medoda diferentelor finite
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COMPLEMENTE DE TEORIA ELASTICITATII SI
PLASTICITATII
TEMA DE CASA Nr.1
METODA DIFERENTELOR FINITE
STUDENT: TANASOIU PETRE-BOGDAN
MASTER INGINERIE GEOTEHNICA, ANUL I
PROFESOR: prof.univ.dr.ing. MIRCEA IEREMIA
1
Metoda Numerica de Calcul a Diferentelor Finite
Sa se determine starea de eforturi unitare care apare in urmatoarea grinda perete si sa
se reprezinte grafic variatia pe inaltimea grinzii in diferite sectiuni a eforturilor unitare: σx, σy,
τxy.
Grinda-perete (diafragma) este din beton armat si este actionata in planul ei median de
un sistem de forte aflate in echilibru.
SCHEMA DE INCARCARE SCHEMA STATICA
Se va aborda rezolvarea numerica a problemei, discretinzandu-se domeniul grinzii-
perete, dupa care se va scrie cate o ecuatie algebrica liniara cu coeficienti constanti in fiecare
nod curent “k” al retelei de calcul folosite. In final se va rezolva un sistem algebric de 15
ecuatii cu 15 necunoscute.
Forma generala matriceala a sistemului de ecuatii algebrice este:
[A](15,15)*{F}(15,1)={B}(15,1), unde:
[A] – Matricea coeficientilor necunoscutelor (depinde de dimensiunile geometrice ale
grinzii-perete si de natura materialului din care e alcatuita grinda);
{F} – Matricea-coloana a functiilor necunoscute pe care urmeaza sa le aflam in fiecare
nod al retelei de calcul;
{B} – Matricea-coloana care depinde de modul de incarcare al grinzii-perete.
Analogia mecanica Aflarea functiei de tensiune pe conturul grinzii-perete si pe extracontur:
Contur: FK=MK , M – momentul incovoietor pe sistemul de baza static determinat;
Extracontur: FK=Fpc+2*a*N, N – forta axiala de pe sistemul de baza static determinat.
2
Diagramele de eforturi pe sistemul de baza
SISTEM DE BAZA DIAGRAMA N DIAGRAMA M
Caroiajul de calcul atasat grinzii Avand in vedere simetriile posibile, domeniul grinzii-perete poate fi discretizat astfel:
3
Definirea functiilor pe contur
F4’=-6 qa
2
F3’=-3.5 qa2
F2’=-2 qa2
F1’=-1.5 qa2
F16’=-6 qa2
F15’=-3 qa2
F14’=-0.75 qa2
F13’=0 qa2
Definirea functiilor pe extracontur
F4V=2*(-3)+F3’= -9.5 qa
2
F3V=2*(-3)+F3= F3-6 qa2
F6V=2*(-3)+F6=F6-6 qa2
F9V=2*(-3)+F9=F9-6 qa2
F12V=2*(-3)+F12=F12-6 qa2
F15V=2*(-3)+F15=F15-6 qa2
F16V=2*(-3)+F15’= -9 qa2
F40=2*(0)+F4’=-6 qa2
F30=F3 qa2
F20=F2 qa2
F10=F1 qa2
F160=2*(0)+F16’=-6 qa2
F150=F15 qa2
F140=F14 qa2
F130=F13 qa2
Reteaua fiind patratica, se va aplica molecula de calcul, in fiecare nod al
retelei:
4
Sistemul de ecuatii se obtine in felul urmator:
(impartit la qa2)
