2
2
(1) Set AB = 2 m and observe that the building's foot-print covers the entire graph.
100 /
Set AB = 7 m and the number of units = 175
100 175 0.005 87.5 /
The stress at point C has drop
E
Average stress in the clay stratum beneath the center 93.8 2
50 48
Average stress in the clay stratum beneath the corner 49
2
( )( )
(2) The differential settlement between the center and a corner of the building is,
0.30 2.70
0.22 2.70 1 0.59
The in-situ stress at mid-clay stratum before the building is built is
c S
the settlement at the center log log 0.33
1 1 0.59 74.3
the settlement at the corner
o sand sand clay at mid clay
C o
the differential settlement is 0.33 0.21
(3) The time required to attain 60% consolidation in th
0.12 1 Solving for the time of settlement in the field ,
0.26 2,500 5 1 1
mm hours day year
t hours days y
*Mohr-01:
Find the angle of internal friction φ of sand.(Revision: Jan-09)
The horizontal stress on a soil particle at a depth of 130 ft is ⅓ of its vertical stress. If the average unit weight of the intervening soil mass above is 116.4 pcf, and the shear stress on the soil particle’s failure plane is 30 psi, what is the soil’s average angle of friction?
Solution:
The vertical stress is the major principal stress σ1 and the horizontal stress is the minor principal stress σ3. The shear stress is related to these two principal stresses via the angle θ, which is the angle to the failure plane with respect to the principal stress σ1. These are related by the equation,
1 3
From this relation, the angle of internal friction is found from θ,
2 90
*Mohr-02:
Simple transformation from principal to general stress state.(Revision: Feb-09)
A soil particle is found to be subjected to a maximum stress of 14.6 kN/m2, and a minimum stress of – 4.18 kN/m2. Find the σ and ττττ on the plane of θ = 50° with respect to the major principal stresses, and also find ττττmax.
(a) The graphical solution,
(b) The calculated solution,
( )
14.6 4.18 14.6 4.18
cos 2 cos 2 50 3.6
*Mohr–03: Find the principal stresses and their orientation.
(Revision: Sept.-08)
Equations for the principal stresses in the elastic half-space shown below for a uniformly loaded strip footing are as follows: σ1 = q/π(α + sin α) and σ3 = q/π(α - sin α).
The direction of the major
principal stress bisects the angle α.
Calculate the vertical stress σy, the horizontal stress σx, and ττττxy at point A if x = 0.75B and y = 0.5B using Mohr’s diagram.
Solution:
*Mohr–04: Find the principal stresses and their orientation.
(Revision: Sept-09)
Given the general stresses at a point in a soil, determine the principal stresses and show them on a properly oriented element.
*Mohr-05:
Find the angle of internal friction.(Revision: Sept-09)
A sample of clean sand was retrieved from 7 m below the surface. The sample had been under a vertical load of 150 kN/m2, a horizontal load of 250 kN/m2, and a shear stress of 86.6 kN/m2. If the angle θθθθ between the vertical stress and the principal stress is 60°°°°, what is the angle of internal friction φφφφ of this sample?
Solution:
σy = 90 lb/ft2 - 40 lb/ft2
τn
σn
*Mohr –
06: Normal and shear stress at a chosen plane.(Revision: Sept-09)
Using a Mohr circle, determine the normal and shear stresses on the plane AB.
Solution:
*Mohr–07:
Find the maximum and minimum stresses on a given plane.(Revision: Feb-10)
Determine the normal and shear stresses on plane AB.
Solution:
2 2
1 2 2 1
3 3
922
400 750 400 750
( 300)
Knowing the principal stresses permits to calculate the stresses on the normal plane,
( ) ( )
1 3 1 3
1 3
922 228 922 228
cos 2 cos2 45 575
**Mohr – 08:
Back figure the failure angle(Revised: Sept-09)
From the stress triangle shown below, find (a) the maximum and minimum principle stresses, (b) the angle alpha, as shown, (c) the angle theta, and (d) the value for the maximum shear stress.
