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```Geol 542
Bodies With Holes /
Stress Perturbations
Around Cracks
Lecture 25
2D Elastic Solutions in Polar Coordinates
We can also take the relations from the Airy stress function:
¶ 2f
s xx = 2
¶y
¶ 2f
s xy = ¶ x¶ y
¶ 2f
s yy = 2
¶x
and relate these to the polar coordinate system to derive general relationships
(see handout):
1 ¶f 1 ¶ 2f
s rr =
+ 2 2
r ¶ r r ¶q
¶ 2f
s qq = 2
¶r
¶ é1 ¶f ù
s rq = - ê
¶ r ë r ¶q úû
Circular Hole in a Biaxial Stress Field
If the principal stresses act along the coordinate axes, we have:
sxxr = s1r
syyr = s2r
sxyr = 0
The remote boundary conditions can
be expressed in polar coordinates as:
s rrr =
1 r
1 r
r
s
+
s
+
s xx - s yyr ) cos2q
(
(
xx
yy )
2
2
s rrq = -
1 r
s xx - s yyr ) sin 2q
(
2
Note, these are simply the Mohr
equations used at r = ∞, where srr 
sii (normal stress on a cubical
element) and srq  sij (i≠j).
Along the hole boundary, we have local boundary conditions for a shear stress
free surface:
srr = srq = 0
(note, sqq ≠ 0)
Circular Hole in a Biaxial Stress Field
The stress function for a circular hole is given by:
f (r,q ) = A ln(r)+ Br 2 + éëCr 2 + Er -2 + F ùûcos2q
where A, B, C, E, and F are constants dictated
by the boundary conditions. Using the Airy
equations in polar coordinates, we get:
s rr = Ar -2 + 2B - éë2C + 6Er -4 + 4Fr -2 ùûcos2q
s qq = -Ar -2 + 2B + éë2C + 6Er -4 ùûcos2q
s rq = éë2C + 6Er -4 - 2Fr -2 ùûsin 2q
These are the general solutions for stress around a circular hole (for any
Circular Hole in a Biaxial Stress Field
Then solve for A, B, C, E, and F using the boundary conditions at r = a and
r = ∞ to get:
4
2ù
é æ a ö2 ù 1
é
æ
ö
æ
ö
1 r
a
a
s rr = (s xx + s yyr ) ê1- ç ÷ ú + (s xxr - s yyr ) ê1+ 3ç ÷ - 4 ç ÷ ú cos2q
èrø
è r ø úû
2
êë è r ø úû 2
êë
4
é æ a ö2 ù 1
é
æaö ù
1 r
r
r
r
s qq = (s xx + s yy ) ê1+ ç ÷ ú - (s xx - s yy ) ê1+ 3ç ÷ ú cos2q
è r ø úû
2
êë è r ø úû 2
êë
4
2
é
æaö
æaö ù
1 r
r
s rq = - (s xx - s yy ) ê1- 3ç ÷ + 2 ç ÷ úsin 2q
èrø
è r ø úû
2
êë
These are the specific solutions for a circular hole with biaxial loading.
Circular Hole in an Isotropic Stress Field
We can use the specific solutions to solve for a circular hole for various remote
boundary conditions and for any spatial location (r, q).
First we consider the case of a uniform remote compression of magnitude –S
(i.e., sxxr = syyr = –S; sxyr =0) and zero stress on the hole boundary (i.e., pf = 0).
Substituting into the specific solution stress
equations in polar coordinates, we get:
é æ a ö2 ù
s rr = -S ê1- ç ÷ ú
êë è r ø úû
s qq
é æ a ö2 ù
= -S ê1+ ç ÷ ú
êë è r ø úû
s rq = 0
So the stress intensity factor is 1+(a/r)2.
