Spring design have?

Spring the daily task of design is to clear spring wire diameter d, the work world number n and other geometry specifications, prompt can consider binding binding compressive strength, bending stiffness and reliability constraints, further also stipulates the relative design parameter values ( Such as volume, net weight, vibration reliability, etc. ) Do * is good. Spring manufacturer to introduce practical design process is: the first according to the standard, regulations and other work, try to choose spring, spring index value C raw materials. Because sb associated with d, assuming that the diameter of the spring wire is usually advance d. Then calculates the value of d, n and other relative geometry specification, if the individual results and design standards, about whole process will be repeated. 

 To consider all the constraint conditions of getting until solution is a feasible plan for this problem. Specific problems, the feasible plan is not *, must from several feasible scheme, usually for better solution. Example 12 - 1 design a narrow cylindrical spiral spring, wire spring model for the ring. Less known * load Fmin = 200 n, larger load Fmax = 500 n, job scheduling, h = 10 mm spring Ⅱ type of job, spring diameter does not exceed 28 mm, * end and tight. Solution: trial ( A) :( 1) Choose spring materials and allowable stress. Using C grade carbon spring stainless steel wire. According to the provisions in diameter, C = 7 the first review, the C = D2 / d = ( D - d) D/d = 3. 5 mm, are found in table 1 to sb = 1570 mpa, known by table 2: t] = 0. 41= 644 mpa。 ( 2) By measuring the spring wire diameter d type K = 1. 21 from the d type 4 or more. 1 mm so, d = 3. 5 mm at the beginning of the numerical value does not conform to the constraints of compressive strength, should be measured again. Trial ( 2) :( 1) Choose the spring like the above raw materials. In order to obtain great I> D value, choose C = 5. 3. By C = ( D - d) / d, d = 4. 4mm。 Look-up table 1 sb = 1520 mpa, by the table 2 know [ t] = 0. 41sb=623MPa。 ( 2) By measuring the spring wire diameter d type K = 1. 29 the type d 3 or higher. 7mm。 Know: I> d=4. 4 mm considering constraints of compressive strength. ( 3) Measure the reasonable work world number n from figure 1 form variables lambda Max: lambda Max = 16. 7mm。

 Table 2, G = 79000 n/was, by type n = 9. 75 take n = 10, considering each side and a circle, the total number of ring n1 = n + 2 = 12. To this, get a considering constraints of compressive strength and bending rigidity of feasible plans, but consider to further reduce the spring size and net weight, carry out the trial again. Trial ( 3) :( 1) Still choose about spring materials, C = 6, for K = 1. 253, d = 4 mm, table 1, sb = 1520 mpa, t] = 0. 41sb=623MPa。 ( 2) Measure the spring wire diameter. D 3 or higher. 91mm。 Know d = 4 mm considering compressive strength standard. ( 3) Measure the reasonable work world number n. By trial, 2) Know, lambda Max = 16. 7 mm, G = 79000 n/was by type n = 6. 11 n = 6. Five laps, still refer to each side and a circle, n1 = n + 2 = 8. 5. That figure is considering constraints of compressive strength and bending rigidity, from the size and net weight, it seems, is also a good solution, the solution of the basic can be clear, the following measure other specifications and reliability strength check. ( 4) Clear shape variable lambda Max, lambda min, lambda lim and specific * to the limit load of the Fmin spring: less number because of the work world by six. 11 to 6. 5, therefore, the spring form variables and * load also has certain change relatively less. By type: lambda min = lambda Max - h=( 17. 77 - 10) mm=7. 77mm( 5) For Spring Festival diameter p, optional ratio H0, spiral Angle gamma and spring wire length L in Fmax effect under the interval of adjacent laps the delta p 0. 1 d = 0。 4 mm, take the delta = 0. 5 mm diameter, no load effect next Spring Festival for p = d + lambda Max/n + delta 1 = ( 4 + 17. 77/6. 5 + 0. 5) mm=7. 23 MMPS base is in ( 1/2 ~ 1/3) The requirement of the D2 category. Grinding inner hole and tight to spring arbitrary take respect H0 ratio of high to width = 52 mm. Spring under no load effect of spiral Angle on the basis of considering the category of gamma = 5 ° ~ 9 °. Spring spring wire length ( 6) Reliability is calculated b = H0 / D2 = 52/24 = 2. 17 choose sides fixed rubber bearing, b = 2. 17' 5. Three, therefore not easy imbalances. ( 7) Drawing the spring characteristic line and * job center.