Temple University, College of Engineering MEE-3305 Materials Laboratory Time: _17:30__ Section: __4___ Group: ___D____________ Spring Lab *Each of the following members has made significant contribution to this lab report and we agree to receive a grade as a group: Name: ____ Yao, Yang-Yi_________________ Signature: _____YY____________ Name: ________________________________ Signature: ________________ Name: ________________________________ Signature: ________________ Name: ______________________________ Signature: ____________________ Date: __________ Grading Rubric Lab objectives 5 points Methods and materials 10 points Experimental results, data processing and analysis 30 points Discussion of the results 30 points Comments: 10/13/2022 1 Conclusions Writing 10 points 15 points Contents Lab Objectives ............................................................................................................................................... 3 Methods and Materials................................................................................................................................. 3 Equipment & Materials: ............................................................................................................................ 3 Testing Methodology: ............................................................................................................................... 3 Experimental Results and Data Analysis ....................................................................................................... 3 Discussion...................................................................................................................................................... 3 Conclusion ..................................................................................................................................................... 3 Appendix ....................................................................................................................................................... 3 Group D 10/13/2022 2 Lab Objectives Methods and Materials Equipment & Materials: Testing Methodology: Experimental Results and Data Analysis Discussion Conclusion Appendix Group D 10/13/2022 3 Electronic Coil Spring Tester - 73505 Monroe, WA ∙ 360 453 2030 ∙ www.longacreracing.com Electronic Coil Spring Tester Instructions • • • • • • • • • • • • • • Push the power button of the Electronic Scale Base on and let it warm up for about 5 minutes. Loosen the thumb knob of the Digital Travel Indicator. Remove the black handled pins and lift the jack slider up high enough that there is room to place the coil spring into the lower spring cup and temporarily re-insert the pins. Place the coil spring into the lower spring cup.  Note: The top spring cup has one side for square end springs (Coil Over) and the other side for pigtail springs. For pigtail remove the bolt from the center of the cup to turn it over. Place the spacer on the bolt before placing the bolt through the spring cup or the bolt will be too long for the pigtail application. Make sure the pigtail is against the step in the cup. Remove the black handled pins and lower the jack slider until the top spring cup touches the coil spring, then raise it to the next available set of holes and re-insert the black handled pins. Tighten the valve at the base of the jack and pump the jack handle until the top and bottom spring cups are touching the coil spring. Slide the Digital Travel Indicator down until it stops and tighten the thumb knob. Zero the Digital Travel Indicator by pressing the 0/On Button. Pump the jack until the Digital Travel Indicator reads 1.0” for preload. This is very important and should be done accurately. You may alternatively use 0.5” (1/2”) depending on the spring quality as long as you always use the same amount. Re-Zero the Digital Travel Indicator. Zero the Electronic Scale by pressing the Zero button. Pump the jack until the Digital Travel Indicator reads 1.0” At this point the Electronic Scale Base shows the spring rate on the display. YOU MUST LOOSEN THE THUMB KNOB ON THE DIGITAL TRAVEL INDICATOR BEFORE RELEASING THE COIL SPRING OR IT WILL BE DAMAGED. DANGER: Compressed coil springs are dangerous. If one comes out of the tester it can cause severe injury or death • NEVER compress a spring beyond the capacity of the tester (3000 lbs) • NEVER compress a coil spring that bows to one side or does not compress straight. • NEVER compress a coil spring that is not seated completely within both spring cups. Longacre Racing Products –Electronic Coil Spring Tester PN 73505 Electronic Coil Spring Tester - 73505 Monroe, WA ∙ 360 453 2030 ∙ www.longacreracing.com Electronic Coil Spring Tester Instructions Longacre Racing Products –Electronic Coil Spring Tester PN 73505 Case Study 2 (CS2) Automobile Valve Spring Learning Objectives After studying this case study, you should be able to do the following: 3. For a steel alloy material in a helical spring, 1. Briefly describe the steps that are used to ascerexplain how cold drawing and shot peening tain whether a particular metal alloy is suitable treatments affect the spring performance and for use in an automobile valve spring. lifetime. 