Draft:S Ghorai. 1 Introduction. 2 Invariance. Constitutive equations. The conservation laws introduced before are. Equations of mass conservation

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1 1 Constitutive equations 1 Introduction The conservation laws introduced before are Equations of mass conservation Conservation of linear momentum Conservation of angular momentum Dρ Dt + ρdiv v = 0 ρ Dv = ρb + div σ dt σ T = σ These equations are valid for all continuum bodies in motion i.e. they do not distinguish between fluids and solids. The total number of unknowns are 10 but we only have 4 equations. Therefore, further conditions, called constitutive laws have to be introduced in order to make a continuum problem in general solvable. Such constitutive laws relates stress tensor to various kinematical quantities. 2 Invariance We require a continuum to be invariant with regards to reference frame, or to have unchanged response when a superposed rigid motion is applied to all the material points of the body. As a result, restrictions are put on the constitutive relations. There are two methods to examine invariance. In material frame indifference, a continuum body s response to applied forces or prescribed motion must be the same as observed from two different reference frames. On the other hand in superposed rigid body motion to the body, the observer maintains the same reference frame. Here each material point has a superposed rigid motion added to it. In this note we consider the second approach. The superposed rigid motion associated with the configuration P + differes from a arbitrary motion x = x(x, t)

2 2 by only superposed rigid motion of the entire body. Thus the material point X of the body which is at x at time t in the configuration P moves to the location x + at time t + t + = t + a in the configuration P + where a is a constant. Thus we have x + = ˆx + (X, t + ) = x + (X, t) Similarly consider the material point Y of the body which at time t in the present configuration P is at y moves to location y + at time t +. Thus we write y + = x + (Y, t) Recalling the inverse relationship X = X(x, t), Y = X(y, t) we write the above expression as x + = x + (X(x, t), t) = x + (x, t) y + = x + (X(y, t), t) = x + (y, t) Since the superposed motion is rigid, the distance between two pairs of material points remains same. Thus y + x + = y x. This gives ( x + (y, t) x + (x, t)) ( x + (y, t) x + (x, t)) = (y x) (y x) or ( x + m(y, t) x + m(x, t))( x + m(y, t) x + m(x, t)) = (y m x m )(y m x m ) This relation is valid for all x, y in the configuration P occupied by the body at time t. Since x and y are independent, we differentiate the above equation consecutively with respect to x and y to obtain [ ] x + 2 m (x, t) ( x + x m(y, t) x + m(x, t)) p [ ] [ ] x + m (x, t) x + m (y, t) = 2(y p x p ) x p x q = δ pq Thus we have [ x + ] T [ (x, t) x + (y, t) = ] 1 x y

3 Thus each side must be tensor of function of time only. Let this tensor be Q T (t). Thus we have [ x + ] (x, t) = Q(t) x Since the above relation is independent of x we must also have [ x + ] (y, t) = Q(t) y 3 Thus we have Q T Q = I which shows that Q is an orthogonal tensor and det(q) = ±1. Since x + = x is a special case of superposed rigid body motion we have det(q) = 1 Thus Q(t) is a proper orthogonal tensor. Integrating we get x + = c(t) + Q(t)x By definition this equation represents superposed rigid body motion. This means the lengths of line elements are preserved and also the angles between two line elements: x + y + 2 = (x + y + ) (x + y + ) cos θ + = x+ y + x + y + = = Q(x y) Q(x y) = x y 2 Q(x y) x y x + z + x + z + Q(x z) x z = x y x y x z x z = cos θ Due to superposed rigid body motion, the area and volume are also preserved. This can be shown as follows. Let F + denote the deformation gradient from reference to superposed configuration. THus Now J + = dv+ dv F + = x+ X = QF = det(f + ) = det(qf ) = det(q)det(f ) = J

