Solutions to Numerical Exercises
Chapters 1-4
Exercise 1.16
Rewriting Eq. (1.1) we have
F   A
The area A is given as 4 cm2, which in SI units becomes 4  10-4 m2. The stress is given as 1
MPa, which is 1  106 Pa (N m-2). Using these values, the above equation yields the axial
force acting on the specimen as
F    A  1  10  4  10 4  400 N
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Exercise 1.17
Rewriting Eq. (1.3) we have
L    L
The strain ε is given as 0.4%, which is equal to 0.004. The original length L is 0.2 m. From
the equation above we then obtain the shortening L as
L    L  0.004  0.2  0.0008m  0.8mm
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Exercise 1.18
The measured strain is 0.12% which is equal to 0.0012. The compressive stress is 60 MPa,
which in standard SI units is 6  107 Pa. From Eq. (1.6) we obtain Young's modulus E as
E
 6  10 7

 5  1010 Pa  50MPa
 0.0012
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Exercise 2.9
From Eq. (2.11) the vertical stress is
 v   r gz  2600  9.81  3000  7.65  10 7 Pa  76.5MPa
This result can be compared with the statement on page 53 that the vertical stress increases by
about 25 MPa for every kilometre in the uppermost part of the crust.
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Exercise 2.10
We rewrite Eq. (2.11) to solve for the depth z and then substitute 1  108 Pa (100 MPa) for
the vertical stress σv and 2700 kg m-3 for the crustal density r. Thus, we have
z
v
1  10 8

 3375.4m  3.4km
 r g 2700  9.81
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Exercise 2.11
The vertical stress is the maximum principal stress and is given as 76.5 MPa (Exercises 2.9),
whereas the minimum principal stress is horizontal and given as 50 MPa. The angle α is 90 −
70 = 20. Thus, from Eq. (2.7) the normal stress on the normal fault at 3 km depth is
n 

1   3
2

1   3
2
cos 2
76.5  50 76.5  50

cos 40  63.3  10.3  53MPa
2
2
Similarly, from Eq. (2.8) the shear stress on the normal fault is (see Fig. 2.6 for the Mohr's
circle construction)

1   3
2
sin 2 
76.5  50
sin 40  13.25  0.643  8.5MPa
2
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Exercise 2.12
The vertical stress is the minimum principal stress and equal to the overburden pressure
(geostatic pressure). To calculate the vertical stress from Eq. (2.11) we need to know the rock
density, which is not given. For a rift zone composed primarily of basaltic lava flows, as in
Fig. 1.6 (referred to in this exercise), the density of the uppermost 100 m is likely to be
between 2500 and 2600 kg m-3. This refers to the in-situ density; the densities of small
laboratory specimens of basalt are generally higher (Appendix E.1). Here we take the density
as 2500 kg m-3. Then, from Eq. (2.11) the vertical stress at 100 m depth is
 v   3   r gz  2500  9.81  100  2.45  10 6 Pa  2.5MPa
We know that the maximum stress is horizontal and given as 20 MPa. The angle α is 90 − 75
= 15. Thus, from Eq. (2.7) the normal stress on the normal fault at 100 m depth is
n 

1   3
2

1   3
2
cos 2
20  2.5 20  2.5

cos 30  11.25  7.58  3.7 MPa
2
2
Similarly, from Eq. (2.8) the shear stress on the normal fault is

1   3
2
sin 2 
20  2.5
sin 30  11.25  0.5  5.6MPa
2
The Mohr's circle construction follows directly from Fig. 2.6.
Notice: in Exercises 2.11 and 2.12, Eqs. (2.5) and (2.6) could also be used, in which case the
angle , rather than α, would be used to define the attitude of the fault plane. The relationship
between these angles is shown in Fig. 2.19 (and in Example 2.7). If Eqs. (2.5) and (2.6) are
used, then the Mohr's construction follows the procedure in Example 2.4.
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Exercise 3.7
Elongation is given by Eq. (3.11) as
 
