This article provides general theoretical data and personal arguments about spring selection. Basically, in the suspension of modern cars, compression coil springs are used, and they will be mainly discussed.
Changing springs to non-standard, basically several goals are pursued:
increase or decrease ground clearance (which is caused by the need to use the car in certain conditions)
increase or decrease the stiffness of the springs (find a compromise between comfort and handling)
It is necessary to set conditions: what ultimately we want to get. Next, we will talk about a specific car - the 2101 VAZ classic, as the first-born of the family of these cars. Now many fans of these cars (to which the author of the article refers) are not happy with much in the behavior of the car on the road. Large rolls in turns, body buildup on longitudinal waves, etc. I hope everyone understands that by replacing springs or shock absorbers alone, a compromise cannot be found between handling and comfort, only comprehensive measures can achieve the desired results. Now we will not touch the shock absorbers, but will set ourselves the goal of lowering the ride height and increasing the rigidity of the system.
So, the main technical characteristic of the VAZ 2101
gross vehicle weight, kg (for sports driving it makes no sense to take it from the factory instructions with 4 passengers and 50 kg in the trunk, it makes sense to load the car with 2 passengers) - 955 kg + 80 kg + 80 kg = 1115 kg
weight per front axle of the equipped vehicle - 430 kg
mass falling on the rear axle of the equipped car - 362 kg
Front suspension coil springs:
free length, mm - 360
the inner diameter of the spring, mm -
bar diameter, mm - 13.13 + 0.05
the number of turns - 8.83
length under load 628 kg, mm - 182
length under load 435 kg, mm (yellow marking of group A) -> 232
Rear suspension coil springs:
bar diameter, mm - 12.3 + 0.5
the inner diameter of the spring, mm - 102.7
the number of working (active) turns - 8
number of turns - 10
spring length without load, mm - 434
length under load 295 kg, mm (yellow marking for group A) -> 273
The easiest way to lower the ride height is to shorten, by cutting one or more turns, a standard spring. Many do this without thinking about the consequences. The question is, what do we get as a result? How will the cut coils affect the springy properties of a spring?
The spring stiffness is inversely proportional to the number of active turns: C = F / S = (G * d ^ 4) / (8 * n * D ^ 3), N / mm
Where: G - shear modulus, for hot-rolled springs: 78,500 N / mm ^ 2, for cold-rolled 81400 N / mm ^ 2,
d is the diameter of the rod, mm,
n is the number of active turns,
D is the average diameter of the rod, mm,
We calculate the reduction of the working turns by 1, 2 for the rear suspension, for comparison, calculate for the standard suspension.
Standard spring stiffness: С0 = (78500 * 12.34) / (8 * 8 * 102.73) = 25.917 N / mm
The rigidity of the cut spring: C1 = (78500 * 12.34) / (8 * 7 * 102.73) = 29.620 N / mm
The rigidity of the cut spring: C2 = (78500 * 12.34) / (8 * 6 * 102.73) = 34.557 N / mm
How will the change in the number of turns affect the deformation of the spring?
X - deformation of the spring, mm,
P - external force, N,
G - shear modulus, for hot-rolled springs: 78,500 N / mm ^ 2, for cold-rolled 81400 N / mm ^ 2,
r is the radius of the rod, mm
n is the number of turns,
R is the radius of the helical axis of the spring, mm,
We will also calculate for the rear suspension of the car:
The deformation of a standard spring from a load of 295 * 9.81 = 2893.95 N (standard load)
Deformation of a spring trimmed for 1 revolution:
Deformation of a spring cut into 2 turns:
To what extent do we have ground clearance? In the case of circumcision, not only the deformation of the spring but also its length changes. When shortening by 1 turn, the length will decrease by a step of a turn - 48.22 mm, by two turns - respectively by two steps of a turn - 96.44 mm.
So with such a load and a free spring length of 434 mm, we get the total spring length:
For a standard spring - 294.427 mm
For cropped one turn - 260.164 mm
For cut into two turns - 225.901 mm
It was found that by changing the number of active turns, the rigidity of the system increases the more the more turns are removed. So shortening the spring by two turns, we will increase its stiffness by 33% and reduce its length under load by 13% with respect to the standard. If we assume that the spring in the rear suspension is upright, then the ground clearance will decrease by the amount of change in the length of the spring under load. That is, in our case, for the rear suspension, it will decrease by 68.5 mm. If you consider that the ground clearance at full load to the rear axle beam is 170 mm, then the ground clearance will decrease by 40%. Nevertheless, a shortening spring should not count on a resource comparable to a standard spring due to increased stresses on the remaining turns. Also, you should not shorten already used springs, it is better to buy new ones - they cost not much. The trimmed end of the spring should be rested on the cup of the lower arm (front suspension), axle (rear suspension), but not in the cup of the body (if something happens to change the lever or digest the cup is much cheaper than repairing the body).
It should be said that with standard shock absorbers, the rear spring can be easily trimmed by 1 turn - there will be a slight preload, the spring does not drop out with the wheel fully hung. If you cut more, or put shorter springs, you will either have to look for a shock absorber with a shorter rod, or move its lower fastening lower by manufacturing the corresponding brackets.
What else you need to know when cutting the springs.
To the ends, the winding pitch of them decreases, forming a platform for support. By cutting the turns this platform decreases (stresses on the cup increase) and it is unevenly located, with some eccentricity - e. That is, in this case we have loading in the form of eccentric compression. The spring is already working differently. The force F shifted relative to the axis by the value of “e” will cause a pure bending moment:
Since the plane of action of the bending pairs may not coincide with any of the main planes of inertia of the spring, in the general case there is a combination of longitudinal compression and pure oblique bending. Imagine a small spring sandwiched between two fingers. It can be simply squeezed and unclenched, and only its length will change, but try to squeeze the same spring resting on your fingers not only its entire area, but only half. The spring will be compressed due to the turns and also due to the curvature. In this case, its characteristics may be completely different.