4 4 n + da + = J + (F + ) T NdA = JQF T NdA = Qnda From this we get (da + ) 2 = (n + da + ) (n + da + ) = (Qnda) (Qnda) = (da) 2 From the last two relations we get n + = Qn From we get QQ T = I QQ T = Q Q T = Q = Q Q T Q = ΩQ Thus Ω is a skew-symmetric tensor since Ω = Q Q T = Ω T If w is the axial vector of Ω then for arbitrary vector a we have Now we calculate the velocity v + as Ωa = w a v + = Dx+ = ċ + Dt Qx + Q Dx = ċ + ΩQx + Qv Dt which using the axial vector can be written as v + = ċ + Ω(x + c) + Qv = ċ + w (x + c) + Qv Let us now calculate the velocity gradient L + and the rate of deformation tensor D + and spin tensor W +. We have Hence L + = v+ x = Ω + Q v x + x x = Ω + + QLQT D + = QDQ T and W + = Ω + QW Q T Now let us derive the expression for density ρ +, stress vector t, stress tensor σ and body forces b +. It is important to emphasize that the balance laws remain form-invariant when expressed relative to P +. The conservation of mass gives ρ + J + = ρ 0

5 5 Conservation of linear and angular momentum respectively gives ρ + v + = ρ + b + + div + σ + and Using J + = J we get (σ + ) T = σ + ρ + = ρ i.e. density remains invariant under superposed rigid body motion. In the equation for linear momentum, we know the expression in the left hand side. To get expression for b + and σ + we make assumption that the normal component of the stress vector remains unchanged under superposed rigid body motion. Thus t + n + = t n Using n + = Qn or n = Q T n + we get [t + Qt(Q T n + )] n + = 0 We can not conclude the quantity inside third bracket is zero for arbitrary n + since the quantity is is function of n +. However using stress tensor we write the above as [σ + QσQ T ] : (n + n + ) = 0 Here we have used double contraction. Since this results hold for arbitrary n + and the coefficient is independent of n + we must have σ + = QσQ T Now we have t + = σ + n + = (QσQ T )(Qn) = Qσn = Qt This implies that t + t + = t t, σ + : σ + = σ : σ Thus the magnitude of stress vector and stress tensor remains unchanged by superposed rigid body motion. Also from the assumption of t + n + = t n

6 we conclude the angle between the stress vector and normal remains unchanged under superposed rigid body motion. To get an expression for b + we note that div + σ + = σ+ ij x + e j i = σ+ ij x k x k x + e j i 6 = Q ip σ pq,k Q jq x k x + i e j = Q ip σ pq,k Q jq Q rk δ ri e j = Q pi σ pq,k Q jq Q ik e j = δ pk Q jq σ pq,k e j = σ kq,k Q jq e j = Qdivσ Hence we get div + σ + = Qdiv σ Thus ρ + v + = ρ + b + + div + σ + becomes ρ v + = ρb + + Qdivσ From the last equation and ρ v = ρb + divσ we get b + = v + + Q(b v) It can be shown the first and second Piola-Kirchoff stress tensors T and S transform according to T + = QT and S + = S Constitutive equations for Viscous fluids: A general viscous fluid is characterized by the constitutive assumption that the Cauchy stress tensor is function of the density ρ, the velocity v and the velocity gradient L. For convenience we break L into D and W and write σ = σ(ρ, v, D, W )

7 7 We shall use invariance under superposed rigid body motion to develop restriction on the form of the above equation. Now we have σ + = σ(ρ +, v +, D +, W + ) We know that σ + = QσQ T where Q is proper orthogonal and function of time only. Thus we get σ(ρ +, v +, D +, W + ) = Q σ(ρ, v, D, W )Q T Using the relations for the other quantities we get σ(ρ, ċ + ΩQx + Qv, QDQ T, Ω + QW Q T ) = Q σ(ρ, v, D, W )Q T Since this relation holds for all superposed rigid motion, we obatin the restriction by considering some special superposed rigid body motion. First consider a rigid body motion in which c 0, Q = I and Ω = 0. Then we have σ(ρ, ċ + v, D, W ) = σ(ρ, v, D, W ) Since we can choose ċ arbitrarily and the right hand side is independent of c, it follows that σ can not depend on the velocity v. Thus σ must be expressed as another function of ρ, D and W only. σ = σ(ρ, D, W ) and due to invariance we must have σ(ρ, QDQ T, Ω + QW Q T ) = Q σ(ρ, D, W )Q T Agin since Ω is arbitrary and the right hand side is idependent of Ω and so the stress tensor is independent of spin tensor W. Thus the constitutive equations for general viscous fluid is given by σ = ˆσ(ρ, D) and this must satisfy ˆσ(ρ, QDQ T ) = Qˆσ(ρ, D)Q T

