Basic Concepts of Strain and Tilt. Evelyn Roeloffs, USGS June 2008

Transcription

1 Basic Concepts of Strain and Tilt Evelyn Roeloffs, USGS June

2 Coordinates Right-handed coordinate system, with positions along the three axes specified by x,y,z. x,y will usually be horizontal, and z will usually be up. We will often use the coordinate system x,y,z=east, (geographic) North, Up. Sometimes we will use 1,2,3 instead of x,y,z to allow equations to be written more compactly. The distinctions among the reference frames used for GPS are not important for the interpretation of strainmeter or tiltmeter data. Page 2 of 24

3 Displacements Displacement of a point is described by the three displacements in these respective directions denoted by u E, u N, and u UP. Page 3 of 24

4 Stress and Strain in One Dimension If L is the length of the rod and the force stretches it by dl, then strain, ε, is the dimensionless quantity ε = dl /L The stress, σ, is the force per unit area, in this case σ = F /πr 2 Since this is a linearly elastic rod, the stress is proportional to the strain: F /πr 2 = E(dL /L) or σ = Eε a) we consider ε positive when the rod gets longer; we consider σ to be positive when the rod is in tension. b) E is a material property called Young s modulus with dimensions of force per unit area, units hpa, MPa, GPa c) The stress σ is present throughout the length of the rod, not just at the end where the force is applied. d) Strain is uniform throughout the length of the rod; displacement increases linearly from the fixed to the free end. Page 4 of 24

5 Typical Values of Young s Modulus Elastic strains are small for many realistic situations. Material Young's modulus (GPa) Strain caused by a 0.1 MPa (apx 14.5 psi) load parallel to axis of a rod Rubber E-3 Spider thread 3.03E-3 Hair 10.01E-3 Brick 14 7E-6=7 microstrain Oak 14 7E-6=7 microstrain Concrete 27 4E-6=4 microstrain Marble 50 2E-6=2 microstrain Aluminum E-6=1.5 microstrain Granite E-6 = 1.4 microstrain Iron E-6=0.5 microstrain Table. Typical values of Young s modulus (from Page 5 of 24

6 Strain in Three Dimensions The strains are the spatial gradients of displacement. If displacement is uniform over a body (ie., every part of the body moves same distance in the same direction), then there is no strain. The body can be picked up, moved, or rotated and there will not necessarily be any strain. The equation for obtaining strain from displacements, in three dimensions, is ε ij = 1 u 2 i / x j + u j / x i [ ] or ε ij = 1 Δu 2 [ i /Δx j + Δu j /Δx i ] for i, j = 1,2,3 (or x, y,z, or East-North-Up). (1) Writing out a few of them: ε zz = 1 u 2 [ z / z + u z / z] = u z / z Δu z /Δz ε xy = 1 u 2 x / y + u y / x [ ] ε NN = 1 u 2 [ N / x N + u N / x N ] = u N / x N Δu N /Δx N ε NE = 1 u 2 [ N / x E + u E / x N ] = 1 u 2 E / x N + u N / x E [ ] = ε EN (a shear strain) Engineering shear strain: γ xy = 2ε xy = [ u x / y + u y / x] Page 6 of 24

7 3D Strain, continued The 3 displacements can each vary in any of 3 directions, so there are 9 i, j combinations. However, there are only 6 independent strains, because if i and j are different, then ε ij = ε ji Example: for the rod, strain in the direction parallel to the rod is the slope of the plot of displacement vs. distance along the rod: ε xx = u x / x = dl /L Page 7 of 24

8 Some basic assumptions: 1) Small region: ie., region in which it is OK to use the value of displacement at a point and its spatial gradients to estimate the displacement Size of region depends on the spatial distribution of displacement. For example, slip on the San Andreas fault produces a displacement field that has a jump across the fault. Can t use displacement and strain at a point SW of the fault to estimate displacement NE, even if points are only 100 m apart. For two points on the same side of the fault, the region might be about 100 m or more. For two sites several km from the fault, the displacement field is much smoother and observations one or more km apart are sufficient. 2) Small deformation: Generally we will be speaking of strains in range (0.1 nanostrain) to 10-4 (100 microstrain). This ranges goes from resolution of GTSM21 borehole strainmeter to approximate strain within 10 s of meters of a M7 fault rupture. 3) Only changes matter: For example, we will consider vertical stress changes caused by atmospheric pressure fluctuations, but we will not explicitly worry about the more or less constant overburden pressure. Page 8 of 24