20*F1-8*(F1’+F2+F4+F2)+2(F2’+F5+ F2’+F5)+1(F1O+F3+F7+F3)=0
20*F1-8*(-1.5+2*F2+F4)+2(-2*2+2*F5)+1(F1+2*F3+F7)=0
20*F1+12-16*F2-8F4+-8+4*F5+F1+2*F3+F7=0
21*F1-16*F2+2*F3-8*F4+4*F5+F7+4.0=0
21*F1-16*F2+2*F3-8*F4+4*F5+F7=-4.0
In mod analog se vor obtine celelalte ecuatii, iar sistemul va rezulta:
21*F1-16*F2+2*F3-8*F4+4*F5+F7=-4.0
22*F2-8*F1-8*F3+2*F4-8*F5+2*F6+F8=0
F1-8*F2+22*F3+2*F5-8*F6+F9=-42.0
4*F2-8*F1+20*F4-16*F5+2*F6-8*F7+4*F8+F10=1.5
2*F1-8*F2+2*F3-8*F4+21*F5-8*F6+2*F7-8*F8+2*F9+F11=8
2*F2-8*F3+F4-8*F5+21*F6+2*F8-8*F9+F12=-14.5
F1-8*F4+4*F5+20*F7-16*F8+2*F9-8*F10+4*F11+F13 =0
F2+2*F4-8*F5+2*F6-8*F7+21*F8-8*F9+2*F10-8*F11+2*F12+F14 =6
F3+2*F5-8*F6+F7-8*F8+21*F9+2*F11-8*F12+F15=-18
F4-8*F7+4*F8+20*F10-16*F11+2*F12-8*F13+4*F14 =0
F5+2*F7-8*F8+2*F9-8*F10+21*F11-8*F12+2*F13-8*F14+2*F15 =6.75
F6+2*F8-8*F9+F10-8*F11+21*F12+2*F14-8*F15 =-15
F7-8*F10+4*F11+21*F13-16*F14+2*F15 =3.0
F8+2*F10-8*F11+2*F12-8*F13+22*F14-8*F15 =6
F9+2*F11-8*F12+F13-8*F14+22*F15=-40.5
Matricea coeficientilor va fi de forma:
A
21
8
1
8
2
0
1
0
0
0
0
0
0
0
0
16
22
8
4
8
2
0
1
0
0
0
0
0
0
0
2
8
22
0
2
8
0
0
1
0
0
0
0
0
0
8
2
0
20
8
1
8
2
0
1
0
0
0
0
0
4
8
2
16
21
8
4
8
2
0
1
0
0
0
0
0
2
8
2
8
21
0
2
8
0
0
1
0
0
0
1
0
0
8
2
0
20
8
1
8
2
0
1
0
0
0
1
0
4
8
2
16
21
8
4
8
2
0
1
0
0
0
1
0
2
8
2
8
21
0
2
8
0
0
1
0
0
0
1
0
0
8
2
0
20
8
1
8
2
0
0
0
0
0
1
0
4
8
2
16
21
8
4
8
2
0
0
0
0
0
1
0
2
8
2
8
21
0
2
8
0
0
0
0
0
0
1
0
0
8
2
0
21
8
1
0
0
0
0
0
0
0
1
0
4
8
2
16
22
8
0
0
0
0
0
0
0
0
1
0
2
8
2
8
22
5
Matricea [A] este o matrice simetrica fata de diagonala principala. Pentru a exprima
mai bine acest fapt, se vor inmulti cu 2 urmatoarele randuri:
Rand: 2, 3, 5, 6, 8, 9, 11, 12, 14,15 rezultand urmatoarea matrice:
Matricea-coloana {B} a termenilor liberi are valoarea:
x2
x2
x2
x2
x2
x2
x2
x2
x2
x2
21 -16 2 -8 4 0 1 0 0 0 0 0 0 0 0
-16 44 -16 4 -16 4 0 2 0 0 0 0 0 0 0
2 -16 44 0 4 -16 0 0 2 0 0 0 0 0 0
-8 4 0 20 -16 2 -8 4 0 1 0 0 0 0 0
4 -16 4 -16 42 -16 4 -16 4 0 2 0 0 0 0
0 4 -16 2 -16 42 0 4 -16 0 0 2 0 0 0
1 0 0 -8 4 0 20 -16 2 -8 4 0 1 0 0
A= 0 2 0 4 -16 4 -16 42 -16 4 -16 4 0 2 0
0 0 2 0 4 -16 2 -16 42 0 4 -16 0 0 2
0 0 0 1 0 0 -8 4 0 20 -16 2 -8 4 0
0 0 0 0 2 0 4 -16 4 -16 42 -16 4 -16 4
0 0 0 0 0 2 0 4 -16 2 -16 42 0 4 -16
0 0 0 0 0 0 1 0 0 -8 4 0 21 -16 2
0 0 0 0 0 0 0 2 0 4 -16 4 -16 44 -16
0 0 0 0 0 0 0 0 2 0 4 -16 2 -16 44
B
4
0
42
1.5
8
14.5
0
6
18
0
6.75
15
3
6
40.5
B
4
0
84
1.5
16
29
0
12
36
0
13.5
30
3
12
81
6
Rezolvarea sistemului se va face folosind metoda matricei inverse, si anume se va
inmulti la stanga fiecare termen cu [A]-1
.