L Ld  Lu

Lu
Lu
Here ε = 0.002 and the extension or change in length of the rod is L = 0.001 m (0.1 cm). We
then rewrite Eq. (3.11) so as to solve for the original length of the rod, Lu, thus
Lu 
L


0.001
 0.5m
0.002
Thus, the original length of the rod was 0.5 m.
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Exercise 3.8
We rewrite Eq. (1.6) and obtain Young's modulus thus
 1  10 7
E 
 5  10 9 Pa  5GPa
 0.002
This is a low laboratory Young's modulus for rocks, but some sedimentary rocks and tuffs
have similar values (Appendix D.1). However, the in-situ Young's modulus of many young
(Holocene) lava flows is similar to this value (cf. page 108 in the book).
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Exercise 3.9
We rewrite Eq. (3.12) to obtain the deformed length Ld as
Ld  S  Lu  1.4  0.1  0.14m
From Eq. (3.12) we get the elongation as
  S  1  1.4  1  0.4
From Eq. (3.13), the quadratic elongation is
L
   d
 Lu
2

  (1   ) 2  S 2  (1.4) 2  1.96

From Eq. (3.19) the natural strain of the fossil is
 N  ln 1     ln S  12 ln   12 ln( 1.96)  0.34
(The result in Example 3.2 should be divided by 2, yielding the correct result: -0.22)
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Exercise 3.10
The stretch of the volcanic rift zone is the difference between its undeformed width, 20000 m
(Lu), the deformed width is 2000 + 0.5 = 20000.5 m (Ld). From Eq. (3.12) the stretch is
S
Ld 20000.5

 1.000025
Lu
20000
From Eq. (3.11) the elongation is
 
L Ld  Lu 20000.5  20000


 0.000025  2.5  10 5
Lu
Lu
20000
Strain rate is calculated from the elongation over time in seconds. The number of seconds in a
year is about 3.15  10 7 , so that the number of seconds in 20 years is about 6.3  10 8 . Since
the elongation (extension) is 2.5  10-5, we get the strain rate from Eq. (3.20) as
 
d 2.5  10 5

 4  10 14 s 1
8
dt
6.3  10
This is a typical geological strain rate and corresponds to about 2.5 centimetres (0.025
metres) per year. This is a common spreading rate - and 'spreading rate' is the other term used
for a strain rate across a rift zone - for a slow-spreading ridge.
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Exercise 4.15
(a) The average density is given, so that we rewrite Eq. (4.45) so as to find the depth z for this
state of stress, namely
z
v
1  10 7

 425m
 r g 2400  9.81
Thus, the rock body with this state of stress would be at a depth of just over 400 m, or similar
to the depth of exposure of the the normal fault in Fig. 2.4.
(b) The principal strains in terms of principal stresses and Young's modulus and Poisson's
ratio follow from Eqs. (4.5-4.7). For the maximum principal compressive strain we have
1 


1
 1   ( 2   3 ) = 1 10 1  10 7  0.25(5  10 6  1  10 5 )  0.0008725  0.087%
E
1  10
For the intermediate principal strain we have
2 


1
 2   ( 3   1 )  1 10 5  10 6  0.25(1  10 5  1  10 7 )  0.0002475  0.025%
E
1  10
For the minimum principal strain we have


1
 3  ( 1   2 )  1 10 1  10 5  0.25(1  10 7  5  10 6 )  0.000365  0.04%
E
1  10
3 
This shows that the minimum principal strain is negative, that is, tensile. Tensile strain (and
stress) at the depth of about 400 m km is close to a typical depth of absolute tension during
rifting events (cf. Chapters 7 and 8).
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Exercise 4.16
We first change the values given into SI values. Thus: the original diameter of the rod is 2
 10-2 m, its original length 0.1 m, the increase in its diameter during loading 3 10-6m, and
the decrease in its length 6 10-5m.
To find Young's modulus, we first need to know the elongation which, from Eq. (3.4), is
obtained as
 yy 
L 6  10 5

 6  10 4
L
0.1
From Eq. (4.1) we then get Young's modulus as
E
 4  10 7

 6.67  1010 Pa  67GPa
4
 6  10
For Poisson's ratio we need, in addition to the elongation (here shortening), the transverse
strain which is obtained from Eqs. (3.3) or (3.5) as
 xx
W  3  10 6


