6. SPRINGS In Mechanics of Deformable Bodies (MDB), springs are usually introduced as an extension topic for Torsion, However, out of the numerous kinds of springs, most discussion and problems are concentrated on helical springs only. ‘In Machine Design I (MD1), discussion about springs is more extensive since design considerations are given more emphasis. Analysis of other kinds of springs include ‘the following: extension springs with binding force, helical springs made of rectangular wires, torsion springs, leaf springs, ete. ‘The first part of the discussion presented here is an excerpt from the pre-requisite subject, Mechanics of Deformable Bodies.This is intended to easily recall the principles underlying the derivation of formulas, particularly about stresses and deformations of springs under the action of loads. A. Helical Springs of Round Wire A belical spring is a good example of a machine ‘component that serves its purpose through considerable deformations, While being deformed axially, the spring wire is subjected to a combination of shear stresses, the seater being a torsional stress on the spring wire. axial load of spring spring wire diameter outside coil diameter inside col diameter mean coil diameter Dy = Do ~ Dy: Dy = Dy + Dy @o+ Dd ae pitch of spring pitch angle Note: All dimensions are measured when F = 0, Machine Design 1: BASAEN, RV Shear Stress Formula, $, spring cross section under “consideration Note: ‘The spring wire is subjected to two shear stresses: torsional shear $,, and direct shear $5 + Torsional Stress in the Spring S,, = Torsional shear stress And, = Qo anes 5 COG) Oy aes Since the ratio of the diameters of spring is commonly used 2, c= (spring index) Then, s,- BFE a Dy (torsional stress in the spring) ‘© Direct Shear Stress in the Spring areas i aries Rare Pw" ar Sa= p57 Dy (Girect shear stress in the spring) 65Resultant Shear Stress in the Spring ‘The shear stresses have a maximum resultant at point ‘A’, xis of sing ‘The total shear stress is then Se= Sut Seo Cbining the wo sia snes inthe spring wicse mimosa cote at “A he Steeratesisthen eon oti eto The vets lie shoe tay calmer en Ge ane Seo th seo col meet te ging is Stpoit Aa shown Dy , AF eeaaeeeee Dy? Dy” ote = aD, Ut 2D, eee = Del * 3e) but : [2+ Ah: cacrtr dicts Ths, axe nD,? wots ‘The above shear stress equation already includes the two shear stresses. However, considering the effect of the curvature of the spring Coil, another factor K, should be included, This is also known as the stress concentration of the spring due to the curvature of the ‘Thus, the corrected total shear stress is then Si RRS 1D,’ where, K, ~ factor for direct shear Ke = factor for eurvature of the coil ‘The two fictors when combined is known as the Wahl’s factor 'K” K=K,K.: Wahl'sfactor) Meche Des NARY In terms of the spring index, C, the Wahl’s factor is defined as K ciety, 2618) ac-4* ¢ ‘Thus, the maximum shear stress in the spring wire i, (exact equation of shear stress) Use this unless otherwise sated in the problem. ‘Neglecting the factor for curvature K., ‘BFC serren = Koo (approximate equation only) Axial Deformation Formula, 8 Note: While the spring absorbs an axial load F which causes an axial deformation 6, the spring wire is under a twisting, ‘moment T which results in an angle of wist 8. ‘Thus, relating the strain energy quantities from axial and angular deformations, where, nmin = Strain energy or elastic potential energy of spring Since the two strain energy quantis (represented by areas) are just equal, 7 Esra = 3 F8|= [Esrana = 37 (8)-(76) But, 1 and 9= 55Length of equivalent straight spring wire Ne ; a \) es ee es Ne ~ Number of active coils ofthe spring ‘ren, equating the two stein energy quantities, 2, Da [(€ 38) dnt) wa 7% [Cero 2 7 3g Pw G [SFD,.