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1
- #1
Omar_Whiz
Mechanical
- May 31, 2019
- 1
thread404-162279
The above thread discusses the issue of finding a helical spring's lateral stiffness.
I have come across this issue myself, and desperately needed a solution.
Now, this is normally a non-standard way of loading a helical spring. Nonetheless, A.M. Wahl in his 1963 book "Mechanical springs" gives an expression for this.
Let us take β to be the spring's flexural rigidity, and γ to be the shearing rigidity. These expressions are given as:
β = (lo.E.G.d^4)/(32.n.r.(2G+E))
γ = (lo.E.d^4)/(64.n.r^3)
where;
lo = spring's free length
E = modulus of elasticity
G = modulus of rigidity
d = wire/coil diameter
n = number of active coils
r = mean coil radius, also equal to the outer diameter minus the wire/coil diameter
Now, Wahl states that the deflection due to a lateral force Q is given as:
δ = (Q.lo^3)/(3.β) + (Q.lo)/γ
And this is due to 1) bending effects, and 2) shearing effects.
If we rearrange the above expression and extract the lateral stiffness, we obtain:
k_lat = (3.β.γ)/((γ.lo^3) + (3.β.lo))
And that is the expression for the lateral stiffness of helical spring.
Extra note: The lateral deflection of the helical spring is effected by axial loads as well. If we have an axial load P, we can find a certain factor C1 which is equal to: C1 = 1/(1 - (P/Pcr)); where Pcr is the critical buckling load of the spring. If the ratio P/Pcr is ≈ 0, then the axial load has no effect on the total lateral deflection, otherwise the total lateral deflection can be found as:
δtot = C1.[(Q.lo^3)/(3.β) + (Q.lo)/γ]
Hope this helps!
P.S. Please check Wahl's 1963 book on "Mechanical springs" as it is the source of inspiration.
The above thread discusses the issue of finding a helical spring's lateral stiffness.
I have come across this issue myself, and desperately needed a solution.
Now, this is normally a non-standard way of loading a helical spring. Nonetheless, A.M. Wahl in his 1963 book "Mechanical springs" gives an expression for this.
Let us take β to be the spring's flexural rigidity, and γ to be the shearing rigidity. These expressions are given as:
β = (lo.E.G.d^4)/(32.n.r.(2G+E))
γ = (lo.E.d^4)/(64.n.r^3)
where;
lo = spring's free length
E = modulus of elasticity
G = modulus of rigidity
d = wire/coil diameter
n = number of active coils
r = mean coil radius, also equal to the outer diameter minus the wire/coil diameter
Now, Wahl states that the deflection due to a lateral force Q is given as:
δ = (Q.lo^3)/(3.β) + (Q.lo)/γ
And this is due to 1) bending effects, and 2) shearing effects.
If we rearrange the above expression and extract the lateral stiffness, we obtain:
k_lat = (3.β.γ)/((γ.lo^3) + (3.β.lo))
And that is the expression for the lateral stiffness of helical spring.
Extra note: The lateral deflection of the helical spring is effected by axial loads as well. If we have an axial load P, we can find a certain factor C1 which is equal to: C1 = 1/(1 - (P/Pcr)); where Pcr is the critical buckling load of the spring. If the ratio P/Pcr is ≈ 0, then the axial load has no effect on the total lateral deflection, otherwise the total lateral deflection can be found as:
δtot = C1.[(Q.lo^3)/(3.β) + (Q.lo)/γ]
Hope this helps!
P.S. Please check Wahl's 1963 book on "Mechanical springs" as it is the source of inspiration.