8 8 Reiner-Rivlin fluid: Since the above relation hold for all proper orthogonal tensor Q, the function ˆσ is called an isotropic tensor function of its argument D. Now using theory of invariants the general form of ˆσ may be expressed as σ = αi + βd + γd 2 where α, β and γ are scaler functions of ρ and the three invariants of D. Newtonian-viscous fluid: This is special case of Reiner-Rivlin fluid in which stress is a linear function of rate of deformation D. Thus σ = p 1 I + λtr(d)i + 2µD where p 1, λ and µ are scaler functions of ρ only. This may be written as where and σ = pi + 2µD p = 1 3 tr(σ) = p 1 D = D 1 3 tr(d)i (λ + 23 µ ) tr(d) This shows that pressure may depend on the rate of volume expansion tr(d). Inviscid fluid: Here stress is independent of the rate of deformation. Hence Now stress vector is σ = p(ρ)i t = σn = pn Thus for inviscid fluid, the stress vector alwaya acts normal to the surface on which it is applied. Theorem: T is an isotropic function of D if and only if T (D) = αi + βd + γd 2 (1) where α, β, γ are scaler functions of principal invariants of D Proof: For T (D) to be an isotropic function of D we must have T (QDQ T ) = QT (D)Q T We consider T and D to be symmetric. Let us denote T = QT Q T, D = QDQ T

9 9 (a) Sufficiency: Since Q is orthogonal, we can prove that the principal invariants of D are isotropic. This is due to the fact that tr( D) = tr(qdq T ) = tr(dq T Q) = tr(d) det( D) = det(qdq T ) = det(q)det(d)det(q T ) = det(d) and from ( D 2 ) = (QDQ T )(QDQ T ) = (QD 2 Q T ) we get tr( D 2 ) = tr(qd 2 Q T ) = tr(d 2 ) Hence α, β and γ are unchanged if D ij is replaced with D ij. Assume (1) holds, then QT (D)Q T = Q(αI + βd + γd 2 )Q T = αi + β D + γqdq T QDQ T = αi + β D + γ D 2 = T ( D) (b) Necessary: Since T is isotrpic we must have T (QDQ T ) = QT (D)Q T Choose the coordinate axes as the principal axes of D. Then in this coordinate we have D [D ij ] = 0 D D 3 and T ij = T ij (D 1, D 2, D 3 ) Choose Q as Then D ij = D ij and [ T ij ] = [Q ij ] = T 11 T 12 T 13 T 12 T 22 T 23 T 13 T 23 T 33

10 10 Using T ij = T ij we get T 12 = T 13 = 0. Similarly we can show that T 23 = 0. Thus both the D and T are diagonal matrix and and hence same principal axes. Thus we have T 11 = T 1 = F (D 1, D 2, D 3 ), T 22 = T 2 = F 2 (D 1, D 2, D 3 ), T 33 = T 3 = F 3 (D 1, D 2, D 3 ) Next choose Q as [Q ij ] = Now we have D T [ D ij ] = 0 D 3 0 [ T ij ] = 0 T D T 1 Thus we have F (D 2, D 3, D 1 ) = F 2 (D 1, D 2, D 3 ) F 2 (D 2, D 3, D 1 ) = F 3 (D 1, D 2, D 3 ) F 3 (D 2, D 3, D 1 ) = F (D 1, D 2, D 3 ) Hence T 1, T 2, T 3 can be expressed as a single function F (D 1, D 2, D 3 as T 1 = F (D 1, D 2, D 3 ), T 2 = F (D 2, D 3, D 1 ), T 3 = F (D 3, D 1, D 2 ) Finally choose Q as Then we have [ D ij ] = [Q ij ] = D D D [ T ij ] = T T T 3 Thus we have F (D 2, D 1, D 3 ) = F (D 2, D 3, D 1 ), F (D 2, D1, D 3 ) = F (D 1, D 2, D 3 )

