spring leaves
A semi-elliptical laminated vehicle spring to carry a load 2000 N, is to consis seven leaves 2,45 mm wide, two of the leaves extending the full length of the spring. The spring is to be 1010 mm in length & attached to the axle by two u-bolts 15 mm apart. Hese bolt hold the central portion of the spring so rigidly. That they may be considered equivalent to a band having a width equal to the distance between bolts. The leaves are to be silicomangenese steel. Assuming an allowed stress of 3500 kg/cm2 .
Detemine: a. thickness of the leaves
b. Deflection of the leaves
c. Diameter of the eye
d. length of the leaves
e. radius to which leaves should be initially bent assume modulus of elasticity as 2,1 X 106 kg/cm2
Detemine: a. thickness of the leaves
b. Deflection of the leaves
c. Diameter of the eye
d. length of the leaves
e. radius to which leaves should be initially bent assume modulus of elasticity as 2,1 X 106 kg/cm2
Answer:
2L = 1010-2/3.15 mm L=1000mm
2W= 2000 N.m W= 1000 N.m
Thickness = 6 W L : ibh2
= 6.1000.1000 : 7bh2
h2 = 6.106 : 7.3500
h2 = 244,8 = 245 : 2,45
h2 = 100
h = 10
2L = 1010-2/3.15 mm L=1000mm
2W= 2000 N.m W= 1000 N.m
Thickness = 6 W L : ibh2
= 6.1000.1000 : 7bh2
h2 = 6.106 : 7.3500
h2 = 244,8 = 245 : 2,45
h2 = 100
h = 10
Deflection
Deflection = 12 W L3 : Ebh3 (2ng + 3nf) E= 210.103 N/cm2
= 12.2000.(1000)3 : 210.103.1000.103 (20)
Deflection = 10
Deflection = 12 W L3 : Ebh3 (2ng + 3nf) E= 210.103 N/cm2
= 12.2000.(1000)3 : 210.103.1000.103 (20)
Deflection = 10
Diameter of the eye
Diameter of the eye = W : cos 450
= 1414 N
Load on the pin = db.bx.pb
Pb = 10 N/mm2 dp = 1414 : 2,45.10
= 58
Length of the leaves
Lp = b + 2 clearance
Lp = 2,45 + 2.2
Lp = 6,45
Diameter of the eye = W : cos 450
= 1414 N
Load on the pin = db.bx.pb
Pb = 10 N/mm2 dp = 1414 : 2,45.10
= 58
Length of the leaves
Lp = b + 2 clearance
Lp = 2,45 + 2.2
Lp = 6,45
Max bending moment
M = load on pin x lp : 4
= 1414.6,45 : 4
= 2281
M = load on pin x lp : 4
= 1414.6,45 : 4
= 2281
stresst = 100 N/mm2
stresst X phi/32 X dp = M
dp = 3(32.M : phi.stresst)
dp = 6,2
stresst X phi/32 X dp = M
dp = 3(32.M : phi.stresst)
dp = 6,2
The required pin diameter is greater from bending consideration. Hence a pin of diameter 20 mm is suggested. For bush wall thickness of 3 mm, the eye diameter will be
6,2+2.3 = 12,2 mm
6,2+2.3 = 12,2 mm
Length of the leaves
L1 = (effective length/i-1) + ineffective length
L1 = (1000/7-1) + 10
L1 = 176 mm
Where ineffective length = 2/3 distance between U-bolts
L2 = (1000/7-2) + 10 = 210 mm
L3 = (1000/7-3) + 10 = 260 mm
L4 = (1000/7-4) + 10 = 343 mm
L5 = (1000/7-5) + 10 = 510 mm
L6 = (1000/7-6) + 10 = 1010 mm
L7 = (1000/7-7) + 10 = 1010 mm
L1 = (effective length/i-1) + ineffective length
L1 = (1000/7-1) + 10
L1 = 176 mm
Where ineffective length = 2/3 distance between U-bolts
L2 = (1000/7-2) + 10 = 210 mm
L3 = (1000/7-3) + 10 = 260 mm
L4 = (1000/7-4) + 10 = 343 mm
L5 = (1000/7-5) + 10 = 510 mm
L6 = (1000/7-6) + 10 = 1010 mm
L7 = (1000/7-7) + 10 = 1010 mm
Radius
R2 = (R – Y)2 + 5502
840 R = 176400 + 302500
R = 478900 : 840 = 570 mm
R2 = (R – Y)2 + 5502
840 R = 176400 + 302500
R = 478900 : 840 = 570 mm
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