Summary

Case Study: Stress Concentration Factors for Complicated Geometries

6.8 Summary

Of the approaches that can be used to determine stress concentration factors, the use of stress concentration charts from the technical literature (of which Figs. 6.2 through 6.8 are examples) represents the most cost-effective method. Especially in the age of the Internet, it is also usually the fastest approach as well. The finite element method has become an extremely widespread approach as well, but it has drawbacks of longer lead time and therefore higher cost. Experimental confirma- tion of designs is also relatively expensive and time consum- ing, but may be necessary to obtain confidence in the safety and suitability of designs.

Figure 6.22: Photoelastic comparison of threaded fastener profiles comparing load distribution. The left image shows a conventional profile where load per tooth varies widely, and the right shows a Spiralock^c profile with more uniform stresses on each tooth. Source: Courtesy of Stanley Engi- neered Fastening - Spiralock.

Key Words

Coulomb-Mohr theory a failure theory for materials with different strengths in tension and compression; identical to the internal friction theory

crack small flaw, always present, that can compromise material strength

distortion-energy theory postulate that failure is caused by elastic energy associated with deformation; identical to von Mises criterion

fracture control maintenance of nominal stress and crack size below critical level

fracture toughness critical value of stress intensity at which crack extension occurs

internal friction theory failure criterion accounting for dif- ference between compressive and tensile strengths of brittle materials; identical to Coulomb-Mohr theory maximum-normal-stress theory theory that yielding will

occur whenever the greatest positive principal stress exceeds the tensile yield strength or whenever the greatest negative principal stress exceeds the compressive yield strength

maximum-shear-stress theory theory that yielding will occur when the largest shear stress exceeds a critical value;

identical to the Tresca criterion

modes of crack propagation principal mechanisms for cracks to enlarge: Mode I, opening through tension;

Mode II, sliding or in-plane shearing; Mode III, tearing modified Mohr theory failure postulate similar to Coulomb-

Mohr theory, except that curve is altered in quadrants II and IV of plane stress plot of principal stresses

stress concentration region where stress raiser is present stress concentration factor factor used to relate actual maxi-

mum stress at discontinuity to nominal stress stress intensity factor stress intensity at crack tip

stress raiser discontinuity that alters stress distribution so as to increase maximum stress

Tresca yield criterion theory that yielding will occur when the largest shear stress exceeds a critical value; identical to maximum-shear-stress theory

von Mises criterion postulate that failure is caused by elastic energy associated with deformation; identical to distortion-energy theory

von Mises stress effective stress based on von Mises criterion

Summary of Equations

Stress concentration factor:

Definition:Kc=Actual maximum stress Average stress

Elliptical hole in plate loaded in tension:Kc= 1 + 2a b Fracture Toughness:KIc=Y σnom√

πa

Failure Prediction for Uniaxial Stress State:ns= σall

σd

Failure Prediction for Multiaxial Stress State:

Maximum shear stress theory (MSST, or Tresca):

ns= Sy

σ1−σ3

von Mises stress:

σe= 1

√2 (σ1

−σ2)2+ (σ3−σ1)2+ (σ3−σ2)2^1/2 Distortion energy theory (DET, or von Mises):ns= Sy

σe

Maximum normal stress theory (MNST):

ns= Sut

σ1 orns= Suc

σ3 , whichever is lower.

Internal Friction Theory (IFT, or Coulomb-Mohr):

Ifσ1>0andσ3<0, σ1

Sut − σ3

Suc

= 1 ns

Ifσ3>0,ns=Sut

σ1

Ifσ1<0,ns=Suc

σ3

Modified Mohr Theory (MMT):

Ifσ1>0andσ3<−Sut, σ1− Sutσ3

Suc−Sut = SucSut

nsSuc−Sut

Ifσ3>−Sut,ns=Sut

σ1

Ifσ1<0,σ3=Suc

References

ASM International (1989)Guide to Selecting Engineering Mate- rials, American Society for Metals.

Bowman, K.J. (2004)Mechanical Behavior of Materials, Wiley.

Coulomb, C.A. (1773) Sur une Application des Regles de max- immis et minimus a quelques problemes de statique relatifs a l’architecture.

Daniels, I.M., and Ishai, O. (2005) Engineering Mechanics of Composite Materials2nd ed., Oxford.

Dowling, N.E. (1993) Mechanical Behavior of Materials, Pear- son.

Doyle, J.F. (2004)Experimental Stress Analysis: Completing the Solution of Partially Specified Problems, Wiley.