Circular Hole in an Isotropic Stress Field
é æ a ö2 ù
s rr = -S ê1- ç ÷ ú
êë è r ø úû
s qq
é æ a ö2 ù
= -S ê1+ ç ÷ ú
êë è r ø úû
At the hole boundary (r = a), srr = srq = 0, and
sqq = -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.
s rq = 0
Circular Hole in an Isotropic Stress Field
Around the hole, principal stresses form
The mean stress (srr + sqq)/2 is constant
everywhere and equal to –S.
The maximum shear stress (srr – sqq)/2 is
equal to S(a/r)2. So the contours
(isochromatics) are concentric around the
hole. Note that despite the isotropic
shear stress. As r  ∞, (a/r) 0, so
ss(max)  0.
Circular Hole With an Internal Fluid Pressure
Next, we consider the case of a uniform remote compression of magnitude –S
(i.e., sxxr = syyr = –S; sxyr =0) and an internal fluid pressure acting on the hole
boundary (i.e., pf = –P = –S). This condition reflects a pressurized borehole, an
oil well, or magma pressure in a cylindrical conduit. Tension is positive.
Substituting into the specific solution stress
equations in polar coordinates, we get:
s rr = -S
s qq = -S
s rq = 0
The result is a homogeneous, isotropic state of
stress. It’s as if the hole isn’t even there.
Circular Hole With an Internal Fluid Pressure
We now consider the case of zero remote stress (i.e., sxxr = syyr = 0; sxyr =0) and
an internal fluid pressure acting on the hole boundary (i.e., pf = –P).
The stress components for this problem are:
æaö
s rr = -P ç ÷
èrø
2
æaö
= Pç ÷
èrø
2
s qq
s rq = 0
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.
Spanish Peaks Dikes
Muller and Pollard, 1977
Remote Stress Plus Internal Fluid Pressure
We now consider the case of isotropic remote stress (i.e., sxxr = syyr = –S; sxyr
=0) and an internal fluid pressure acting on the hole boundary (i.e., pf = –P),
where P ≠ S.
The stress components for this problem are:
2
é
æaö ù
s rr = -S - ê( P - S ) ç ÷ ú
è r ø úû
êë
2
é
æaö ù
s qq = -S + ê( P - S ) ç ÷ ú
è r ø úû
êë
s rq = 0
If P = S, this result reduces to the equations derived previously.
≠
The specific solutions for a circular hole can also be used for the boundary
For example, the circumferential stress component can be solved at r = a
to show:
s qq = ( SH + Sh ) - 2 ( SH - Sh ) cos2q
Hence, at q = 0, p: sqq = 3Sh – SH.
At q = p/2, 3p/2: sqq = 3SH – Sh.
(i.e., as described previously)
Circular Hole in a Biaxial Stress Field
Around the hole, principal stresses are
perturbed.
Stress Around Elliptical Holes
A similar approach can be applied to the problem of stresses around
elliptical holes.
e.g., dikes, sills, veins, joints, Griffith flaws
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
Stress Around Elliptical Holes
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 x (xi) and h (eta).
The transformation equations are:
x = c cosh x cos h
y = c sinh x sin h
where 2c is the focal separation (see figure).
Stress Around Elliptical Holes
The stress components are:
sxx
shh
sxh
which act on any particular element in this coordinate system.
It is the shh component that tells us of the circumferential stress acting along
the hole boundary, and always acts along lines of constant h.
Stress Around Elliptical Holes
As with the circular hole, solutions are found by specifying the boundary
conditions both at infinity (remote) and on the hole boundary, designated
at x = xo.
The semi-major and semi-minor axes are given by:
a = c cosh xo
and
b = c sinh xo
As xo 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).
Elliptical Hole in an Isotropic Tension
Boundary conditions:
Uniform remote tension of magnitude S (i.e., sxxr = syyr = S; sxyr =0) and zero
stress on the hole boundary (i.e., pf = sxx = sxh = 0).
