4 Stress Path

 

Stress Paths

As you know, states of stress, point in equilibrium can be represented by a Mohr circle in a τ-σ  coordinate system. Sometimes it is convenient to represent that state of stress by a  stress point, which has the coordinates (σ (σ1 - σ3)/2 and (σ (σ1 + σ3)/2 shown in Fig.1. For many situations in geotechnical engineering, assume σ1 and σ3 act on vertical and horizontal planes, so the coordinates of the stress point become ( σv - σh)/2 and ( σv + σh)/2, or simply q and p, respectively; or

Fig.1 A Mohr circle of stress and corresponding stress point. q = ( σv - σh)/2

(1)

 p = ( σv + σh)/2

(2)

Both q and  p could, of course, be defined in terms of the principal stresses. By convention, q is considered positive when σv > σh ; otherwise it is negative. We often want to show successive states of stress which a test specimen or a typical element in the field undergoes during loading or unloading. A diagram showing the successive states with a series of Mohr circles could be used (fig.2a), but it might be confusing, especially if the stress path were complicated. Therefore it is simpler to show only the locus of the stress points. This locus is called the  stress path, and it is plotted on what we call a p-q diagram (Fig.2b). Note that both p and q could be defined either in terms of total stresses or effective stresses. As before, a  prime mark is used to indicate effective stresses. So from Eqs.1 and 2 the effective stress equation, we know that q' = q while while p'  p' = p - u, where u is the excess hydrostatic or pore water pressure. Very often in geotechnical engineering practice, if you understand the complete stress path of your problem, you are well along the way towards the solution that problem. A simple case to illustrate stress paths is the common triaxial test which s  3  remains fixed as we increase s  1. Some Mohr circles for this test are shown in Fig. 2a along with their stress points. The corresponding stress path shown in Fig. 2b is a straight line at an angle of 45° from the horizontal  because the stress point represents the state of stress on the plane oriented 45° from the principal  planes. (Note that this is the plane of maximum shear stress.)

 

Some examples of stress paths are shown in Figs.3 and 4. In Fig. 3 the initial conditions are σv = σh an equal-all-around or hydro- static state of stress: Those in Fig. 4, where the initial vertical stress is not the same as the initial horizontal stress, represent a non-hydrostatic state of stress. You should verify that each stress path in Figs. 3 and 4 has in fact the direction as indicated in the figures. It is often convenient to consider stress consider  stress ratios. We defined a lateral stress ratio K, ratio  K, which is the ratio of horizontal to vertical stress, K = σh/σv  In terms of effective stresses, this ratio is

(3)

K = σ’h/σ’v 

(4)

where  Ko is called the coefficient of lateral earth pressure at rest conditions of no lateral strain. Finally, we can define a ratio Kf ratio Kf for stress conditions at failure. K f f = σ’hf  /  /σ σ’vf    

(5)

where σ’hf  = the horizontal effective stress at failure, and σ’vf  = the vertical effective stress at failure.

σ1 increasing

σ

σ3 (a) q

45ο

p (b)

 

Fig. 2 (a) Successive Mohr circles; (b) stress path for constant s3 and increasing s1, (after Lambe and Whitman, 1969).

 

 ∆σv σv

σv  ∆σh σh

σh

 σh  ∆σh

 σh  σv  σv

 

 ∆σv

 

 

q

D

C 45o

E

B

45o

18.4o

p

0  A F -q

 

Fig.3 Different stress paths for initially hydrostatic stress conditions (after Lambe and Whitman, 1969). Usually  Kf is defined in terms of effective stresses, but it may just as well be in terms of total stresses. Constant stress ratios appear as straight lines on a  p-q diagram (Fig.5). These lines could also be stress path for initial conditions of σv = σh = 0 with loadings of  K equal to a const (that is, constant σh / σv). Other initial conditions are, of course, possible such as those shown in Figs. 3 and 4.

 

 

 ∆σv σv

σ v

 ∆σh σh

σh

 σh  ∆σh

 σh  σv  σv

 

 ∆σv

 

Fig. 4 Different stress paths for initially non-hydrostatic stress conditions (after Lambe and Whitman, 1969).  Note that (q/p) = tan β = (1-K)/(1+K)

(6)

 K = (1-tanβ (1-tanβ )/(1+tan  )/(1+tanβ β )

(7)  (7) 

or in terms of K of K

where β is the slope of the line of constant  K when when K  K <  K  f . At failure, the slope of the K  the  K  f . line is indicated by the symbol ψ. Note also that for any point where you know  p and q, (for example  point A  point  A in Fig. 5), σh and σv  can readily be found graphically; that is, lines at 45° from the stress  point intersect the σ-axis at σh and σv. Finally, there is no reason why σv  must always be greater than σh. It usually is, but there are many important situations in geotechnical engineering where σh > σv .In these cases, by convention q is negative and K> and K> 1, as show in. Fig.5

 

q Kf  (compression) −ψ K = 0.5 = K0 for normally consolidated clays

β

σv > ση

K1

σv > ση

−ψ

Kf (extension) -q

 

