Spring

[JRM]

Can anyone help me with this??

A spring has a free length of 14.5mm and has 9 coils. When it is installed in a mechanism it has an installed length of 9mm and the load exerted by the spring is 1kg. If the spring were removed from the mechanism and shortened to a free length of 10mm by cutting off 4.5mm; what force would it exert when reinstalled in the mechanism?

Cheers

 

[israelkk]

You have to give info such as the the wire diameter and what are the types of the spring end. Is the spring ends are squared and ground, squared only, etc. Is the 9 coils are the total number of coils or the active coils? What is the spring outside diameter?

 

[JRM]

Wire and sprind diameters are unknown.
9 is the total number of coils.

 

[desertfox]

Hi JRM

israelkk is right, we need more detail of the spring however to give you an idea of what force you will get
I have calculated the stiffness of the existing spring and
the estimated stiffness of the spring with 4.5mm cut off.

assumed 9 active coils

stiffness of existing spring = 1x9.81/(14.5-9)
= 1.7836 N/mm = k

stiffness of a compression spring is also given by:-

G*d^4/(8*D^3*n)=k
where G= modulus of rigidity
D= mean diameter of spring
n= number of active coils (assumed)

therefore G*d^4/(8*D^3*n)= 1.7836

now n in the formula will = 9 initially and so we can calculate the value of G*d^4/(8*D^3)which will equal 9*1.7836= 16.0524

this figure will be constant even when you cut the spring down so looking at the existing spring before modification we can say that:-

16.0524/9 =1.7836N/mm

I estimate you will lose about 2.79 coils when you cut off the spring, call it 3 coils for ease now the new stiffness of the spring we be

16.0524/6 = 2.6754N/mm

now assuming your new free length is 10mm you are only compressing the spring by 1mm so the new force is 2.6754N
which equates to about 0.2727kg.

A further word of caution if you cut a spring down as you intend you need to check the stresses within the spring itself as you may exceed its design limits.

regards desertfox.

 

[TheTick]

Actually, he does have enough info.

Given constant wire diameter, pitch, and mean diameter, the spring rate for an open-end, unground compression spring varies inversely to the number of coils.

Where k=spring rate (force/distance)
c=coils
f=final
o=initial

k(f)/k(o) = c(o)/c(f)

If a spring has a rate of 10 N/mm and 10 free coils, that same spring "cut in half" would have a rate of 20 N/mm.

Be careful about using kg as a force equivalent.

"Great ideas need landing gear as well as wings."--C. D. Jackson

 

[Cockroach]

"TheTick" is correct, there is enough information to obtain a closed form solution set. However, I don't think the right approach has been discussed yet. Using a model of similtanity, previous issues of spring stiffness, wire diameter, etc can be overcome.

Spring load is a function of geometry, i.e. mean coil diameter D, wire diameter d, number of active coils N, deflection or spring travel x and the material property of stiffness modulus G. Load can be expressed mathematically as:

P = (G d^4 x) / (8 D^3 N)

We have been given properties of the spring as free length L=14.5 mm, N=9 and P=1.0 kg which is 9.81 N, the proper dimensional unit for load. Furthermore, installation length is 9 mm which means spring compression is 5.5 mm. Call these characteristics SPRING 1.

In the second case, the free length has been shortened to 10 mm. Call this case SPRING 2. Taking the ratio of loads between SPRING 1 & 2,

P1/P2 = [(G d^4 x1)/(8 D^3 N1)][(8 D^2 N2)/(G d^4 x2)]

P1/P2 = (x1/x2)(N2/N1) {1}

Spring wire diameter cancels out and is not required to solve for P2. Note also that material property is not a factor. The ratio of spring load is fully dependent on the ratio of deflection and number of active coils.

Clearly, chopping off free length alters the coil population, pitch remaining the same. Defining pitch (p)as the ratio of coils (N) to free length (L), then

p = N/L such that p1/p2 = (N1/L1)(L2/N2) = (N1/N2)(L2/L1)

p1 = p2 so that p1/p2 = 1 implying N1/N2 = L1/L2 {2}

Substituting equation {2} into equation {1} results as:

P1/P2 = (x1/x2)(L2/L1)

Therefore the ratio of spring load between the two cases is dependent on the ratio of spring deflection and ratio of spring free length. We indeed have enough information to solve the problem!

x1 = 14.5 - 9.0 mm = 5.5 mm
x2 = 10.0 - 9.0 mm = 1.0 mm
P1 = 9.81 N

9.81 N / P2 = (4.5mm/1.0mm)(10.0mm/14.5mm) = 3.10345
P2 = 3.161 N = 0.71 lbf

I suggest that the spring load with 4.5 mm of material removed shall be 3.2 N, approximately 1/3 of the original load.

Hope this helps everyone out.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[Cockroach]

I was looking at DesertFox solution and wondering why we had a slight difference even though the approach is sort of the same. In doing so, I discovered a numerical substitional error, 14.5 - 9 mm = 5.5 mm and not 4.5 mm.

9.81 N/P2=(5.5mm/1.0mm)(10.0mm/14.5mm) = 3.79310
P2 = 2.586 N = 0.581 lbf

Therefore P2 = 2.586 N which is 0.264 kg mass equivalent. Essentially DesertFox and I have the same answer.

My apologies for the slight confusion.



Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[JRM]

Thanks very much you guys.

I used the ticks method to come up with the answer of 0.264.

However the full explanation from Mr Hueston has also helped to clarify things.

Once again thanks to everyone.

All the best for the new year.