Stress Preliminaries

Invariants of the Cauchy Stress

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Invariants of the deviatoric stress tensor

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Stress space• The Cauchy stress tensor has six independent

components.

• Hence to represent a stress state geometrically, we need a six dimensional space with the six independent components as its coordinate axes.

• To describe the multiaxial failure envelope for general anisotropic materials, we have to work in the six dimensional stress space.

• This is because for anisotropic materials, the orientation of the principal stress is as important as the magnitude of the principal stress.

Stress space• However when the material is isotropic, the situation

simplifies considerably.

• Since the material properties are the same in any direction, only the magnitude of the principal stresses play any role in describing the failure surface.

• Therefore we only need a three-dimensional stress space using the three principal stresses as the coordinate axes.

• This stress space is called the Haigh-Westergaard stress space.

Haigh Westergaard stress space• In this space, every point with coordinates

represents a stress state with these principal stresses.

• Two stress states having the same principal stresses but different principal directions is not distinguishable in this space.

• Recall that each stress state can be split into a hydrostatic and deviatoric component. This decomposition can be geometrically represented in a convenient manner in the principal stress space.

),σ,σ(σ 321

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Hydrostatic axis• Recall we defined the octahedral planes to be families of 8

planes each of whose normals made equal angles with each of the principal directions.

• In the principal stress space therefore, the normal to each octahedral plane would form equal angles with the

axes.

• We consider one of these 8 normals, specifically the normal with direction cosines i.e. the normal

that lies in the positive octant.

• All stress states that lie on this diagonal satisfy the conditions

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1(

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The π plane• In other words along this line all stress states represent a

state of pure hydrostatic pressure. Hence this line is called the hydrostatic axis in the stress space.

• The planes normal to this axis are of course the octahedral planes with equations where C is the perpendicular distance from the origin to the plane

• The octahedral plane passing through the origin i.e. with C=0 is known as the π plane.

• Each of these planes contain all possible stress states with the same hydrostatic pressure C. Thus the π plane contains all possible stress states with zero hydrostatic pressures

3321 C

3

Projection along hydrostatic axis• Any stress state , represented by an unique

vector in the principal stress space, can be decomposed into two components, one of which lie along the hydrostatic axis, and another on one of the family of octahedral planes normal to the hydrostatic axis.

• The projected component along the hydrostatic axis gives the hydrostatic pressure part of the stress:

• The stress vector left after projecting out the hydrostatic component of the stress, which lies on an octahedral plane, must be a purely deviatoric stress state.

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The π plane through the origin

1

2

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devs

Projection on deviatoric plane• The length of the projection vector on the deviatoric plane,

is given by:

• To determine the orientation of this vector on the deviatoric plane, we project the axes on the deviatoric plane, denoting them as

• The unit vector along the axis is denoted as . It makes an angle of with the axis and equal angles with the axes.

• Thus this unit vector has direction cosines of

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Projection on deviatoric plane• The projection of the vector in the direction is given

by:

• The angle θ is known as the Lode angle.

• The Lode angle θ, the length of the projection of the stress vector in the deviatoric plane, r, and the length of its projection on the hydrostatic axis, denoted by ξ (equal to

) yield another parametrization of the stress state.

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Haigh Westergaard coordinates• This is a cylindrical coordinate system and each stress

point is represented by its (r, θ, z) = (r, θ, ξ) components. The (r, θ, ξ) components can all be expressed in terms of

the invariants of σ and S.

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Lode angles for common stress states

• One can calculate the Lode angles for common stress states encountered during laboratory testing.

• We will consider three such stress states: the case of uniaxial tension in the presence of hydrostatic pressure, pure shear in the presence of hydrostatic pressure and uniaxial compression in the presence of hydrostatic pressure.

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Uniaxial tension + h.s. pressure

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Pure shear under hydrostatic pressure

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Triaxial stress states• Triaxial stress states encountered from laboratory testing

on cylinders have two principal stresses equal.

• The third principal stress, aligned along the axis of the cylinder is usually different.

• Hence all such stress states lie in the plane, if the loading axis is assumed to be aligned with the

direction.

• Such planes are called rendulic planes. In the rendulic plane two curves, described as meridians are of interest.

• Along one of the curves, is always higher (more compressive) than

21 3

321

Compressive and Tensile meridians

• Along the other meridian, is always lower (less compressive) than

• This curve is known as the tensile meridian.

321

21

3Rendulic Plane

compressive meridian

tensile meridian

Multiaxial failure: Biaxial compression

• The micromechanics of failure in uniaxial compression can be extended to the multiaxial case.

