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Unformatted text preview: bud21932_ch05_204-256 07/8/06 6:38 PM Page 204 CONFIRMING PAGES PART 2 Failure Prevention bud21932_ch05_204-256 07/8/06 6:38 PM Page 205 CONFIRMING PAGES 5 Chapter Outline Static Strength Failure Theories 5–1 5–2 5–3 5–4 5–5 5–6 5–7 5–8 5–9 5–10 5–11 5–12 5–13 5–14 208 Failures Resulting from Static Loading Stress Concentration 211 209 Maximum-Shear-Stress Theory for Ductile Materials Distortion-Energy Theory for Ductile Materials Coulomb-Mohr Theory for Ductile Materials Failure of Ductile Materials Summary 222 213 219 211 Maximum-Normal-Stress Theory for Brittle Materials Modifications of the Mohr Theory for Brittle Materials Failure of Brittle Materials Summary Selection of Failure Criteria Stochastic Analysis 230 231 229 226 227 Introduction to Fracture Mechanics 240 246 Important Design Equations 205 bud21932_ch05_204-256 07/8/06 6:38 PM Page 206 CONFIRMING PAGES 206 Mechanical Engineering Design In Chap. 1 we learned that strength is a property or characteristic of a mechanical element. This property results from the material identity, the treatment and processing incidental to creating its geometry, and the loading, and it is at the controlling or critical location. In addition to considering the strength of a single part, we must be cognizant that the strengths of the mass-produced parts will all be somewhat different from the others in the collection or ensemble because of variations in dimensions, machining, forming, and composition. Descriptors of strength are necessarily statistical in nature, involving parameters such as mean, standard deviations, and distributional identification. A static load is a stationary force or couple applied to a member. To be stationary, the force or couple must be unchanging in magnitude, point or points of application, and direction. A static load can produce axial tension or compression, a shear load, a bending load, a torsional load, or any combination of these. To be considered static, the load cannot change in any manner. In this chapter we consider the relations between strength and static loading in order to make the decisions concerning material and its treatment, fabrication, and geometry for satisfying the requirements of functionality, safety, reliability, competitiveness, usability, manufacturability, and marketability. How far we go down this list is related to the scope of the examples. “Failure” is the first word in the chapter title. Failure can mean a part has separated into two or more pieces; has become permanently distorted, thus ruining its geometry; has had its reliability downgraded; or has had its function compromised, whatever the reason. A designer speaking of failure can mean any or all of these possibilities. In this chapter our attention is focused on the predictability of permanent distortion or separation. In strength-sensitive situations the designer must separate mean stress and mean strength at the critical location sufficiently to accomplish his or her purposes. Figures 5–1 to 5–5 are photographs of several failed parts. The photographs exemplify the need of the designer to be well-versed in failure prevention. Toward this end we shall consider one-, two-, and three-dimensional stress states, with and without stress concentrations, for both ductile and brittle materials. Figure 5–1 (a) Failure of a truck drive-shaft spline due to corrosion fatigue. Note that it was necessary to use clear tape to hold the pieces in place. (b) Direct end view of failure. bud21932_ch05_204-256 07/8/06 6:38 PM Page 207 CONFIRMING PAGES Failures Resulting from Static Loading 207 Figure 5–2 Impact failure of a lawnmower blade driver hub. The blade impacted a surveying pipe marker. Figure 5–3 Failure of an overhead-pulley retaining bolt on a weightlifting machine. A manufacturing error caused a gap that forced the bolt to take the entire moment load. Figure 5–4 Chain test fixture that failed in one cycle. To alleviate complaints of excessive wear, the manufacturer decided to case-harden the material. (a) Two halves showing fracture; this is an excellent example of brittle fracture initiated by stress concentration. (b) Enlarged view of one portion to show cracks induced by stress concentration at the support-pin holes. bud21932_ch05_204-256 07/8/06 6:38 PM Page 208 CONFIRMING PAGES 208 Mechanical Engineering Design Figure 5–5 Valve-spring failure caused by spring surge in an oversped engine. The fractures exhibit the classic 45◦ shear failure. 5–1 Static Strength Ideally, in designing any machine element, the engineer should have available the results of a great many strength tests of the particular material chosen. These tests should be made on specimens having the same heat treatment, surface finish, and size as the element the engineer proposes to design; and the tests should be made under exactly the same loading conditions as the part will experience in service. This means that if the part is to experience a bending load, it should be tested with a bending load. If it is to be subjected to combined bending and torsion, it should be tested under combined bending and torsion. If it is made of heat-treated AISI 1040 steel drawn at 500◦ C with a ground finish, the specimens tested should be of the same material prepared in the same manner. Such tests will provide very useful and precise information. Whenever such data are available for design purposes, the engineer can be assured of doing the best possible job of engineering. The cost of gathering such extensive data prior to design is justified if failure of the part may endanger human life or if the part is manufactured in sufficiently large quantities. Refrigerators and other appliances, for example, have very good reliabilities because the parts are made in such large quantities that they can be thoroughly tested in advance of manufacture. The cost of making these tests is very low when it is divided by the total number of parts manufactured. You can now appreciate the following four design categories: 1 Failure of the part would endanger human life, or the part is made in extremely large quantities; consequently, an elaborate testing program is justified during design. The part is made in large enough quantities that a moderate series of tests is feasible. The part is made in such small quantities that testing is not justified at all; or the design must be completed so rapidly that there is not enough time for testing. 2 3 bud21932_ch05_204-256 07/8/06 6:38 PM Page 209 CONFIRMING PAGES Failures Resulting from Static Loading 209 4 The part has already been designed, manufactured, and tested and found to be unsatisfactory. Analysis is required to understand why the part is unsatisfactory and what to do to improve it. More often than not it is necessary to design using only published values of yield strength, ultimate strength, percentage reduction in area, and percentage elongation, such as those listed in Appendix A. How can one use such meager data to design against both static and dynamic loads, two- and three-dimensional stress states, high and low temperatures, and very large and very small parts? These and similar questions will be addressed in this chapter and those to follow, but think how much better it would be to have data available that duplicate the actual design situation. 5–2 Stress Concentration Stress concentration (see Sec. 3–13) is a highly localized effect. In some instances it may be due to a surface scratch. If the material is ductile and the load static, the design load may cause yielding in the critical location in the notch. This yielding can involve strain strengthening of the material and an increase in yield strength at the small critical notch location. Since the loads are static and the material is ductile, that part can carry the loads satisfactorily with no general yielding. In these cases the designer sets the geometric (theoretical) stress concentration factor K t to unity. The rationale can be expressed as follows. The worst-case scenario is that of an idealized non–strain-strengthening material shown in Fig. 5–6. The stress-strain curve rises linearly to the yield strength Sy , then proceeds at constant stress, which is equal to Sy . Consider a filleted rectangular bar as depicted in Fig. A–15–5, where the crosssection area of the small shank is 1 in2. If the material is ductile, with a yield point of 40 kpsi, and the theoretical stress-concentration factor (SCF) K t is 2, • A load of 20 kip induces a tensile stress of 20 kpsi in the shank as depicted at point A in Fig. 5–6. At the critical location in the fillet the stress is 40 kpsi, and the SCF is K = σmax /σnom = 40/20 = 2 . Figure 5–6 An idealized stress-strain curve. The dashed line depicts a strain-strengthening material. Tensile stress , kpsi 50 C D B E Sy A 0 Tensile strain, bud21932_ch05_204-256 07/8/06 6:38 PM Page 210 CONFIRMING PAGES 210 Mechanical Engineering Design • A load of 30 kip induces a tensile stress of 30 kpsi in the shank at point B. The fillet stress is still 40 kpsi (point D), and the SCF K = σmax /σnom = Sy /σ = 40/30 = 1.33 . • At a load of 40 kip the induced tensile stress (point C) is 40 kpsi in the shank. At the critical location in the fillet, the stress (at point E) is 40 kpsi. The SCF K = σmax /σnom = Sy /σ = 40/40 = 1 . For materials that strain-strengthen, the critical location in the notch has a higher Sy . The shank area is at a stress level a little below 40 kpsi, is carrying load, and is very near its failure-by-general-yielding condition. This is the reason designers do not apply K t in static loading of a ductile material loaded elastically, instead setting K t = 1. When using this rule for ductile materials with static loads, be careful to assure yourself that the material is not susceptible to brittle fracture (see Sec. 5–12) in the environment of use. The usual definition of geometric (theoretical) stress-concentration factor for normal stress K t and shear stress K ts is σmax = K t σnom τmax = K ts τnom (a) (b) Since your attention is on the stress-concentration factor, and the definition of σnom or τnom is given in the graph caption or from a computer program, be sure the value of nominal stress is appropriate for the section carrying the load. Brittle materials do not exhibit a plastic range. A brittle material “feels” the stress concentration factor K t or K ts , which is applied by using Eq. (a) or (b). An exception to this rule is a brittle material that inherently contains microdiscontinuity stress concentration, worse than the macrodiscontinuity that the designer has in mind. Sand molding introduces sand particles, air, and water vapor bubbles. The grain structure of cast iron contains graphite flakes (with little strength), which are literally cracks introduced during the solidification process. When a tensile test on a cast iron is performed, the strength reported in the literature includes this stress concentration. In such cases K t or K ts need not be applied. An important source of stress-concentration factors is R. E. Peterson, who compiled them from his own work and that of others.1 Peterson developed the style of presentation in which the stress-concentration factor K t is multiplied by the nominal stress σnom to estimate the magnitude of the largest stress in the locality. His approximations were based on photoelastic studies of two-dimensional strips (Hartman and Levan, 1951; Wilson and White, 1973), with some limited data from three-dimensional photoelastic tests of Hartman and Levan. A contoured graph was included in the presentation of each case. Filleted shafts in tension were based on two-dimensional strips. Table A–15 provides many charts for the theoretical stress-concentration factors for several fundamental load conditions and geometry. Additional charts are also available from Peterson.2 Finite element analysis (FEA) can also be applied to obtain stress-concentration factors. Improvements on K t and K ts for filleted shafts were reported by Tipton, Sorem, and Rolovic.3 1 R. E. Peterson, “Design Factors for Stress Concentration,” Machine Design, vol. 23, no. 2, February 1951; no. 3, March 1951; no. 5, May 1951; no. 6, June 1951; no. 7, July 1951. 