5. Soil physics article.pdf

*Corresponding Author: Paul T E Cusack, Email: St-michael@hotmail.com
RESEARCH ARTICLE
Available Online at www.ajms.in
Asian Journal of Mathematical Sciences 2017; 1(3): 145-169
Universal Soil Mechanics
*
Paul T E Cusack1
*1
BScE, DULE, 1641 Sandy Point Rd., Saint John, NB E2K 5E, Canada
Received on: 25/03/2017, Revised on: 12/04/2017, Accepted on: 18/05/2017
ABSTRACT
Here is a paper that continues on with Astrotheology by the same author. It provides solutions to the
Kalonite Universe using well established Soil Mechnics principles used now in Soil Physics. In addition,
an explanation for Black Holes is provided as well as a calculation on moisture content of the universal
material. This Kalonite is sometimes called the Ether or perhaps Dark Matter. There is no vacume in the
universe, thus particles can communicate from on end to the other.
Keywords: Soil Mechanics, Kalonite, Foundations, Retaining Walls, Embankments, Water Drawdown
INTRODUCTION
I’ve introduced that the universe can be thought of as a giant ellipsoid that is being compressed by gravity
to such an expent that the material that makes it u forms mass. This material I’ve dubbed “the Ether”.
Really, it is modelled best by Kalonite Clay particles that I’ve referenced in my paper on Astrotheology.
Since this material is like a kalonite clay, the science of Soil Mechanics, once the exclusive domine of
Civil Engineers, is now a branch of Physics, namely, Soil Physics. I present here some spot calculations
on the Universal Soil Physics which can be widely understood by Soil Physists and civil Engineers alike.
There is no further line of reasoning in this particular paper; just spot calculations, so that the reader may
examine them in no particular rder.
KALONITE
Normally consolidated cohesive soil:
Ko=0.44 +0.42 (PI %/100)
Ko=0.44 +).42 (40/100)
K o=1.724~ sqrt 3
K o=sigma h / sigma v=sqrt 3
K o=tan theta
From B M das Geotechnical Engineering Handbook,
cu/ sigma=0.0037 PI +0.11
PI=40%
cu/sigma=0.115~c^2 (speed of light)
cu/sigma'=0.22
=2c^2
cu/sqrt 2 /sigma=c
0.1427=sigma*c
1-sigma c=0.857 0.858=E
c=K/sigma=SPEED OF LIGHT
Continuing,
tan ^-1(cuz)=22.9 degrees = 1314 rads 0.858
And,

Paul T E Cusack et al.Universal Soil Mechanics
146
© 2017, AJMS. All Rights Reserved.
c *cos phi=P/A *c (cos 22.5 degrees)=1318 0.868
E=2q=2c^2=8N 60
c=2 sqrt N60
N 60/d^2=60
D =N60/60=0.4290/ 60=0.0072 1388 0.862
N60 =42= very dense soil.
The Universe is a very dense soil.
CLAY HIGH PLASTICITY EQUATION
LI=w-PL]/ [LL-PL]
1.125=10% -40%]/[LL -40%
-30/-1.125=26.667=LL-40%
c*10/ E min=F =G -|D|SOIL MECHANICS SHEAR STRENGTH
Shear=1/2 (p1-p2)sin 2 alpha
alpha=60 degrees
p1-p2=0.4330
0.4233-0.2=p1-p2
R-dM/dt=p1-p2
P=F*sin 60 degrees
In tri-axial forces on a soil, P=compressive force, p1 and p2 are the lateral forces
Again , I won't go through all the math (they don't pay me enough!),
M/c+ dM/dt=Shear
Integrate
Integral Shear=Moment (we know from Structural Engineering.)
M/2c^2+M=Moment=1-sin 1
M(1/(2c^2-1)=0,1585 1/2c^2-1 =0.0353-.1585
1/c^2 -1=0.1232 1/c^2=1.1232
c^2=0.8903~9
c=3
But t=s
Mass *t=Moment
Mt=1-sin1
f=6 *Mass
6d=ratio of volume of a sphere to area of a circle of radius 1
LIQUID LIMIT
LI=w -wp/ PI
PI=wl -wp
=10%-90%=80%
LI=90%/80%=1.125=E min
A=PI/ % clay
0.75=80%/% clay
% clay=1.0667 =THE UNIVERSE IS A FLUID WHEN COMPRESSED
1-1.0667=0.667
1/ 0.667=15% 0.85
SPECIFIC GRAVITY AND PERMITTIVITY
Gs=Ms/[Vs * rho w] 4.486/19860/1
-0.00002259
CLAY particles~0.0002 mm
0.0002/0.0002259=0.8854=eo=permittivity of "free space.
