Design method of tension spring and compression spring

The task of spring design is to determine the diameter d of the spring wire, the number of working turns n and other geometric dimensions, so as to meet the strength constraints, stiffness constraints and stability constraints, and further require corresponding design indicators (such as volume, weight, vibration Stability, etc.) achieve the best.
The specific design steps are: first try to select the spring material and spring index C according to the working conditions and requirements. Since sb is related to d, the diameter d of the spring wire is often assumed in advance. Next, calculate the values ​​of d and n and the corresponding other geometric dimensions. If the results obtained do not meet the design conditions, the above process must be repeated. Until the solution that satisfies all constraints is obtained, it is a feasible solution to this problem. In actual problems, feasible solutions are not unique, and it is often necessary to find a better solution from multiple feasible solutions.

Example 12-1 designs a cylindrical helical compression spring with a circular cross-section. It is known that the minimum load Fmin=200N, the maximum load Fmax=500N, the working stroke h=10mm, the spring type II works, the outer diameter of the spring should not exceed 28mm, and the ends should be ground tightly.

solution:

Trial calculation (1):

(1) Choose spring material and allowable stress. Choose C grade carbon spring steel wire.

According to the requirements of outer diameter, the primary selection is C=7, d=3.5mm from C=D2/d=(Dd)/d, sb=1570MPa from Table 1, and from Table 2: [t]=0.41sb= 644MPa.

(2) Calculate the spring wire diameter d

From the formula, K=1.21

From the formula, d≥4.1mm

It can be seen that the initial calculated value of d=3.5mm does not meet the strength constraints and should be recalculated.

Trial calculation (two):

(1) Select the spring material as above. To obtain a larger value of I>d, choose C=5.3.

From C=(D-d)/d, d=4.4mm.

Look up Table 1 to get sb=1520MPa, and from Table 2 we know [t]=0.41sb=623MPa.

(2) Calculate the spring wire diameter d

From the formula, K=1.29

From the formula, d≥3.7mm.

It can be seen that: I>d=4.4mm satisfies the strength constraint.

(3) Calculate the effective number of working circles n

Determine the amount of deformation λmax from Figure 1: λmax=16.7mm.

Look up table 2, G=79000N/mm2,

From the formula, n=9.75

Take n=10, and consider that both ends are tied tightly, then the total number of turns n1=n+2=12. So far, a feasible solution that satisfies the constraints of strength and stiffness has been obtained, but considering the further reduction of spring dimensions and weight, the trial calculation is performed again.

Trial calculation (3):

(1) Still choose the above spring material, take C=6, get K=1.253, d=4mm, look up Table 1, get sb=1520MPa, [t]=0.41sb=623MPa.

(2) Calculate the diameter of the spring wire. Get d≥3.91mm. Knowing that d=4mm meets the strength condition.

(3) Calculate the effective number of working circles n. From the trial calculation (2), λmax=16.7mm, G=79000N/mm2

From the formula, n=6.11

Take n=6.5 circles, still refer to the two ends and tighten one circle, n1=n+2=8.5.

This calculation result not only satisfies the constraints of strength and rigidity, but also a better solution in terms of dimensions and weight. This solution can be preliminarily determined, and then other dimensions will be calculated and checked for stability.

(4) Determine the deformation λmax, λmin, λlim and the actual minimum load Fmin

The limit load of the spring is:

Because the number of working turns has been changed from 6.11 to 6.5, the spring deformation and minimum load have also changed accordingly.

From the formula:

λmin=λmax-h=(17.77-10)mm=7.77mm

(5) Find the pitch p of the spring, the free height H0, the helix angle γ and the unfolded length of the spring wire L

Under the action of Fmax, the distance between two adjacent turns δ≥0.1d=0.4mm, and δ=0.5mm, the spring pitch under no load is

p=d+λmax/n+δ1 = (4+17.77/6.5+0.5)mm=7.23mm

p basically conforms to the specified range of (1/2~1/3)D2.

The free height of the spring with the end face tightly ground is

Take the standard value H0=52mm.

The helix angle of the spring under no load is

It basically satisfies the range of γ=5°~9°.

Unfolded length of spring wire

(6) Stability calculation

b=H0/D2=52/24=2.17

The fixed support at both ends is adopted, b=2.17<5.3, so it will not lose stability.

(7) Draw the spring characteristic line and the working drawing of the part.