099 27 41,9 1,551852 246 1603 1796 1897 833 2756 2563 2462 1197 40 20785 26 42,9 1,65 236 1645 1827 1922 793 2799 2618 2522 1157 Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 50 44793 34,45 34,45 1 465 1837 2227 2433 1681 3504 3114 2908 2156 50 41175 33 35,9 1,087879 428 1869 2228 2417 1529 3482 3123 2934 2004 50 38845 32 36,9 1,153125 406 1895 2234 2412 1436 3477 3139 2960 1911 50 36638 31 37,9 1,222581 385 1926 2245 2414 1353 3482 3163 2995 1828 50 34545 30 38,9 1,296667 367 1961 2262 2421 1277 3497 3195 3037 1752 50 32556 29 39,9 1,375862 350 2000 2284 2434 1208 3521 3237 3087 1683 50 30665 28 40,9 1,460714 335 2044 2312 2453 1145 3555 3288 3147 1620 50 28864 27 41,9 1,551852 321 2094 2346 2478 1088 3600 3348 3216 1563 50 27147 26 42,9 1,65 309 2149 2386 2510 1036 3656 3419 3294 1511
1300 Nedadgående ventil (0 dwell) 1200 1100 1000 [N] 900 800 FKN1 FKN2 FKN3 FV1 700 600 500 400 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 L 1 /L 2
Opadgående ventil (0 dwell) 900 800 700 600 [N] 500 FKN1 FKN2 FKN3 FV1 400 300 200 1 1,1 1,2 1,3 1,4 1,5 1,6 1,7 L 1 /L 2
Kraftligevægt: 6.000 RPM Stødstang nedadgående Ventil opadgående Stødstang opadgående Ventil nedadgående [grader] [m/s 2 ] [mm] [mm] [N/mm] [N] [N] [N] [N] [N] [N] [N] [N] Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 0 8574 34,45 34,45 1 89 352 426 466 322 671 596 557 413 0 7881 33 35,9 1,087879 82 358 426 463 293 666 598 562 384 0 7435 32 36,9 1,153125 78 363 428 462 275 666 601 567 366 0 7013 31 37,9 1,222581 74 369 430 462 259 667 605 573 350 0 6612 30 38,9 1,296667 70 375 433 463 244 669 612 581 335 0 6231 29 39,9 1,375862 67 383 437 466 231 674 620 591 322 0 5869 28 40,9 1,460714 64 391 443 469 219 681 629 602 310 0 5525 27 41,9 1,551852 62 401 449 474 208 689 641 616 299 0 5196 26 42,9 1,65 59 411 457 481 198 700 654 631 289 Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 10 10203 34,45 34,45 1 106 418 507 554 383 798 709 663 491 10 9379 33 35,9 1,087879 98 426 507 551 348 793 711 668 456 10 8848 32 36,9 1,153125 92 432 509 550 327 792 715 674 435 10 8346 31 37,9 1,222581 88 439 512 550 308 793 720 682 416 10 7869 30 38,9 1,296667 84 447 515 551 291 797 728 692 399 10 7416 29 39,9 1,375862 80 456 520 554 275 802 737 703 383 10 6985 28 40,9 1,460714 76 466 527 559 261 810 749 717 369 10 6575 27 41,9 1,551852 73 477 534 565 248 820 763 733 356 10 6184 26 42,9 1,65 70 490 543 572 236 833 779 751 344 Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 20 12346 34,45 34,45 1 128 506 614 671 463 966 858 802 594 20 11349 33 35,9 1,087879 118 515 614 666 421 960 861 809 552 20 10707 32 36,9 1,153125 112 522 616 665 396 958 865 816 527 20 10098 31 37,9 1,222581 106 531 619 665 373 960 872 825 504 20 9521 30 38,9 1,296667 101 541 624 667 352 964 881 837 483 20 8973 29 39,9 1,375862 97 551 630 671 333 970 892 851 464 20 8452 28 40,9 1,460714 92 564 637 676 316 980 906 867 447 20 7956 27 41,9 1,551852 89 577 647 683 300 992 923 886 431 20 7482 26 42,9 1,65 85 592 658 692 286 1008 942 908 417 Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 30 15242 34,45 34,45 1 158 625 758 828 572 1193 1060 990 734 30 14011 33 35,9 1,087879 146 636 758 822 520 1185 1063 998 682 30 13218 32 36,9 1,153125 138 645 760 821 489 1183 1068 1007 650 30 12467 31 37,9 1,222581 131 655 764 821 460 1185 1076 1019 622 30 11755 30 38,9 1,296667 125 667 770 824 434 1190 1087 1033 596 30 11078 29 39,9 1,375862 119 681 777 828 411 1198 1101 1051 573 30 10435 28 40,9 1,460714 114 696 787 835 390 1210 1119 1071 551 30 9822 27 41,9 1,551852 109 713 798 843 370 1225 1139 1094 532 30 9238 26 42,9 1,65 105 731 812 854 353 1244 1163 1121 514 Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 40 19291 34,45 34,45 1 200 791 959 1048 724 1509 1341 1253 928 40 17732 33 35,9 1,087879 184 805 959 1041 658 1500 1345 1263 863 40 16729 32 36,9 1,153125 175 816 962 1039 619 1498 1352 1275 823 40 15779 31 37,9 1,222581 166 829 967 1039 583 1500 1362 1290 787 40 14877 30 38,9 1,296667 158 845 974 1043 550 1506 1376 1308 755 40 14021 29 39,9 1,375862 151 861 984 1048 520 1516 1394 1330 725 40 13206 28 40,9 1,460714 144 881 996 1056 493 1531 1416 1355 698 40 12431 27 41,9 1,551852 138 902 1010 1067 469 1550 1442 1385 673 40 11691 26 42,9 1,65 133 926 1027 1081 446 1574 1473 1419 651 Dwell Aknast L1 L2 L2/L1 Kfjeder FKN1 FKN2 FKN3 FV1 FKN1 FKN2 FKN3 FV1 50 25196 34,45 34,45 1 262 1033 1253 1369 945 1971 1752 1636 1213 50 23161 33 35,9 1,087879 241 1051 1253 1359 860 1959 1757 1650 1127 50 21850 32 36,9 1,153125 228 1066 1257 1357 808 1956 1766 1665 1075 50 20609 31 37,9 1,222581 217 1083 1263 1358 761 1959 1779 1685 1028 50 19431 30 38,9 1,296667 206 1103 1272 1362 718 1967 1797 1708 985 50 18313 29 39,9 1,375862 197 1125 1285 1369 679 1980 1821 1737 947 50 17249 28 40,9 1,460714 189 1150 1300 1380 644 2000 1849 1770 912 50 16236 27 41,9 1,551852 181 1178 1319 1394 612 2025 1883 1809 879 50 15270 26 42,9 1,65 174 1209 1342 1412 583 2056 1923 1853 850
Bilag 5 Spændingsmoment i topstykke ˆ Minimum tilspændingsmoment ˆ Maksimum tilspændingsmoment ˆ Topstykke - udbøjning ˆ Bolt/Møtrik - udbøjning ˆ Pindbolt - spændingsberegning ˆ Sikkerhedsfaktor for bolten ˆ Toppakning - brud 27
restart : Minimum tilspændingsmoment Kraften Fv er en gættet værdi for bolttilspænding.tilspændingsmomentet regnes ved at benytte formlerne i "Maskinelement bogen s. 116 (6.16)" Enheden er [Nm] µ værdien ligger typisk imellem 0.10-0.19, dette afhænger af om der er tørt eller olieret mellem kontaktfladerne for henholdsvis pindbolt og møtrik p max d 80$10 5 d boring d 0.025 d pitch d 0.00535 F total d 3.1415 4 F min d F total 4 F v d 1500 2 $d boring $p max 8000000 0.025 0.00535 3920. 980. 1500 (1) (2) (3) (4) (5) (6)
θd 29.95$3.1415 180 βd 3.4$3.1415 180 αd 60$3.1415 180 µd 0.19 r n d 0.004 0.523 0.0594 1.04 0.19 0.004 (7) (8) (9) (10) (11) T min d F v $ d pitch 2 $ cos θ $tan β Cµ cos θ Kµ$tan β Cr n $µ 2.28 (12)
Maximum tilspændingsmoment Da bolten stort set er defekt efter flydespændingen er opnået, bruges denne værdi for at beregne max kraften, således kan max tilspændingsmomentet beregnes. Enheden er [Nm] F max dσ y $A s σ y A s (13) T max d F max $ d pitch 2 $ cos θ $tan β Cµ cos θ Kµ$tan β Cr n $µ 0.00152 σ y A s (14)
