Übung zu Globale Geophysik I 05 Answers
Übung zu Globale Geophysik I: Wednesday, 16:00 – 18:00, Theresienstr. 41, Room C419
Lecturer: Heather McCreadie
Companion class to Globale Geophysik I: Monday, 12:00 – 14:00, Theresienstr. 41, Room C419
Lecturer: Rocco Malservisi
1. Define stress (in words, no symbols allowed).
5
2
1
1
1
From page 425 Fowler, C.M.R., (1990) The solid Earth: An Introduction to Global Geophysics,
Cambridge University Press, (reprint 1992) ISBN: 0-521-38590-3 [+1]
[ /10]
2. Define strain (in words, no symbols allowed).
From page 426 Fowler, C.M.R., (1990) The solid Earth: An Introduction to Global Geophysics,
Cambridge University Press, (reprint 1992) ISBN: 0-521-38590-3 [+1]
OR
The change in length, volume or shape due to the application of a deforming force is called
"Strain". Strain has no unit because it is a ratio of two similar physical quantities.
From http://www.citycollegiate.com/matterb.htm
If you have another source and another way of writing you would also get full marks.
[10]
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Übung zu Globale Geophysik I 05 Answers
3. Define hooks law: use a plot if you need to (in words, no symbols allowed).
hint: Make sure you are defining it for rocks.
From page 429 Fowler, C.M.R., (1990) The solid Earth: An Introduction to Global Geophysics,
Cambridge University Press, (reprint 1992) ISBN: 0-521-38590-3
If you plotted a function then you should have plotted a straight line with a gradient
dependent only upon the material you have used. Because the theory hold for perfectly
elastic materials only this plot should go through zero. If it doesn’t then you don’t have a
perfectly elastic material.
[10]
4. According to Hook's law, if stress is increased, what happens to the ratio of stress to
strain?
It doesn’t change.
Hooke’s law states that stress/strain = constant.
How eloquent :-)
Extra marks if you discussed elasticity (+10)
[10]
5. How many parameters does an isotropic medium have?
According to page 429 Fowler 1 , an isotropic material has only two parameters. In the case
of Fowler they have used the Lamé constants .one of which corresponds to the shear
modulus.
[5]
For extra marks: The Lamé constants have almost the same value for rocks so they are
often approximated to the same value (termed Poisson’s relation). They are always positive
and have the SI units of Nm-2 .One Nm-2 is a Pascal (Pa). This corresponds to a Poisson
ratio of 0.25.
[+5]
Fowler, C.M.R., (1990) The solid Earth: An Introduction to Global Geophysics,
Cambridge University Press, (reprint 1992) ISBN: 0-521-38590-3
1
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Übung zu Globale Geophysik I 05 Answers
6. Write balance of forces on a cube in terms of strain not stress only in the y-direction.
Hint: look at the equations of motion.
In Fowler Appendix 2 this was done for x. Following this example for the y-direction.
Figure from page 431 and equation from page 432, Fowler, C.M.R., (1990) The solid
Earth: An Introduction to Global Geophysics, Cambridge University Press, (reprint 1992)
ISBN: 0-521-38590-3
Where ρ is the density of the parallelepiped; t is time; v is change of length in the ydirection. The Lamé constants are given by λ and μ. Δ is the 3-D equation of velocity and
nabla squared is the Laplacian operator.
[5]
The equation is only valid for an homogenous, isotropic, perfectly elastic medium, assuming
infinitesimal strain and no body forces.
