Soil Mechanics Lab: Direct Shear Test

Experiment: Direct Shear Test (Cohesion & Friction Angle)

1. OBJECTIVE

The primary objectives of this experiment are:

• To determine the shear strength parameters, namely the cohesion (c) and the angle of internal friction (φ), for a given soil sample.
• To study the relationship between shear stress and normal stress for the soil under controlled laboratory conditions.
• To plot the failure envelope (Mohr-Coulomb failure envelope) for the soil based on the test results.
• To understand the shear behavior of soil when subjected to direct shearing forces.

2. APPARATUS

• Direct shear box apparatus (including shear box, porous stones, and loading frame)
• Proving ring to measure shear force
• Dial gauges to measure horizontal and vertical displacement
• Loading yoke and weights for applying normal stress
• Spatula and straightedge for preparing the sample
• Weighing balance

3. THEORY

The shear strength of a soil is its capacity to resist forces that cause the internal structure of the soil to slide against itself. This is a critical property in geotechnical engineering, as it dictates the stability of structures such as foundations, slopes, and retaining walls. The shear strength of a soil is not a single value but is dependent on the normal stress applied to the plane of shearing.

The relationship between shear strength and normal stress is defined by the Mohr-Coulomb failure criterion, which is expressed as:

where:

• τ (tau) is the shear stress at the point of failure
• c is the cohesion of the soil, which represents the shear strength when the normal stress is zero
• σ (sigma) is the normal stress acting on the failure plane
• φ (phi) is the angle of internal friction, which represents the friction between the soil particles

Factors contributing shear strength:

1. Structural resistance: It is due to the interlocking of the particles
2. Frictional resistance: resisting force to the sliding between the particles along the failure plane
3. Cohesion: cohesive force between the particles along the failure plane

4. PROCEDURE

1. The soil sample was prepared and placed into the shear box, ensuring it was level and had a uniform density. The initial dimensions and weight of the sample were recorded.
2. The shear box was assembled and placed in the loading frame.
3. A specific normal load (5 kg, 10 kg, or 15 kg) was applied to the sample via the loading yoke and allowed to consolidate.
4. The dial gauges for measuring horizontal shear displacement and vertical displacement were set to zero.
5. A horizontal shearing force was applied at a constant rate of strain.
6. Readings from the proving ring (for shear force) and the horizontal displacement dial gauge were taken at regular intervals of displacement.
7. The test was continued until the shear force reading peaked and then either remained constant or began to decrease, indicating that the sample had failed in shear. The maximum shear force was recorded.
8. The procedure was repeated for at least two additional identical specimens under different normal loads (10 kg and 15 kg).

5. OBSERVATION AND CALCULATIONS

Given Data:

• Area of mould (A): 36 cm²
• Calibration Factor of Proving Ring: 0.3 kg/division

Observations:

Calculations:

• For Normal Load = 5 kg:
  • Normal Stress (σ) = 5 kg / 36 cm² = 0.139 kg/cm²
  • Maximum Proving Ring Reading = 48 divisions
  • Maximum Shear Force = 48 div × 0.3 kg/div = 14.4 kg
  • Shear Stress (τ) = 14.4 kg / 36 cm² = 0.4 kg/cm²
• For Normal Load = 10 kg:
  • Normal Stress (σ) = 10 kg / 36 cm² = 0.278 kg/cm²
  • Maximum Proving Ring Reading = 56 divisions
  • Maximum Shear Force = 56 div × 0.3 kg/div = 16.8 kg
  • Shear Stress (τ) = 16.8 kg / 36 cm² = 0.467 kg/cm²
• For Normal Load = 15 kg:
  • Normal Stress (σ) = 15 kg / 36 cm² = 0.417 kg/cm²
  • Maximum Proving Ring Reading = 94 divisions
  • Maximum Shear Force = 94 div × 0.3 kg/div = 28.2 kg
  • Shear Stress (τ) = 28.2 kg / 36 cm² = 0.783 kg/cm²

Summary of Results:

Graphical Analysis:

A graph of Shear Stress (τ) on the y-axis versus Normal Stress (σ) on the x-axis was plotted using the data above. A best-fit line (the Mohr-Coulomb failure envelope) was drawn through the data points.

• The equation of the best-fit line is approximately: τ = 1.37σ + 0.208

From this equation:

• Cohesion (c): The y-intercept gives the cohesion, c = 0.208 kg/cm².
• Angle of Internal Friction (φ): The slope of the line is equal to tan(φ).
  • tan(φ) = 1.37
  • φ = arctan(1.37) ≈ 53.9 degrees

6. DISCUSSION AND CONCLUSION

The direct shear test was successfully conducted to determine the shear strength parameters of the provided soil sample. The results indicate that the soil is a c-φ soil, meaning it possesses both cohesion and internal friction. The cohesion (c) was determined to be 0.208 kg/cm², and the angle of internal friction (φ) was found to be 53.9 degrees.

The value of cohesion suggests the presence of fine-grained particles like clay or silt, which contribute to the cohesive bonds within the soil matrix. The relatively high angle of internal friction indicates a significant granular component, suggesting a well-graded soil with interlocking particles that provide frictional resistance.

It is important to acknowledge the limitations of the direct shear test. The test forces failure to occur along a predetermined horizontal plane, which may not necessarily be the weakest plane within the soil sample. Additionally, stress concentrations can occur at the edges of the shear box, potentially affecting the results. Despite these limitations, the direct shear test provides valuable data for many geotechnical design applications.

The results obtained are specific to the sample tested and the conditions of the test (e.g., moisture content, density). These parameters are fundamental for analyzing the stability of slopes, calculating the bearing capacity of foundations, and designing earth-retaining structures.