• For example in biaxial compression, considering a micromechanics model comprising rigid particles, it is clear that if the compressive stress is sufficiently high, it would counteract the wedge splitting forces caused by the

vertical stress

• As a result of the confinement, the local splitting forces in the direction are reduced.

• Consequently a higher vertical stress must be applied to counteract the effect of and cause failure.

2

1

1

12

Biaxial compression• Thus the failure stress in biaxial compression must be

higher than in uniaxial compression.

• However the increase in failure load is only marginal. The reason for this is that the 3 direction is still unconfined.

• Splitting forces in the 3 direction are unopposed – therefore cracking is still relatively easy.

• Thus particle heterogeniety, in combination with a simple tensile/shear criterion for fracture, can be used to construct the biaxial failure envelope.

Biaxial compression: Experiments• The biaxial failure envelope for concrete was first plotted

by Kupfer.

• Specimens of sizes 200 x 200 x 50 mm were tested under biaxial compression, with specially designed plattens with brushes to reduce friction.

• The tests were performed on concretes with different uniaxial strengths: 19.1, 31.1 and 59.4 MPa

• The shape of the biaxial curve was found to be about the same for all these different mixes.

The Biaxial failure envelope

Biaxial compression: Maximum increase

• Under equibiaxial stress, , the strength increases by about 15-20% compared to the uniaxial compressive strength.

• However the largest increase in strength is not for equibiaxial loading. The largest increase occurs when the ratio of the principal stresses is about .5 i.e. the compressive stress applied in one direction is half the compressive stress in the orthogonal direction.

• In the tension-compression regime, the compressive strength decreases sharply even with a small tensile component.

21

Biaxial Tension Compression• Tension compression tests are harder to perform – also the

data shows considerably more scatter than the results of biaxial compression tests.

• However, unlike in the biaxial compression regime, it is clear that concrete quality has a greater influence in the biaxial tension-compression regime.

• For higher quality concretes, i.e. concretes with higher uniaxial compressive strengths, the rate of decline in compressive strength with tensile stress is higher.

• This may seem counter-intuitive, since it is known that tensile strength in concrete increases with compressive strength.

Biaxial Tension Compression• But recall that empirical relationships predict:

• Again, a microstructural explanation for this is lacking. Increase in compressive strength implies reduction in porosity, increased hydration in cement etc.

• All of these appear to have an effect on the tensile strength as well, but to a significantly lesser (hence the square root) extent.

• It seems to point to certain factors that are conducive to tensile cracking that are not really affected the increase in compressive strength.

cc

t

ct

ff

f

1

Effect of restraints on biaxial strength

• One such factor may be the weak planes that naturally result from the casting process. However a clear identification of these factors is lacking.

• The tests by Kupfer were carried out using loading plattens that were designed to reduce friction.

• Some researchers have studied the effect of frictional plattens on the biaxial failure envelope.

• The results are as expected: unrestrained testing (frictionless plattens) provide a lower bound to concrete compressive strength.

Effect of load paths on biaxial strength

• Other researchers have investigated the influence of load paths on biaxial strength.

• For instance if the total load is applied in a monotonically increasing fashion, or alternatively, if there is some intermediate unloading prior to reloading, will the biaxial strength be different?

• The answer appears to be that the biaxial strength does not depend on the load path unless there is appreciable damage sustained to the specimen during the previous

loadings.

Triaxial tests on concrete• Triaxial testing on concrete started with the tests by

Richart in the late 1920s and they quickly revealed the very different behavior of confined and unconfined concrete.

• However Richart’s tests and many subsequent tests that followed were not truly triaxial in nature, as they were performed on concrete cylinders.

• Fluid pressure was applied to the circumference of the cylinder in a pressure cell and axial loading was applied simultaneously in a compression machine.

Cylindrical testing vs. true triaxial loading.

• Since two of the principal stresses are always the same – the tests are not truly triaxial in nature.

• The full triaxial stress space can only be explored by a triaxial loading device.

• In such situations, cubes rather than cylinders are used.

• But cubes have the disadvantage that they are susceptible to stress concentrations along the edges and along the corners.

• Moreover platten effects play a larger role in cubes than in cylinders.

Cylindrical testing vs. true triaxial loading.

• True triaxial loading devices are expensive and are not readily available.

• Because of this, as well as the problems attendant to testing with cubes, triaxial tests on cylinders subjected to fluid pressures are still the most commonly adopted triaxial test.

• Since two of the stresses are always equal, only a single plane in the triaxial stress space can be investigated with cylinder tests.

Cylindrical testing: results for stress states on a single plane

1

2

3

3

3

p = 1 = 2

stress states only on this plane