1. Lecture 25 Bodies With Holes / Stress Perturbations Around Cracks Geol 542 Textbook reading: p.313-318; 354-357

2. We can also take the relations from the Airy stress function: 2D Elastic Solutions in Polar Coordinates and relate these to the polar coordinate system to derive general relationships (see handout):

3. If the principal stresses act along the coordinate axes, we have:  xx r =  1 r  yy r =  2 r  xy r = 0 Circular Hole in a Biaxial Stress Field The remote boundary conditions can be expressed in polar coordinates as: Note, these are simply the Mohr equations used at r = ∞, where  rr   ii (normal stress on a cubical element) and  r    ij (i≠j). Along the hole boundary, we have local boundary conditions for a shear stress free surface:  rr =  r  = 0(note,   ≠ 0)

4. The stress function for a circular hole is given by: Circular Hole in a Biaxial Stress Field where A, B, C, E, and F are constants dictated by the boundary conditions. Using the Airy equations in polar coordinates, we get: These are the general solutions for stress around a circular hole (for any loading condition).

5. Then solve for A, B, C, E, and F using the boundary conditions at r = a and r = ∞ to get: Circular Hole in a Biaxial Stress Field These are the specific solutions for a circular hole with biaxial loading.

6. We can use the specific solutions to solve for a circular hole for various remote boundary conditions and for any spatial location (r,  ). First we consider the case of a uniform remote compression of magnitude – S (i.e.,  xx r =  yy r = – S;  xy r =0) and zero stress on the hole boundary (i.e., p f = 0). Circular Hole in an Isotropic Stress Field Substituting into the specific solution stress equations in polar coordinates, we get: So the stress intensity factor is 1+(a/r) 2.

7. Circular Hole in an Isotropic Stress Field At the hole boundary (r = a),  rr =  r  = 0, and   = -2S everywhere (i.e., circumferential compression). So there is a stress concentration factor of 2, independent of hole size. This becomes important if 2S is greater than the uniaxial compressive strength of the rock.

8. Circular Hole in an Isotropic Stress Field Around the hole, principal stresses form radial and concentric stress trajectories. The mean stress (  rr +   )/2 is constant everywhere and equal to –S. The maximum shear stress (  rr –   )/2 is equal to S(a/r) 2. So the contours (isochromatics) are concentric around the hole. Note that despite the isotropic loading, the hole perturbation creates shear stress. As r  ∞, (a/r)  0, so  s(max)  0.

9. Next, we consider the case of a uniform remote compression of magnitude – S (i.e.,  xx r =  yy r = – S;  xy r =0) and an internal fluid pressure acting on the hole boundary (i.e., p f = – P = – S). This condition reflects a pressurized borehole, an oil well, or magma pressure in a cylindrical conduit. Tension is positive. Circular Hole With an Internal Fluid Pressure Substituting into the specific solution stress equations in polar coordinates, we get: The result is a homogeneous, isotropic state of stress. It’s as if the hole isn’t even there.

10. We now consider the case of zero remote stress (i.e.,  xx r =  yy r = 0 ;  xy r =0) and an internal fluid pressure acting on the hole boundary (i.e., p f = – P). Circular Hole With an Internal Fluid Pressure The stress components for this problem are: The result is a tension all around the hole equal in magnitude to the fluid pressure inside the hole (i.e., a stress concentration factor of -1). If this tension exceeds the tensile strength of the rock, hydrofracturing may occur.

11. Spanish Peaks Dikes Muller and Pollard, 1977

12. We now consider the case of isotropic remote stress (i.e.,  xx r =  yy r = – S;  xy r =0) and an internal fluid pressure acting on the hole boundary (i.e., p f = – P), where P ≠ S. Remote Stress Plus Internal Fluid Pressure The stress components for this problem are: If P = S, this result reduces to the equations derived previously. ≠

13. The specific solutions for a circular hole can also be used for the boundary conditions of biaxial loading. Biaxial Loading For example, the circumferential stress component can be solved at r = a to show: Hence, at  = 0,  :   = 3S h – S H. At  =  /2, 3  /2:   = 3S H – S h. (i.e., as described previously)

14. Circular Hole in a Biaxial Stress Field Around the hole, principal stresses are perturbed.

15. A similar approach can be applied to the problem of stresses around elliptical holes. e.g., dikes, sills, veins, joints, Griffith flaws Stress Around Elliptical Holes If the hole is oriented with long axes parallel to the x and y coordinate axes, respectively, the hole boundary is defined by: (x/a) 2 + (y/b) 2 = 1

16. Just as it was more useful to use a polar coordinate system for circular holes, it is appropriate to use an elliptical curvilinear system for elliptical holes, with components  (xi) and  (eta). Stress Around Elliptical Holes The transformation equations are: x = c cosh  cos  y = c sinh  sin  where 2c is the focal separation (see figure).