11 11 Thus we have αi + βd 1 + γd1 2 = F (D 1, D 2, D 3 ) αi + βd 2 + γd2 2 = F (D 2, D 3, D 1 ) αi + βd 3 + γd3 2 = F (D 3, D 1, D 2 ) These give solutions of α, β, γ as functions of D 1, D 2, D 3. Also since F (D 1, D 2, D 3 ) has symmetry, the above solutions are unaltered if any pair of D 1, D 2 and D 3 are interchanged. Hence α, β and γ are symmetric functions of D 1, D 2, D 3. Hence α, β, γ can be expressed as functions of D 1 + D 2 + D 3 = I 1 D 1 D 2 + D 2 D 3 + D 3 D 1 = I 2 D 1 D 2 D 3 = I 3 Also we have T D T 2 0 = α β 0 D T D 3 + γ which together with the relations above proof the necessity. 3 Conservation of energy D D D 2 3 It is sometimes useful to consider the mechanical energy equation which is not derived from the equation of balance of energy, rather it is obtained from the equation of linear momentum using making use of the equation of mass conservation. Taking scaler product of the velocity vector and equation of the linear momentum we get 1 2 ρ D Dt (v iv i ) = ρb i v i + σ ij,j v i The integral form of the above equation is obtained by integrating the above equation over volume with surface as 1 D ρv v dv = ρb v dv + 2 Dt t v ds (σ : D) dv (2) where we have use mass conservation principle, divergence theorem, L = D + W and σ : W = 0 due to symmetric σ and skew-symmetric W. Let us introduce the

12 following notations K 1 ρv v dv = Kinetic energy in 2 R c t v ds = rate of work done by surface forces on the boundary ρb v dv = rate of work done by body forces in the material R b 12 The scalar σ : D is the rate of work by stresses per unit volume of the body and is called stress power. We can hypothesize the existence of a scalar called the specific internal energy, u = u(x, t) (internal energy per unit mass). The internal energy for the material body will be ρu dv Heat may enter the body through the surface by a heat flux vector q. In addition, heat may enter the body as specific heat supply, r = r(x, t): the heat generating within the body per unit mass. The balance of energy statement is a statement that the rate of increase of internal energy and kinetic energy of the body is equal to the rate of work by body forces, contact forces plus the energy entering the body per unit time due to heading. Thus D ( ρ u + v v ) dv = ρb v dv + Dt 2 t v ds + ρr dv q n ds Using mechanical energy equation (2) and divergence theorem we get D ρu dv = div q dv + ρr dv + (σ : D) dv Dt or using mass conservation principle this gives the energy balance equation as ρ Du Dt = div q + ρr + σ : D 4 Infinitesimal strain We consider the coordinate system concident. The displacement field u is given by u = x X which can be expressed using material coordinate as u(x, t) = x(x, t) X or u i (X, t) = x i (X, t) X i

13 13 and Eulerian coordinate as From thus we get u(x, t) = x X(x, t) or u A (x, t) = x A X A (x, t) F ia = x i,a = u i,a + δ ia F 1 Ai = X A,i = δ Ai u A,i Now and similarly 2E AB = F ia F ib δ AB = u i,b δ ia + u i,a δ ib + u i,a u i,b 2e ij = δ ij F 1 Ai F 1 Aj = u A,jδ Ai + u A,i δ Aj u A.i u A,j Now for infinitesimal displacement gradient approximation we have E AB = 1 2 (u i,bδ ia + u i,a δ ib ) 2e ij = 1 2 (u A,jδ Ai + u A,i δ Aj ) For infinitesimal dispalcement gradient we have u A = u A x i = u A,i (u i,b + δ ib ) = U A,i δ ib x B x i x B Thus u 1 = u 1 X 1 x 1 and it is immaterial whether we take derivatives of u with respect to x A or x i. Hence E AB = 1 2 (U A,iδ ib + U B,j δja) = 1 2 (U j,i + U i,j )δjaδib = e ij δ ia δ jb Thus the difference between two stain measures disappear and we introduce a single notation for strain tensor ε where ε ij = 1 ( ui + u ) j = 1 2 X j X i 2 ( ui + u ) j = 1 x j x i 2 (u i,j + u j,i ) Also from J = 1 6 ɛ ijkɛabcx i,a x j,b x k,c = 1 6 ɛ ijkɛabc(δ ia + u i,a )(δ jb + u j,b )(δ kc + u k,c ) 1 + u k,k

