CHAPTER 12 SPRINGS Summary Close-coiled springs (a)Under axial load w Maximum shear stress set up in the material of the spring 2WR 8WD =Tmax= nd3 Total deflection of the spring for n turns 4WR3n 8WD3n =6= Gr4= Gd4 where r is the radius of the wire and R the mean radius of the spring coils. W Gd4 i.e. Spring rate=δ-8nD (b)Under axial torque T 4T32T Maximum bending stress set up omax nd3 8TRn 64TDn Wind-up angle =0= Er4 Ed4 nEd4 .∴.Torque per turn= 0/2π32Dn The stress formulae given in (a)and(b)may be modified in practice by the addition of'Wahl' correction factors. Open-coiled springs (a)Under axial load W Deflection 6=2nnWR3 sec a cos2a sin2a GJ+EI Angular rotation 2mnWR2 sina [11 297
CHAPTER 12 SPRINGS Summary Close-coiled springs (a) Under axial load W Maximum shear stress set up in the material of the spring 2WR 8WD - Tmax= __ = - xr3 xd3 - Total deflection of the spring for n turns 4WR3n 8WD3n Gr4 Gd4 - -6=- =- where r is the radius of the wire and R the mean radius of the spring coils. i.e. W Gd4 Spring rate = - = ~ 6 8nD3 (b) Under axial torque T 4T 32T Maximum bending stress set up = omax = - = __ xr3 xd3 8TRn 64TDn Wind-up angle = e = - E ___ Er4 Ed4 T xEd4 0/2x 32Dn :. Torque per turn = ~ - ~ The stress formulae given in (a) and (b) may be modified in practice by the addition of ‘Wahl’ correction factors. Open-coiled springs (a) Under axial load W cosza sin’a Deflection 6 = 2xn WR3 sec a Angular rotation 0 = 2xn WRz sin a [t - - - :I] 297
298 Mechanics of Materials (b)Under axial torque T sin2a cos2a Wind-up angle 0 2mnRT sec a GJ where a is the helix angle of the spring. 1 Axial deflection 8=2nnTR2 sina GJ EI Springs in series Stiffness S= SS2 (S1+S2) Springs in parallel Stiffness S=S+S2 Leaf or carriage springs (a)Semi-elliptic Under a central load W: 3WL maximum bending stress 2nbt2 3WL3 deflectionδ= 8Enbt3 where L is the length of spring,b is the breadth of each plate,t is the thickness of each plate, and n is the number of plates. 8Enbt3 Proof load W,33 where 6 is the initial central "deflection". 4tE Proof or limiting stress (b)Quarter-elliptic 6WL Maximum bending stress nbt2 6WL3 Deflection 6 Enbt3
298 Mechanics of Materials (b) Under axial torque T sin’a cos2a Wind-up angle 8 = 2mRT sec a where a is the helix angle of the spring. Axial deflection 6 = 2nnTR’ sina - -- [;.l Ell] Springs in series Springs in parallel Stiffness S = SI + S, Leaf or carriage springs (a) Semi-elliptic Under a central load W 3 WL maximum bending stress = - 2nbtz 3 WL3 deflection 6 = - 8Enbt3 where L is the length of spring, b is the breadth of each plate, t is the thickness of each plate, and n is the number of plates. 8Enbt3 Proof load Wp = ~ 3L3 where 6, is the initial central “deflection”. 4tE Proof or limiting stress 0, = Lz6p (b) Quarter-elliptic 6WL Maximum bending stress = - nbt’ 6 WL3 Deflection 6 = - Enbt3
§12.1 Springs 299 Plane spiral springs 6Ma Maximum bending stress RBt2 or,assuming a =2R, 12M maximum bending stress Bt2 ML wind-up angle 0= EI where M is the applied moment to the spring spindle,R is the radius of spring from spindle to pin,a is the maximum dimension of the spring from the pin,B is the breadth of the material of the spring,t is the thickness of the material of the spring,L is equal to(mn)(a+b),and b is the diameter of the spindle. Introduction Springs are energy-absorbing units whose function it is to store energy and to release it slowly or rapidly depending on the particular application.In motor vehicle applications the springs act as buffers between the vehicle itself and the external forces applied through the wheels by uneven road conditions.In such cases the shock loads are converted into strain energy of the spring and the resulting effect on the vehicle body is much reduced.In some cases springs are merely used as positioning devices whose function it is to return mechanisms to their original positions after some external force has been removed. From a design point of view "good"springs store and release energy but do not significantly absorb it.Should they do so then they will be prone to failure. Throughout this chapter reference will be made to strain energy formulae derived in Chapter 11 and it is suggested that the reader should become familiar with the equations involved. 