Lesson 03

22
Μηχανικές Διαμορφώσεις & Σχεδιασμός Καλουπιών

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καλουπια καλουπια καλουπια καλουπια καλουπια

Transcript of Lesson 03

& & & : i. ii. iii. iv. v. vi. vii. & : s l PB =P = = l = s = s l 8 . 0 PB = 5% & : D/s 2: B

D/s = 1.51.9: B 25% D/s = 1.01.4: B 90% D/s = 0.70.9: B 150% 1 & : s x PP=HAP = P = x = s = = : & ( , ) ... & : P = = % P = 100P P=E 2 & 6 . 22kN 2.5mm. 1 : 2 : (%) ( 2) 2 : 132kN kN 6 22 P 6= = =oP% 20 = akN 33 kN 132 25 . 0 P= = =Eo P & 30% & ... & : P = P = P = P P 1.3 PP P 0.3 P P+ = + + = & & ... & : = P = = P = l = P = % s = l P100s P 3 . 1 El P P 3 . 1 E + = + = 3 & 0.2%C : & 1 : 2kp/mm 0 4 = : (D/s> 2 ) kp 18396 P kp 3 ) 1 . 25 1 . 25 1 . 103 ( 40 s ) l l (l s l P3 2 1 B B= + + = + + = = : tn 30 kp 29315 P kp ) 18396 (1.3 P 1.3 PK K~ = = =mm 1 . 103 mm ]484 ) 8 25 ( 2 ) 8 30 ( 2 [ l1= + + =tmm 1 . 125 mm 8 l l3 2= = = t / & , 7mm. 19% . : H H : 180mm 1mm & 1 : 2kp/mm 45 = 10 =kp 2545 P kp10025447 10100P P = ==E E : 2 : : (D/s>2 ) kp 25447 Pkp 1 180 45 s D s l PB B= = = = 1 2 & : 3 : : : kp 35626 Pkp ) 2545 25447 (1.3 P P 1.3 PK K= + = + =kpm 74 . 27 kpm ) 10 7 2545 10 3 25447 3 . 1 (l P100s P 3 . 1 l P P) 3 . 1 ( E3 4 = + == + = + = kpm 146 E m kp ) 19 . 0 / 74 . 27 ( E E 0.19 = = E = 30 = 3 0.3%C, , 50 mm 2.5 mm. & 50 & mm 1 . 0 x = 3 22 =mm 55 . 0 mm1005 . 2 22100s === o ... & 2kp/mm 8 4 =kp 6000 P 5 . 2 50 48 s D s l PB B= = = =(D/s>2 ) 1 kp 7 . 307 P kp55 . 0 5 . 21 . 0 6000 s x PP = ==HA HA 1 & 0.3%..C. 2.5mm. . 3mm.H 2mm. , 7.5 mm.) ) . ) . ) 20% .