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Universal Structural Mechanics | OMICS International
ISSN: 2476-2296
Fluid Mechanics: Open Access
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Universal Structural Mechanics

Paul TE Cusack*

Independent Researcher, BSc E, DULE, 1641 Sandy Point Rd, Saint John, NB, Canada E2K 5E8, Canada

*Corresponding Author:
Cusack PTE
Independent Researcher, BSc E, DULE
1641 Sandy Point Rd, Saint John
NB, Canada E2K 5E8, Canada
Tel: (506) 214-3313
E-mail: [email protected]

Received date: May 30, 2017; Accepted date: August 16, 2017; Published date: August 24, 2017

Citation: Cusack PTE (2017) Universal Structural Mechanics. Fluid Mech Open Acc 4: 173. doi: 10.4172/2476-2296.1000173

Copyright: © 2017 Cusack PTE. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

Visit for more related articles at Fluid Mechanics: Open Access

Abstract

In this paper, we provide some interesting spot calculations from the Theory of Elastic Stability. Certain key constants of Cusack’s Astrotheology are derived from basic structural mechanics’ formula. No attempts for proofs are made. The reader is referred to Timoshenko and Gere’s classic book on stability. Proton Mass was determined to be within the margin of error.

Keywords

Elastic stability; Astrotheology; Moment; Euler’s critical load; Golden mean parabola

Introduction

Consider the beam column. Using our knowledge of Physical Constants derived in Astrotheology math, we can derive other variable, such as energy and time. The Super force is sin t. we begin with a column loaded with that Super force load. Then we move on to a beam column loaded laterally as well as axially. We end with Euler’s Critical load which allows us to derive the maximum mass in the periodic table of the elements (Figure 1).

fluid-mechanics-beam-column-showing

Figure 1: Beam column showing k.

Refer to Figure 1 showing cusack’s modulus k=cuz=0.4233.

y=sin θ

y’=-cos θ

0.4233=cuz=-cos θ

θ=64.95°=113.37 rads

=1/0.882~1/ε0

And,

θ=64.95°=1/0.1539=1/ (1-sin 1.0085 rads)

1.0085= Mass of H+

Continuing,

0-2π sin θ = |-cos θ |

-(-1) -(1)=2=-dM/dt

Δt/dt=1 rad/ 1 sec.

sin 1=0.8415=1/118=1/M (Periodic Table of the elements)

EI d³y/dx³+ P dy/dx=-V [1] pg. 2

(0.4233)I(G)+ 2.667(0.8415)=-1.617

I=0, 4936

1/I=Y=0.202

I=1/Y where Y=e-t cos (2πt) & t=1

EI d²M/dx²=-M [1].

(0.4233) (1/Y)(0.8415)=M= 1758=1.00735 ~Mass H+ (Figure 2).

fluid-mechanics-beam-column-showing

Figure 2: Beam column showing Q.

d²y/dx²+k²y=-Qc x/ [EIl] [1]

Aside:

k²=P/[EI]

=2.7667/(1/0.4233)(1/Y)

=127.6

-2.667c/[0.4233 × (1/Y) l] x

6.67 +127.6=127.6 cx/l

cx/l=522.5

From the Golden mean parabola with roots -0.618 & 1.618

(-0.618) x/l=522.5

x/l=118.3=Mass M

118.3=x/(1.618+0.618)

= x/224

x=52.91=1.083 rads

522.5=cx/l

522=c ×52.91) /224

c=221

G+ 127.6= 2.667 c x /[(0.4233)(1/Y)(l)

Y=202

6.67-1.27=2.667/ 0.4233 × x/ 0.202

2.667/0.4233=6.3=l

Y(G-ρ)=c(1-c)

Y(G-ρ)=c-c²

Golden Mean parabola

c²-c+Y(G-ρ)=0

Y(G-ρ)=1

0.202(6.67-1.27)=1

24.5-1=23.6=Ln π.

