1ME311 Machine DesignW Dornfeld14Nov2013Fairfield UniversitySchool of EngineeringLecture 10: Springs(Chapter 17)Compression SpringsHamrock Page 445A Free Body Diagram of a coil spring (cutting through anywhere on the coil) shows that there must be torsion on the coil to balance the load.Coil springs have these features:Wire diameter, d or dwireCoil Mean diameter, D or DmeanCoil Inner Diameter, ID = D – dCoil Outer Diameter, OD = D + dSpring Index, dDC =You can think of the OD as the Mean Diameter plus twice the wire radius, andD + 2r = D + d2Spring IndexHamrock Eqn. 17.7It is seldom practical to make a spring with an Index, C, less than 3 or greater than 12.A small Index means a large curvature, and a large index means a small curvature.C = 3C = 12Springs with C in the range of 5 to 10 are preferred. Springs with small C are hard to manufacture, and have large stress concentrations due to the tight curvature.Stress In SpringsHamrock Page 445+=34max,82)32(dPDdTdJTrtorsππτ===2max,4dPAPdirπτ==+=DddDPtot2183max,πτPD/2PTorsional ShearDirect ShearTotal Shear3Stress In SpringsHamrock Page 445/6The term in parentheses is a constant, so this can be rewritten:+=DddDPtot2183max,πτCCCDdKd5.021121+=+=+=Torsion DirectIf the Spring Index, C, ranges from 3 to 12, then the Direct Shear is from 1/6thto 1/24thof the Torsional Shear.3max,8dPDKdtotπτ=[Eqn. 17.8]Use this for STATICloading.Where Kd, the Transverse Shear Factor, isSpring Stress ExerciseA spring made from 0.1 in. music wire has an outside diameter of 1 inch. If it has a load of 25 Lbs applied to it, what is the maximum shear stress?4Curvature EffectsHamrock Page 446CCCKw615.04414+−−=CurvatureDirectFor cyclic (dynamic) loading, we include this curvature effect, and write:where3max,8dPKDwtotπτ=[Eqn. 17.10]Use this for CYCLICloading.Adding the effect of curvature drives up the stress at the InnerRadius.Remember curved beams & their stress distribution?Shear FactorsKdKwKb0.80.91.01.11.21.31.41.51.61.73 4 5 6 7 8 9 10 11 12Spring Index, CK3424−+=CCKbAnother Cyclic correction factor is the Bergsträsser factor, KbIt is simpler and very close to KwClearly, the lower the Index, the higher the curvature, and the higher the max shear.5Spring MaterialsmputdAS =There are a very limited number of materials commonly used for making springs, listed in Table 17.1.The allowable shear yield of these materials Ssy = 0.40 Sut, and the Sut varies with wire size!The variation is:where Ap and m come from Table 17.2.Caution: Use Ap in psi with d in inches, and Ap in MPa with d in mm.A plot of Sut versus wire diameter for the material in Table 17.2 is shown on the next two slides, first in Metric and then in English units.Hamrock Page 442[Eqn. 17.3]How much factor of safety did our spring have?[Eqn. 17.2]Min. Ultimate Tensile Strengths of Common Spring WiresMusic WireMusic WireChrome SiliconChrome VanadiumOil-TemperedHard Drawn1000150020002500300035000.1 1.0 10.0 100.0Wire Diameter (mm)Sut (MPa)From Hamrock, Table 17.2 & Eqn. 17.26Min. Ultimate Tensile Strengths of Common Spring WiresMusic WireChrome SiliconChrome VanadiumOil-TemperedHard Drawn1001502002503003504000.010 0.100 1.000Wire Diameter (in)Sut (KSI)From Hamrock, Table 17.2 & Eqn. 17.2Spring DeflectionCastigliano’s theorem gives spring deflection as+=235.018CGdNPCaδ[Eqn. 17.17]Because C is usually between 3 and 12, the second term would be between 0.056 and 0.0035, and the equation is often shortened toGdNPCa38=δ[Eqn. 17.15]thereby ignoring between 0.35% and 5.6% of the deflection.I suggest that you generally use Eqn. 17.17, dropping the second term only if C is large.In both equations, P is the applied load on the spring, G is thematerial Shear Modulus , and Na is the number of active coils.)1(2ν+=E7Active CoilsThe number of active coils depends on how the ends of the coils are finished.[~Table 17.3](Lfree– 2d)/Nad x NtotNtot- 22Squared & Ground(Lfree– 3d)/Nad(Ntot+ 1)Ntot- 22SquaredLfree/(Na+1)d x NtotNtot- 11Plain & Ground(Lfree– d)/Nad(Ntot+ 1)Ntot0PlainPitch*LsolidNactiveNendEnds* Pitch is ONLY measured when spring is UNLOADED! (L = Lfree)PlainPlain & GroundSquared & GroundSquaredSpring StiffnessThe stiffness, k, is the force per deflection. )5.01(823CNCGdka+=[Eqn. 17.19][Eqn. 17.18]If C is large, this can be reduced toaaNDGdNCGdk34388==Note: k, δ, and τ are functions of d, D, Na, G, and P, but NOT of pitch and therefore not Lfree.8Spring ExampleMy Garage Door spring (yes, it’s an extension spring, but it is close enough).dwire= 0.155 in.OD = 1.400 in.Ntot= 160 turnslblkForceinlbNCGdka3.97)3.265.62)(687.2(/687.2)160()032.8(8)155.0)(105.11(8363=−=∆==×==Lfree= 26.3 in.Lextended= 62.5 in.G = 11.5 x 106psi032.8155.0245.1.245.1155.0400.1====−=−=dDCindODDmean0078.051.645.0032.85.05.022===CSo we missed <1%ksidPKDCCCKww97.87)155.0()3.97)(183.1)(245.1(88183.1615.0441433====+−−=ππτSpring Force and DeflectionAt Free LengthAt Solid LengthP = k x ∆LP = k x ∆LA change in spring length is always accompanied by a change in spring force = k x ∆L, up until the spring bottoms out.P = 0Two things to (almost) always count on.9Spring BucklingLike all long, skinny things with a load on them, springs can buckle.Buckling is related to the Free Length of the spring, and to the End Conditions.To make it stable, you can “guide” it on the inside or outside.Spring VibrationSprings can vibrate longitudinally (or surge) just like a Slinky:The frequency isHzGDdNfanρπ3222=Here G is the Shear Modulus, and ρis the mass density (or weight density divided by g).To avoid resonances, avoid cyclic loading a spring near integral multiples of fn.For steel springs where G and ρare constants, this can be simplified to:)(000,353)(900,1322mminDanddHzNDdfinchesinDanddHzNDdfnn==[Eqn. 17.20]For the garage door spring, fn=8.7Hz.10Fatigue / Cyclic Loading of Helical Springs• Helical springs are NEVER used as both compression and extensionsprings (Hamrock, top of Section 17.3.7).• Therefore, loading is never fully reversing, so we will use the modified Goodman diagram instead of an S-N plot.1. Get the steady (mean) and alternating loads, Pmeanand Palt.2. Compute the mean and alternating shears, using KWahlfor BOTH:3. FOS against yielding:4. FOS against fatigue (Infinite life):38dPDKmeanWmeanπτ=max4.0τττUTmeanaltSysSSn =+=altSEsSnτ=38dPDKaltWaltπτ=Design