The solution to the circumferential stress on the
hole boundary is given by:
s hh (x = x o ) =
2S sinh 2x o
cosh 2x o - cos2h
The maximum values occur at the crack tips where h = 0, p, so cos 2h = 1.
This can be solved to show:
s hh(max) = 2S
a
b
The stress concentration factor is thus 2a/b (i.e., hole shape is important).
e.g., if a = 5b, then shh(max) = 10S.
Elliptical Hole in an Isotropic Tension
The minimum values occur along the crack
edges where h = p/2, 3p/2, so cos 2h = -1.
This can be solved to show:
s hh(min) = 2S
b
a
The stress diminution factor is thus 2b/a. So if a = 5b, then shh(min) = (2/5)S.
We can reduce our solution to two special cases:
(1) Circular hole:
(2) Crack:
a=b
b0

shh(max) = shh(min) = 2S

shh(max) = ∞
Infinite stresses are predicted at the crack tip. This is referred to as a stress
singularity in linear elastic fracture mechanics.
Pressurized Elliptical Hole with Zero Remote Stress
Boundary conditions:
Zero remote stress (i.e., sxxr = syyr = sxyr =0) and an internal pressure on the
hole boundary (i.e., pf = sxx = –P; sxh= 0).
In this case we get:
é a ù
s hh(max) = P ê2 -1ú
ë b û
h = 0, p
For the special case of a crack-like hole, a>>b, so the stress concentration
factor becomes ~2a/b. Also:
é b ù
s hh(min) = P ê2 -1ú
ë a û
h = p/2, 3p/2
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.
Elliptical Hole with Orthogonal Uniaxial Tension
Boundary conditions:
Uniaxial remote tension parallel to minor axis b (i.e., sxxr = 0; syyr = S; sxyr =0)
and an zero pressure on the hole boundary (i.e., pf = sxx = sxh= 0).
The general solution is:
é sinh 2x (1+ e-2xo ) ù
o
s hh (x = x o ) = Se2xo ê
-1ú
êë cosh 2x o - cos2h úû
We thus get the same result determine by Inglis:
é a ù
s hh(max) = S ê2 +1ú
ë b û
h = 0, p
and
s hh(min) = -S
h = p/2, 3p/2
So shh(min) is independent of hole shape. If a = 5b, shh(max) = 11S and shh(min) = -S.
If a = b, shh(max) = 3S and shh(min) = -S (as we determined earlier in polar coords).
For a crack, a>>b, so shh(max) = 2Sa/b () and shh(min) = -S.
Plots of tangential stress around two elliptical holes (a/b = 2 and 4) with three
Elliptical Hole with Stress at an Angle to Crack
S1
Boundary conditions:
y
S2
S2 at b to x-axis
b
x
S1 at b+p/2 to x-axis
(S1>S2 OR S2>S1)
S2
The general solution is:
s hh(max) =
S1
2ab ( S1 + S2 ) + ( S1 - S2 ) éë( a + b2 ) cos2 ( b - h ) - ( a 2 - b 2 ) cos2 b ùû
a 2 + b2 - ( a 2 - b 2 ) cos2h
This is the equation that Griffith solved with respect to h to develop his
compressive stress failure criterion. So we’ve already examined an
application of this.
Elliptical Hole with Stress at an Angle to Crack
Jaeger and Cook, 1969
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:
é a ù
s hh = T ê2 +1ú
ë b
û
So the stress scales as 2a/b. From the geometry of an ellipse, the radius of
curvature at the end of the ellipse r = b2/a. Substituting b = √ra into the above
equation, we get:
s hh
éaù
» 2s ê ú
ërû
1
2
where s is the remote stress acting perpendicular to the crack.
So as r is decreased, shh gets bigger = bad!
Solutions for Holes with Other Shapes
Obert and Duvall, 1967
Note: even for rounded corners, the stress concentrations are greatest at the corners.
Solutions for Holes with Other Shapes
Obert and Duvall, 1967
A reminder of why it matters…
THE END!
```