4 Stress and strain. 4.1 Force. Chapter. Force components z. Force z. F z. F x x

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1 Chapter 4 In this chapter we introduce the concepts of stress and strain which are crucial for the understanding of glacier flow. Any calculation of deformation rates and flow velocities involves stresses and a flow law which relates them to the strain rates. It is also very useful to take now a look at Appendi C Vectors and Tensors. The notation introduced there will be used below. 4.1 Force There are two different kinds of forces: body forces and surface forces. Body forces act on each volume of mass, independent on the surrounding material. The gravity force is the body force that causes glaciers to flow. It eerts on each volume of ice a force that is proportional to the mass within that volume. Other eamples of body forces are inertia forces such as the centrifugal force. Surface forces arise from the action of one body on another across the surface of contact between them. A typical eample is the force eerted from the glacier to its base, and vice versa. Moreover, across any internal surface of arbitrary orientation that divides a block of material into two, one side of the block applies a surface force on the other side. Body forces Surface forces A force F (typeset in bold face) is a vector quantity and may be divided into its components along perpendicular directions. Vectors and tensors are eplained in Appendi B. Force z Force components z F F z F 33

2 Chapter Stress on A force that acts on a surface is called a stress (or pressure when it is compressive). The intensity of the force depends on the area of the surface over which the force is distributed. It is called a traction and is commonly represented in terms of its components perpendicular and parallel to a surface. Traction Traction components e stress In order to satisfy the requirement of mechanical equilibrium, any surface must have a pair of equal and opposite tractions acting on opposite sides of the surface. This pair of tractions defines the surface stress vector Σ which is defined by the total force eerted on the surface divided by the surface area Σ = F A. Stress and traction are measured in units of force per unit area [ ] F = N A m = Pa, and the derived units Pa = 1 bar = 0.1 MPa. Surface stress Surface stress components Σ (top) σ (top) s σ (top) n Σ (bot) σ (bot) n σ (bot) s l stress stress It is convenient to resolve the surface stress into components, one perpendicular to the surface and two others parallel to the surface at right angles. These components are called normal stress and shear stress and are denoted by σ n and σ s1 resp. σ s2. The condition of mechanical equilibrium implies F (top) + F (bot) = 0, and therefore F (top) A + F(bot) A = 0, Σ (top) + Σ (bot) = 0. (4.1) 34

3 Physics of Glaciers I HS 2013 Equation (4.1) asserts that the tractions on top and bottom of the surface are equal and opposite. The same must be true for the components of the surface stress σ (top) n = σ n (bot) and σ s (top) = σ s (bot). (4.2) A pair of normal stresses that point towards each other is called a compressive stress (negative sign), a pair pointing away from each other is called a etensive stress (positive sign). Stress equilibrium For ease of presentation we consider two dimensions only (analogous relations hold in three dimensions). The stresses acting on a small volume of material eert forces and moments that must balance for a mechanical equilibrium. σ (top) zz Σ (top) z σ (top) z Σ (rt) σ (lft) σ (lft) z dz σ (rt) Σ (lft) d σ (bot) z σ (rt) z Σ (bot) z σ (bot) zz From the balance of normal and shear tractions (Eq. 4.2) we obtain the relations σ (rt) σ z (rt) = σ (lft) = σ (lft) z σ zz (top) σ z (top) = σ (bot) zz = σ (bot) z. (4.3) We also require that all moments with respect to the body center are balanced (otherwise the body would rotate). This involves only the shear components, since the moments of all normal components are zero. Denoting the surface areas A and A z, and taking the moment anti-clockwise, we obtain σ (top) z A z dz + σ (bot) z A z dz σ z (lft) A d σ z (rt) A d =! 0. (4.4) 35