Topstykke udbøjning Stivheden for topstykket regnes først, derefter kan udbøjningen beregnes ved en given kraft Fm. "Maskinelement bogen s. 120 (6.31)" Enheden er [m] F m d 1500 E m d 71$10 9 d h d 0.006 d w d 0.010 d A d 0.014 l m d 0.035 C 1 d 0.79670 C 2 d 0.63816 1500 71000000000 0.006 0.010 0.014 0.035 0.79670 0.63816 (15) (16) (17) (18) (19) (20) (21) (22) C 2 $d h k m d E m $d h $C 1 $e l m 3.81 10 8 (23) δ m d F m k m 0.00000394 (24)
Bolt/møtrik udbøjning Tilsvarende regnes boltstivheden, sådan at udbøjningen kan beregnes. Kraften der påføres er pr. bolt. "Maskinelement bogen s. 120 (6.30)" Enheden er [m] F b d 1500 E b d 206$10 9 d norm d 0.006 d core d 0.004773 l s d 0.095 l k d 0.035 l t d 0.010 A n d 3.1415$d 2 norm 4 A i d 3.1415 4 2 $d norm A 3 d 3.1415$d 2 core 4 1500 206000000000 0.006 0.004773 0.095 0.035 0.010 0.0000283 0.0000283 0.0000179 (25) (26) (27) (28) (29) (30) (31) (32) (33) (34) (35) f T1 d 0.5$d norm E b $A 3 L TN1 d 0.4$d norm f N1 d L TN1 E b $A n 8.15 10-10 0.0024 (36) (37)
f TN1 d f T1 Cf N1 f T2 d 0.5$d norm E b $A 3 L TN2 d 0.5$d norm f N2 d L TN2 E b $A n f TN2 d f T2 Cf N2 f t d l t E b $A 3 f i d l s E b $A i f bolt d f TN1 Cf i Cf t Cf TN2 k b d 1 f bolt δ b d F b k b 4.13 10-10 1.23 10-9 8.15 10-10 0.0030 5.16 10-10 1.33 10-9 2.71 10-9 1.63 10-8 2.15 10-8 4.65 10 7 0.0000323 (38) (39) (40) (41) (42) (43) (44) (45) (46) (47) (48)
Pindbolt - spændingsberegning For at sikre at pindbolten holder under drift, skal spændingerne i pindbolten kendes. Dette beregnes og derefter sammenlignes dette med materialedata for pindbolten. As værdien er taget ud fra en tabel i "Maskinelement bogen afsnittet Threaded Fasterners s. 109-127 " σe formel er taget fra "Fundamentals of Machine Elements, af Bernard Hamrock, Bo Jacobson Steven Schmid 1999 s.265 (7.7)" Stagboltens materialedata er følgende : Stål, E-modul 206.8 GPa, m6 8.8 De beregnede spændinger har enheden [MPa] A s d 20.1 Φd k b k b Ck m F AU d 1600 F AL d 600 F A dφ$ F AU KF AL 2 20.1 0.109 1600 600 54.5 (49) (50) (51) (52) (53) σ a d F A A s σ ut d 800 σ y d 0.8$σ ut F m d F v CF A 2.71 800 640. 1550. (54) (55) (56) (57) σ m d F m A s 77.1 (58)
σ i d F v A s 74.6 (59) σ e d 0.45$σ ut 360. (60)
Sikkerhedsfaktoren for Pindbolten Goodman sikkerhedsfaktoren beregnes ved brug af formelen fra "Maskinelement bogen s. 88(4.40)", der antages konstant forespænding, værdien er givet ved Fv. Kærvfølsomhedsfaktoren er taget fra en tabel i bogen "Machine Design" s.910 se vedlagt bilag. K f d 3 3 (61) n fg d σ e $ σ ut Kσ i σ e $ σ m Kσ i Cσ ut $σ a $K f 35.2 (62)
Toppakningen - brud For at sikre toppakningen ikke bliver skudt ud under drift, beregnes kraften der virker på toppakningen, og sammenlignes med friktionskraften samt max trykket for motoren. Der kigges på en kvart cirkel, og trykket antages til at trykke på pakningens areal med p_max. h er pakningen højde. rd0.0125 o d 3.1415 4 h d 0.001 A p d o$h $r 0.0125 0.00981 0.001 0.00000981 (63) (64) (65) (66) µ s d 0.15 f klem d 600 f frik d f klem $µ s f kammer d p max $A p 0.15 600 90.0 78.5 (67) (68) (69) (70)
Bilag 6 Spændingsmoment i krumtap ˆ Tilspændingsmoment - krumtap ˆ Tandhjul - moment 38
restart p pitch d 0.002 d pitch d 0.014701 d washer d 0.024 d hole d 0.016 r n d d washer Cd hole 4 µd 0.19 α rad d 60$3.1415 180 Tilspændingsmoment Krumtap 0.0020 0.0147 0.0240 0.0160 0.0100 0.1900 1.0472 (1) (2) (3) (4) (5) (6) (7) βdarctan p pitch 3.1415$d pitch 0.0433 (8) θd arctan cos β $tan r ydre d 0.032 α rad 2 0.5232 (9) r indre d 0.023 0.0320 (10) 0.0230 (11) r knast d r ydre Cr indre 2 F KN1 d 1225 0.0275 (12) T knast d F KN1 $r knast 1225 (13) F v d F a 33.6875 (14) F a (15) µ knast d 0.1
ηd2 F a d η$t knast µ knast $r knast 0.1000 2 24500.0000 (16) (17) (18) TdF v $ d pitch 2 $ cos θ $tan β Cµ cos θ Kµ$tan β Cr n $µ 94.3027 (19)
P tryk d 65$10 5 θ tryk d 15$3.1415 180 r slag d 0.025 d stempel d 0.025 A stempel d 3.1415 4 2 $d stempel r krumtap d sin θ tryk $r slag T tanhjul d P tryk $A stempel $r krumtap Tandhjul moment 6500000 0.2618 0.0250 0.0250 0.0005 0.0065 20.6440 (20) (21) (22) (23) (24) (25) (26)
Bilag 7 Reaktionskræfter i lejer 42
restart : Reaktionskræfter i lejer Den påførte kraft på leje A F A d Vector 0,K4000 N, 0 0 K4000 N Reaktionskræfterne på leje A findes umiddelbart: R A d Vector 0, 4000 N, 0 0 (1) Reaktionskræfterne på leje B R B d Vector R BX, R BY, R BZ 0 4000 N 0 (2) R BX R BY (3) Reaktionskræfterne på leje C R C d Vector R CX, R CY, 0 R BZ
R CX R CY (4) Afstanden mellem leje B og leje A, udtykt i vektor koordinater: S BA d Vector 0, 25$10 K3 m,k34.5$10 K3 m 0 1 40 0 m K0.03450000000 m Afstanden mellem leje B og leje C, udtykt i vektor koordinater: S BC d Vector 0, 0, 39$10 K3 m 0 0 39 1000 Afstanden mellem leje C og leje A, udtykt i vektor koordinater: S CA d Vector 0, 25$10 K3 m,k73.5$10 K3 m 1 40 0 m m K0.07350000000 m Afstanden mellem leje C og leje B, udtykt i vektor koordinater: S CB d Vector 0, 0,K39$10 K3 m (5) (6) (7) Der tages momentligevægt om leje B: M B d S BA # F A CS BC #R C 0 0 K 39 1000 m (8) K138.0000000 m N K 39 1000 39 1000 Momentligevægten løses for R[CY] og R[CX]: 0 m R CX m R CY (9)
eq 1 d M B 1 : solve eq 1 = 0, R CY : assign R CY, evalf % : R CY K3538.461538 N eq 2 d M B 2 : solve eq 2 = 0, R CX : assign R CX, evalf % : R CX Der tages momentligevægt om leje C: M C d S CA #F A CS CB #R B 0. K294.0000000 m N C 39 1000 K 39 1000 m R BX Momentligevægten løses for R[BY] og R[BX]: eq 3 d M C 1 : solve eq 3 = 0, R BY : assign R BY, evalf % : R BY 7538.461538 N eq 4 d M C 2 : solve eq 4 = 0, R BX : assign R BX, evalf % : R BX R BZ d M C 3 Nu kendes reaktionskræfterne i lejer B og C: R B 0 0. 0 m R BY (10) (11) (12) (13) (14) (15) R C 0. 7538.461538 N 0 0. K3538.461538 N 0 (16) (17)
Bilag 8 Feder-not beregning 46
