Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from ©The McGraw-Hil 219 Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition 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 a'=Sy. -S. Pure shear load line (a=-0s=T) DE ---MSS Equation(5-13)is a rotated ellipse in the oA.og plane,as shown in Fig.5-9 with '=Sy.The dotted lines in the figure represent the MSS theory,which can be seen to be more restrictive,hence,more conservative. Using xyz components of three-dimensional stress,the von Mises stress can be written as g'= 万[a,-,P+a,-P+a:-aP+66+最+】p (5-14④ and for plane stress, a'=(a2-aa,+a+3r)2 (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 conser- vative.Consider an isolated element in which the normal stresses on each surface are equal to the hydrostatic stress.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 trans- formations the octahedral shear stress is given by ta=3[a1-022+(o2-0)2+(a3-o1)2]2 (5-16) The three-dimensional equations for DE and MSS can be plotted relative to three-dimensional. 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. SFor a derivation,see Arthur P.Boresi,op.cit.,pp.36-37
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGraw−Hill 219 Companies, 2008 Failures Resulting from Static Loading 215 Equation (5–13) is a rotated ellipse in the σA, σB 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 σ = 1 √2 (σx − σy ) 2 + (σy − σz) 2 + (σz − σx ) 2 + 6 τ 2 xy + τ 2 yz + τ 2 zx 1/2 (5–14) and for plane stress, σ = σ2 x − σxσy + σ2 y + 3τ 2 xy 1/2 (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 3 (σ1 − σ2) 2 + (σ2 − σ3) 2 + (σ3 − σ1) 21/2 (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. 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 Sy Sy B A DE MSS Pure shear load line (A B )
20 Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hil Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition 216 Mechanical Engineering Design Figure 5-10 Octahedral surfaces. Under the name of the octahedral-shear-stress theory,failure is assumed to occur when- ever 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 o =Sy and o2=03=0.From Eq.(5-16)the octahedral shear stress under this condition is (5-17刀 When,for the general stress case,Eq.(5-16)is equal or greater than Eq.(5-17),yield is predicted.This reduces to 1-2+@2-a2+(a3-m2]2 (5-18) 2 ≥S, 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/2A,and not the hydrostatic stresses which are P/3A.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 (5-19 n 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 the- ory 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 txy, where for plane stress ox=oy =0.For yield,Eq.(5-11)with Eq.(5-15)gives (3i )2=Sy or ty= S=0.577S, (5-201
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading 220 © The McGraw−Hill Companies, 2008 216 Mechanical Engineering Design 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 τoct = √2 3 Sy (5–17) 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/2A, and not the hydrostatic stresses which are P/3A. 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 τxy , where for plane stress σx = σy = 0. For yield, Eq. (5–11) with Eq. (5–15) gives 3τ 2 xy 1/2 = Sy or τxy = Sy √3 = 0.577Sy (5–20) Figure 5–10 Octahedral surfaces. 2 1 3 oct av
Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw--Hill 21 Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition Failures Resulting from Static Loading 217 Thus,the shear yield strength predicted by the distortion-energy theory is Sy=0.577S (5-211 which as stated earlier,is about 15 percent greater than the 0.5 S,predicted by the MSS theory.For pure shear,txy the principal stresses from Eq.(3-13)are A =-0g =txy. The load line for this case is in the third quadrant at an angle of 45 from the oA.o axes shown in Fig.5-9. EXAMPLE 5-1 A hot-rolled steel has a yield strength of Sy=Syc=100 kpsi and a true strain at fracture of sf=0.55.Estimate the factor of safety for the following principal stress states: (a)70,70,0kpsi (b)30.70.0kpsi. (c)0,70.-30kpsi. (d)0,-30,-70kpsi. (e)30,30,30kpsi. Solution Sinces>0.05 and Sye and Sy are equal,the material is ductile and the distortion- energy (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 A=a1=70,og =02=70.03=0 kpsi. DE From Eq.(5-13), a'=[702-70(70)+702]/2=70kpsi Answer n=S、100 =70=1.43 MSS Case 1,using Eq.(5-4)with a factor of safety, Answer n=S=100 =1.43 70 (b)The ordered principal stresses are oA=o1=70,og =02=30,03=0 kpsi. DE g'=[702-70(30)+302/2=60.8kpsi Answer n= S100 。=60.8 =1.64 MSS Case 1.using Eq.(5-4), Answer n=S100 A 70 =1.43
