MAE 323: Chapter 4. Plane Stress and Plane Strain. The Stress Equilibrium Equation

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The Stress Equilibrium Equation As we mentioned in Chapter 2, using the Galerkin formulation and a choice of shape functions, we can derive a discretized form of most differential equations. In Structural Mechanics, the governing differential equation will most often be the stress equilibrium equation. We will therefore derive the two dimensional version of it here (see course handouts for the three dimensional version) The version we will show is quite standard and was first offered b Cauch in 1823 b considering the pressure that eists on a plane of arbitrar orientation within a differential element of material (a continuum). In this paper, Cauch first introduced the notion of a stress tensor.* *Cauch was actuall not the first person to derive an equilibrium epression for the elastic continuum. This honor usuall goes to Navier (1821), but his method was based on a molecular theor of isotropic solids which wasn t quite correct. For a fascinating account, see: S. P. Timoshenko, Histor of Strength of Materials. Dover, 1983. 1

The Stress Equilibrium Equation The stress tensor and surface traction 2 T 2S2 T n 1 n n S 1 T 1 F T T L The static equilibrium on a infinitesimal wedge ma be found b considering the pressure on an arbitrar plane The eternal pressure ma be thought of as the component of traction normal to the plane The traction is defined b: T = lim A 0 ( F) A Where A is the surface area of the plane*, Fis an arbitrar eternal force acting normal to the plane *in our case, this is just L 1= L 2

The Stress Equilibrium Equation The stress tensor and surface traction 2 T 2S2 T S n n F T n T L In the most general sense, an arbitrar continuousl varing force, Fon a surface ma be split into a component normal (the pressure) and tangential to the surface In Cauch s analsis, this is done on all three sides of the wedge: the inclined side and the two sides of the wedge parallel to the and -aes, which for the time being are numbered 1 and 2 1 S 1 T 1 In what follows, we will assume that the force, Facts normal to the surface (simpl for clarit), so that Sis zero 3

The Stress Equilibrium Equation The stress tensor and surface traction 2 T 2S2 T n 1 n S 1 n F T T L F T L T L cosθ S Lsinθ = 0 2 1 T = T cosθ + S F 2 1 sinθ T L T Lsinθ S L cosθ = 0 1 2 T = T sinθ + S 1 2 cosθ T 1 4

The Stress Equilibrium Equation The stress tensor and surface traction Now, sum moments about point P T M P 2 T 2S2 2 1 P 1 S 1 S + S = 0 S = S 2 1 So, we see that the tangential pressures on the two sides parallel to the and aes are alwas equal T 1 We ll use this fact to swap the S 1 and S 2 terms in the previous summation 5

The Stress Equilibrium Equation The stress tensor and surface traction Now, we know that 2 T 2S2 T n n n F T T L n n = cosθ = sinθ So the traction equilibria ma epressed as (after swapping the S 2 and S 1 terms): T = T n + S n 2 2 1 S 1 T = T n + S n 1 1 Or, as a matri equation*: T 1 T T = T S S 2 2 T 1 1 n n (1) 6

The Stress Equilibrium Equation The stress tensor and surface traction The matri of normal and tangential pressures is known as the Cauch or infinitesimal stress tensor. It s two-dimensional form is shown below. It gets this name because it is onl valid for small increments of stress and corresponding strain T S 2 2 S1 T = 1 As we saw in slide 5: = So, the matri is also smmetric The preceding derivation could have been done in three dimensions following the same procedure, but on a plane cut through a 3D infinitesimal cube (a tetrahedron) 7

The Stress Equilibrium Equation The stress tensor and surface traction A diagram like the one below could be used to follow the same procedure as was done in 2D, but in all three dimensions, ielding the full 3 3 Cauch stress tensor T T 2 T z 3 T 1 z z = z z z zz z z = = = z z 8

The Stress Equilibrium Equation Net,we d like to derive the stress equilibrium differential equation itself, which forms the basis for most structural problems. Again, we ll work just in two dimensions to keep things simple. This time, we ll use an infinitesimal element with all four sides as shown Here, b is a bod force per unit volume b + + b + + 9