Feder-not beregning Uendelig levetid forekommer hvis den beregnede spænding er under 48% af brudspændingen, gælder kun stål. restart σ brud d 600$10 6 σ y d 340$10 6 Md 20.64 h d 0.007 b d 0.008 ld 0.009 rd0.0115 A shear d b$l 600000000 340000000 20.64 0.008 0.009 0.0115 0.000072 (1) (2) (3) (4) (5) (6) (7) A bearing d h 2 $l 0.00003150000000 (8) Fd M r 1794.782609 τd F A shear η shear d σ y 3 $τ evalf % σ b d F A bearing 2.492753624 10 7 4.546511625 3 7.874789134 5.697722568 10 7 (9) (10) (11) (12) (13)
η b d σ y σ b 5.967296511 (14)
Bilag 9 Bjælkeberegning & Eulers søjletilfælde ˆ Udbøjning i bjælke ˆ Euler - stødstang 49
Udbøjning bjælke restart b d 0.0075 h d 0.007 Ed2.1$10 11 a d 1 ld 0.037 Fd1100 0.0075 0.0070 2.1000 10 11 1 0.0370 1100 (1) (2) (3) (4) (5) (6) yd h 2 0.0035 Jd 1 12 $b$h3 TdF$l vd F$l3 3$E$J $a3 2.1438 10-10 40.7000 0.0004 (7) (8) (9) (10) σd T J $y 6.6449 10 8 (11)
restart Ed2.1$10 11 rd0.0022 Ad3.1415$r 2 Jd 3.1415 4 ld 0.160 $r 4 Euler stødstang 2.1000 10 11 0.0022 0.0000 1.8398 10-11 0.1600 (1) (2) (3) (4) (5) P k d 3.14152 $E$J l 2 1489.4340 (6) σ k d P k A 9.7958 10 7 (7)
Bilag 10 Spændinger i kontaktader ˆ Kontaktspændinger mellem stødstang og vippearm ˆ Kontaktspændinger mellem stødstang og knastfølger ˆ Kontaktspændinger mellem knast og knastfølger 52
restart : Kontaktspændinger mellem stødstang og vippearm I følgende regnestykke står a for kuglen i justeringsskruen på vippearmen (convex), mens b står for koppen på toppen af stødstangen (concave). r ax d 0.00125 m r ay d r ax r bx dk0.001275 m r by d r bx solve solve 0.00125 m 0.00125 m K0.001275 m K0.001275 m 1 = 1 C 1, R R x r ax r x : R x d % bx 0.06375000004 m 1 = 1 C 1, R R y r ay r y : R y d % by 0.06375000004 m (1) (2) (3) (4)
solve 1 R effective = 1 R x C 1 R y, R effective : R effective d % R d d R effective $ 1 R x K 1 R y 0.03187500002 m (5) 0. (6) α r d R y R x 1.000000000 (7) Hert s antagelser (1881): 1. Materialerne er homogen og deres kritiske spændinger er ikke overskredet. 2. Ingen tangent kræfter er til stede mellem de to solide. 3. Kontakten er begrænset til en lille del af overfladen således at kontaktregionens dimensioner er små i forhold til ellipsoidernes radier. 4. De to solide er i hvile og ligevægt. Den normal påførte kraft er: w z d 1094 N 1094 N Da begge to kontakt flader er sfæriske, bliver ellipticity parameteren lige med 1: kd1 1 Det elliptiske integral af første orden er givet ved: Fd π 2 1K 1K 1 k 2 $sin 2 x K 1 2 dx evalf % (8) (9) 0 1 2 π 1.570796327 (10) Det elliptiske integral af anden orden er givet ved: εd π 2 1K 1K 1 k 2 $sin 2 x 1 2 dx evalf % 0 E a d 2.1$10 11 Pa E b d E a 1 2 π 1.570796327 2.100000000 10 11 Pa (11)
νd 0.3 E effective d 1Kν 2 E a 2 C 1Kν2 E b 2.100000000 10 11 Pa 0.3 2.307692308 10 11 Pa Kontakt areal diameterne er beregnet: Diameter y d 2$ simplify units 6$k 2 $ε$w z $R effective π$e effective 1 3 evalf % 0.0008455056904 6 1/3 N m Pa 0.001536385802 N m Pa 1/3 1/3 (12) (13) (14) (15) Diameter x d 2$ 6$ε$w z $R effective π$k$e effective 0.001536385802 m 1 3 evalf % (16) simplify units 0.0008455056904 6 1/3 N m Pa 0.001536385802 N m Pa 1/3 1/3 (17) 0.001536385802 m Den maksimal elastiske udbøjning er givet ved: δ m d F$ simplify units 9 2$ε$R effective $ w z π$k$e effective 2 1 3 evalf % 0.000002803450480 π 9 1/3 2 2/3 N 2 0.000009256789531 π 3 Pa 2 m N 2 Pa 2 m 1/3 1/3 (18) (19) Kontaktarealet er: 0.000009256789531 m (20)
(18) 2 $ π 4 : evalf % 0.000001853917704 m 2 Maksimum trykket indenfor kontaktarealet findes til: 6$w z p m d π$diameter x $Diameter y simplify units 1.530327040 10 9 N 6 1/3 π N m Pa 2/3 (21) (22) evalf % 1.530327040 10 9 6 1/3 Pa π 8.851525593 10 8 Pa (23) (24)
restart : Kontaktspændinger mellem stødstang og knastfølger I følgende regnestykke står a for den nedste del af stødstangen (convex), mens b står for koppen som er udgravet i knastfølgeren (concave). r ax d 0.0015 m r ay d r ax r bx dk0.00153 m r by d r bx solve solve 0.0015 m 0.0015 m K0.00153 m K0.00153 m 1 = 1 C 1, R R x r ax r x : R x d % bx 0.07649999956 m 1 = 1 C 1, R R y r ay r y : R y d % by 0.07649999956 m (1) (2) (3) (4)
solve 1 R effective = 1 R x C 1 R y, R effective : R effective d % R d d R effective $ 1 R x K 1 R y 0. (5) (6) α r d R y R x 1.000000000 (7) Hert s antagelser (1881): 1. Materialerne er homogen og deres kritiske spændinger er ikke overskredet. 2. Ingen tangent kræfter er til stede mellem de to solide. 3. Kontakten er begrænset til en lille del af overfladen således at kontaktregionens dimensioner er små i forhold til ellipsoidernes radier. 4. De to solide er i hvile og ligevægt. Den normal påførte kraft er: w z d 1139 N 1139 N Da begge to kontakt flader er sfæriske, bliver ellipticity parameteren lige med 1: kd1 1 Det elliptiske integral af første orden er givet ved: Fd 0 π 2 1K 1K 1 k 2 $sin 2 x K 1 2 dx evalf % 1 2 π 1.570796327 Det elliptiske integral af anden orden er givet ved: εd π 2 1K 1K 1 k 2 $sin 2 x 1 2 dx evalf % (8) (9) (10) 0 E a d 2.1$10 11 Pa E b d E a 1 2 π 1.570796327 2.100000000 10 11 Pa 2.100000000 10 11 Pa (11) (12)
νd 0.3 E effective d 1Kν 2 E a 2 C 1Kν 0.3 2.307692308 10 11 Pa Kontakt areal diameterne er beregnet: Diameter y d 2$ simplify units E b 2 6$k 2 $ε$w z $R effective π$e effective 1 3 evalf % 0.0009106379600 6 1/3 N m Pa 0.001654738990 N m Pa 1/3 1/3 (13) (14) (15) Diameter x d 2$ 6$ε$w z $R effective π$k$e effective 0.001654738990 m 1 3 evalf % (16) simplify units 0.0009106379600 6 1/3 N m Pa 0.001654738990 N m Pa 1/3 1/3 (17) 0.001654738990 m Den maksimal elastiske udbøjning er givet ved: δ m d F$ simplify units 9 2$ε$R effective $ w z π$k$e effective 2 1 3 evalf % 0.000002710004900 π 9 1/3 2 2/3 N 2 0.000008948239024 π 3 Pa 2 m N 2 Pa 2 m 1/3 1/3 (18) (19) Kontaktarealet er: (18) 2 $ π 4 : evalf % 0.000008948239024 m (20)
0.000002150546719 m 2 Maksimum trykket indenfor kontaktarealet findes til: 6$w z p m d π$diameter x $Diameter y simplify units 1.373511260 10 9 N 6 1/3 π N m Pa 2/3 (21) (22) evalf % 1.373511260 10 9 6 1/3 Pa π 7.944491441 10 8 Pa (23) (24)
restart : Kontaktspændinger mellem knast og knastfølger Rullediameteren på knastfølgeren er: R ax d 0.014 m 0.014 m Diameteren på knasten hvor maksimum accelerationen sker: R bx d 0.03217805991 m 0.03217805991 m Referenceradius i x retningen findes: 1 solve = 1 C 1, R R x R ax R x : R x d % bx 0.009755560101 m Kraft pr.bredde, E-moduler og Poissons forhold indtastes: (1) (2) (3) w z d 1244 N 0.008 m E a d 210$109 Pa E b d E a νd 0.3
