3. 3
Compression springs
It can be used as a working drawing in
order to save drafting time, with the
appropriate dimensions and details
added.
4. 4
Flat springs
▪No standard drawing to cover this
type of spring
▪Flat springs are usually made from
high-carbon steel in the annealed
condition, and are subsequently heat
treated
5. 5
Spiral-Flat springs
▪The spring consists of a strip of steel
spirally wound and capable of storing
energy in the form of torque.
▪If the spring is close wound and fitted
in a housing then the illustrations in
(c) and (d) are applicable.
9. Applications of Springs
To store and return energy,
To apply and maintain a definite force,
as in relief valves
To isolate vibrations, as in automobile.
To indicate and/or control load, as in a
scale
To return or displace a component, as in
a brake pedal or engine valve.
11. Helical Coil
Figure 17.2: Helical coil. (a) Coiled wire showing applied force; (b) coiled wire with
section showing torsional and direct (vertical) shear acting on the wire.
12. HelicalCompression Springs
Helical coil. (a) Straight wire before coiling; (b) coiled wire
showing transverse (or direct) shear; (c) coiled wire showing
Shear Stresses on Wire and Coil
13. Shear Stresses
Torsional Shear Stress
where D is the mean coil diameter and d is the wire diameter
Transverse Shear Stress
Combined Torsional and Transverse Shear Stress
The spring index, which is a measure of coil curvature, is
where Kd is the transverse shear factor, given by
15. Deflection of Helical
Compression Springs
the shear strain due to torsional loading is
The deflection due to torsional loading is:
Where C = spring index = D/d
Na = Number of active coils ?
G = Shear modulus of elasticity
16. Deflection of Helical Springs
The deflection-force relations are quite easily obtained by using Castigliano’s theorem.
where N=Na =number of active coils. Then using Castigliano’s theorem,
The spring rate, also called the scale of the spring, is k = F/y, and so
Since C = D/d
17. Compression Spring Ends
Four end types commonly used in compression springs. (a) Plain; (b) plain and ground; (c) squared; (d) squared and ground.
18. Figure 17.5 shows four types of ends commonly used in
compression springs.
The plain ends are less expensive than squared and
ground ends, for example, but are not as uniformly
loaded and therefore more susceptible to fatigue failure.
Thus, a spring can have an active number of coils that is
different from the total.
It is difficult to identify just how many coils should be
considered end coils, as this can vary with spring index,
solid length, and specific manufacturing parameters.
However, an average number based on experimental
results is used in Table 17.3 and is useful for designers.
Figure 17.5a shows plain ends that have a noninterrupted
helicoid; the spring rates for the ends are the same as if
they were not cut from a longer coil.
Figure 17.5b shows a plain end that has been ground.
In Fig. 17.5c, a spring with plain ends that are squared
(or closed) is obtained by deforming the ends to a 0 ◦
helix angle.
Figure 17.5d shows squared
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20. 18 April 2021
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20
Various lengths and forces applicable to helical compression springs. (a) Unloaded; (b)
under initial load; (c) under operating load; (d) under solid load.
21. Stability
we learned that a column will buckle when the load becomes too large.
Similarly, compression coil springs may buckle when the deflection becomes too
large.
The critical deflection is given by the equation
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MACHINE DESIGN 2
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The quantity λeff) is the effective slenderness ratio and is given by the equation
22. 18 April 2021
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22 The end-condition constant α is given by:
C′1 and C′2 are dimensionless elastic constants defined by the equations
25. Spring Materials
Characteristics of spring materials include
• High Strength
• Low loss coefficient (fractional energy loss per stress-strain cycle)
26. Spring Material Properties
Table 17.2: Coefficients used in Eq. (17.2) for selected spring materials.
Size range Exponent, Constant, Ap
Material in. mm m ksi MPa
Music wirea 0.004-0.250 0.10-6.5 0.146 196 2170
Oil-tempered wireb 0.020-0.500 0.50-12 0.186 149 1880
Hard-drawn wirec 0.028-0.500 0.70-12 0.192 136 1750
Chromium vanadiumd 0.032-0.437 0.80-12 0.167 169 2000
Chromium silicone 0.063-0.375 1.6-10 0.112 202 2000
302 stainless steel 0.013-0.10 0.33-2.5 0.146 169 1867
0.10-0.20 2.5-5 0.263 128 2065
0.20-0.40 5-10 0.478 90 2911
Phosphor-bronzef 0.004-0.022 0.1-0.6 0 145 1000
0.022-0.075 0.6-2 0.028 121 913
0.075-0.30 2-7.5 0.064 110 932
a Surface is smooth and free from defects and has a bright, lustrous f
in
i s
h.
b Surface has a slight heat-treating scale that must be removed before plating.
c Surface is smooth and bright with no visible marks.
d Aircraft-quality tempered wire; can also be obtained annealed.
e Tempered to Rockwell C49 but may also be obtained untempered.
f SAE CA510, tempered to Rockwell B92-B98.
27. Strength of Spring Materials
Coefficients used in the above Equation for five spring materials.
28. Cyclic Loading
28
310 for unpeened springs
465 for peened springs
se
se
S MPa
S MPa
= →
= →
“Goodman Criterion” for ∞-‐life:
29. EXAMPLE 1
An overflow valve, shown in sketch b, has a piston diameter of 15 mm and a slit
length of 5 mm. The spring has mean coil diameter D = 10 mm and wire diameter
d = 2 mm. The valve should open at 1 bar pressure and be totally open at 3 bar
pressure when the spring is fully compressed. Calculate the number of active coils,
the free length, and the pitch of the spring. The shear modulus for the spring
material G = 80 GPa. The spring ends are squared and ground. Determine the
maximum shear stress for this geometry.
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30. SOLUTION
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The piston area is
When the pressure is one bar (0.1 MPa), the valve starts to open.
The force associated with this pressure is (1.767 × 10−4 m2 )(0.1 MPa) = 17.7 N.
The stiffness of the spring
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for squared and ground ends,
We know that at 53.0 N, the spring is at the solid length of 0.0492 m. When this load is removed, the
deflection is δ = P/k = 53. N/7060 N/m = 0.00751 m. Therefore the free length is
the pitch is
