Mechanics.Of.Materials.Saouma Episode 13 pps

Draft 19.4 Plastic Yield Conditions (Classical Models) 5 12 In terms of the principal stresses, those invariants can be simplified into I 1 = σ 1 + σ 2 + σ 3 (19.24) I 2 = −(σ 1 σ 2 + σ 2 σ 3 + σ 3 σ 1 ) (19.25) I 3 = σ 1 σ 2 σ 3 (19.26) 13 Similarly, J 1 = s 1 + s 2 + s 3 (19.27) J 2 = −(s 1 s 2 + s 2 s 3 + s 3 s 1 ) (19.28) J 3 = s 1 s 2 s 3 (19.29) 19.4.1.2 Physical Interpretations of Stress Invariants 14 If we consider a plane which makes equal angles with respect to each of the principal-stress directions, π plane, or octahedral plane, the normal to this plane is given by n = 1 √ 3    1 1 1    (19.30) The vector of traction on this plane is t oct = 1 √ 3    σ 1 σ 2 σ 3    (19.31) and the normal component of the stress on the octahedral plane is given by σ oct = t oct ·n = σ 1 + σ 2 + σ 3 3 = 1 3 I 1 = σ hyd (19.32) or σ oct = 1 3 I 1 (19.33) 15 Finally, the octahedral shear stress is obtained from τ 2 oct = |t oct | 2 − σ 2 oct = σ 2 1 3 + σ 2 2 3 + σ 2 3 3 − (σ 1 + σ 2 + σ 3 ) 2 9 (19.34) Upon algebraic manipulation, it can be shown that 9τ 2 oct =(σ 1 − σ 2 ) 2 +(σ 2 − σ 3 ) 2 +(σ 1 − σ 3 ) 2 =6J 2 (19.35) or τ oct =  2 3 J 2 (19.36) and finally, the direction of the octahedral shear stress is given by cos 3θ = √ 2 J 3 τ 3 oct (19.37) Victor Saouma Mechanics of Materials II Draft 6 3D PLASTICITY 16 The elastic strain energy (total) per unit volume can be decomposed into two parts U = U 1 + U 2 (19.38) where U 1 = 1 − 2ν E I 2 1 Dilational energy (19.39-a) U 2 = 1+ν E J 2 Distortional energy (19.39-b) 19.4.1.3 Geometric Representation of Stress States Adapted from (Chen and Zhang 1990) 17 Using the three principal stresses σ 1 , σ 2 ,andσ 3 , as the coordinates, a three-dimensional stress space can be constructed. This stress representation is known as the Haigh-Westergaard stress space, Fig. 19.4. 3 1 Cos J 2 −1 (s ,s ,s ) 3 1 2 3 1 σ N(p,p,p) ρ ξ O σ 2 σ Hydrostatic axis deviatoric plane P( , , ) σ σ σ 1 2 3 Figure 19.4: Haigh-Westergaard Stress Space 18 The decomposition of a stress state into a hydrostatic, pδ ij and deviatoric s ij stress components can be geometrically represented in this space. Considering an arbitrary stress state OP starting from O(0, 0, 0) and ending at P(σ 1 ,σ 2 ,σ 3 ), the vector OP can be decomposed into two components ON and NP. The former is along the direction of the unit vector (1 √ 3, 1/ √ 3, 1/ √ 3), and NP⊥ON. 19 Vector ON represents the hydrostatic component of the stress state, and axis Oξ is called the hy- drostatic axis ξ, and every point on this axis has σ 1 = σ 2 = σ 3 = p,or ξ = √ 3p (19.40) Victor Saouma Mechanics of Materials II Draft 19.4 Plastic Yield Conditions (Classical Models) 7 20 Vector NP represents the deviatoric component of the stress state (s 1 ,s 2 ,s 3 ) and is perpendicular to the ξ axis. Any plane perpendicular to the hydrostatic axis is called the deviatoric plane and is expressed as 1 √ 3 (σ 1 + σ 2 + σ 3 )=ξ (19.41) and the particular plane which passes through the origin is called the π plane and is represented by ξ = 0. Any plane containing the hydrostatic axis is called a meridian plane. The vector NP lies in a meridian plane and has ρ =  s 2 1 + s 2 2 + s 2 3 =  2J 2 (19.42) 21 The projection of NP and the coordinate axes σ i on a deviatoric plane is shown in Fig. 19.5. The σ ’ 2 σ 3 ’ 120 0 120 0 120 0 P’ θ N’ σ ’ 1 Figure 19.5: Stress on a Deviatoric Plane projection of N  P  of NP on this plane makes an angle θ with the axis