[A]*{F}={B}
[A]-1
*[A]*{F}=[A]-1
*{B}
{F}=[A]-1
*{B}
Efectuand calculele, va rezulta matricea {F}:
x qa2
A1
0.1092
0.0556
0.0168
0.0924
0.0606
0.0214
0.0589
0.0426
0.0159
0.0297
0.0222
0.0083
0.0094
0.007
0.0024
0.0556
0.062
0.0226
0.0603
0.0551
0.0228
0.0422
0.0357
0.0148
0.022
0.0179
0.0072
0.007
0.0055
0.002
0.0168
0.0226
0.0353
0.0207
0.0223
0.0209
0.015
0.0142
0.0096
0.0078
0.0068
0.0037
0.0024
0.002
0.0009
0.0924
0.0603
0.0207
0.2134
0.1319
0.0464
0.1621
0.1138
0.0426
0.089
0.0658
0.0249
0.0297
0.022
0.0078
0.0606
0.0551
0.0223
0.1319
0.1267
0.0505
0.1136
0.0988
0.0414
0.0658
0.0545
0.0225
0.0222
0.0179
0.0068
0.0214
0.0228
0.0209
0.0464
0.0505
0.054
0.0422
0.0412
0.0313
0.0249
0.0225
0.0135
0.0083
0.0072
0.0037
0.0589
0.0422
0.015
0.1621
0.1136
0.0422
0.2521
0.1621
0.0589
0.1621
0.1136
0.0422
0.0589
0.0422
0.015
0.0426
0.0357
0.0142
0.1138
0.0988
0.0412
0.1621
0.1512
0.0611
0.1138
0.0988
0.0412
0.0426
0.0357
0.0142
0.0159
0.0148
0.0096
0.0426
0.0414
0.0313
0.0589
0.0611
0.0593
0.0426
0.0414
0.0313
0.0159
0.0148
0.0096
0.0297
0.022
0.0078
0.089
0.0658
0.0249
0.1621
0.1138
0.0426
0.2134
0.1319
0.0464
0.0924
0.0603
0.0207
0.0222
0.0179
0.0068
0.0658
0.0545
0.0225
0.1136
0.0988
0.0414
0.1319
0.1267
0.0505
0.0606
0.0551
0.0223
0.0083
0.0072
0.0037
0.0249
0.0225
0.0135
0.0422
0.0412
0.0313
0.0464
0.0505
0.054
0.0214
0.0228
0.0209
0.0094
0.007
0.0024
0.0297
0.0222
0.0083
0.0589
0.0426
0.0159
0.0924
0.0606
0.0214
0.1092
0.0556
0.0168
0.007
0.0055
0.002
0.022
0.0179
0.0072
0.0422
0.0357
0.0148
0.0603
0.0551
0.0228
0.0556
