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In these Lecture notes, Professor has tried to illustrate the following points : Stress II, Geologists, Blocks, Equal Force, Smaller Block, Intuitively, Stresses, Vector, Shear, Perpendicular
Typology: Study notes
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Geologists are not using forces. Instead they use stresses! Why stress?
Consider two blocks of different size that are subjected to an equal force F~. Intuitively, the smaller block is going to deform a lot more than the larger block.
For this reason it makes sense to work with stresses rather than with forces.
STRESS = FORCE / AREA
Like force, the stress is a vector. The units of stress (in mks) are Pascal, where:
Pascal = Newton / m^2
A stress of 1Pa is very small! For example, the load due to 1m of water is about 10^4 Pa (why?). For this reason, geologists use MPa, which is 10^6 Pascals.
Other stress units are bars and atm:
1MPa = 10^6 Pa = 10bars = 9.8692atm
Consider a small cubic element of rock extracted from the earth. The stresses acting on this element may be visualized as follows:
The first index indicates the plane in question (recall that a plane is indicated by the direction of its outward normal). The second index indicates the direction of the stress. Docsity.com
Sign Convention
Unfortunately, the sign convention adopted by geologists is different than that adopted by engineers.
In Engineering, the stress is positive if it acts in the positive direction on the positive plane. In other words, the stress is positive in tension, and negative in compression.
In Geology, the stress is positive if it acts in the negative direction on the positive plane. In other words, the stress is negative in tension, and positive in compression. Why is that? This is because the stresses in the earth are compressive (although locally, tensional stresses are also possible). Docsity.com
The Symmetry of the Stress Tensor
The torque has to be zero, otherwise the block rotates. In the example below, the condition of zero torque may be written as:
−(σxyδy)δx/2 + (σyxδx)δy/2 = 0
And since δxδy 6 = 0, we get that σxy = σyx, and the number of independent stress components is equal to 3.
Similarly, in 3D we get that: σxy = σyx, σxz = σzx, and σyz = σzy, and the number of independent stress components is equal to 6.
The force components are:
f 1 x = −σxxδAnx f 1 y = −σxyδAnx f 2 x = −σyxδAny f 2 y = −σyyδAny
Summing the force components and setting these sums to zero we have: ∑ fx = txδA − σxxδAnx − σyxδAny = 0 ∑ fy = tyδA − σxyδAnx − σyyδAny = 0
Rearranging:
tx = σxxnx + σyxny ty = σxynx + σyyny
This is equivalent to:
tj = σij ni
where t is the traction acting on n .Docsity.com
Principal Stresses
We have learned that the stress tensor is symmetric. A property of symmetric matrices is that they may be diagonalized. The transformation from the non-diagonal to the diagonal tensor requires transformation of the coordinate system. The axes of the new coordinate system are the principal axes, and the diagonal elements of the tensor are referred to as the principal stresses.
σ ij∗ =
σ 1 0 0 0 σ 2 0 0 0 σ 3
Expressing forces in terms of stresses leads to:
σN = σ 1 cos^2 θ + σ 3 sin^2 θ
σS = (σ 1 − σ 3 ) sin θ cos θ
Substituting these identities:
sin^2 θ = (1 − cos 2θ)/ 2 sin θ cos θ = sin 2θ/ 2
gives:
σN = σ^1 + 2 σ^3 + σ^1 − 2 σ^3 cos 2θ
σS = σ^1 − 2 σ^3 sin 2θ
The above equation defines a circle with a center on the horizontal axes at (σ 1 + σ 3 )/2, and a radius that is equal to (σ 1 − σ 3 )/ 2