Reference E-modulen findes: 2 1Kν 2 : E effective d % C 1Kν2 E a E b 1.555000000 10 5 N m 210000000000 Pa 210000000000 Pa 0.3 2.307692308 10 11 Pa (4) (5) w z E effective $R x simplify units 0.00006907172179 N m 2 Pa (6) Den dimensionsløse belastning findes: W effective d % Halv-kontaktbredde er defineret ved b d R x $ 8$ % π 1 2 : evalf % 0.00006907172179 0.00006907172179 0.0001293815671 m Den maksimal deformation for en rektangulær sammentræf: δ m d 2$W effective $R x 2$π $ ln K1 : evalf % π W effective 0.000004469171942 m Den maksimal tryk for en rektangulær sammentræf: p m d E effective $ evalf % W effective 2$π 1 2 9.589546025 10 8 Pa 2 7.651350716 10 8 Pa 1 π (7) (8) (9) (10) (11) (12)
Bilag 11 Vibrationsberegninger ˆ Vibrationsanalyse - stødstang ˆ Vibrationsanalyse - ventilfjeder 63
Vibrationsanalyse Stødstang restart E AISI1080 d 1.72$10 11 ρd7.8$10 3 ld 0.160 1.720000000 10 11 7800.0 0.160 (1) (2) (3) cd E AISI1080 ρ ω n d 3.1415$c l 4695.879263 92200.65439 (4) (5) fd ω n 2$3.1415 14674.62269 rpmdf$60 8.804773614 10 5 (6) (7)
kd121094 md 0.012 Fjeder 121094 0.012 (8) (9) ω nf d k m 3176.659672 (10) f f d ω nf 2$3.1415 505.5960006 rpm f d f f $60 30335.76004 (11) (12)
Bilag 12 Levetid af lejer på krumtappen ˆ NKI 20/16 ˆ 6305 ˆ NK 14/20 66
restart : with CurveFitting : Estimeret levetid for NKI 20/16 Indtast omdrejningshastighed i RPM: n d 4000 4000 Indtast den ønskede levetid i timer: ReqLifetime hours d 100 100 L req d ReqLifetime hours * 60 * n 24000000 Indtast den radial, tangential og den aksial kraft der virker på lejen: F r d 4000 F t d 0 F a d 0 4000 0 sqrt F r ^2CF t ^2 : F r total d % 4000 Indtast navnet på lejen og dennes principielle værdier: Bearingd NKI2016 CC d 15400 C 0 d 24500 f 0 d 12 NKI2016 0 15400 24500 12 Indtast p værdi: (kugleleje = 3... rulleleje = 10/3) p d 10 3 10 3 Indtast clearance værdi: X d 0.56 0.56 Værdiet nedenfor burde være det første gæt på et værdi for C når leje skal vælges: L req / 10^6 ^ 1 / 3 * F r total : C first_try d evalf % 11537.99656 xpoints d 0.172, 0.345, 0.689, 1.03, 1.38, 2.07, 3.45, 5.17, 6.89 : epoints d 0.19, 0.22, 0.26, 0.28, 0.30, 0.34, 0.38, 0.42, 0.44 : Ypoints d 2.30, 1.99, 1.71, 1.55, 1.45, 1.31, 1.15, 1.04, 1.00 : (1) (2) (3) (4) (5) (6) (7) (8) (9) h d x / piecewise x!.345,.159c.181 * xk.257 * xk.172 ^3, x!.689,.165c.158 * xk.133 * x K.345 ^2 C0.35e-1 * xk.345 ^3, x! 1.030,.206 C0.79e-1 * xk0.98e-1 * xk.689 ^2 C.114 * xk.689 ^3, x! 1.380,.226C0.52e-1 * xc0.19e-1 * xk1.030 ^2K0.13e-1 * xk1.030 ^3, x! 2.070,.216 C0.61e-1 * x C0.6e-2 * xk1.380 ^2K0.14e-1 * xk1.380 ^3, x! 3.450,.240 C0.49e-1 * xk0.23e-1 * xk2.070 ^2 C0.7e-2 * xk2.070 ^3, x! 5.170,.304 C0.22e-1 * x C0.4e-2 * xk3.450 ^2K0.2e-2 * xk3.450 ^3,.324 C0.19e-1 * xk0.6e-2 * xk5.170 ^2 C0.1e-2
* xk5.170 ^3 : g d y / piecewise y!.345, 2.637K1.957 * yc5.508 * yk.172 ^3, y!.689, 2.494K1.462 * y C2.859 * yk.345 ^2K2.832 * yk.345 ^3, y! 1.030, 2.055K.501 * yk0.64e-1 * yk.689 ^2 C.459 * yk.689 ^3, y! 1.380, 1.946K.384 * yc.406 * yk1.030 ^2K.355 * yk1.030 ^3, y! 2.070, 1.768K.231 * y C0.33e-1 * yk1.380 ^2 C0.10e-1 * yk1.380 ^3, y! 3.450, 1.662 K.170 * y C0.55e-1 * yk2.070 ^2K0.11e-1 * yk2.070 ^3, y! 5.170, 1.436K0.83e-1 * y C0.9e-2 * yk3.450 ^2 C0.1e-2 * yk3.450 ^3, 1.253K0.41e-1 * y C0.16e-1 * yk5.170 ^2K0.3e-2 * y K5.170 ^3 : f 0 * F a / C 0 : evalf % 0. (10) F a / F r total : evalf % 0. e d h f 0 * F a / C 0 0.1603077311 if F a / F r total! e then Pd F r total else Pd X * F r total CY* F a end if 4000 Y d g f 0 * F a / C 0 2.608972828 L req / 10^6 ^ 1 / 3 * P : C req d evalf % 11537.99656 L 10 d CC / P ^p : evalf % 89.44081851 L 10 * 10^6 / n * 60 : Lifetime hours d evalf % 372.6700772 if CC! C req then print "Lejen ", Bearing, " opfylder ikke kravet om levetid!" else print "Lejen ", Bearing, " opfylder kravet om levetid!" end if print "Den har levetid på ", Lifetime hours, " stunder." "Lejen ", NKI2016, " opfylder kravet om levetid!" "Den har levetid på ", 372.6700772, " stunder." (11) (12) (13) (14) (15) (16) (17) (18)
restart : with CurveFitting : Estimeret levetid for 6305 Indtast omdrejningshastighed i RPM: n d 4000 4000 Indtast den ønskede levetid i timer: ReqLifetime hours d 100 100 L req d ReqLifetime hours * 60 * n 24000000 Indtast den radial, tangential og den aksial kraft der virker på lejen: F r d 7500 F t d 0 F a d 0 7500 0 sqrt F r ^2CF t ^2 : F r total d % 7500 Indtast navnet på lejen og dennes principielle værdier: Bearingd 6305 CC d 23400 C 0 d 11600 f 0 d 12 6305 0 23400 11600 12 Indtast p værdi: (kugleleje = 3... rulleleje = 10/3) p d 3 3 Indtast clearance værdi: X d 0.56 0.56 Værdiet nedenfor burde være det første gæt på et værdi for C når leje skal vælges: L req / 10^6 ^ 1 / 3 * F r total : C first_try d evalf % 21633.74356 xpoints d 0.172, 0.345, 0.689, 1.03, 1.38, 2.07, 3.45, 5.17, 6.89 : epoints d 0.19, 0.22, 0.26, 0.28, 0.30, 0.34, 0.38, 0.42, 0.44 : Ypoints d 2.30, 1.99, 1.71, 1.55, 1.45, 1.31, 1.15, 1.04, 1.00 : (1) (2) (3) (4) (5) (6) (7) (8) (9) h d x / piecewise x!.345,.159c.181 * xk.257 * xk.172 ^3, x!.689,.165c.158 * xk.133 * x K.345 ^2 C0.35e-1 * xk.345 ^3, x! 1.030,.206 C0.79e-1 * xk0.98e-1 * xk.689 ^2 C.114 * xk.689 ^3, x! 1.380,.226C0.52e-1 * xc0.19e-1 * xk1.030 ^2K0.13e-1 * xk1.030 ^3, x! 2.070,.216 C0.61e-1 * x C0.6e-2 * xk1.380 ^2K0.14e-1 * xk1.380 ^3, x! 3.450,.240 C0.49e-1 * xk0.23e-1 * xk2.070 ^2 C0.7e-2 * xk2.070 ^3, x! 5.170,.304 C0.22e-1 * x C0.4e-2 * xk3.450 ^2K0.2e-2 * xk3.450 ^3,.324 C0.19e-1 * xk0.6e-2 * xk5.170 ^2 C0.1e-2 * xk5.170 ^3 : g d y / piecewise y!.345, 2.637K1.957 * yc5.508 * yk.172 ^3, y!.689, 2.494K1.462 * y
