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 linear springs: k=F/y F nonlinear springs: k k y dF dy Parallel Series 1 ktotal 1 1 1 k1 k 2 k3 ktotal=k1+k2+k3 Types of Springs Helical: Compression Extension Torsion More Springs Washer Springs: Beams: Power springs: Helical Compression Springs d D Lf p Nt diameter of wire mean coil diameter free length pitch Total coils may also need: Do and Di Length Terminology minimum of 10-15% clash allowance Free Length Lf Assembled Length Max Working Load Bottomed Out L L L a m s End Conditions Plain Plain Ground Na= Active Coils Square Square Ground Stresses in Helical Springs F Spring Index 2C 1 max K s , where K s 2C d 3 8FD T F F T F C=D/d Curvature Stress Inner part of spring is a stress concentration (see Chapter 4) Kw includes both the direct shear factor and the stress concentration factor 4C 1 0.615 max K w , where K w 4C 4 C d 3 8FD under static loading, local yielding eliminates stress concentration, so use Ks under dynamic loading, failure happens below Sy: use Ks for mean, Kw for alternating Spring Deflection y 3 8 FD N 4 d G a Spring Rate y 3 8 FD N a 4 d G k=F/y k 4 d G 3 8D N a 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 Sut ultimate tensile strength Figure 13-3 Table 13-4 with Sut=Adb Sys torsional yield strength Table 13-6 – a function of Sut 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 Ks, not Kw (stress concentration relieved) Load Reversal with Springs Shot Peening What type of failure would this be most effective against? What are You Designing? Given Find F, y k, y k F + d, C, D*, Lf*, Na*, clash allowance ()**, material** design variables Such that: Safety factor is > 1 Spring will not buckle Spring will fit in hole, over pin, within vertical space * - often can calculate from given ** - often given/defined Static Spring Flow Chart if GIVEN F,y, then find k; If GIVEN k, y, then find F N a, d, C D, Ks, Kw material strengths material STRESSES DEFLECTION Ns=Sys/ Lf, yshut, Fshut for shut spring if possible if not, for max working load Three things to know: • effect of d • shortcut to finding d • how to check buckling ITERATE? CHECK buckling, Nshut, Di, Do Nshut=Sys/shut Static Design: Wire Diameter max K s 8FD y d 3 8 FD 3 N a d 4G Based on Ns=Ssy/ and above equation for : 1 ( 2b ) 8 N s C 0.5Fwork 1 Finitial d K m A Three things to know: • effect of d • shortcut to finding d • how to check buckling use Table 13-2 to select standard d near calculated d Km=Sys/Sut *maintain units (in. or mm) for A, b **see Example 13-3A on MathCad CD Buckling S .R. Lf D yinit y working y Lf 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 Sus ultimate shear strength Sfw´ torsional fatigue strength Sus0.67 Sut Table 13-7 -- function of Sut, # of cycles repeated, room temp, 50% reliability, no corrosion Sew´ 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 a Sfs 0.5 Sfw Sfw, Sew are for torsional strengths, so von Mises not used C S fs 0.5 B S fwSus A 0.5 Sfw Sus Sus 0.5 S fw m Fatigue Safety Factor a Fi=Fmin Fa=(Fmax-Fmin)/2 Fm=(Fmax+Fmin)/2 Sfs N fs Sa a mload 0.5 Sfw Sa a mgood 0.5 Sfw a,load = a,good at intersection i m Sus m S fs Sus i N fs S fs m i Sus a …on page 828 What are you Designing? Given Find Fmax,Fmin, y k, y k F + d, C, D*, Lf*, Na*, clash allowance ()**, material** design variables 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 * - often can calculate from Given ** - often given/defined Fatigue Spring Design Strategy if GIVEN F,y, then find k; If GIVEN k, y, then find F N a, d, C D, Ks, Kw material strengths STRESSES S fs Sus i N fs S fs m i Sus a DEFLECTION Lf, yshut, Fshut material ITERATE? Two things to know: • shortcut to finding d • how to check frequency CHECK buckling, frequency, Nshut, Di, Do Nshut=Sys/shut 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: 1 ( 2b ) b 8CN fs N fs 1 Ad d K s Fmin 1.34 1 K w Fa K s Fm N fs S fw 0.67A use Table 13-2 to select standard d near calculated d Two things to know: • shortcut to finding d • how to check frequency *maintain units (in. or mm) for A, b **see Example 13-4A on MathCad CD Natural Frequency: Surge Surge == longitudinal resonance for fixed/fixed end conditions: 1 kg fn 2 Wa (Hz) ideally, fn will be at least 13x more than fforcing… it should definitely be multiple times bigger Two things to know: • shortcut to finding d • how to check frequency …see pages 814-815 for more Review of Design Strategy ITERATIVE USING d EQUATION Find Loading Select C, d Find Loading Select C, safety factor Find stresses Determine material properties Find safety factor Solve for d, pick standard d Find stresses Determine material properties Check safety factor Strategy Review Continued Find spring constant, Na, Nt Find FSHUT (must find lengths and y’s to do this) Find static shut shear stress and safety factor Check Buckling Check Surge Check Di, Do 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 max K s 8FD , use K w for a d 3 surge check is same as before Lb=d(Na+1) However, no peening, no setting, no concern about buckling Difference 1: Initial Force force F “preloading” Fi deflection y F=Fi+ky F Fi d 4G k y 8D 3 N a Difference 1a: Deflection 3 8( F Fi ) D N a y 4 d G Difference 2: Initial Stress take initial stress as the average stress between these lines, then find Fi Difference 3: Ends!: Bending a Kb 16DF Kb C1 3 d 4F 2 d 4C12 C1 1 4C1 (C1 1) 2 R1 2 D C d 2d S e ( Sut min ) N fb S e ( mean min ) Sut alt Se S es 0.67 standard end Difference 3a: Ends: Torsion 4C2 1 max K w2 , K w2 4C2 4 d 3 8FD C2=2R2/d pick a value >4 Materials – Same Sys, Sfw, Sew – same for body Sys, Sfw, Sew – see Tables 13-10 and 1311 for ends Sut 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 1 MLw rev , 2 EI Lw length of wire DN a rev,roundwire 10.8 MDN a k d 4E M rev Stresses Compressive is Max – Use for Static – Inside of Coil imax K bi M max c 32M max K bi I d 3 4C 2 C 1 K bi 4C (C 1) For Fatigue – Slightly lower Outside Tensile Stress – Outside of Coil omax K bo 32M max d 3 omin K bo 4C 2 C 1 K bo 4C (C 1) 32M min d 3 Materials see Tables 13-13 and 13-14, page 850 follow book on Sewb=Sew/0.577… for now Strategy Select C, d M K • fit over pin (if there is one) • don’t exceed stresses Helical Springs Compression Nomenclature Stress Deflection and Spring Constant Static Design Fatigue Design Extension Torsion