0.062
0.0226
0.0024
0.002
0.0009
0.0078
0.0068
0.0037
0.015
0.0142
0.0096
0.0207
0.0223
0.0209
0.0168
0.0226
0.0353
F
1.4523
1.9645
3.4893
1.3272
1.8668
3.4519
1.1054
1.6902
3.3802
0.7634
1.4128
3.2642
0.3296
1.0497
3.1126
7
Verificarea ecuatiilor de conditie:
[A]*{F}={B}
Ecuatia 1: (simplificand qa2)
21*F1-16*F2+2*F3-8*F4+4*F5+F7=
=21*(-1.4523) - 16*(-1.9645) + 2*(-3.4893) - 8*(-1.3272) + 4*(-1.8668) +
0*(-3.4519) + 1*(-1.1054) + 0*(-1.6902) + 0*(-3.3802) + 0*(-0.7634)+ 0*(-1.4128)+
0*(-3.2642)+ 0*(-0.3296)+ 0*(-1.0497) + 0*(-3.1126) = -3.999999999999999999882
Ecuatia 7: (simplificand qa2)
F1-8*F4+4*F5+20*F7-16*F8+2*F9-8*F10+4*F11+F13 =
=1*(-1.4523) + 0*(-1.9645)+ 0*(-3.4893) - 8*(-1.3272) + 4*(-1.8668) +
0*(-3.4519) + 20*(-1.1054) -16*(-1.6902) + 2*(-3.3802) -8*(-0.7634)+ 4*(-1.4128)+
0*(-3.2642)+1*(-0.3296)+ 0*(-1.0497) + 0*(-3.1126) = -0.007
Ecuatia 10: (simplificand qa2)
F5+2*F7-8*F8+2*F9-8*F10+21*F11-8*F12+2*F13-8*F14+2*F15 =
=0*(-1.4523) + 0*(-1.9645) + 0*(-3.4893) + 0*(-1.3272) + 1*(-1.8668) +
0*(-3.4519) + 2*(-1.1054) -8*(-1.6902) + 2*(-3.3802) -8*(-0.7634)+ 21*(-1.4128)-
8*(-3.2642)+2*(-0.3296)-8*(-1.0497) +2*(-3.1126) = 6.7488
Ecuatia 13: (simplificand qa2)
F7-8*F10+4*F11+21*F13-16*F14+2*F15 =
=0*(-1.4523) + 0*(-1.9645) + 0*(-3.4893) + 0*(-1.3272) + 0*(-1.8668) +
0*(-3.4519) + 1*(-1.1054) +0*(-1.6902) + 0*(-3.3802) -8*(-0.7634)+ 4*(-1.4128)-
0*(-3.2642)+21*(-0.3296)-16*(-1.0497) +2*(-3.1126) = 3.0
Calculul Tensiunilor in interiorul grinzii-perete Tensiuni normale : σx
(σx)k =𝐹𝑙−2𝐹𝑘+𝐹𝑗