12.1.Close-coiled helical spring subjected to axial load W (a)Maximum stress A close-coiled helical spring is,as the name suggests,constructed from wire in the form of a helix,each turn being so close to the adjacent turn that,for the purposes of derivation of formulae,the helix angle is considered to be so small that it may be neglected,i.e.each turn may be considered to lie in a horizontal plane if the central axis of the spring is vertical. Discussion throughout the subsequent section on both close-coiled and open-coiled springs will be limited to those constructed from wire of circular cross-section and of constant coil diameter. Consider,therefore,one half-turn of a close-coiled helical spring shown in Fig.12.1.Every cross-section will be subjected to a torque WR tending to twist the section,a bending moment tending to alter the curvature of the coils and a shear force w.Stresses set up owing to the shear force are usually insignificant and with close-coiled springs the bending stresses
412.1 Springs 299 Plane spiral springs 6Ma Maximum bending stress = - RBt2 or, assuming a = 2R, 12M maximum bending stress = - Bt2 ML wind-up angle 8 = - EZ where M is the applied moment to the spring spindle, R is the radius of spring from spindle to pin, a is the maximum dimension of the spring from the pin, B is the breadth of the material of the spring, t is the thickness of the material of the spring, L is equal to $ (m) (a + b), and b is the diameter of the spindle. Introduction Springs are energy-absorbing units whose function it is to store energy and to release it slowly or rapidly depending on the particular application. In motor vehicle applications the springs act as buffers between the vehicle itself and the external forces applied through the wheels by uneven road conditions. In such cases the shock loads are converted into strain energy of the spring and the resulting effect on the vehicle body is much reduced. In some cases springs are merely used as positioning devices whose function it is to return mechanisms to their original positions after some external force has been removed. From a design point of view “good springs store and release energy but do not significantly absorb it. Should they do so then they will be prone to failure. Throughout this chapter reference will be made to strain energy formulae derived in Chapter 11 and it is suggested that the reader should become familiar with the equations involved. 12.1. Close-coiled helical spring subjected to axial load W (a) Maximum stress A close-coiled helical spring is, as the name suggests, constructed from wire in the form of a helix, each turn being so close to the adjacent turn that, for the purposes of derivation of formulae, the helix angle is considered to be so small that it may be neglected, i.e. each turn may be considered to lie in a horizontal plane if the central axis of the spring is vertical. Discussion throughout the subsequent section on both close-coiled and open-coiled springs will be limited to those constructed from wire of circular cross-section and of constant coil diameter. Consider, therefore, one half-turn of a closecoiled helical spring shown in Fig. 12.1. Every cross-section will be subjected to a torque WR tending to twist the section, a bending moment tending to alter the curvature of the coils and a shear force W. Stresses set up owing to the shear force are usually insignificant and with close-coiled springs the bending stresses
