Design method of tension spring and compression spring

Design method of tension spring and compression spring:
The task of spring design is to determine the diameter of spring wire D, the number of working coils N and other geometric dimensions, so as to meet the strength constraints, stiffness constraints and stability constraints, and further require the corresponding design indicators (such as volume, weight, vibration stability, etc.) to achieve the best
The specific design steps are as follows: first, according to the working conditions and requirements, try to select the spring material and spring index C. Since sb is related to D, it is often necessary to assume the diameter D of the spring wire in advance. Next, calculate the values of D, N and other corresponding geometric dimensions. If the results obtained do not conform to the design conditions, The above process needs to be repeated until a solution satisfying all constraints is obtained. In practical problems, the feasible scheme is not unique, and it is often necessary to obtain the optimal solution from multiple feasible schemes
Example 12-1 a cylindrical spiral compression spring is designed. The spring wire section is circular. The minimum load Fmin = 200N, the maximum load Fmax = 500N, the working stroke H = 10mm. For class II operation of spring, the outer diameter of the spring shall not exceed 28mm, and the end shall be tightly ground
Solution:
Trial calculation (1)
(1) Select spring material and allowable stress. Select C grade carbon spring steel wire
According to the requirement of outer diameter, C = 7, d = 3.5mm from C = D2 / D = (D-D) / D, sb = 1570mpa from table 1, and from table 2: [t] = 0.41sb = 644mpa
(2) Calculate the diameter of spring wire D
From the equation, k = 1.21
From the formula, D ≥ 4.1 mm
Therefore, the initial value of D = 3.5mm does not meet the strength constraint condition and should be recalculated
Trial calculation (2)
(1) 3. D = C > 3
From C = (D-D) / D, d = 4.4mm
According to table 1, sb = 1520mpa, from table 2 we can know that [t] = 0.41sb = 623mpa
(2) Calculate the diameter of spring wire D
From the equation, k = 1.29
From the formula, D ≥ 3.7mm
It can be seen that I > d = 4.4mm meets the strength constraint condition
(3) Calculate the number of effective working cycles n
According to figure 1, the deformation is determined as λ Max: λ max = 16.7mm
According to table 2, g = 79000n / mm2,
From the formula, n = 9.75
Taking n = 10, considering that both ends of the spring are tightly connected, the total number of turns is N1 = n + 2 = 12. So far, a feasible scheme meeting the constraint conditions of strength and stiffness is obtained. However, considering further reducing the spring’s overall size and weight, the trial calculation is carried out again
Trial calculation (3)
(1) Still select 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) The diameter of spring wire is calculated. D ≥ 3.91mm. It is known that d = 4mm meets the strength condition
(3) According to trial calculation (2), λ max = 16.7mm, g = 79000n / mm2
From the formula, n = 6.11
Take n = 6.5 turns, still refer to both ends and tighten one circle, N1 = n + 2 = 8.5
This result satisfies the constraint conditions of strength and stiffness, and is a better solution in terms of overall dimension and weight. This solution can be initially determined, and other dimensions can be calculated and checked for stability
(4) The deformation values λ max, λ min, λ Lim and the actual minimum load Fmin are determined
The ultimate load of the spring is as follows:
Because the number of working turns is changed from 6.11 to 6.5, the deformation and minimum load of the spring are also changed accordingly
According to the formula:
λmin=λmax-h=(17.77-10)mm=7.77mm
(5) Calculate the spring pitch P, free height H0, helix angle γ and spring wire development length L
Under the action of Fmax, if the distance between two adjacent coils is greater than or equal to 0.1D = 0.4mm, if δ = 0.5mm, the pitch of spring under no load is
p=d+λmax/n+δ1 =(4+17.77/6.5+0.5)mm=7.23mm
P is basically in the range of (1 / 2 ~ 1 / 3) d 2
The free height of the spring with the end face tightly ground is
Take the standard value H0 = 52mm
The helix angle of spring without load is
The range of γ = 5 ° to 9 ° is basically satisfied
Development length of spring wire
(6) Stability calculation
b=H0/D2=52/24=2.17
Using fixed support at both ends, B = 2.17 < 5.3, so it will not lose stability