*N. Pa] BFEN, D, (axial deformation formula) Spring Constant, load _F. ‘Teformation = 5 Aso, aren, Dy, CDy aen, Note: It is also referred to as spring rate, spring seale, spring modulus, spring gradient, or stiffness of spring. Bit the deformation is constant ‘Thus, Watb+e) = Fila) + Path) Deformations: Deformations: br by + bs r= total deformation Spring under impact load 1, Froma Potential Energy Source Spring Arrangements 1, Springs arranged in parallel a) Equal length Ww Forces: Wakes Deformations: 60 ~ 5:= 65 = constant where ‘W = total load F,= force exerted by the outer spring = force exerted by the inner spring '8,= deformation of the outer spring, {5)= deformation ofthe inner spring h = 1 erga eh) | mero = = a2. From a Kinetic Energy Source “Energy balance: KE ~ Strain energy g P= impact load of spring k= spring constant Other Important Considerations for Helical Springs: 1. Pitch angle of spring, a [2] ow lial Note: For more accurate results of the stress equation, the pitch angle should not exceed 127, < 12, 2. Spring wire diameters are based on stardard gage numbers. Table AT1S of DME Faires lists the standard sizes. ‘Wire Diameter, in| Gage | W&M | B&S No. | Ferrous | Nonferrous 70 | aor 0 | oasis 50-0505. 0 [03958 | 00. 3.0 [025 [nao 2003310 | 0365 ‘0 [03065 35 1_[07930_[- 0289 1 2 [38830356 3 [-o2is7 | 0229. [02255] —n208 5 [e200 | 9.831 [01920 | 062. [arr — oie 5 [01920 | 0.138 S| oass [outa 1] -01350_| 0:10 | a1205—| 0091 12] 01055] ost 13—[-o0o15_—| nora 14 | 0.0800 | 0.065 15 | -a07a0 | o0s7 16 | 0.0625 | nasi 17] asso | 005 1=_L_o0a7s_[ 080 —L ‘Machine Dssin 1: DASAEN, RY, 3. Recommended values for the spring index, C: To avoid problems in coiling the the spring, the sprig index should not be less than 5: € > 5. From Machinery’s Handbook, best proportioned springs use a spring index of C~ 7 through 9. 4. Compression springs have the following lengths FL = freelength CL = compressed tengsn 3 = deformation of spring SH ~ solid height Scoua~ deformation at solid height at a= z= F 5. Approximate free lengths and solid heights for different spring ends. (AT16 DME Faires) ‘Tye of | Free [Total | Sold Ends" | Leng | Cotls_| Height Plain DNo#D. | Ne) DaNNe+Dn Piainground | PN. | oN | “Dut ‘Squared PNe+3Dw | Net2 | DeNe+3Du ‘Squared and | pNe+2De | Net2 | DeNe+2De ground lL Example 1: ‘A helical compression spring is used f0 support an axial oad of SOON. The spring is made of 8mm diameter steel spring wire that is would to form an outside coil diameter of Sémm and a pitch of 24mm. The spring consists of 10 coils, with squared and gracund ends. Determine the following: a) Maximum stress induced in the spring ) Axial deformation of spring in mm ©) Spring rate 4) Pitch angle of the spring if its free length is thrice its ‘wire diameter ©) Length of straight spring wire used to coil the given spring 4) Deformation of spring if itis to be compressed to its solid height. 