determine the spring constant of a helical spring that has the following characteristics:?

Question

Material: Aluminum with shear modulus 4 x 10 to the sixth power
Number of turns - 10
Mean coil diameter - 10 in
Wire diameter - 1 in
free length - 15 in

Answer 1

The spring stiffness formula is g*d^4/(8*N*D^3), where g is shear modulus, d is wire diameter, N is number of turns and D is coil diameter. Assuming your d value is in pound-inch-compatible units, the spring constant is 50 lb/inch. Derivation of the formula is in the ref. under "Close-coiled springs", bottom of page 3. Note their formula gives deflection del (Greek "d") as a function of W (weight or force) and R (coil radius). In those terms the spring constant is W/del, and their 64R^3 is our 8D^3. If you wish to account for the helix angle (~.15 rad) you should use the more complex formula on the same page that accounts for the change in coil diameter with stretching.