Trading wire size for length, diameter, or cycle life: Now we are really going to save you some money, if you just recall your high school algebra class (and I don't mean that cute cheerleader who sat next to you). If you further understand the role of the 4th power of the spring wire size (letter d in the formulas above) in the numerator of the spring rate formula, and how to increase or decrease d to compensate for changes in length, diameter, and cycle life, then you're qualified for elite spring calculations. Matching springs is a matter of equating the 4th power of the proportion in wire size change to the proportion of change in the diameter or length or the product of both diameter and length. However, it is usually best to only increase wire size when substituting a spring, since this does not derate the cycle life. If you observe that the formula for bending stress is proportionate to the inverse 3rd power of the diameter, then physically a proportionate increase in wire size will result in a dramatic increase in cycle life of the 3rd power of that proportion. Trade-off example: Yawn with me while we ponder my original spring once more. Let's say I was in a fit of engineering mania, and wanted to replace my spring having a 0.2253 inch diameter wire (d = 0.2253) with a 0.262 wire version (d = 0.262). How much longer is the spring with equal torque rate, assuming we use the same coil diameter? The proportion of this change is 0.262/0.2253 = 1.163, and the 4th power of that is 1.83. This means the length must increase by a factor of 1.83 (again, not counting dead coils). Recalling that the length in Example 1 was 102 non-dead coils, the heavier wire spring must be about 1.83*102 = 187 coils, which when adding 5 dead coils and multiplying by the wire size to get the overall length, is (187+5)*0.262 = 50 inches, versus 24 inches in the original. So using this heavier wire more than doubles the length (and thus the mass and thus the cost). While the cost about doubles, the stress goes down by the inverse 3rd power of the wire size proportion, or 1/(1.163**3) = 0.64. Sress is favorably, non-linearly related to cycle lifetime (halving the stress more than doubles the lifetime), so this decreased stress should more than double the expected lifetime of the spring. While the up-front cost is more, the true cost of an amortized lifetime is much less. In short, per cycle it is cheaper. Ah, the wonders of engineering calculations! Conclusion: Observe that the stress formula (and thus the cycle lifetime) depends only on wire diameter (d) for equal torques. Thus the only way to improve cycle lifetime is to use heavier wire. For equal torques, heavier wire size, due to the exponents in the formulas, increases cycle lifetime much faster than it increases mass (and thus cost), physically speaking.
The "safety issue" trick: Another tip-off is the use of language like "safety issue". This is meant to trump any objections you might have to a costly repair bill. Don't be manipulated by the suggestion that you are risking disaster if you don't buy something expensive. Even if you think the risk is genuine, get another estimate, and tell the second repairman you are skeptical; every technician loves to prove the competition made a mistake.
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If your door feels heavy, it is likely that your springs have started to wear down and are no longer capable of bearing the weight that they once did. Now, don’t worry, just because a spring is starting to lose its strength doesn’t mean it will snap at any moment. However, simultaneously, a weak spring isn’t any safer to try and repair on your own.