Normal stress causes normal strain σ 22
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1 Normal stress causes normal strain blue box = before yellow box = after x 2 = Eɛ 22 ɛ 22 = E x 3 x 1 force is acting on the x2 face force is acting in the x2 direction Why do I draw this with equal stresses acting in opposite directions on either end of the box? We are ignoring other forces for now - so assume it s a massless box. x 2 x 3 x 1 You cannot apply a stress on just one side - the block will accelerate infinitely. There must be force balance. To get force balance, you must have stress acting on the top and the bottom surfaces, with equal force vectors in opposite directions. Think of what is happening inside the box. If it is tension then normal stress is positive. If compression (force vectors pointing into the box) then normal stress is negative.
2 x area = Force (up) If the box had mass and gravitational force was parallel to x2 and down, then for force balance stress on the bottom would have to be (less than / greater than) stress at the top. x 2 gravity force (down) x 3 x 1 x area = Force (down) Recall: A normal stress actually causes three normal strains x 2 x 3 x 1 ɛ 33 = ɛ 11 = ν E ν is Poissonʼs ratio (unitless) ν Typical is 0.25.
3 ɛ 33 = σ 33 E νσ 11 E ν E ɛ 11 = σ 11 E ν E νσ 33 E ɛ 22 = E νσ 11 E νσ 33 E If you have three equal normal stresses... σ 11 = = σ 33 = σ then ɛ 11 = ɛ 22 = ɛ 33 = ɛ ɛ = σ(1 2ν) and % volume change = 3ɛ When the normal stresses are equal to each other, there are no shear stresses anywhere (and normal stress = -1 x pressure) Shear stress ( sometimes called ) causes shear strain τ τ x 2 Simple shear τ 21 = 2Gɛ 21 ɛ 21 = τ 21 2G x 1 G is the Shear Modulus In rocks, G is also very, very large. typical value is 3 x Pascals ONE shear stress causes ONE shear strain (easy!)
4 When you are standing on a flat surface, what is the normal stress you exert on the ground? What is the shear stress? How could you exert a non-zero shear stress on the ground? Lithostatic pressure in the Earth... P = ρgh ρ =2500kg/m 3 g =10m/s 2 h = depth (m) Moho (crust-mantle boundary) is 30 km down Cascadia SZ (80 km below us) Deep ocean floor = 4 km (EQ works for water or ice sheets too with correct density) ρ h
5 In a fluid normal stresses are -P ( hydrostatic pressure ) shear stresses are 0 [ ] [ σ1 0 σ 0 = 0 σ 2 0 σ ] = [ P 0 0 P ] Below the Earth s upper crust normal stresses are close to -P ( lithostatic pressure ). Shear stresses are MUCH smaller. [ ] [ σ1 0 σ 0 0 σ 2 0 σ ] = [ P 0 0 P ] ρ h σ 11 σ 11 In the brittle upper crust, the mean of the normal stresses ( ) equals -P. σ The shear stresses and the (normal stresses minus their average) make up the deviatoric stress matrix [ ] [ ] [ ] σ11 σ 12 σ11 σ = 12 σ 0 σ 21 σ 21 0 σ DEV At a typical hypocentre = 7 km σ =? Deviatoric stresses are smaller. How much smaller? It is a subject of debate! Only the deviatoric stresses contribute to shear stress (which drives faulting!)
6 Rocks in the upper crust are brittle: this limits the maximum deviatoric stresses Differential stress low P Differential stress medium P Differential stress high P rocks are stronger in compression than in tension! However, even at high pressures their strength has a limit: 100ʼs of MPa. low P high P
7 σ = ( ) σ1 0 0 σ 2 σ 2 σ 1 If you know the principal stresses it is easy to solve for normal and shear stresses on any fault plane (using basic trigonometry). σ n τ You can very easily estimate the differential stress. It canʼt exceed a few hundred MPa or the rock will break. Coulomb failure criterion: The fault can slip if: τ = µσ n τ = µ s σ n Byerleeʼs Law (1978) Does rock type matter? This works at T less than C: hot rocks creep
8 Coulomb failure criterion: Right before an earthquake, The fault can slip if: At 7 km depth what is τ? τ = µσ n σ n = ρgh µ =0.7 ρ =2500kg/m 3 We can also estimate τ from shear strain accumulated between earthquakes if we know shear modulus G: t =200yr ɛ 12 = /yr ɛ 12 = ɛ 12 t G =30GP a = Pa τ 12 =2Gɛ 12 τ =123MPa from τ = µσ n τ =6MPa from τ =2Gɛ 12 Which answer is right? How could each answer be mistaken?
9 For many faults, earthquake shear stress drop is much smaller than background shear stress shear stress earthquakes! 123 MPa 117 MPa 6 MPa time For many faults, earthquake shear stress drop is much smaller than background shear stress τ = µ s σ n shear stress earthquakes! τ = µ d σ n τ shear stress drop is much less than the background stress time
10 Based on measurements of stresses in California crust, the shear stress acting on the SAF is very small. A large earthquake may release most of the accumulated stress. Cascadia SZ fault may be the same way. This is probably not typical of most faults. shear stress τ earthquakes! time τ = µ s σ n τ = µ d σ n The fault strength debate rages on... µ =0.1! µ =0.7! Consensus seems to be that major faults are weak and minor faults are strong... more later!
When you are standing on a flat surface, what is the normal stress you exert on the ground? What is the shear stress?
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