9 Units of Strain Strain is dimensionless but is often referred to as if it had units A 1% change in volume is a volumetric strain of 0.01 = 10,000 microstrain = 10,000 parts-per-million (ppm) A 1-mm change in a 1-km baseline is a linear strain of 10-6 =1 microstrain 1 microstrain is sometimes written 1 µε 1 nanostrain=10-9 =0.001 microstrain, also called 1 part-per-billion (ppb) 1 nanostrain is sometimes written 1 nε We will mostly be considering strains that range from about 0.1 nanostrain (an approximate lower bound on resolution for borehole strainmeters) to 1000 microstrain. 10 microstrain is a ballpark value for the coseismic strain within 5 km of a M7 earthquake. Page 9 of 24

10 Sign Conventions for Strain In mathematical descriptions, increases of length or volume are always considered to be positive strains. There is also an unambiguous sign convention for shear strain based on its defining equation. ε xy = 1 u 2 [ x / y + u y / x] Page 10 of 24

11 Example: Strain Near Transition from Creeping to Locked on a Strike-Slip Fault The strain-displacement equations unambiguously define a sign convention for shear strain. However, an arbitrary sign convention is sometimes used in reporting data. It s always a good idea to check. If fault creep decreases from 30 mm/year to 0 over a 30 km reach of the fault, then ε yy increases by 1 microstrain/year. If the creeping zone is 100 m wide, then at the creeping end ε xy changes by -300 microstrain/year. Page 11 of 24

12 Matrix Notation It s convenient to represent strain using a matrix: ε 11 ε 12 ε 13 ε xx ε xy ε xz ε EE ε EN ε EZ ε 3 3 = ε 12 ε 22 ε 23 or ε 3 3 = ε xy ε yy ε yz or ε 3 3 = ε EN ε NN ε NZ ε 13 ε 23 ε 33 ε xz ε yz ε zz ε EZ ε NZ ε ZZ The term strain tensor is often used. This is because strain can be expressed in different coordinate systems ( transformed ) according to certain rules. We will often work in 2 horizontal dimensions, where ε 2 2 = ε xx ε xy ε xy ε yy The three independent strain components in 2D are usually written in vector form: ε xx ε yy ε xy 3 1 or ε xx + ε yy ε xx ε yy 2ε xy 3 1 Page 12 of 24

13 Form of the Horizontal Strain Tensor that We Will Use in this Workshop We will mostly use the form on the right and refer to its entries as follows: ε xx + ε yy Areal Strain ε a ε xx ε yy Differential Extension γ 1 2ε xy Engineering Shear Strain γ 2 Dashed line = initial shape Solid line = deformed shape All of the diagrams show positive strains. Page 13 of 24

14 Stress Stress has dimensions of force per unit area. Inside a deforming body, forces act on every small planar region. Force has magnitude and direction, and 3 components: one perpendicular to the surface, and two parallel to the surface, at right angles to each other. σ xx σ xy σ xz σ 3 3 = σ xy σ yy σ yz σ xz σ yz σ zz There can be 3 directions of force on each plane, and 3 orthogonal planes, each perpendicular to one of the coordinate axes, giving rise to 9 components of stress. We do not distinguish between σ ij and σ ji, because stresses arise from the forces that do not move (accelerate) the body. Mathematically, the requirement that σ ij = σ ji results from the equilibrium equations (see any elasticity text). Intuitively, the picture can be used to visualize what would happen if these shear stresses were not equal. Because σ ij = σ ji, there are only 6 independent stress components in 3 dimensions, similar to the situation for strain. Page 14 of 24

