Closed Coiled Helical Spring under axial pull:
Consider one half turn of helical spring as shown
Every cross section would be under torsion F*R
thus, maximum stress by torsion theory:
\[\tau_{max} = \frac{16 T}{\pi d^3}\]
\[\boxed{\tau_{max} = \frac{8FD}{\pi d^3}}\]
where D = 2*R = diameter of coil
d = diameter of spring wire
If one cross section twists angle θ relative to other then:
\[θ = \frac{TL}{GJ} = \frac{FR(\pi R)}{GJ}\]
\[\delta' = Rθ = \frac{\pi F R^3}{GJ}\]
Total deflection \( \delta = 2N* \delta' = \frac{2N F \pi R^3}{GJ} = \frac{8NF D^3}{G d^4}\)
Spring Stiffness(K) :
\[\boxed{K = \frac{F}{\delta} = \frac{G d^4}{8 N D^3}}\]
where, N : number of turns
G : Modulus of rigidity
NOTE: Stiffness is inversely proportional to the number of coils, therefore when a spring cut into half its stiffness becomes double for each part.
Wahl's correction factor( $k_w$ )
The simple torsion theory gives inappropriate results, thus for accurate result multiply by Wahl's factor.
\[\tau_{max} = \frac{k_w 8FD}{\pi d^3}\]
Wahl's correction factor is defined as:
\[k_w = \frac{4C-1}{4C-4} + \frac{0.615}{C}\]
where C is spring index = D/d
Springs in series:
When different stiffness springs joined such that shares common load, said to be in series.
Total deflection \( \delta = \frac{F}{K_{eq}} = \delta_1 + \delta_2 + ..... + \delta_n = \frac{F}{K_1} + \frac{F}{K_2} + .....+ \frac{F}{K_n}\)
\[\boxed{\frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2} + .....+ \frac{1}{K_n}}\]
Springs in parallel:
When different stiffness springs joined such that shares common deflection, said to be in parallel.
\[ \delta = \frac{F}{K_{eq}} = \delta_1 = \delta_2 = ..... = \delta_n = \frac{F_1}{K_1} = \frac{F_2}{K_2} = .....= \frac{F_n}{K_n}\]
Total load \(F = F_1 + F_2 + ..... + F_n\)
\[\boxed{K_{eq} = K_1 + K_2 + ...... + K_3}\]
Spiral Spring
Consider a rectangular cross sectional (breadth B and thickness t) spiral spring is given by following equation:
\[ r = \frac{b}{2} + \frac{a-b}{4 \pi N}θ \]
Springs will be subjected to uniform bending due to action of central moment which tends to reduce the radius of curvature at all points.
When winding couple M is applied to the spindle, resisting force F will be set up at the pin such that
M = F * R
Now, Consider two small elements of length dl at distance x each side of center.
For small deflection change in slope is = $\frac{M dl}{EI}$.
for left side element change in slope,
\[dθ_1 = \frac{F(R+x) dl}{EI}\]
for right side element change in the slope,
\[dθ_2 = \frac{F(R-x) dl}{EI}\]
sum of these two slopes = $dθ_1 + dθ_2 = \frac{2FR dl}{EI}$
Total angle of twist = $\frac{1}{2} \int_{0}^{L} \frac{2FR}{EI}dl = \frac{FRL}{EI} = \frac{ML}{EI}$
Length of spiral $ L = \frac{\pi N}{2}(a+b) $
We get wind up angle of spiral spring is :
\[θ = \frac{M}{EI} \Bigg[ \frac{\pi N}{2} (a+b) \Bigg] \]
Maximum Bending moment = F*a
Maximum bending stress = $\frac{M y}{I}$
\[ \sigma_{max} = \frac{F a(t/2)}{Bt^3/12}\]
\[ \sigma_{max} = \frac{6 F a}{Bt^2}\]
