Mechanical Properties - Stresses & Strains

Transcription

1 Mechanical Properties - Stresses & Strains Types of Deformation : Elasic Plastic Anelastic Elastic deformation is defined as instantaneous recoverable deformation Hooke's law : For tensile loading, σ = E ε where σ is stress defined as the load per unit area : σ = P/A o, N/m 2, Pa and strain is given by the change in length per unit length ε = l l, % o The proportional constant E is the Young's modulus or modulus of Elasticity : E ~ 10x10 6 psi [68.9 GPa] for metals [varying from 10x10 6 psi for Al, 30x10 6 for Fe and 59x106 for W]. Poisson's Ratio [ν] : ratio of lateral contraction to longitudinal elongation ν = - ε x / ε z = - ε y / ε z [for isotropic materials]; in general, ν ~ 0.3 Thus the total contractile strains is less than the expansion along the tensile axis thereby resulting in a slight increase in the volume of the material under stress - this is known as Elastic Dilation. Modulus Of Rigidity or Shear Modulus [G] : G = τ / γ ; G is the shear modulus and is related to E andν, G = E / 2(1+ν ). Bulk Modulus [κ] : the change in volume to the original volume is proportional to the hydrostatic pressure [σ hyd ] : V/V = β σ hyd, where β is the compressibility. The inverse of the compressibility is the bulk modulus [κ] : κ = 1/β. κ is also related to E and ν : κ = E / 3(1-2ν ) = 2G(1+ν ) / 3(1-2ν ). Thus one can evaluate the various elastic moduli from one or more experimentally evaluated constants. Note that the elastic moduli are related to the interatomic bonding and thus decrease [slightly] with increasing temperature. Any change in the crystal structure, for example following a phase change [polymorphism], one notes a distinct change in the elastic moduli. KL Murty page 1 MAT 450

2 Stress - Strain Curve Definitions Nominal (engineering) vs True S = P A, e = l o l o Proportional limit (PL) Yield strength (S y ) 0.2% offset; (S LY ) Tensile strength (TS or UTS or S UTS ) Fracture strength (S F ) Uniform elongation (e u ) Total elongation (ductility) (e t or e f in 2 ) Necking strain (e n = e t - e u ) Reduction in area (ductility) (RA) Volume increases (Elastic Dilation) σ = P A, ε = ln ( l l o ) σ = S (1+e) & ε = ln (1+e) true Yield stress (σ y ) 0.2% offset true Tensile strength (TS or UTS or σ UTS ) true Fracture strength (σ F ) true Uniform strain (ε u ) true Total elongation (ductility) (ε t or e f in 2 ) true Necking strain (ε n = ε t - ε u ) Volume is conserved A o l o = Al Energy to fracture Elastic (σ = E ε ) Plastic (σ = K εn ) Resilience (U el = σ2 2E ) units (J/m3 ) Toughness (J = K εn+1 n+1 ) KL Murty page 2 MAT 450

3 σ - ε curves smooth (SSs & fcc) with yield point (steels & bcc) Plastic deformation is rate dependent Rate Effects d ln σ d ln ε (generally at high temperatures) : σ f( εd ) = A εd m, m = SRS = ( ) m ~ 0 at low temperatures m max = 1 T,ε m e t vs n e u Group Work : left of the instructor : (1) Derive relation between σ and S : right of the instructor : (2) Derive relation between ε and e : all (3) Show that ε u = n KL Murty page 3 MAT 450

4 Concept of Stress σ = F A lim 0 A Force extended on reference section by remaining sections body in equilibrium Normal and Shear Stresses σ N = F A cosθ σ shear = τ = F A sinθ τ x = τ y = recall RSS (τ RSS ) : KL Murty page 4 MAT 450

5 F, Force is a vector (1st rank tensor) Stress - Strain Relationships Elastic Behavior while σ ij should be specified with 2 directions : plane normal and force direction - acts on plane perpendicular to i along j direction z Sign Convention (Fig. 2.2) σ z τ zx τ zy τ yz τ yx σ y y Tension : +ive Compression : -ve τ ij (or σ ij ) is +ive if both i and j are +ive or ive τ ij (or σ ij ) is - ive if one of i and jis - ive x σ xx σ xy σ xz σ= σ yx σ yy σ yz no net moment σ ij = σ ji or only 6 components or σ zx σ zy σ zz σ xx σ xy σ xz σ x τ xy τ xz σ= σ xy σ yy σ yz ; book notation σ y τ yz σ xz σ yz σ zz σ z Values of σ ij depend on the choice of reference axes (see 2-D example 2.3) {can determine these components using tensor transformations, Mohr circle, etc.} First, we look at 3 important examples of Stress States 1. Plane Stress (p.20) 2. Hydrostatic & Deviatoric Stresses (p.46) 3. Principal Stresses (various sections such as 2.14) KL Murty page 5 MAT 450

