Helical spring is a vital element for many machines and most of the time helical compression spring is used rather than helical tension spring. This article will talk specifically about design of helical compression spring. More over, we will not consider buckling for spring design, non circular spring wire and non linear springs here.
Some important basic things about spring design
Most of the times, when you will be asked to design a helical spring practically, you will already have data like loading conditions, movements and material properties (like, shear modulus, maximum permissible shear stress etc.) available with you. Also, you will know the sizes of wire available in market. You have to find out spring diameter, spring constant, number of active coils, free length of the spring etc. Also you have to validate that whether stress develop in your designed spring is not exceeding maximum permissible tensional stress. Let’s explain the basic terminologies used in spring design:
· Spring constant (K): It is the ratio of applied force on spring (F) to the change of shape (delta) for the applied force.
· Spring index (C): It is the ratio of mean spring diameter to wire diameter of the spring.
· Number of active coil (Na): This is the number of coils participate in resisting the external applied force.
· Free length (Lf): This is the length of spring without applying any load.
We will try to understand design approach for helical compression spring through the typical example below:
Maximum working load (Fmax) = 500 N
Minimum working load (Fmin) = 100 N
Length of spring at maximum load (Lmax) = 50 MM
Length of spring at minimum load (Lmin) = 70 MM
Shear modulus of material (G) = 77500 Mpa
Permissible torsional shear stress (Tmax) = 725 Mpa
Available wire diameter (d) = 5 to 20 mm
- First of all we will try to find out approximate mean diameter (D) of the spring from the following equation:
T = (8DF/pi*d^3)……………eqn.1
We will start with d=5 MM, T=Tmax and F=Fmax.
And will get, D=71.14 MM.
- Calculate Spring constant (K) and free length of the spring (Lf) by solving the following equations:
Fmax = K (Lf – Lmax)…………….eqn.2
Fmin = K (Lf – Lmin)……………..eqn.3
You will get, k=20 and Lf = 75.
At this point of time, please note the relation between Tmax and Fmax shown in eqn.1 is approximate and for correcting it we need to multiply by Wahl correction factor (W) and it will become:
T = W*(8DF/pi*d^3)…………..eqn.4
- Calculate spring index by using following equation:
Take, D=70 (nearest round figure of 71.14, which we found from eqn.1)
You will get, C=14
- Calculate Wahl Factor by using following equation:
W= (4C-1)/(4C-4) + (0.615/C)…..eqn.6
You will get, W=1.10162
- Use eqn.4 to calculate T by using C=14 and W=1.10162. You will get
Now, note from the given input data that, Permissible torsional shear stress (Tmax) is less than T calculated from step 5.
- Repeat from Step3 to Step5 by taking a lower value of D (say 65) until you will get T value less than Tmax. You will get final D value as 63 MM.
- Calculate Number of active coil (Na) from the following equation:
You will get Na=1.15, you can consider Na=1 or 2
I have developed a Helical Compression Spring Design Calculator, where you can get the design output by putting input details.
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