Spring Constants: Does Changing Diameter Affect Constant?

[simplyphysics]

Homework Statement

If you change the diameter of a spring does it affect the spring constant? Assume the spring length, if fully extended, was kept constant.

Homework Equations

F=kx; F=mg

The Attempt at a Solution

From my research the spring constant (k) is measured by applying a mass and determining the vertical displacement (x) of the spring. The spring force is just equal but opposite to that of the force of gravity. Using Hooke's law the spring constant of a specific spring can be determined.
I know that the spring constant tells us the stiffness of the spring. I'm not sure how this would change if we were to use the same spring, unravelling it and then creating a larger diameter but lowering the number of coils. I think this would be difficult to test experimentally as the spring state would definitely be compromised. For some reason my mind is fuzzing out and I cannot think of whether changing the number of coils would affect the spring constant or stretch. I think maybe the spring would not stretch as far, keeping the mass applied the same, the spring constants would be different...but I'm still not sure. Help?

 

[haruspex]

You may need to think about how a spring works. Consider a very short section of the wire. As the load is applied, what deformation (primarily) occurs?

 

[vijay kumar]

k = Gd4/[8nD3]

Where:

k = constant, pounds of load per inch of deflection
G = modulus of rigidity of spring material, pounds per square inch
d = wire diameter, inches
n = number of active coils, which is the number of coils subjected to flexure (always less than the total number of coils)
D = mean coil diameter, inches = Outer Diameter - Wire Diameter

 

[Suraj M]

Don't springs work on the concept of shearing stress -rigidity modulus?
changing the number of coils would change the natural length of the spring.
αR=θL
so required shear angle = θ
θ α R/L
and also F/A α θ
I don't know if this is even right!
Sorry if this could not help you.