15 Types of Stress and Stress States The three stress components with two equal subscripts are referred to as normal stresses. They apply tension or compression along a specified coordinate axis. They act parallel to the normal to the face of the cube. Stress components with unequal subscripts are shear stresses. They apply equal and opposite forces on opposing faces of a cube of material, acting parallel to those faces. ( ) is called the mean stress, or sometimes the average The average of the three normal stresses 1 σ 3 xx + σ yy + σ zz stress. It is frequently written σ kk /3, using the repeated index to imply the sum. A simple and important stress state is that in which the three normal stresses are equal. This is referred to as hydrostatic or isotropic stress. Pressure in a liquid is equivalent to a hydrostatic stress state. See Engelder (1993) for discussions of stress states relevant to the lithosphere. Page 15 of 24

16 Stress-strain Relations The way in which stress and strain are coupled for a particular material is described by constitutive equations. For a linearly elastic medium, the constitutive relations basically say that strain is proportional to stress, in 3 dimensions. An isotropic material is one which has the same mechanical properties in all directions. The constitutive equations for normal stresses and strains, in an isotropic linearly elastic material, are: 2Gε xx = σ xx ν ( 1+ ν σ + σ + σ xx yy zz), 2Gε yy = σ yy ν ( 1+ ν σ + σ + σ xx yy zz), 2Gε zz = σ zz ν 1+ ν σ + σ + σ xx yy zz and the constitutive equations for shear stresses and strains are: 2Gε xy = σ xy, 2Gε xz = σ xz, 2Gε yz = σ yz (2d,e,f) All 6 constitutive equations can be written compactly as: 2Gε ij = σ ij ν 1+ ν σ kkδ ij (another way to write 2a-f) using the summation convention and δ ij =1 for i = j, and 0 for i j. ( ) (2a,b,c) Two material properties appear: G, the shear modulus (with units of force per unit area); and ν, the Poisson ratio (dimensionless). More elastic moduli are needed if the material is not isotropic, but that situation is beyond the scope of this short course. Page 16 of 24

17 Relationships Among Elastic Moduli Although only two material properties are needed to relate stress and strain in an isotropic elastic material, there are many ways to represent these properties. The inter-relationships useful in this short course are: Young s modulus, E Bulk modulus, K Shear modulus, G E 3(1 2ν)K 2G(1+ ν) K E 3(1 2ν) 2(1+ ν)g 3(1 2ν) G E 2(1+ ν) 3(1 2ν)K 2(1+ ν) Table. Relationships between elastic moduli We will use K and G frequently. They are examples of moduli, with dimensions of force/unit area. The Poisson ratio couples extension in one direction to contraction in the perpendicular directions. It is always >0 and <0.5, taking on the upper limit of 0.5 for liquids. The Poisson ratio is dimensionless and is not a modulus. To see the role of the bulk modulus, add up the equations relating the three normal stresses and strains: ( ε 11 + ε 22 + ε 33 ) = 3(1 2ν) σ 11 + σ 22 + σ 33 = 1 σ 11 + σ 22 + σ 33 2G(1+ ν) 3 K 3 The bulk modulus is the coefficient of proportionality between volumetric strain (ie., expansion or contraction), and mean stress change (for example, pressure). Page 17 of 24

18 Typical Values of Elastic Moduli and Poisson ratio Material Bulk Modulus (GPa) Shear modulus (GPa) Poisson ratio Air (0 C, x MPa) Water (25 C) Polycrystalline Ice Plagioclase Quartz Crystal Calcite Olivine Berea Sandstone Westerly Granite Tennessee Marble Table. Typical values of bulk modulus, shear modulus, and Poisson ratio Page 18 of 24

19 Sizes of relevant and/or familiar stresses and strains: Material immediately adjacent to a fault that experiences a seismic stress drop of 10 MPa undergoes strains order of 1000 microstrain. Barometric pressure drop during a typical storm is 20 millibars = 0.02 bars = MPa = 2 kpa =20 hpa Pressure underwater increases at rate of 2.3 psi/foot = 1 bar/10 meters. = 1MPa/100 meters; pressure difference from bottom to top of 10-foot-deep pool is 23 psi = apx 0.3 bar = 0.03 MPa. Page 19 of 24