14 5 Elasticity, Hooke s Law and Strain energy 14 A material is said to be elastic if it posseses a homogeneous stress free state and if in a finite neighbourhood of this state there exits a one to one correspondence between the stress tensor and the strain tensor. A material is said to be hyperelastic if hyperelastic if it posseses a strain energy function per unit volume W such that W is a function of the current deformation only and is not a function of deformation rate or history of deformation and σ ij = W ε ij Now we simplify this expression using linearized theory. We have using Taylor series exoansion Thus W = W(0) + W(0) ε ij W(0) ε ij ε kl + ε ij 2 ε ij ε kl σ ij = W(0) W(0) ε kl + ε ij 2 ε ij ε kl It is standard practice to assume that the residual stress is zero in the unstrained stress i.e. σ ij = 0 when ε ij = 0. Thus by retaining only the linear term we obtain the constitutive relations for the linear elastic material as Now due to we have σ ij = 1 2 W(0) ε kl = C ijkl ε kl or σ = Cε 2 ε ij ε kl 2 W(0) = 2 W(0) ε ij ε kl ε kl ε ij C ijkl = C klij Also due to symmetry of σ and ε we have C ijkl = C jikl = C ijlk The coefficient C ijkl has 3 4 = 81 coefficients. However due to the above symmetry the number of distinct components of is 21. The tensor character of C can be shown by considering the transformation laws in orthogonal coordinate. Let us consider σ ij = C ijpm ε pm

15 15 But Thus comparing the above two we get σ ij = a iq a js σ qs = a iq a js C qskl ε kl = a iq a js C qskl a pk a ml ε pm C ijpm = a iq a js a pk a ml C qskl The relation σ ij = C ijkl ε kl or σ = Cε is called generalized Hooke s law. Here we assume that adiabatic and isothermal condition. Thus the coefficients at the most depend on position only. 6 Hooke s law for isotropic media Using the most general form of isotropic fourth order tensor we write C as C ijkl = λδ ij δ km + µ(δ ik δ jm + δ im δ jk ) + β(δ ik δ jm δ im δ jk ) Interchanging i and j and using symmetry we get β = 0. Thus which reduces to σ ij = [λδ ij δ km + µ(δ ik δ jm + δ im δ jk )] ε kl σ ij = λδ ij ε kk + 2µε ij The relation between stress and strain may also be written as Let σ and ε be the devetoric stress i.e. ε ij = 1 + ν E σ ij ν E δ ijσ kk σ = σ 1 3 tr(σ)i ε = ε 1 3 tr(ε)i Using this we can write stress-strain relation for linear isotropic solid as σ ii = 3Kε ii σ ij = 2µε ij

16 Thus we can five elstic parameters: λ, Lame s constant; µ shear modulas; E, Young s modulas; K, the bulk modulas and ν, Poisson s relation. These constant are related. For examle E = µ(3λ + 2µ), K = λ + 2 λ + µ 3 µ, ν = λ 2(λ + µ) 7 Field Equations for linear isotropic problem 16 Problems in elasticity can be divided into two categories: elastostatic and elastodynamics. The difference is that in elastodynamics problem the term ρ v is not zero in the linear momentum equations. And boundary conditions require more specification. Due to infinitesimal deformation ρ 0 = Jρ ρ. Hence ρ is known. 7.1 Elastostatic problem a. Equilibrium equation (3 equations) σ ij,j + ρb i = 0 b. Strain displacement relations (6 equations) c. Hooke s law (6 equations) or 2ε ij = u i,j + u j,i σ ij = λδ ij ε kk + 2µε ij ε ij = 1 ( ) λ σ ij 2µ 3λ + 2µ δ ijσ kk d. Sometimes the compatiability equations (6 equations) ε ij,kl + ε kl,ij ε ik,jl ε jl,ik = 0 This is actually 81 equations but due to redundancy, the 81 equations reduces to 6 equations. These equations are obtained from by taking (i, j, kl) as (1, 1, 2, 2), (1, 1, 2, 3), (2, 2, 3, 3), (2, 2, 1, 3), (3, 3, 1, 1), (3, 3, 1, 2)

