Chapter 7. Yielding and Fracture under Combined Stresses
Transcript of Chapter 7. Yielding and Fracture under Combined Stresses
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Mechanical Strengths and Behavior of Solids
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Chapter 7. Yielding and Fracture under Combined Stresses
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Contents
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1 Introduction
2 General Form of Failure Criteria
3 Maximum Normal Stress Fracture Criterion
4 Maximum Shear Stress Yield Criterion
5 Octahedral Shear Stress Yield Criterion
6 Discussion of the Basic Failure Criteria
7 Coulomb-Mohr Fracture Criterion
8 Modified Mohr fracture Criterion
9 Brittle Versus Ductile Behavior
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7.1 Introduction
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β’ A failure criterion is required to predict the safe limits for use of a material under combined stresses.
Figure 7.1 Yield strengths for a ductile metal under various states of stress: (a) uniaxial tension, (b) tension with transverse compression, (c) biaxial tension, and (d) hydrostatic compression.
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7.2 General Form of Failure Criteria
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β’ A mathematical function of the principal normal stresses:ππ ππ1,ππ2,ππ3 = οΏ½ππ
Failure occurs when οΏ½ππ = ππππ, where ππππ is the failure strength of the material.Failure is not expected if οΏ½ππ is less than ππππ.
β’ The safety factor against failure:
ππ =πππποΏ½ππ
β Safety factor describes the structural capacity of a system beyond expected loads or applied loads.
β In other words, the applied stresses can be increased by a factor X before failure occurs.
οΏ½ππ = ππππ (at failure)
οΏ½ππ < ππππ (no failure)
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7.3 Maximum Normal Stress Fracture Criterion (1)
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β’ A maximum normal stress fracture criterion:
For plane stress, as shown in Fig. 7.2(a), the box is the region that satisfies
MAX( ππ1 , ππ2 ) β€ πππ’π’
Figure 7.2 Failure locus for the maximum normal stress fracture criterion for plane stress.
πππ’π’ = MAX( ππ1 , ππ2 , ππ3 ) (at fracture)
Safety factor
ππ =π·π·πππ·π·
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7.3 Maximum Normal Stress Fracture Criterion (2)
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For the general case, where all three principal normal stresses are nonzero, the failure surface is illustrated as Fig. 7.3.
Figure 7.3 Three-dimensional failure surface for the maximum normal stress fracture criterion.
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7.4 Maximum Shear Stress Yield Criterion (1)
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β’ Yielding of ductile materials occurs when the maximum shear stress on any plane reaches a critical value ππππ (ππππ = ππππππππ).
β’ Maximum shear stress yield criterion (Tresca criterion)
ππππ = MAX ππ1, ππ2, ππ3 = MAX(ππ2 β ππ3
2,ππ1 β ππ3
2,ππ1 β ππ2
2)
In uniaxial tension test,ππ1 = ππππ, ππ2 = ππ3 = 0
Substitution of these values into the yield criterion gives
ππππ =ππππ2
Thus,
ππππ2 = MAX(
ππ1 β ππ22 ,
ππ2 β ππ32 ,
ππ3 β ππ12 )
Let the effective stress be οΏ½ππππ, the safety factor is then
ππ =πππποΏ½ππππ
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7.4 Maximum Shear Stress Yield Criterion (2)
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Figure 7.4 The plane of maximum shear in a uniaxial tension test.
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7.4 Maximum Shear Stress Yield Criterion (3)
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For plane stress, a maximum shear stress yield criterion:
ππππ = MAX ππ1 β ππ2 , ππ2 , ππ1Boundaries:
ππ1 β ππ2 = Β±ππππ, ππ2 = Β±ππππ, ππ1 = Β±ππππ
Figure 7.5 Failure locus for the maximum shear stress yield criterion for plane stress.