20 Rotating Coordinates We ll need to switch back and forth between coordinate systems that have different horizontal orientations (the vertical direction remains the same in these notes; See any text on elasticity for the 3D formulas). ε x'x' ε y'y' ε x'y' 3 1 = 1 1+ cos2θ 1 cos2θ 2sin2θ 1 cos2θ 1+ cos2θ 2sin2θ 2 2sin2θ 2sin2θ 2cos2θ ε xx ε yy 3 3 ε xy 3 1 ε x'x' + ε y'y' ε x'x' ε y'y' 2ε x'y' = 0 cos2φ sin2φ 0 sin2φ cos2φ 3 3 ε xx + ε yy ' ε xx ε yy 2ε xy 3 1 (3a,b) The method for converting stresses to a rotated coordinate system is exactly the same as that for converting strains. Page 20 of 24

21 Invariants of Strain and Stress ε x'x' + ε y'y' ε x'x' ε y'y' 2ε x'y' = 0 cos2φ sin2φ 0 sin2φ cos2φ 3 3 ε xx + ε yy ' ε xx ε yy 2ε xy 3 1 (3b again) Inspecting equation (3b) shows that : ε x'x' + ε y'y' = ε xx + ε yy for any value of φ (4) This is also true of the volumetric strain in 3D, and of mean stresses. In other words, the areal and volumetric strain, and the mean stress, are the same (invariant) no matter what Cartesian coordinates they are expressed in. 2 In 2D there is one other invariant of strain, which is ε xy ε xx ε yy Principal Strains and Principal Stresses By setting ε x'y' =0 in equation (3b), φ can be found such that the shear stresses (and strains) vanish. In this coordinate system, the e stress and strain tensors have only diagonal elements, called principle stresses and strains., related by: ε I 0 0 1/ E ν / E ν / E σ I ε II 0 = ν / E 1/ E ν / E 0 σ II ε III ν / E ν / E 1/ E 0 0 σ III (5) Near the Earth s surface, the vertical stress is usually a principle stress, so the other two principle stresses are in the horizontal plane. Page 21 of 24

22 Reducing to Two Dimensions using the Plane Stress Assumption Plane stress: one of the principle stresses is zero, so that only the stress components in one plane are nonzero. This is the situation when the vertical stress is unchanging (for example, it equals the overburden) and can therefore be assumed to be zero. Shear strains in vertical planes also vanish, so the only nonzero strains are ε xx,ε yy, ε zz, andε xy. However, we can express ε zz in terms of ε xx and ε yy. To see this, write out the constitutive equations for the normal strains: 2Gε zz = ν 1+ ν (σ xx + σ yy ) 2G(ε xx + ε yy ) = 1 ν 1+ ν (σ xx + σ yy ) Comparing these two equations shows that ε zz = ν 1 ν (ε xx + ε yy ) (6) ε zz is of opposite sign from ε xx + ε yy ; when there is contraction in the horizontal plane, the vertical strain is extensional. Consequently, for plane stress the volumetric strain is smaller than the areal strain: 1 2ν ε xx + ε yy + ε zz = 1 ν (ε + ε xx yy ) for plane stress (7) Page 22 of 24

23 Tilt Tilt is defined simply as the change in inclination with respect to either the horizontal or the vertical. Tilt is conceptually different from strain - it does not necessarily entail deformation. That is, displacements can be distributed spatially such that there is no strain, but there are nonzero tilts. Tilt can be measured using vertical sensors in boreholes or horizontal sensors on the earth s surface. As for borehole strainmeters, tiltmeters in boreholes are usually considered to be at the surface for mathematical purposes. Two tiltmeter components are needed to characterize the change in inclination of a plane. Tilt is a vector. Figure. (A) Tilt without strain. (B) Tilt with strain. If one component is oriented East-West and the other North-South, tilts are related to displacements by: tilt EW = u x / z = u z / x and tilt NS = u y / z = u z / y Page 23 of 24

24 Borehole Strainmeters as Elastic Inclusions Page 24 of 24