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Figure 1: Kaolinite Microscopic Particles. Figure 2 Kaolinite Particle [ DAS., PRINCIPLES
OF GEOTECNICAL ENGINEERING]
KALONITE
1-=0.1420=0.858
MOHR'S CIRCLE- FAILURE PLANE = 60 DEGREES
p1=p2N p1=p2(3)
p1=p3+2c p2(3)=p3 *2c p2/p3=2c/3 =0.666c
p2=p3*0.666c
p1/3=p3*0.666c
p1=p3*2c
p1=p2(3)=p3 +2c
p3*2c=p3+2c
2c=1+2c
1=1/2c+1
0=2c
c=0
Figure 3: Mohr’s Circle Failure Envelope Figure 4: Mohr’s Circle Rupture Plane
[SOIL MECHANICS IN ENGINEERING PRACTICE]
GRAPH THEORY OF CONSOLIDATION
Time Factor=1.4159~1.4
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C1=60 DEGREES / 59 KN / DEGREE=1.0169 2.9972=c LIGHT
C2=0.4233DEGREES /59=0.7175 1394 86 FIRM CEILING
C3= 70 DEGREES/59 =1.1864 0.8429 = sin 1 MASS / EARTH
v=1/k *1/ gamma w* dp/dt {SOIL MECHANICS IN ENGINEERING PRACTICE, K. TERZACHI]
v=1/R *1/ rho * F/dt
0.8415=1/R*1/0.127* 26/.5/ 1.0472 rads
R=0.4223~cuz
TIME FACTOR
T=cv */ H^2 * t
du/dt=cv d2u/dt2
cv=1
So,
T= 1/ 0.86^2)* (1.0472)
T=14159
T=Pi-3
1- 0.1416= 0.858=E
Pi is expressed exactly as
Pi=T+3
Pi=cv/ H^2 *t t=60 degrees
c+T=E
Figure 5: Kalonite Particle Size Figure 6: Particle Envelope Failure Plane
Figure 7: Triaxle Failure of Soil Specimen Figure 8: Coffer dam
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delta h=h1/Nd
sqrt[sqrt[Pi]]=32/18
=s
s^0.333=1.1=1/c^2
0.1334 (c^2)=12=|Dm|
sc^2=|Dm|=1/c^2
E=1/c^4 E=1/c^2*1/c^2=ts0.333
HYDROSTATIC PRESSURE=SPEED OF LIGHT
a^2 x 135 x delta h x gamma w =8x15x(32)(,81) =3.01 =c
COFFER DAM
dv/dx+dv/dz=0
dv/dx=dv/dz
dx=dz
Integrate
x=z
t=s
t=1/c^2 s(0.1334)^1/3=1/c^2
HYDRAULIC GRADIENT i
i=delta h/a
i=32/0.84=4
i=det |A-Lambda I|
12/4=3=c=sc^2/2=sc^2/(delta h/a sc^2=|Dm|
sc^2-c=a/ deltah
c(sc-1)=a/delta h
sc^2 *a/ deltah=c
Figure 9: Net
11 DIMENSIONS
30% *0.86=2.58
100%=2.58* 100/30=7.740
1/7.74=0.1292
1292 *1/1118=1.1550
0.8444=sin 1
V=IR =7.74(0.1334)=1.03 rads = 59 degrees
sin 59 degres=0.8572=0.858
1/0.1239=8.07=81
Figure 10: Earth Dam
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x=1/2 {q/k- k/q *z^2]
x=1/2[26.667/0.4233-0.4233/26.667 (0.86)^2]
x=3.149
x=Pi
[H+dH/dt] * gamma >Pi
0.86+dH/dt](0.127)>Pi
dH/dt=209.4
Recall 209.4/0.4233=201.6
dH/dt/R=Y
Y=e^-tcos (2Pit)
INTEGRAL dH/dt=H +INTEGRAL RY
H=R* e^-t sin (2Pi t)
0.86/0.4233=e^t sin (2Pi t)
203.17=e^tsin (2Pit)
Ln 203.17=Ln (sin (2Pit)
0.7089=sin (2Pi t)
45.14=2Pit
45 degrees=360 degress +t
t=3.148
E=t=Pi
Figure 11: Earthquake model
F=sin t =sin (0.2018) =0.858 =E
t=Y
F=E
F=1/Y
1/F=P.E. + K.E.
1/ (Ma)=Mgh +1/2Mv^2
1/a=ah+1/2v^2
v=a
1=a^2h+1/2a^2
h=1
1=a^2+1/2a^2 1=3/2 a^2
a= sqrt 2/sqrt 3
a=0.816
THIS IS 618 IN REVERSE. THE LAWS OF PHYSICS WORK IN REVERSE.
ALSO IT IS 81
1/81=0.01234567
[GEOTECHNICAL EARTHQUAKE ENGINEERING HANDBOOK, DAY]
q=0.4[1-r] gamma BN
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r=0.85 from chart
26.667=0.4(1-0.85)(0.127) (1)N
N=35.5 degrees=0.6196 rads
For our universe:
THE CUSACK "GOLDEN MEAN" EARTHQUAKE EQUATION:
G.M. = 1.618=N=(2.5 F)/[(1-sin 1)gamma]
From Groundwater movement:
v is proportional to i
V=Ci
0.8415=C (-0.1364)
C=0.618 = 1-G.M.
C= sqrt (-1)
GROUNDWATER HYDROLOGY
C/A INTEGRAL dt=INTEGRAL dS/S
Area of ellipse=0.7854 L x H=0.7854 *24*3=0.5654
t=1/sqrt 3
S=0.87 (from above =frost pemetration)
C/[sqrt 3*0.5654)=Ln (0;.87)
C=- 0.1364 0.864
voids e / C=Mass M
26.5%/ 0.1364=1.94~2=dM/dt
S[=]m/kg
C[=] kg/m
S=1/C
CS=1
C*0.87=1
C=1.151/SA ]*t=1
t=SA
1/sqrt 3=A (0.87)
A=1/2
A=0.7854 x L x H =1/sqrt Pi
1/2=1/sqrt Pi
0.884~0.885 PERMEABILITY
Figure 12: Gap Graded Soil
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Figure 13: Consolidation vs. Depth {Geotechnical Engineering Handbook, Das]
Z=0.86
T=0.265% U=36%
Permittivity =0.885Permittivity * U=0.885*3.6=31.8 Hz=freq=1/Period T
The universe is compressed to match the human perception which evolved to be so.