Fish, J., and Belytschko, T. (2007)A First Course in Finite Ele- ments, Wiley.

Hughes, T.J.R. (2000)The Finite Element Method, Dover.

Kalpakjian, S., and Schmid, S.R. (2014)Manufacturing Engi- neering and Technology, 7th ed., Pearson.

Kaw, A.K. (1997)Mechanics of Composite Materials, CRC Press.

Murphy, G. (1964)Advanced Mechanics of Materials, McGraw- Hill.

Pilkey, W.D., and Pilkey, D.F. (2008)Peterson’s Stress Concen- tration Factors, 3rd ed., Wiley.

Reddy, J.N. (1996) Mechanics of Laminated Composite Plates, CRC Press.

Tresca, H. (1868) “Mem. prenetes par divers savants,” vol. 59, p. 754,Comptes Rendus Acad. Sci..

Young, W.C., and Budynas, R. (2001) Roark’s Formulas for Stress and Strain, 7th ed., McGraw-Hill.

Questions

6.1 What is a stress concentration?

6.2 What is a stress concentration factor?

6.3 Define the term “crack” as relates to material fracture.

6.4 List the modes of crack growth.

6.5 What is fracture toughness?

6.6 Is a material with a high fracture toughness a ductile material? Explain.

6.7 What is the maximum shear stress theory?

6.8 Do the distortion energy criterion and maximum shear stress criterion give very different results? Explain.

6.9 Without the use of equations, define the von Mises stress.

6.10 What is a yield locus?

6.11 What is the MNST and the IFT?

6.12 For what materials is the IFT most useful?

Qualitative Problems

6.13 A round bar has a fillet withr/d= 0.15 andD/d= 1.5.

The bar transmits both bending moment and torque. A new construction is considered to make the shaft stiffer and stronger by making it equally thick on each side of the fillet or groove. Determine whether that is a good idea.

6.14 Are stress concentrations more severe for tension, bending, or torsion for a bar with a groove?

6.15 A diamond cutting tool is used to make a shallow but sharp groove in a glass plate. The glass is struck with a hard rubber mallet to produce a well-defined cut at the location of the groove. Referring to the discussions in this chapter, explain the phenomena that are important in this application.

6.16 Explain why the approach used to cut glass will not work for a ductile aluminum.

6.17 A plate of titanium is diffusion bonded to a plate of aluminum. There are no holes, grooves, fillets, or notches in either member, and the two plates are then exposed to uniform tension. Sketch the stress distribution through the thickness of the plates. Is there a stress concentration? Explain.

6.18 Give three examples of each mode of crack growth de- scribed in Section 6.4.

6.19 List and briefly explain the variables that can influence the fracture toughness of a material.

6.20 Some magnesium alloys have a lower strength in compression than in tension. What failure criterion would you use for such a material? Produce an equivalent sketch as Fig. 6.11 for this failure criterion.

6.21 If a material strain hardens, what effect does plastic deformation have on the yield locus?

6.22 Figs. 6.2 through 6.8 do not show stress concentrations for compressive loadings. Why not?

6.23 Review Fig. 6.4 and plot the stress concentration factor as a function ofr/hfor the case where the grooves are semicircular.

6.24 Figure 7.3 shows an R.R. Moore test specimen for fatigue tests. Explain why this specimen has its shape, and estimate the stress concentration factor.

Quantitative Problems

6.25 Complete the following table:

Geometry r

Shape Loading (mm) (mm) K^c

Plate w/ hole Tension b=50,d=5 — Plate w/ hole Bending b=50,d=5 — Plate w/ fillet Tension H=100,h=50 2.0

Bar w/ fillet Tension D=50,d=25 2.0

Bar w/ groove Torsion D=10,d=9.09 0.9

6.26 Given that the stress concentration factor is 3.81 for a machine element made of steel with a modulus of elasticity of 207 GPa, find the stress concentration factor for an identical machine element made of aluminum in- stead of steel. The modulus of elasticity for aluminum is 69 GPa.

6.27 A flat part with constant thicknessbis loaded in tension as shown in Fig. 6.3a. The height changes from 50 to 87 mm with a radiusr = 4.0mm. Find how much lower a load can be transmitted through the bar if the height increases from 50 to 100 mm and the radius increases from 4.0 to 10 mm.Ans.10% higher load.

6.28 A flat steel plate axially loaded as shown in Sketchahas two holes for electric cables. The holes are situated be- side each other and each has a diameterd. To make it possible to draw more cables, the two holes are replaced with one hole having twice the diameter2d, as shown in Sketchb. Assume that the ratio of diameter to width is d/b = 0.2for the two-hole plate. Which plate will fail first?

