To measure the frictional resistance between wood and steel, a wooden block and a steel block can be stacked vertically and placed on a rough surface as shown in Figure 5.2. A constant vertical loadF is applied on top of the steel block. Then a gradually increasing lateral load is applied to the steel block until it starts sliding against the wooden block. Sliding indicates that shear failure has occurred between
F
A T
Rough Surface
σ = F/A
σ = F/A τf= Tf/A
τ = T/A
φ = tan−1µ = tan−1(τf/σ)
τ = T/A
(a) (b)
Steel
Wood
φ σ
FIGURE 5.2 Shear test concept.
the two blocks. Thus, the applied shear stress at failure (τf = Tf/A) is equal to the shear resistance (or shear strength) between the blocks, whereTf is the applied lateral force at failure andA is the cross-sectional area of the block. In reference to Figure 5.2b, the friction factor between the two materials can be calculated as µ = tanφ = τf/σ, where, φ is the friction angle between the two blocks and σ is the normal stress applied (= F/A). In this testing arrangement, termed direct shear, the shear strength at a predetermined shear plane is measured at a constant normal stress.
The direct shear test for soils uses the same concept as that illustrated above.
The soil is placed in a “split” shear box that consists of two halves, as shown in Figure 5.3. The box has a square cross section and can accommodate a soil
T Soil
Actuator
Actuator Vertical Displacement
Shear Displacement
F= constant
σ = F/A τ = T/A
Shear Box Displacement
Transducer
Disp. Transducer
FIGURE 5.3 Direct shear test apparatus.
specimen 10 cm× 10 cm in plane and 2.5 cm in height. The upper half of the shear box is not allowed to move laterally, whereas the bottom half can slide laterally by the action of a horizontal actuator. After placing the soil inside the box, a loading plate is seated on top of the soil and a vertical load is applied using a vertical actuator. A gradually increasing lateral displacement (at a constant rate) is then applied to the bottom half of the box, via the horizontal actuator, to generate shear stresses within the soil. The shear force,T , is measured using a load cell attached to the piston of the horizontal actuator. The horizontal displacement (shear displacement) of the bottom half of the shear box is measured during shearing using a horizontal displacement transducer. Also, the vertical displacement of the loading plate is measured during shearing using a vertical displacement transducer.
Figure 5.4 is a photo of a direct shear apparatus.
The results of a direct shear test are plotted in the shear displacement versus shear stress plane such as the one shown in Figure 5.5a. The vertical displacement of the loading plate is plotted against the shear displacement as shown in Figure 5.5b.
The test results shown in Figure 5.5a and 5.5b are typical for loose sands. Note that the shear stress is calculated by dividing the measured shear force by the cross-sectional area of the soil specimen. Figure 5.5a shows that the shear stress increases in a nonlinear manner as the shear displacement increases. The shear
FIGURE 5.4 Direct shear test apparatus. (Courtesy of Geocomp)
Compression Dilation
Vertical Displacement
Shear Displacement
Compression Dilation
Vertical Displacement
Shear Displacement
(b) (d )
Shear Displacement τf
Shear Displacement τf
Ultimate Strength Ultimate Strength
Peak Strength
(c) (a)
τ τ
FIGURE 5.5 Typical direct shear test results on (a, b) loose sand and (c, d) dense sand.
stress approaches failure at τ = τf, where τf is the shear strength of the soil.
Figure 5.5b shows a downward displacement of the loading plate, indicating soil compression. Loose sand consists of loosely packed grains with large voids in between. During shearing, some of the voids in the shear zone and its vicinity will collapse, causing the soil specimen to compress (downward plate displacement).
The behavior of dense sand during shearing is quite different from that of loose sand even though the two sands are assumed to be identical in terms of their gradation and specific gravity—they differ only in relative density. In the shear displacement versus shear stress plane (Figure 5.5c) the dense sand exhibits greater strength, or peak strength, at an early stage during shearing. After reaching peak strength, the shear stress decreases as the shear displacement increases until reach-ing an ultimate strength that is approximately the same as the ultimate strength of the loose sand (Figure 5.5a). Figure 5.5d shows the direct shear test results for dense sand in the shear displacement versus vertical displacement plane. In this figure, downward displacement of the loading plate indicates soil compression, and upward displacement of the plate indicates soil expansion (dilation). A pecu-liar characteristic is noted in the figure. At an early stage of shearing, the sand compresses slightly and then starts to dilate until a later stage of shearing, when it levels out as shown in the figure. Dense sand consists of densely packed grains with small voids in between. During shearing, some of the grains will slide and roll on top of other densely packed particles in the shear zone and its vicinity, causing the soil specimen to dilate (upward plate displacement).
The results of a direct shear test provide the shear strength (τf) of the soil at a specific normal stress (σ). The direct shear test is repeated several times on identical soil specimens using different normal stresses. Typical direct shear test results for sand using three different normal stresses are shown in Figure 5.6a. The shear strength of the soil at different normal stresses can be determined from the figure as indicated by points 1, 2, and 3. The at-failure tests results (points 1, 2,
Shear Displacement (mm)
τ (kPa)
1 2
3
(a) σ′ = 10 kPa
σ′= 30 kPa σ′ = 50 kPa
0 0 10 20 30 40 50
1 2 3 4 5 6
φ′ ≈ 37°
σ′ (kPa)
τ (kPa)
τ =σ′ tanφ′
Mohr–Coulomb Failure Criterion for Sand
1
2
3
(b) 0
10 20 30 40 50
0 10 20 30 40 50 60
FIGURE 5.6 Determination of the Mohr–Coulomb failure criterion for sand (direct shear test).
and 3) are presented in the normal stress versus shear stress plane in Figure 5.6b.