T=1.781-0.933 Ln (100-U)
T=R *Moment T= cuz *{1-sin 1] =0.865
T=Pi/4 U^2 =Pi/4 (036%)^2 = 0.1018
1-1018=c^2
1-T=c^2
MAXIMUM COMPACTION
In the sin and cos curves, they meet at 1 rad or 0.1 =w/S
Max compaction =26.5%
So, 0.8415%*26.5%=0.3149=Pi Energy Density
Figure 14: Consolidation
From Accounting, we have the formula
A-L=O.E.
A=L+O.E.
O.E.=A-L
A=PQ
O.E.=PQ-L
O.E. +i=PQ-L +i
O.E.=PQ-L +i]/i
O.E.=PQ/i=net sales/i
O.E. (cap rate)=PQ +L
From Soil Mechanics
e^2/(1+e) where e=max compression=26.%
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(0.26)^2/(1+0.26) =0.0537
0.0537=PQ-L]/O/E.
5.37% O.E.=PQ-L
=net sales
O.E.=net sales/ 5.37%
O.E.=13.64 (net sales)
sin 60 degrees =0.864 = net sales
Now i Eigenvector= (0.26)^3/[1.26) =1.395~1.4
40% compression cf Re=403
E.V=1 .395=Re=Inertia/ Viscous
1.395/0.403=0.3136
1/0.3136=0.6538
Ratio of Sphere to Area = s^1/3=6D=6*2r
2R*6=4/3R'12R-4/3R=0
R=32/3R =10.666R
Ln 10.666- Ln R=-2.5671
R=0.6554 cf 0.6538
ds/dt=s^2/2=0.6538
s^2=1.3076
s=1.1435 ~115
PQ/11435=(0.221)(1.580)/0.1.1435=0.3999~0.4
=40% compression=Re
1/Compression=251=T
sin 60 degress = 0.866=net sales
1/sin 60=1.1547 1+0.1546/2=1+0.0774
8 % debt
8% equity
d3y/dt3=40& comptesssion d2y/dt2=0.1395y
dy/dt=0.1395/2 y^2 y=4.6375
e^3/(1+e)=6.94 7-7e-e^3=0
e=21.6 (0.618)=1.335=s
s=1.1s^2 c=1.0493=5% 0.0537(0.05494)-26.5% = MAX COMPRESSSION
CLAY CURVE [R. F. Craig Soil Mechanics]
The universe lies between Low and High Plasticity Clay.
This then is the diagram of the universe. As the dry universe became more and more compacted, the dry
density went up and the water content went down until the length of the hypotenuse(Pythagoras) equals
base e 2.718
CUSACK’S FINAL 13 EQUATIONS:
[e/10]^2+ [1-sin t]^2=0.884^2=PERMITTIVITY
0.1580^2+0.221^2=2.67^2=F=BALANCED FORCE
(0.221+0.05)/10)^2+0.1580)^2= 0.4277^2=cuz=RESISTANCE TO MASS
Sqrt 2^2+(1-SIN 1)^2=4.486^2=VISIBLE MASS
(0.252*0.271)^2+(sin 1)^2=118 CHEMICAL ELEMENTS
(0.271)^2+(0.8415)^2=0.8415^2=a=v FORCE AND MOMENTUM
(0.271/10)^2+(1-sin 1)^2=0.397^2=t TIME
(1-sin 1)^2=251^2=T=PERIOD
(0.217+0.05)^2+(1-SIN 1)^2=1.2448^2 = E min=MINIMUM ENERGY
(0.221)^^2 +(0.8415)^2=13^2 = SPACE
(1-sin1)^2(10/81)^2 =0.858=ENERGY
{[(0,396)0.271)]^2]}/ (1/SIN 1))^2=C^2 SPEED OF LIGHT
( 0.221)^2+(0.1580)^2=2.718^2=e=y=y'=CUSACK UNIVERSAL EQUATION
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Figure 15: Dry Density Curves
G=2.69 ZERO AIR VOIDS.
q=Q/B 26.667=Q/0.525
Q=s=0.1334
THE FOUR PILLARS UPON WHICH THE EARTH SITS IS:
Q=qA+2Pi r f D
0.1334=26.667 (1)+2Pi(1)fD
26.5=Compression=2Pi fD
fD=0.4223~R=uz
Four pillars
So, 4 * 0.4233=1.6392
59.2 kN/ degree C*300 degreesC=1.77=sqrt Pi
1/ delta F=sqrt Pi
y=Ln F y'=1/F y'=0.3046~305
y=Ln (F)=Ln (26.667)=3.28334
y'=1/F=3046
y'=sin^-1(0.3046)=1.77=Ln f
y'=sin (Ln F)
theta=188.12 degrees
cos (188.12)=0.99~1
A right triangle hypotenuse 1/c^2 , leg =304.6
cos theta= 0,3046/ 1/c^2 =-29.25 degrees
friction angle
Pile force=F cos theta=26.667 cos 188=23.2667
Ln 23.2667=Pi=P
1/0.1127=0.887=PERMITIVITY
f fricton= Nx
So
Nx(d)/4=0.4223
N=1.0435=59./79degrees~60 degrees
sin 59.79 degrees=0.8462
Force=gamma b cos theta
26.667=(0.127)(0.525)cos theta
F=G cos theta
c=SPEED OF LIGHT = c
F=Gc
cos 66.4 degrees=3/ hyp
hyp=s=0.1334
Figure 16: Retaining Wall Failure Plane
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F=26.667~26.5 (recall) sin60 degrees=0.866 sin 57.29 degrees=0.8415
F sin theta=222.9
t^2-t-1=0 (0.229)^2-(0.229)-1=1/0.2016~201.8=Y
So God first created Energy, then time, then space
dE/dt
s=|E||t| cos theta
1/2=(2t-1)(t)=cos theta = Light
theta = 60 degrees, the failure or "flow plane."