The Stress Equilibrium Equation Appling Newton s Third Law in the -direction gives: F = + + + b = 0 + + Rearranging and dividing b ields: + + + + b = 0 Taking the limit as 0, 0: + + b = 0 b + + b + + 10

The Stress Equilibrium Equation Similarl, repeating the previous three steps in the -direction ields: + + b = 0 And, once again, even though we won t go thru the steps, we will simpl point out that the full three dimensional equations can be obtained in a similar manner, considering a three-dimensional cube element instead of a square. Without showing the derivation, the equations are: z + + + b = 0 z z + + + b = 0 z z z zz + + + bz = 0 z 11

The Stress Equilibrium Equation Regardless of spatial dimension, the stress equilibrium can be written compactl in indicial notation as: ij, j bi 0 + = Where, as usual, the Einstein summation convention implies summation over repeated indices and the comma in the subscript means derivitativewith respect to the coordinate variable implied b the inde following the comma 12

Small Strains The stress equilibrium equations derived previousl give us onl half the picture. To full solve for the response of a bounded continuum (a structure), we need a wa to relate stresses and forces to strains and displacements, respectivel. So, before we look at that, we will introduce the common notion of small, or infinitesimal strains From the work we ve alread done, we know that in a structural problem, we relate forces to displacements through Hooke s Law (remember it kept coming back again and again). Further, we saw that in reduced continuum solutions, we had solutions over elements (in two dimensions) of the form: u = f (, ) = N (, ) u i= 1 v = g(, ) = N (, ) v n n i= 1 i i i i 13

Small Strains Just as the displacement field is the natural counterpart to the eternal forces in a structure, the strain field is the natural counterpart to the stress field in a structure. So, in two spatial dimensions, we can epect four strain components (three independent ones due to the smmetr in slide 7) We will designate the strain tensor as. And, in three dimensions: = z = z z z zz 14

Small Strains Although it s a little difficult to conve graphicall, an infinitesimal element undergoing nonzero,, and z represents a small perturbation from an equilibrium configuration (the dashed square), such that the actual lengths, and don t change. With this in mind, the strain components are defined as: u u u +,, = lim = 0 u, v, + u, v +, + v v v, +, = lim = 0 θ v v v = +,, lim = tanθ 0 θ u, v, θ u, v +, 15

Small Strains And finall: u u u =, +, lim = tanθ 0 θ The engineering shear strain is given b: γ u v = + u, v, + u, v +, + The tensor shear strain is ½ the engineering shear strain: 1 u v = 2 + θ u, v, θ u, v +, 16

Small Strains Thus, it should be clear that γ = γ and = Note that the engineering strain represents the TOTAL shear angle, whereas the tensor strain represents an average Once again, without going thru the details, we ll point out that in three dimensions, there are three additional unique strain components (letting w stand for displacement in the z-direction) zz z z v v w z z,, z+ z,, z = lim = 0 1 u w 1 = + = γ 2 z 2 1 v w 1 = γ 2 + = z 2 z z 17

Generalized Hooke s Law As we mentioned earlier, in order to full solve the response of a bounded continuum to an eternal load, we need a constitutive relation between the stress and strain components. This is understood as a propert of the continuum (different materials will respond differentl over the same bounded continuum under a given loading regime) So, we re looking for something of the form =C, where Cplas the role of K in Hooke s law Unfortunatel, a thorough discussion of this mathematical object is beond the scope of this course*. However, we ll hit the introductor highlights. We ll begin b noting that in its most general form, Cis a fourth rank tensor, and in indicial notation, the relation between stress and strain is given b: ij = Cijkl kl (2a) *We suggest the interested student begin b looking up Hooke s Law in Wikipedia, for a prett good introduction 18

