AGEN 304 Lecture 8 - Shear Stress and Failure Theories
AGEN 304 Lecture 8 - Shear Stress and Failure Theories
AGEN 304 Lecture 8 - Shear Stress and Failure Theories
Effectives stress:
• no flow, downward and upward conditions
Shear strength:
– Definition
– Components
– Coloumb’s law
– Mohr theory
– Mohr-Coulomb theory
– Mohr-coulomb failure criterion & Mohr circle
Lecture 8: Shear strength and failure theories
How can we make the earth material with the Mohr circle above fail?
Failure theories: Mohr-Coulomb criterion
• When the circle touches the Mohr-Coulomb failure
criterion/envelope (the straight line) shear failure occurs.
• There are three ways for the circle reach the straight line
to reach failure:
• 1)Increase σ1;
• 2)Decrease σ3;
• The previous lecture essentially dealt with Mohr’s circle of stresses for a
material in equilibrium.
• If at any point within a soil mass the shear stress becomes equal to the
shear strength of the soil failure will occur at that point.
= c + n tan
• The plot of normal and shear stresses corresponding to failure will give
what is termed a strength envelope or the failure criterion straight line.
This line of failure shear and normal stress combinations intersects the
shear axis (n = 0) at the value of cohesion c and has a slope to the
horizontal axis
Failure theories
• The strength limit (strength envelope) forms two straight lines in the
graph above making angles of with the n axis and intersecting the -
axis at c.
• It is therefore possible for the soil to obey Coulomb’s law and Mohr’s
description of equilibrium stresses simultaneously.
Failure theories
Relationship between f and the angle of internal friction :
• 2f = 900 +
• Note:
• (1) That f the angle of the soil failure plane to the principal
stress 1 is independent of cohesion.
• The shear stress is increased until the specimen fails along that plane.
Figure 1 A typical direct shear box apparatus for measuring soil shear strength
Disadvantages
• Drainage conditions cannot be controlled (so only total
normal stress can be determined because pore water
pressure cannot).
• The cylindrical soil specimen is enclosed within a thin rubber membrane and
placed in a triaxial cell filled with a fluid.
• Pressure is then applied to the fluid in the cell to induce a compressive stress
б3 on the specimen.
Disadvantages:
Typical averaged drained shear test results for sand at increasing normal stress levels
Shear stress
Find
• (1) By graphical means the value of apparent cohesion and angle of internal
friction of the soil.
• (2) Check your solutions using the equations relating 1 and 3 above
• (3) Calculate f, the angle of the failure plane to the principal major stress 1
Shear strength: triaxial test
Solution:
• Calculate the Major and Minor principal stresses from the data
provided
• First the cell pressure is equivalent to the minor stress (σ3) since
it acts horizontally on the sample.
• Therefore the major principal stress (σ1) at each failure state and
conditions is the sum of the axial load (additional vertical
pressure) and the cell pressure
1. Tabulated stresses
Test 3 Cell pressure 1 Total vertical 1+3 1-3
(kN/m2) pressure(kN/m2) (kN/m2)
1 50 134 184 84
2 150 284 434 134
3 250 436 686 186
2. Construct Mohr’s circles:
For each test using the circle radius and circle centre values
calculated from