Fig. 5 Different constant stress ratios and examples of stress paths, starting from σv = σh = 0 (after Lambe and Whitman, 1969).  Now we shall describe some stress paths which are important in geotechnical engineering. When soils are deposited in a sedimentary environment like a lake or the sea, there is a gradual  build up of overburden stress as additional material is deposited from above. As this stress increases, the sediments consolidate and decrease in volume. If the area of deposition is relatively large compared with the thickness of the deposit, then it seems reasonable that the compression is essentially one dimensional. In this case the stress ratio would be constant and equal to  Ko, and the stress path during sedimentation and consolidation would be similar to path  AB in Fig. 6. Typical values of Ko of  Ko for granular materials range from about 0.4 to 0.6, whereas for normally consolidated clays Ko clays  Ko can be a little less than 0.5 up to 0.8 or 0.9. A good average value is about 0.5. When a sample of the soil is taken, stress decrease occurs because the overburden stress σ’v0 has to be removed to get at the sample. The stress path follows approximately line BC line  BC in Fig.6, and the soil specimen ends up someplace on the hydrostatic ( σh  = σv ) or  K = I axis. This stress path and its relation to the strength of clays will be discussed later. If, instead of sampling, the overburden stress was decreased by erosion or some other geologic process, an unloading stress path similar to  BC would be followed. If the vertical stress continued to be removed, the path could extend to a point well below the p-axis. The soil would then be over consolidated and Ko and Ko would be greater than 1.0. Sometimes in engineering practice a test specimen is reconsolidated in the laboratory under  Ko conditions so as to reinstate the estimated in situ stresses. Such conditions are shown in Fig.4 and at  point A  point  A in Fig.7. After consolidation, the loading (or unloading) path followed to failure depends on

 

the field loading conditions one wishes to model. Four common field conditions and the laboratory stress paths which model them are shown in Fig. 7. Note that these stress paths are for drained loading (discussed in the next chapter) in which there is no excess pore water pressure; therefore total stresses equal effective stresses and the total stress path (TSP) for a given loading is identical to the effective stress path (ESP).

Fig. 6 Stress paths during sedimentation and sampling of normally consolidated clay, where Ko < 1

Symbol Geotechnical Engineering example AC: Axial Compression Foundation Foundation loading-increase σv, σh constant LE: Lateral Extension Active earth pressure-decrease σh, σv constant AE: Axial Extension Unloading (excavation)- decrease σv, σh constant LC: Lateral Compression Passive earth pressure- increase σh, σv constant Fig. 7 Stress paths during drained loadings on normally consolidated clays and sand (after Lambe,1967).

 

As suggested by Eq.5, we are often interested in conditions at failure, and it is useful to know the relationship between the  Kf line and the Mohr-Coulomb failure envelope. Consider the two Mohr circles shown in Fig.10. The circle on the left, drawn for illustrative purposes only, represents failure in terms of the  p-q diagram. The identical circle on the right is the same failure circle on the Mohr τ-σ diagram. To establish the slopes of the two lines and their intercepts, several Mohr circles and stress paths, determined over a range of stresses, were used. The equation of the  Kf line is q f  = a + p f  tan ψ   where a = the intercept on the q-axis, in stress units, and ψ= the angle of the Kf the  Kf line with respect to the horizontal, in degrees. The equation of the Mohr-Coulomb failure envelope is

(8)

τ   ff  ff  = σff  tan φ + c  From the geometries of the two circles, it can be shown that sinφ sinφ = tan ψ  and C = a/cos φ 

(9) (10)

So, from a p-q a p-q diagram the shear strength parameters f and c may readily be computed.

Fig.8 Relationship between the K, the K, line and the Mohr-Coulomb failure envelope. Another useful aspect of the  p-q diagram is that it may be used to show both total and effective stress paths on the same diagram. We said before that for drained loading, the total stress  path (TSP) and the effective stress path (ESP) were identical. This is because the pore water  pressure induced by loading was approximately equal to zero at all times during shear. However, in general, during undrained loading the TSP is not equal to the ESP because excess pore water  pressure develops. For axial compression (AC) loading of a normally consolidated clay (Ko < 1)  positive excess pore water pressure ∆u develops. Therefore the ESP lies the left of the TSP because σ' = σ- ∆u. At any point during the loading the pore water pressure ∆u may be scaled off any horizontal line between the TSP and ESP, as shown in Fig.9.

Fig. 9 Stress paths during undrained axial compression loading of a normally consolidated clay.

 

If a clay is overconsolidated (Ko > 1), then negative  pore w: pressure (- ∆u) develops  because the clay tends to expand during shear, but it can't. (Remember: we are talking about undrained loading in which. no volume change is allowed.) For AC loading on an overconsolidated clay, stress paths like those shown in Fig.10 will develop. Similarly, we can plot total and effective stress paths for other types of loadings, unloadings, for both normally and overconsolidated soils. In most practical situations in geotechnical engineering, there exists static ground water table; thus an initial pore water pressure u0 , is acting on the element in question. So there are really three stress paths we should consider, the ESP, the TSP, and the (T – u o)SP. These three paths shown in Fig. 11 for a normally consolidated clay with an initial I water pressure uo undergoing AC loading. Note that as long as the ground water table remains at the same elevation,  elevation,  uo  does not affect either the ESP or the conditions at failure.

Fig. 10 Stress paths during axial compression of a heavily overconsolidated clay

Fig. 11 ESP, TSP and (T-uo)SP for a normally consolidated clay