2. Explain how the tensile strength and fatigue strength of a particular metal affect the materials selection process for an automobile valve spring. CS2.1 MECHANICS OF SPRING DEFORMATION The basic function of a spring is to store mechanical energy as it is initially elastically deformed and then recoup this energy at a later time as the spring recoils. In this section, helical springs that are used in mattresses and in retractable pens and as suspension springs in automobiles are discussed. A stress analysis will be conducted on this type of spring, and the results will then be applied to a valve spring that is used in automobile engines. Consider the helical spring shown in Figure CS2.1, which has been constructed of wire having a circular cross section of diameter d; the coil center-to-center diameter is denoted as D. The application of a compressive force F causes a twisting force, or moment, denoted T, as shown in the figure. A combination of shear stresses results, the sum of which, t, is t 8FD Kw pd 3 (CS2.1) where Kw is a force-independent constant that is a function of the Dd ratio: Kw  1.60 a D 0.140 b d F d T (CS2.2) Figure CS2.1 Schematic diagram of a helical spring showing the twisting moment T that results from the compressive force F. (Adapted from K. Edwards and P. McKee, Fundamentals of Mechanical Component Design. Copyright © 1991 by McGraw-Hill, Inc. Reproduced with permission of The McGraw-Hill Companies.) D F • CS2.1 CS2.2 • Case Study 2 / Automobile Valve Spring F ␦c D 2 D 2 D 2 (a) (b) Figure CS2.2 Schematic diagrams of one coil of a helical spring, (a) prior to being compressed, and (b) showing the deflection dc produced from the compressive force F. (Adapted from K. Edwards and P. McKee, Fundamentals of Mechanical Component Design. Copyright © 1991 by McGraw-Hill, Inc. Reproduced with permission of The McGraw-Hill Companies.) In response to the force F, the coiled spring will experience deflection, which will be assumed to be totally elastic. The amount of deflection per coil of spring, dc, as indicated in Figure CS2.2, is given by the expression dc  8FD3 d 4G (CS2.3) where G is the shear modulus of the material from which the spring is constructed. Furthermore, dc may be computed from the total spring deflection, ds, and the number of effective spring coils, Nc, as dc  ds Nc (CS2.4) Now, solving for F in Equation CS2.3 gives F d 4dcG 8D3 (CS2.5) and substituting for F in Equation CS2.1 leads to t Condition for nonpermanent spring deformation—shear yield strength and its relationship to shear modulus, number of effective coils, and spring and wire diameters dcGd pD 2 Kw (CS2.6) Under normal circumstances, it is desired that a spring experience no permanent deformation upon loading; this means that the right-hand side of Equation CS2.6 must be less than the shear yield strength ty of the spring material or that ty 7 dcGd pD2 Kw (CS2.7) CS2.2 VALVE SPRING DESIGN AND MATERIAL REQUIREMENTS We shall now apply the results of the preceding section to an automobile valve spring. A cutaway schematic diagram of an automobile engine showing these springs is presented in Figure CS2.3. Functionally, springs of this type permit both intake and exhaust valves to alternately open and close as the engine is in operation. Rotation of the camshaft causes a valve to open and its spring to be compressed, so CS2.2 Valve Spring Design and Material Requirements • CS2.3 Cam Camshaft Figure CS2.3 Cutaway drawing of a section of an automobile engine in which various components, including valves and valve springs, are shown. Valve spring Exhaust valve Intake valve Piston Crankshaft that the load on the spring is increased. The stored energy in the spring then forces the valve to close as the camshaft continues its rotation. This process occurs for each valve for each engine cycle, and over the lifetime of the engine, it occurs many millions of times. Furthermore, during normal engine operation, the temperature of the springs is approximately 80°C (175°F). A photograph of a typical valve spring is shown in Figure CS2.4. The spring has a total length of 1.67 in. (42 mm), is constructed of wire having a diameter d of 0.170 in. Figure CS2.4 Photograph of a typical automobile valve spring. CS2.4• Case Study 2 / Automobile Valve Spring (4.3 mm), has six coils (only four of which are active), and has a center-to-center diameter D of 1.062 in. (27 mm). Furthermore, when installed and when a valve is completely closed, its spring is compressed a total of 