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. prof.univ.dr.ing. MIRCEA IEREMIA
0
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 =F pc+2*a*N, N – forta axiala de pe sistemul de baza static determinat.
1
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:
2
Definirea functiilor pe contur F4’=-6 qa2 F3’=-3.5 qa2 F2’=-2 qa2 F1’=-1.5 qa2 F16’=-6 qa2 F15’=-3 qa2 F14’=-0.75 qa 2 F13’=0 qa2
Definirea functiilor pe extracontur 2
F4V=2*(-3)+F3’= -9.5 qa 2 F3V=2*(-3)+F3= F3-6 qa 2 F6V=2*(-3)+F6=F6-6 qa 2 F9V=2*(-3)+F9=F9-6 qa 2 F12V=2*(-3)+F12=F12-6 qa 2 F15V=2*(-3)+F15=F15-6 qa 2 F16V=2*(-3)+F15’= -9 qa 2 F40=2*(0)+F4’=-6 qa 2 F30=F3 qa 2 F20=F2 qa F10=F1 qa2 2 F160=2*(0)+F16’=-6 qa F150=F15 qa 2 F140=F14 qa 2 F130=F13 qa 2
Reteaua fiind patratica, se va aplica molecula de calcul , in fiecare nod al retelei:
3
Sistemul de ecuatii se obtine in felul urmator: 2 (impartit la qa ) 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:
21 16
2
8
8 22 8 2 1 8 22 0 8 4 0 20 2 8 2 8 0 2 8 1 1 0 0 8 A 0 1 0 2 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0
0
0
0
1
0
0
0
0
0
0
0
8
2
0
1
0
0
0
0
0
0
2
8
0
0
1
0
0
0
0
0
16
2
8
4
0
1
0
0
0
0
21
8
2
8
2
0
1
0
0
0
8
21
0
2
8
0
0
1
0
0
4
0
20
16
2
8
4
0
1
0
8
2
8
21
8
2
8
2
0
1
2
8
1
8
21
0
2
8
0
0
0
0
8
4
0
20
16
2
8
4
1
0
2
8
2
8
21
8
2
8
0
1
0
2
8
1
8
21
0
2
0
0
1
0
0
8
4
0
21
16
0
0
0
1
0
2
8
2
8
22
0 0 0 0 0 0 1 0 2 8 2 8
0
0
0
0
1
0
2
8
1
8
22
4
0
4
0
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:
A=
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
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
44
Matricea-coloana {B} a termenilor liberi are valoarea:
4
4 0 42 1. 5 8 14.5 0 B 6 18 0 6.75 15 3 6 40.5
0 84 1. 5 16 29 0 B 12 36 0 13.5 30 3 12
x2 x2
x2 x2 x2 x2
x2 x2 x2 x2
81
5
-16
2
Rezolvarea sistemului se va face folosind metoda matricei inverse, si anume se va inmulti la stanga fiecare termen cu [A]-1. [A]*{F}={B} -1 -1 [A] *[A]*{F}=[A] *{B} {F}=[A]-1*{B}
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 1 A 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
Efectuand calculele, va rezulta matricea {F}:
1.4523 1.9645 3.4893 1.3272 1.8668 3.4519 1.1054 F 1.6902 3.3802 0.7634 1.4128 3.2642 0.3296 1.0497
x qa2
3.1126 6
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 2
Ecuatia 7: (simplificand qa ) 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 2
Ecuatia 10: (simplificand qa ) 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 2