2 3 Walter D. Pilkey, Peterson’s Stress Concentration Factors, 2nd ed, John Wiley & Sons, New York, 1997. S. M. Tipton, J. R. Sorem Jr., and R. D. Rolovic, “Updated Stress-Concentration Factors for Filleted Shafts in Bending and Tension,” Trans. ASME, Journal of Mechanical Design, vol. 118, September 1996, pp. 321–327. bud21932_ch05_204-256 07/8/06 6:38 PM Page 211 CONFIRMING PAGES Failures Resulting from Static Loading 211 5–3 Failure Theories Section 5–1 illustrated some ways that loss of function is manifested. Events such as distortion, permanent set, cracking, and rupturing are among the ways that a machine element fails. Testing machines appeared in the 1700s, and specimens were pulled, bent, and twisted in simple loading processes. If the failure mechanism is simple, then simple tests can give clues. Just what is simple? The tension test is uniaxial (that’s simple) and elongations are largest in the axial direction, so strains can be measured and stresses inferred up to “failure.” Just what is important: a critical stress, a critical strain, a critical energy? In the next several sections, we shall show failure theories that have helped answer some of these questions. Unfortunately, there is no universal theory of failure for the general case of material properties and stress state. Instead, over the years several hypotheses have been formulated and tested, leading to today’s accepted practices. Being accepted, we will characterize these “practices” as theories as most designers do. Structural metal behavior is typically classified as being ductile or brittle, although under special situations, a material normally considered ductile can fail in a brittle manner (see Sec. 5–12). Ductile materials are normally classified such that ε f ≥ 0.05 and have an identifiable yield strength that is often the same in compression as in tension ( Syt = Syc = Sy ). Brittle materials, ε f < 0.05, do not exhibit an identifiable yield strength, and are typically classified by ultimate tensile and compressive strengths, Sut and Suc , respectively (where Suc is given as a positive quantity). The generally accepted theories are: Ductile materials (yield criteria) • Maximum shear stress (MSS), Sec. 5–4 • Distortion energy (DE), Sec. 5–5 • Ductile Coulomb-Mohr (DCM), Sec. 5–6 Brittle materials (fracture criteria) • Maximum normal stress (MNS), Sec. 5–8 • Brittle Coulomb-Mohr (BCM), Sec. 5–9 • Modified Mohr (MM), Sec. 5–9 It would be inviting if we had one universally accepted theory for each material type, but for one reason or another, they are all used. Later, we will provide rationales for selecting a particular theory. First, we will describe the bases of these theories and apply them to some examples. 5–4 Maximum-Shear-Stress Theory for Ductile Materials The maximum-shear-stress theory predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tensiontest specimen of the same material when that specimen begins to yield. The MSS theory is also referred to as the Tresca or Guest theory. Many theories are postulated on the basis of the consequences seen from tensile tests. As a strip of a ductile material is subjected to tension, slip lines (called Lüder lines) form at approximately 45° with the axis of the strip. These slip lines are the bud21932_ch05_204-256 07/8/06 6:39 PM Page 212 CONFIRMING PAGES 212 Mechanical Engineering Design beginning of yield, and when loaded to fracture, fracture lines are also seen at angles approximately 45° with the axis of tension. Since the shear stress is maximum at 45° from the axis of tension, it makes sense to think that this is the mechanism of failure. It will be shown in the next section, that there is a little more going on than this. However, it turns out the MSS theory is an acceptable but conservative predictor of failure; and since engineers are conservative by nature, it is quite often used. Recall that for simple tensile stress, σ = P / A , and the maximum shear stress occurs on a surface 45° from the tensile surface with a magnitude of τmax = σ/2. So the maximum shear stress at yield is τmax = Sy /2. For a general state of stress, three principal stresses can be determined and ordered such that σ1 ≥ σ2 ≥ σ3 . The maximum shear stress is then τmax = (σ1 − σ3 )/2 (see Fig. 3–12). Thus, for a general state of stress, the maximum-shear-stress theory predicts yielding when τmax = Sy σ1 − σ3 ≥ 2 2 or σ1 − σ3 ≥ Sy (5–1) Note that this implies that the yield strength in shear is given by Ssy = 0.5 Sy (5–2) which, as we will see later is about 15 percent low (conservative). For design purposes, Eq. (5–1) can be modified to incorporate a factor of safety, n. Thus, τmax = Sy 2n or σ1 − σ3 = Sy n (5–3) Plane stress problems are very common where one of the principal stresses is zero, and the other two, σ A and σ B , are determined from Eq. (3–13). Assuming that σ A ≥ σ B , there are three cases to consider in using Eq. (5–1) for plane stress: Case 1: σ A ≥ σ B ≥ 0. For this case, σ1 = σ A and σ3 = 0. Equation (5–1) reduces to a yield condition of σ A ≥ Sy Case 2: σ A ≥ 0 ≥ σ B . Here, σ1 = σ A and σ3 = σ B , and Eq. (5–1) becomes σ A − σ B ≥ Sy σ B ≤ − Sy (5–5) (5–4) Case 3: 0 ≥ σ A ≥ σ B . For this case, σ1 = 0 and σ3 = σ B , and Eq. (5–1) gives (5–6) Equations (5–4) to (5–6) are represented in Fig. 5–7 by the three lines indicated in the σ A , σ B plane. The remaining unmarked lines are cases for σ B ≥ σ A , which are not normally used. Equations (5–4) to (5–6) can also be converted to design equations by substituting equality for the equal to or greater sign and dividing Sy by n. Note that the first part of Eq. (5-3), τmax = Sy /2n , is sufficient for design purposes provided the designer is careful in determining τmax . For plane stress, Eq. (3–14) does not always predict τmax . However, consider the special case when one normal stress is zero in the plane, say σx and τx y have values and σ y = 0. It can be easily shown that this is a Case 2 problem, and the shear stress determined by Eq. (3–14) is τmax . Shaft design problems typically fall into this category where a normal stress exists from bending and/or axial loading, and a shear stress arises from torsion. bud21932_ch05_204-256 07/8/06 6:39 PM Page 213 CONFIRMING PAGES Failures Resulting from Static Loading 213 Figure 5–7 The maximum-shear-stress (MSS) theory for plane stress, where σ A and σ B are the two nonzero principal stresses. – Sy B Sy Case 1 Sy A Case 2 – Sy Case 3 5–5 Distortion-Energy Theory for Ductile Materials The distortion-energy theory predicts that yielding occurs when the distortion strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension or compression of the same material. The distortion-energy (DE) theory originated from the observation that ductile materials stressed hydrostatically exhibited yield strengths greatly in excess of the values given by the simple tension test. Therefore it was postulated that yielding was not a simple tensile or compressive phenomenon at all, but, rather, that it was related somehow to the angular distortion of the stressed element. To develop the theory, note, in Fig. 5–8a, the unit volume subjected to any three-dimensional stress state designated by the stresses σ1 , σ2 , and σ3 . The stress state shown in Fig. 5–8b is one of hydrostatic tension due to the stresses σav acting in each of the same principal directions as in Fig. 5–8a. The formula for σav is simply σav = σ1 + σ2 + σ3 3 (a) Thus the element in Fig. 5–8b undergoes pure volume change, that is, no angular distortion. If we regard σav as a component of σ1 , σ2 , and σ3 , then this component can be 2 av 2– av = 1 av + 3– 1– av 3 1 > 2 > av 3 av (a) Triaxial stresses (b) Hydrostatic component (c) Distortional component Figure 5–8 (a) Element with triaxial stresses; this element undergoes both volume change and angular distortion. (b) Element under hydrostatic tension undergoes only volume change. (c) Element has angular distortion without volume change. bud21932_ch05_204-256 07/8/06 6:39 PM Page 214 CONFIRMING PAGES 214 Mechanical Engineering Design subtracted from them, resulting in the stress state shown in Fig. 5–8c. This element is subjected to pure angular distortion, that is, no volume change. The strain energy per unit volume for simple tension is u = 1 σ . For the element 2 of Fig. 5–8a the strain energy per unit volume is u = 1 [ 1 σ1 + 2 σ2 + 3 σ3 ] . 2 Substituting Eq. (3–19) for the principal strains gives u= 1 2 2 σ 2 + σ2 + σ3 − 2ν(σ1 σ2 + σ2 σ3 + σ3 σ1 ) 2E 1 (b) The strain energy for producing only volume change u v can be obtained by substituting σav for σ1 , σ2 , and σ3 in Eq. (b). The result is uv = 2 3σav (1 − 2ν) 2E (c) If we now substitute the square of Eq. (a) in Eq. (c) and simplify the expression, we get uv = 1 − 2ν 2 2 2 σ1 + σ2 + σ3 + 2σ1 σ2 + 2σ2 σ3 + 2σ3 σ1 6E (5–7) Then the distortion energy is obtained by subtracting Eq. (5–7) from Eq. (b). This gives ud = u − uv = 1+ν 3E (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 2 (5–8) Note that the distortion energy is zero if σ1 = σ2 = σ3 . For the simple tensile test, at yield, σ1 = Sy and σ2 = σ3 = 0, and from Eq. (5–8) the distortion energy is ud = 1+ν 2 S 3E y (5–9) So for the general state of stress given by Eq. (5–8), yield is predicted if Eq. (5–8) equals or exceeds Eq. (5–9). This gives (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 2 1/2 ≥ Sy (5–10) If we had a simple case of tension σ , then yield would occur when σ ≥ Sy . Thus, the left of Eq. (5–10) can be thought of as a single, equivalent, or effective stress for the entire general state of stress given by σ1 , σ2 , and σ3 . This effective stress is usually called the von Mises stress, σ , named after Dr. R. von Mises, who contributed to the theory. Thus Eq. (5–10), for yield, can be written as σ ≥ Sy where the von Mises stress is σ= (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 2 1/2 (5–11) (5–12) For plane stress, let σ A and σ B be the two nonzero principal stresses. Then from Eq. (5–12), we get 2 2 σ = σ A − σ AσB + σB 1/2 (5–13) bud21932_ch05_204-256 07/8/06 6:39 PM Page 215 CONFIRMING PAGES Failures Resulting from Static Loading 215 Figure 5–9 The distortion-energy (DE) theory for plane stress states. This is a plot of points obtained from Eq. (5–13) with σ = Sy . – Sy Sy B Sy A Pure shear load line ( – Sy DE MSS A B ) Equation (5–13) is a rotated ellipse in the σ A , σ B plane, as shown in Fig. 5–9 with σ = Sy . The dotted lines in the figure represent the MSS theory, which can be seen to be more restrictive, hence, more conservative.4 Using xyz components of three-dimensional stress, the von Mises stress can be written as 1 2 2 2 σ = √ (σx − σ y )2 + (σ y − σz )2 + (σz − σx )2 + 6 τx y + τ yz + τzx 2 and for plane stress, 2 2 2 σ = σx − σx σ y + σ y + 3τx y 1/2 1/2 (5–14) (5–15) The distortion-energy theory is also called: • The von Mises or von Mises–Hencky theory • The shear-energy theory • The octahedral-shear-stress theory Understanding octahedral shear stress will shed some light on why the MSS is conservative. Consider an isolated element in which the normal stresses on each surface are equal to the hydrostatic stress σav . There are eight surfaces symmetric to the principal directions that contain this stress. This forms an octahedron as shown in Fig. 5–10. The shear stresses on these surfaces are equal and are called the octahedral shear stresses (Fig. 5–10 has only one of the octahedral surfaces labeled). Through coordinate transformations the octahedral shear stress is given by5 τoct = 1 (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 3 1/2 (5–16) 4 The three-dimensional equations for DE and MSS can be plotted relative to three-dimensional σ1 , σ2 , σ3 , coordinate axes. The failure surface for DE is a circular cylinder with an axis inclined at 45° from each principal stress axis, whereas the surface for MSS is a hexagon inscribed within the cylinder. See Arthur P. Boresi and Richard J. Schmidt, Advanced Mechanics of Materials, 6th ed., John Wiley & Sons, New York, 2003, Sec. 4.4. 