0.24 in. (6.1 mm), which, from Equation CS2.4, gives an installed deflection per coil dic of dic  0.24 in.  0.060 in. /coil 11.5 mm/coil2 4 coils The cam lift is 0.30 in. (7.6 mm), which means that when the cam completely opens a valve, the spring experiences a maximum total deflection equal to the sum of the valve lift and the compressed deflection, namely, 0.30 in.  0.24 in.  0.54 in. (13.7 mm). Hence, the maximum deflection per coil, dmc, is dmc  0.54 in.  0.135 in. /coil 13.4 mm/coil2 4 coils Thus, we have available all of the parameters in Equation CS2.7 (taking dic  dmc), except for ty, the required shear yield strength of the spring material. However, the material parameter of interest is not really ty, inasmuch as the spring is continually stress cycled as the valve opens and closes during engine operation; this necessitates designing against the possibility of failure by fatigue rather than against the possibility of yielding. This fatigue complication is handled by choosing a metal alloy that has a fatigue limit (Figure 8.19a of Introduction; Figure 9.25a of Fundamentals) that is greater than the cyclic stress amplitude to which the spring will be subjected. For this reason, steel alloys, which have fatigue limits, are normally employed for valve springs. When using steel alloys in spring design, two assumptions may be made if the stress cycle is reversed [if tm  0, where tm is the mean stress, or, equivalently, if tmax  tmin, in accordance with Equation 8.14 of Introduction (Equation 9.15 of Fundamentals) and as noted in Figure CS2.5]. The first of these assumptions is that the fatigue limit of the alloy (expressed as stress amplitude) is 45,000 psi (310 MPa), the threshold of which occurs at about 106 cycles. Secondly, for torsion and on the basis of experimental data, it has been found that the fatigue strength at 103 cycles is 0.67TS, where TS is the tensile strength of the material (as measured from a pure tension test). The S–N fatigue diagram (i.e., stress amplitude versus logarithm of the number of cycles to failure) for these alloys is shown in Figure CS2.6. Now let us estimate the number of cycles to which a typical valve spring may be subjected in order to determine whether it is permissible to operate within the fatigue limit regime of Figure CS2.6 (i.e., if the number of cycles exceeds 106). For the sake of argument, assume that the automobile in which the spring is mounted travels a minimum of 100,000 miles (161,000 km), at an average speed of 40 mph (64.4 km/h), with Figure CS2.5 Stress versus time for a reversed cycle in shear. Stress ␶max 0 ␶min Time CS2.3 One Commonly Employed Steel Alloy • CS2.5 Figure CS2.6 Shear stress amplitude versus logarithm of the number of cycles to fatigue failure for typical ferrous alloys. Stress amplitude, S 0.67TS 45,000 psi 103 105 107 109 Cycles to failure, N (logarithmic scale) Goodman’s law— computation of the nonzero-mean-stress fatigue limit for a material using tensile strength and zeromean-stress fatigue limit values an average engine speed of 3000 rpm (rev/min). The total time it takes the automobile to travel this distance is 2500 h (100,000 mi/40 mph), or 150,000 min. At 3000 rpm, the total number of revolutions is (3000 rev/min)(150,000 min)  4.5  108 rev, and because there are 2 rev/cycle, the total number of cycles is 2.25  108.This result means that we may use the fatigue limit as the design stress, inasmuch as the limit cycle threshold has been exceeded for the 100,000-mile distance of travel (i.e., since 2.25  108 cycles  106 cycles). Furthermore, this problem is complicated by the fact that the stress cycle is not completely reversed (i.e., tm  0), inasmuch as between minimum and maximum deflections the spring remains in compression; thus, the 45,000-psi (310-MPa) fatigue limit is not valid. What we would now like to do is first make an appropriate extrapolation of the fatigue limit for this tm  0 case and then compute and compare with this limit the actual stress amplitude for the spring. If the stress amplitude is significantly below the extrapolated limit, then the spring design is satisfactory. A reasonable extrapolation of the fatigue limit for this tm  0 situation may be made using the following expression (termed Goodman’s law): tal  te a 1  tm b 0.67TS (CS2.8) where tal is the fatigue limit for the mean stress tm; te is the fatigue limit for tm  0 [i.e., 45,000 psi (310 MPa)]; and, again, TS is the tensile strength of the alloy. To determine the new fatigue limit tal from this expression necessitates the computation of both the tensile strength of the alloy and the mean stress for the spring. CS2.3 ONE COMMONLY EMPLOYED STEEL ALLOY For an ASTM 232 steel wire, dependence of tensile strength on drawn wire diameter One common spring alloy is an ASTM 232 chrome–vanadium steel, having a composition of 0.48–0.53 wt% C, 