σ  1 . cos 3θ = 3 √ 3 2 J 3 J 3/2 2 (19.43) 22 The three new variables ξ,ρ and θ can all be expressed in terms of the principal stresses through their invariants. Hence, the general state of stress can be expressed either in terms of (σ 1 ,σ 2 ,σ 3 ), or (ξ, ρ, θ). For 0 ≤ θ ≤ π/3, and σ 1 ≥ σ 2 ≥ σ 3 ,wehave    σ 1 σ 2 σ 3    =    p p p    + 2 √ 3  J 2    cos θ cos(θ −2π/3) cos(θ +2π/3)    (19.44-a) =    ξ ξ ξ    +  2 3 ρ    cos θ cos(θ −2π/3) cos(θ +2π/3)    (19.44-b) (19.44-c) 19.4.2 Hydrostatic Pressure Independent Models Adapted from (Chen and Zhang 1990) 23 For hydrostatic pressure independent yield surfaces (such as for steel), their meridians are straigth lines parallel to the hydrostatic axis. Hence, shearing stress must be the major cause of yielding for Victor Saouma Mechanics of Materials II Draft 8 3D PLASTICITY this type of materials. Since it is the magnitude of the shear stress that is important, and not its direction, it follows that the elastic-plastic behavior in tension and in compression should be equivalent for hydorstatic-pressure independent materials (such as steel). Hence, the cross-sectional shapes for this kind of materials will have six-fold symmetry, and ρ t = ρ c . 19.4.2.1 Tresca 24 Tresca criterion postulates that yielding occurs when the maximum shear stress reaches a limiting value k. max  1 2 |σ 1 − σ 2 |, 1 2 |σ 2 − σ 3 |, 1 2 |σ 3 − σ 1 |  = k (19.45) from uniaxial tension test, we determine that k = σ y /2 and from pure shear test k = τ y . Hence, in Tresca, tensile strength and shear strength are related by σ y =2τ y (19.46) 25 Tresca’s criterion can also be represented as 2  J 2 sin  θ + π 3  − σ y =0 for 0≤ θ ≤ π 3 (19.47) 26 Tresca is on, Fig. 19.6: Planeπ σ σ 3 σ 1 2 ξ ρ ρ 0 ξ ρ σ 1 ’ ρ 0 σ y σ 2 −σ y −σ y σ 1 σ τ σ 3 σ ’ 2 ’ Figure 19.6: Tresca Criterion • σ 1 σ 2 σ 3 space represented by an infinitly long regular hexagonal cylinder. Victor Saouma Mechanics of Materials II Draft 19.4 Plastic Yield Conditions (Classical Models) 9 • π (Deviatoric) Plane, the yield criterion is ρ =2  J 2 = σ y √ 2sin  θ + π 3  for 0 ≤ θ ≤ π 3 (19.48) a regular hexagon with six singular corners. • Meridian plane: a straight line parallel to the ξ axis. • σ 1 σ 2 sub-space (with σ 3 = 0) an irregular hexagon. Note that in the σ 1 ≥ 0,σ 2 ≤ 0 the yield criterion is σ 1 − σ 2 = σ y (19.49) • στ sub-space (with σ 3 = 0) is an ellipse  σ σ y  2 +  τ τ y  2 = 1 (19.50) 27 The Tresca criterion is the first one proposed, used mostly for elastic-plastic problems. However, because of the singular corners, it causes numerous problems in numerical analysis. 19.4.2.2 von Mises 28 There are two different physical interpretation for the von Mises criteria postulate: 1. Material will yield when the distorsional (shear) energy reaches the same critical value as for yield as in uniaxial tension. F (J 2 )=J 2 − k 2 = 0 (19.51) =  (σ 1 − σ 2 ) 2 +(σ 2 − σ 3 ) 2 +(σ 1 − σ 3 ) 2 2 − σ y = 0 (19.52) 2. ρ, or the octahedral shear stress (CHECK) τ oct , the distance of the corresponding stress point from the hydrostatic axis, ξ is constant and equal to: ρ 0 = τ y √ 2 (19.53) 29 Using Eq. 19.52, and from the uniaxial test, k is equal to k = σ y / √ 3, and from pure shear test k = τ y . Hence, in von Mises, tensile strength and shear strength are related by σ y = √ 3τ y (19.54) Hence, we can rewrite Eq. 19.51 as f(J 2 )=J 2 − σ 2 y 3 = 0 (19.55) 30 von Mises is on, Fig. ??: • σ 1 σ 2 σ 3 space represented by an infinitly long regular circular