Figure 17: Passive Moment on Retaining Wall
PLASTICITY INDEX
PI =40 [B. M. DAS GEOTECHNICAL ENGINEERING HANDBOOK]
ACTIVITY OF CLAY
A=PI/ %CLAY
ACTIVE CLAY A=0.75
0.75=40/%CLAY
=53.33%
P.E..=1.22.1=45.3%
K.E. =54.75-53.33=1.42 1420 0.858
E=Mc^2
0.858=M*9
M=10.48
10.48-4.486=6
6=6D
d=1
r=1/2=t Friday, 5 June 2015
SOIL PHASE AND THE PERMITTIVITY EQUATIONS
w=Se/ Gs
Gs=F=2.667
e=26.6% (recall)
w/S=0.1=10%
s=|E||t|cos 60
=1/2
t^2-t-1=0
2.24^2-2.24-1=0.0200
delta s/delta t=1/2]/ [0.020]=0.0999=0.1=10%
So, Mw/S=delta s/delta tM=permittivity /2
Permittivity=2Mw/S
Permittivity=c^2
Permittivity=Eigenvalue^2
Sqrt Permittivity=c
Sqrt c=Eigenvector
Sqrt Eigen value=Eigenvector
Eigen vector=1.73~0.175=10 degrees=w/S
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Eigenvector=w/S=water content %/ Saturation %
0.173=w/S
w=17.3%
w/n=0.173/0.265=0.6604
n/w=1.51 ~1.5
delta s/ delta t=c
Permittivity= 2Mc
K.E.=1/2Mc^2
Permittivity/K.E.=4/c=s space
So s=|E||t|cos theta
|E||t|cos theta= 4/c
c=4/ (1)(1)*cos theta
3=4/cos theta
cos theta=0.1334
theta=1.437 rads
1437-1=0.856=P.E.
P.E.=Mgh
0.856=M a(0.86)
Ma=0.9957~1
F=1 (recall)
Since F=0.2667 or 0.265 26.5%
c/26.5=3/26.5=1/ Permittivity
THE SPEED OF LIGHT IS ACTUALLY THE INVERSE OF THE PERMITTIVITY IF F=1
PERMITIVITY *EIGENVECTOR=0.885*1.73=1.531 1531 0.8469=sin 1
EIGENVECTOR=sin t/ Permittivity
EIGENVECTOR^2=c
c=( sin t)^2/ Permittivity ^2
c^2=sin t/ Permittivity
cos 60 (1-sin 1)=0.4330> 0.4233=cuz
Resistance to Mass formation is overcome.
26/667 *cos 60*K=0334 (0.866)=0.885=Permittivity
2-sin 1=Permittivity
dM/dt-F=Permittivity
dM/dt-Ma=Permittivity
Integrate
M-M^2/2 *a'=Permittivity
M^2-2/C1M-2/C1 Permittivity=0
M^2-1/2M-0.44=0
QUADRATIC
M=0.606, 6.85
RMS=2.708=base e
Figure 18: Time Required For Consolidation
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TIME REQUIRED FOR CONSOLIDATION:
(1+t)^11=0.8411
Ln (1+t)=Ln (0,8411)/11
t=0.1560 *2Pi of a cycle= 0.098~1
And,
(1+t)^11=26.5% Compression
11*Ln (1+t)=Ln 0.265
t=0.886~ PERMITIVITY
K+qH 0.44+0.42(PI%/100)+ qH
PI=40% H=0.86 K+qh =0.608+1.529
=0.1589 cf 0.1585 =1-sin 1
RETAING WALLS IN SOIL MECHANICS:
Ka=[1-sin theta]/[1+sin theta]
But Ka=p/(gamma h)
1-sin theta=cos theta= speed of light
1+sin theta = Moment = Fs
So,
F/s^2)/ (gamma)*=cos theta
26.667 (0.1334)]=[0.127]=cos theta
cos theat = 0.1574
theta =80.9 ~81 THE UNIVERSAL NUMBER
80.9 DEGREES = SQRT 2 RADS
Figure 19: Pile Bearing Capacity
Volume of Universe=4/3Piabc=4/3Pi *3*24*66=19905
19905*0.127 /4 piles supporting universe=632 load /pile
Bearing Capacity of toe:
q=0.5 Nq *tan phi =0.5 (100)(tan 35.5 degrees) [Nq comes from graph in Hicks IBID]
=35.66
Ff=632-35.66=596.78~600=6D
Ff=N bar/50
596.78*50=29839 tons / ft ^2
x 2000 lbs/ton / 2.2 lbs/kg=271057=base e
271057/271828=1.0028
26.667/1.0028=265=void ratio
2./71828/26.667=0.1019
1-0.1019=8.981=c^2
K 0=1-sin Phi =Moment
=0.8415
K = sigma h / sigma 0 =F/ (E/S.A. of ellipsoid) =1=M (Pi.e) =M=1.1575
0.865
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M4.486/0.865=0.257~26% Compression
0.864-0.885=0.0006
ratio SphereVolume to Ara=6D=6 (2R)=12r
12R=0.006
R=2=E (30-60-90 triangle)
R=
26.666 *(.8415)=Pi/A
1/A=7.1430
A=0.14 0.86
Now,
0.8415=M (0.8415)/(Pi/10312 =2.71412
2.7412/2.71828=1.008 (Hydrogen M)
MASS GAP:
dtheta/dt=M theta
Integral dtheta/dt dt = Integral M theta
theta = M theta ^2/2
2 theta=M
theta=Pi/2
But we know theta=60 degrees as retaining wall is at that angle where passive resistance balances active
pressure.