4 Chapter 4 Using A = A z, d = dz and Equation (4.3) we obtain 2σ z 2σ z! = 0, which is equivalent to σ z = σ z. The stress state induced by [Σ, Σ z ] (four numbers) is therefore fully described by the three components σ, σ z = σ z, and σ zz. (4.5) An analogous relation holds in three dimensions where the stress tensor components are Stress tensor σ, σ yy, σ zz σ y = σ y, σ z = σ z, σ yz = σ zy. (4.6) The stress components in Equation (4.5) form a two-dimensional tensor of second order [ ] ( ) Σ σ σ σ = [σ ij ] = = z. (4.7) Σ z σ z Since σ z = σ z, the stress tensor is symmetric, that is σ zz σ = [σ ij ] = [σ ji ] = σ T. (4.8) Using similar arguments in the three-dimensional case, it is possible to show that the stress tensor is also symmetric and has the general form Σ σ σ y σ z σ = [σ ij ] = Σ y = σ y σ yy σ yz. (4.9) Σ z σ z σ yz σ zz Notice that only si components are independent, since balance of moments (Eq. 4.6) leads to σ y = σ y, σ yz = σ zy and σ z = σ z. For the stress vector Σ and the unit normal ˆn on an arbitrary surface the following important relation holds Σ i (ˆn) = σ ˆn = σ ik n k. (4.10) If the stress tensor is known in one coordinate system K, it can be calculated in any other system K. The transformation formula is the same as in Equation (C.25) where R = [α] is an arbitrary rotation. σ ij = α ip α jq σ pq or σ = RσR T. (4.11) The above transformation eplains, why the shear stress components change their value by moving from a vertically aligned to a tilted coordinate system. 36

5 Physics of Glaciers I HS 2013 Eample The components of the stress tensor are σ = [σ ij ] = Find the traction on a plane defined by F () = = 0. Also determine the angle θ between the stress vector Σ and the surface normal ˆn. Solution: The unit normal on the surface is F 1 ˆn = F = F 3 and the traction on the surface is Σ(ˆn) = σˆn = = The angle θ is cos θ = Σ(ˆn) ˆn Σ(ˆn) = θ = 56. Stress invariants From the eamples in section C.2 we know how to calculate quantities that are independent of the orientation of the coordinate aes. For a second order tensor in three dimensions three invariants can be constructed. The first is I σ = 1 3 σ ii = 1 3 trσ = 1 3 (σ + σ yy + σ zz ), (4.12) and is also called the mean stress σ m. For incompressible materials like glacier ice, the isotropic mean stress does not contribute to deformation. It is therefore useful to characterize the stress state by the stress deviator. The deviatoric stress tensor is that part of the stress tensor which is etra from the isotropic stress state mean stress deviator σ (d) ij := σ ij σ m δ ij = σ ij 1 3 σ iiδ ij. (4.13) The second invariant of the deviatoric stress tensor is defined by second invaria 37

6 Chapter 4 (II σ (d)) 2 = 1 2 σ(d) ij σ(d) ij = 1 2 (σ(d) ) 2 = 1 2 ( (σ (d) ) 2 + (σ (d) yy ) 2 + (σ (d) zz ) 2 + 2(σ (d) y ) 2 + 2(σ (d) z ) 2 + 2(σ (d) yz ) 2). (4.14) invariant It is also called the octahedral stress or the effective shear stress, and is often denoted by τ or σ e. It will be important for the formulation of the ice flow law. The third invariant III σ (d) is the determinant of the deviatoric stress tensor It is seldom used in glaciology. Principal stresses III σ (d) = det(σ (d) ij ) = 1 3 σ(d) ij σ(d) jk σ(d) ki. (4.15) A face Fˆn with unit normal ˆn is free of shear forces, if the stress vector Σ(ˆn) is parallel to ˆn. In this case the vectors Σ(ˆn) and ˆn differ only by a numerical factor so that we can write Σ(ˆn) = σ ˆn = λˆn. (4.16) The proportionality constant λ is an eigenvalue and the vector ˆn an eigenvector of the tensor σ. An eigenvector of the stress tensor always fulfills equation (4.16). It therefore follows that an eigenvector of σ defines the orientation of a face without shear stresses. Furthermore, the eigenvalue is the normal stress on this face. A short reminder of some properties of symmetric tensors: All eigenvalues are real numbers. Two eigenvectors that belong to different eigenvalues are perpendicular to each other. There eists at least one coordinate system in which the representation of the tensor has only nonzero values on the main diagonal. For the Cauchy stress tensor, a coordinate system can always be found in which the tensor is purely diagonal. The three eigenvectors of σ, designated with s (1), s (2) and s (3), are perpendicular to each other and define a orthogonal coordinate system. In this coordinate system σ has the form λ σ = 0 λ 2 0. (4.17) 0 0 λ 3 Since σ s (i) = λ i s (i) (no summation convention!), the eigenvalues λ 1, λ 2 and λ 3 are the normal stresses. No tangential stresses act on the faces with unit normal s (i). 38