Quantitative Problems 155 d

(a)

2d b

(b)

Sketchesaandb, for Problem 6.28

6.29 A 5-mm thick, 100-mm wide AISI 1020 steel rectangular plate has a central elliptical hole 8 mm in length transverse to the applied stress and 2 mm in diameter along the stress. Determine the applied load that causes yielding at the edge of the hole.Ans.92.43 kN.

6.30 A 250-mm wide plate loaded in tension contains a 50 mm long, 10 mm wide slot. Estimate the stress concentration by:

(a) Approximating the slot as an ellipse that is in- scribed within the slot.

(b) Obtaining the stress concentration at the edge of the slot by taking a section through the slot and approximating the geometry as a rectangular plate with a groove.

6.31 A machine has three circular shafts, each with fillets giv- ing stress concentrations. The ratio of fillet radius to shaft diameter is 0.1 for all three shafts. One of the shafts transmits a tensile force, one transmits a bending torque, and one transmits torsion. Because they are stressed exactly to the stress limit (n^s = 1), a design change is proposed doubling the notch radii to get a safety factor greater than 1. How large will the safety factors be for the three shafts if the diameter ratio is 2 (D/d= 2)?Ans.

ns,tension= 1.21,ns,bending= 1.19,ns,torsion= 1.17.

6.32 A material with a yield strength of Sy = 350MPa is subjected to the stress state shown in Sketchc. What is the factor of safety based on the maximum shear stress and distortion energy theories? Ans. For MSST,ns = 11.67.

70 MPa

20 MPa 50 MPa

85 MPa

Sketchc, for Problems 6.32 and 6.33

6.33 A material subjected to the stresses shown in Sketchcis known to be yielding. What stress would cause yielding in uniaxial tension?Ans.For MSST,Sy= 30MPa.

6.34 The shaft shown in Sketchdis subjected to tensile, torsional, and bending loads. Determine the principal stresses at the location of stress concentration. Ans.

σ1= 53.0MPa,σ2= 0,σ3=−12.27MPa.

150 mm

120 mm D = 45 mm

d = 30 mm r = 3 mm

100 N-m 1000 N 500 N

Sketchd, for Problem 6.34

6.35 A steel plate with dimensions shown in Sketcheis subjected to 150-kN tensile force and 300-N-m bending moment. The plate is made of AISI 1080 steel, quenched and tempered at 800^◦C. A hole is to be punched in the center of the plate. What is the maximum diameter of the hole for a safety factor of 1.5?Ans.d= 170mm.

d M

M P 235 mm

5 mm

1000 mm

Sketche, for Problem 6.35

6.36 A Plexiglas plate with dimensions 1 m×1 m×1 cm is loaded by a nominal tensile stress of 55 MPa in one direction. The plate contains a small crack perpendicular to the load direction. At this stress level a safety factor of 2 against crack propagation is obtained. Find how much larger the crack can get before it grows catastrophically.

Ans.a2= 4a1.

6.37 A pressurized container is made of AISI 4340 steel. The wall thickness is such that the tensile stress in the material is 1000 MPa. The dimensionless geometry correction factorY = 1for the given geometry. Find how big the largest crack can be without failure if the steel is tempered

(a) At 260^◦C.Ans.1.592 mm.

(b) At 425^◦C.Ans.4.863 mm.

6.38 Two tensile test rods are made of AISI 4340 steel tempered at 260^◦C and aluminum alloy 2024-T351. The dimensionless geometry correction factor Y = 1. Find how high a stress each rod can sustain if there is a crack of 1-mm half-length in each of them. Ans. AISI 4340:

σ= 892MPa.

6.39 A plate made of titanium alloy Ti-6Al-4V has the dimensionless correction factor Y = 1. How large can the largest crack in the material be if it still should be possible to plastically deform the plate in tension?Ans.1.488 mm.

6.40 Consider the 2014-T4 aluminum plate shown in Sketch f. What is the largest tensile stress this plate can with- stand? What is the largest center crack that can exist while ensuring that the material will still yield? Ans.

σ= 115.7MPa,lc= 2.644mm.

σ 20 σ

10 10

Sketchf, for Problem 6.40. All dimensions are in millimeters.

6.41 For a plate with length 50 mm, width 25 mm, and thickness 5 mm, what is the ratio of allowable stress for a center-cracked specimen compared to an edge-cracked specimen? Assume the crack is 10 mm long for both cases.Ans.σedge/σcenter= 0.607.