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGraw−Hill 221 Companies, 2008 Failures Resulting from Static Loading 217 Thus, the shear yield strength predicted by the distortion-energy theory is Ssy = 0.577Sy (5–21) which as stated earlier, is about 15 percent greater than the 0.5 Sy predicted by the MSS theory. For pure shear, τxy the principal stresses from Eq. (3–13) are σA = −σB = τxy . The load line for this case is in the third quadrant at an angle of 45o from the σA, σB 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 εf = 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. Solution Since εf > 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 σA = σ1 = 70, σB = σ2 = 70, σ3 = 0 kpsi. DE From Eq. (5–13), σ = [702 − 70(70) + 702 ] 1/2 = 70 kpsi Answer n = Sy σ = 100 70 = 1.43 MSS Case 1, using Eq. (5–4) with a factor of safety, Answer n = Sy σA = 100 70 = 1.43 (b) The ordered principal stresses are σA = σ1 = 70, σB = σ2 = 30, σ3 = 0 kpsi. DE σ = [702 − 70(30) + 302 ] 1/2 = 60.8 kpsi Answer n = Sy σ = 100 60.8 = 1.64 MSS Case 1, using Eq. (5–4), Answer n = Sy σA = 100 70 = 1.43
Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from ©The McGraw-Hil Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition 218 Mechanical Engineering Design (c)The ordered principal stresses are oa =o1=70.02=0,og=03 =-30 kpsi. DE g=[702-70(-30)+(-30)2]/2=88.9kpsi Answer n= S_100 -889=1.1B MSS Case 2,using Eq.(5-5), n=-Sy 100 Answer 0A-0g70-(-30=1.00 (d)The ordered principal stresses are o1=0,A =02=-30.og=03 =-70 kpsi. DE a'=[(-70)2-(-70)(-30)+(-30)2]/2=60.8kpsi Answer - =1.64 MSS Case 3,using Eq.(5-6), ns- 100 Answer =--70=1.43 OB (e)The ordered principal stresses are o=30,o2=30,o3 =30 kpsi DE From Eq.(5-12). a' 「30-302+(30-30)2+(30-30)271/2 =0 kpsi Answer n= S,100 0=0→0 MSS From Eq.(5-3). 100 Answer Sy n= 1-0=30-30→0 A tabular summary of the factors of safety is included for comparisons (a) (b) (c) (d) (e) DE 1.43 1.64 1.13 1.64 00 MSS 1.43 1.43 1.00 1.43 0 Since the MSS theory is on or within the boundary of the DE theory,it will always pre- dict 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 oA,og plane are shown in Fig.5-11.Case (e)is not plane stress.Note that the load line for case (a)is
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading 222 © The McGraw−Hill Companies, 2008 218 Mechanical Engineering Design (c) The ordered principal stresses are σA = σ1 = 70, σ2 = 0, σB = σ3 = −30 kpsi. DE σ = [702 − 70(−30) + (−30) 2 ] 1/2 = 88.9 kpsi Answer n = Sy σ = 100 88.9 = 1.13 MSS Case 2, using Eq. (5–5), Answer n = Sy σA − σB = 100 70 − (−30) = 1.00 (d) The ordered principal stresses are σ1 = 0, σA = σ2 = −30, σB = σ3 = −70 kpsi. DE σ = [(−70) 2 − (−70)(−30) + (−30) 2 ] 1/2 = 60.8 kpsi Answer n = Sy σ = 100 60.8 = 1.64 MSS Case 3, using Eq. (5–6), Answer n = − Sy σB = − 100 −70 = 1.43 (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 1/2 = 0 kpsi Answer n = Sy σ = 100 0 → ∞ MSS From Eq. (5–3), Answer n = Sy σ1 − σ3 = 100 30 − 30 → ∞ A tabular summary of the factors of safety is included for comparisons. (a) (b) (c) (d) (e) DE 1.43 1.64 1.13 1.64 ∞ MSS 1.43 1.43 1.00 1.43 ∞ 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 σA, σB plane are shown in Fig. 5–11. Case (e) is not plane stress. Note that the load line for case (a) is
Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hill 23 Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition Failures Resulting from Static Loading 219 Figure 5-11 (a) Load lines for Example 5-1. 0 (c) 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 pre- dict 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 Sthan 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 o 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 interal-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 a12>o3.The largest circle connects o and o3.as shown in Fig.5-13.The centers of the circles in Fig.5-13 are C.C2.and C3.Triangles OB,C are similar,therefore BaC2 -BC1=B3C3 -BiC 0C2-0C1-0C3-0C1
Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGraw−Hill 223 Companies, 2008 Failures Resulting from Static Loading 219 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 OBi Ci are similar, therefore B2C2 − B1C1 OC2 − OC1 = B3C3 − B1C1 OC3 − OC1 Figure 5–11 Load lines for Example 5–1. –Sy –Sy Sy Sy B A (a) (b) (c) (d) DE MSS Load lines A B the only plane stress case given in which the two theories agree, thus giving the same factor of safety.