5 For a derivation, see Arthur P. Boresi, op. cit., pp. 36–37. bud21932_ch05_204-256 07/8/06 6:39 PM Page 216 CONFIRMING PAGES 216 Mechanical Engineering Design 2 Figure 5–10 Octahedral surfaces. av oct 1 3 Under the name of the octahedral-shear-stress theory, failure is assumed to occur whenever the octahedral shear stress for any stress state equals or exceeds the octahedral shear stress for the simple tension-test specimen at failure. As before, on the basis of the tensile test results, yield occurs when σ1 = Sy and σ2 = σ3 = 0. From Eq. (5–16) the octahedral shear stress under this condition is √ 2 τoct = Sy (5–17) 3 When, for the general stress case, Eq. (5–16) is equal or greater than Eq. (5–17), yield is predicted. This reduces to (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 2 1/2 ≥ Sy (5–18) which is identical to Eq. (5–10), verifying that the maximum-octahedral-shear-stress theory is equivalent to the distortion-energy theory. The model for the MSS theory ignores the contribution of the normal stresses on the 45° surfaces of the tensile specimen. However, these stresses are P /2 A , and not the hydrostatic stresses which are P /3 A . Herein lies the difference between the MSS and DE theories. The mathematical manipulation involved in describing the DE theory might tend to obscure the real value and usefulness of the result. The equations given allow the most complicated stress situation to be represented by a single quantity, the von Mises stress, which then can be compared against the yield strength of the material through Eq. (5–11). This equation can be expressed as a design equation by σ= Sy n (5–19) The distortion-energy theory predicts no failure under hydrostatic stress and agrees well with all data for ductile behavior. Hence, it is the most widely used theory for ductile materials and is recommended for design problems unless otherwise specified. One final note concerns the shear yield strength. Consider a case of pure shear τx y , where for plane stress σx = σ y = 0. For yield, Eq. (5–11) with Eq. (5–15) gives 2 3τx y 1/2 = Sy or Sy τx y = √ = 0.577 Sy 3 (5–20) bud21932_ch05_204-256 07/8/06 6:39 PM Page 217 CONFIRMING PAGES Failures Resulting from Static Loading 217 Thus, the shear yield strength predicted by the distortion-energy theory is Ssy = 0.577 Sy (5–21) which as stated earlier, is about 15 percent greater than the 0.5 Sy predicted by the MSS theory. For pure shear, τx y the principal stresses from Eq. (3–13) are σ A = −σ B = τx y . The load line for this case is in the third quadrant at an angle of 45o from the σ A , σ B axes shown in Fig. 5–9. EXAMPLE 5–1 A hot-rolled steel has a yield strength of Syt = Syc = 100 kpsi and a true strain at fracture of ε f = 0.55. Estimate the factor of safety for the following principal stress states: (a) 70, 70, 0 kpsi. (b) 30, 70, 0 kpsi. (c) 0, 70, −30 kpsi. (d ) 0, −30, −70 kpsi. (e) 30, 30, 30 kpsi. Since ε f > 0.05 and Syc and Syt are equal, the material is ductile and the distortionenergy (DE) theory applies. The maximum-shear-stress (MSS) theory will also be applied and compared to the DE results. Note that cases a to d are plane stress states. (a) The ordered principal stresses are σ A = σ1 = 70, σ B = σ2 = 70, σ3 = 0 kpsi. DE From Eq. (5–13), σ = [702 − 70(70) + 702 ]1/2 = 70 kpsi Solution Answer n= Sy 100 = = 1.43 σ 70 MSS Case 1, using Eq. (5–4) with a factor of safety, Answer n= Sy 100 = = 1.43 σA 70 (b) The ordered principal stresses are σ A = σ1 = 70, σ B = σ2 = 30, σ3 = 0 kpsi. DE Answer σ = [702 − 70(30) + 302 ]1/2 = 60.8 kpsi n= MSS Case 1, using Eq. (5–4), Answer n= Sy 100 = = 1.43 σA 70 Sy 100 = 1.64 = σ 60.8 bud21932_ch05_204-256 07/8/06 6:39 PM Page 218 CONFIRMING PAGES 218 Mechanical Engineering Design (c) The ordered principal stresses are σ A = σ1 = 70, σ2 = 0, σ B = σ3 = −30 kpsi. DE Answer σ = [702 − 70(−30) + (−30)2 ]1/2 = 88.9 kpsi n= MSS Case 2, using Eq. (5–5), Answer n= Sy 100 = = 1.00 σ A − σB 70 − (−30) Sy 100 = = 1.13 σ 88.9 (d ) The ordered principal stresses are σ1 = 0, σ A = σ2 = −30, σ B = σ3 = −70 kpsi. DE Answer σ = [(−70)2 − (−70)(−30) + (−30)2 ]1/2 = 60.8 kpsi n= MSS Case 3, using Eq. (5–6), Answer n=− Sy 100 =− = 1.43 σB −70 Sy 100 = = 1.64 σ 60.8 (e) The ordered principal stresses are σ1 = 30, σ2 = 30, σ3 = 30 kpsi DE From Eq. (5–12), σ= (30 − 30)2 + (30 − 30)2 + (30 − 30)2 2 n= MSS From Eq. (5–3), Answer n= Sy 100 →∞ = σ1 − σ3 30 − 30 Sy 100 = →∞ σ 0 1/2 = 0 kpsi Answer A tabular summary of the factors of safety is included for comparisons. (a) DE MSS 1.43 1.43 (b) 1.64 1.43 (c) 1.13 1.00 (d ) 1.64 1.43 (e) ∞ ∞ Since the MSS theory is on or within the boundary of the DE theory, it will always predict a factor of safety equal to or less than the DE theory, as can be seen in the table. For each case, except case (e), the coordinates and load lines in the σ A , σ B plane are shown in Fig. 5–11. Case (e) is not plane stress. Note that the load line for case (a) is bud21932_ch05_204-256 07/8/06 6:39 PM Page 219 CONFIRMING PAGES Failures Resulting from Static Loading 219 Figure 5–11 Load lines for Example 5–1. Sy B (a) B (b) – Sy A Sy A (c) – Sy (d ) DE MSS Load lines the only plane stress case given in which the two theories agree, thus giving the same factor of safety. 5–6 Coulomb-Mohr Theory for Ductile Materials Not all materials have compressive strengths equal to their corresponding tensile values. For example, the yield strength of magnesium alloys in compression may be as little as 50 percent of their yield strength in tension. The ultimate strength of gray cast irons in compression varies from 3 to 4 times greater than the ultimate tensile strength. So, in this section, we are primarily interested in those theories that can be used to predict failure for materials whose strengths in tension and compression are not equal. Historically, the Mohr theory of failure dates to 1900, a date that is relevant to its presentation. There were no computers, just slide rules, compasses, and French curves. Graphical procedures, common then, are still useful today for visualization. The idea of Mohr is based on three “simple” tests: tension, compression, and shear, to yielding if the material can yield, or to rupture. It is easier to define shear yield strength as Ssy than it is to test for it. The practical difficulties aside, Mohr’s hypothesis was to use the results of tensile, compressive, and torsional shear tests to construct the three circles of Fig. 5–12 defining a failure envelope, depicted as line ABCDE in the figure, above the σ axis. The failure envelope need not be straight. The argument amounted to the three Mohr circles describing the stress state in a body (see Fig. 3–12) growing during loading until one of them became tangent to the failure envelope, thereby defining failure. Was the form of the failure envelope straight, circular, or quadratic? A compass or a French curve defined the failure envelope. A variation of Mohr’s theory, called the Coulomb-Mohr theory or the internal-friction theory, assumes that the boundary BCD in Fig. 5–12 is straight. With this assumption only the tensile and compressive strengths are necessary. Consider the conventional ordering of the principal stresses such that σ1 ≥ σ2 ≥ σ3 . The largest circle connects σ1 and σ3 , as shown in Fig. 5–13. The centers of the circles in Fig. 5–13 are C1, C2, and C3. Triangles OBiCi are similar, therefore B3 C3 − B1 C1 B2 C2 − B1 C1 = OC2 − OC1 OC3 − OC1 bud21932_ch05_204-256 07/8/06 6:39 PM Page 220 CONFIRMING PAGES 220 Mechanical Engineering Design Figure 5–12 Three Mohr circles, one for the uniaxial compression test, one for the test in pure shear, and one for the uniaxial tension test, are used to define failure by the Mohr hypothesis. The strengths Sc and S t are the compressive and tensile strengths, respectively; they can be used for yield or ultimate strength. A B C D E –Sc St Figure 5–13 Mohr’s largest circle for a general state of stress. Coulomb-Mohr failure line B3 B2 B1 –Sc 3 C3 C2 1 C1 St O or σ1 − σ3 St Sc St − − 2 2=2 2 St σ1 + σ3 Sc St − + 2 2 2 2 Cross-multiplying and simplifying reduces this equation to σ1 σ3 − =1 St Sc (5–22) where either yield strength or ultimate strength can be used. For plane stress, when the two nonzero principal stresses are σ A ≥ σ B , we have a situation similar to the three cases given for the MSS theory, Eqs. (5–4) to (5–6). That is, Case 1: σ A ≥ σ B ≥ 0. For this case, σ1 = σ A and σ3 = 0. Equation (5–22) reduces to a failure condition of σ A ≥ St σB σA − ≥1 St Sc σ B ≤ − Sc (5–23) Case 2: σ A ≥ 0 ≥ σ B . Here, σ1 = σ A and σ3 = σ B , and Eq. (5–22) becomes (5–24) Case 3: 0 ≥ σ A ≥ σ B . For this case, σ1 = 0 and σ3 = σ B , and Eq. (5–22) gives (5–25) bud21932_ch05_204-256 07/8/06 6:39 PM Page 221 CONFIRMING PAGES Failures Resulting from Static Loading 221 Figure 5–14 Plot of the Coulomb-Mohr theory of failure for plane stress states. B St – Sc St A –Sc A plot of these cases, together with the normally unused cases corresponding to σ B ≥ σ A , is shown in Fig. 5–14. For design equations, incorporating the factor of safety n, divide all strengths by n. For example, Eq. (5–22) as a design equation can be written as σ3 1 σ1 − = St Sc n (5–26) Since for the Coulomb-Mohr theory we do not need the torsional shear strength circle we can deduce it from Eq. (5–22). For pure shear τ, σ1 = −σ3 = τ . The torsional yield strength occurs when τmax = Ssy . Substituting σ1 = −σ3 = Ssy into Eq. (5–22) and simplifying gives Ssy = Syt Syc Syt + Syc (5–27) EXAMPLE 5–2 A 25-mm-diameter shaft is statically torqued to 230 N · m. It is made of cast 195-T6 aluminum, with a yield strength in tension of 160 MPa and a yield strength in compression of 170 MPa. It is machined to final diameter. Estimate the factor of safety of the shaft. The maximum shear stress is given by τ= 16(230) 16T = 3 πd π 25 10−3 3 Solution = 75 106 N/m2 = 75 MPa The two nonzero principal stresses are 75 and −75 MPa, making the ordered principal stresses σ1 = 75, σ2 = 0 , and σ3 = −75 MPa. From Eq. (5–26), for yield, Answer n= 1 1 = 1.10 = σ1 / Syt − σ3 / Syc 75/160 − (−75)/170 Syt Syc 160(170) = 82.4 MPa = Syt + Syc 160 + 170 Alternatively, from Eq. (5–27), Ssy = bud21932_ch05_204-256 07/8/06 6:39 PM Page 222 CONFIRMING PAGES 222 Mechanical Engineering Design and τmax = 75 MPa . Thus, Answer n= Ssy 82.4 = = 1.10 τmax 75 5–7 Failure of Ductile Materials Summary Having studied some of the various theories of failure, we shall now evaluate them and show how they are applied in design and analysis. In this section we limit our studies to materials and parts that are known to fail in a ductile manner. Materials that fail in a brittle manner will be considered separately because these require different failure theories. To help decide on appropriate and workable theories of failure, Marin6 collected data from many sources. Some of the data points used to select failure theories for ductile materials are shown in Fig. 5–15.7 Mann also collected many data for copper and nickel alloys; if shown, the data points for these would be mingled with those already diagrammed. Figure 5–15 shows that either the maximum-shear-stress theory or the distortion-energy theory is acceptable for design and analysis of materials that would Figure 5–15 Experimental data superposed on failure theories. (From Fig. 7.11, p. 257, Mechanical Behavior of Materials, 2nd ed., N. E. Dowling, Prentice Hall, Englewood Cliffs, N.J., 1999. Modified to show only ductile failures.) –1.0 0 2 /Sc Oct. shear Yielding ( Sc = Sy ) Ni-Cr-Mo steel AISI 1023 steel 2024-T4 Al 3S-H Al 1.0 Max. shear 1 1.0 /Sc –1.0 6 Joseph Marin was one of the pioneers in the collection, development, and dissemination of material on the failure of engineering elements. He has published many books and papers on the subject. Here the reference used is Joseph Marin, Engineering Materials, Prentice-Hall, Englewood Cliffs, N.J., 1952. (See pp. 156 and 157 for some data points used here.) Note that some data in Fig. 5–15 are displayed along the top horizontal boundary where σ B ≥ σ A . This is often done with failure data to thin out congested data points by plotting on the mirror image of the line σB = σ A . 