Assignment – Mohr’s Circle for CIVL4230 Advanced Soil Mechanics

Running head: ASSIGNMENT – MOHR’S CIRCLE
CIVL4230 Advanced soil mechanics
Assignment – Mohr’s Circle
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ASSIGNMENT – MOHR’S CIRCLE 2
Question 1
(a) The magnitude of the principal stresses
From the initial stresses in figure 1, the Mohr Circle is as shown below
From the graph,
Max Principal Stress = 44.14 kPa
Min Principal Stress =15.86 kPa
Max Shear Stress, τmax = 14.142kPa
(b) The direction of the principal stresses is at angle of 22.5 as shown in the diagram⁰

ASSIGNMENT – MOHR’S CIRCLE 3
(c) The stresses on the horizontal plane H-H.
If the stresses shown in the figure are the principal stresses, then the stresses in the H-H plane
before rotation are:
σx= 35.0kPa
σy= 25.0kPa
τxy = 8.66kPa
Using the equations for stress at a point:
σavg = (σx+σy)/2 = (σ1+ σ2)/2 = 30.000
R =τmax =σ1−σ2/2= √ [(σx−σavg) 2 + τxy2] = 14.142
σ1= σavg + R =44.142
σ2 = σavg − R= 15.858
θp = tan −1 [τxy/ (σx−σavg)] 2 = 22.5
θτmax = θp−π/4 = -22.5
Also:

ASSIGNMENT – MOHR’S CIRCLE 4
σx = σavg + R + cos(2⋅θp) = 40.0
σy = σavg – R + cos (2⋅θp) = 20.0
τxy = R + sin (2⋅θp) = 10.0
Question 2
In the Fig. 2, a load q per unit area acting on a strip of infinite length and of constant width 2a.
The vertical stress at any arbitrary point P due to a line load can be expressed as
Applying the principle of superposition, the total stress σz, at point P due to a strip load
distributed over a width 2a may be written as
Where b = a
A non-dimensional value of σz/q are always given in form of graphs and expressed more
conveniently as
where β = α , α and δ are angles as shown in figure 2.Thus the principal stresses σ1 and σ3 at
any point P may be obtained from the equations
σ1 = q/π (α + Sin α)
σ3 = q/π (α - Sin α)
The direction of the principal stresses is positive as the load is acting downwards
Question 3
a)
Increment shear stress, Δτxy = q/π {Sinα Sin (α + 2δ)}
Given that:
q= 200 kPa
Width = 10m
Water depth = 2.0m
Bulk unit weight of the soil = 19kN/m3
Distance below = 5m
Area a= B x W