300 Mechanics of Materials §12.2 R W Fig.12.1.Close-coiled helical spring subjected to axial load W. are found to be negligible compared with the torsional stresses.Thus the maximum stress in the spring material may be determined to a good approximation using the torsion theory. Tr WRr tma=万-元r产2 2WR 8WD i.e. maximum stress 元r3 nd3 (12.1) (b)Deflection Again,for one half-turn,if one cross-section twists through an angle 6 relative to the other, then from the torsion theory 0= TL_WR(πR)22WR2 G -X- G Gr4 2WR3 But 8=R0=1 Gr total deflection =2n= 4WR3n 8WD3n Gr4 Gd (12.2) W Gd Spring rate=57=8nD3 12.2.Close-coiled helical spring subjected to axial torque T (a)Maximum stress In this case the material of the spring is subjected to pure bending which tends to reduce the radius R of the coils(Fig.12.2).The bending moment is constant throughout the spring and equal to the applied axial torque T.The maximum stress may thus be determined from the bending theory My Tr O max= 1πr4/4
300 Mechanics of Materials Q 12.2 i w Fig. 12.1. Closecoiled helical spring subjected to axial load W. are found to be negligible compared with the torsional stresses. Thus the maximum stress in the spring material may be determined to a good approximation using the torsion theory. Tr WRr Tmax= - = __ J xr412 i.e. 2WR 8WD maximum stress = - = - zr3 nd3 (12.1) (b) Dejection Again, for one half-turn, if one cross-section twists through an angle 8 relative to the other, then from the torsion theory TL WR(nR) 2 2WR2 e=-= x-=- GJ G xr4 Gr4 But .. 4WR'n - 8WD3n total deflection 6 = 2nd' = ~ - ~ Gr4 Gd4 (1 2.2) W Gd4 Spring rate = - = - 6' 0n~3 12.2. Close-coiled helical spring subjected to axial torque T (a) Maximum stress In this case the material of the spring is subjected to pure bending which tends to reduce the radius R of the coils (Fig. 12.2). The bending moment is constant throughout the spring and equal to the applied axial torque T. The maximum stress may thus be determined from the bending theory
§12.3 Springs 301 4T32T i.e. maximum bending stress= nd3 (12.3) Fig.12.2.Close-coiled helical spring subjected to axial torque T. (b)Deflection (wind-up angle) Under the action of an axial torque the deflection of the spring becomes the "wind-up angle"of the spring,i.e.the angle through which one end turns relative to the other.This will be equal to the total change of slope along the wire,which,according to Mohr's area-moment theorem (see $5.7),is the area of the M/EI diagram between the ends. 0= MdL TL EI EI 0 where L total length of the wire =2Rn. 0=T2πRm、4 4 8TRn i.e. wind-up angle 0 = Er4 (12.4) N.B.The stress formulae derived above are slightly inaccurate in practice,particularly for small D/d ratios,since they ignore the higher stress produced on the inside of the coil due to the high curvature of the wire."Wahl"correction factors are therefore introduced-see page 307. 12.3.Open-coiled helical spring subjected to axial load W (a)Deflection In an open-coiled spring the coils are no longer so close together that the effect of the helix angle a can be neglected and the spring is subjected to comparable bending and twisting effects.The axial load Wcan now be considered as a direct load Wacting on the spring at the mean radius R,together with a couple WR about AB(Fig.12.3).This couple has a component about AX of WR cos a tending to twist the section,and a component about AY
912.3 Springs i.e. 4T 32T maximum bending stress = ~ - __ nr3 - nd3 301 (12.3) Fig. 12.2. Closecoiled helical spring subjected to axial torque T. (b) Defection (wind-up angle) Under the action of an axial torque the deflection of the spring becomes the “wind-up angle” of the spring, i.e. the angle through which one end turns relative to the other. This will be equal to the total change of slope along the wire, which, according to Mohr’s area-moment theorem (see 9 5.7), is the area of the M/EZ diagram between the ends. L .. e=jF== MdL TL 0 where L = total length of the wire = 2nRn. .. e=- XE nr4 T2nRn 4 i.e. 8T Rn wind-up angle 8 = ~ Er4 (12.4) N.B. The stress formulae derived above are slightly inaccurate in practice, particularly for small D/d ratios, since they ignore the higher stress produced on the inside of the coil due to the high curvature of the wire. “Wahl” correction factors are therefore introduced - see page 307. 