𝛥𝑦 2 , unde: Fj, Fk, Fl sunt valorile functiilor necunoscute, luate pe verticala
: 𝛥𝑦=a
(σx)1’ =𝐹10 −2𝐹1 ′+𝐹1
𝑎2 = −1.4523 −2∗ −1.5 −1.4523 𝑞𝑎2
𝑎2 = 0.095 q
(σx)2’ =𝐹20 −2𝐹2 ′+𝐹2
𝑎2 = −1.9645−2∗ −2 −1.9645 𝑞𝑎2
𝑎2 = 0.071 q
(σx)3’ =𝐹30 −2𝐹3 ′+𝐹3
𝑎2 = −3.4893−2∗ −3.5 −3.4893 𝑞𝑎2
𝑎2 = 0.022 q
(σx)1 =𝐹1 ′−2𝐹1+𝐹4
𝑎2 = −1.5−2∗ −1.4523 − 1.3272 𝑞𝑎2
𝑎2 = 0.077 q
(σx)2 =𝐹2 ′−2𝐹2+𝐹5
𝑎2 = −2−2∗ −1.9645 −1.8668 𝑞𝑎2
𝑎2 = 0.062 q
(σx)3 =𝐹3 ′−2𝐹3+𝐹6
𝑎2 = −3.5−2∗ −3.4893 − 3.4519 𝑞𝑎2
𝑎2 = 0.027 q
8
(σx)4 =𝐹1−2𝐹4+𝐹7
𝑎2 = −1.4523 −2∗ − 1.3272 −1.1054 𝑞𝑎2
𝑎2 = 0.097 q
(σx)5 =𝐹2−2𝐹5+𝐹8
𝑎2 = −1.9645−2∗ −1.8668 −1.6902 𝑞𝑎2
𝑎2 = 0.079 q
(σx)6 =𝐹3−2𝐹6+𝐹9
𝑎2 = −3.4893−2∗ − 3.4519 −3.3802 𝑞𝑎2
𝑎2 = 0.034 q
(σx)7 =𝐹4−2𝐹7+𝐹10
𝑎2 = − 1.3272 −2∗ −1.1054 −0.7634 𝑞𝑎2
𝑎2 = 0.120 q
(σx)8 =𝐹5−2𝐹8+𝐹11
𝑎2 = −1.8668 −2∗ −1.6902 −1.4128 𝑞𝑎2
𝑎2 = 0.081 q
(σx)9 =𝐹6−2𝐹9+𝐹12
𝑎2 = − 3.4519−2∗ −3.3802 −3.2642 𝑞𝑎2
𝑎2 = 0.044 q
(σx)10 =𝐹7−2𝐹10 +𝐹13
𝑎2 = −1.1054 −2∗ −0.7634 −0.3296 𝑞𝑎2
𝑎2 = 0.092 q
(σx)11 =𝐹8−2𝐹11 +𝐹14
𝑎2 = −1.6902−2∗ −1.4128 −1.0497 𝑞𝑎2
𝑎2 = 0.086 q
(σx)12 =𝐹9−2𝐹12 +𝐹15
𝑎2 = −3.3802 −2∗ −3.2642 −3.1126 𝑞𝑎2
𝑎2 = 0.036 q
(σx)13 =𝐹10−2𝐹13 +𝐹13 ′
𝑎2 = −0.7634−2∗ −0.3296 + 0 𝑞𝑎2
𝑎2 = -0.104 q
(σx)14 =𝐹11−2𝐹14 +𝐹14 ′
𝑎2 = −1.4128−2∗ −1.0497 −0.75 𝑞𝑎2
𝑎2 = -0.063 q
(σx)15 =𝐹12−2𝐹15 +𝐹15 ′
𝑎2 = −3.2642−2∗ −3.1126 −3 𝑞𝑎2
𝑎2 = -0.039 q
(σx)13’ =𝐹13−2𝐹13 ′+𝐹13𝑂
𝑎2 = −0.3296−2∗ 0 −0.3296 𝑞𝑎2
𝑎2 = -0.659 q
(σx)14’ =𝐹14−2𝐹14 ′+𝐹14𝑂
𝑎2 = −1.0497−2∗ −0.75 −1.0497 𝑞𝑎2
𝑎2 = -0.599 q
(σx)15’ =𝐹15−2𝐹15 ′+𝐹15𝑂
𝑎2 = −3.1126 −2∗ −3 −3.1126 𝑞𝑎2
𝑎2 = -0.225 q
9
Tensiuni normale : σy
(σy)k =𝐹𝑘+1−2𝐹𝑘+𝐹𝑘−1