So theta =Pi/2*1/1.5=Pi/3=60 degrees
M theta=F=ks=omega R
4.486*(1.5) *(Pi/3)=7.04
1/7.04=0.1419=[1-sin phi]=K
1-0.1419=0.858=E=sin (1)=sin t
There is no Mass M until the minimum Energy reaches this value. thus the Mass gap.
ASIDE: R=cuz=Pi-e
M-(2.7412/1.008 =)=cuz
M=Area of circle=Pi (R^2)
R=t=1
M=Pi
So R=cuz-=M-F/S.A.
Pi-26.667/ (10312) S.A. of ellipsoid 1 x 8 x 22 = 3 x 24 x 66 LY^3
Figure 20: Shallow Footing Bearing Capacity
BEARING CAPACITY FOR SHALLOW FOOTING:
q=cN c +gamma D N q+ 1/2 Gamma B N gamma
q=26.5 c=0 gamma=0.127 D=1.8638
N gamma=4Pp/[gamma * B^2] cos (30-30)
=19.1496 /B^2
Q=gamma D 26.667=0.127 D
D=209.9
209/9/0.4233=201.6=YSolving for B=0.0525
5 1/4 =21/4=7*3/4
Jesus / Paul = 1/3
7/4=0.175=35/2=Holy Spirit /2
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Jesus / Paul =H.S /2
2 Jesus/ HS=Paul
2 God / God = God
[FOUNDATION ENGINEERING, P. C Varghese]
ULTIMATE BEARING CAPACITY
q=cN + gamma (D N d)+ 0.5 B gamma N gamma
=0.127 (0.1334)(50)+0.5(1)(0.127) (50) 402.2=Re
SETTLEMENT
N=40 B1 delta=8.1 or 81
1/81=0.01234567/m
(1^2+8^2) * 22^2=2.249 x 3^3=609.35*891=0.1334=s
MASS GAP BY BOUSSINESQ'S EQUATION FROM SOIL MECHANICS
sigma= 3/2 *[Q/Pi][z^3/R^3]
Let Q=F, and Lambda = frequency
sigma=1.5 F Lambda
Differentiate
1.5 d /dt (dF/ =1.5 dM/dt* d2a/dt2
=1.5 (2)(0.8415)
=252=T=Period
sigma=T
F/A=T=252
26.667/A=252
A=0.1058
But for he ellipsoid universe 3 x 8 x 66
A=0.78 x L x H =0.7854 x 3 x 24 =56.54
1/A=sqrt Pi
sqrt Pi/ 0.1058=1/6=5.98~6
Volume of Sphere / Area of Circle=6
UNIVERSAL COEFFICIENT OF THERMAL EXPANSION
alpha=1/V (dV/dt)
=sqrt 3 (E/e)
=0.5467 cf SILVER =054
(0.8456)^2=1.0875~1.09
9%=Water
13012/0.919=1431 0857
Tension between expansion and contraction:
SILVER/WATER=Volumetric/ Linear =54/9=6 6D=Vol/Area
0.5467/0.0875=6.248~2Pi =Circumference
Figure 21: Embankment Moment at toe of Slope
FROST HEAVE
ds/dt=0.87=E rate of frost penetration
dh/dt=0.175=1 degree
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Temperature drop=T=2.718
dT/dt=e
ds/dt=E
MFL> internal angle of friction
(4.486)(a)(L=sin theta
a=0.8415
L=4.359
1/L=t=2.24
E=4. 532/c^2=0.5=s
Figure 22: Earthquake Force Contours on Embankment
W=rho * Volume
=0,127 *13012 =1652
c u/ sigma'=0.22 for clays
c u = 0.22(1.17) =2574
Wd=cu 1652 d =0.2574
d=0.1558
0.8442
sin^-1 (0.8442)= 57.59 degrees ~1 radian
t^2-t-1=0.26667
t^2-t-0.7333=0
t= 1.4916, 0.4916
RMS= 1
Figure 23 Retaining Wall Soil Wedge
K a c = 0.333
K a cos delta
from graph
0.6=1/3 cos delta
delta=78.46 degrees
cos delta= 26.667/ P
cos 78.64 degrees /26.666= P
P=0,1334=s= space
The weight of the universal clay forces space to form.