12.3. Open-coiled helical spring subjected to axial load W (a) Defection In an opencoiled spring the coils are no longer so close together that the effect of the helix angle a can be neglected and the spring is subjected to comparable bending and twisting effects. The axial load Wcan now be considered as a direct load Wacting on the spring at the mean radius R, together with a couple WR about AB (Fig. 12.3). This couple has a component about AX of WR cos a tending to twist the section, and a component about AY
302 Mechanics of Materials §12.3 M-WR sina A a Wsin a x a Wcos a T'=WR cos a Fig.12.3.Open-coiled helical spring. of WR sin a tending to reduce the curvature of the coils,i.e.a bending effect.Once again the shearing effect of W across the spring section is neglected as being very small in comparison with the other effects. Thus T'=WR cosa and M'=WR sina Now,the total strain energy,neglecting shear, T2L M2L U=2GJ+2EI (see $11.3 and 11.4) =L(WR cosa)L(WR sina) 2GJ 2EI (12.5) and this must equal the total work done W. ww[2+] From the helix form of Fig.12.4 2πRn=Lcosa L=2πRn sec a cos2a sin2a deflection 8=2an WR3 seca G+ (12.6)
302 Mechanics of Materials $12.3 1’ W Fig. 12.3. Opencoiled helical spring. of WR sin u tending to reduce the curvature of the coils, i.e. a bending effect. Once again the shearing effect of W across the spring section is neglected as being very small in comparison with the other effects. Thus T’ = WRcosa and M = WRsina Now, the total strain energy, neglecting shear, u=-- *’ +- M2 (see $5 11.3 and 11.4) 2GJ 2EI L ( WR cos a)’ 2GJ L ( WR sin a)2 - -I 2EI LW‘R’ cosza sin’a - --[-+-I 2 GJ EI and this must equal the total work done $ W6. .. 2 [ GJ +I] From the helix form of Fig. 12.4 LW~R’ cos2u sin’u $WS=- __ 2nRn = L cos a .. L = 2nRn sec u [ cos’a ~ sin2a] .. deflection S = 2an WR3 seca - GJ EI (12.5) (12.6)
§12.3 Springs 303 ◆-2TRn=Lc05a Fig12.4. Since the stiffness of a spring S is normally defined as the value of W required to produce unit deflection, stiffness W S=W=2πnR3seca 16 cos2a sin2a GJ+ 十 (12.7) EI Alternatively,the deflection in the direction of w is given by Castigliano's theorem (see§11.11)as aU 8= =LWR2 cos2 a sin2 GJ +EI and with L=2πRn sec a 6=2πnWR3sec [+] (12.8) This is the same equation as obtained previously and illustrates the flexibility and ease of application of Castigliano's energy theorem. (b)Maximum stress The material of the spring is subjected to combined bending and torsion,the maximum stresses in each mode of loading being determined from the appropriate theory. From the bending theory My 0= with M=WR sin a and from the torsion theory Tr with T=WR cosa The principal stresses at any point can then be obtained analytically or graphically using the procedures described in 13.4. (c)Angular rotation Consider an imaginary axial torque Tapplied to the spring,together with W producing an angular rotation of one end of the spring relative to the other
$12.3 Springs 303 +2nRn=Lcos a ---/ Fig. 12.4. . .. Since the stiffness of a spring S is normally defined as the value of --’ w requirea to proauce W stiffness S = - 6 unit deflection, .. - 16 = - = 2nnR3 sec a [‘e+-] sin2 a sw GJ EZ Alternatively, the deflection in the direction of W is given by (see $ 11.11) as au a LW~R~ cos2a sin’a 6=--=-. ___ aw aw[ 2 (r+r)] = LWRz[--Gj-+y] cosza sin’u and with L = 2nRn sec CI cos’a sin’cc 6 = 2nn WR3 sec a (1 2.7) Castigliano’s theorem (12.8) This is the same equation as obtained previously and illustrates the flexibility and ease of application of Castigliano’s energy