Given:
Weight of rail carrier (W) = 200 kN = 200,000 N
Velocity (v) = 5 km/h = 36005×1000=1.389m/s
Number of springs (n) = 4
Spring index (C) = dD=5
Deflection per spring (δ) = 220 mm = 0.22 m
Maximum allowable shear stress (τ) = 400 MPa = 400×106Pa
Modulus of rigidity (G) = 80,000 MPa = 80×109Pa
Ends of the spring are squared and ground.
Step 1: Calculate the Kinetic Energy Absorbed by the Springs
The kinetic energy of the rail carrier is absorbed by the 4 springs:
KE=21mv2
First, find the mass (m):
m=gW=9.81200,000=20,387.36kg
Now, calculate KE:
KE=21×20,387.36×(1.389)2=19,684.5J
This energy is absorbed by 4 springs:
KEper spring=419,684.5=4,921.13J
Step 2: Relate Energy to Spring Parameters
The energy stored in a spring is:
E=21Pδ
Where P is the load on the spring and δ is the deflection.
4,921.13=21×P×0.22P=0.224,921.13×2=44,737.5N
Maximum load on each spring (P) = 44.74 kN
Step 3: Find Wire Diameter (d)
The shear stress in the spring is given by:
τ=Kπd28PC
Where K is Wahl's correction factor:
K=4C−44C−1+C0.615=4×5−44×5−1+50.615=1.3105
Now, substitute τ=400×106, P=44,737.5, C=5:
400×106=1.3105×πd28×44,737.5×5d2=π×400×1061.3105×8×44,737.5×5d2=0.001865d=0.001865=0.0432m=43.2mm
Wire diameter (d) = 43.2 mm
Step 4: Find Coil Diameter (D)
Given C=dD=5:
D=5d=5×43.2=216mm
Coil diameter (D) = 216 mm
Step 5: Find Number of Turns (N)
The deflection formula for a spring is:
δ=Gd48PD3N
Given δ=0.22m, P=44,737.5N, D=0.216m, G=80×109Pa, d=0.0432m:
0.22=80×109×(0.0432)48×44,737.5×(0.216)3N0.22=80×109×3.49×10−68×44,737.5×0.01008N0.22=279.2×1033,612.6NN=3,612.60.22×279.2×103=17turns
Number of active turns (N) = 17
Step 6: Find Free Length (Lf)
For squared and ground ends, the free length is:
Lf=(N+2)d+δ+0.15δLf=(17+2)×43.2+220+0.15×220Lf=19×43.2+220+33=820.8+220+33=1,073.8mm
Free length (Lf) = 1,074 mm
Final Answers:
Maximum load on each spring (P) = 44.74 kN
Wire diameter (d) = 43.2 mm
Coil diameter (D) = 216 mm
Number of active turns (N) = 17
Free length of the spring (Lf) = 1,074 mm
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