7 Physics of Glaciers I HS 2013 principal stress The eigenvalues λ i are called principal stress and the eigenvectors s (i) principal aes. The eigenvalues can be found by solving the problem σ s = λs which, written in components, reads σ ij n j s j λs i = 0 or (σ ij λδ ij )s j = 0. The trivial solution is s j = 0. The requirement for a non-trivial solution is det(σ ij λδ ij ) = 0. This equation leads to a polynomial of third order in λ, which can be written as λ 3 I 1 λ 2 + I 2 λ I 3 = 0, where use has been made of the following invariants of the stress tensor I 1 := trσ = σ ii, I 2 := 1 2 (σ iiσ jj σ ij σ ij ), I 3 := det(σ). Note: these invariants are different from the ones used before, but can be combined to yield the same forms. 39

8 Chapter Deformation A rigid body motion (translation, rotation) induces no change of the body shape. The strain of a body is the change in size and shape that the body has eperienced during deformation. The strain is homogeneous if the changes in size and shape are proportionately identical for each small part of the body and for the body as a whole. The strain is inhomogeneous if the changes in size and shape of small parts of the body are different from place to place: straight lines become curved, planes become curved surfaces, and parallel planes and lines do not remain parallel after deformation. h Linear strain The stretch s n of a material line segment is defined as the ratio of the deformed length l f to its undeformed length l o s n := l f l o. (4.18) sion strain The etension e n of a material line segment is the ratio of change in length l to its initial length l o e n := l f l o = l = s n 1. (4.19) l o l o (Note the sign convention: a positive etension is lengthening, a negative etension is shortening the body.) The above definition gives the average etension after a length change. Going to very small etension increments, one defines the strain ε ε := dl l, (4.20) that is, the ratio of the infinitesimal current etension increment dl with respect to the current length l. To obtain the finite strain of the etension from l o to l f we have to integrate Equation (4.20) with respect to l (the reference length l is increasing with increasing etension) ε := lf l o 1 l ( ) lf dl = ln = ln(s n ). (4.21) l o For obvious reasons ε is also called logarithmic strain. 40

9 Physics of Glaciers I HS 2013 Strain We now consider the deformation of an arbitrary body by studying the relative displacement of three neighboring points P, P, P in the body. If they are transformed to the points Q, Q, Q in the deformed configuration, the change in area and angles of the triangle is completely determined if we know the change in length of the sides. a 3, 3 P P P (a 1, a 2, a 3 ) Q Q Q ( 1, 2, 3 ) a 1, 1 a 2, 2 Consider an infinitesimal line element connecting the point P (a 1, a 2, a 3 ) to a neighboring point P (a 1 + da 1, a 2 + da 2, a 3 + da 3 ). The square of the length ds o of P P in the original configuration is given by ds 2 o = da da da 2 3 = da i da i. When P and P are deformed to the points Q( 1, 2, 3 ) and Q ( 1 + d 1, 2 + d 2, 3 + d 3 ), respectively, the square of the length ds of the new element QQ is ds 2 = d d d 2 3 = d i d i. We may epress the transformation from the a coordinate system into the coordinate system and its inverse by the epressions i = i (a 1, a 2, a 3 ) and a i = a i ( 1, 2, 3 ). (4.22) Therefore, using the Kronecker delta, we can write (with an arbitrary but convenient choice of inde labels) ds 2 o = δ kl da k da l = δ kl a k i a l j d i d j, ds 2 = δ ij d i d j = δ ij i a k j a l da k da l. (4.23) 41

10 Chapter 4 The difference between the squares of the length elements may be written as ( ) ds 2 ds 2 i j o = δ ij δ kl da k da l, (4.24) a k a l or as ds 2 ds 2 o = ( δ ij δ kl a k i ) a l d i d j. (4.25) j tensor We define the strain tensor in two variants Green - St. Venant E kl = 1 ( ) i k δ ij δ kl, (4.26) 2 a j a l Cauchy e ij = 1 ( ) a k a l δ ij δ kl, (4.27) 2 i j so that (remember that inde names are arbitrary) ds 2 ds 2 o = 2E ij da i da j, (4.28) ds 2 ds 2 o = 2e ij d i d j. (4.29) The Green strain tensor E ij is the strain with reference to the original, undeformed state and is often referred to as Lagrangian. We will mainly use the Cauchy strain tensor which is defined with respect to the momentaneous configuration. It is often referred to as Eulerian. E ij and e ij are tensors in the coordinate systems {a i } and { i }, respectively. Obviously both are symmetric E ij = E ji, e ij = e ji. (4.30) An immediate consequence of Equations (4.28) and (4.29) is that ds 2 ds 2 o = 0 implies E ij = e ij = 0 and vice versa. Therefore, a deformation in which the length of every line element remains unchanged is a rigid-body motion (translation or rotation). Strain components If we introduce the displacement vector u with the components then we can write u p = p a p p a i = u p a i + δ pi, a p i = δ pi u p i, 42

