Post on 05-May-2023
One dimensional elastic rod
β’ ππ!" = #""
with #"#$= π
β’ Elastic constitutive law:β’ πΌπ!" = π (elastic stiffness πΌ = 200 [GPa] )β’ Assuming πΌ is constant, the above constitutive
law can be expressed as:
β’ ππ!" = %πΌππ
One dimensional elastic rod
β’ ππ!" = #""
with #"#$= π
β’ ππ!" = %πΌππ
β’ π = π' + ππ‘
β’ %πΌππ = #"
"β β«'
( %πΌππ = β«"!
" #""β %
πΌπ = ln "
"!
π = πΌ lnππ!
One dimensional elastic rodππ!" = #"
"with #"
#$= π
ππ!" = πΌ ππβ’ π = π' + ππ‘
1πΌππ =
πππ‘π' + ππ‘
β 6'
( 1πΌππ = π6
'
$ ππ‘π' + ππ‘
β1πΌπ = ln π' + ππ‘ β ln π'
π = πΌ lnπ! + ππ‘π!
Confirm following results:β’ π' = 20 ππ
β’ #"#$= 0.001 ππ
β’ πΌ = 200 πΊππ (equivalently, 200000 πππ)β’ Loading for 30 seconds
Now that you have stress, you could estimate the force as well.
β’ Letβs say, the thickness is 1 [mm] and the width is 12.5 [mm]; The cross-sectional area amounts to 12.5 [mm^2]
πΉ = π΄ β π = π΄ πΌ lnπ! + ππ‘π!
π΄ β π = ππ(πππ = 10)*π ( 10+ππ( = π
One dimensional Newtonian rod
β’ ππ = #""
with #"#$= π
β’ Newtonian fluidβs constitutive law:
β’ π #)#$= π
β’ Assuming π is constant (Newtonian fluid), the above constitutive law can be expressed as:
β’ π #)#$= π β π
"###$= π β *
"#"#$= π
L vs π?
π£ = cmms
ππ£ππ‘ = 0
π π‘ = 0 = π!
The cross-sectional area amounts to 12.5 [mm^2]
One dimensional Newtonian rod
ππππππ‘= π
ππ' + ππ‘
π = π
WaterβsviscosityisknownasΞ· = 1.3Γ10"# [Pa β s] at 10Β°πΆππππππ‘ =
ππ β π ππ
πππ = ππ
Newtonian fluid (water) under tensionπ! = 5. ππππππ‘ = 0.1
πππ ππ
π = 1.3Γ10"# ππ β π
The cross-sectional area amounts to 12.5 [mm^2]
Newtonian fluid (water) under compression
π! = 5. ππππππ‘ = β0.1
πππ ππ
π = 1.3Γ10"# ππ β π
Letβs compare two different speed of compression.
π! = 50 πππ = 1.3Γ10"# ππ β π
ππππ‘ = β0.1
πππ ππ
ππππ‘ = β1.0
πππ ππ
A reasonable consideration (*VERY* phenomenological constitutive model)
Newtonian fluidβs constitutive law:π "#"$= π
non-Newtonian fluid
π(π)ππππ‘= π
withπ π = π! + πΌπ
Continuedβ’ ππ = #"
"with #"
#$= π
πππππ‘= π β π' + πΌπ
πππππ‘
= π
βπ!πππππ‘+πΌππππππ‘= π β
π!ππ+πΌπππ= π
βπ!ππ+πΌπππ= π β
π!ππ= π 1 β
πΌππ
βπ!ππ β πΌπ
= πOkay, analytical solution was easily found.
A reasonable consideration (*quite* demanding phenomenological
constitutive model)
Newtonian fluidβs constitutive law:
πππππ‘= π
non-Newtonian fluid
π(π)ππππ‘= π
with a viscosity that is exponentially related with stress?
π π = π! expππΌ
Continuedβ’ ππ = #"
"with #"
#$= π
π πππππ‘= π β π' exp
ππΌ
πππππ‘
= π
I gave up looking for the analytical solution of π(π)β¦
Newton-Raphson method
Howtosolve?
πP expQR
STSU = π
Ourgoalistofindacertainπ thatgivesπ π = 0
Example NRβ’ π π₯ = π¦ = cos π₯ + 3 sin π₯ + 3 exp(β2π₯) = 0μ ν΄(μ¦, y=0 μΌλμ xκ°)λ₯Ό μ°Ύμ보μ.