sigma a '=gamma z Ka
K a= cos alpha { cos alpha -sqrt [cos ^2 alpha - cos ^2 phi]/[ cos alpha + sqrt { cos ^2 Aalpha - cos ^2
phi]
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=1* 1-sqrt 0.5]/ [1+sqrt 0.25]
=0.5/1.5
=0.3333
sigma a '=(0.127)(0.8415)( 0.333)=0.356 1/0.356=28.70
Ln 28.70= 3.3348
1/Ln 28.07=2.999=c
c u/ sigma a '=0.22
c u= (0.22)( 2.747)=0.0604
[SOIL MECHANICS, R F CRAIG]
sigma1 ' / sigma 2 '=[tan ^2 (45+phi/2)][1-dv/dx]
sqrt 3 (0.8415) =1.4575
0.854
(0.8542 (26.667)=22.78
1/22.78=0.0439~0.44% (recall)
22.78/26.667=0.8542
58.67 degrees
cos 58.67degrees=0.5199~0.52
K.E.=1/2 Mv^2 =1/2(56)(0.8415)^2 =19827 ~196 Infinity
Figure 24: Retaining Wall Rate of Failure
0.127(0.86)=0.1092
gamma H=(H _+ dH/dt)(F+dF/dt) 0.1092=(0.8415)(26.667) =2.224
t=2.24
t^2-t-1=1.77=sqrt Pi
sqrt Pi/ 0.1092=0.1623~1.618
w/C= 22.1% /36%=1629 Gs= gamma d / gamma w
2.69=gamma d/ 0.221
gamma d=0.5945~0,5959 (recall)
=0.6D =dVol/ of sphered Area of circle
Now 0.86/sqrt 3=201.4~201.8=Y=e^-t (cos 2Pit)
P=M v =4.486(0.86)=1/26.5=1/n
n^3/[1+n]=1/253=1/ T=t
F=Ma F=(5.6)( 0.8415)=4.7125
1/F=2.122
2.122*2=0.4233=cuz
This is the essence of my entire theory.
3.618=2+1.618
=dM/dt +t
dE/dc^2+1/E 2.25+1/E =t+t=2t
t=1.809~1.81 ( Recall )
ASIDE:
E=Mc^2 E/c^2=M
dc/dt*dE/dt=dM/dt
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2/c *2t-1=2
t=9.4=2.25
Figure 25: Embankment Slump Failure
The Universal Earthquake was a 6.309 according to my calculations.
So, 26.667/6.309=4.2284.233=cuz=Magnitude Scaling factor
The Energy dissipated for a 6.309 earthquake =0.1834
The Seismic Moment = 0.3585
v=d/t
t=d/v
t=0.1334/0.8415 =0.1585 =-sn 1 =MOMENT
Figure 26: Tunnel Formula Figure 27: Flattened Cigar Universal Figure 28: Ground Water Table
Ellipsoid
As the silver cigar of the universe descended into the universal fluid, when the depth and thus pressure
became just right, the silver liquified. this liquification lead to a universal earthquake precipitated. the
shear failure occurs at the reynolds number 403 at a distance or frequency =31.8. mass is wave forms that
fit through the screens to allow the human mid which is tuned to 31.8 hz to pick up on the signal.
PORE WATER PRESSURE
r=u/[gamma/h]
0.85=u/[ 0.127/0.1334]
u=0.805~81
1/81 =0.01234567 is the resonant frequency that leads to failure in the universal silver.
Figure 29: Water Migration
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LIQUIFICATION
F/A=26.667 / (Pi (R)^2=0.8488
sin 8488=58 degrees
Minuscus
58 degrees *0.75=43.56
1/F=0.23
Ln 23=1/0.85
Figure 30: Saturated Sand Figure 31: Water Drawdown
k=q[Ln(r2/r1]/[Pi (h1^2-h2^2] k=0.885 r1/r2=Pi h1=0 h2=Pi
q=0.862
or
k=1.22q/[sw ho]
=0.8415=sin 1
[ELEMENTS OF SOIL MECHANICS 6TH EDITION G.N. SMITH]
Ln (0.4233 )=-0.8587
Sin^-1(0.8597)=59.28 degrees ~60 degrees
PI=0.73(LL -20%) =40%=0.73(LL-20%) LL=74.79% GAP GRADED UNIVERSAL SOIL
Now SATURATION RATE
Gamma= gamma w[ Gs +e Sr]/[1+e
Sr=0.127 =gamma
Desity=saturation rate
0.125/0.0127=10
S=1/2.781=0.3679
Ln (0.3679)=1 Derivative and algebra, 1/e^E=Gs
1/x=Gs’
E^E=Gs
Ln Gs=2.06
Finally, from Earthquake Magnitudes:
M=Fs=mew AD F=Mew s’ ’26.667=0.4233 x
X=62.99
1/63=0.1587=1-sin 1=Moment
R/[1-sin 1]=1/M -{1-sin 60 degres}/{1-sin 1 rad}
R=1/M-dVol/dArea
R/RM =1/M -1/M - dVol/dArea
1/M=1/M - dVol /dArea
dVol/dArea=0
Vol =Area
4/3 Pi R^3=Pi R^2
R=3/4=0.75
Vol=sqrt Pi
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=sqrt F
F^2=M^2 * a^2= K.E.=0.,5959=F^2
K.E.=F^2
(0.2667)^2=1-0.89=0.86
Posted by Paul T E Cusack at 18:15 No comments: Email ThisBlogThis!Share to TwitterShare to
FacebookShare to Pinterest Since F=Pi
1/F-(gamma d^2 - w^2)=R=cuz
R=1/F -G s
R=1/M {1-sin 1}- { 1-sin 60 degrees}
RM=1-sin 1
cuz(M)=Moment
cuz=d
1-sin 1=M cuz
HYDROMETER
This is an important piece of apparatus from Soil Mechanics
.[B. M Das SOIL MECHANICS LABORATORY MANUAL]
L=47.34-{47.34-22.4}/[0.127]* hydrometer reading
47.34=149.0x hyd reading
hydro reading = 31.77~ 31.8 =1/ Pi= frequency
One of the most important equation in Soil Mechanics is
c u/ sigma '=0.22
This is the 22.1% finer that lubricates the embankment failure that is our universe. How does this figure
increase until failure? It may be some kind of mechanical weathering (friction breakdown).