cylinder. • π (Deviatoric) Plane, the yield criterion is ρ =  2 3 σ y (19.56) a circle. Victor Saouma Mechanics of Materials II Draft 10 3D PLASTICITY π Plane 1 σ 3 σ 2 σ ξ ρ ρ 0 ξ ρ σ 1 ’ ρ 0 σ τ σ 2 σ ’ 2 σ ’ 3 σ 1 x y Figure 19.7: von Mises Criterion • Meridian plane: a straight line parallel to the ξ axis. • σ 1 σ 2 sub-space (with σ 3 = 0) an ellipse σ 2 1 + σ 2 2 − σ 1 σ 2 = σ 2 y (19.57) • στ sub-space (with σ 3 = 0) is an ellipse  σ σ y  2 +  τ τ y  2 = 1 (19.58-a) Note that whereas this equation is similar to the corresponding one for Tresca, Eq. 19.50, the difference is in the relationships between σ y and τ y . 19.4.3 Hydrostatic Pressure Dependent Models 31 Pressure sensitive frictional materials (such as soil, rock, concrete) need to consider the effects of both the first and second stress invariants. frictional materials such as concrete. 32 The cross-sections of a yield surface are the intersection curves between the yield surface and the deviatoric plane (ρ, θ) which is perpendicular to the hydrostatic axis ξ and with ξ = constant. The cross-sectional shapes of this yield surface will have threefold symmetry, Fig. 19.8. 33 The meridians of a yield surface are the intersection curves between the surface and a meridian plane (ξ,ρ) which contains the hydrostatic axis. The meridian plane with θ = 0 is the tensile meridian,and passes through the uniaxial tensile yield point. The meridian plane with θ = π/3isthecompressive meridian and passes through the uniaxial compression yield point. 34 The radius of a yield surface on the tensile meridian is ρ t , and on the compressive meridian is ρ c . Victor Saouma Mechanics of Materials II Draft 19.4 Plastic Yield Conditions (Classical Models) 11 σ ’ 1 σ ’ 2 σ 3 ’ ρ ρ ρ c t θ DEVIATORIC PLANE ρ ξ MERIDIAN PLANE Tensile Meridian Compressive Meridian θ=0 θ=π/3 Figure 19.8: Pressure Dependent Yield Surfaces 19.4.3.1 Rankine 35 The Rankine criterion postulates that yielding occur when the maximum principal stress reaches the tensile strength. σ 1 = σ y ; σ 2 = σ y ; σ 3 = σ y ; (19.59) 36 Rankine is on, Fig. 19.9: • σ 1 σ 2 σ 3 space represented by XXX • π (Deviatoric) Plane, the yield criterion is  ρ t = 1 √ 2 ( √ 3σ Y − ξ) ρ c = √ 2( √ 3σ y − ξ) (19.60) a regular triangle. • Meridian plane: Two straight lines which intersect the ξ axis ξ y = √ 3σ y • σ 1 σ 2 sub-space (with σ 3 = 0) two straight lines.  σ +1 = σ y σ 2 = σ y (19.61) • στ sub-space (with σ 3 = 0) is a parabola σ σ y +  τ σ y  2 = 1 (19.62-a) 19.4.3.2 Mohr-Coulomb 37 The Mohr-Coulomb criteria can be considered as an extension of the Tresca criterion. The maximum shear stress is a constant plus a function of the normal stress acting on the same plane. |τ| = c − σ tan φ (19.63) Victor Saouma Mechanics of Materials II Draft 12 3D PLASTICITY c Cot π Plane 3 1 σ 2 σ ξ ρ σ Φ σ 1 ’ σ 2 σ t σ t σ t σ ’ 2 σ ’ 3 σ 1 ξ ρ ρ ρ cy ty θ=0 θ=π/3 σ τ Figure 19.9: Rankine Criterion where c is the cohesion, and φ the angle of internal friction. 38 Both c and φ are material properties which can be calibrated from uniaxial tensile and uniaxail compressive tests.  