12) Spring's design stress if itis made of ASTM A313, stainless steel wire. Use equal spring index and standard spring wire diameter closest to 8mm (but greater), The spring is to be used for only about 100,000 times during its life. cySolution: ‘This is abasic example that applies the useful formulas ‘about torsion in MDB. Design considerations are introduced in the last part of the problem. Owsbmn Nr 10 cots , 69 B ge D0 +5emm New rk New @ setve exis (adss*? : : Fem RAED, + Cem + Das Sr RA Dw Dw oF se8 cre we seol pete fea te ke 42525 S5= 12535 -B6500N) (6) Ww (8)* rn S52 9951 MPa ) 672 From SeREGING "5G Geygpy# 1-5 « 10° psi ad 279865 MPa Ne* ® ecils 86.6500 N).(695 6) TR 26B BL, Comm) Se 10-40 mm cone? KLE + -S00N. 445 7M F | Bama Toh Machin GAYA, pitch dogle i S45 2 Fa er Dew tare rom -4 Mt tafe) Pt 3 ow (given? Dz € Dw 7608) 248 Has] 593. Agi ae 4-048") cere? tr + (ie) Chron) be Length of teat Ne 40 atts Mcoits (Crome PF Lrcoits fOr aby? Ga Ltcoit = 152-64 mm Lys (10) (152-64) ui 526-4 mam) {£9 Ssu0 © From FLESH + S300 wtene: FL pNe + 2Dw Sis Owne + 2bw Ssno* FL-SH + Ne (p-Dw) #8 (24-8) J ram arse Saoui0 + 128 een (g)Ssd+2 CMATL: ASTM A313, 51 FRom Atty FOR Dw: 6mm xm __ 0+ 315in Bam FROM ATAS, ose gage H2-0, Dw=0-331" Sed Frs (Suonesea) * Fs [85] FacToR FOR THE TYPE OF SERVICE Light service £ oP 40 104 Himes aves Fmediam etevice + 104 $e 10% severe secvices mare than 10° Fines 1: BASAEN, RV.WHERE: Suvrented * (0186 Ow < THUS, For 400.600 times service is average Ssd 20-26 (Sud * 0-26 fee] ge the service Dwr Ssd +026 | 47000 , pet (an35n) (Sed « 39 660 Example 2: ‘A spring isto be used asa relief valve fora boiler. “The spring has te following data: D,= 13mm D,=115mm FL =203mmNr~9.5 coils (with sq. and gr. ends) G=80GPa ‘The valve seat is 32 mm_and the relief valve should ‘open once the steam pressure reaches 1.4 MPa, Determine the following: 8) Length to which the spring should be intially ‘compressed inside the valve to hold the given Cm Ose boay pressure. 1b) Stress on the spring atthe given setting, and ‘when the spring is compressed to its solid height. Solutio Dw 18m Do 115m crBe + 7-846 Neenr-2 Nes 75 coals 86 82mm Fs 1126 N Cinitiat gorce compressing the spring (set © vam) ] ace? FROM FLeCL +5 WHERE 6:BE.C* Ne OW B= B(1126 NDG7-846)°(75)_ 80.200 Nrm® (15era) 8 31-38 som Dm = o- Dw + 102m Nre 45am, 818 gd Machine Design |: BASAEN, RV THus, OLS FL+§ = 203- 3% GL 16 me (b) Ss #7 @ initial setting | Sseaua=? Sore BEC 6 beagth @ set-up Pena se 4188 Ss u1eg £01126) (7-866) ao ‘SS 156-1 MPs AtS0, Ssoucg + FL~SH* CFL-(Dw et 200)) 5 208-C18(7-5) #2 (13) Gsotrd | 19-5 c+ $26 « Feoia B88 “Ie Feoitd » 2853 N Sssuttd 2654. (158-1) Example ‘A spring sto absorb an impact load froma 12-kg rigid body wihich isto be dropped at a height of 15m from the unsupported end of the spring. Ihe spring gradient is 35 Nim, determine the following: 2) Maximum deflection ofthe spring 1) Velocity of the body at an instant when the spring's deformation is 80mm 7Solution: a) Maximum deformation, 5 =? or we 12 C4.80N #1500 men (ange? FROM winrels 4 96 2 WHERE! KEE ; PERE = wOhrs]e4 Ks sacar) [1500 + 6] +4 (95)C61% jem = Nem 52 103.4 mm Example 4: ‘A erate of mass 1800 kg at a speed of 1.2 mis is Drought to rest by two helical steel springs. In stopping the crate, the springs are compressed by 200mm. Take the spring index to he 6 and the allawahle stress ta he 360 MPa ‘on cach spring. Determine the following: a) the maximum Toad on each spring, b) the minimum spring wire diameter necessary, and ¢) the number of active coils. Use G= 80 GPa Solution: 2) Maximum load on each spring, ae na WF WWE oO Dee Klose "ACL pts ot V0 ere Bena The Po ie 1.2528 chine Desion I: BASAEN, RV. G2) Ps? max Lead absorved by each Spriog OKeE CSD S.t. £m WDE Pr& 5 BUT roar (1800 kg CHEM m2 = (209 (0-2 $00 Kg IAM aE =A. (20) Co-2m