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12 Cards in this Set
- Front
- Back
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Stress Invariants
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A general stress state existing at a material may be expressed as,
σx , σy , σz , τxy , τ yz , τ zx This stress state maybe considered equivalent to 3 perpendicular principal stresses acting on planes of zero shear stress. σ1 ; σ2 ; σ3 Compressive stresses – positive sign convention adopted. Here, compression = +ve and tension = -ve |
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Space diagonal and its coordinates
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The line passing through the origin and equally inclined to the principal axes is called the space diagonal (or hydrostatic axis) and has the following directional cosines,
(1/sqrt 3, 1/sqrt 3, 1/sqrt 3) |
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π-plane
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any planes perpendicular to Space axis is called the π-plane (sometimes called the Octahedral plane).
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Octahedral plane
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any planes perpendicular to Space axis is called the π-plane (sometimes called the Octahedral plane).
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hydrostatic axis and its coordinates
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The line passing through the origin and equally inclined to the principal axes is called the space diagonal (or hydrostatic axis) and has the following directional cosines,
(1/sqrt 3, 1/sqrt 3, 1/sqrt 3) |
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expressing a point P in principal stress space with co-ordinates , what is the length of the OP , which o is the origin
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By Pythagoras the length OP is given by,
OP = (σ12 + σ22 + σ32)0.5 which : A general stress state existing at a material may be expressed as, σx , σy , σz , τxy , τ yz , τ zx This stress state maybe considered equivalent to 3 perpendicular principal stresses acting on planes of zero shear stress. σ1 ; σ2 ; σ3 Compressive stresses – positive sign convention adopted. σm & J2 are symmetrical function of the principal stresses and are therefore invariants. However, we require a third invariant to find the exact position of P, i.e. Angular position in the π-plane; This is called the lode angle ϴ. |
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expressing a point P in principal stress space with co-ordinates , what is the length of the OP , which o is the origin
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The point P can also be arrived at by travelling a distance OA along the space diagonal and a distance AP along the π-plane in which P lies.
Therefore, OP2 = OA2 + AP2 The lengths OA + AP and be expressed as, Length OA= σm * sqrt 3 Which : σm = 1/3 ( σ1 + σ2 + σ3) Length AP =sqrt 2 * sqrt J2 which J2 = t2 = Tua 2 = Second devatoric stress invariant = 1/6 [ (σ1- σ2)^2 + (σ2- σ3)^2 + (σ3- σ1)^2 ] = J2 |
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Second devatoric stress invarian
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J2 = t2 = Tua 2 = Second devatoric stress invariant =
1/6 [ (σ1- σ2)^2 + (σ2- σ3)^2 + (σ3- σ1)^2 ] = J2 Note : σm & J2 are symmetrical function of the principal stresses and are therefore invariants. Which : σm = 1/3 ( σ1 + σ2 + σ3) |
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Load angle , ϴ
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The point P can also be arrived at by travelling a distance OA along the space diagonal and a distance AP along the π-plane in which P lies.
Therefore, OP2 = OA2 + AP2 The lengths OA + AP and be expressed as, Length OA= σm * sqrt 3 Which : σm = 1/3 ( σ1 + σ2 + σ3) Length AP =sqrt 2 * sqrt J2 which J2 = t2 = Tua 2 = Second devatoric stress invariant = 1/6 [ (σ1- σ2)^2 + (σ2- σ3)^2 + (σ3- σ1)^2 ] = J2 σm & J2 are symmetrical function of the principal stresses and are therefore invariants. ------------------------------------------------------------- However, we require a third invariant to find the exact position of P, i.e. Angular position in the π-plane; This is called the lode angle ϴ. ----------------------------- Projections of the positive and negative principal stress axis on the π-plane can be seen to subdivide into six sectors (symmetrical), each subtending 60° at the origin. If we let σ1 be the most compressive and σ3 the least, then only one sector will satisfy this, therefore any stress state existing in the soil will lie in the sector subtending 60° at the space diagonal and between + σ1 and – σ3 SO Axis (assuming a compression positive sign convention) The Lode angle ϴ varies between ± 30° in this sector, - Ж ϴ = -30° <==> Corresponds to the positive principal axis - Ж ϴ = +30° <==> Corresponds to the positive principal axis - ϴ = 0° <==>Corresponds to pure shear σ2 = 0.5 * (σ2 + σ2 ) which Ж sign depends on convention In triaxial conditions σ2 = σ3 and ϴ = -30° corresponds to triaxial compression, whereas ϴ = +30° corresponds to triaxial extension. |
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each angle corresponds to which stress - strain state of the soil
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Axis (assuming a compression positive sign convention)
The Lode angle ϴ varies between ± 30° in this sector, - Ж ϴ = -30° <==> Corresponds to the positive principal axis - Ж ϴ = +30° <==> Corresponds to the positive principal axis - ϴ = 0° <==>Corresponds to pure shear σ2 = 0.5 * (σ2 + σ2 ) which Ж sign depends on convention In triaxial conditions σ2 = σ3 and ϴ = -30° corresponds to triaxial compression, whereas ϴ = +30° corresponds to triaxial extension. |
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Which invariants fully define the stress state in the soil:
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( Sqrt 3*σm , sqrt J2*Sqrt 2, ϴ ) = ( S, t , ϴ )
Which : -- S = Isotropic Stress = Sqrt 3 * σm & σm = 1/3 ( σ1 + σ2 + σ3) -- t = Devatoric Stress = sqrt J2 * Sqrt 2 & J2 = t2 = Tua 2 = Second devatoric stress invariant = 1/6 [ (σ1- σ2)^2 + (σ2- σ3)^2 + (σ3- σ1)^2 ] = J2 -- ϴ = Load Angle ------------- Which in our present sign convention S = -ve tension S = +ve compression t = always positive ϴ = ± 30° or ± π/6 |
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Define the soil stress state in triaxial compression.
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Note that in TC , σ2 = σ3 , so some of the following only applies for TC.
In triaxial compression, i.e. σ2 = σ3 and ϴ = -π/6 - σ1 = σm + 2 * ( sqrt ( J2/3) ) - σ2 = σ3= σm - sqrt ( J2/3) Often we use the following notation for S & t in triaxial compression, p= σm = S/ sqrt 3 = mean stress q= σ1 - σ3 = sqrt 3 * sqrt J2 = +t * sqrt (3/2) = devatoric stress = difference in stress |