Generalized Hooke s Law In three spatial dimensions, the tensor Chas 81 (3 3 3 3) components! However, if we consider the stress and strain tensors to be smmetric (as we do for small linear small strain problems), the corresponding entries in Cwill also be smmetric. This double smmetr allows us to reduce the rank of Cb two and the rank of and b 1. This rank-reduction is ver convenient for doing calculations and is known as Voigt notation*. Thus, (2) ma be written as: = C i ij j (2b) Note that with this rank reduction, the 2 nd rank stress and strain tensors are replaced with the equivalent stress and strain vectors: zz = z z zz = γ γ z γ z *see http://en.wikipedia.org/wiki/voigt_notation 19

Hooke s Law for isotropic materials This is done to compensate for the loss of a factor of two in the shear terms. Note in particular that the tensor shear strain components are replaced b engineering shear strains! when epressing the internal strain energ densit in the equivalent vector form: V 1 = ij ji 2 1 T V = 2 So, we see that in three spatial dimensions, C has 36 (6 6) terms (21 independent due to smmetr). If we further require that the continuum is isotropic, then we re left with onl two independent variables! (3) 1 ν ν ν 0 0 0 ν 1 ν ν 0 0 0 zz E ν ν 1 ν 0 0 0 zz = (1 + ν )(1 2 ν ) 0 0 0 (1 2 ν ) / 2 0 0 γ z 0 0 0 0 (1 2 ν ) / 2 0 γ z zz 0 0 0 0 0 (1 2 ν ) / 2 γ z (4) 20

Hooke s Law for isotropic materials So, equation (4) gives us the constitutive relation for an isotropic material in three dimensions, such that C in equation (2b) is: 1 ν ν ν 0 0 0 ν 1 ν ν 0 0 0 E ν ν 1 ν 0 0 0 C = (1 + ν )(1 2 ν ) 0 0 0 (1 2 ν ) / 2 0 0 0 0 0 0 (1 2 ν ) / 2 0 0 0 0 0 0 (1 2 ν ) / 2 If we had some other tpe of material (perhaps orthotropic, or transversel isotropic for eample), we would simpl replace Cwith the appropriate constitutive relation. An other material will have more non-zero terms and more independent variables. The minimum number of independent (scalar) material properties for a continuum in at least two spatial dimensions is two. We have given equation (4) in a particular form, where Eis the Elastic Modulus and ν is Poisson s Ratio 21

The Elastic Modulus and Poisson s Ratio The elastic modulus, E(also known as Young s Modulus) measures the material s tensile strain under a uniform tensile stress., while Poisson s Ratio gives a measure of how much the material contracts in a direction transverse to the uniform tensile stress (this is known as Poisson s Effect). This ma be seen b taking the inverse of equation (4): 1 ν ν 0 0 0 ν 1 ν 0 0 0 zz 1 ν ν 1 0 0 0 zz = γ E 0 0 0 2(1 + ν ) 0 0 γ z 0 0 0 0 2(1 + ν ) 0 z γ zz 0 0 0 0 0 2(1 + ν ) z Now, appl a uniaial stress, = { } 0 0 0 0 0 T 1 ν ν 0 0 0 ν 1 ν 0 0 0 0 zz 1 ν ν 1 0 0 0 0 = γ E 0 0 0 2(1 + ν ) 0 0 0 γ z 0 0 0 0 2(1 + ν ) 0 0 γ zz 0 0 0 0 0 2(1 + ν ) 0 22 (5)

The Elastic Modulus and Poisson s Ratio This results in the equations: zz 1 = E ν = E ν = E (6a) (6b) (6c) The fundamental linear relationship between stress and strain is embodied in (6a), and we ma in fact definethe elastic modulus Eunder a uniaial stress, from (6a) as: E = (7) Furthermore, (6b) and (6c) state that there are two transverse components of compressive strain (the sign of the strain is opposite that of the stress) resulting from the uniaial stress field. And, if we substitute =E into (6b) and (6c) to solve for ν, we obtain: zz ν = = (8) 23

The Elastic Modulus and Poisson s Ratio In other words, Poisson s ratio ma be thought of as a ratio of transverse to aial strain (or contraction) under a uniaial stress as shown in the figure below. It is worth repeating: This a material propert. Different materials will ehibit different ratios of contraction under a given tensile load (the values ma even be negative or zero, but the must never eceed 0.5, in which case the material is perfectl incompressible) : zz z 24