D=0.04 from hydrometer log scale 20% finer.
Vol/Area=0.6 D=0.6(0.04)=0.24
=[4/3 Pi R^3]/ Pi R^2
=0.2667 R =F*R =moment
F=26.667 R=1
1-20%=80%
80%*4.486=5.6 Dark Matter
K.E.=1/2Mv^2=1/2 (5.6)(0.8415)^2
=1`.9827 ~2
P.E=Mc^2 5.6(9) =5.04 ~5
2^2+5^2]/base e=1.98~2
1/1.98=0.5 ~2
dVol/ Area'=2PiR/2Pi R=1
=2R
2R=0.6D =0.24
R=12 =|Dm|
Vol=4/3 Pi 12^3=7.238 1382 0.862
Area=Pi (12)^2=45.23
Vol/A'=0.016
A'/Vol-=62.5 (recall)
Ln (|Dm|/T/E.)=Pi
HYDROMETEER
G s= gamma d/ gamma w
2For clay,
2.69=0.127/ gam,ma w
gamma w=0.0472
K.E.=1/2 Mv^2 1/2 (4.486)(0.8415)^2 =0.5959~0.6
0.5959/0.0472=12.65=100% +26.5%
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0.5959/0.265=2.2487
t^2-t-1=1.0808~1,.81 (recall)
v=[gamma s - gamma w / [18 {Re} ]*D^2
0.8415= 0.127-0.0472/[18(0.403) ]*D^2
D^2=71.8 ~72
D=8.474 cf 0.8415
1/D=118 elements
Area=Pi R^2=Pi *(D/2)^2
=112.7~113
1/A=0.886=PERMITIVITY
0.06=0.6D
D=0.1
D=A sqrt L/sqrt tt minutes)=0.1
sqrt L=4.7239
dP/dt=dM/dt L=2(v)=2(L/sqrt1)=0.4233
2117*2=0.4234 ~0.4233
The rate of change of the momentum = R=cuz
dP/dt=cuz
So where does the Universal Force come from?
Of course,
F=gamma H
=0.127 (86)=10.9122
26.667/10.92=2.4416
t^2-t-1=0
(2.44)^2-2.44-1=251.97=Period T
BLACK HOLES ARE THE PORTALTO THE SPIRITUAL WORLD.
54.82 Light E 54.3 Mass E
109.12 *c^2=87.5
Figure 32: Water Drawdown inside and outside universe
[B M DAS PRI NCIPLES OF GEOTECHNICAL ENGINEERING]
c u/ sigma ' =0.22
sigma'=0.22/c u minimum=0.22/ 1.618
=0.864
N=1.0435
gamma=G s w/[1+w *Gs]
0.127=2.70 w/[1+0.211* 2.7]
w=13.32=s
w= 1-sin 1=Moment
The Passive force K acts at- 9.568 degrees from horizontal.
F-Compression =26.667-26.5= -0.167
cos^-1(0.167)=80.38~81
cos (^-1)delta F)=81
delta F~10 degrees
1/ delta F=1.77=sqrt Pi
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Figure 33: Vane Shear Test [DAS PRI NCIPLES OF GEOTECHNICAL ENGINEERING]
BENDING MOMENT DIAGRAM
INTEGRAL [1-sin 1] dL
L-cos 1= 1-cos 1= 0.1585=1/p=P max
[k Terzaghi, 2 ed]
p=gamma *z* 1/N
p=(0.127)(0.86)(1/sqrt 3)
1/p=0.1585 0.8415=sin1
So,
CUSACK RANKIN BACKFILL PRESSURE EQUATION:
1/ [1-sin1]=gmma (z)1/N
=1/ Moment
C0ontinuing.
INTEGRAL P =1/2 gamma (H^2) 1/N
INTEGRAL P= 0.0271
INTEGRAL P=base e
Derivative
Total Pressure
P= e^x
cuz=Pi -P
Pi=F/A
A=8377
R=5.164
Figure 34: Black Holes and the Cathode Model
BLACK HOLES
Light can't escape from a black hole so long as there it is a cathode.
v=D SE/[4PinL]
0.8415=D(0,866)(0.866)/[4Pi(3.36) (12]
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D=dielectric=0.001759~176=t=1 rad
frm above recall:
0.59kN/(m^2 degrees C)*300 degrees=590
26.667/590=452=P.E
.0.001759/88.5=500=1/2=s
s=1/2
dtheta/dt=M+c/|det|A-lambda I|
c=|D|[Pi/6 -4.486/10]
COEFFICIENT OF PERMEABILITY
{B M Das]
x2^2-x1^2/t=2k/nS(h-hc)
(x2^2-x1^2)/t=2 (0.4233)/[0.26/(1.26)(100%) *(0.866)
(2x^2) =3.55(0.396)=1.4058~1.41sqrt 2
x=0.8384
0.8384/18=0.0466
46.6%=P.E.