6.42 A cylindrical pressure vessel constructed from 4340 steel tempered at 260^◦C has an outer diameter of 250 mm, a wall thickness of 2.5 mm, and an internal pressure of 4 MPa. What crack can be tolerated in the axial direction before fracture occurs?Ans.lc= 39.8mm.

6.43 A 100-mm wide, 200-mm long plate made of titanium alloy Ti-6Al-4V has a single edge crack. How large can the crack be if it still should be possible to plastically deform the plate in tension? What if the plate is very long?Ans.lc= 8.79mm.

6.44 A polymethylmethacrylate (KIc= 3MPa√m) model of a gear has a 1-mm half-length crack formed in its fillet curve (where the tensile stress is maximum). The model is loaded until the crack starts to propagate. Assume thatY = 1.5. How much higher a load can a gear made of AISI 4340 steel tempered to 425^◦C carry with the same crack and the same geometry? Ans. σsteel/σplexiglass = 29.13.

6.45 A pressure vessel made of aluminum alloy 2024-T351 is manufactured for a safety factor of 3.0 guarding against yielding. The material contains cracks through the wall thickness with a crack half-length less than 3 mm. Y = 1. Find the safety factor when considering crack propagation.Ans.ns= 3.17.

6.46 The clamping screws holding the top lid of a nuclear reactor are made of AISI 4340 steel tempered at 260^◦C.

They are stressed to a maximum level of 1250 MPa during a pressurization test before starting the reactor.

Find the safety factor guarding against yielding and the safety factor guarding against crack propagation if the initial cracks in the material have Y = 1 and a = 1 mm. Also, find the safety factor if the same material is used, but tempered at 425^◦C.Ans. AISI 4340 tempered at 260^◦C:ns= 0.714.

6.47 A glass tube used in a pressure vessel is made of aluminum oxide (sapphire) to make it possible to apply 30- MPa pressure and still have a safety factor of 2 guarding against fracture. For a soda-lime glass of the same geometry only 7.5-MPa pressure can be allowed if a safety

factor of 2 is to be maintained. Find the size of the cracks the glass tube can tolerate at 7.5 MPa pressure and a safety factor of 2. Y = 1for both tubes. Ans.Sapphire:

lc<75.2µm, glass:lc<65.6µm.

6.48 A stress optic model used for demonstrating the stress concentrations at the ends of a crack is made of polymethylmethacrylate. An artificially made crack 50 mm long is perpendicular to the loading direction. Y = 1.

Calculate the highest tensile stress that can be applied to the model without propagating the crack. Ans. σnom= 3.57MPa.

6.49 A passengerless airplane requires wings that are lightweight and resistant to fracture with cracks up to 2 mm in length. The dimensionless geometry correction factorY is usually 1.5 for a safety factor of 2.

(a) What is the appropriate alloy for this application?

Ans. Either Aluminum 2020-T351 or Alloy steel 4340 tempered at 425^◦C.

(b) IfY is increased to 4.5, what kind of alloy from Table 6.1 should be used?Ans.Al 2020-T351.

6.50 The anchoring of the cables carrying a suspension bridge are made of cylindrical AISI 1080 steel bars, quenched and tempered at 800^◦C, and are 200 mm in diameter. The force transmitted from the cable to the steel bar is 3.5 MN. Calculate the safety factor against yielding.Ans.ns= 3.41.

6.51 The arm of a crane has two steel plates connected with a rivet that transfers the force in pure shear. The rivet is made of annealed AISI 1040 steel and has a circular cross section with a diameter of 25 mm. The load on the rivet is 15 kN. Calculate the safety factor according to the von Mises criterion.Ans.ns= 5.67.

6.52 An aluminum alloy yields at a stress of 50 MPa in uniaxial tension. If this material is subjected to the stresses σ1 = 25MPa,σ2= 15MPa, andσ3 =−26MPa, will it yield? Explain.

6.53 A material with a yield stress of 70 MPa is subjected to principal (normal) stresses ofσ1 = σ, σ2 = 0, and σ3=−σ/2. What is the value ofσwhen the metal yields according to the von Mises criterion? What ifσ2=σ/3?

Ans.σ= 52.91MPa.

6.54 Assume that a material with a uniaxial yield strengthSy

yields under a stress state of principal stressesσ1, σ2, andσ3, whereσ1 ≥ σ2 ≥σ3. Show that the superpo- sition of a hydrostatic stress pon this system (such as placing the specimen in a chamber pressurized with a liquid) does not affect yielding. In other words, the material will still yield according to yield criteria.