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7.4 Maximum Shear Stress Yield Criterion (4)
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For the general case, boundaries:ππ1 β ππ2 = Β±ππππ, ππ2 β ππ3 = Β±ππππ, ππ1 β ππ3 = Β±ππππ
Figure 7.6 Three-dimensional failure surface for the maximum shear stress yield criterion.
Safety factor
ππ =π·π·π·πππ·π·π·
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7.5 Octahedral Shear Stress Yield Criterion (1)
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β’ Yielding of ductile materials occurs when the maximum shear stress on octahedral plane reaches a critical value ππβππ (ππβ = ππβππ).
β’ Octahedral Plane: For the special case where πΌπΌ = π½π½ = πΎπΎ, the oblique plane which intersects the principal axes at equal distances from the origin
Figure 6.15 Octahedral plane shown relative to the principal normal stress axes (a), and the octahedron formed by the similar such planes in all octants (b).
πΌπΌ = cosβ113
= 54.7Β°
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7.5 Octahedral Shear Stress Yield Criterion (2)
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β’ Octahedral shear stress yield criterion (von Mises or distortion energy criterion):
ππβππ =13
ππ1 β ππ2 2 + ππ2 β ππ3 2 + ππ3 β ππ1 2
In uniaxial tension with ππ1 = ππππ, and ππ2 = ππ3 = 0,
ππβππ =2
3ππππ
The same result can be obtained from Mohrβs circle that the normal stress is
ππβ =ππ1 + ππ2 + ππ3
3=ππ13
Figure 7.7 The plane of octahedral shear in a uniaxial tension test.
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7.5 Octahedral Shear Stress Yield Criterion (3)
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β’ The yield criterion expressed in terms of the uniaxial yield strength:
For plane stress,
ππππ =12
ππ1 β ππ2 2 + ππ22 + ππ12
Manipulation givesππππ2 = ππ12 β ππ1ππ2 + ππ22
Which is the equation of an ellipse, as shown in Fig. 7.8.
ππππ =12
ππ1 β ππ2 2 + ππ2 β ππ3 2 + ππ3 β ππ1 2 (at yielding)
Figure 7.8 Failure locus for the octahedral shear stress yield criterion for plane stress, and comparison with the maximum shear criterion.
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7.5 Octahedral Shear Stress Yield Criterion (4)
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For the general case, boundaries:
max. shear
oct. shear
Figure 7.9 Three-dimensional failure surface for the octahedral shear stress yield criterion.
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7.6 Comparison of Failure Criteria
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β’ Ductile materialsβ Tresca and von Mises criteria are widely used to predict yielding of ductile materials.β Both of these indicate that hydrostatic stress does not affect yielding.β Because of the small difference between the two, a choice between the two is not a
matter of major importance.
β’ Brittle materialsβ They are usually follow the normal stress criterion.
Figure 7.11 Plane stress failure loci for three criteria. These are compared with biaxial yield data for ductile steels and aluminum alloys, and also with biaxial fracture data for gray cast iron. (The steel data are from [Lessells 40] and [Davis 45], the aluminum data from [Naghdi 58] and [Marin 40], and the cast iron data from [Coffin 50] and [Grassi 49].)
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7.6 Yield Criteria for Anisotropic and Pressure-Sensitive Materials
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β’ The anisotropic yield criterion described in Hill:π»π» ππππ β ππππ 2 + πΉπΉ ππππ β ππππ 2 + πΊπΊ ππππ β ππππ 2 + 2ππππππππ2 + 2πΏπΏππππππ2 + 2ππππππππ2 = 1
π»π» + πΊπΊ =1ππππππ2
, π»π» + πΉπΉ =1ππππππ2
, πΉπΉ + πΊπΊ =1ππππππ2
2ππ =1
ππππππππ2 , 2πΏπΏ =1
ππππππππ2 , 2ππ =1
ππππππππ2
In case of polymers, yield strengths in compression is often higher than those in tension.
Figure 7.12 Biaxial yield data for various polymers compared with a modified octahedral shear stress theory. (Data from [Raghava 72].)