6 Plane Stress (p. 20) - stresses are zero in one of the primary directions (or 2-D stress state) - Examples : 1. Thin sheet with loaded in the plane (stresses are zero along the thickness direction) 2. Pressurized thin cylinder (stresses along r or thickness direction are zero for cylinders when wall-thickess is about 1/10 th of diameter) : σ θ = Pr, σz = t Pr with σr 0 2t Principal Stresses (p. 22) Principal Planes are the planes on which maximum normal stresses act with no shear stresses and these stresses are the Principal Stresses Designate σ 1, σ 2, σ 3 implies no shear stresses or σ ij = σ ij δ ij Proof is clear from Mohr s circle representation (see text Fig.2.6) for 2-D Note : τ max σ1 σ3 = 2 Hydrostatic and Deviatoric Stresses (p. 46) Total stress tensor can be divided into two components (Fig. 2-18) Hydrostatic or mean stress tensor (σ m ) involving only pure tension or compression & Deviatoric stress tensor (σ ' ij ) representing pure shear with no normal components σkk σ m = 3 ' σ ij = σij + σ1 + σ2 + σ3 = δijσkk KL Murty page 6 MAT 450

7 Given the stress state: σ ij = Example on Hydrostatic and Deviatoric Stresses a. Find the hydrostatic part of the stresses , b. Find the deviatoric part of the stresses Ans. (a) σ hyd ij = σ m δ ij = 1 3 (σ 11+σ 22 +σ 33 ) δ ij where σ m = 1 3 ( ) = 30 so that σ = hyd ij (b) By definition, σ dev ij = σ ij - σ hyd ij = = Note that the mean hydrostatic stress for σ dev ij = (σ dev 11 +σdev 22 +σdev 33 ) = 0, as expected. KL Murty page 7 MAT 450

8 Principal Stresses (p. 22) Principal Planes are the planes on which maximum normal stresses act with no shear stresses and these stresses are the Principal Stresses Designate σ 1, σ 2, σ 3 implies no shear stresses or σ ij = σ ij δ ij Proof is clear from Mohr s circle representation (see text Fig.2.6) for 2-D For 3-D such an analogy is not useful and these are determined from the roots of σ of the determinant (cubic in σ) : σ 11 σ σ 12 σ 13 σ 21 σ 22 σ σ 23 σ 31 σ 32 σ 33 σ = 0 ; Expand the determinant 0 = σ 3 - (σ 11 + σ 22 + σ 33 ) σ 2 + (σ 11 σ 22 +σ 22 σ 33 +σ 33 σ 11 -σ σ σ 2 31 )σ - (σ 11 σ 22 σ 33 +2σ 12 σ 23 σ 31 -σ 11 σ σ 22 σ σ 33 σ 2 12 ) σ 3 I 1 σ 2 I 2 σ Ι 3 = 0 where I s are invariants of the stress tensor (Eqs. on p.28 of Text) : I 1 = (σ 11 + σ 22 + σ 33 ) I 2 = -(σ 11 σ 22 +σ 22 σ 33 +σ 33 σ 11 -σ σ σ 2 31 ) I 3 = σ 11 σ 22 σ 33 +2σ 12 σ 23 σ 31 -σ 11 σ σ 22 σ σ 33 σ 2 12 [text - x,y,z 1,2,3] (a) Uniaxial stress : (b) Biaxial stress : (c) Hydrostatic pressure : (d) Pure shear : σ σ σ p p p 0 σ 0 σ Special Stress States KL Murty page 8 MAT 450

9 Normal and Shear Stresses on a Given Plane [Cut-Surface Method] Given σ ij in reference system ; 3 n S ˆ n is the unit vector normal to the plane = n 1 n 2 n 3 ˆ m is the unit vector in the plane = m 1 m 2 m 3 σ ij m 2 σ N = normal stress along n τ = shear stress along m = Note : nˆ. mˆ = 0; n + n + n 1 and m m m 2 3 = 1 note : if n = 1, 2, 5 ˆ n = 1 30, 2 30, 5 30 ˆ n is a unit vector where = 30 so that n n n 2 3 = Find the stress vector (S ) {the stress vector : the vector force per unit area acting on the cut }: S = σ. ˆ n S 1 S 2 S 3 = σ 11 σ 12 σ 13 σ 21 σ 22 σ 23 σ 31 σ 32 σ 33 n 1 n 2 n 3 3 S i = σ ik n k ; i.e. S 1 =σ 11 n 1 +σ 12 n 2 +σ 13 n 3 ; etc. k=1 2. σ N and τ follow as : σ N = S. ˆ n = S 1 n 1 + S 2 n 2 + S 3 n 3 τ = S. ˆ m = S 1 m 1 + S 2 m 2 + S 3 m 3 and τ max occurs when n, S and m are in the same plane, or from Fig. S 2 =σ 2 Ν + τ2 max cut plane n S m KL Murty page 9 MAT 450