17. The stress components are:       which act on any particular element in this coordinate system. Stress Around Elliptical Holes It is the   component that tells us of the circumferential stress acting along the hole boundary, and always acts along lines of constant .

18. As with the circular hole, solutions are found by specifying the boundary conditions both at infinity (remote) and on the hole boundary, designated at  =  o. Stress Around Elliptical Holes The semi-major and semi-minor axes are given by: a = c cosh  o and b = c sinh  o As  o  0, a  c, and b  0. This produces a pair of straight lines connecting the foci and is the special case of a crack (cf. Griffith’s approximation).

19. Boundary conditions: Uniform remote tension of magnitude S (i.e.,  xx r =  yy r = S;  xy r =0) and zero stress on the hole boundary (i.e., p f =   =   = 0). The solution to the circumferential stress on the hole boundary is given by: The maximum values occur at the crack tips where  = 0, , so cos 2  = 1. This can be solved to show: Elliptical Hole in an Isotropic Tension The stress concentration factor is thus 2a/b (i.e., hole shape is important). e.g., if a = 5b, then   (max) = 10S.

20. The minimum values occur along the crack edges where  =  /2, 3  /2, so cos 2  = -1. This can be solved to show: Elliptical Hole in an Isotropic Tension The stress diminution factor is thus 2b/a. So if a = 5b, then   (min) = (2/5)S. We can reduce our solution to two special cases: (1)Circular hole: a = b    (max) =   (min) = 2S (2)Crack:b  0    (max) = ∞ Infinite stresses are predicted at the crack tip. This is referred to as a stress singularity in linear elastic fracture mechanics.

21. Pressurized Elliptical Hole with Zero Remote Stress Boundary conditions: Zero remote stress (i.e.,  xx r =  yy r =  xy r =0) and an internal pressure on the hole boundary (i.e., p f =   = – P;   = 0). In this case we get: For the special case of a crack-like hole, a>>b, so the stress concentration factor becomes ~2a/b. Also: If a>2b, this is a compressive stress, and in the limit a>>b, the stress approaches –P. In other words, a pressure acting on a flat surface induces a compressive stress of the same magnitude parallel to the surface.  = 0,   =  /2, 3  /2

22. Elliptical Hole with Orthogonal Uniaxial Tension Boundary conditions: Uniaxial remote tension parallel to minor axis b (i.e.,  xx r = 0;  yy r = S;  xy r =0) and an zero pressure on the hole boundary (i.e., p f =   =   = 0). The general solution is: We thus get the same result determine by Inglis: So   (min) is independent of hole shape. If a = 5b,   (max) = 11S and   (min) = -S. If a = b,   (max) = 3S and   (min) = -S (as we determined earlier in polar coords). For a crack, a>>b, so   (max) = 2Sa/b (  ∞ ) and   (min) = -S.  = 0,  and  =  /2, 3  /2

23. Elliptical Hole with Various Loadings Plots of tangential stress around two elliptical holes (a/b = 2 and 4) with three loading configurations:

24. Elliptical Hole with Stress at an Angle to Crack Boundary conditions: S 2 at  to x-axis S 1 at  +  /2 to x-axis (S 1 >S 2 OR S 2 >S 1 ) The general solution is: This is the equation that Griffith solved with respect to  to develop his compressive stress failure criterion. So we’ve already examined an application of this. x y S2S2 S2S2 S1S1 S1S1 

25. Elliptical Hole with Stress at an Angle to Crack Jaeger and Cook, 1969

26. Solutions for Holes with Other Shapes Analytical methods for determining solutions for holes with other shapes were introduced by Greenspan (1944) and are reviewed in the book “Rock mechanics and the design of structures in rock” by Obert & Duvall (1967). One of the most important considerations when addressing holes in rock is the effect of sharp corners on stress concentration. The sharper a corner, the greater the concentration of stress. We can explain this by re-examining the elliptical hole problem. For a uniaxial tension T applied orthogonal to the long axis of an elliptical hole, the circumferential stress at the tip is: So the stress scales as 2a/b. From the geometry of an ellipse, the radius of curvature at the end of the ellipse  = b 2 /a. Substituting b = √  a into the above equation, we get: where  is the remote stress acting perpendicular to the crack. So as  is decreased,   gets bigger = bad!

27. Solutions for Holes with Other Shapes Obert and Duvall, 1967 Note: even for rounded corners, the stress concentrations are greatest at the corners.

28. Solutions for Holes with Other Shapes Obert and Duvall, 1967

29. A reminder of why it matters… THE END!