Consider,

y=Q sin k(l-c)/[P k sin (kl)] sin k(l-x) -Q (l-c)(l-x)/ Pl

Let

Q=P

k=√ρ

y=sin 59.65/[√ 127.6 ×sin 253] - 522.531

y=-0.868

y=sin 60°

Y=0.868=sin 60°

Y=√3/2

Y=t/[dM/dt]

dM/dt=t/Y

Y=E= dM/dt × t

But E=Mc²

E²/2=M t²/2

E=[√M]t

E=√4.482 × π

E=6.65~G

EI d²y/dt²=-Py [1]

y=y'

EI=P

0.4233(I)=P.

I=2.667/0.4233=6.3=1/0.1585=1/[1-sin 1]

I=1/ moment

Moment=1/I

F × d=Y=1/ 6.3=1585

2.667d=0.1585

d=0.594~0.6

Circ. =Area

2πR=πR²

R=2

Circ =2π(0.6)=1/F=Area

E=Work ×t=Fdt

F(1/F)(t)=1

t=1=E (Figure 3).

fluid-mechanics-eccentric-load-column

Figure 3: Eccentric load on column.

EI d²y/dx²=P(δ-y) [1]

(0.4233)(1/Y)(0.8415)=2.667(δ-4/3)

1.50 (δ-4/3)=1

1.5δ-2.00=1

1.5δ=3

δ=2=dM/dt

M=1.77=√π

d=1/F=Y

√π=δ-y

=δ-1/F

δ=√π+1/F

119.6 ×943]/ 127.6=88.3=ε0

χu≫1 when u=≫0

χu≫∞ when u≫π/2

Mmax=-EI y'= M0 sec u

(0.4233)(1/Y)(0.8415)=M0 sec (π/2)

M0=1.7582=1.00762 = Mass H+

M0=Fd

175=2.667(1/F)

F=6.59~1/152

F~G

Finally,

Pcritical= 4π4EI/l² [1]

2.667=4π4 =(0.4233)(1/0.202)/ l²

I=1.77=√π

=√t critical=√(1.618+0.618)=1.4953~1.50=Mass Gap.

√π=1.2533²

Rigidity:

Aside:

EI=F

(0.4233)I=2.667

I=6.3

α=2EI/l

=2(0.4233)(6.3)/(1/2)

=126~ρ

Critical Load:

ψ(u)=-3Ib/[4Il]

b=l

ψ(u)=3/4=1/s

α/ψ(u)=ρ/(1/s)=ρs=Mass M

Proton Mass: Refer to Figure 4.

fluid-mechanics-proton-mass-column

Figure 4: Proton mass column.

ρk/ EI

=127(0.4233)/2π

=53.759/2π

=938.27

=Proton Mass (Figure 4).

The rigidity of the universe is the density, and the critical load is the Mass.

EI d4y/dx4 + P d3y/dx3 + q/g d2y2/dx2=0 [1]

y=y’’ etc.

EI+P+q/G=0

EI+F=G[dM/dx].

Mass-Gravity Equation [2]

dM/dx=[EI +F]/G=[(0.4233)(6.3) + 2.6678]/ 6.67=0.799~0.8α 8

1/8=1.25=Emin

M=1/c4=1/81=0.12345679 Eight digits

Now Integrate wrt x or s:

∫dM/dx dx= 1/G [∫EI dx + ∫F dx]

M=1/G × E 4s5/5 +∫sin 60°

M=1/6.67× [(0)(0.4233)(4/3)5 /5 -cos 60°]

=0.75=3/4

M =1/s

Ms=Y

Ms=Fd×t

dM/dt ×ds/dt =dF/dt ×dd/dt+ dt/dt +C1

2(.8415)=(0.5)(0.8415)(1)+C1

1683=0.4208 +C1

C1=126=ρ

1683-42=1641

1641 + 1/c=1641 +33.3333=1674

= 1671.6+0.901

=Mp+ +e.

Conclusion

In this paper, we provided some spot calculations using formula from elasticity stability theory. The beam column and the eccentricity loaded column can be used as a model for certain universal calculations. The rigidity of the universe is the density, and the critical load is the Mass. Perhaps those interested could find more in stability theory to explain Cusack’s model of the Universe.

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