Plane Stress When analzing bounded continua in two spatial dimensions, one can make one of two assumptions. The first one is that all stress components in the third dimension (let s call it z) are zero. In other words: We plug this assumption into (4) : zz = z = z = 0 1 ν ν 0 0 0 ν 1 ν 0 0 0 zz 1 ν ν 1 0 0 0 0 = γ E 0 0 0 2(1 + ν ) 0 0 γ z 0 0 0 0 2(1 + ν ) 0 0 γ zz 0 0 0 0 0 2(1 + ν ) 0 These components are zero to ield: ν 1 = ν + E γ 2(1 ν ) + and: ν E ( ) (9) zz = + (10) 25

Plane Stress Note that separation of the z-component strain reaction (due purel to Poisson s Effect) in (10) is integral to the plane stress assumption. We omit it from the constitutive relation and appl it as a bookkeeping eercise after calculating the in-plane reactions Re-writing (10) as a matri equation: 1 ν 0 1 ν 1 0 = E γ 0 0 2(1 ν ) + (11) And, after inversion (using Mathematica ): 1 ν 0 E ν 1 0 = 2 1 ν 0 0 (1 ν ) / 2 γ (12) And remember our bookkeeping: ν = + E ( ) zz 26

Plane Stress Note that inverting a truncated version of (5) did NOT ield a corresponding truncated version of (4)! This is ver important and distinguishes plane stress from plane strain from a purel mathematical standpoint. The assumptions of plane stress are an approimation. In real structures, the outof-plane (z) strain would in fact be accompanied b a corresponding out-of-plane stress as required b (4). However, the approimation is good under certain circumstances which we will discuss later 27

Plane Strain The other 2D assumption we could make is that all the strain components in the z-direction are zero: zz = z = z = This is a different kind of assumption because it can be eplicitl enforced(it ma be stated as a boundar condition on the LHS of the governing differential equation), whereas the plane stress assumption arises from an observed approimation due to the absence of loads, gradients, and boundar conditions in the z-direction. Simpl plug the above assumptions into (4): 1 ν ν ν 0 0 0 ν 1 ν ν 0 0 0 zz E ν ν 1 ν 0 0 0 0 = (1 + ν )(1 2 ν ) 0 0 0 (1 2 ν ) / 2 0 0 γ z 0 0 0 0 (1 2 ν ) / 2 0 0 zz 0 0 0 0 0 (1 2 ν ) / 2 0 0 28

Plane Strain This ields: And: 1 ν ν 0 1 ν 1 ν 0 = (1 ν )(1 2 ν ) + 0 0 (1 2 ν ) / 2 γ = Eν zz ( ) (1 + ν )(1 2 ν ) + (13) Where we again remove the out-of-plane reaction purel due to Poisson s Effect and calculate it separatel. As we mentioned before, this condition can be eplicitl enforced in a structure. Specificall, If we fi the z-displacements at the etreme z-locations (the ends of the structure) and ensure no load gradients in the z-direction, then the plane strain assumption should hold at locations far from the ends. 29

Plane Strain (aismmetric version) The planar aismmetricformulation is another planar stress/strain formulation, but we do not include it as a separate categor because it is actuall equivalent to the plane strain formulation in clindrical coordinates (all properties of plane strain hold equall true of aismmetricelements). Switching to a clindrical reference frame does introduce some interesting differences, however. all that is required to convert the plane strain formulation to an aismmetric one is replace the - coordinate with the clindrical z, the -coordinate with r, and z coordinate with θ. This produces a 4 4 constitutive matri and 4 1 stress/strain vector: z Eν = θ θθ ( rr zz ) (1 + ν )(1 2 ν ) r + 30

Plane Strain (aismmetric version) r 1 ν ν ν 0 r θ E ν 1 ν ν 0 θ = ( 1 )( 1 2 ) z + ν ν ν ν 1 ν 0 z rz 0 0 0 1/ 2 ν γ rz z Eν = θ θθ ( rr zz ) (1 + ν )(1 2 ν ) r + 31