C2.859 * yk.345 ^2K2.832 * yk.345 ^3, y! 1.030, 2.055K.501 * yk0.64e-1 * yk.689 ^2 C.459 * yk.689 ^3, y! 1.380, 1.946K.384 * yc.406 * yk1.030 ^2K.355 * yk1.030 ^3, y! 2.070, 1.768K.231 * y C0.33e-1 * yk1.380 ^2 C0.10e-1 * yk1.380 ^3, y! 3.450, 1.662 K.170 * y C0.55e-1 * yk2.070 ^2K0.11e-1 * yk2.070 ^3, y! 5.170, 1.436K0.83e-1 * y C0.9e-2 * yk3.450 ^2 C0.1e-2 * yk3.450 ^3, 1.253K0.41e-1 * y C0.16e-1 * yk5.170 ^2K0.3e-2 * y K5.170 ^3 : f 0 * F a / C 0 : evalf % 0. (10) F a / F r total : evalf % 0. e d h f 0 * F a / C 0 0.1603077311 if F a / F r total! e then Pd F r total else Pd X * F r total CY* F a end if 7500 Y d g f 0 * F a / C 0 2.608972828 L req / 10^6 ^ 1 / 3 * P : C req d evalf % 21633.74356 L 10 d CC / P ^p : evalf % 30.37132800 L 10 * 10^6 / n * 60 : Lifetime hours d evalf % 126.5472000 if CC! C req then print "Lejen ", Bearing, " opfylder ikke kravet om levetid!" else print "Lejen ", Bearing, " opfylder kravet om levetid!" end if print "Den har levetid på ", Lifetime hours, " stunder." "Lejen ", 6305, " opfylder kravet om levetid!" "Den har levetid på ", 126.5472000, " stunder." (11) (12) (13) (14) (15) (16) (17) (18)
restart : with CurveFitting : Estimeret levetid for NK 14/20 Indtast omdrejningshastighed i RPM: n d 4000 4000 Indtast den ønskede levetid i timer: ReqLifetime hours d 100 100 L req d ReqLifetime hours * 60 * n 24000000 Indtast den radial, tangential og den aksial kraft der virker på lejen: F r d 3500 F t d 0 F a d 0 3500 0 sqrt F r ^2CF t ^2 : F r total d % 3500 Indtast navnet på lejen og dennes principielle værdier: Bearingd NK1420 CC d 12800 C 0 d 16600 f 0 d 12 NK1420 0 12800 16600 12 Indtast p værdi: (kugleleje = 3... rulleleje = 10/3) p d 10 3 10 3 Indtast clearance værdi: X d 0.56 0.56 Værdiet nedenfor burde være det første gæt på et værdi for C når leje skal vælges: L req / 10^6 ^ 1 / 3 * F r total : C first_try d evalf % 10095.74699 xpoints d 0.172, 0.345, 0.689, 1.03, 1.38, 2.07, 3.45, 5.17, 6.89 : epoints d 0.19, 0.22, 0.26, 0.28, 0.30, 0.34, 0.38, 0.42, 0.44 : Ypoints d 2.30, 1.99, 1.71, 1.55, 1.45, 1.31, 1.15, 1.04, 1.00 : (1) (2) (3) (4) (5) (6) (7) (8) (9) h d x / piecewise x!.345,.159c.181 * xk.257 * xk.172 ^3, x!.689,.165c.158 * xk.133 * x K.345 ^2 C0.35e-1 * xk.345 ^3, x! 1.030,.206 C0.79e-1 * xk0.98e-1 * xk.689 ^2 C.114 * xk.689 ^3, x! 1.380,.226C0.52e-1 * xc0.19e-1 * xk1.030 ^2K0.13e-1 * xk1.030 ^3, x! 2.070,.216 C0.61e-1 * x C0.6e-2 * xk1.380 ^2K0.14e-1 * xk1.380 ^3, x! 3.450,.240 C0.49e-1 * xk0.23e-1 * xk2.070 ^2 C0.7e-2 * xk2.070 ^3, x! 5.170,.304 C0.22e-1 * x C0.4e-2 * xk3.450 ^2K0.2e-2 * xk3.450 ^3,.324 C0.19e-1 * xk0.6e-2 * xk5.170 ^2 C0.1e-2
* xk5.170 ^3 : g d y / piecewise y!.345, 2.637K1.957 * yc5.508 * yk.172 ^3, y!.689, 2.494K1.462 * y C2.859 * yk.345 ^2K2.832 * yk.345 ^3, y! 1.030, 2.055K.501 * yk0.64e-1 * yk.689 ^2 C.459 * yk.689 ^3, y! 1.380, 1.946K.384 * yc.406 * yk1.030 ^2K.355 * yk1.030 ^3, y! 2.070, 1.768K.231 * y C0.33e-1 * yk1.380 ^2 C0.10e-1 * yk1.380 ^3, y! 3.450, 1.662 K.170 * y C0.55e-1 * yk2.070 ^2K0.11e-1 * yk2.070 ^3, y! 5.170, 1.436K0.83e-1 * y C0.9e-2 * yk3.450 ^2 C0.1e-2 * yk3.450 ^3, 1.253K0.41e-1 * y C0.16e-1 * yk5.170 ^2K0.3e-2 * y K5.170 ^3 : f 0 * F a / C 0 : evalf % 0. (10) F a / F r total : evalf % 0. e d h f 0 * F a / C 0 0.1603077311 if F a / F r total! e then Pd F r total else Pd X * F r total CY* F a end if 3500 Y d g f 0 * F a / C 0 2.608972828 L req / 10^6 ^ 1 / 3 * P : C req d evalf % 10095.74699 L 10 d CC / P ^p : evalf % 75.35980244 L 10 * 10^6 / n * 60 : Lifetime hours d evalf % 313.9991768 if CC! C req then print "Lejen ", Bearing, " opfylder ikke kravet om levetid!" else print "Lejen ", Bearing, " opfylder kravet om levetid!" end if print "Den har levetid på ", Lifetime hours, " stunder." "Lejen ", NK1420, " opfylder kravet om levetid!" "Den har levetid på ", 313.9991768, " stunder." (11) (12) (13) (14) (15) (16) (17) (18)
Bilag 13 Smøring ˆ Minimum olie-lm tykkelse 73
restart : Minimum olie-film tykkelse Knast-knastfølger Den følgende formel kan bruges for at finde den minimal oliefilmtykkelse: h min d 1.806$ w z K0.128 η 0 $u 0.694 $ξ 0.568 $R x 0.434 hvor wz er den normal påførte kraft N w z d 1225 Der bruges data fra en mineral gear olie. 1.806 ξ 0.568 R x 0.434 w z η 0 u 0.088832 1225 η0 er den absolut viskositet for den specifikke olietype ved p=0 og konstant temperature N $ η 0 d 0.01 0.01 u er overfladens hastighed i x-retningen, (ua+ub)/2 u d 2 ξ er tryk-viskositet koefficienten for den specifikke olietype ξd 2$10 K8 1 50000000 Rx er den effektive radius i x-retningen m : R x d 0.0007574257425 0.0007574257425 Minimum olie-film tykkelse findes til at være m h min 2 m s : m 2 N : 0.000001798916748 Minimum olie-film tykkelse, udtykt ved micrometre: % $1000000 1.798916748 Den krævede ruhed, Rq, for at opnå elastohydronynamisk smørring (3 < λ < 10) findes [micrometre]: % solve = 3, R q 2 $R q 0.4240087438 s m 2 : (1) (2) (3) (4) (5) (6) (7) (8) (9)
Bilag 14 ANSYS-Vippearm 75
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 1 of 12 11-02-2010 Project Author Michael Oftedal og Helgi Gudbjartsson Subject Udvikling af ventilsystem til 2-takts forbrændingsmotor Prepared for Afgangsprojekt First Saved Friday, January 22, 2010 Last Saved Tuesday, January 26, 2010 Product Version 11.0 SP1 Release
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 2 of 12 11-02-2010 Contents Model Geometry VIPPERARM_2 Mesh Refinement Static Structural Analysis Settings Loads Solution Material Data Units Model Geometry Solution Information Results Convergence Max Equivalent Stress Results Max Shear Stress Results Low alloy steel, AISI 4340 TABLE 1 Unit System Metric (mm, kg, N, C, s, mv, ma) Angle Degrees Rotational Velocity rad/s TABLE 2 Model > Geometry Object Name Geometry State Fully Defined Definition E:\Users\Mike\Documents\Skolen\Dtu\7 semester\bachelor projekt_12_01_10\proe Source ANSYS\Workbench Vippearm 22_01_10\vipperarm_2.prt.1 Type ProEngineer Length Unit Millimeters Element Control Program Controlled Display Style Part Color Bounding Box Length X 76,336 mm Length Y 17,54 mm Length Z 10, mm Properties Volume 5119,3 mm³ Mass 3,9931e-002 kg Statistics Bodies 1 Active Bodies 1 Nodes 258765
