Tarea-2-Geodinamica

Homework 2: Geodynamics 1,2

Luis Blanco , Edgar Tarazona1, Santiago Cuartas1, Carlos Valencia1,3 Departamento de Geociencias 2 Departamento de Ingeniería civil y ambiental 3 Departamento de Ingeniería mecánica 1

Universidad de los Andes, Bogotá, Colombia 31-Agosto-2016

2.2 A mountain range has an elevation of 5 km. Assuming that ρc = 2800 kg

ρm

= 3300 kg m−3,

m−3 , and that the reference or normal continental crust has a

thickness of 35 km, determine the thickness of the continental crust beneath the mountain range. Assume that hydrostatic equilibrium is applicable

Solution: Using Airy’s hypothesis of isostasy:

( ∆ h+hc +r ) ρc =h c ρc +r ρm ∆ h ρc +h c ρ + r ρ c =hc ρc + r ρ c c

∆ h ρc =r ( ρm− ρc ) 2800 kg ) 3 ∆ h ρc m r= = =28000 m=28 km ρm −ρc ( 3300−2800 ) kg /m3 5000 m(

2.3 There is observational evidence from the continents that the sea level in the Cretaceous was 200 m higher than today. After a few thousand years, however, the seawater is in isostatic equilibrium with the ocean basins. What was the corresponding increase in the depth of the ocean basins? Take ρw = 1000 kg m−3 and the density of the displaced mantle to be ρm = 3300 kg m−3.

Solution: ρw ( 200+h o+ hm ) + ρ oc ho= ρw h o + ρoc h oc + ρm hm

ρm ( 200+h m )=ρ m h m hm =

ρw∗200 200∗1000 = =86.95 m ρ m−ρw 2300

Rta=286.95 m

2.6 A simple model for a continental mountain belt is the crustal compression model illustrated in Figure 2–6. A section of the continental crust of width w0 is compressed to a width wmb. The compression factor β is defined by ρm − ρs α β= w0 . wmb Show that the height of the mountain belt h is given by h=h (ρm −ρcc)(β−1). Assumingβ=2, hcc =35km,ρm =3300kgm−3,andρcc =2800kgm−3, determine the height of the mountain belt h and the thickness of the crustal root b.

Solution:

β=

wo wmb

w o h cc =w mb (h+ hcc +b) h+h cc + b=β hcc b=hcc ( β−1 )−h

( h+ hoc +b ) ρc =ρc hcc + ρm b

h ρc +b ρc =ρ m b hρc =b ρ m−b ρc

h=

h=

h=

h+

b ( ρm−ρ c ) ρc

[ h cc ( β−1 ) −h ] ( ρm −ρc ) ρc

h cc ( β−1 ) ( ρm −ρc ) h ( ρm− ρc ) − ρc ρc

h ( ρm −ρc ) ρc

=hcc

( ρm−ρ c ) ρc

ρ ¿ ( ¿ m−ρc ¿) /ρ c ( ρ m−ρc ) 1+¿=hcc ( β−1 ) ρc h¿

h

ρm ( ρ m− ρ c ) =hcc ( β−1 ) ρc ρc

( )

h=hcc

( ρm−ρc ) ρm

( β −1 )

Now we have:

β =2 hoc =¿ 35 km ρm=¿ 3300

kg /m3

ρcc =¿ 2800

kg /m

3

( β−1 )

h=35 km

( 3300−2800 ) ( 2−1 ) 3300

h=5.303 km

b=35 ( 2−1 )−5.303 b=29.697

2.10 Consider a block of rock with a height of 1 m and horizontal dimensions of 2 m. The density of the rock is 2750 kg m−3. If the coefficient of friction is 0.8, what force is required to push the rock on a horizontal surface?