*Mohr – 09: find the
Principle pressure using Mohr(Revised Sept-09)
The temporary excavation shown below is braced with a steel tube strut. Every morning, a misguided foreman tightens the screw mechanism on the strut “just to be safe”. The stress on a soil particle at point A, just behind the wall, has been measured with a pressure sensor installed by the Engineer. It now measures 40 kN/m2. If the potential failure planes in the soil behind the wall sustain 60°°°° angles with respect to the vertical wall, estimate the normal and shear stresses at that point A along a potential failure plane.
Solution:
At point A:
σv = hγ = (1.25 m) (16 kN/m3) = 20 kN/m2
∴∴
∴∴ σv is the minor principal stress at A,
Since θθθθ = 60° is with respect to the major principal stress (σ1) plane, then σv=σ 3
∴∴
∴∴ σθ = (σ1+σ3)/2 + (σ1-σ3)/2 cos 2θ = (40+20)/2 + (40-20)/2 cos 120°
∴
∴
∴
∴ σθ = 25 kN/m2
and τθ = (σ1-σ3)/2 sin 2θ = (40-20)/2 sin 120°
∴
∴
∴
∴ τθ = 8.7 kN/m2
*Mohr –10: Relation between θ and φ.
(Revised Sept-09)
For a clean sand, prove that θθθθ = 45°°°°+ φφφφ/2 using Mohr’s circle.
Solution:
For sand c=0.
By inspection in ∆OAB (180°-2θ)+90°+φ=180°
∴∴
∴∴ 2θ=90°+φ
∴
∴
∴
∴ θθθθ = 45°°°° + φφφφ/2
A failure test on a clean sand (i.e. c=0) shows that σσσσ1=11.5 ksf and σσσσ3=3.2 ksf at failure. Find the angle φφφφ for this sand.
*Mohr – 11:
Normal and shear stresses on any plane.(Revised Sept-09)
Determine the normal and shear stresses on the plane AB.
Solution.
001 = (300+125)/2 = 212.5 psf 0102 = (300-125)/2 = 87.5 psf 01B = 87.52+552 = 103 psf σ1 = 212.5 + 103 = 315.5 psf σ3 = 212.5 – 103 = 109.5 psf πC0102 = tan-1 (55/87.5) = 32°
σn = 212.5 – 103 cos (32+40) = 181 psf τn = 103 sin (32+40) = 98 psf
*Mohr–12:
Derive the transformation from general to principal stresses.(Revised Sept-09)
o) Derive the equation that transforms a general state of stress to the principal state of stress.
(Hint: Use Mohr’s circle for a graphical solution).
p) Determine the value of the major principal stress.
q) Determine the angle θθθθ between the major principal stress and the state of stress shown in the figure above.
*Mohr – 13:
Tensile stress from the Mohr-Coulomb failure envelope.(Revised Sept-09)
A soil sample has been tested and when plotted developed the Mohr-Coulomb envelope of failure shown below. Find (1) axial stresses at failure, (2) the normal and shear stresses on the failure plane, (3) the angle of failure with respect to the principal axis, and (4) the soil tensile strength.
Solution:
σ1f = qu = 300 Pa σf = σθ = 86 Pa τf = τθ = 136 Pa
θ = 115°/2 = 57.5°
qu = -123 Pa
+ σ + τ
25 °
96 kN/m2
(1.) σ1f = qu + σ
(4.) qu
+ τ
σ1
2θ = 115°
(3.) θ (2.) σf = σθ
τf = τθ
**Mohr–14:
Maximum principal stress from triaxial test.(Revised Sept-09)
A dry sample of sand was tested in a triaxial test. The angle of internal friction was found to be 36°. If the minor principal stress was 300 kPa, at what value of maximum principal stress will the sample fail? The same test was then performed on a clay sample that had the same φ, and cohesion of 12 kPa. What was the new maximum principal stress?
Solution:
a) Failure will occur when the Mohr circle becomes tangent to the failure envelopes.
*Mohr – 15:
Derive the general formula for horizontal stress.(Revised Sept-09)
Derive the general formula of the horizontal stress as a function of the vertical stress, cohesion and the angle of internal friction.
Solution:
According to the Mohr’s circle properties:
2