[5]
If you derived the equation from scratch: (+10)
[+10]
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Übung zu Globale Geophysik I 05 Answers
7. Write your answer to 6 in terms of Young’s modulus and Poisson’s Ratio.
Fowler (appendix 2) gives the definition of Young’s modulus and Poison’s ratio in terms of
Lames constants. As shown below
‘
[5]
Rewriting the equation in terms of Young’s modulus (E) and Poisson’s ratio (σ) and
inserting these into the answer for (6) yields
Eσ
λ=
(1 + σ )(1 − 2σ )
μ=
E
2 (1 + σ )
⎞ δΔ ⎛
⎞ 2
∂ 2υ ⎛
Eσ
E
E
=
+
+
∇υ
⎜
⎟
⎜
⎟
∂t 2 ⎝⎜ (1 + σ )(1 − 2σ ) 2 (1 + σ ) ⎠⎟ δ t ⎝⎜ 2 (1 + σ ) ⎟⎠
Rearranging
ρ
ρ
⎛
⎞
E
1
∂ 2υ
δΔ
=
+ ∇ 2υ ⎟⎟
⎜⎜
2
2 (1 + σ ) ⎝ (1 − 2σ ) δ t
∂t
⎠
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Page 4 of 6
Übung zu Globale Geophysik I 05 Answers
8. Why is hooks law useful for seismic research? Hint: Think of speed….
The P-wave speed (Vp) and the S-wave speed (Vs) for a perfectly elastic, homogeneous,
isotropic material, are given on page 78 of Fowler 2 rewriting them in terms of the Lamé
constant and the and the shear modulus:
4
K+ μ
3 = λ + 2μ
V =
ρ
p
Vs =
Vp
Vs
=
ρ
μ
ρ
0.5 − σ
1−σ
Where K is the bulk modulus and the other symbols have been defined in previous
questions.
Therefore the constants derived from Hook’s law allow us to determine the velocities of
seismic waves.
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9. A slab of granite is deformed. The Poisson’s ratio for a granite rock was determined to
be 0.2, Young’s Modulus 200 MPa and the density was 2.75 g/cm3. What can you tell
me about the granite?
Convert everything to SI units
Density is 2.75g/cm3 t o convert in kg/m3 divide by (1000) to convert into kg and then
divide by (10-2)3 to convert in to m3. Therefore Density is 2.75x103 kg/m3
Young’s Modulus is 200 MPa. = 200 x106 Pa. As 1Pa=1Nm-2 it is already in SI units.
Poisson’s Ratio is dimensionless.
[5]
Using the equations in the previous questions (see working below) the Vp was found to be
284.2 m/s, the Vs was found to be 174.1/s and the calculated ratio Vs/Vp was 0.6. This ratio
is not the same as the determined ratio (174.1/284.2=0.78). Why? The given Young’s
modulus is too low for granite. The rock is also not an homogenous, isotropic, perfectly
elastic medium and we have assumed infinitesimal strain and no body forces on the rock
which is unrealistic.
[5]
2
Fowler, C.M.R., (1990) The solid Earth: An Introduction to Global Geophysics,
Cambridge University Press, (reprint 1992) ISBN: 0-521-38590-3
Page 5 of 6
Übung zu Globale Geophysik I 05 Answers
Vp =
Eσ
E
+2
(1 + σ )(1 − 2σ ) 2 (1 + σ )
λ + 2μ
=
ρ
ρ
=
(1 − σ ) ⎞
E⎛
⎜⎜
⎟
ρ ⎝ (1 + σ )(1 − 2σ ) ⎟⎠
⎞
⎞
1 − 0.2 )
(
2 ×108 ⎛
2 × 105 ⎛
0.8
=
=
⎜
⎟
⎜
⎟
2.75 ×10+3 ⎜⎝ (1 + 0.2 )(1 − 2(0.2) ) ⎟⎠
2.75 ⎜⎝ (1.2 )(1 − (0.4) ) ⎟⎠
=
Vs =
2 ×105 ⎛ 0.8 ⎞
1.6 ×105
=
= 284.2m / s
⎜
⎟
2.75 ⎝ 1.2 × 0.6 ⎠
2.75 × 0.72
μ
=
ρ
E
2 (1 + σ )
ρ
=
2 ×108
2 (1 + 0.2 )
=
2.75 ×103
108
105
=
= 174.1m / s
(1.2 ) × 2.75 ×103 1.2 × 2.75
Vs
0.5 − σ
0.5 − 0.2
0.3
=
=
=
= 0.6
1−σ
1 − 0.2
0.8
Vp
10]
Some Parameters for question 9 taken from
http://www.liv.ac.uk/seismic/links/info.html
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