Mechanical Properties - Stresses & Strains

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

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

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

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

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

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 = 3 1 3 δijσkk KL Murty page 6 MAT 450

Given the stress state: σ ij = Example on Hydrostatic and Deviatoric Stresses 80 20 50 a. Find the hydrostatic part of the stresses. 20 40 30, b. Find the deviatoric part of the stresses. 50 30 50 Ans. (a) σ hyd ij = σ m δ ij = 1 3 (σ 11+σ 22 +σ 33 ) δ ij where σ m = 1 3 (80-40 +50) = 30 so that σ = 0 0 30 30 0 0 hyd ij 0 30 0. (b) By definition, σ dev ij = σ ij - σ hyd ij = 80 20 50 20 40 30-50 30 50 30 0 0 0 30 0 = 0 0 30 50 20-50 20-70 30-50 30 20 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

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 2 12 -σ 2 23 -σ 2 31 )σ - (σ 11 σ 22 σ 33 +2σ 12 σ 23 σ 31 -σ 11 σ 2 23 -σ 22 σ 2 31 -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 2 12 -σ 2 23 -σ 2 31 ) I 3 = σ 11 σ 22 σ 33 +2σ 12 σ 23 σ 31 -σ 11 σ 2 23 -σ 22 σ 2 31 -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 0 0 0 0 0 0 0 0 σ 1 0 0 0 σ 2 0 0 0 0 p 0 0 0 p 0 0 0 p 0 σ 0 σ 0 0 0 0 0 Special Stress States KL Murty page 8 MAT 450

Normal and Shear Stresses on a Given Plane [Cut-Surface Method] Given σ ij in reference system 1 2 3 ; 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 1 2 2 2 1 2 3 = Note : nˆ. mˆ = 0; n + n + n 1 and m 2 1 + m 2 2 + m 2 3 = 1 note : if n = 1, 2, 5 ˆ n = 1 30, 2 30, 5 30 ˆ n is a unit vector 2 2 2 where 1 + 2 + 5 = 30 so that n 2 1 + n 2 2 + n 2 3 = 1. 1. 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

Normal and Shear Stresses on a Given Plane [Cut-Surface Method] EXAMPLE 8 2 5 A stress state in a given reference frame is (MPa): σ ij = 2 4 3 5 3 6 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, -2 10 and ˆ m = 0, 1 2, 1 2. 1 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 = -3.53 MPa and S 3 = -4.13 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)] = - 5.42 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 2 2 + S 2 3 = (8)2 + (-4.13)2 + (-5.42)2 = 93.5. From the figure, note that S 2 =σ 2 Ν + τ2 max or τ max = S 2 - σ 2 Ν = 93.5 - (3.96) 2 = 8.82 MPa. cut plane n S m KL Murty page 10 MAT 450

Given σ ij = Example : Stress Tensor Transformations 10 5 0 5 20 0 0 0 0 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 ( 1 2 )(5) + 0 + a xy a yx σ yx + a xy a yy σ yy + a xy a yz σ yz + 1 2 (- 1 2 )(5) + 12 ( 1 2 )(20) + 0 + a xz a yx σ zx + a xz a yy σ zy + a xz a yz σ zz + 0 + 0 + 0 = 5 MPa b. Show that σ ' x = 20 MPa and σ' y = 10 MPa. Thus σ ' ij = 20 5 0 5 10 0 0 0 0. Note that σ x + σ y = σ ' x + σ' y (= 30 MPa) KL Murty page 11 MAT 450

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. 2.11. KL Murty page 12 MAT 450