theorem. (b) Maximum stress The material of the spring is subjected to combined bending and torsion, the maximum From the bending theory stresses in each mode of loading being determined from the appropriate theory. MY I 0 = - with M = WRsina and from the torsion theory Tr J 5 = - with T = WRcosa The principal stresses at any point can then be obtained analytically or graphically using the procedures described in Q 13.4. (c) Angular rotation Consider an imaginary axial torque Tapplied to the spring, together with W producing an angular rotation 8 of one end of the spring relative to the other
304 Mechanics of Materials §12.4 The combined twisting moment on the spring cross-section is then T=WR cosa+Tsin a and the combined bending moment M=Tcos a-WR sin a The total strain energy of the system is then T2L M2L U=2GJ2EI (WR cosa+Tsin )L (Tcosa-WR sin)L 2GJ 2EI Now from Castigliano's theorem the angle of twist in the direction of the axial torque T is aU given bynd sinceTall terms includingTmay be ord. 02WR cos asinaL(-2WR sin acosa)L 2GJ 2EI 「11 =WRLcosa sina 0=2πnWR2sina 117 i.e. GJEI」 (12.9) 12.4.Open-coiled helical spring subjected to axial torque T (a)Wind-up angle When an axial torque T is applied to an open-coiled helical spring it has components as shown in Fig.12.5,i.e.a torsional component T sin g about AX and a flexural (bending) component T cos a about AY,the latter tending to increase the curvature of the coils. ①Tcos a Tsin a Fig.12.5.Open-coiled helical spring subjected to axial torque T
304 Mechanics of Materials $12.4 The combined twisting moment on the spring cross-section is then - T= WRcosu+Tsinu and the combined bending moment M =Tcosu- WRsinu The total strain energy of the system is then - - T~L M~L u=-- +- 2GJ 2EI ( WR cos u + Tsin u)’L 2GJ (Tcos u - WR sin u)’ L 2EI + - Now from Castigliano’s theorem the angle of twist in the direction of the axial torque T is given by 0 = - and since T = 0 all terms including T may be ignored. au aT .. i.e. 2WRcosusinuL (-2WRsinucosa)L 2EI + 2GJ e= = WRLcosusinu --- [iJ ;I] 0 = 2xnWR’sina [A A] ( 12.9) 12.4. Open-coiled helical spring subjected to axial torque T (a) Wind-up angle When an axial torque Tis applied to an open-coiled helical spring it has components as shown in Fig. 12.5, i.e. a torsional component T sin u about AX and a flexural (bending) component T cos u about AY the latter tending to increase the curvature of the coils. Fig, 12.5. Opencoiled helical spring subjected to axial torque T
s12.5 Springs 305 As for the close-coiled spring the total strain energy is given by strain energy U=GJ2EI T2L M2L =L(T'sin2+Tcosa)‖ GJ T2L「sin2x,cos2x 2GJ EI (12.10) and this is equal to the work done by T,namely,T0,where 0 is the angle turned through by one end relative to the other,i.e.the wind-up angle of the spring. 70=T2L and,with L =2Rn sec a as before, wind-up angle 0=2mnRT sec a sin2 a cos2a GJ+EI (12.11) (b)Maximum stress The maximum stress in the spring material will be found by the procedure outlined in 12.3(b)with a bending moment of T'cos a and a torque of T sin a applied to the section. (c）Axial defection Assuming an imaginary axial load W applied to the spring the total strain energy is given by eqn.(11.5)as U=(WRcos+Tsin a)L (Tcosa-WRsina)L 2GJ 2EI Now from Castigliano's theorem the deflection in the direction of W is given by 6= aU ow 「1 =TRLcos a sin GJEI when W=0 deflection 8=2anTR2 sin a 11 GJ EI (12.12) 12.5.Springs in series If two springs of different stiffness are joined end-on and carry a common load W,they are said to be connected in series and the combined stiffness and deflection are given by the following equations