17 17 The equations are 2 ε 11 x ε 22 x ε 33 x ε 22 x ε 33 x ε 11 x 2 3 = 2 2 ε 12 x 1 x 2, = 2 2 ε 23 x 2 x 3, = 2 2 ε 13 x 1 x 3, 2 ε 11 x 2 x 3 = 2 ε 22 x 1 x 3 = 2 ε 33 x 1 x 2 = ( ε 23 + ε 13 + ε ) 12 x 1 x 1 x 2 x 3 ( ε23 ε 13 + ε ) 12 x 2 x 1 x 2 x 3 ( ε23 + ε 13 ε ) 12 x 3 x 1 x 2 x 3 Thus there are 15 equations not counting compatiability. The total number of unknowns are also 15: 6 for stress, 6 for strain and 3 for displacement. Body forces are assumed to be known. The conditions to be specifide on the boundary may be of the three form Displacement specified on the boundary Tractions (stress vector) specified everywhere A mix of traction and displacement on the boundary The field equations are linear in the unknown. Thus if σij, 1 ε 1 ij and u 1 i are solutions with body force b 1 i and surface tractions t 1 i and σij, 2 ε 2 ij and u 2 i are solutions with body force b 2 i and surface tractions t 2 i then σ ij = σij 1 + σij, 2 ε ij = ε 1 ij + ε 2 ij, quad u i = u 1 i + u 2 i offers solution for the situation where b i = b 1 i + b 2 i and t i = t 1 i + t 2 i. This is principle of superposition. For problems where boundary conditions are given in terms of displacement, it is convenient to eliminate stress and strain unknowns from the field equations and so as to state the problem solely in terms of the displacement components. The equations are µu i,jj + (λ + µ)u j,ji + ρb i = 0 which are called Navier s equations. Thus solution from this are substituted to get strain and then we obtain stress. When the boundary conditions are given in terms of traction vector, we eliminate displacement and starin components from the field equations. To do this we

18 combined Hooke s law for stain interms of stress, the compatiability equations and the equilibrium equations to arrive at σ ij,kk ν σ kk,ij + ρ(b i,j + b j,i ) + ν 1 ν δ ijρb k,k = 0 which are known as Beltrami-Michell stress equations of compatibility. Combined with the equilibrium equations σ ij,j + ρb i = 0 18 the Beltrami-Michell equations make the statement of the problem in terms of stress. We derive the Beltrami-Michell stress equations of compatibility as follows. Using compatiability equations for starin we get (1 + ν)(σ ij,kl + σ kl,ij σ ik,jl σ jl,ik ) = ν(δ ij σ pp,kl + δ kl σ pp,ij δ ik σ pp,jl δ jl σ pp,ik ) Now there are 3 4 = 81 equations in total. Out of these only 6 are independent. They are obtained using the following steps: first contract k and l. This gives (1 + ν)(σ ij,kk + σ pp,ij σ ik,jk σ jk,ik ) = ν(δ ij σ pp,kk + σ pp,ij ) Using linear momentum equations this gives σ ij,kk + σ pp,ij + ρ(b i,j + b j,i ) = Now contracting i and j we get Thus the equations becomes σ pp,kk = 1 + ν 1 ν ρb i,i ν 1 + ν (δ ijσ pp,kk + σ pp,ij ) σ ij,kk ν σ pp,ij + ρ(b i,j + b j,i ) + ν 1 ν δ ijρb k,k = Elastodynamic problem Here the equilibrium equations are replaced by σ ij,j + ρb i = ρü i Here all the field quantities are functions of time as well as coordinates. Thus displacement field is u i = u i (x, t). The solution satisfy not only the boundary conditions which may be function of time as in u i = u i (x, t) on boundary, ( denotes prescribed value)

19 19 or t (n) i = t (n) i (x, t) on boundary but also initial conditions, which usually are taken as u i = u i (x, 0) and u i = u i (x, 0) The governing equations for displacement in elastodynamics problem is µu i,jj + (λ + µ)u j,ji + ρb i = ρü i which are also called Navier s equations. 8 Plane elasticity In a number of engineering application, the geometry and loading pattern lead to a essentially two dimensional form of the equations of elasticity. In plane stress problem the geometry of the body is that of a thin plate with one dimension very much snaller than the other two. The loading in this case is in the plane of the plate and is assumed to be uniform across the plate thickness. Plane strain problems involves bodies having one dimension very much larger than the other two and the loads act perpendicular to and distributed uniformly with respect to this large dimension. 8.1 Plane stress For the plane stress problem, let the x 3 axis is perpendicular to the thin plate. In this case σ 13 = σ 23 = σ 33 = 0. The equilibrium equations reduces to σ 11,1 + σ 12,2 + ρb 1 = 0 σ 12,1 + σ 22,2 + ρb 2 = 0 The strain-displacement relation becomes ε 11 = u 1,1, ε 22 = u 2,2, 2ε 12 = u 1,2 + u 2,1 The compatiability equations for strain are given by ε 11,22 + ε 22,11 = 2ε 12,12