𝛥𝑥 2 , unde: Fk+1, Fk, Fk-1 sunt valorile functiilor necunoscute, luate pe
orizontala
: 𝛥𝑥=a
(σy)1’ =𝐹2 ′−2𝐹1 ′+𝐹2 ′
𝑎2 = −2−2∗ −1.5 −2 𝑞𝑎2
𝑎2 = -1.000 q
(σy)2’ =𝐹3 ′−2𝐹2 ′+𝐹1 ′
𝑎2 = −3.5−2∗ −2 −1.5 𝑞𝑎2
𝑎2 = -1.000 q
(σy)3’ =𝐹4 ′−2𝐹3 ′+𝐹2 ′
𝑎2 = −6−2∗ −3.5 −2 𝑞𝑎2
𝑎2 = -1.000 q
(σy)1 =𝐹2−2𝐹1+𝐹2
𝑎2 = −1.9645−2∗ −1.4523 −1.9645 𝑞𝑎2
𝑎2 = -1.024 q
(σy)2 =𝐹3−2𝐹2+𝐹1
𝑎2 = −3.4893−2∗ −1.9645 −1.4523 𝑞𝑎2
𝑎2 = -1.013 q
(σy)3 =𝐹4 ′−2𝐹3+𝐹2
𝑎2 = −6−2∗ −3.4893 −1.9645 𝑞𝑎2
𝑎2 = -0.986 q
(σy)4 =𝐹5−2𝐹4+𝐹5
𝑎2 = −1.8668 −2∗ − 1.3272 −1.8668 𝑞𝑎2
𝑎2 = -1.079 q
(σy)5 =𝐹6−2𝐹5+𝐹4
𝑎2 = − 3.4519−2∗ −1.8668 − 1.3272 𝑞𝑎2
𝑎2 = -1.045 q
(σy)6 =𝐹4 ′−2𝐹6+𝐹5
𝑎2 = −6−2∗ − 3.4519 −1.8668 𝑞𝑎2
𝑎2 = -0.963 q
(σy)7 =𝐹8−2𝐹7+𝐹8
𝑎2 = −1.6902−2∗ −1.1054 + −1.6902 𝑞𝑎2
𝑎2 = -1.170 q
(σy)8 =𝐹9−2𝐹8+𝐹7
𝑎2 = −3.3802 −2∗ −1.6902 −1.1054 𝑞𝑎2
𝑎2 = -1.033 q
(σy)9 =𝐹4 ′−2𝐹9+𝐹8
𝑎2 = −6−2∗ −3.3802 +−1.6902 𝑞𝑎2
𝑎2 = -0.930 q
(σy)10 =𝐹11−2𝐹10 +𝐹11
𝑎2 = −1.4128 −2∗ −0.7634 −1.4128 𝑞𝑎2
𝑎2 = -1.300 q
(σy)11 =𝐹12−2𝐹11 +𝐹10
𝑎2 = −3.2642 −2∗ −1.4128 −0.7634 𝑞𝑎2
𝑎2 = -1.202 q
(σy)12 =𝐹4 ′−2𝐹12 +𝐹11
𝑎2 = −6−2∗ −3.2642 −1.4128 𝑞𝑎2
𝑎2 = -0.884 q
(σy)13 =𝐹14−2𝐹13 +𝐹14
𝑎2 = −1.0497−2∗ −0.3296 −1.0497 𝑞𝑎2
𝑎2 = -1.440 q
(σy)14 =𝐹15−2𝐹14 +𝐹13
𝑎2 = −3.1126 −2∗ −1.0497 −0.3296 𝑞𝑎2
𝑎2 = -1.343 q
10
(σy)15 =𝐹4 ′−2𝐹15 +𝐹14
𝑎2 = −6−2∗ −3.1126 −1.0497 𝑞𝑎2
𝑎2 = -0.825
(σy)13’ =𝐹14 ′−2𝐹13 ′+𝐹14 ′
𝑎2 = −0.75−2∗ 0 −0.75 𝑞𝑎2
𝑎2 = -1.500 q
(σy)14’ =𝐹15 ′−2𝐹14 ′+𝐹13 ′
𝑎2 = −3−2∗ −0.75 −0 𝑞𝑎2
𝑎2 = -1.500 q
(σy)15’ =𝐹16 ′−2𝐹15 ′+𝐹14 ′
𝑎2 = −6−2∗ −3 −0.75 𝑞𝑎2
𝑎2 = -0.75 q
Tensiuni tangentiale : τxy