 1.5  10 4
2
W
2  10
We have to use minus in front of the increase in diameter, because strictly extension is
negative whereas compression or, here, shortening is positive. Poisson's ratio is the negative
of the lateral extension over the axial shortening, or
 
 xx
 1.5  10 4

 0.25
 yy
6  10  4
Here the shortening is here parallel with the y-axis and the extension parallel with the x-axis,
in contrast to that in Example 4.2.
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Exercise 4.17
There are several ways to find Young's modulus E and Poisson's ratio ν when the shear
modulus G (15.7 GPa) and the bulk modulus K (29 GPa) are known using the various
relations among the elastic constants in Appendix D.2. Here we do as follows. We obtain
Young's modulus using the relation
E
9 KG
9  29  15.7

 40GPa
3K  G 3  29  15.7
Similarly, we obtain Poisson's ratio from the relation
v
3K  2G 3  29  2  15.7

 0.27
6 K  2G 6  29  2  15.7
These are both very reasonable laboratory values for many solid rocks (Appendix D.1).
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Exercise 4.18
The theoretical horizontal stress, in relation to the vertical stress, when the horizontal strains
are assumed zero can be obtained from Eqs. (4.51) or (4.53). Using Eq. (4.53), and
remembering that m = 1/ν and thus 1/0.25 = 4 in this case, we have
h 
v
m 1

v
1
 v
4 1 3
At a depth of 2 km, and with a crustal density of 2500 kg m-3, the vertical stress is, from Eq.
(4.45) obtained as
 v   r gz  2500  9.81  2000  4.9  10 7 Pa  49MPa
The horizontal stress is 1/3 of the vertical stress, or about 16.4 MPa, and the stress difference
is thus 49  16.3  32.7MPa . This stress difference is many times larger than the in-situ
tensile strength, so that long before this difference could be reached (through reduction in the
horizontal stress from its initial lithostatic value equal to the vertical stress) a dyke or an
inclined sheet would be injected and bring the state of stress again close to lithostatic at the
boundary of a fluid magma chamber.
Notice that in this exercise, Young's modulus, although given, is not needed. Normally,
however, when Poisson's ratio is known and given, Young's modulus is also known and thus
given here for completeness.
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Exercise 4.19
From Eq. (4.69) we obtain the strain energy density, that is, strain energy per unit volume, as
U0 
2
2E

(2  10 6 Pa) 2
 50 Jm 3
10
8  10 Pa
This is the strain energy per unit volume. The volume of the specimen is its area times its
height. The area is A = πR2, where R = 0.01 m (1 cm). Thus, A  0.0314 m2. The height of
the specimen is 0.1 m (10 cm), so that its volume is
V  A  H  0.0314  0.1  0.003m 3
It follows that the total strain energy stored in the the specimen is
U = 50  0.003= 0.15 J
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Exercise 4.20
We have the following information: diameter of the bar is 4 cm = 4  10-2 m (and its radius,
R, is thus 2  10-2 m), length of the bar is 2 m, and the applied tensile force is 80 kN. Since
stress is force per unit area, we first find the cross-sectional area thus
A  R 2   (2  10 2 ) 2  1.26  10 3 m 2
The axial stress on the bar, from Eq. (1.1), is thus
F
 8  10 4
 
 6.35  10 7 Pa  63.5MPa
3
A 1.26  10
The corresponding strain or elongation, from Eq. (4.1), is


E

 6.35  10 7
 8  10 4
10
8  10
The strain (elongation) has a negative sign since it is tensile, that is, the bar has increased its
length. The strain energy per unit volume of bar (strain energy density) is, from Eq. (4.69),
obtained as
U0 
2
2E

(6.35  10 7 ) 2
 2.5  10 4 Jm 3
11
1.6  10
The volume of the bar is its area times its height, that is
V  A  H  1.25  10 3  2  2.52  10 3 m 3
It follows that the total strain energy stored in the bar is
U = 2.5  104  2.52  10-3 = 63 J
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