20 20 Hooke s law can be written as σ 11 = λ(ε 11 + ε 22 + ε 33 ) + 2µε 11, Eε 11 = σ 11 νσ 22 σ 22 = λ(ε 11 + ε 22 + ε 33 ) + 2µε 22, Eε 22 = σ 22 νσ 11 0 = λ(ε 11 + ε 22 + ε 33 ) + 2µε 33, Eε 33 = ν(σ 11 + σ 22 ) σ 12 = E 1 + ν ε 12, Eε 12 = (1 + ν)σ 12 We may eliminate ε 33 from the equations to get equations in terms of ε 11, ε 22 or σ 11, σ 22. Note that ε zz is not in general zero and that u 1, u 2, u 3 may be functions x 3 also. Thus plane stress is not truly a two dimensional problem. In terms of the displacement components u i (i = 1, 2) the Navier s equation for elastostatic problem becomes 8.2 Plane strain E 2(1 + ν) u E i,jj + 2(1 ν) u j,ji + ρb i = 0, (i, j = 1, 2) For the plane strain situation we assume that u 3 = 0 and the remaining two displacemnt components are functions of x 1, x 2 only: u i = u i (x 1, x 2 ) (i = 1, 2) This leads to the condition of plane strain since ε 13 = ε 23 = ε 33 = 0 The equilibrium equations, strain-displacement relations and compatiability equations remain the same as the plane stress problem. The Hooke s law may be written as along with σ 11 = σ 22 = σ 12 = σ 33 = E (1 + ν)(1 2ν) [(1 ν)ε 11 + νε 22 ] E (1 + ν)(1 2ν) [νε 11 + (1 ν)ε 22 ] E 1 + ν ε 12 Eν (1 + ν)(1 2ν) (ε 11 + ε 22 )

21 We can invert these equation to get strain in terms of stress also. In general σ 33 is not zero but is a function of x, y only. Therefore some end forces and couples are require to maintain state of plane strain. For elastostatic problem the appropriate Navier equation can be written as E 2(1 + ν) u E i,jj + 2(1 + ν)(1 2ν) u j,ji + ρb i = 0 (i, j = 1, 2) 21 9 Airy Stress function If we assume that the stress components are functions of x 1, x 2 only for the plane stress. Further we also assume that the variation of displacement components with x 3 is negligible. In this case the compatiability equation reduces to ε 11,22 + ε 22,11 = 2ε 12,12 For the plane strain problem the compatiability equation remains the same. Now for the plane stress and plane strain problem the equilibrium equations are given by σ 11,1 + σ 12,2 + ρb 1 = 0 σ 12,1 + σ 22,2 + ρb 2 = 0 Let ρb i = V i then the above equation reduces to (σ 11 + V ),1 + σ 12,2 = 0 σ 12,1 + (σ 22 + V ),2 = 0 Let us choose a function φ(x 1, x 2 ) so that σ 11 = φ,22 V, σ 22 = φ,11 V, σ 12 = φ,12 Then the equilibrium equations are satisfied automatically. Now σ 33 = βν(σ 11 + σ 22 ) with β = 0 for plane stress and β = 1 for plane strain. Hence for Hooke s law ε ij = 1 + ν E σ ij ν E σ kkδ ij

22 22 we get ε 11 = 1 + ν E σ 11 ν E (1 + βν)(σ 11 + σ 22 ) ε 22 = 1 + ν E σ 22 ν E (1 + βν)(σ 11 + σ 22 ) ε 12 = 1 + ν E σ 12 Using these values in the strain compatiability equation we get ( ) 1 + ν 2 σ σ 22 νe ( 2 E x 2 2 x (1 + βν) x 2 2 x 2 1 ) (σ 11 + σ 22 ) ν E Substituting the value of σ ij in terms of φ and simplifying we get 4 φ φ + 4 φ = 1 ( ) βν2 2 V + 2 V x 4 1 x 2 1 x 2 2 x ν 2βν 2 x 2 1 x 2 2 Thus we have where C(ν) = 4 φ = C(ν)(V,11 + V,22 ) 1 ν 1 2ν 1 1 ν Plane strain Plane stress 2 σ 12 x 1 x 2 = 0