ν΄: y=0 λ§μ‘±νλμ μμμ x μ’ν
Newton Raphson Method - Algorithmβ’ 1. Guess x value and letβs name it as π₯! where the subscript 0 means βinitialβ.
β’ 2. Obtain new guess π₯$ by following the below tasks.
β’ Estimate π π₯! and "#"$
. In case "#"$
is a function of π₯. For the first attempt, use π₯!.
β’ Obtain the next guess π₯% by drawing a tangent line at the point of (π₯!, π π₯! ) and obtain its intercept with x-axis. You can do it by defining the line function derived from the tangent line, i.e.,
π =ππππ₯ π₯! Γ π β π₯! + π(π₯!)
Find the intercept of the line with x-axis, i.e., π¦ = 0, which gives π₯$:
0 =ππππ₯ π₯! Γ π₯$ β π₯! + π π₯! β π₯$ β π₯! = β
π π₯!ππππ₯ π₯!
β π₯$ = π₯! βπ π₯!ππππ₯ π₯!
β’ 3. We are using this intercept as the new π₯.
And repeat 2-1/2-2 steps until π π₯% β 0.
μ΄νμ΄μ§λ₯ΌνμΈνμΈμ:https://youngung.github.io/nr_example/
NR summary
β’ π₯%&' = π₯% β( )!"#"$ )!
β’ Repeat the above until π π₯% < tolerance
β’Of course, you can do it manually, step-by-step. Usually, people make computer do the repetitive and tedious tasks.
Howtosolve?
πP expQR
STSU = π
π π = π β π! Μπ expππΌ
ππ πππ = 1 β
π! ΜππΌ exp
ππΌ
LetβsconsiderSTSU is given as Μπ
Now, if you have a reasonable guess on π (say, π!), letβs estimate next guess π$ and so on.
π%&$ = π% βπ π%ππππ₯ π%
Continuedβ’ ππ = #"
"with #"
#$= π
π πππππ‘= π β π' exp
ππΌ
πππππ‘
= π
I gave up looking for the analytical solution of π(π)β¦
We might be able to use NR method to solve the above in combination with Euler method!
Euler + Newton-Raphson
πP expππΌ
πππππ‘
= π β πP expππΌ
ΞππΞπ‘
= π β 0 = π β πP expππΌ
ΞππΞπ‘
π(ijk) = π(i) + Ξπ
π‘(ijk) = π‘(i) + Ξπ‘
π‘P = 0πP = 0
Ξπ‘ is, as usual, fixed as constant
Note that &'&( = π. If we apply Euler approximation,
ΞπΞπ‘ = π β Ξπ = πΞπ‘
β 0 = π β πP expππΌ
cΞπ‘πΞπ‘
β 0 = π β πP expππΌππ
We apply the Newton-Raphson method to as below function:
β π π(%) = π(%) β π! expπ(%)πΌ
ππ(%)
βππ πππ = 1 β
π!πΌ exp
π(%)πΌ
ππ(%)
Euler + Newton-Raphson
π£ =ππππ‘ = cos(π‘) β 0.5
ππ£ππ‘
= sin π‘
π π‘ = 0 = π!
The cross-sectional area amounts to 12.5 [mm^2]
Euler + Newton-Raphson
πP expππΌ
πππππ‘
= π β πP expππΌ
ΞππΞπ‘
= π β 0 = π β πP expππΌ
ΞππΞπ‘
π(ijk) = π(i) + Ξπ
π‘(ijk) = π‘(i) + Ξπ‘
π‘P = 0πP = 0
Ξπ‘ is, as usual, fixed as constant
Note that &'&( = cos(π‘) β 0.5. If we apply Euler approximation,ΞπΞπ‘ = cos(π‘) β 0.5 β Ξπ = cos(π‘) β 0.5 Ξπ‘
β 0 = π β π! expππΌ
cos(π‘) β 0.5 Ξπ‘πΞπ‘
β 0 = π β π! expππΌcos(π‘) β 0.5
π
We apply the Newton-Raphson method to as below function:
β π π(%) = π(%) β π! expπ(%)πΌ
cos π‘(%) β 0.5π(%)
βππ πππ = 1 β
π!πΌ exp
π(%)πΌ
cos π‘(%) β 0.5π(%)