GG 450 March 19, 2008. Stress and Strain Elastic Constants. Deformation of solids: We will define a set of parameters that provide models for the deformation of materials that will be used to describe seismic waves and what they tell us about the materials that they pass through.
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Stress and Strain
Shear strains: stains along one axis change with distance in the other axis:
In the figure above, the strain (u) in the x direction changes with y, but not with x. Shear strains are changes in shape, or angles.
NOTE: The tangent of u/y= strain in the x direction divided by y.
There are normal stresses that cause dilatational strains and tangential stresses that cause shear strains. ELASTIC CONSTANTS relate stress to strain through 3-dimensional versions of Hooke's Law, our constituative relationship.
Consider the deformation of water under a stress in the x direction. The resulting strains involve changes in all three directions as the water is pushed away, not just in the axis in which the force is directed.
If =0, then the other dimensions do not change in response to stress, and volume change is maximum. If = 0.5, then the volume does not change at all. For fluids, 0.5, while for slinky 0. For most solid rocks, = 0.1-0.25.
Note in the above arguments that we assume that the solid is isotropic, with no changes in elastic constants with direction. This is a MAJOR assumption, but the alternative - allowing elastic constants to change with direction (anisotropy), results in difficult math still being worked out.
For analysis of an isotropic solid, only 2 elastic constants are needed. E (Young's modulus), (Poisson's ratio), and G (rigidity) are important to seismology, as are the compressibility, , and bulk modulus, : where =1/, and is the ratio of change in pressure divided by the resulting change in volume.
q is the displacement of the any particle along the x axis , thus
dq/dt is the particle velocityof that point in the material, and
d2q/dt2 is the particle acceleration.
V is called the propagation velocity, or the wave velocity, or phase velocity. It is a measure of how fast the wave moves through the earth.
which describes a wave with amplitude A traveling in the +x direction with a velocity V, and k (wave number) is the number of cycles (times 2π) per distance.
Is it really a solution to this equation? Let’s try it out…
A bit more work shows that two types of waves are possible - compressional (or p) waves generated by dilational strains, and
shear (or s) waves generated by shear strains.
Near a boundary, such as the earth’s surface, other types of waves - surface waves- are generated.
Surface waves come in two types, Rayleigh waves, with particle motions like ocean waves, and
Love waves, with only horizontal particle motion.
Rayleigh waves are a combination of compressional and shear energy, while Love waves are only shear energy.
In terms of other elastic constants (the two velocities ARE elastic constants):
where is the shear modulus, k is the bulk modulus and is the density.
Note that the compressional velocity is always greater than the shear velocity, and that the shear velocity is independent of the bulk modulus.
In a uniform homogeneous isotropic solid where u is the displacement.
Where is the shear velocity and u is the displacement in a direction perpendicular to the direction the wave is moving in.
Here’s a good problem from Shearer’s Intro. SEISMOLOGY, page 34, # 3.5. In this problem, the wave equation is solved to show a shear wave traveling along a 100 km long bar (km/s). Use finite differences to write a small program to propagate a plane wave in a solid.
Using distance increments of dx=1 km and time increments of dt=0.1 sec, and assuming a source at u=50 km for 0 < t < 5 sec.
Apply a stress-free boundary condition at x=0 (this is a “free surface”, like the surface of the earth), and a fixed boundary condition at x=100 km (this condition says the bar is not allowed to move at this point).
The first derivative in digital form (OF ANY WELL-BEHAVED FUNCTION!) is given by:
Don’t believe it?: Google “difference calculus”. Or try it with a sine wave.
The second derivatives can be approximated by the following:
where the subscript is a change in x and u1, u2, and u3 are time increments.
Here’s the program in Matlab:
% WAVE EQN for bar using finite differences
% problem 3.6 page 34 of Shearer Seismology
beta=4.; % shear velocity
dt=0.1; % time incriment
dx=1.; % displacement incriment
tlen=5.; % time that the source is active (seconds)
% initialize variables
u1=zeros(1,101); % displacement at time j-1 (previous time step)
u2=zeros(1,101); % displacement at time j (current time step)
u3=zeros(1,101); % displacement at time j+1 (next time step)
for time=[1:dt:33]; %increment time for 33 seconds
for i=[2:100]; %calculate new displacement at each point for next time step
rhs=beta^2*(u2(i+1)-2.*u2(i)+u2(i-1))/dx^2; % right-hand-side of wave eqn
u3(i)=dt^2*rhs+2.*u2(i)-u1(i); % new displacement (u3)
% apply boundary conditions
u3(1)=1*u3(2); % du/dx=0|x=1, so: u3(1)=u3(2); boundary condition at x=1: stress-free, so
u3(101)=0.; % boundary condition at x=101: bar fixed, so u3(x=101)==0
u3(51)=sin(3.14159*time/tlen)^2; % source-time function for plane wave
% u3(1)=sin(3.14159*time/10); % another source function
for i=[1:101]; % for each x, propagate the displacement
plot(u2); % plot the current displacement
axis([1 101 -2 2]); % set axes for plot
pause % pause at each time step
This program displays the expected wave properties when a wave reflects from a fixed boundary and from a free surface.