σ t = 2c cos φ 1+sin φ σ c = 2c cos φ 1−sin φ (19.64) 39 In terms of invariants, the Mohr-Coulomb criteria can be expressed as: 1 3 I 1 sin φ +  J 2 sin  θ + π 3  +  J 2 3 cos  θ + π 3  sin φ − c cos φ =0 for 0≤ θ ≤ π 3 (19.65) 40 Mohr-Coulomb is on, Fig. 19.10: • σ 1 σ 2 σ 3 space represented by a conical yield surface whose normal section at any point is an irregular hexagon. • π (Deviatoric) Plane, the cross-section of the surface is an irregular hexagon. • Meridian plane: Two straight lines which intersect the ξ axis ξ y =2 √ 3c/ tan φ, and the two characteristic lengths of the surface on the deviatoric and meridian planes are  ρ t = 2 √ 6c cos φ−2 √ 2ξ sin φ 3+sin φ ρ c = 2 √ 6c cos φ−2 √ 2ξ sin φ 3−sin φ (19.66) Victor Saouma Mechanics of Materials II Draft 19.4 Plastic Yield Conditions (Classical Models) 13 c Cot π Plane 3 2 1 σ σ ξ ρ σ Φ σ 1 ’ ρ ty ρ cy σ t σ t σ 1 σ 2 −σ c σ t −σ c σ ’ 2 σ ’ 3 ξ ρ σ θ=0 θ=π/3 ρ cy ρ ty τ Figure 19.10: Mohr-Coulomb Criterion • σ 1 σ 2 sub-space (with σ 3 = 0) the surface is an irregular hexagon. In the quarter σ 1 ≥ 0,σ 2 ≤ 0, of the plane, the criterios is mσ 1 − σ 2 = σ c (19.67) where m = σ c σ t = 1+sinφ 1 − sin φ (19.68) • στ sub-space (with σ 3 = 0) is an ellipse  σ + m−1 2m σ c m+1 2m σ c  2 +  τ √ m 2m σ c  2 (19.69-a) 19.4.3.3 Drucker-Prager 41 The Drucker-Prager postulates is a simple extension of the von Mises criterion to include the effect of hydrostatic pressure on the yielding of the materials through I 1 F (I 1 ,J 2 )=αI 1 + J 2 − k (19.70) The strength parameters α and k can be determined from the uni axial tension and compression tests  σ t = √ 3k 1+ √ 3α σ c = √ 3k 1− √ 3α (19.71) Victor Saouma Mechanics of Materials II Draft 14 3D PLASTICITY or  α = m−1 √ 3(m+1) k = 2σ c √ 3(m+1) (19.72) 42 Drucker-Prager is on, Fig. 19.11: c Cot π Plane 1 σ 3 Φ 2 σ σ ξ ρ σ 1 ’ ρ 0 σ 2 σ 1 σ t σ ’ 3 σ ’ 2 ξ ρ y x σ τ −σ c Figure 19.11: Drucker-Prager Criterion • σ 1 σ 2 σ 3 space represented by a circular cone. • π (Deviatoric) Plane, the cross-section of the surface is a circle of radius ρ. ρ = √ 2k − √ 6αξ (19.73) • Meridian plane: The meridians of the surface are straight lines which intersect with the ξ axis at ξ y = k/ √ 3α. • σ 1 σ 2 sub-space (with σ 3 = 0) the surface is an ellipse  x + 6 √ 2kα 1−12α 2 √ 6k 1−12α 2  2 +   y √ 2k √ 1−12α 2   2 (19.74) where x = 1 √ 2 (σ 1 + σ 2 ) (19.75-a) y = 1 √ 2 (σ 2 − σ 1 ) (19.75-b) Victor Saouma Mechanics of Materials II [...]... 2 ˙ σ = [[K − G]I 2 ⊗ I 2 + Gep ] : ˙ 3 (19.112) (19. 113) (19.114) (19.115) (19.116) where the elastoplastic material tensor is 2 Eep = [[K − G]I 2 ⊗ I 2 + Gep ] 3 19.9.1 (19.117) Isotropic Hardening/Softening(J2 − plasticity) In isotropic hardening/softening the yield surface may shrink (softening) or expand (hardening) uniformly (see figure 19 .13) 66 s3 Fn+1= 0 for Hp > 0 (Hardening) Fn = 0 for Hp... surface may shrink (softening) or expand (hardening) uniformly (see figure 19 .13) 66 s3 Fn+1= 0 for Hp > 0 (Hardening) Fn = 0 for Hp = 0 (Perfectly Plastic) Fn+1= 0 for Hp < 0 (Softening) s1 s2 Figure 19 .13: Isotropic Hardening/Softening 1 Yield function for linear strain hardening/softening: F (s, Victor Saouma p ef f ) = 1 o 1 s : s − (σy + Ep 2 3 p 2 ef f ) =0 (19.118) Mechanics of Materials II Draft . stress must be the major cause of yielding for Victor Saouma Mechanics of Materials II Draft 8 3D PLASTICITY this type of materials. Since it is the magnitude of the shear stress that is important,. extension of the Tresca criterion. The maximum shear stress is a constant plus a function of the normal stress acting on the same plane. |τ| = c − σ tan φ (19.63) Victor Saouma Mechanics of Materials. σ 3 ) 2 =6J 2 (19.35) or τ oct =  2 3 J 2 (19.36) and finally, the direction of the octahedral shear stress is given by cos 3θ = √ 2 J 3 τ 3 oct (19.37) Victor Saouma Mechanics of Materials II Draft 6 3D PLASTICITY 16 The elastic