7 bud21932_ch05_204-256 07/8/06 6:39 PM Page 223 CONFIRMING PAGES Failures Resulting from Static Loading 223 fail in a ductile manner. You may wish to plot other theories using a red or blue pencil on Fig. 5–15 to show why they are not acceptable or are not used. The selection of one or the other of these two theories is something that you, the engineer, must decide. For design purposes the maximum-shear-stress theory is easy, quick to use, and conservative. If the problem is to learn why a part failed, then the distortion-energy theory may be the best to use; Fig. 5–15 shows that the plot of the distortion-energy theory passes closer to the central area of the data points, and thus is generally a better predictor of failure. For ductile materials with unequal yield strengths, Syt in tension and Syc in compression, the Mohr theory is the best available. However, the theory requires the results from three separate modes of tests, graphical construction of the failure locus, and fitting the largest Mohr’s circle to the failure locus. The alternative to this is to use the Coulomb-Mohr theory, which requires only the tensile and compressive yield strengths and is easily dealt with in equation form. EXAMPLE 5–3 This example illustrates the use of a failure theory to determine the strength of a mechanical element or component. The example may also clear up any confusion existing between the phrases strength of a machine part, strength of a material, and strength of a part at a point. A certain force F applied at D near the end of the 15-in lever shown in Fig. 5–16, which is quite similar to a socket wrench, results in certain stresses in the cantilevered bar OABC. This bar (OABC) is of AISI 1035 steel, forged and heat-treated so that it has a minimum (ASTM) yield strength of 81 kpsi. We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find this force. y 2 in Figure 5–16 O A 12 in z 1 1 -in D. 2 1 8 B -in R. 1-in D. 2 in C 15 in F 1 1 -in D. 2 x D bud21932_ch05_204-256 07/8/06 6:39 PM Page 224 CONFIRMING PAGES 224 Mechanical Engineering Design Solution We will assume that lever DC is strong enough and hence not a part of the problem. A 1035 steel, heat-treated, will have a reduction in area of 50 percent or more and hence is a ductile material at normal temperatures. This also means that stress concentration at shoulder A need not be considered. A stress element at A on the top surface will be subjected to a tensile bending stress and a torsional stress. This point, on the 1-in-diameter section, is the weakest section, and governs the strength of the assembly. The two stresses are σx = τzx = M 32 M 32(14 F ) = = = 142.6 F 3 I /c πd π(13 ) Tr 16T 16(15 F ) = = = 76.4 F 3 J πd π(13 ) 1/2 Employing the distortion-energy theory, we find, from Eq. (5–15), that 2 2 σ = σx + 3τzx = [(142.6 F )2 + 3(76.4 F )2 ] 1/2 = 194.5 F Equating the von Mises stress to Sy , we solve for F and get Answer F= Sy 81 000 = = 416 lbf 194.5 194.5 In this example the strength of the material at point A is Sy = 81 kpsi. The strength of the assembly or component is F = 416 lbf. Let us see how to apply the MSS theory. For a point undergoing plane stress with only one non-zero normal stress and one shear stress, the two nonzero principal stresses σ A and σ B will have opposite signs and hence fit case 2 for the MSS theory. From Eq. (3–13), σ A − σB = 2 σx 2 2 2 + τzx 1/2 2 2 = σx + 4τzx 1/2 For case 2 of the MSS theory, Eq. (5–5) applies and hence 2 2 σx + 4τzx 2 1/2 = Sy [(142.6 F ) + 4(76.4 F )2 ]1/2 = 209.0 F = 81 000 F = 388 lbf which is about 7 percent less than found for the DE theory. As stated earlier, the MSS theory is more conservative than the DE theory. EXAMPLE 5–4 The cantilevered tube shown in Fig. 5–17 is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size tube from Table A–8 using a design factor n d = 4. The bending load is F = 1.75 kN, the axial tension is P = 9.0 kN, and the torsion is T = 72 N · m. What is the realized factor of safety? bud21932_ch05_204-256 07/8/06 6:39 PM Page 225 CONFIRMING PAGES Failures Resulting from Static Loading 225 Figure 5–17 y 12 0m m F z P T x Solution Since the maximum bending moment is M = 120 F , the normal stress, for an element on the top surface of the tube at the origin, is σx = P Mc 9 120(1.75)(do /2) 9 105do + =+ =+ A I A I A I (1) where, if millimeters are used for the area properties, the stress is in gigapascals. The torsional stress at the same point is τzx = Tr 72(do /2) 36do = = J J J (2) For accuracy, we choose the distortion-energy theory as the design basis. The von Mises stress, as in the previous example, is 2 2 σ = σx + 3τzx 1/2 (3) On the basis of the given design factor, the goal for σ is σ≤ Sy 0.276 = = 0.0690 GPa nd 4 (4) where we have used gigapascals in this relation to agree with Eqs. (1) and (2). Programming Eqs. (1) to (3) on a spreadsheet and entering metric sizes from Table A–8 reveals that a 42- × 5-mm tube is satisfactory. The von Mises stress is found to be σ = 0.06043 GPa for this size. Thus the realized factor of safety is Answer n= Sy 0.276 = = 4.57 σ 0.06043 For the next size smaller, a 42- × 4-mm tube, σ = 0.07105 GPa giving a factor of safety of n= Sy 0.276 = = 3.88 σ 0.07105 bud21932_ch05_204-256 07/8/06 6:39 PM Page 226 CONFIRMING PAGES 226 Mechanical Engineering Design 5–8 Maximum-Normal-Stress Theory for Brittle Materials The maximum-normal-stress (MNS) theory states that failure occurs whenever one of the three principal stresses equals or exceeds the strength. Again we arrange the principal stresses for a general stress state in the ordered form σ1 ≥ σ2 ≥ σ3 . This theory then predicts that failure occurs whenever σ1 ≥ Sut or σ3 ≤ − Suc (5–28) where Sut and Suc are the ultimate tensile and compressive strengths, respectively, given as positive quantities. For plane stress, with the principal stresses given by Eq. (3–13), with σ A ≥ σ B , Eq. (5–28) can be written as σ A ≥ Sut or σ B ≤ − Suc (5–29) which is plotted in Fig. 5–18a. As before, the failure criteria equations can be converted to design equations. We can consider two sets of equations for load lines where σ A ≥ σ B as Figure 5–18 (a) Graph of maximum-normalstress (MNS) theory of failure for plane stress states. Stress states that plot inside the failure locus are safe. (b) Load line plot. B Sut –Suc Sut A – Suc (a) B Load line 1 O Sut A Load line 2 – Suc Load line 4 (b) Load line 3 bud21932_ch05_204-256 07/8/06 6:39 PM Page 227 CONFIRMING PAGES Failures Resulting from Static Loading 227 σA = Sut n σ A ≥ σB ≥ 0 σ A ≥ 0 ≥ σB and and σB Suc ≤ σA Sut σB Suc > σA Sut Load line 1 Load line 2 Load line 3 Load line 4 (5–30b) (5–30a) σB = − Suc n σ A ≥ 0 ≥ σB 0 ≥ σ A ≥ σB where the load lines are shown in Fig. 5–18b. Before we comment any further on the MNS theory we will explore some modifications to the Mohr theory for brittle materials. 5–9 Modifications of the Mohr Theory for Brittle Materials We will discuss two modifications of the Mohr theory for brittle materials: the BrittleCoulomb-Mohr (BCM) theory and the modified Mohr (MM) theory. The equations provided for the theories will be restricted to plane stress and be of the design type incorporating the factor of safety. The Coulomb-Mohr theory was discussed earlier in Sec. 5–6 with Eqs. (5–23) to (5–25). Written as design equations for a brittle material, they are: Brittle-Coulomb-Mohr σA = Sut n σ A ≥ σB ≥ 0 σ A ≥ 0 ≥ σB 0 ≥ σ A ≥ σB (5–31a) (5–31b) σA σB 1 − = Sut Suc n σB = − Suc n (5–31c) On the basis of observed data for the fourth quadrant, the modified Mohr theory expands the fourth quadrant as shown in Fig. 5–19. Modified Mohr σA = Sut n σ A ≥ σB ≥ 0 σ A ≥ 0 ≥ σB σB 1 ( Suc − Sut ) σ A − = Suc Sut Suc n σB = − Suc n and σB ≤1 σA and σB >1 σA (5–32a) σ A ≥ 0 ≥ σB 0 ≥ σ A ≥ σB (5–32b) (5–32c) Data are still outside this extended region. The straight line introduced by the modified Mohr theory, for σ A ≥ 0 ≥ σ B and |σ B /σ A | > 1, can be replaced by a parabolic relation bud21932_ch05_204-256 07/8/06 6:39 PM Page 228 CONFIRMING PAGES 228 Mechanical Engineering Design B, Figure 5–19 Biaxial fracture data of gray cast iron compared with various failure criteria. (Dowling, N. E., Mechanical Behavior of Materials, 2/e, 1999, p. 261. Reprinted by permission of Pearson Education, Inc., Upper Saddle River, New Jersey.) max. normal MPa 300 Sut - Mo hr mo – Suc –700 d. M ohr Cou lomb Sut 0 300 –300 A, MPa – Sut n io rs To –300 Gray cast-iron data – Suc –700 which can more closely represent some of the data.8 However, this introduces a nonlinear equation for the sake of a minor correction, and will not be presented here. 8 See J. E. Shigley, C. R. Mischke, R. G. Budynas, Mechanical Engineering Design, 7th ed., McGraw-Hill, New York, 2004, p. 275. EXAMPLE 5–5 Consider the wrench in Ex. 5–3, Fig. 5–16, as made of cast iron, machined to dimension. The force F required to fracture this part can be regarded as the strength of the component part. If the material is ASTM grade 30 cast iron, find the force F with (a) Coulomb-Mohr failure model. (b) Modified Mohr failure model. We assume that the lever DC is strong enough, and not part of the problem. Since grade 30 cast iron is a brittle material and cast iron, the stress-concentration factors K t and K ts are set to unity. From Table A–24, the tensile ultimate strength is 31 kpsi and the compressive ultimate strength is 109 kpsi. The stress element at A on the top surface will be subjected to a tensile bending stress and a torsional stress. This location, on the 1-indiameter section fillet, is the weakest location, and it governs the strength of the assembly. The normal stress σx and the shear stress at A are given by σx = K t τx y = K ts M 32 M 32(14 F ) = Kt = (1) = 142.6 F I /c π d3 π(1)3 Tr 16T 16(15 F ) = K ts = (1) = 76.4 F 3 J πd π(1)3 142.6 F − 0 2 2 Solution From Eq. (3–13) the nonzero principal stresses σ A and σ B are σ A, σB = 142.6 F + 0 ± 2 + (76.4 F )2 = 175.8 F , −33.2 F bud21932_ch05_204-256 07/8/06 6:39 PM Page 229 CONFIRMING PAGES Failures Resulting from Static Loading 229 This puts us in the fourth-quadrant of the σ A , σ B plane. (a) For BCM, Eq. (5–31b) applies with n = 1 for failure. σB 175.8 F (−33.2 F ) σA − = − =1 Sut Suc 31(103 ) 109(103 ) Solving for F yields Answer F = 167 lbf (b) For MM, the slope of the load line is |σ B /σ A | = 33.2/175.8 = 0.189 < 1. Obviously, Eq. (5–32a) applies. 