TUGAS 3 HELICAL TORSION SPRINGS

Posted: 19th October 2012 by Musa H. R. Situmorang in ELEMEN MESIN 1

Problem:

Design a concentric spring for an air craft engine valve to exert a maximum force of 5000 N under a deflection of 40 mm.

Both the springs have same free length, solid length and are subjected to equal maximum shear strees of 850 MPa.

The spring index for both the springs is 6. [ Ans.d1 = 8 mm ; d2 = 6 mm ; n = 4 ]

Answer:

  • Load each spring

P1 : P2 = (d1 : d2)2 P1 : P2

= (8 : 6)2

= 1,69 mm2
P1 + P2 = 5000 (1,69 . P2) + P2

= 5000

2P2 = 2958,6
P2 = 1479,3

P1 = 3520,7

 

  • Mean diameter

k1 = k2 = (4C – 1) : (4C – 4) + 0,615 : C

= (4.6 – 1) : (4.6 – 4) + 0,615 : 6

= 1,2525

  • Max shear stress 1

850 = k1 (8.P1.C) : ( phi.d12) 850

= 1,2525 ( 8.(3520,7).6 : (phi..d12)
d1 = 8.9 mm
D1 = C.d1

= 6.8,9

= 53,4 mm

  • Max shear stress 2

850 = k2 (8.P2.C) : ( phi.d12)

850 = 1,2525 ( 8.(1479,3).6 : (phi.d22)
d2 = 5,7 mm
D2 = Cd2

= 6.5,7

= 34,2 mm

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