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 3 of 12 11-02-2010 Elements 175945 Preferences Import Solid Bodies Yes Import Surface Bodies Yes Import Line Bodies Yes Parameter Processing Yes Personal Parameter Key DS CAD Attribute Transfer No Named Selection Processing No Material Properties Transfer No CAD Associativity Yes Import Coordinate Systems No Reader Save Part File No Import Using Instances Yes Do Smart Update No Attach File Via Temp File No Analysis Type 3-D Mixed Import Resolution None Enclosure and Symmetry Processing Yes TABLE 3 Model > Geometry > Parts Object Name VIPPERARM_2 State Meshed Graphics Properties Visible Yes Transparency 1 Definition Suppressed No Material Low alloy steel, AISI 4340 Stiffness Behavior Flexible Nonlinear Material Effects No Bounding Box Length X 76,336 mm Length Y 17,54 mm Length Z 10, mm Properties Volume 5119,3 mm³ Mass 3,9931e-002 kg Centroid X -4,8408 mm Centroid Y 3,6124 mm Centroid Z 2,0415e-006 mm Moment of Inertia Ip1 0,94458 kgmm² Moment of Inertia Ip2 16,509 kgmm² Moment of Inertia Ip3 17,079 kgmm² Statistics Nodes 258765 Elements 175945
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 4 of 12 11-02-2010 Mesh TABLE 4 Model > Mesh Object Name Mesh State Solved Defaults Physics Preference Mechanical Relevance 0 Advanced Relevance Center Coarse Element Size Default Shape Checking Standard Mechanical Solid Element Midside Nodes Program Controlled Straight Sided Elements No Initial Size Seed Active Assembly Smoothing Low Transition Fast Statistics Nodes 258765 Elements 175945 TABLE 5 Model > Mesh > Mesh Controls Object Name Refinement State Fully Defined Scope Scoping Method Geometry Selection Geometry 150 Faces Definition Suppressed No Refinement 1 FIGURE 1 Model > Mesh > Figure
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 5 of 12 11-02-2010 Static Structural TABLE 6 Model > Analysis Object Name Static Structural State Fully Defined Definition Physics Type Structural Analysis Type Static Structural Options Reference Temp 22, C TABLE 7 Model > Static Structural > Analysis Settings Object Name Analysis Settings State Fully Defined Step Controls Number Of Steps 1, Current Step Number 1, Step End Time 1, s Auto Time Stepping Program Controlled Solver Controls Solver Type Program Controlled Weak Springs Program Controlled
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 6 of 12 11-02-2010 Large Deflection Inertia Relief Force Convergence Moment Convergence Displacement Convergence Rotation Convergence Line Search Calculate Stress Calculate Strain Calculate Results At Solver Files Directory Future Analysis Save ANSYS db Delete Unneeded Files Nonlinear Solution Off Off Nonlinear Controls Program Controlled Program Controlled Program Controlled Program Controlled Program Controlled Output Controls Yes Yes All Time Points Analysis Data Management E:\Users\Mike\Documents\Skolen\Dtu\7 semester\bachelor projekt_12_01_10\proe ANSYS\Workbench Vippearm 22_01_10\vipperarm_2 Simulation Files\Static Structural\ None No Yes No FIGURE 2 Model > Static Structural > Figure
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 7 of 12 11-02-2010 TABLE 8 Model > Static Structural > Loads Object Name Fixed Support Force Force 2 State Fully Defined Scope Scoping Method Geometry Selection Geometry 6 Faces 1 Face Definition Type Fixed Support Force Suppressed No Define By Components X Component 0, N (ramped) Y Component 1094, N (ramped) 532, N (ramped) Z Component 0, N (ramped) FIGURE 3 Model > Static Structural > Force FIGURE 4 Model > Static Structural > Force 2
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 8 of 12 11-02-2010 Solution TABLE 9 Model > Static Structural > Solution Object Name Solution State Solved Adaptive Mesh Refinement Max Refinement Loops 7, Refinement Depth 3, TABLE 10 Model > Static Structural > Solution > Solution Information Object Name Solution Information State Solved Solution Information Solution Output Solver Output Newton-Raphson Residuals 0 Update Interval 2,5 s Display Points All TABLE 11 Model > Static Structural > Solution > Results Object Name Equivalent Stress Maximum Shear Stress Total Deformation State Solved Scope Geometry All Bodies Definition Type Equivalent (von-mises) Stress Maximum Shear Stress Total Deformation Display Time End Time Results Minimum 0,33319 MPa 0,19046 MPa 0, mm Maximum 142,47 MPa 73,346 MPa 5,5381e-002 mm
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 9 of 12 11-02-2010 Information Time 1, s Load Step 1 Substep 1 Iteration Number 1 TABLE 12 Model > Static Structural > Solution > Equivalent Stress > Convergences Object Name Convergence State Solved Definition Type Maximum Allowable Change 1, % Results Last Change 0,27079 % Converged Yes FIGURE 5 Model > Static Structural > Solution > Equivalent Stress > Convergence Model > Static Structural > Solution > Equivalent Stress > Convergence Equivalent Stress (MPa) Change (%) Nodes Elements 1 140,01 258765 175945 2 142,08 1,4714 404915 281098 3 142,47 0,27079 1237907 882901 FIGURE 6 Model > Static Structural > Solution > Equivalent Stress > Figure
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 10 of 12 11-02-2010 FIGURE 7 Model > Static Structural > Solution > Equivalent Stress > Figure 2
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 11 of 12 11-02-2010 TABLE 13 Model > Static Structural > Solution > Stress Safety Tools Object Name Max Equivalent Stress State Solved Definition Theory Max Equivalent Stress Stress Limit Type Tensile Yield Per Material TABLE 14 Model > Static Structural > Solution > Max Equivalent Stress > Results Object Name Safety Factor Safety Margin State Solved Scope Geometry All Bodies Definition Type Safety Factor Safety Margin Display Time End Time Results Minimum 5,4048 4,4048 Information Time 1, s Load Step 1 Substep 1 Iteration Number 1 TABLE 15 Model > Static Structural > Solution > Stress Safety Tools
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 12 of 12 11-02-2010 Material Data Low alloy steel, AISI 4340 Object Name Max Shear Stress State Solved Definition Theory Max Shear Stress Factor 0,5 Stress Limit Type Tensile Yield Per Material TABLE 16 Model > Static Structural > Solution > Max Shear Stress > Results Object Name Safety Factor Safety Margin State Solved Scope Geometry All Bodies Definition Type Safety Factor Safety Margin Display Time End Time Results Minimum 5,2491 4,2491 Information Time 1, s Load Step 1 Substep 1 Iteration Number 1 TABLE 17 Low alloy steel, AISI 4340 > Constants Structural Young's Modulus 2,05e+005 MPa Poisson's Ratio 0,285 Density 7,8e-006 kg/mm³ Thermal Expansion 0, 1/ C Tensile Yield Strength 770, MPa Compressive Yield Strength 770, MPa Tensile Ultimate Strength 865, MPa Compressive Ultimate Strength 940, MPa