412.5 Springs 305 As for the close-coiled spring the total strain energy is given by T~L M~L strain energy U = __ +- 2GJ 2EI (1 2.10) 1 (Tsinu)’ (TCOS~)~ El + and this is equal to the work done by T, namely, 4 TQ, where 6 is the angle turned through by one end relative to the other, i.e. the wind-up angle of the spring. iTQ =fT2L[7+EI] sin’a cos2 a and, with L = 2zRn sec a as before, sinZa cos’a wind-up angle 0 = 2anRTsec a (12.11) (b) Maximum stress The maximum stress in the spring material will be found by the procedure outlined in tj 12.3(b) with a bending moment of Tcos u and a torque of T sin a applied to the section. (c) Axial deflection Assuming an imaginary axial load W applied to the spring the total strain energy is given by eqn. (1 1.5) as ( WR cos a + Tsin u)’L 2GJ (Tcos u - WR sin a)’ L 2EI U= + Now from Castigliano’s theorem the deflection in the direction of W is given by au aw a=-- =TRLcosusina [jJ --- d,] when W=O deflection 6 = 2xnTR’sin a - - - [iJ A] 12.5. Springs in series (1 2.12) If two springs of different stiffness are joined end-on and carry a common load W, they are said to be connected in series and the combined stiffness and deflection are given by the following equations
306 Mechanics of Materials 12.6 W WW Deflection =+a 117 -W Ls+s」 (12.13) 11.1 ss+52 and stiffness S= SiS2 (12.14) S1+S2 12.6.Springs in parallel If two springs are joined in such a way that they have a common deflection o they are said to be connected in parallel.In this case the load carried is shared between the two springs and total load W=W1+W2 (1) Now wW W2 8= S-S S (12.15) so that =Sw S2W S and W2= Substituting in eqn.(1) W=SW S2W S+ S W「 9+, i.e. combined stiffness S=S+S2 (12.16) 12.7.Limitations of the simple theory Whilst the simple torsion theory can be applied successfully to bars with small curvature without significant error the theory becomes progressively more inappropriate as the curvatures increase and become high as in most helical springs.The stress and deflection equations derived in the preceding sections,are,therefore,slightly inaccurate in practice, particularly for small D/d ratios.For accurate assessment of stresses and deflections account should be taken of the infuence of curvature and slope by applying factors due to Wahlt and Ancker and Goodiert.These are discussed in Roark and Young$where the more accurate tA.M.Wahl,Mechanical Springs,2nd edn.(McGraw-Hill,New York 1963). C.J.Ancker (Jr)and J.N.Goodier,"Pitch and curvature correction for helical springs",ASME J.Appl.Mech.. 25(4).Dec.1958. $R.J.Roark and W.C.Young.Formulas for Stress and Strain,5th edn.(McGraw-Hill,Kogakusha,1965)
306 Mechanics of Materials $12.6 .. and W ww Deflection = - = 6, +6, = -+- S Sl s2 =w -+- a 111 _- +- s-s, s, SlS, Sl + sz stiffness S = ~ (12.1 3) (1 2.14) 12.6. Springs in parallel If two springs are joined in such a way that they have a common deflection 6 they are said to be connected in parallel. In this case the load carried is shared between the two springs and total load W = W, + W, (1) Now so that Substituting in eqn. (1) i.e. SlW and W, = - s2w S w, =- S s,w s2w W=-- +- S S =‘Y[s,+s,] S combined stiffness S = S, + Sz (12.15) (1 2.16) 12.7. Limitations of the simple theory Whilst the simple torsion theory can be applied successfully to bars with small curvature without significant error the theory becomes progressively more inappropriate as the curvatures increase and become high as in most helical springs. The stress and deflection equations derived in the preceding sections, are, therefore, slightly inaccurate in practice, particularly for small D/d ratios. For accurate assessment of stresses and deflections account should be taken of the influence of curvature and slope by applying factors due to Wahlt and Ancker and GoodierS. These are discussed in Roark and Young§ where the more accurate t A. M. Wahl, Mechanical Springs, 2nd edn. (McGraw-Hill, New York 1963). $ C. J. Ancker (Jr) and J. N. Goodier, “Pitch and curvature correction for helical springs”, ASME J, Appl. Mech., 25(4), Dec. 1958. R. J. Roark and W. C. Young, Formulasfor Stress and Strain, 5th edn. (McGraw-Hill, Kogakusha, 1965)