11 Physics of Glaciers I HS 2013 and the strain tensors reduce to the simpler form E ij = 1 [ ( ) ( ) ] up uq δ pq + δ pi + δ qj δ ij 2 a i a j = 1 [ uj + u i + u ] q u p 2 a i a j a i a j and e ij = 1 2 = 1 2 ( [δ ij δ pq u p + δ pi i [ uj + u i u ] q u p i j i j ) ( u )] q + δ qj j We now write out the components for e (the epressions for E are completely analogous), and use the more conventional variable names, y, z instead of 1, 2, 3, and u, v, w instead of u 1, u 2, u 3 (notice that we use here u, v, w to designate displacements). This leads to nine terms of the general form [ ( u e = u 1 ) e y = 1 [ u 2 y + v ( ) 2 v + ( u u y + v ( ) ] 2 w, v y + w )] w. (4.31) y If the components of displacement u i are such that their first derivatives are very small and the squares and products of the derivatives of u i are negligible, then e ij reduces to Cauchy s infinitesimal strain tensor ε ij = 1 ( ui + u ) j. (4.32) 2 j i In unabridged notation it reads ε = u, ε y = 1 2 ε yy = v y, ε z = 1 2 ε zz = w z, ε yz = 1 2 ( u y + v ) = ε y, ( u z + w ) = ε z, (4.33) ( v z + w ) = ε zy. y In the case of infinitesimal displacement, the distinction between the Lagrangian and Eulerian tensor disappears, since it is unimportant whether the derivatives of the displacements are calculated at the position of a point before or after deformation. Four common cases are shown in Figure

12 Chapter 4 z u u+ u d z u u+ u d Case 1: u > 0, w = 0 Case 2: u < 0, w = 0 z Slope to vertical = u z z Slope = w Case 3: u z > 0, w > 0 Case 4: u z > 0, u = w = 0 Figure 4.1: Different strain states: uniaial etension (case 1), uniaial compression (case 2), shear (case 3) and simple shear (case 4). Rotation Consider the infinitesimal displacement field u i ( 1, 2, 3 ). We then can form the cartesian tensor ω ij = 1 ( uj u ) i, (4.34) 2 i j which is antisymmetric, i.e. ω ij = ω ji. (4.35) Therefore the rotation tensor ω ij has only three independent components ω 12, ω 23 and ω 31 because ω 11 = ω 22 = ω 33 = 0. We can therefore write any relative movement of two points as the sum of a rotation and a deformation. Consider a point P with coordinates i and a point P in the neighborhood with coordinates i +d i. The relative displacement of P with respect 44

13 Physics of Glaciers I HS 2013 to P is This can be rewritten as du i = 1 ( ui + u ) j d j j i 2 Strain rate du i = u i j d j. (4.36) ( ui u ) j d j = (ε ij + ω ij ) d j. (4.37) j i For the study of glacier flow, we are concerned with the velocity field v(, y, z), which describes the velocity of every particle of the body. At every point (, y, z), the velocity field is epressed by the components (from now on u, v and w are used to denote the components of the velocity vector) or by v i ( 1, 2, 3 ) in inde notation. u(, y, z), v(, y, z), w(, y, z), Note: we reuse the letter u, v, w to designate velocity components instead of displacement components. Since the velocity is just the change in time of the infinitesimal displacement, the equations from section (4.3) apply unaltered. Instead of the infinitesimal strain tensor, we now look at the strain rate tensor ε ij := 1 ( vi + v ) j. (4.38) 2 j i strain rate The only change with respect to Eq. (4.32) is the dot, the use of velocity v i instead of displacement u i. Remember that the dot is part of the symbol used to designate strain rate and does not indicate a time derivative. Notice that other authors (e.g. K. Hutter) use the symbol D ij. 45