0.80–1.10 wt% Cr, a minimum of 0.15 wt% V, and the balance being Fe. Spring wire is normally cold drawn (Section 11.4 of Introduction; Section 14.2 of Fundamentals) to the desired diameter; consequently, tensile strength will increase with the amount of drawing (i.e., with decreasing diameter). For this alloy, it has been experimentally verified that, for the diameter d in inches, the tensile strength is TS 1psi2  169,0001d2 0.167 (CS2.9) CS2.6 • Case Study 2 / Automobile Valve Spring Since d  0.170 in. for this spring, TS (psi)  (169,000)(0.170 in.)0.167  227,200 psi (1570 MPa) Computation of the mean stress tm is made using Equation 8.14 of Introduction (Equation 9.15 of Fundamentals), modified to the shear stress situation as follows: tm  tmin  tmax 2 (CS2.10) It now becomes necessary to determine the minimum and maximum shear stresses for the spring, using Equation CS2.6. The value of tmin may be calculated from Equations CS2.6 and CS2.2, inasmuch as the minimum dc is known 1i.e., dic  0.060 in.2 . A shear modulus of 11.5  106 psi (79 GPa) will be assumed for the steel; this is the room-temperature value, which is also valid at the 80°C service temperature. Thus, tmin is just tmin  dicGd D 0.140 K  c 1.60 a b d w d pD2 pD2 dicGd  c 10.060 in.2 111.5  106 psi2 10.170 in.2 p11.062 in.2 2 (CS2.11a) d c 1.60 a 1.062 in. 0.140 b d 0.170 in.  41,000 psi (280 MPa) Now tmax may be determined taking dc  dmc  0.135 in. as follows: tmax  dmcGd  c pD2 c 1.60 a D 0.140 b d d (CS2.11b) 10.135 in.2 111.5  106 psi2 10.170 in.2 p11.062 in.2 2 d c 1.60 a 1.062 in. 0.140 b d 0.170 in.  92,200 psi (635 MPa) Now, from Equation CS2.10, tmin  tmax 2 41,000 psi  92,200 psi   66,600 psi 1460 MPa 2 2 The variation of shear stress with time for this valve spring is noted in Figure CS2.7; the time axis is not scaled, inasmuch as the time scale will depend on engine speed. Our next objective is to determine the fatigue limit amplitude 1tal 2 for this tm  66,600 psi (460 MPa) using Equation CS2.8 and for te and TS values of 45,000 psi (310 MPa) and 227,200 psi (1570 MPa), respectively. Thus, tm  tal  te c 1  tm d 0.67TS  145,000 psi2 c 1  66,600 psi d 10.672 1227,200 psi2  25,300 psi (175 MPa) CS2.3 One Commonly Employed Steel Alloy • CS2.7 Figure CS2.7 Shear stress versus time for an automobile valve spring. 100 Stress (103 psi) 80 60 ␶aa = 25,600 psi ␶max = 92,200 psi 40 ␶m = 66,600 psi ␶min = 41,000 psi 20 0 Time Now let us determine the actual stress amplitude taa for the valve spring using Equation 8.16 of Introduction (Equation 9.17 of Fundamentals), modified to the shear stress condition: taa   tmax  tmin 2 (CS2.12) 92, 200 psi  41, 000 psi  25,600 psi 1177 MPa 2 2 Thus, the actual stress amplitude is slightly greater than the fatigue limit, which means that this spring design is marginal. The fatigue limit of this alloy may be increased to greater than 25,300 psi (175 MPa) by shot peening, a procedure described in Section 8.10 of Introduction (Section 9.13 of Fundamentals). Shot peening involves the introduction of residual compressive surface stresses by plastically deforming outer surface regions; small and very hard particles are projected onto the surface at high velocities. This is an automated procedure commonly used to improve the fatigue resistance of valve springs; in fact, the spring shown in Figure CS2.4 has been shot peened, which accounts for its rough surface texture. Shot peening has been observed to increase the fatigue limit of steel alloys in excess of 50% and, in addition, to significantly reduce the degree of scatter of fatigue data. This spring design, including shot peening, may be satisfactory; however, its adequacy should be verified by experimental testing. The testing procedure is relatively complicated and, consequently, will not be discussed in detail. In essence, it involves performing a relatively large number of fatigue tests (on the order of 1000) on this shot-peened ASTM 232 steel, in shear, using a mean stress of 66,600 psi (460 MPa) and a stress amplitude of 25,600 psi (177 MPa), and for 106 cycles. On the basis of the number of failures, an estimate of the survival probability can be made. For the sake of argument, let us assume that this probability turns out to be 0.99999; this means that one spring in 100,000 produced will fail. Suppose that you are employed by one of the large automobile companies that manufactures on the order of 1 million cars per year, and that the engine powering each automobile is a six-cylinder one. Since for each cylinder there are two valves, and thus two valve springs, a total of 12 million springs would be produced every year. For the preceding survival probability rate, the total number of spring failures CS2.8 • Case Study 2 / Automobile Valve Spring would be approximately 120, which also corresponds to 120 engine failures. As a practical matter, one