Outline  Spring Functions & Types  Helical Springs  Compression  Extension  Torsional

The Function(s) of Springs Most fundamentally: to STORE ENERGY Many springs can also: push pull twist

Some Review F y k linear springs: k=F/y nonlinear springs: Parallel k total =k 1 +k 2 +k 3 Series

Types of Springs Helical: Compression Extension Torsion

More Springs Washer Springs: Beams: Power springs:

Helical Compression Springs ddiameter of wire Dmean coil diameter L f free length ppitch N t Total coils may also need: D o and D i

Length Terminology Free Length Assembled Length Max Working Load Bottomed Out minimum of 10-15% clash allowance LfLf LaLa LmLm LsLs

End Conditions Plain Square Plain Ground Square Ground N a = Active Coils

F F F F Stresses in Helical Springs Spring IndexC=D/d F F T T

Curvature Stress  under static loading, local yielding eliminates stress concentration, so use K s  under dynamic loading, failure happens below S y : use K s for mean, K w for alternating Inner part of spring is a stress concentration (see Chapter 4) K w includes both the direct shear factor and the stress concentration factor

Spring Deflection

Spring Rate k=F/y

Helical Springs  Compression  Nomenclature  Stress  Deflection and Spring Constant  Static Design  Fatigue Design  Extension  Torsion

Static Spring Design  Inherently iterative  Some values must be set to calculate stresses, deflections, etc.  Truly Design  there is not one “correct” answer  must synthesize (a little bit) in addition to analyze

Material Properties  S ut ultimate tensile strength  Figure 13-3  Table 13-4 with S ut =Ad b  S ys torsional yield strength  Table 13-6 – a function of S ut and set

Spring/Material Treatments  Setting  overstress material in same direction as applied load »increase static load capacity 45-65% »increase energy storage by 100%  use K s, not K w (stress concentration relieved)  Load Reversal with Springs  Shot Peening  What type of failure would this be most effective against?

What are You Designing? Given F, y k, y Find k F Such that: Safety factor is > 1 Spring will not buckle Spring will fit in hole, over pin, within vertical space d, C, D *, L f *, N a *, clash allowance (  ) **, material ** design variables + * - often can calculate from given ** - often given/defined

Static Spring Flow Chart STRESSES N s =S ys /  for shut spring if possible if not, for max working load if GIVEN F,y, then find k; If GIVEN k, y, then find F D, K s, K w material strengths DEFLECTION L f, y shut, F shut Three things to know: effect of d shortcut to finding d how to check buckling ITERATE? d, C material N a,  CHECK buckling, N shut, D i, D o N shut =S ys /  shut

Static Design: Wire Diameter Three things to know: effect of d shortcut to finding d how to check buckling **see Example 13-3A on MathCad CD *maintain units (in. or mm) for A, b use Table 13-2 to select standard d near calculated d Based on N s =S sy /  and above equation for  : K m =S ys /S ut

Buckling Three things to know: effect of d shortcut to finding d how to check buckling

Helical Springs  Compression  Nomenclature  Stress  Deflection and Spring Constant  Static Design  Fatigue Design  Extension  Torsion

Material Properties  S us ultimate shear strength  S us  0.67 S ut  S fw´ torsional fatigue strength  Table function of S ut, # of cycles  repeated, room temp, 50% reliability, no corrosion  S ew´ torsional endurance limit  for steel, d < 10mm  see page 816 (=45 ksi if unpeened, =67.5 ksi if peened)  repeated, room temp., 50% reliability, no corrosion

Modified Goodman for Springs  S fw, S ew are for torsional strengths, so von Mises not used 0.5 S fw aa mm Repeated S fs C B S us A

Fatigue Safety Factor 0.5 S fw aa mm S fs S us load line SaSa aa ii mm m load m good  a,load =  a,good at intersection …on page 828 F i =F min F a =(F max -F min )/2 F m =(F max +F min )/2

What are you Designing? Given F max,F min,  y k,  y Find k F Such that: Fatigue Safety Factor is > 1 Shut Static Safety Factor is > 1 Spring will not buckle Spring is well below natural frequency Spring will fit in hole, over pin, within vertical space d, C, D *, L f *, N a *, clash allowance (  ) **, material ** design variables + * - often can calculate from Given ** - often given/defined

Fatigue Spring Design Strategy d, C if GIVEN F,y, then find k; If GIVEN k, y, then find F N shut =S ys /  shut CHECK buckling, frequency, N shut, D i, D o D, K s, K w material strengths DEFLECTION L f, y shut, F shut Two things to know: shortcut to finding d how to check frequency ITERATE? material N a,  STRESSES

Fatigue Design:Wire Diameter as before, you can iterate to find d, or you can use an equation derived from relationships that we already know: use Table 13-2 to select standard d near calculated d Two things to know: shortcut to finding d how to check frequency **see Example 13-4A on MathCad CD *maintain units (in. or mm) for A, b

Natural Frequency: Surge Two things to know: shortcut to finding d how to check frequency Surge == longitudinal resonance for fixed/fixed end conditions: (Hz) …see pages for more ideally, f n will be at least 13x more than f forcing … it should definitely be multiple times bigger

Review of Design Strategy Find Loading Select C, d Find stresses Determine material properties Find safety factor Solve for d, pick standard d Find stresses Determine material properties Check safety factor Find Loading Select C, safety factor ITERATIVEUSING d EQUATION

Strategy Review Continued Find spring constant, N a, N t Find F SHUT (must find lengths and y’s to do this) Find static shut shear stress and safety factor Check Buckling Check Surge Check D i, D o if pin to fit over, hole to fit in

Consider the Following:

Helical Springs  Compression  Nomenclature  Stress  Deflection and Spring Constant  Static Design  Fatigue Design  Extension  Torsion

Extension Springs As before, 4 < C < 12 However, no peening, no setting, no concern about buckling surge check is same as before L b =d(N a +1)

Difference 1: Initial Force deflection y force F FiFi “preloading” F=F i +ky

Difference 1a: Deflection

Difference 2: Initial Stress take initial stress as the average stress between these lines, then find F i

Difference 3: Ends!: Bending standard end

Difference 3a: Ends: Torsion C 2 =2R 2 /d pick a value >4

Materials  S ut – Same  S ys, S fw, S ew – same for body  S ys, S fw, S ew – see Tables and for ends

Strategy similar to compression + end stresses - buckling

Helical Springs  Compression  Nomenclature  Stress  Deflection and Spring Constant  Static Design  Fatigue Design  Extension  Torsion

Torsion Springs close-wound, always load to close Deflection & Spring Rate

Stresses Compressive is Max – Use for Static – Inside of Coil For Fatigue – Slightly lower Outside Tensile Stress – Outside of Coil

Materials see Tables and 13-14, page 850 follow book on S ewb =S ew /0.577… for now

Strategy MK  fit over pin (if there is one) don’t exceed stresses Select C, d

Helical Springs  Compression  Nomenclature  Stress  Deflection and Spring Constant  Static Design  Fatigue Design  Extension  Torsion