(τxy)k = 𝐹𝑙−1+𝐹𝑗 +1 −(𝐹𝑙+1+𝐹𝑗−1)
4∗𝛥𝑥 ∗𝛥𝑦 , unde: Fj-1, Fj+1, Fl-1, Fl+1 sunt valorile functiilor
necunoscute in jurul punctului k
: 𝛥𝑦 = 𝛥𝑥=a
(τxy)1’ = 𝐹2𝑂+𝐹2 −(𝐹2𝑂+𝐹2)
4∗𝑎2 =0 q
(τxy)2’ = 𝐹3𝑂+𝐹1 −(𝐹1𝑂+𝐹3)
4∗𝑎2 = −3.4893−1.4523 −(−1.4523 −3.4893 )
4∗𝑎2 =0 q
(τxy)3’ = 𝐹4𝑂+𝐹2 −(𝐹2𝑂+𝐹4 ′)
4∗𝑎2 = −6−1.9645 −(−1.9645−6)
4∗𝑎2 =0 q
(τxy)1 = 𝐹2 ′+𝐹5 −(𝐹2 ′+𝐹5)
4∗𝑎2 =0 q
(τxy)2 = 𝐹3 ′+𝐹4 −(𝐹1 ′+𝐹6)
4∗𝑎2 = −3.5− 1.3272 −(−1.5− 3.4519 )
4∗𝑎2 =0.031 q
(τxy)3 = 𝐹4 ′+𝐹5 −(𝐹2 ′+𝐹4 ′)
4∗𝑎2 = −6−1.8668 −(−2− 6)
4∗𝑎2 =0.033 q
(τxy)4 = 𝐹2+𝐹8 −(𝐹2+𝐹8)
4∗𝑎2 =0 q
(τxy)5 = 𝐹3+𝐹7 −(𝐹1+𝐹9)
4∗𝑎2 = −3.4893−1.1054 −(−1.4523 −3.3802 )
4∗𝑎2 =0.060 q
(τxy)6 = 𝐹4 ′+𝐹8 −(𝐹2+𝐹4 ′)
4∗𝑎2 = −6−1.6902 −(−1.9645−6)
4∗𝑎2 =0.069 q
11
(τxy)7 = 𝐹5+𝐹11 −(𝐹5+𝐹11 )
4∗𝑎2 =0 q
(τxy)8 = 𝐹6+𝐹10 −(𝐹4+𝐹12 )
4∗𝑎2 = − 3.4519−0.7634 −(− 1.3272 −3.2642 )
4∗𝑎2 =0.094 q
(τxy)9 = 𝐹4 ′+𝐹11 −(𝐹5+𝐹4 ′)
4∗𝑎2 = −6−1.4128 −(−1.8668 −6)
4∗𝑎2 =0.114 q
(τxy)10 = 𝐹8+𝐹14 −(𝐹8+𝐹14 )
4∗𝑎2 =0 q
(τxy)11 = 𝐹9+𝐹13 −(𝐹7+𝐹15 )
4∗𝑎2 = −3.3802 −0.3296 −(−1.1054 −3.1126 )
4∗𝑎2 =0.127 q
(τxy)12 = 𝐹4′+𝐹14 −(𝐹8+𝐹4 ′)
4∗𝑎2 = −6−1.0497 −(−1.6902−6)
4∗𝑎2 =0.160 q
(τxy)13 = 𝐹11 +𝐹14 ′ −(𝐹11 +𝐹14 ′)
4∗𝑎2 =0 q
(τxy)14 = 𝐹12 +𝐹13 ′ −(𝐹10 +𝐹15 ′)
4∗𝑎2 = −3.2642 +0 −(−0.7634 −3)
4∗𝑎2 =0.125 q
(τxy)15 = 𝐹4′+𝐹14 ′ −(𝐹11 +𝐹16 ′)
4∗𝑎2 = −6−0.75 −(−1.4128 −6)
4∗𝑎2 =0.166 q
(τxy)13’ = 𝐹14 +𝐹14𝑂 −(𝐹14 +𝐹14𝑂 )
4∗𝑎2 =0 q
(τxy)14’ = 𝐹15 +𝐹13𝑂 −(𝐹13 +𝐹15𝑂 )
4∗𝑎2 = −3.1126 −0.3296 −(−0.3296−3.1126 )
4∗𝑎2 =0 q
(τxy)15’ = 𝐹4 ′+𝐹14𝑂 −(𝐹14 +𝐹16𝑂 )
4∗𝑎2 = −6−1.0497 −(−1.0497−6)
4∗𝑎2 =0 q
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