EXAMPLE 13-3ADesign of a Helical Compression Spring for Static Loading: An Alternate ApproachProblemDesign a compression spring for a static load over a known deflec- tion with a factor of safety against yielding at shut height of at least 1.1.Unitsksi 103psi.GivenMinimum forceFinit100 lbf.Maximum forceFwork150 lbf.Working deflection∆y 0.75 in.AssumptionsUse the least expensive, unpeened, cold-drawn spring wire (ASTM A227) since the loads are static.Shear modulusG 11.5 106.psi.SolutionSee Mathcad file EX13-03A.1 We will derive a design equation for this problem that will yield a value for the wire diameter that is a function of two parameters, spring index C and the ratio, α, of the clash allowance to the working deflection. To start, we write the equation for the factor of safety against yielding at shut heightτshutSysNs(a)From the given data we have a desired value for the spring ratekFworkFinit∆yk 66.667lbfin=(b)But, from equations 13.5 and 13.7, the spring rate is given askdG.8 C3.Na.(c)Eliminating k from equations b and c and solving for the number of active coils, Na, we haveNaG∆y.d.8 C3.FworkFinit.(d)Combining equations 13.5, 13.7, and 13.8b, the stress at shut height isEX13-03A.MCD 4/28/2005 - 4:42 PM 1/5τshut8 k.C 0.5().πd2.yshut.(e)where the shut height yshut isyshutywork∆yclashFworkkα∆y.(f)andα∆yclash∆ySubstituting equation f into equation e,τshut8 k.C 0.5().πd2.Fworkkα∆y..(g)From equation 13.3 and Table 13-6, the torsional yield strength of the wire isSysKmA.db.(h)and Km is the reduction factor taken from Table 13-6, expressed as as a decimal fraction.Substituting equations g and h into a and solving for d yields our design equationd8 Ns.C 0.5().Fwork1α().αFinit..πKm.A.12 b(i)Once we choose a material for the wire, the only unknowns in this equation are the parameters C (spring index) and α (clash allowance to working deflection ratio).2 Assume a spring index of 8 and a clash allowance of 15% of the working deflection, thenSpring indexC 8(j)Clash allowance ratioα0.153 From Tables 13-4 and 13-6 for A227 wire we haveA 141.04 ksi.b 0.1822 Km0.60(k)4 Using these values and equation i we can solve for the required wire diameter. In order to compare this solution with Example 13-3, let Ns1.24 EX13-03A.MCD 4/28/2005 - 4:42 PM 2/5d8 Ns.C 0.5().Fwork1α().αFinit..πKm.A.in2.12 bin.d 0.192 in=This is a preferred diameter as given in Table 13-2, so we will accept it. Notice that the term in the large square brackets has units of in2. In order to raise this term to a fractional exponent, we must make it dimensionless by dividing by in2 and then multiplying the result by in.5Calculate the mean coil diameter D from equation 13.5 for d 0.192 in. .Mean coil diameterDCd.D 1.536 in=(l)6 Find the direct shear factor Ks and use it to calculate the shear stress in the coil at the larger force.Direct shear factorKs10.5CKs1.063=(mStress at FworkτworkKs8 Fwork.D.πd3..τwork88.1 ksi=(n)7 Find the ultimate tensile strength of this wire material from equation 13.3 and Table 13-4 and use it to find the torsional yield strength from Table 13-6, assuming that the set has been removed and using the low end of the recommended range.Ultimate tensilestrengthSutAdinb.Sut190.5 ksi=(o)Shear yieldstrengthSysKmSut.Sys114.3 ksi=(p)8 Find the safety factor against yielding at this working deflection from equation 13.14.Safety factor atworking deflectionNsSysτworkNs1.30=(q)9 To achieve the desired spring rate, the number of active coils must satisfy equation 13.7, solving for Na yields:Number of activecoilsNad4G.8 D3.k.Na8.086= Na8(r)EX13-03A.MCD 4/28/2005 - 4:42 PM 3/5Note that we round it to the nearest 1/4 coil as the manufacturing tolerance cannot achieve better than that accuracy. Having rounded the number of active coils, we must now calculate the spring rate using equation 13.7:Correctedspring ratekd4G.8 D3.Na.k 67.38lbfin=(s)10 Assume squared and ground ends making the total number of coils, from Figure 13-9:Total coilsNtNa2 Nt10=(t)10 The shut height can now be determined.Shut heightLsdNt.Ls1.920 in=(u)11 The initial deflection to reach the smaller of the two loads isInitial deflectionyinitFinitkyinit1.484 in=(v)12 Calculate the clash allowance:Clash allowance∆yclashα∆y.∆yclash0.112 in=(w13 The free length (see Figure 13-8) can now be found fromLfLs∆yclash∆y yinitLf4.267 in=(x)14 To check for buckling, two ratios need to be calculated, Lf/D and ymax/Lf.Slenderness ratiosrLfDsr 2.778=(y)Deflection ratioy'yinit∆yLfy' 0.524=Take these two values to Figure 13-14 and find that their coordinates are safely within the zones that are stable against buckling for either end-condition case.15 The inside and outside coil diameters areInside coil diaDiDd Di1.344 in=(z)Outside coil diaDoDd Do1.728 in=16 The smallest hole and largest pin that should be used with this spring areSmallest holeholeminDo0.05D.holemin1.80in=EX13-03A.MCD 4/28/2005 - 4:42 PM 4/5holeminDo0.05Dholemin1.80inLargest pinpinmaxDi0.05 D.pinmax1.27 in=(aa)17 The total weight of the spring isWeight densityρ0.28 lbf.in3.WeightWtπ2d2.D.Nt.ρ.4Wt0.39 lbf=(ab)18 We now have a complete design specification for this A227-wire spring:Wire diameterd 0.192 in=Outside diameterDo1.728 in=Total coilsNt10=ends squared and groundFree lengthLf4.267 in=(ac)EX13-03A.MCD 4/28/2005 - 4:42 PM
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