Bilag 15 ANSYS-krumtap 88
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 1 of 12 11-02-2010 Project Author Michael Oftedal & Helgi Guðbjartsson Subject Udvikling af ventilsystem til 2-takts forbrændingsmotor Prepared for Afgangsprojekt First Saved Wednesday, January 20, 2010 Last Saved Saturday, January 23, 2010 Product Version 11.0 SP1 Release
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 2 of 12 11-02-2010 Contents Model Geometry KRUMTAPPEN_SUBMODEL Mesh Mesh Controls Static Structural Analysis Settings Loads Solution Solution Information Results Convergence Max Equivalent Stress Results Max Shear Stress Results Material Data GGG-70 Units Model Geometry TABLE 1 Unit System Metric (mm, kg, N, C, s, mv, ma) Angle Degrees Rotational Velocity rad/s TABLE 2 Model > Geometry Object Name Geometry State Fully Defined Definition Source E:\Users\Mike\Documents\Skolen\Dtu\7 semester\bachelor projekt_12_01_10\proe ANSYS\19_01_10\Workbench Krumtap Submodel\krumtappen_submodel.prt.2 Type ProEngineer Length Unit Millimeters Element Control Program Controlled Display Style Part Color Bounding Box Length X 34, mm Length Y 63,023 mm Length Z 71,3 mm Properties Volume 48317 mm³ Mass 0,34547 kg Statistics Bodies 1 Active Bodies 1 Nodes 73072
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 3 of 12 11-02-2010 Elements 49285 Preferences Import Solid Bodies Yes Import Surface Bodies Yes Import Line Bodies Yes Parameter Processing Yes Personal Parameter Key DS CAD Attribute Transfer No Named Selection Processing No Material Properties Transfer No CAD Associativity Yes Import Coordinate Systems No Reader Save Part File No Import Using Instances Yes Do Smart Update No Attach File Via Temp File No Analysis Type 3-D Mixed Import Resolution None Enclosure and Symmetry Yes Processing TABLE 3 Model > Geometry > Parts Object Name KRUMTAPPEN_SUBMODEL State Meshed Graphics Properties Visible Yes Transparency 1 Definition Suppressed No Material GGG-70 Stiffness Behavior Flexible Nonlinear Material Effects No Bounding Box Length X 34, mm Length Y 63,023 mm Length Z 71,3 mm Properties Volume 48317 mm³ Mass 0,34547 kg Centroid X 6,0405e-005 mm Centroid Y 6,0806 mm Centroid Z 4,9654 mm Moment of Inertia Ip1 167,01 kgmm² Moment of Inertia Ip2 147,99 kgmm² Moment of Inertia Ip3 59,967 kgmm²
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 4 of 12 11-02-2010 Statistics Nodes 73072 Elements 49285 Mesh TABLE 4 Model > Mesh Object Name Mesh State Solved Defaults Physics Preference Mechanical Relevance 0 Advanced Relevance Center Fine Element Size Default Shape Checking Standard Mechanical Solid Element Midside Nodes Program Controlled Straight Sided Elements No Initial Size Seed Active Assembly Smoothing High Transition Program Controlled Statistics Nodes 73072 Elements 49285 TABLE 5 Model > Mesh > Mesh Controls Object Name Face Sizing Refinement State Fully Defined Scope Scoping Method Geometry Selection Geometry 31 Faces 10 Faces Definition Suppressed No Type Element Size Element Size 10, mm Edge Behavior Curv/Proximity Refinement Refinement 1 FIGURE 1 Model > Mesh > Figure
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 5 of 12 11-02-2010 Static Structural TABLE 6 Model > Analysis Object Name Static Structural State Fully Defined Definition Physics Type Structural Analysis Type Static Structural Options Reference Temp 22, C TABLE 7 Model > Static Structural > Analysis Settings Object Name Analysis Settings State Fully Defined Step Controls Number Of Steps 1, Current Step Number 1, Step End Time 1, s Auto Time Stepping Program Controlled Solver Controls Solver Type Program Controlled Weak Springs Program Controlled
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 6 of 12 11-02-2010 Large Deflection Inertia Relief Force Convergence Moment Convergence Displacement Convergence Rotation Convergence Line Search Calculate Stress Calculate Strain Calculate Results At Solver Files Directory Future Analysis Save ANSYS db Delete Unneeded Files Nonlinear Solution Off Off Nonlinear Controls Program Controlled Program Controlled Program Controlled Program Controlled Program Controlled Output Controls Yes Yes All Time Points Analysis Data Management E:\Users\Mike\Documents\Skolen\Dtu\7 semester\bachelor projekt_12_01_10\proe ANSYS\Workbench Krumtap 22_01_10\krumtappen Simulation Files\Static Structural\ Prestressed analysis Yes Yes No FIGURE 2 Model > Static Structural > Figure
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 7 of 12 11-02-2010 TABLE 8 Model > Static Structural > Loads Object Name Bearing Load Fixed Support State Fully Defined Scope Scoping Method Geometry Selection Geometry 2 Faces 1 Face Definition Define By Components Type Bearing Load Fixed Support X Component 0, N Y Component -4000, N Z Component 0, N Suppressed No FIGURE 3 Model > Static Structural > Bearing Load Solution TABLE 9 Model > Static Structural > Solution Object Name Solution State Solved Adaptive Mesh Refinement Max Refinement Loops 7, Refinement Depth 2, TABLE 10 Model > Static Structural > Solution > Solution Information Object Name Solution Information State Solved Solution Information
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 8 of 12 11-02-2010 Solution Output Solver Output Newton-Raphson Residuals 0 Update Interval 2,5 s Display Points All TABLE 11 Model > Static Structural > Solution > Results Object Name Equivalent Stress Maximum Shear Stress Total Deformation State Solved Scope Geometry All Bodies Definition Type Equivalent (von-mises) Stress Maximum Shear Stress Total Deformation Display Time End Time Results Minimum 2,2622e-002 MPa 1,2601e-002 MPa 0, mm Maximum 227,78 MPa 126,03 MPa 0,18153 mm Information Time 1, s Load Step 1 Substep 1 Iteration Number 1 TABLE 12 Model > Static Structural > Solution > Equivalent Stress > Convergences Object Name Convergence State Solved Definition Type Maximum Allowable Change 1, % Results Last Change 0,59497 % Converged Yes FIGURE 4 Model > Static Structural > Solution > Equivalent Stress > Convergence