175.8 F σA = =1 Sut 31(103 ) Answer F = 176 lbf As one would expect from inspection of Fig. 5–19, Coulomb-Mohr is more conservative. 5–10 Failure of Brittle Materials Summary We have identified failure or strength of brittle materials that conform to the usual meaning of the word brittle, relating to those materials whose true strain at fracture is 0.05 or less. We also have to be aware of normally ductile materials that for some reason may develop a brittle fracture or crack if used below the transition temperature. Figure 5–20 shows data for a nominal grade 30 cast iron taken under biaxial Figure 5–20 A plot of experimental data points obtained from tests on cast iron. Shown also are the graphs of three failure theories of possible usefulness for brittle materials. Note points A, B, C, and D. To avoid congestion in the first quadrant, points have been plotted for σ A > σ B as well as for the opposite sense. (Source of data: Charles F. Walton (ed.), Iron Castings Handbook, Iron Founders’ Society, 1971, pp. 215, 216, Cleveland, Ohio.) Modified Mohr –Sut B 30 Sut –120 – Suc –90 –60 –30 30 Sut A ASTM No. 30 C.I. Sut = 31 kpsi, Suc = 109 kpsi Coulomb-Mohr –30 – Sut B B A A = –1 –60 Maximum-normal-stress –90 B C D A –120 – Suc –150 bud21932_ch05_204-256 07/8/06 6:39 PM Page 230 CONFIRMING PAGES 230 Mechanical Engineering Design stress conditions, with several brittle failure hypotheses shown, superposed. We note the following: • In the first quadrant the data appear on both sides and along the failure curves of maximum-normal-stress, Coulomb-Mohr, and modified Mohr. All failure curves are the same, and data fit well. • In the fourth quadrant the modified Mohr theory represents the data best. • In the third quadrant the points A, B, C, and D are too few to make any suggestion concerning a fracture locus. 5–11 Selection of Failure Criteria For ductile behavior the preferred criterion is the distortion-energy theory, although some designers also apply the maximum-shear-stress theory because of its simplicity and conservative nature. In the rare case when Syt = Syc , the ductile Coulomb-Mohr method is employed. For brittle behavior, the original Mohr hypothesis, constructed with tensile, compression, and torsion tests, with a curved failure locus is the best hypothesis we have. However, the difficulty of applying it without a computer leads engineers to choose modifications, namely, Coulomb Mohr, or modified Mohr. Figure 5–21 provides a summary flowchart for the selection of an effective procedure for analyzing or predicting failures from static loading for brittle or ductile behavior. Figure 5–21 Failure theory selection flowchart. Brittle behavior Ductile behavior < 0.05 f ≥ 0.05 No Conservative? Yes No · Syt = Syc? Yes Mod. Mohr (MM) Eq. (5-32) Brittle Coulomb-Mohr Ductile Coulomb-Mohr (BCM) (DCM) Eq. (5-31) Eq. (5-26) No Conservative? Yes Distortion-energy (DE) Eqs. (5-15) and (5-19) Maximum shear stress (MSS) Eq. (5-3) bud21932_ch05_204-256 07/8/06 6:39 PM Page 231 CONFIRMING PAGES Failures Resulting from Static Loading 231 5–12 Introduction to Fracture Mechanics The idea that cracks exist in parts even before service begins, and that cracks can grow during service, has led to the descriptive phrase “damage-tolerant design.” The focus of this philosophy is on crack growth until it becomes critical, and the part is removed from service. The analysis tool is linear elastic fracture mechanics (LEFM). Inspection and maintenance are essential in the decision to retire parts before cracks reach catastrophic size. Where human safety is concerned, periodic inspections for cracks are mandated by codes and government ordinance. We shall now briefly examine some of the basic ideas and vocabulary needed for the potential of the approach to be appreciated. The intent here is to make the reader aware of the dangers associated with the sudden brittle fracture of so-called ductile materials. The topic is much too extensive to include in detail here and the reader is urged to read further on this complex subject.9 The use of elastic stress-concentration factors provides an indication of the average load required on a part for the onset of plastic deformation, or yielding; these factors are also useful for analysis of the loads on a part that will cause fatigue fracture. However, stress-concentration factors are limited to structures for which all dimensions are precisely known, particularly the radius of curvature in regions of high stress concentration. When there exists a crack, flaw, inclusion, or defect of unknown small radius in a part, the elastic stress-concentration factor approaches infinity as the root radius approaches zero, thus rendering the stress-concentration factor approach useless. Furthermore, even if the radius of curvature of the flaw tip is known, the high local stresses there will lead to local plastic deformation surrounded by a region of elastic deformation. Elastic stress-concentration factors are no longer valid for this situation, so analysis from the point of view of stress-concentration factors does not lead to criteria useful for design when very sharp cracks are present. By combining analysis of the gross elastic changes in a structure or part that occur as a sharp brittle crack grows with measurements of the energy required to produce new fracture surfaces, it is possible to calculate the average stress (if no crack were present) that will cause crack growth in a part. Such calculation is possible only for parts with cracks for which the elastic analysis has been completed, and for materials that crack in a relatively brittle manner and for which the fracture energy has been carefully measured. The term relatively brittle is rigorously defined in the test procedures,10 but it means, roughly, fracture without yielding occurring throughout the fractured cross section. Thus glass, hard steels, strong aluminum alloys, and even low-carbon steel below the ductile-to-brittle transition temperature can be analyzed in this way. Fortunately, ductile materials blunt sharp cracks, as we have previously discovered, so that fracture occurs at average stresses of the order of the yield strength, and the designer is prepared References on brittle fracture include: H. Tada and P. C. Paris, The Stress Analysis of Cracks Handbook, 2nd ed., Paris Productions, St. Louis, 1985. D. Broek, Elementary Engineering Fracture Mechanics, 4th ed., Martinus Nijhoff, London, 1985. D. Broek, The Practical Use of Fracture Mechanics, Kluwar Academic Pub., London, 1988. David K. Felbeck and Anthony G. Atkins, Strength and Fracture of Engineering Solids, Prentice-Hall, Englewood Cliffs, N.J., 1984. Kåre Hellan, Introduction to Fracture Mechanics, McGraw-Hill, New York, 1984. 10 9 BS 5447:1977 and ASTM E399-78. bud21932_ch05_204-256 07/8/06 8:01 PM Page 232 CONFIRMING PAGES 232 Mechanical Engineering Design for this condition. The middle ground of materials that lie between “relatively brittle” and “ductile” is now being actively analyzed, but exact design criteria for these materials are not yet available. Quasi-Static Fracture Many of us have had the experience of observing brittle fracture, whether it is the breaking of a cast-iron specimen in a tensile test or the twist fracture of a piece of blackboard chalk. It happens so rapidly that we think of it as instantaneous, that is, the cross section simply parting. Fewer of us have skated on a frozen pond in the spring, with no one near us, heard a cracking noise, and stopped to observe. The noise is due to cracking. The cracks move slowly enough for us to see them run. The phenomenon is not instantaneous, since some time is necessary to feed the crack energy from the stress field to the crack for propagation. Quantifying these things is important to understanding the phenomenon “in the small.” In the large, a static crack may be stable and will not propagate. Some level of loading can render the crack unstable, and the crack propagates to fracture. The foundation of fracture mechanics was first established by Griffith in 1921 using the stress field calculations for an elliptical flaw in a plate developed by Inglis in 1913. For the infinite plate loaded by an applied uniaxial stress σ in Fig. 5–22, the maximum stress occurs at (±a, 0) and is given by (σ y )max = 1 + 2 a σ b (5–33) Note that when a = b , the ellipse becomes a circle and Eq. (5–33) gives a stress concentration factor of 3. This agrees with the well-known result for an infinite plate with a circular hole (see Table A–15–1). For a fine crack, b/a → 0, and Eq. (5–34) predicts that (σ y )max → ∞. However, on a microscopic level, an infinitely sharp crack is a hypothetical abstraction that is physically impossible, and when plastic deformation occurs, the stress will be finite at the crack tip. Griffith showed that the crack growth occurs when the energy release rate from applied loading is greater than the rate of energy for crack growth. Crack growth can be stable or unstable. Unstable crack growth occurs when the rate of change of the energy release rate relative to the crack length is equal to or greater than the rate of change of the crack growth rate of energy. Griffith’s experimental work was restricted to brittle materials, namely glass, which pretty much confirmed his surface energy hypothesis. However, for ductile materials, the energy needed to perform plastic work at the crack tip is found to be much more crucial than surface energy. Figure 5–22 y b x a bud21932_ch05_204-256 07/8/06 8:01 PM Page 233 CONFIRMING PAGES Failures Resulting from Static Loading 233 Figure 5–23 Crack propagation modes. Mode I Mode II Mode III Crack Modes and the Stress Intensity Factor Three distinct modes of crack propagation exist, as shown in Fig. 5–23. A tensile stress field gives rise to mode I, the opening crack propagation mode, as shown in Fig. 5–23a. This mode is the most common in practice. Mode II is the sliding mode, is due to in-plane shear, and can be seen in Fig. 5–23b. Mode III is the tearing mode, which arises from out-of-plane shear, as shown in Fig. 5–23c. Combinations of these modes can also occur. Since mode I is the most common and important mode, the remainder of this section will consider only this mode. Consider a mode I crack of length 2a in the infinite plate of Fig. 5–24. By using complex stress functions, it has been shown that the stress field on a d x dy element in the vicinity of the crack tip is given by σx = σ σy = σ τx y = σ a θ θ 3θ cos 1 − sin sin 2r 2 2 2 a θ θ 3θ cos 1 + sin sin 2r 2 2 2 a θ θ 3θ sin cos cos 2r 2 2 2 0 ν(σx + σ y ) (for plane stress) (for plane strain) (5–34a) (5–34b) (5–34c) σz = (5–34d) Figure 5–24 Mode I crack model. y dx dy r a x bud21932_ch05_204-256 07/8/06 6:39 PM Page 234 CONFIRMING PAGES 234 Mechanical Engineering Design The stress σ y near the tip, with θ = 0, is σ y |θ =0 = σ a 2r (a) As with the elliptical crack, we see that σ y |θ =0 → ∞ as r → 0, and again the concept of an √ infinite stress concentration at the crack tip is inappropriate. The quantity √ σ y |θ =0 2r = σ a , however, does remain constant as r → 0. It is common practice to define a factor K called the stress intensity factor given by √ K = σ πa (b) √ √ where the units are MPa m or kpsi in. Since we are dealing with a mode I crack, Eq. (b) is written as √ KI = σ πa (5–35) The stress intensity factor is not to be confused with the static stress concentration factors K t and K ts defined in Secs. 3–13 and 5–2. Thus Eqs. (5–34) can be rewritten as KI θ θ 3θ σx = √ cos 1 − sin sin 2 2 2 2π r KI θ θ 3θ σy = √ cos 1 + sin sin 2 2 2 2π r KI θ θ 3θ τx y = √ sin cos cos 2 2 2 2π r σz = 0 ν(σx + σ y ) (for plane stress) (for plane strain) (5–36a) (5–36b) (5–36c) (5–36d) The stress intensity factor is a function of geometry, size and shape of the crack, and the type of loading. For various load and geometric configurations, Eq. (5–35) can be written as √ K I = β π a (5–37) where β is the stress intensity modification factor. Tables for β are available in the literature for basic configurations.11 Figures 5–25 to 5–30 present a few examples of β for mode I crack propagation. Fracture Toughness When the magnitude of the mode I stress intensity factor reaches a critical value, K I c crack propagation initiates. The critical stress intensity factor K I c is a material property that depends on the material, crack mode, processing of the material, temperature, 11 See, for example: H. Tada and P. C. Paris, The Stress Analysis of Cracks Handbook, 2nd ed., Paris Productions, St. Louis, 1985. G. C. Sib, Handbook of Stress Intensity Factors for Researchers and Engineers, Institute of Fracture and Solid Mechanics, Lehigh University, Bethlehem, Pa., 1973. Y. Murakami, ed., Stress Intensity Factors Handbook, Pergamon Press, Oxford, U.K., 1987. W. D. Pilkey, Formulas for Stress, Strain, and Structural Matrices, 2nd ed. John Wiley& Sons, New York, 2005. bud21932_ch05_204-256 07/8/06 6:39 PM Page 235 CONFIRMING PAGES Figure 5–25 Off-center crack in a plate in longitudinal tension; solid curves are for the crack tip at A; dashed curves are for the tip at B. 2.2 AA 2.0 2a A 1.8 d 2b B A 1.6 0.4 1.4 d b = 1.0 0.2 B B 0.4 1.2 0.2 1.0 0 0.2 0.4 a d ratio 0.6 0.8 Figure 5–26 Plate loaded in longitudinal tension with a crack at the edge; for the solid curve there are no constraints to bending; the dashed curve was obtained with bending constraints added. 