1=0.466=53.42%=K.E.
P.E.=46.6% x 0.858=0.3996~0.4=t
Re=vDrho/mew [Advanced Soil Mechanics, B. M. Das]
0.403=(0.8415)(12.666)(0.127)/mew
mew=0.2979 coefficient of viscosity.
1/mew=3.3569m*sec/gm (recall)
KELVIN TEMPERATURE
273-300 DEGREES =26 DEGREES CELCIUS
26 DEGREES/26 % COMPACTION MAX.=1/100
n=0.01
k=e^3/[1+e] 0.01^3/1.01
9.901
1/k=spring constant=101
Ln 101=86.17
0.862
26 degrees/Vol Ellipsoid = 26/19860=0.1309 0.879=E rho Energy density
E=Mc^2
0.879/c^2=26.94=RM=0.4233 *4.486
CLAY TEMPERATURE
Change in shear with temperature
dSu/dT=0.59 kN/(m^2 degree celcius) [B/. M. Das]
F=1.96-0.59=1.370 0.1370 0.863
Temperature cooling =K=e^-t=0.863
273 K -86.3=186.7
=Ln (0.863)=0.1473 0.853
delta H=dK/dt=e^-t
e^-t= 186.7
s= t=5.2295 rads = 2.9962 degrees = c
As the universe cooled the shear plane in the universal kaolinite became at the point of rupture on the
360-299.62=60.38 degrees~60 degrees.
PERMEABILITY IN SAND
k=e^3/[1+e]
k=25%/[1.26] =0.01395
q=kiA 1.334=0.01395(c)(A
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A=PiR^2
R^2=1.01
R=1=t=E=1/t=1/1
The P.E. for the coffer dam is:
P.E.=mgh=Mah =4.486*2(0.8415)(0.866) =6.53 1/P.E=0.1529=K.E.=1/2 M/c^2
K.E.=1/2 (2^4.86)(9 =40.3=0.403=Re
So where forward meetsbackwards:
0.866=0.668
s=0.1333=0.3331
0.668+0.3331=0.99=HOLY SPIRIT
9*9*9=729=73 =JESUS
0.1372 -1=0.863
363
1/0.363=002717=e^1
Figure 35: Hydrometer Test Table [Das Soil Mechanics Laboratory Manual]
UNIVERSAL SOIL MECHANICS
COMPRESSIBILITY INDEX=C=0.4233
C=0.007(Lw_10%)
0.4233=0.007(Lw-10%)
Lw=50.47% of dry weight
M=4.486* (1/0.5)=8,9720=c^2
c=2.9953 E=c^2 *c^2=c^4=81
t=1/E=1/81=0.012345679
According to the preeminent soil mechanics, the maximum compressibility is 26%
n=e/1-e
=0.26/1.26 =206
E=Mc^2
=4.486(8.9893)
=T
SOIL MECHANICS HYDRAULICS
ip=gamma w*h/l=2.6(1)=2.6 v=K/n *ip =0.8415 =K *2.6/0.6412
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Aside Re=0.402=mew/[rho *v*L]
Re=I.F./ V/F.=0.402(0.127)(0.8415)(6)
mew=0,2578/402=0.6412
K=207.5
K/k0=207.5/0.885=1/ 0.4265~1/R=1/cuz
K=207.5/cuz=49
F=ks=49Pi
0.1539 0.8461
E=0.8461=F/sec= F=0.335
1/F=c=2.98~3 VELOCITY THROUGH VOIDS IS THE SPEED OF LIGHT.
POROSITY:
[Advanced Soil Mechanics B. Das]
So I end it off where I began, with soil mechanics. "God made man from dust." Actually He made Man
from the universe and the universe is made from dust.
s=1.334 s^1/3=1.1006=1/c^2=t
F=26.667=ks
=26.667=k(1.1006) k=24.229
k/k0=k/R=k/cuz=57.29
=57.29 degrees = 1 radian
So when E reaches 0.8415, the force squeezed out all the voids and the universe is buoyant. This occurs
where F=P or a=v=0.8415
CONCLUSION
We see that Soli Mechanics or Soil Physics plays a large part in helping Engineers and Scientist
understand the Universe.
REFERENCES
1. Craig, R. F., Soil Mechanics, CRC Press, New York,. 2002.
2. Das, B. M. Advanced Soil Mechanics.,Taylor & Francis, New York,. 2008.
3. Soil Mechanics Laboratory Manual , 8 ed. Oxford U.P. New York,. 2002
4. Principles of Geotechnical Engineering. Centgage Learning., 2009.
5. Geotechnical Engineering Handbook,J. Ross Pub., USA, 201
6. Day, R. W. ,,Geotechnical Engineering Earthquake Handbook., McGraw Hill USA, 200
7. Hicks., T.G. Handbook of Civil Engineering Calculations, McGraw Hill, New York., 2000
8. Smith, G. N., Elements of Soil Mechanics 6TH Ed., Oxford, England, 1991
9. Terzaghi, K. , Soil Mechanics in Engineering Practice,Wiley., 1948.
10. Varghese, P. C., Foundation EngineeringPHI., New Delhi, 2005
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