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 9 of 12 11-02-2010 Model > Static Structural > Solution > Equivalent Stress > Convergence Equivalent Stress (MPa) Change (%) Nodes Elements 1 214,14 73072 49285 2 222,84 3,985 135146 93886 3 226,43 1,5971 238965 168726 4 227,78 0,59497 618489 442851 FIGURE 5 Model > Static Structural > Solution > Equivalent Stress > Figure Von Mises stress
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 10 of 12 11-02-2010 FIGURE 6 Model > Static Structural > Solution > Equivalent Stress > Figure 2
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 11 of 12 11-02-2010 TABLE 13 Model > Static Structural > Solution > Stress Safety Tools Object Name Max Equivalent Stress State Solved Definition Theory Max Equivalent Stress Stress Limit Type Tensile Yield Per Material TABLE 14 Model > Static Structural > Solution > Max Equivalent Stress > Results Object Name Safety Factor Safety Margin State Solved Scope Geometry All Bodies Definition Type Safety Factor Safety Margin Display Time End Time Results Minimum 1,8263 0,8263 Information Time 1, s Load Step 1 Substep 1 Iteration Number 1 TABLE 15 Model > Static Structural > Solution > Stress Safety Tools
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 12 of 12 11-02-2010 Material Data GGG-70 Object Name Max Shear Stress State Solved Definition Theory Max Shear Stress Factor 0,5 Stress Limit Type Tensile Yield Per Material TABLE 16 Model > Static Structural > Solution > Max Shear Stress > Results Object Name Safety Factor Safety Margin State Solved Scope Geometry All Bodies Definition Type Safety Factor Safety Margin Display Time End Time 1, s Results Minimum 1,6504 0,65042 Information Time 1, s Load Step 1 Substep 1 Iteration Number 1 TABLE 17 GGG-70 > Constants Structural Young's Modulus 1,72e+005 MPa Poisson's Ratio 0,27 Density 7,15e-006 kg/mm³ Tensile Yield Strength 416, MPa Compressive Yield Strength 425, MPa Tensile Ultimate Strength 700, MPa Compressive Ultimate Strength 480, MPa
Bilag 16 ANSYS-stødstang 101
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 1 of 15 11-02-2010 Project Author Michael Oftedal & Helgi Guðbjartsson Subject Udvikling af ventilsystem til 2-takts forbrændingsmotor Prepared for Afgangsprojekt First Saved Saturday, January 23, 2010 Last Saved Thursday, February 11, 2010 Product Version 11.0 SP1 Release
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 2 of 15 11-02-2010 Contents Model Geometry STOEDSTANG Mesh Refinement Static Structural Analysis Settings Loads Solution Solution Information Results Max Equivalent Stress Results Max Shear Stress Results Linear Buckling Initial Condition Analysis Settings Solution Solution Information Total Deformation Material Data Carbon steel, AISI 1080 annealed Units Model Geometry TABLE 1 Unit System Metric (mm, kg, N, C, s, mv, ma) Angle Degrees Rotational Velocity rad/s Object Name State Source Type Length Unit Element Control Display Style Length X Length Y Length Z Volume TABLE 2 Model > Geometry Geometry Fully Defined Definition E:\Users\Mike\Documents\Skolen\Dtu\7 semester\bachelor projekt_12_01_10\proe ANSYS\Workbench Stødstang 23_01_10\STOEDSTANG.PRT.1 ProEngineer Millimeters Program Controlled Part Color Bounding Box 159,48 mm 5,4706 mm 5,4706 mm Properties 2535,2 mm³
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 3 of 15 11-02-2010 Mass 1,9775e-002 kg Statistics Bodies 1 Active Bodies 1 Nodes 18728 Elements 11570 Preferences Import Solid Bodies Yes Import Surface Bodies Yes Import Line Bodies Yes Parameter Processing Yes Personal Parameter Key DS CAD Attribute Transfer No Named Selection Processing No Material Properties Transfer No CAD Associativity Yes Import Coordinate Systems No Reader Save Part File No Import Using Instances Yes Do Smart Update No Attach File Via Temp File No Analysis Type 3-D Mixed Import Resolution None Enclosure and Symmetry Processing Yes TABLE 3 Model > Geometry > Parts Object Name STOEDSTANG State Meshed Graphics Properties Visible Yes Transparency 1 Definition Suppressed No Material Carbon steel, AISI 1080 annealed Stiffness Behavior Flexible Nonlinear Material Effects Yes Bounding Box Length X 159,48 mm Length Y 5,4706 mm Length Z 5,4706 mm Properties Volume 2535,2 mm³ Mass 1,9775e-002 kg Centroid X 80,471 mm Centroid Y 8,4759e-017 mm Centroid Z 2,5831e-017 mm Moment of Inertia Ip1 4,8271e-002 kgmm² Moment of Inertia Ip2 40,978 kgmm²
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 4 of 15 11-02-2010 Moment of Inertia Ip3 40,978 kgmm² Statistics Nodes 18728 Elements 11570 Mesh TABLE 4 Model > Mesh Object Name Mesh State Solved Defaults Physics Preference Mechanical Relevance 0 Advanced Relevance Center Coarse Element Size Default Shape Checking Standard Mechanical Solid Element Midside Nodes Program Controlled Straight Sided Elements No Initial Size Seed Active Assembly Smoothing Low Transition Fast Statistics Nodes 18728 Elements 11570 TABLE 5 Model > Mesh > Mesh Controls Object Name Refinement State Fully Defined Scope Scoping Method Geometry Selection Geometry 10 Faces Definition Suppressed No Refinement 2 FIGURE 1 Model > Mesh > Figure
Project file://c:\users\michael\appdata\roaming\ansys\v110\simulation_report\simulatio... Page 5 of 15 11-02-2010 Static Structural TABLE 6 Model > Analysis Object Name Static Structural State Fully Defined Definition Physics Type Structural Analysis Type Static Structural Options Reference Temp 22, C TABLE 7 Model > Static Structural > Analysis Settings Object Name Analysis Settings State Fully Defined Step Controls Number Of Steps 1, Current Step Number 1, Step End Time 1, s Auto Time Stepping Program Controlled Solver Controls Solver Type Program Controlled Weak Springs Program Controlled