6.55 A machine element is loaded so that the principal normal stresses at the critical location for a biaxial stress state areσ1 = 100MPa andσ2 = −50MPa. The material is ductile with a yield strength of 300 MPa. Find the safety factor according to

(a) The maximum-shear-stress theory (MSST) Ans.

ns= 2.00.

(b) The distortion-energy theory (DET) Ans. ns = 2.27.

Quantitative Problems 157 6.56 A bolt is tightened, subjecting its shank to a tensile

stress of 80 MPa and a torsional shear stress of 50 MPa at a critical point. All of the other stresses are zero.

Find the safety factor at the critical point by the DET and the MSST. The material is high-carbon steel (AISI 1080, quenched and tempered at 800^◦C). Will the bolt fail because of the static loading? Ans. ns,DET = 3.22, ns,MSST= 2.97.

6.57 A torque is applied to a piece of chalk used in a class- room until the chalk cracks. Using the maximum- normal-stress theory (MNST) and assuming the tensile strength of the chalk to be small relative to its compressive strength, determine the angle of the cross section at which the chalk cracks.Ans.45^◦.

6.58 A cantilevered bar 500 mm long with square cross section has 25-mm sides. Three perpendicular forces are applied to its free end, a 1000 N force is applied in thex- direction, a 100 N force is applied in theydirection, and an equivalent force of 100 N is applied in thez-direction.

Calculate the equivalent stress at the clamped end of the bar by using the DET when the sides of the square cross section are parallel with they- and z-directions. Ans.

σe= 40MPa.

6.59 A shaft transmitting torque from a gearbox to the rear axle of a truck is unbalanced, so that a centrifugal load of 500 N acts at the middle of the 3-m-long shaft. The annealed AISI 1040 tubular steel shaft has an outer diameter of 70 mm and an inner diameter of 58 mm. Simul- taneously, the shaft transmits a torque of 3000 N-m. Use the DET to determine the safety factor guarding against yielding.Ans.ns= 2.39.

6.60 The right-angle-cantilevered bracket used in Problem 5.30, Sketchw, has a concentrated force of 1000 N and a torque of 300 N-m. Calculate the safety factor. Use the DET and neglect transverse shear. Assume that the bracket is made of annealed AISI 1040 steel and use the following values: a = 0.5m,b = 0.3m,d = 0.035m, E= 205GPa, andν= 0.3.Ans.ns= 1.76.

6.61 A 100-mm-diameter shaft is subjected to a 10-kN-m steady bending moment, an 8-kN-m steady torque, and a 120-kN axial force. The yield strength of the shaft material is 600 MPa. Use the MSST and the DET to determine the safety factors for the various types of loading.

Ans.ns= 4.21.

6.62 Use the MSST and the DET to determine the safety factor for 2024-T4 aluminum alloy for each of the following stress states:

(a) σx = 10MPa,σy = −60MPa. Ans. ns,MSST = 4.64.

(b) σx=σy=τxy=−30MPa.Ans.ns,DET= 5.42.

ns,MSST= 1.67.

(d) σx = 2σy =−112MPa, andτxy = 33MPa. Ans.

ns,DET= 2.31.

6.63 Four different stress elements, each made of the same material, are loaded as shown in Sketchesg,h,i, andj.

Use the MSST and the DET to determine which element is the most critical.Ans.Sketchgis most critical.

21 MPa

7.5 MPa

28.5 MPa

30 MPa

10 MPa

30 MPa (g) (h)

(i) (j)

Sketchesg,h,i, andj, for Problem 6.63 6.64 The rod shown in Sketchk is made of annealed AISI

1040 steel and has two 90^◦ bends. Use the MSST and the DET to determine the minimum rod diameter for a safety factor of 2 at the most critical section.Ans.d= 35 mm.

300 mm

750 mm

y' x y

x' 50 mm

1500 N 100 N

800 N

Sketchk, for Problem 6.64

6.65 The shaft shown in Sketchlis made of AISI 1020 steel, quenched and tempered at 870^◦C. Determine the most critical section by using the MSST and the DET. Dimen- sions of the various diameters shown in Sketch j are d= 30mm,D= 45mm, andd2= 40mm.

40 mm 40 mm

40 mm

d r = 6 mm r = 9 mm

T = 500 N-m 100 kN 10 kN d₂

Sketchl, for Problem 6.65

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