Having cut the spring to ensure uniform bearing on the cups, you can release the last turn. To do this, heat it to 600-650 C (the color of red-hot brown-red) and allow to cool in air. Here a problem arises connected with the fact that: you need to heat to a fairly high temperature, you need to heat only one turn (local heating). To such a temperature local heating can only be provided with a gas burner, and then, the probability of spoiling the part is very high.
Rigidity how to determine?
First, let's fix what is a spring? This is a mandatory component of the suspension, which is presented in the role of a specific elastic element. It provides mitigation of strokes, shocks, including during sudden braking and starts. The meaning of the spring is to quickly return the wheel to its original "position", after hitting an obstacle.
Too stiff detail significantly affects the handling of the car, especially on rough roads. However, the advantage of increased rigidity is greater safety when driving at high speeds. That is, it does not allow the body to swing as much as that of too soft. The latter cope with almost all the pits without discomfort for the driver, but with such springs it is difficult to enter into bends.
Remember, there are several key factors that affect stiffness. Knowing them, you can independently determine the type of your own element mounted on the suspension. So:
1. The diameter of the rod itself. Remember the important pattern, the thicker the rod, the more rigid the part.
2. The diameter of the spring on the outside. The larger the diameter, the lower the actual stiffness.
3. Form. There are several main types: conical, cylindrical, “barrel-shaped”. Each variety has its own characteristics and characteristics. There are also combined.
4. The number of turns. The pattern is this: the more turns, the less stiffness will be.
The stiffness is determined quite simply. In most cases, the manufacturer independently affixes a marking, according to which it is clear to which class the product belongs.
Remember that marking in yellow indicates a length of up to 240 mm. But, basically all the indicators that will be required to calculate the stiffness are on the product.
If no marking is found, the indicator can be calculated as follows. So, prepare the scales (ordinary floor), a wooden bar, a ruler, the product itself. You need to lay the bar on the scale, but remember that the width of the board should be greater than the diameter of the spring. Next, take the second board and press it on top and measure the length of the product, naturally, without taking into account the boards. Alone or using a special press, the spring must be compressed to a certain level. This is usually 40 mm. Record information from the balance. Next, having the initial length in the expanded position and compressed, we calculate the difference. Next, you need the weight obtained after compression, divided by the difference, thereby obtaining a stiffness indicator.
There are more complicated methods of calculation, but they are not worth talking about, because you will need at least two more values that can be deduced, knowing Hooke's formulas and laws, as well as the theory of proportionality. For an ordinary driver, this method of calculus is superfluous, you can find out much easier.
Resource, and which is better to choose?
As a rule, among the most popular queries in search engines, what is the "mileage" of the springs, and which ones are better to choose? It is difficult to judge which ones are better, because each motorist has his own preferences in this regard. Someone loves a fast ride and for him the rigidity of the safety requirements, while someone prefers comfort. All individually, moreover, today there are a huge number of manufacturers who can offer universal springs, including the so-called adjustable ones.
What, in principle, should be understood as adjustable? It only seems to you that everything is simple, and yet they can still differ among themselves. There are two types:
• With an adjustable "nut", which is screwed onto the cylinder and allows you to both increase and decrease the stiffness depending on the "wrap".
• With adjustable spacer, in principle, many of these two points do not even share
What is the resource? It is difficult to answer in the affirmative about the term of operation, because everything depends on the quality of roads directly. Somewhere, the springs serve 100,000 km, but somewhere they do not "pass" 10,000 km. In general, it can be noted that the greater the stiffness, the more durable, and vice versa. The average "mileage" rarely exceeds 50,000 km.
Faults and their symptoms
As such, there are not so many malfunctions, in fact, the constructive part of this element does not represent anything complicated. These types of problems are divided:
• Broken bar at the edges.
• Trite "tired" metal.
In general, that's all, there is nothing more to add. Either the spring sank, due to frequent overloads, or the coil burst, that's all.
What are the symptoms of malfunctions? Here you can note a wide range of features:
1. Decrease in clearance (clearance).
2. The appearance of vibration or knocking.
3. The valency of the car. That is, the situation when the car "pecks" when braking and starting.
4. "Breakdowns" of the suspension. When hitting bumps, holes, "humps", there is a contact of metal elements, for example, turns between themselves. In serviceable products, this should not be.
5. Big rolls in the corners.
6. The discrepancy in the height of the front and rear.
The reasons for all this are varied.:
• Depreciation due to old age.
• Incorrect operation (transportation of large loads).
- January 26, 2015, 12:35
- January 26, 2015, 16:41
- January 26, 2015, 22:32
- January 27, 2015, 19:00
- January 28, 2015 00:56
- January 28, 2015 00:57
I believe. But
1) Nitric acid is a volatile acid. Therefore, its traces simply evaporate. Stupidly rinse well - rolls, as if it sounded.
2) But it is quite logical that the workplace and part were washed in a weak solution of soda ash.
3) An oxide film is formed on the surface (there are many solutions for chemical burnishing on nitric acid), the film is uniform and strong. After lubricating the spring with any lubricant, this film behaves appropriately.
4) Oddly enough, etching is uniform. It is checked not on one spring - there are no complaints.
Again, such a method has served me faithfully for more than two years. It seems so far nothing has fallen apart, has not broken off, has not rusted.
Maybe it's magic?