7.0 6.0 h a b 5.0 h 4.0 3.0 h b = 0.5 1.0 2.0 1.0 0 0.2 0.4 a b ratio 0.6 0.8 235 bud21932_ch05_204-256 07/8/06 6:39 PM Page 236 CONFIRMING PAGES 236 Mechanical Engineering Design Figure 5–27 Beams of rectangular cross section having an edge crack. 2.0 M a F 1.8 a F 2 1.6 l l h F 2 h M Pure bending 1.4 l =4 h 1.2 l =2 h 1.0 0 0.2 0.4 a h ratio 0.6 0.8 Figure 5–28 Plate in tension containing a circular hole with two cracks. 3 2a 2 2b r r = 0.25 b r = 0.5 b 1r =0 b 0 0 0.2 0.4 a b ratio 0.6 0.8 loading rate, and the state of stress at the crack site (such as plane stress versus plane strain). The critical stress intensity factor K I c is also called the fracture toughness of the material. The fracture toughness for plane strain is normally lower than that for plane stress. For this reason, the term K I c is typically defined as the mode I, plane strain fracture toughness. Fracture toughness K I c for engineering metals lies in the range √ 20 ≤ K I √≤ 200 MPa · m ; for engineering polymers and ceramics, 1 ≤ K I c ≤ c 5 MPa · m. For a 4340 steel, where the yield strength due√ heat treatment ranges to from 800 to 1600 MPa, K I c decreases from 190 to 40 MPa · m. bud21932_ch05_204-256 07/8/06 6:39 PM Page 237 CONFIRMING PAGES Failures Resulting from Static Loading 237 Figure 5–29 A cylinder loading in axial tension having a radial crack of depth a extending completely around the circumference of the cylinder. 4.0 3.0 a a ri ro = 0 0.1 0.4 2.0 ri ro 0.8 1.0 0 0.2 0.4 a (ro – ri ) ratio 0.6 0.8 Figure 5–30 Cylinder subjected to internal pressure p, having a radial crack in the longitudinal direction of depth a. Use Eq. (4–51) for the tangential stress at r = r 0 . 3.4 a 3.0 ri ro 2.6 pi 2.2 1.8 ri ro = 0.9 0.75 0.35 1.4 1.0 0 0.2 0.4 a (ro – ri ) ratio 0.6 0.8 Table 5–1 gives some approximate typical room-temperature values of K I c for several materials. As previously noted, the fracture toughness depends on many factors and the table is meant only to convey some typical magnitudes of K I c . For an actual application, it is recommended that the material specified for the application be certified using standard test procedures [see the American Society for Testing and Materials (ASTM) standard E399]. bud21932_ch05_204-256 07/8/06 6:39 PM Page 238 CONFIRMING PAGES 238 Mechanical Engineering Design Table 5–1 Values of KIc for Some Engineering Materials at Room Temperature Material Aluminum 2024 7075 7178 Titanium Ti-6AL-4V Ti-6AL-4V Steel 4340 4340 52100 √ K Ic, MPa m 26 24 33 115 55 99 60 14 Sy, MPa 455 495 490 910 1035 860 1515 2070 One of the first problems facing the designer is that of deciding whether the conditions exist, or not, for a brittle fracture. Low-temperature operation, that is, operation below room temperature, is a key indicator that brittle fracture is a possible failure mode. Tables of transition temperatures for various materials have not been published, possibly because of the wide variation in values, even for a single material. Thus, in many situations, laboratory testing may give the only clue to the possibility of a brittle fracture. Another key indicator of the possibility of fracture is the ratio of the yield strength to the ultimate strength. A high ratio of Sy / Su indicates there is only a small ability to absorb energy in the plastic region and hence there is a likelihood of brittle fracture. The strength-to-stress ratio K I c / K I can be used as a factor of safety as n= KIc KI (5–38) EXAMPLE 5–6 A steel ship deck plate is 30 mm thick and 12 m wide. It is loaded with a nominal uniaxial tensile stress of 50 MPa. It is operated below its ductile-to-brittle transition temperature with K I c equal to 28.3 MPa. If a 65-mm-long central transverse crack is present, estimate the tensile stress at which catastrophic failure will occur. Compare this stress with the yield strength of 240 MPa for this steel. For Fig. 5–25, with d = b, 2a = 65 mm and 2b = 12 m , so that d /b = 1 and a /d = 65/12(103 ) = 0.00542. Since a/d is so small, β = 1, so that √ √ K I = σ π a = 50 π (32.5 × 10−3 ) = 16.0 MPa m From Eq. (5–38), n= KIc 28.3 = = 1.77 KI 16.0 Solution bud21932_ch05_204-256 07/8/06 6:39 PM Page 239 CONFIRMING PAGES Failures Resulting from Static Loading 239 The stress at which catastrophic failure occurs is Answer σc = KIc 28.3 σ= (50) = 88.4 MPa KI 16.0 The yield strength is 240 MPa, and catastrophic failure occurs at 88.4/240 = 0.37, or at 37 percent of yield. The factor of safety in this circumstance is K I c / K I = 28.3/16 = 1.77 and not 240/50 = 4.8. EXAMPLE 5–7 A plate of width 1.4 m and length 2.8 m is required to support a tensile force in the 2.8-m direction of 4.0 MN. Inspection procedures will detect only through-thickness edge cracks larger than 2.7 mm. The two Ti-6AL-4V alloys in Table 5–1 are being considered for this application, for which the safety factor must be 1.3 and minimum weight is important. Which alloy should be used? (a) We elect first to estimate the thickness required to resist yielding. Since σ = P /wt , we have t = P /wσ. For the weaker alloy, we have, from Table 5–1, Sy = 910 MPa. Thus, σall = Thus t= P 4.0(10)3 = = 4.08 mm or greater wσall 1.4(700) Sy 910 = = 700 MPa n 1.3 Solution For the stronger alloy, we have, from Table 5–1, σall = and so the thickness is Answer t= 4.0(10)3 P = = 3.59 mm or greater wσall 1.4(796) 1035 = 796 MPa 1.3 (b) Now let us find the thickness required to prevent crack growth. Using Fig. 5–26, we have a 2.7 = = 0.001 93 b 1.4(103 ) √ . Corresponding to these ratios we find from Fig. 5–26 that β = 1.1, and K I = 1.1σ π a . √ KIc 115 103 KIc n= = σ= √, √ KI 1.1σ π a 1.1n π a 2.8/2 h = =1 b 1.4 bud21932_ch05_204-256 07/8/06 6:39 PM Page 240 CONFIRMING PAGES 240 Mechanical Engineering Design √ From Table 5–1, K I c = 115 MPa m for the weaker of the two alloys. Solving for σ with n = 1 gives the fracture stress σ= 115 1.1 π (2.7 × 10−3 ) = 1135 MPa which is greater than the yield strength of 910 MPa, and so yield strength is the basis for the geometry decision. For the stronger alloy Sy = 1035 MPa, with n = 1 the fracture stress is σ= 55 KIc = 542.9 MPa = nKI 1(1.1) π (2.7 × 10−3 ) P 4.0(103 ) = = 6.84 mm or greater wσall 1.4(542.9/1.3) which is less than the yield strength of 1035 MPa. The thickness t is t= This example shows that the fracture toughness K I c limits the geometry when the stronger alloy is used, and so a thickness of 6.84 mm or larger is required. When the weaker alloy is used the geometry is limited by the yield strength, giving a thickness of only 4.08 mm or greater. Thus the weaker alloy leads to a thinner and lighter weight choice since the failure modes differ. 5–13 Stochastic Analysis12 Reliability is the probability that machine systems and components will perform their intended function satisfactorily without failure. Up to this point, discussion in this chapter has been restricted to deterministic relations between static stress, strength, and the design factor. Stress and strength, however, are statistical in nature and very much tied to the reliability of the stressed component. Consider the probability density functions for stress and strength, and S, shown in Fig. 5–31a. The mean values of stress and strength are μσ and μ S , respectively. Here, the “average” factor of safety is n= ¯ μS μσ (a) The margin of safety for any value of stress σ and strength S is defined as m = S−σ (b) ¯ The average part will have a margin of safety of m = μ S − μσ . However, for the overlap of the distributions shown by the shaded area in Fig. 5–31a, the stress exceeds the strength, the margin of safety is negative, and these parts are expected to fail. This shaded area is called the interference of and S. Figure 5–31b shows the distribution of m, which obviously depends on the distributions of stress and strength. The reliability that a part will perform without failure, R, is the area of the margin of safety distribution for m > 0. The interference is the area 12 Review Chap. 20 before reading this section. bud21932_ch05_204-256 07/8/06 6:40 PM Page 241 CONFIRMING PAGES Failures Resulting from Static Loading 241 Figure 5–31 Plot of density functions showing how the interference of S and is used to obtain the stress margin m. (a) Stress and strength distributions. (b) Distribution of interference; the reliability R is the area of the density function for m greater than zero; the interference is the area (1 − R). f (s), f ( ) S s Stress (a) f (m) m (1 – R) R – 0 m Stress margin (b) + 1 − R where parts are expected to fail. We next consider some typical cases involving stress-strength interference. Normal-Normal Case ˆ ˆ Consider the normal distributions, S = N(μ S , σ S ) and = N(μσ , σσ ) . The stress margin is m = S − , and will be normally distributed because the addition or subˆ traction of normals is normal. Thus m = N(μm , σm ) . Reliability is the probability p that m > 0. That is, R = p( S > σ ) = p( S − σ > 0) = p(m > 0) (5–39) To find the chance that m > 0 we form the z variable of m and substitute m = 0 [See ˆ ˆ2 ˆ2 Eq. (20–16)]. Noting that μm = μ S − μσ and σm = (σ S + σσ )1/2 , we write z= m − μm 0 − μm μm μ S − μσ = =− =− 1/2 σm ˆ σm ˆ σm ˆ σ S + σσ ˆ2 ˆ2 (5–40) Equation (5–40) is called the normal coupling equation. The reliability associated with z is given by R= x ∞ 1 u2 du = 1 − F = 1 − √ exp − 2 2π (z ) (5–41) The body of Table A–10 gives R when z > 0 and (1 − R = F ) when z ≤ 0. Noting that n = μ S /μσ , square both sides of Eq. (5–40), and introduce C S and Cσ where ¯ Cs = σs /μs and Cσ = σσ /μσ . Solve the resulting quadratic for n to obtain ˆ ˆ ¯ n= ¯ 1± 2 1 − 1 − z2C S 2 1 − z2C S 2 1 − z 2 Cσ (5–42) The plus sign is associated with R > 0.5, and the minus sign with R < 0.5. bud21932_ch05_204-256 07/8/06 6:40 PM Page 242 CONFIRMING PAGES 242 Mechanical Engineering Design Lognormal–Lognormal Case ˆ ˆ Consider the lognormal distributions S = LN(μ S , σ S ) and = LN(μσ , σσ ). If we interfere their companion normals using Eqs. (20–18) and (20–19), we obtain 2 μln S = ln μ S − ln 1 + C S (strength) σln S = ˆ and 2 μln σ = ln μσ − ln 1 + Cσ 2 ln 1 + C S (stress) σln σ = ˆ 2 ln 1 + Cσ Using Eq. (5–40) for interfering normal distributions gives μS μσ 2 1 + Cσ 2 1 + CS 2 1 + Cσ z=− μln S − μln σ σln S + σln σ ˆ2 ˆ2 1/2 ln =− 2 ln 1 + C S (5–43) The reliability R is expressed by Eq. (5–41). The design factor n is the random variable that is the quotient of S/ . The quotient of lognormals is lognormal, so pursuing the z variable of the lognormal n, we note μn = μS μσ Cn = 2 2 C S + Cσ 2 1 + Cσ σn = Cn μn ˆ ˆ The companion normal to n = LN(μn , σn ) , from Eqs. (20–18) and (20–19), has a mean and standard deviation of 2 μ y = ln μn − ln 1 + Cn σy = ˆ 2 ln 1 + Cn The z variable for the companion normal y distribution is z= y − μy σy ˆ ¯ Failure will occur when the stress is greater than the strength, when n < 1, or when y < 0. z= 2 2 ln μn / 1 + Cn ln μn − ln 1 + Cn μy 0 − μy =− =− =− ˙ σy ˆ σy 2 2 ln 1 + Cn ln 1 + Cn (5–44) Solving for μn gives Cn . 