P= 6460 N (b) Ow: Feow, Sst q BPC Ss SEE, SD Sd 12525 SAGSHON(G) = Ow [De BSE mm] (Ne? FROM OFFN EQN 6+ Beco Ne Ow 3eo.N, + 6+ 80050 MPa 200 + (6480) ¢6)* Ne 80000 (18-56) New 26-52 caits Example §: ‘Two concentric springs of unequal lengths are loaded as shown. Before the 2 KN load is applied, the outer ng is longer by 20mm. Other data ae given in the table Weak (No toad) Curae with W) Spring | Material [No] Dy De (em)_| (am i G=s0GPa fiz; 8 36 2 G=socra | 8 | 10 90 Determine the maximum stress and deformation of ccach spring if they are to be axially loaded with 2 kN vets b2525 2 Kae 184 n are creSolution: DEFORMATIONS 62 +1420 —+@ » Forces FoeFt= WwW Fat Fr © 2000 —+@® Y RATES koe ke Gow Beene 1+ 80009 (8) 2-86 8 (ertti2) S12 En, oe 2790000 (10); Kar 15.49 BCE) (8) $a: fe Bai THUS, EAN @ BECOMES Ege Eset ate Ba fe 27 © soins © & @: Fae 56 N Fie gaan, Hence, Si b44 Sie 2 Bie 27-55 mem | 62: 47-35 men FoR MAX STRESSES + aa atec Se oa 1 162. -Btedar (er Sei * 1ses -B.cedane Seas s1pq BL156)(6) WC) Sees 276-8 MPa shine Design: BASAEN, RV. Example 6: ‘An eccentric circular cam is assembled as shown at low position, the force between the cam and follower is 50. Athigh position, the comesponding, fovve is 80 « The spring is connected to maintain contact between the cam and follower, with the following data c= 5 =60 Calculate the following: a) spring wire diameter Dy and, b) number of active coils Nc. forte spring. Solution: 1 HIGH Posi Travel = 2¢e) #2" Mion od [eme+o1* Low Posi St LEM +01 Fe (ts) = 60 F2+240lb TION Fi (4/3) #5001) Fi 0 150 te R(8) Dw? Sse STK GEC > Sed 1 Ft Fuax #2 ge K wos, , teaig S12aaM) Dw 0294 in 1 a S10: 18 bo be chosen, Select gage no: Ow 0.3065" Co) Nee? FROM OEF'N EON 6 BEI Ne, ‘ew Nee eFct BY SIMILAR , 3 wy Ta SUBING VALUES INTO EON © Dw. BE BeSNE ag ss xtot fo-zeay , 240-150 eer) Ned 2h Nes 813 coils Design of a Helical Spring Characteristics Problems are usually open-ended, assumptions are often necessary. ‘The requirements may involve material selection & determination of necessary spring dimensions. ‘Solution must satisfy all problem requirements and all conditions of good design as advised by the textbook, and other sources. Spring serviee Light service: less than 108 eycles Machine Average service: 10 to 10 cycles Severe service: beyond 10* cycles Some Spring Materials ‘Typical spring materials have a carbon content of 0.5% ot greater. Carbon Steels: Hard Drawn Wire Me for Hard drawn spring wire (cheapest) sic Wire ‘Most commonly used and most reliable material general application up to a wire diameter of inch. Oil tempered spring wire Allloy S Used as an alternative to music wire, but less reliable Steel Selected for a particular application in which carbon steels are no Tonger satisfactory (high endurance, high toughness and corrosion resistance may be desired). Example of spring materials made of alloy steel: Cr,V, Si Mn, Stainless steel Example of Non-Ferrous materials inchude copper and nickel based. “Monel = Cu + Ni Inconel = Cr + Ni +Fe Example of Design of a Helical Spring: ‘Avelical spring ist fitaboutan 2 in diameter rod, ‘A mainmam load of 8 tb sto compress the spring by 12 in, The fre lenth it be 22 tn ores. The sping so bbe used only for about 50,000 times during its lif, but it istobe subjected to 2 relatively high temperature and corrosive atmosphere while in operation. spring Select a spring material and specify the necessary diameter dimensions Dy,Dg, Ne, FL, ete. Meet ail condition used by the textbook. FL S275" B