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7.6 Fracture in Brittle Materials
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β’ In brittle materials, deviation from maximum normal stress criterion occur if the normal stress having the largest absolute value is compressive.
Figure 7.13 Biaxial fracture data of gray cast iron compared with two fracture criteria. (Data from [Grassi 49].)
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7.7 Coulomb-Mohr Fracture Criterion (1)
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β’ Hypothesis: Fracture occurs when a critical combination of shear and normal stress acts on the plane.
β’ Mathematical function:ππ + ππππ = ππππ
Figure 7.14 CoulombβMohr fracture criterion as related to Mohrβs circle, and predicted fracture planes.
Stresses on the plane of fracture
More compressive, more friction
Necessary to increase the πππ· to cause fracture
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7.7 Coulomb-Mohr Fracture Criterion (2)
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Assume ππ1 β₯ ππ2 β₯ ππ3 (with signs considered).From the largest Mohrβs circle at the point (ππβ², πππ·),
ππβ² =ππ1 + ππ3
2+
ππ1 β ππ32
cos2ππππ , πππ· =ππ1 β ππ3
2sin2ππππ
cos2ππππ = sinππ, sin2ππππ = cosππ, ππ = tanππ =sinππcosππ
, sin2ππ + πππππ π 2ππ = 1
After manipulation,ππ1 β ππ3 + ππ1 + ππ3 sinππ = 2ππππcosππππ1 β ππ3 + ππ ππ1 + ππ3 = 2ππππ 1 βππ2
ππ1 β ππ3 + ππ ππ1 + ππ3 = πππ’π’ππβ² (1 βππ)In simple compression, ππ3 = πππ’π’ππβ² ,ππ1 = ππ2 = 0.Therefore, the expected strength πππ’π’ππβ² (negative) is
πππ’π’ππβ² = β2ππππ1 + ππ1 βππ
In simple tension, ππ1 = πππ’π’π‘π‘β² ,ππ2 = ππ3 = 0.Therefore, the expected strength πππ’π’π‘π‘β² is
πππ’π’π‘π‘β² = 2ππππ1 βππ1 + ππ
(ππ = sinππ)
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Figure 7.15 Fracture planes predicted by the CoulombβMohr criterion for uniaxial tests in tension and compression.
7.7 Coulomb-Mohr Fracture Criterion (3)
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In simple torsion, ππ1 = βππ3 = πππ’π’β² ,ππ2 = 0.Therefore, the fracture strength πππ’π’β² is
πππ’π’β² = ππ 1 βππ2
Figure 7.16 Pure torsion and the fracture planes predicted by the CoulombβMohr criterion.
7.7 Coulomb-Mohr Fracture Criterion (4)
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If the assumption ππ1 β₯ ππ2 β₯ ππ3 is dropped,ππ1 β ππ2 + ππ ππ1 + ππ2 = πππ’π’ππβ² (1 βππ)ππ2 β ππ3 + ππ ππ2 + ππ3 = πππ’π’ππβ² (1 βππ)ππ3 β ππ1 + ππ ππ3 + ππ1 = πππ’π’ππβ² (1 βππ)
For plane stress with ππ3 = 0,ππ1 β ππ2 + ππ ππ1 + ππ2 = πππ’π’ππβ² (1 βππ)
ππ2 + ππ ππ2 = πππ’π’ππβ² (1 βππ)ππ1 + ππ ππ1 = πππ’π’ππβ² (1 βππ)
And relationship between the fracture strengths in tension and compression is
πππ’π’π‘π‘β² = πππ’π’ππβ²1 βππ1 + ππ
7.7 Coulomb-Mohr Fracture Criterion (5)
Figure 7.17 Failure locus for the CoulombβMohr fracture criterion for plane stress.