2 2 ¯ μn = n = exp −z ln 1 + Cn + ln 1 + Cn = exp Cn − z + 2 (5–45) bud21932_ch05_204-256 07/8/06 6:40 PM Page 243 CONFIRMING PAGES Failures Resulting from Static Loading 243 Equations (5–42) and (5–45) are remarkable for several reasons: ¯ • They relate design factor n to the reliability goal (through z) and the coefficients of variation of strength and stress. • They are not functions of the means of stress and strength. • They estimate the design factor necessary to achieve the reliability goal before decisions involving means are made. The C S depends slightly on the particular material. The Cσ has the coefficient of variation (COV) of the load, and that is generally given. EXAMPLE 5–8 A round cold-drawn 1018 steel rod has an 0.2 percent yield strength S y = N(78.4, 5.90) kpsi and is to be subjected to a static axial load of P = N(50, 4.1) kip. What value of ¯ the design factor n corresponds to a reliability of 0.999 against yielding (z = −3.09)? Determine the corresponding diameter of the rod. C S = 5.90/78.4 = 0.0753 , and P 4P = A π d2 Since the COV of the diameter is an order of magnitude less than the COV of the load or strength, the diameter is treated deterministically: 4.1 = 0.082 Cσ = C P = 50 From Eq. (5–42), = n 1 1 Solution [1 ( 3.09) (0.0753 )][1 1 ( 3.09) (0.0753 ) 2 2 2 2 ( 3.09) (0.082 )] 2 2 1.416 The diameter is found deterministically: Answer d= ¯ 4P = ¯¯ π Sy /n 4(50 000) = 1.072 in π(78 400)/1.416 Check S y = N(78.4, 5.90) kpsi, P = N(50, 4.1) kip, and d = 1.072 in. Then π d2 π(1.0722 ) = = 0.9026 in2 4 4 ¯ P (50 000) σ= ¯ = = 55 400 psi A 0.9026 4.1 C P = Cσ = = 0.082 50 A= σσ = Cσ σ = 0.082(55 400) = 4540 psi ˆ ¯ σ S = 5.90 kpsi ˆ From Eq. (5–40) 78.4 − 55.4 = −3.09 (5.902 + 4.542 )1/2 From Appendix Table A–10, R = (−3.09) = 0.999 . z=− bud21932_ch05_204-256 07/8/06 6:40 PM Page 244 CONFIRMING PAGES 244 Mechanical Engineering Design EXAMPLE 5–9 Solution Rework Ex. 5–8 with lognormally distributed stress and strength. C S = 5.90/78.4 = 0.0753 , and Cσ = C P = 4.1/50 = 0.082 . Then = Cn = 4P P = A π d2 2 2 C S + Cσ = 2 1 + Cσ 0.07532 + 0.0822 = 0.1110 1 + 0.0822 From Table A–10, z = −3.09. From Eq. (5–45), n = exp − (−3.09) ln(1 + 0.1112 ) + ln 1 + 0.1112 = 1.416 ¯ d= ¯ 4P = ¯¯ π Sy /n 4(50 000) = 1.0723 in π(78 400)/1.416 Check S y = LN(78.4, 5.90), P = LN (50, 4.1) kip. Then π(1.07232 ) π d2 = = 0.9031 4 4 ¯ P 50 000 σ= ¯ = = 55 365 psi A 0.9031 4.1 Cσ = C P = = 0.082 50 A= σσ = Cσ μσ = 0.082(55 367) = 4540 psi ˆ From Eq. (5–43), ⎛ ln ⎝ z=− 78.4 55.365 1 + 0.082 ⎠ 1 + 0.07532 2 ⎞ ln[(1 + 0.07532 )(1 + 0.0822 )] = −3.1343 Appendix Table A–10 gives R = 0.99950. Interference—General In the previous segments, we employed interference theory to estimate reliability when the distributions are both normal and when they are both lognormal. Sometimes, however, it turns out that the strength has, say, a Weibull distribution while the stress is distributed lognormally. In fact, stresses are quite likely to have a lognormal distribution, because the multiplication of variates that are normally distributed produces a result that approaches lognormal. What all this means is that we must expect to encounter interference problems involving mixed distributions and we need a general method to handle the problem. It is quite likely that we will use interference theory for problems involving distributions other than strength and stress. For this reason we employ the subscript 1 to bud21932_ch05_204-256 07/8/06 6:40 PM Page 245 CONFIRMING PAGES Failures Resulting from Static Loading 245 Figure 5–32 (a) PDF of the strength distribution; (b) PDF of the load-induced stress distribution. f1(S ) dF1(x) = f1(x) dx (a) x f2( ) dx S Cursor F2(x) R2(x) (b) designate the strength distribution and the subscript 2 to designate the stress distribution. Figure 5–32 shows these two distributions aligned so that a single cursor x can be used to identify points on both distributions. We can now write Probability that stress is less than strength = dp(σ < x ) = d R = F2 (x ) d F1 (x ) By substituting 1 − R2 for F2 and −d R1 for d F1 , we have d R = −[1 − R2 (x )] d R1 (x ) The reliability for all possible locations of the cursor is obtained by integrating x from −∞ to ∞; but this corresponds to an integration from 1 to 0 on the reliability R1 . Therefore R=− 1 0 [1 − R2 (x )] d R1 (x ) which can be written R =1− 0 1 R2 d R1 (5–46) where R1 ( x ) = x ∞ x ∞ f1( S) d S (5–47) R2 ( x ) = f 2 (σ ) d σ (5–48) bud21932_ch05_204-256 07/8/06 6:40 PM Page 246 CONFIRMING PAGES 246 Mechanical Engineering Design 1 1 R2 R2 R1 (a) 1 R1 (b) 1 Figure 5–33 Curve shapes of the R 1 R 2 plot. In each case the shaded area is equal to 1 − R and is obtained by numerical integration. (a) Typical curve for asymptotic distributions; (b) curve shape obtained from lower truncated distributions such as the Weibull. For the usual distributions encountered, plots of R1 versus R2 appear as shown in Fig. 5–33. Both of the cases shown are amenable to numerical integration and computer solution. When the reliability is high, the bulk of the integration area is under the right-hand spike of Fig. 5–33a. 5–14 Important Design Equations The following equations and their locations are provided as a summary. Maximum Shear Theory p. 212 τmax = Sy σ1 − σ3 = 2 2n (5–3) Distortion-Energy Theory Von Mises stress, p. 214 σ= (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 2 1/2 (5–12) 1/2 1 2 2 2 p. 215 σ = √ (σx − σ y )2 + (σ y − σz )2 + (σz − σx )2 + 6(τx y + τ yz + τzx ) 2 (5–14) Plane stress, p. 214 2 2 σ = (σ A − σ A σ B + σ B )1/2 (5–13) (5–15) p. 215 2 2 2 σ = (σx − σx σ y + σ y + 3τx y )1/2 Yield design equation, p. 216 σ= Shear yield strength, p. 217 Ssy = 0.577 Sy (5–21) Sy n (5–19) bud21932_ch05_204-256 07/8/06 6:40 PM Page 247 CONFIRMING PAGES Failures Resulting from Static Loading 247 Coulomb-Mohr Theory p. 221 σ1 σ3 1 − = St Sc n (5–26) where St is tensile yield (ductile) or ultimate tensile (brittle), and St is compressive yield (ductile) or ultimate compressive (brittle) strengths. Maximum-Normal-Stress Theory p. 226 σ1 = Sut n or σ3 = − Suc n (5–30) Modified Mohr (Plane Stress) Use maximum-normal-stress equations, or p. 227 ( Suc − Sut )σ A σB 1 − = Suc Sut Suc n σ A ≥ 0 ≥ σB and σB >1 σA (5–32b) Failure Theory Flowchart Fig. 5–21, p. 230 Brittle behavior Ductile behavior < 0.05 f ≥ 0.05 No Conservative? Yes No · Syt = Syc? Yes Mod. Mohr (MM) Eq. (5-32) Brittle Coulomb-Mohr Ductile Coulomb-Mohr (BCM) (DCM) Eq. (5-31) Eq. (5-26) No Conservative? Yes Distortion-energy (DE) Eqs. (5-15) and (5-19) Maximum shear stress (MSS) Eq. (5-3) Fracture Mechanics p. 234 √ K I = βσ π a (5–37) where β is found in Figs. 5–25 to 5–30 (pp. 235 to 237) bud21932_ch05_204-256 07/8/06 6:40 PM Page 248 CONFIRMING PAGES 248 Mechanical Engineering Design p. 238 n= KIc KI (5–38) where K I c is found in Table 5–1 (p. 238) Stochastic Analysis ¯ Mean factor of safety defined as n = μ S /μσ (μ S and μσ are mean strength and stress, respectively) Normal-Normal Case p. 241 n= 1± 2 2 1 − (1 − z 2 Cs )(1 − z 2 Cσ ) 2C 2 1−z s (5–42) ˆ ˆ where z can be found in Table A–10, C S = σ S /μ S , and Cσ = σσ /μσ . Lognormal-Lognormal Case p. 242 where Cn = 2 2 C S + Cσ 2 1 + Cσ Cn . 2 2 n = exp −z ln(1 + Cn ) + ln 1 + Cn = exp Cn −z + 2 (5–45) (See other definitions in normal-normal case.) PROBLEMS 5–1 A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 50 kpsi. Using the distortion-energy and maximum-shear-stress theories determine the factors of safety for the following plane stress states: (a) σx = 12 kpsi, σ y = 6 kpsi (b) σx = 12 kpsi, τx y = −8 kpsi (c) σx = −6 kpsi, σ y = −10 kpsi, τx y = −5 kpsi (d ) σx = 12 kpsi, σ y = 4 kpsi, τx y = 1 kpsi Repeat Prob. 5–1 for: (a) σ A = 12 kpsi, σ B = 12 kpsi (b) σ A = 12 kpsi, σ B = 6 kpsi (c) σ A = 12 kpsi, σ B = −12 kpsi (d ) σ A = −6 kpsi, σ B = −12 kpsi Repeat Prob. 5–1 for a bar of AISI 1020 cold-drawn steel and: (a) σx = 180 MPa, σ y = 100 MPa (b) σx = 180 MPa, τx y = 100 MPa (c) σx = −160 MPa, τx y = 100 MPa (d ) τx y = 150 MPa Repeat Prob. 5–1 for a bar of AISI 1018 hot-rolled steel and: (a) σ A = 100 MPa, σ B = 80 MPa (b) σ A = 100 MPa, σ B = 10 MPa (c) σ A = 100 MPa, σ B = −80 MPa (d ) σ A = −80 MPa, σ B = −100 MPa 5–2 5–3 5–4 bud21932_ch05_204-256 07/8/06 6:40 PM Page 249 CONFIRMING PAGES Failures Resulting from Static Loading 249 5–5 5–6 5–7 Repeat Prob. 5–3 by first plotting the failure loci in the σ A , σ B plane to scale; then, for each stress state, plot the load line and by graphical measurement estimate the factors of safety. Repeat Prob. 5–4 by first plotting the failure loci in the σ A , σ B plane to scale; then, for each stress state, plot the load line and by graphical measurement estimate the factors of safety. An ASTM cast iron has minimum ultimate strengths of 30 kpsi in tension and 100 kpsi in compression. Find the factors of safety using the MNS, BCM, and MM theories for each of the following stress states. Plot the failure diagrams in the σ A , σ B plane to scale and locate the coordinates of each stress state. (a) σx = 20 kpsi, σ y = 6 kpsi (b) σx = 12 kpsi, τx y = −8 kpsi (c) σx = −6 kpsi, σ y = −10 kpsi, τx y = −5 kpsi (d ) σx = −12 kpsi, τx y = 8 kpsi For Prob. 5–7, case (d ), estimate the factors of safety from the three theories by graphical measurements of the load line. Among the decisions a designer must make is selection of the failure criteria that is applicable to the material and its static loading. A 1020 hot-rolled steel has the following properties: Sy = 42 kpsi, Sut = 66.2 kpsi, and true strain at fracture ε f = 0.90. Plot the failure locus and, for the static stress states at the critical locations listed below, plot the load line and estimate the factor of safety analytically and graphically. (a) σx = 9 kpsi, σ y = −5 kpsi. (b) σx = 12 kpsi, τx y = 3 kpsi ccw. (c) σx = −4 kpsi, σ y = −9 kpsi, τx y = 5 kpsi cw. (d) σx = 11 kpsi, σ y = 4 kpsi, τx y = 1 kpsi cw. A 4142 steel Q&T at 80◦ F exhibits Syt = 235 kpsi, Syc = 275 kpsi, and ε f = 0.06. Choose and plot the failure locus and, for the static stresses at the critical locations, which are 10 times those in Prob. 5–9, plot the load lines and estimate the factors of safety analytically and graphically. For grade 20 cast iron, Table A–24 gives Sut = 22 kpsi, Suc = 83 kpsi. Choose and plot the failure locus and, for the static loadings inducing the stresses at the critical locations of Prob. 5–9, plot the load lines and estimate the factors of safety analytically and graphically. A cast aluminum 195-T6 has an ultimate strength in tension of Sut = 36 kpsi and ultimate strength in compression of Suc = 35 kpsi, and it exhibits a true strain at fracture ε f = 0.045. Choose and plot the failure locus and, for the static loading inducing the stresses at the critical locations of Prob. 