Mechanical Springs Shigley Ch 10 Lecture 21

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1 Mechanical Springs Shigley Ch 10 Lecture 21

2 Why do we need springs? Flexibility of structure Storing and releasing of energy Mechanical Springs Standard spring types: Wire springs -Helical springs of round or square wire, made to deflect under Compression, tension or torsion -Can be used as a straight piece of wire for stiffer spring types (torsion bar suspensions springs) Flat springs -Cantilever, elliptical, wound clock type (spiral), Flat Spring Washers (usually called Belleville springs), leaf springs, etc Special shape springs also exist for custom cases Most springs can be Linear or non linear, depending on geometry

3 Helical compression spring

4 Flat-leaf spring

5 Belleville spring

6 Torsion spring

7 Stresses in Helical Spring τ max τ = Maximum shear equation Torsion Shear force Tr F = + J A 8FD 4F + 3 πd πd 2

8 We define a term called the Spring index as C = D d C-Usually ranges between 4 and 12 Next define a term K s =shear stress correction factor 2 C +1 K s = 2C Then the stress equation becomes τ = K s 8FD 3 πd

9 Spring design consideration Use round spring wire where possible, if space is limited use nested round springs Springs with round wire are made in large quantities, so for cost avoid square/other shape sections

10 Valve spring in action, showing nested round springs (Grey, red and white)

11 The Curvature effect Helical springs are not straight but curved Taking the curvature into account replace K s with K B (Bergsträsserfactor) K B 4C + 2 = 4 C 3 2 C +1 K s = 2C Thus the largest shear stress is now given by 8FD τ = K B 3 πd

12 Deflection of Helical Springs Strain energy for a helical spring is given by adding the torsional-and shear-energy components U 2 T l 2GJ 2 F l + 2AG Substituting T, L, A and J gives = 4F d D G 2F d DN G Where N=N a is the number of active coils N U = + 4 2

13 From Castigliano s theorem the total deflection Spring rate or also called scale of the spring is given by G d N FD C G d N FD G d FDN G d N FD F U y = + = = D N G d y F k =

14 Compression Springs, end conditions End bent down to form 0 helix angle End ground flat to form flat mounting surface for spring

15 End Conditions may differ check with manufacturer or count and measure them

16 Stability As long columns, long springs can buckle with high loads 1 Where λ = eff α D L o y cr 1 Effective slenderness ratio α End condition constant from table 10-2 ' 2 = ' C 2 LoC1 1 2 λeff C ' 1 = E 2( E G) C ' 2 2π 2 ( E G) = 2 G+ E Elastic constants

17 Table 10-2 End condition Spring supported between flat parallel surfaces One end on flat surface perpendicular to axis (fixed) other end hinged Both ends hinged 1 One end clamped other end free 2 Constant α

18 Absolute stability is given by L 0 < πd α 2( E 2G + G E ) 1 2 For steels L0 < D α Squared and ground ends α = 0.5 thus L < D

19 Spring Materials Hot-or cold-working processes, depending on size, spring index and properties desired In general Pre Hardenedmaterial should not to be used if D/d < 4 or d > 6mm Table 10-3 shows most commonly used spring materials

20 Table 10-3

21 Spring Materials Graph of tensile strength vs. wire diameter is a straight line when plotted on log-log paper using the following equation: S = ut d A m Table 10-4 gives values for m and A

22 Table 10-4

23 Spring Calculations Torsional yield strength is needed in designing springs but because tensile strength is easy to determine the torsional yield strength is usually estimated from that From DET the torsional yield strength is given by S = sy S y The yield strength is between 60 and 90 percent of the ultimate tensile strength for steel, this approach results in the range 0.35 S S ut sy S ut Table 10-5 gives the range of S sy and table 10-6 gives the maximum percentage of tensile strength

24 Table 10-5

25 Set removal or presetting Make spring longer than required, close/press to solid length to introduce residual stresses (10 to 30 percent of length is reduced this way) Stress at solid height is between 1.1 and 1.3 the torsional yield strength Make spring stronger by inducing residual stresses opposite those induced in service Not good for fatigue applications

Chapter 5: Mechanical Springs

ENT 345 MECHANICAL COMPONENT DESIGN Chapter 5: Mechanical Springs IR. DR. KHAIRUL SALLEH BASARUDDIN School of Mechatronic Engineering Universiti Malaysia Perlis Email : khsalleh@unimap.edu.my http://khairulsalleh.unimap.edu.my/