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7.7 Coulomb-Mohr Fracture Criterion (6)
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For the general case, fracture criterion represents six planes as shown in Fig. 7.18.The surface forms a vertex along the line ππ1 = ππ2 = ππ3 at the point
ππ1 = ππ2 = ππ3 = πππ’π’ππβ²1 βππ2ππ
Figure 7.18 Three-dimensional failure surface for the CoulombβMohr fracture criterion.
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7.7 Coulomb-Mohr Fracture Criterion (7)
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β’ Effective stress for the C-M criterionπΆπΆ12 =
11 βππ
ππ1 β ππ2 + ππ ππ1 + ππ2
πΆπΆ23 =1
1 βππ[ ππ2 β ππ3 + ππ ππ2 + ππ3 ]
πΆπΆ31 =1
1 βππ[ ππ3 β ππ1 + ππ ππ3 + ππ1 ]
Hence, effective stress isοΏ½πππΆπΆπΆπΆ = MAX(πΆπΆ12,πΆπΆ23,πΆπΆ31)
And safety factor against fracture is
πππΆπΆπΆπΆ =πππ’π’ππβ²
οΏ½πππΆπΆπΆπΆβ’ Discussion
β The fracture planes predicted for both tension and torsion are incorrect. Brittle materials generally fail on planes normal to the maximum tension stress in both cases, not on the planes predicted by the C-M theory, as in Figs. 7.15 and 7.16.
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7.8 Modified Mohr Fracture Criterion (1)
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β’ Modified Mohr (M-M) fracture criterion: Combination of the C-M criterion and the maximum normal stress fracture criterion
Figure 7.20 The modified Mohr (M-M) fracture criterion, formed by the maximum normal stress criterion truncating the CoulombβMohr (CβM) criterion.
ππππ = β πππ’π’ππβ² + πππ’π’π‘π‘1 + ππ1 βππ
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7.8 Modified Mohr Fracture Criterion (2)
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β’ In three dimensions, the M-M failure locus is similar to Fig. 7.21.β’ The three positive faces of the normal stress cube correspond to
ππ1 = πππ’π’π‘π‘ , ππ2 = πππ’π’π‘π‘ , ππ3 = πππ’π’π‘π‘β’ Three requirements for M-M criterion
β The slope of the C-M failure envelope (specified by ππ,ππ,ππππ , or ππ)β The intercept ππππ of the C-M failure envelopeβ The ultimate tensile strength πππ’π’π‘π‘
β’ For the maximum normal stress components,οΏ½ππππππ = MAX ππ1,ππ2,ππ3 , ππππππ =
πππ’π’π‘π‘οΏ½ππππππ
β’ Safety factor for the M-M criterion:πππΆπΆπΆπΆ = MIN XCM, XNP or
1πππΆπΆπΆπΆ
= MAX(οΏ½πππΆπΆπΆπΆπππ’π’ππβ²
,οΏ½πππππππππ’π’π‘π‘
)
Figure 7.21 Three-dimensional failure surface for the modified Mohr fracture criterion. The CoulombβMohr surface is truncated by three faces of the maximum normal stress cube.
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7.9 Brittle Versus Ductile Behavior (1)
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Figure 7.22 Stressβstrain data for limestone cylinders tested under axial compression with various hydrostatic pressures ranging from one to 10,000 atmospheres. The applied compressive stress plotted is the stress in the pressurized laboratory, that is, the compression in excess of pressure. (Adapted from [Griggs 36]; used with permission; Β© 1936 The University of Chicago Press.)
Figure 7.23 Effect of pressures ranging from one to 26,500 atmospheres on the tensile behavior of a steel, specifically AISI 1045 with HRC = 40. Stress in the pressurized laboratory is plotted. (Data from [Bridgman 52] pp. 47β61.)
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7.9 Brittle Versus Ductile Behavior (2)
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Figure 7.24 Relationships of the limiting surfaces for yielding and fracture for materials that usually behave in a ductile manner, and also for materials that usually behave in a brittle manner.