10 Normal and Shear Stresses on a Given Plane [Cut-Surface Method] EXAMPLE A stress state in a given reference frame is (MPa): σ ij = Assume that the stresses are independent of position (uniform stress state). A plane "cut" is made through the body such that the normal to the cut is n = 2, 2, -2. a. What is the normal stress σ N on the plane? b. What is the shear stress τ along the direction m = 0, 1, 1 in the plane? c. What is the maximum shear stress in the plane (consider all directions in the plane)? Answer : n ˆ = 2 10, 2 10, and ˆ m = 0, 1 2, S 1 =σ 11 n 1 + σ 12 n 2 + σ 13 n 3 = [8 ( 2 ) + 2 (2) + (-5) (-2)] = 8 MPa, 10 Similarly, S 2 = MPa and S 3 = MPa. 1 (a) σ N = S 1 n 1 +S 2 n 2 +S 3 n 3 = [8 ( 2 ) + (-3.53) (2) + (-4.13) (-2)] = 3.96 MPa 10 (b) τ = S. ˆ m = S 1 m 1 + S 2 m 2 + S 3 m 3 = 1 2 [8 (0) + (-3.53) (1) + (-4.13) (1)] = MPa. note: - sign means the shear stress acts in the - ˆ m direction (c) To find the maximum shear stress in the plane : Since τ = S. m ˆ, the maximum projection of S along m ˆ will occur when these 2 vectors are coplanar (containing n ˆ also) - see Fig. below: S 2 =S 2 1 +S S 2 3 = (8)2 + (-4.13)2 + (-5.42)2 = From the figure, note that S 2 =σ 2 Ν + τ2 max or τ max = S 2 - σ 2 Ν = (3.96) 2 = 8.82 MPa. cut plane n S m KL Murty page 10 MAT 450

11 Given σ ij = Example : Stress Tensor Transformations wrt x,y,z axes. New axes (x',y',z') are rotated 45 around z-axis. Need to find σ ' ij. σ' ij = a ik a jn σ kn -y z, z' 45 o -x 45 o y' y x x' a. Calculate τ ' xy. x y z x' 45 o 45 o 90 o θ ij : y' 135 o 45 o 90 o z' 90 o 90 o 0 o τ ' xy = a xx a yx σ xx + a xx a yy σ xy + a xx a yz σ xz = 12 (- 1 2 )(10) ( 1 2 )(5) a xy a yx σ yx + a xy a yy σ yy + a xy a yz σ yz (- 1 2 )(5) + 12 ( 1 2 )(20) a xz a yx σ zx + a xz a yy σ zy + a xz a yz σ zz = 5 MPa b. Show that σ ' x = 20 MPa and σ' y = 10 MPa. Thus σ ' ij = Note that σ x + σ y = σ ' x + σ' y (= 30 MPa) KL Murty page 11 MAT 450

12 Example 2 : Same as above but in 2-D for a general case x', y' rotation by θ x y θ ij : x' θ 90-θ y' 90+θ θ θ θ y' y a ij = cosθ sinθ y σ y sin θ cosθ x x' B τ yx τ xy A x σ x σ y' B A σ x' τ xy τ yx θ τ ' xy (or τ' xy ) = a xx a yx σ xx + a xx a yy τ xy + a xy a yx τ yx + a xy a yy σ yy = - sinθ cosθ σ xx + cos y θ τ xy - sin y θ τ xy + sinθ cosθ σ yy = σ yy - σ xx 2 sin2θ + τ xy cos2θ Eq. 2.7 Similarly find σ ' x = σ x + σ y 2 + σ x - σ y 2 cos2θ + τ xy sin2θ (Eq 2.5) and σ ' y = σ x + σ y 2 - σ x - σ y 2 cos2θ - τ xy sin2θ ( Eq 2.6) & Fig. 2.4 shows variation of these 3 stresses with θ. Note : σ x + σ y = σ ' x + σ' y as should be since I 1 is invariant i.e., sum of normal stresses on mutually perpendicular planes is invariant. same thing can be done using Mohr's circle representation (easier for 2-D case) If τ ' xy = 0, principal planes and stresses: tan2θ = 2τ xy σ x - σ y (Eq. 2.8) whereas maximum shear stresses when tan2θ = - σ x - σ y 2τ xy (Eq.2.10) & τ max given by Eq KL Murty page 12 MAT 450