would have to weigh the cost of replacing these 120 engines against the cost of a spring redesign. Redesign options would involve taking measures to reduce the shear stresses on the spring by altering the parameters in Equations CS2.2 and CS2.6. This would include either (1) increasing the coil diameter D, which would also necessitate increasing the wire diameter d, or (2) increasing the number of coils Nc. SUMMARY A stress analysis was performed on a helical spring, which was then extended to an automobile valve spring. It was noted that the possibility of fatigue failure was crucial to the performance of this spring application. The shear stress amplitude was computed, the magnitude of which was almost identical to the calculated fatigue limit for a chrome–vanadium steel that is commonly used for valve springs. It was noted that the fatigue limit of valve springs is often enhanced by shot peening. Finally, a procedure was suggested for assessing the economic feasibility of this spring design, incorporating the shot-peened chrome–vanadium steel. REFERENCES Juvinall, R. C., and K. M. Marshek, Fundamentals of Machine Component Design, 4th edition, Chapter 12, Wiley, Hoboken, NJ, 2005. Shigley, J., C. Mischke, and R. Budynas, Mechanical Engineering Design, 7th edition, Chapter 10, McGraw-Hill, New York, 2004. SPREADSHEET PROBLEM CS2.1SS Generate a spreadsheet that allows the user to specify the number of effective coils (N), the spring coil-to-coil diameter (D), the wire cross-section diameter (d), the fatigue limit (for tm  0) 1te 2 , the tensile strength (TS), and the shear modulus (G), and calculates the fatigue limit (for tm  0) 1tal 2 as well as the actual stress amplitude 1taa 2 for an automobile valve spring. Incorporate into this routine values cited for installed and maximum deflections per coil (i.e., dic  0.24 in. and dmc  0.54 in.2 . DESIGN PROBLEMS CS2.D1 A spring having a center-to-center diameter of 20 mm (0.8 in.) is to be constructed of cold-drawn and annealed 316 stainless steel wire that is 2.5 mm (0.10 in.) in diameter; this spring design calls for eight coils. (a) What is the maximum tensile load that may be applied, such that the total spring deflection will be no more than 6.5 mm (0.26 in.)? (b) What is the maximum tensile load that may be applied without any permanent deformation of the spring wire? Assume that the shear yield strength is 0.6sy, where sy is the yield strength in tension. CS2.D2 You have been asked to select a material for a spring that is to be stressed in tension. It is to consist of ten coils, and the coil-to-coil diameter called for is 15 mm; furthermore, the diameter of the spring Design Problems • CS2.9 wire must be 2.0 mm. Upon application of a tensile force of 35 N, the spring is to experience a deflection of no more than 12 mm and not plastically deform. (a) From the materials included in the database in Appendix B (of Introduction and Fundamentals), make a list of candidate materials that meet the preceding criteria. Assume that the shear yield strength is 0.6sy, where sy is the yield strength in tension, and that the shear modulus is equal to 0.4E, E being the modulus of elasticity. (b) Now, from this list of candidate materials, select the one you would use for this spring application. In addition to the preceding criteria, the material must be relatively corrosion resistant and, of course, capable of being fabricated into wire form. Justify your decision. CS2.D3 A spring having seven coils and a coil-to-coil diameter of 0.5 in. is to be made of colddrawn steel wire. When a tensile load of 15 lbf is applied, the spring is to deflect no more than 0.60 in. The cold-drawing operation will, of course, increase the shear yield strength of the wire, and it has been observed that ty (in ksi) depends on wire diameter d (in in.) according to 63 (CS2.13) d 0.2 If the shear modulus for this steel is 11.5  106 psi, calculate the minimum wire diameter required, such that the spring will not plastically deform when subjected to the preceding load. CS2.D4 A helical spring is to be constructed from a 4340 steel. The design calls for five coils, a coil-to-coil diameter of 12 mm, and a wire diameter of 3 mm. Furthermore, a maximum total deflection of 5.0 mm is possible without any plastic deformation. Specify a heat treatment for this 4340 steel wire in order for the spring to meet the preceding criteria. Assume a shear modulus of 80 GPa for this steel alloy and that ty  0.6sy. [Note: Heat treatment of the 4340 steel is discussed in Section 10.8 of Introduction (Section 11.8 of Fundamentals).] ty  All in mm Spring 1: L: 205.74mm ID: 63.99 OD: 205.74 Diameter: 7.50 Pitch: W/1/14/8/ 22.25 Spring 2: L: 210.82 ID: 49.62 OD: 62.62 Spring 31 L:203.20 ID, 49.08 OD: 62.72 Dia : 6.71 Pitch: 24.23 Dia 17.19 Pitch: 23.04
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