Solution: ∆ σ xx=fgρL ∆ σ xx =0.8(9.81 m/s 2)(2750 kg /m3 )(1 m) ∆ σ xx=215.82 Pa ∆ σ xx =

F A

F=∆ σ xx A=21582 Pa ( 2 m∗2 m ) F=8.63 ×104 N

2.11 Consider a rock mass resting on an inclined bedding plane as shown in Figure 2–12. By balancing the forces acting on the block parallel to the inclined plane, show that the tangential force per unit area σx′y′ on the plane supporting the block is ρghsinθ (ρ is the density and h is the thickness of the block). Show that the sliding condition is

θ = tan−1(f). Solution:

σ yy =ρgh

∑ F x =0 ∑ F x =−σ xy +σ yy sinθ=0 σ xy =ρghsinθ

∑ F y =0 ∑ F y =σ yy cosθ F N =ρghcosθ σ xy =f F N =fρghcosθ The friction force is proportional to the normal force. Then:

ρghsinθ=fρghcosθ f =tanθ

θ=tan−1 f

2.12 The pressure ph of fluids (water) in the pores of rocks reduces the effective normal stress pressing the surfaces together along a fault. Modify Equation (2–25) to incorporate this effect. SOLUTION:

σ xX =f ρC gL where:

f = coefficient of friction. L=Thrust sheet length.

The Water pore pressure can be defined as:

Ps=g w h w where:

gw =unit weight of water. hw =depth below the water table.

We define then the new effective normal stress: σ 'xx =σ xX −P s=f ρC gL σ 'xx =σ xx=f ρC gL+ P s '

σ xx =σ xx=f ρC gL+ g w h w

2.18 Consider a simple two-layer model of a planet consisting of a core of density ρc and radius b surrounded by a mantle of density ρm and thickness a − b. SOLUTION: F g=

G∗M 1 M 2 r

2

a g=

a g=

G∗M 1 r2

G∗( ρ C V c ) r

2

(

G∗ ρC =

r

4 3 πr 3

3

2

b≤r≤a

Now for r (¿ ¿ 3−b3 ) 4 ρm π ¿ 3 ¿ ¿ a g=G ¿ r (¿ ¿ 3−b3 ) ρ m¿ ¿ ¿ 4 ¿G π ¿ 3

r (¿ ¿3−b 3) ρm ¿ ¿ (+( ρC b3 )¿¿ r2 ) ¿ 4 ¿G π ¿ 3

(

ρm r 3− ρm b 3+ ( ρC b3 ) 4 ¿G π 3 r2 ρ b (¿ ¿ c− ρm)+ ρ m r 3 3

2

r ¿ 4 ¿G π ¿ 3

)= 4 G π r ρ

)

C

b 3 ( ρc −ρm ) 4 ¿ G π ρ m r +[ ] 3 r2

(

)

P, is defined by:

dp =−ρg dr so, d p=∫ −ρg dr

For

b≤r≤a : 3

(

r

)

b ( ρc −ρm ) 4 p=∫ −ρg dr=−∫ ρm πG r ρm + dr 2 3 r a r b3 ( ρ c − ρ m ) 4 ¿− ρm πG ∫ r ρm + dr 3 r2 a

(

)

3

−2 2 4 ρm πG b ( ρc − ρm ) ¿ ρm πG r 2 + 3 3 r

r

|

a

a ¿ 4 1 1 (¿ 2−r ¿¿ 2)+ ρm πG b3 ( ρc − ρm ) − 3 r a ¿ 2 2 ¿ ρm πG ¿ 3

(

For

)

b≤r≤a : r

−4 p=∫ −ρg dr= πGρc 2∫ r dr 3 b

r

¿

|

−2 2 πGρc 2 r 2 = πGρ c2 (b 2−r 2 ) 3 3 b

Taking into account, the previous calculation of P(r) for mantle and evaluating it in

|❑ ❑

2.20 The measured horizontal principal stresses at a depth of 200 m are given in Table 2–1 as a function of distance from the San Andreas fault. What are the values of maximum shear stress at each distance? SOLUTION:

The maximum shear stress is half the difference of the principal stresses: σ 1 (¿¿ xy )max= (σ 1−σ 2) 2 ¿ Distance (km) 2 4 22 34

Maximum Shear Stress (MPa) 0.5 3 5 5.5