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How can I find stiffness of a steel coil spring used in dump trucks?

Here is the data of the question.

Length of spring unloaded = 220mm

Outer Diameter = 27mm

Inner Diameter = 15mm

Wire Diameter = 6mm

Material is steel, G= 77.2 GPa

Spring must bear load of 15kN

Formula which I used for the problem:

enter image description here

G= Modulus of rigidity

d= Wire diameter

D= Mean Diameter

N= No. of turns

By using simple mechanics, I calculated stiffness and it is found 168.8 kN/m. Solution is attached.

enter image description here

Main problem is I do not know N (No. of turns) and I assumed it 8 by doing some research (comparative study). Can I find no. of turns through some table or data books or may be from unloaded length of the spring? Kindly resolve this problem. Thanks.

Ibrahim Omer
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2 Answers2

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The values that you have provided don't make too much sense, as you might have already noted in the comments below your question. But I am going to assume you are just doing as example, so the values are not the actual ones but just hypothetical.

Besides, the stiffness of the spring (longitudinal or torsional) changes with the amount of load; i.e. the stiffness is not constant. Say, if the spring deflects 1 mm for a load of 1000 N, or rotates 1 deg for a torque of 1000 N.mm, then it won't be 2 mm for 2000 N or 2 deg for 2000 N.mm. This implies that the stiffness is not constant, infact for any practically possible structure present in the world. You can only use that formula for extremely small deflections (for longitudinal spring) and rotations (for torsional spring).

Furthermore, the number of turns is an input and not an output, so it needs to be inputted if you want to figure out the spring's stiffness. Number of turns are usually decided when we need to achieve a certain level of stiffness for the spring. It can be anything, 4 or 8 or 40 or 120 etc.

So what you can do is that you can decide on the amount of longitudinal deformation you want (if the spring is longitudinal), like say you only want it to deform to 222 mm from 220 mm for this loading condition. Then divide this load by by 2 mm (i.e. 222 - 220), and that is the linear stiffness for the spring (again note that it might not be the same if the load changes since stiffness is non-linear in reality). Use this stiffness value in the equation you have already posted in your question, and then find out the number of turns required on the spring, that will give you a deformation of 2 mm.

Rameez Ul Haq
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  • The unloaded length is 220mm so why would it get longer to 222mm? The loaded length will be shorter. – Solar Mike Feb 23 '22 at 20:26
  • 'Say' means for example. He may do it for a lesser length. Linear stiffness won't change in tension or compression. The deformation is still 2 mm, which is an absolute value. – Rameez Ul Haq Feb 23 '22 at 20:28
  • So define the unloaded length as given in the question, since you seem to be the expert ( definition of expert)… – Solar Mike Feb 23 '22 at 20:31
  • Its not about being expert, its about reaching to a conclusion when you have a problem given. You see something, and you have to figure out the inputs necessary to generate the output. Or vice versa, decide on the output you want and find the inputs needed for it. And unloaded length means the length of the spring without any load; i.e. initial length. Say the spring deforms to 218 mm (as per requirement) from 220 mm, the deformation is same from the initial length i.e. still 2 mm. And absolute value of deformation should be used for the stiffness. – Rameez Ul Haq Feb 23 '22 at 20:37
  • No, springs can, and do, have variable rates - which could be possible since you assume things not in the question... – Solar Mike Feb 23 '22 at 20:48
  • I already mentioned springs are not linear in reality in my answer; their stiffness change upon the application of force and its magnitude. You wanna go in details? lol. – Rameez Ul Haq Feb 23 '22 at 20:50
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You have some freedom to choose the number of turns (N). If the spring operates in compression, one constraint that limits the solution is the spring must be able to compress the distance F/k.

For example, if the spring has N=220/6=36.6 turns, the wires would be touching and the spring cannot compress at all. If there are 16 turns, the spring can compress by 220 mm - 6 mm*16 = 124 mm. However, 16 turns give a spring constant of k=8.44E4 N/m. To support a 15000 N load requires compressing the spring 15000/8.44E4=177.7 mm. This would not be possible with 16 turns.

Another consideration to investigate is whether the spring will buckling.

JohnHoltz
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