5–9, plot the load lines and estimate the factors of safety analytically and graphically. An ASTM cast iron, grade 30 (see Table A–24), carries static loading resulting in the stress state listed below at the critical locations. Choose the appropriate failure locus, plot it and the load lines, and estimate the factors of safety analytically and graphically. (a) σ A = 20 kpsi, σ B = 20 kpsi. (b) τx y = 15 kpsi. (c) σ A = σ B = −80 kpsi. (d ) σ A = 15 kpsi, σ B = −25 kpsi. This problem illustrates that the factor of safety for a machine element depends on the particular point selected for analysis. Here you are to compute factors of safety, based upon the distortion-energy theory, for stress elements at A and B of the member shown in the figure. This bar is made of AISI 1006 cold-drawn steel and is loaded by the forces F = 0.55 kN, P = 8.0 kN, and T = 30 N · m. 5–8 5–9 5–10 5–11 5–12 5–13 5–14 bud21932_ch05_204-256 07/8/06 6:40 PM Page 250 CONFIRMING PAGES 250 Mechanical Engineering Design y 10 0m A B m Problem 5–14 z F 20-mm D. P x T 5–15 The figure shows a crank loaded by a force F = 190 lbf which causes twisting and bending of the 3 -in-diameter shaft fixed to a support at the origin of the reference system. In actuality, the 4 support may be an inertia which we wish to rotate, but for the purposes of a strength analysis we can consider this to be a statics problem. The material of the shaft AB is hot-rolled AISI 1018 steel (Table A–20). Using the maximum-shear-stress theory, find the factor of safety based on the stress at point A. y 1 in A Problem 5–15 z C dia. B 1 4 F 1 -in 2 3 -in 4 in dia. 1 14 in 4 in 5 in x 5–16 5–17* 5–18 Solve Prob. 5–15 using the distortion energy theory. If you have solved Prob. 5–15, compare the results and discuss the difference. Design the lever arm CD of Fig. 5–16 by specifying a suitable size and material. A spherical pressure vessel is formed of 18-gauge (0.05-in) cold-drawn AISI 1018 sheet steel. If the vessel has a diameter of 8 in, estimate the pressure necessary to initiate yielding. What is the estimated bursting pressure? *The asterisk indicates a problem that may not have a unique result or may be a particularly challenging problem. bud21932_ch05_204-256 07/8/06 6:40 PM Page 251 CONFIRMING PAGES Failures Resulting from Static Loading 251 5–19 This problem illustrates that the strength of a machine part can sometimes be measured in units other than those of force or moment. For example, the maximum speed that a flywheel can reach without yielding or fracturing is a measure of its strength. In this problem you have a rotating ring made of hot-forged AISI 1020 steel; the ring has a 6-in inside diameter and a 10-in outside diameter and is 1.5 in thick. What speed in revolutions per minute would cause the ring to yield? At what radius would yielding begin? [Note: The maximum radial stress occurs at r = (ro ri )1/2 ; see Eq. (3–55).] A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. This cylinder has a 3 1 -in OD, a 0.065-in wall thickness, and ν = 0.334. The purchase order spec2 ifies a minimum yield strength of 46 kpsi. What is the factor of safety if the pressure-release valve is set at 500 psi? A cold-drawn AISI 1015 steel tube is 300 mm OD by 200 mm ID and is to be subjected to an external pressure caused by a shrink fit. What maximum pressure would cause the material of the tube to yield? What speed would cause fracture of the ring of Prob. 5–19 if it were made of grade 30 cast iron? The figure shows a shaft mounted in bearings at A and D and having pulleys at B and C. The forces shown acting on the pulley surfaces represent the belt tensions. The shaft is to be made of ASTM grade 25 cast iron using a design factor n d = 2.8 . What diameter should be used for the shaft? x 5–20 5–21 5–22 5–23 300 lbf y 6-in D. 50 lbf 27 lbf 360 lbf D C 6 in Problem 5–23 8-in D. 8 in z A 8 in B 5–24 By modern standards, the shaft design of Prob. 5–23 is poor because it is so long. Suppose it is redesigned by halving the length dimensions. Using the same material and design factor as in Prob. 5–23, find the new shaft diameter. The gear forces shown act in planes parallel to the yz plane. The force on gear A is 300 lbf. Consider the bearings at O and B to be simple supports. For a static analysis and a factor of safety of 3.5, use distortion energy to determine the minimum safe diameter of the shaft. Consider the material to have a yield strength of 60 kpsi. Repeat Prob. 5–25 using maximum-shear-stress. The figure is a schematic drawing of a countershaft that supports two V-belt pulleys. For each pulley, the belt tensions are parallel. For pulley A consider the loose belt tension is 15 percent of the tension on the tight side. A cold-drawn UNS G10180 steel shaft of uniform diameter is to be selected for this application. For a static analysis with a factor of safety of 3.0, determine the minimum preferred size diameter. Use the distortion-energy theory. 5–25 5–26 5–27 bud21932_ch05_204-256 07/8/06 6:40 PM Page 252 CONFIRMING PAGES 252 Mechanical Engineering Design y 20 in O 16 in FC 10 in 20° Problem 5–25 z Gear A 24-in D. A B C FA 20° Gear C 10-in D. x y 300 45° O T2 Problem 5–27 Dimensions in millimeters 250 Dia. A 50 N B C 300 Dia. z 400 T1 150 x 270 N 5–28 5–29 Repeat Prob. 5–27 using maximum shear stress. The clevis pin shown in the figure is 12 mm in diameter and has the dimensions a = 12 mm and b = 18 mm. The pin is machined from AISI 1018 hot-rolled steel (Table A–20) and is to be loaded to no more than 4.4 kN. Determine whether or not the assumed loading of figure c yields a factor of safety any different from that of figure d. Use the maximum-shear-stress theory. Repeat Prob. 5–29, but this time use the distortion-energy theory. A split-ring clamp-type shaft collar is shown in the figure. The collar is 2 in OD by 1 in ID by 1 2 in wide. The screw is designated as 1 -28 UNF. The relation between the screw tightening torque 4 T, the nominal screw diameter d, and the tension in the screw Fi is approximately T = 0.2 Fi d . The shaft is sized to obtain a close running fit. Find the axial holding force Fx of the collar as a function of the coefficient of friction and the screw torque. 5–30 5–31 bud21932_ch05_204-256 07/8/06 6:40 PM Page 253 CONFIRMING PAGES Failures Resulting from Static Loading 253 F (b) b 2 a+b a F a (c) b Problem 5–29 d b (a) a+b (d ) A Problem 5–31 5–32 Suppose the collar of Prob. 5–31 is tightened by using a screw torque of 190 lbf · in. The collar material is AISI 1040 steel heat-treated to a minimum tensile yield strength of 63 kpsi. (a) Estimate the tension in the screw. (b) By relating the tangential stress to the hoop tension, find the internal pressure of the shaft on the ring. (c) Find the tangential and radial stresses in the ring at the inner surface. (d ) Determine the maximum shear stress and the von Mises stress. (e) What are the factors of safety based on the maximum-shear-stress hypothesis and the distortionenergy theory? In Prob. 5–31, the role of the screw was to induce the hoop tension that produces the clamping. The screw should be placed so that no moment is induced in the ring. Just where should the screw be located? A tube has another tube shrunk over it. The specifications are: Inner Member ID OD 1.000 ± 0.002 in 2.000 ± 0.0004 in 5–33 5–34 Outer Member 1.999 ± 0.0004 in 3.000 ± 0.004 in Both tubes are made of a plain carbon steel. (a) Find the nominal shrink-fit pressure and the von Mises stresses at the fit surface. (b) If the inner tube is changed to solid shafting with the same outside dimensions, find the nominal shrink-fit pressure and the von Mises stresses at the fit surface. bud21932_ch05_204-256 07/8/06 6:40 PM Page 254 CONFIRMING PAGES 254 Mechanical Engineering Design 5–35 Steel tubes with a Young’s modulus of 207 GPa have the specifications: Inner Tube ID OD 25 ± 0.050 mm 50 ± 0.010 mm Outer Tube 49.98 ± 0.010 mm 75 ± 0.10 mm These are shrink-fitted together. Find the nominal shrink-fit pressure and the von Mises stress in each body at the fit surface. 5–36 5–37 Repeat Prob. 5–35 for maximum shrink-fit conditions. A 2-in-diameter solid steel shaft has a gear with ASTM grade 20 cast-iron hub ( E = 14.5 Mpsi) shrink-fitted to it. The specifications for the shaft are 2.000 + 0.0000 − 0.0004 in 1 The hole in the hub is sized at 1.999 ± 0.0004 in with an OD of 4.00 ± 32 in. Using the midrange values and the modified Mohr theory, estimate the factor of safety guarding against fracture in the gear hub due to the shrink fit. 5–38 Two steel tubes are shrink-fitted together where the nominal diameters are 1.50, 1.75, and 2.00 in. Careful measurement before fitting revealed that the diametral interference between the tubes to be 0.00246 in. After the fit, the assembly is subjected to a torque of 8000 lbf · in and a bending-moment of 6000 lbf · in. Assuming no slipping between the cylinders, analyze the outer cylinder at the inner and outer radius. Determine the factor of safety using distortion energy with Sy = 60 kpsi. Repeat Prob. 5–38 for the inner tube. For Eqs. (5–36) show that the principal stresses are given by θ θ KI 1 + sin cos σ1 = √ 2 2 2π r θ KI θ σ2 = √ cos 1 − sin 2 2 2π r ⎧ (plane stress) ⎨0 ⎩ 2 θ ν K I cos πr 2 (plane strain) 5–39 5–40 σ3 = 5–41 Use the results of Prob. 5–40 for plane strain near the tip with θ = 0 and ν = 1 . If the yield 3 strength of the plate is Sy , what is σ1 when yield occurs? (a) Use the distortion-energy theory. (b) Use the maximum-shear-stress theory. Using Mohr’s circles, explain your answer. A plate 4 in wide, 8 in long, and 0.5 in thick is loaded in tension in the direction of the length. The plate contains a crack as√ shown in Fig. 5–26 with the crack length of 0.625 in. The material is steel with K I c = 70 kpsi · in , and Sy = 160 kpsi. Determine the maximum possible load that can be applied before the plate (a) yields, and (b) has uncontrollable crack growth. A cylinder subjected to internal pressure pi has an outer diameter of 350 mm and a 25-mm wall √ thickness. For the cylinder material, K I c = 80 MPa · m, Sy = 1200 MPa, and Sut = 1350 MPa. 5–42 5–43 bud21932_ch05_204-256 07/8/06 6:40 PM Page 255 CONFIRMING PAGES Failures Resulting from Static Loading 255 If the cylinder contains a radial crack in the longitudinal direction of depth 12.5 mm determine the pressure that will cause uncontrollable crack growth. 5–44 A carbon steel collar of length 1 in is to be machined to inside and outside diameters, respectively, of Di = 0.750 ± 0.0004 in Do = 1.125 ± 0.002 in This collar is to be shrink-fitted to a hollow steel shaft having inside and outside diameters, respectively, of di = 0.375 ± 0.002 in do = 0.752 ± 0.0004 in These tolerances are assumed to have a normal distribution, to be centered in the spread interval, and to have a total spread of ±4 standard deviations. Determine the means and the standard deviations of the tangential stress components for both cylinders at the interface. 5–45 5–46 Suppose the collar of Prob. 5–44 has a yield strength of S y = N(95.5, 6.59) kpsi. What is the probability that the material will not yield? A carbon steel tube has an outside diameter of 1 in and a wall thickness of 1 in. The tube is to 8 carry an internal hydraulic pressure given as p = N(6000, 500) psi. The material of the tube has a yield strength of S y = N(50, 4.1) kpsi. Find the reliability using thin-wall theory. bud21932_ch05_204-256 07/8/06 6:40 PM Page 256 CONFIRMING PAGES ...
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This note was uploaded on 03/22/2010 for the course MEEN ISEN 302 taught by Professor Kim during the Spring '10 term at Texas A&M University–Commerce.

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