The three data points in Figure 5.6b are best fitted with a straight line. This straight line is the Mohr–Coulomb failure criterion. The slope of this line is the internal friction angle of the soil, φ, and its intercept with the shear stress axis is the cohesion intercept,c. The parameterscandφ are the strength parameters of the soil. They are unique parameters for a given soil. For a sandy soilc is zero, that is, the Mohr–Coulomb failure criterion passes through the origin, soτf = σtanφ. In Figure 5.6b the internal friction angle can be calculated from the slope of the Mohr-Coulomb failure criterion:φ= tan−1(τf/σ) ≈ 37◦.
Figure 5.7a presents typical direct shear test results for clay under three differ-ent normal stresses. The Mohr–Coulomb failure criterion for this clay is shown in Figure 5.7b. The figure shows that this soil has a cohesion intercept of approxi-mately 9 kPa and an internal friction angle of approxiapproxi-mately 26.5◦. With the help of Figure 5.7b, one can define the cohesion intercept (or cohesion) as the shear strength of the soil at zero normal stress (or zero confining pressure). This means that clays have some shear strength even when they are not subjected to confining pressure. It also means that sands do not have any shear strength without confining pressure. That is why we cannot make shapes out of dry sand, although we can certainly make shapes out of clay. For this reason sands and gravels are called cohesionless, whereas clays are called cohesive.
Note that we used effective stresses in the discussion above when describing the Mohr–Coulomb failure criterion and the shear strength parameters of the soil. This is because the shear strength of soil is dependent on effective stresses rather than total stresses (recall the effective-stress principle). Also, this means that when we conduct a direct shear test on wet or saturated soils, we have to facilitate drainage while shearing the soil specimen to prevent the development of excess pore water pressures in the soil. When the soil is saturated, the shear stress must be applied very slowly to prevent the development of excess pore water pressure. That way the total stress is equal to the effective stress because the pore water pressure is kept equal to zero:σ= σ − u = σ − 0 = σ.
When measuring the shear strength of a soil using a direct shear test, one needs to duplicate the field conditions of the soil being tested. Take, for example, the case of the slope shown in Figure 5.1. To measure the shear strength at point A, we can estimate the normal stressσat that point. This requires knowledge of the in situ unit weight of the soil and the location of the groundwater table. In the lab-oratory we can reconstitute a soil specimen in the direct shear box, aiming for the same in situ soil density. The soil sample can be submerged in a water basin (not shown in Figure 5.3) to simulate field conditions. A constant vertical stress equal to the normal stressσ calculated at point A is then applied to the soil specimen.
The shearing stage should not start until equilibrium within the soil specimen is achieved. This means that if the soil is a clayey soil, we have to wait until the excess pore water pressure generated as a result of stress application is dissipated.
Shearing can then be applied until failure. The shear stress applied at failure should reflect the true shear strength of the soil at point A.
1 2
3
(a) σ′ = 10 kPa
σ′ = 30 kPa σ′ = 50 kPa
τ (kPa)
Shear Displacement (mm) 0
0 10 20 40
30 50 60
1 2 3 4 5 6
c′ ≈ 9 kPa
σ′ (kPa)
τ (kPa)
τ = c′ +σ′ tan φ′
Mohr–Coulomb Failure Criterion for Clay
(b) 1
2
3
0 0 10 20 40
30 50 60
10 20 30 40 50 60
φ′ ≈ 26.5°
FIGURE 5.7 Determination of the Mohr–Coulomb failure criterion for clay (direct shear test).
Example 5.1 A dry sand sample is subjected to a normal stressσ= 20 kPa in a direct shear test (Figure 5.3). Calculate the shear force at failure if the soil sample is 10 cm× 10 cm in plane and 2.5 in height. The strength parameters of the sand arec= 0 and φ= 38◦.
σ′ = 1000 kPa τ = 550 kPa
Rocky Wedge
Inclined Plane
FIGURE 5.8 Is the wedge going to slide?
SOLUTION: From the Mohr–Coulomb failure criterion we can calculate the shear strength of the sand atσ= 20 kPa:
τf = c+ σtanφ= 0 + 20 × tan 38◦ = 15.6 kPa
At failure, the applied shear stress is equal to the shear strength of the soil (i.e., τ = τf); therefore, the shear force at failure is:
T = τ × A = 15.6 × 0.1 × 0.1 = 0.156 kN = 156 N.
Example 5.2 An intact rocky wedge is situated on an inclined plane as shown in Figure 5.8. Due to the self-weight of the wedge, a normal stress of 1000 kPa and a shear stress of 550 kPa are applied to the inclined plane. Determine the safety of the wedge against sliding given that the friction angle between the wedge and the inclined plane is 30◦.
SOLUTION: The shear strength offered at the wedge–plane interface can be cal-culated using the Mohr–Coulomb failure criterion withσ= 1000 kPa:
τf = c+ σtanφ= 0 + 1000 × tan 30◦ = 577.35 kPa
To avoid sliding, the applied shear stress must be less than the shear strength offered by the wedge–plane interface (i.e.,τ < τf). Luckily, τ is 550 kPa, which is smaller thanτf = 577.35 kPa. Therefore, the wedge is safe against sliding.