Ecuatia 13: (simplificand qa ) 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
−2 + , unde: F j, Fk , Fl sunt valorile functiilor necunoscute, luate pe verticala 2
(σx)k =
: =a
10 −2 1 ′ +1 −1.4523 −2∗−1.5 −1.4523 2 = = 0.095 q 2 2
(σx)1’ =
20 −2 2 ′ +2 −1.9645 −2 ∗−2−1.9645 2 (σx)2’ = = = 0.071 q 2 2 30 −2 3 ′ +3 −3.4893 −2 ∗−3.5 −3.4893 2 = = 0.022 q 2 2
(σx)3’ =
1 ′−21 +4 −1.5 −2 ∗−1.4523 − 1.3272 2 = = 0.077 q 2 2
(σx)1 =
2 ′−22 +5 −2 −2 ∗−1.9645 −1.8668 2 = = 0.062 q 2 2
(σx)2 =
3 ′−23 +6 −3.5 −2 ∗−3.4893 − 3.4519 2 = = 0.027 q 2 2
(σx)3 =
7
1 −2 4 +7 −1.4523 −2∗− 1.3272 −1.1054 2 (σx)4 = = = 0.097 q 2 2 2 −2 5 +8 −1.9645 −2∗−1.8668 −1.6902 2 = = 0.079 q 2 2
(σx)5 =
3 −2 6 +9 −3.4893 −2∗− 3.4519 −3.3802 2 = = 0.034 q 2 2
(σx)6 =
4 −2 7 +10 − 1.3272 −2 ∗−1.1054 −0.7634 2 = = 0.120 q 2 2
(σx)7 =
5 −2 8 +11 −1.8668 −2 ∗−1.6902 −1.4128 2 = = 0.081 q 2 2
(σx)8 =
6 −2 9 +12 − 3.4519 −2 ∗−3.3802 −3.2642 2 (σx)9 = = = 0.044 q 2 2 7 −210 +13 −1.1054 −2 ∗−0.7634 −0.3296 2 = = 0.092 q 2 2
(σx)10 =
8 −211 +14 −1.6902 −2 ∗−1.4128 −1.0497 2 = = 0.086 q 2 2
(σx)11 =
9 −212 +15 −3.3802 −2 ∗−3.2642 −3.1126 2 (σx)12 = = = 0.036 q 2 2 10 −2 13 +13 ′ −0.7634 −2 ∗−0.3296 + 0 2 = = -0.104 q 2 2
(σx)13 =
11 −2 14 +14 ′ −1.4128 −2 ∗−1.0497 −0.75 2 = = -0.063 q 2 2
(σx)14 =
12 −2 15 +15 ′ −3.2642 −2 ∗−3.1126 −3 2 = = -0.039 q 2 2
(σx)15 =
13 −2 13 ′ +13 −0.3296 −2∗0−0.3296 2 = = -0.659 q 2 2
(σx)13’ =
14 −2 14 ′ +14 −1.0497 −2∗−0.75−1.0497 2 = = -0.599 q 2 2
(σx)14’ =
15 −2 15 ′ +15 −3.1126 −2∗−3−3.1126 2 = = -0.225 q 2 2
(σx)15’ =
8
Tensiuni normale : σy
+1 −2 +− 1 , unde: F k+1, Fk , Fk-1 sunt valorile functiilor necunoscute, luate pe 2
(σy)k = orizontala
: =a 2 ′−21 ′ +2 ′ −2 −2∗−1.5−2 2 = = -1.000 q (σy)1’ = 2 2 3 ′−22 ′ +1 ′ −3.5 −2 ∗−2−1.5 2 = = -1.000 q 2 2
(σy)2’ =
4 ′−23 ′ +2 ′ −6 −2∗−3.5−2 2 = = -1.000 q 2 2
(σy)3’ =
2 −2 1 +2 −1.9645 −2∗−1.4523 −1.9645 2 (σy)1 = = = -1.024 q 2 2 3 −2 2 +1 −3.4893 −2∗−1.9645 −1.4523 2 = = -1.013 q 2 2
(σy)2 =
4 ′−23 +2 −6 −2 ∗−3.4893 −1.9645 2 = = -0.986 q 2 2
(σy)3 =
5 −2 4 +5 −1.8668 −2∗− 1.3272 −1.8668 2 = = -1.079 q 2 2
(σy)4 =
6 −2 5 +4 − 3.4519 −2∗−1.8668 − 1.3272 2 = = -1.045 q 2 2
(σy)5 =
4 ′−26 +5 −6 −2 ∗− 3.4519 −1.8668 2 = = -0.963 q 2 2
(σy)6 =
8 −2 7 +8 −1.6902 −2∗−1.1054 + −1.6902 2 = = -1.170 q 2 2
(σy)7 =
9 −2 8 +7 −3.3802 −2∗−1.6902 −1.1054 2 = = -1.033 q 2 2
(σy)8 =
4 ′−29 +8 −6 −2 ∗−3.3802 +−1.6902 2 (σy)9 = = = -0.930 q 2 2 11 −2 10 +11 −1.4128 −2∗−0.7634 −1.4128 2 = = -1.300 q 2 2
(σy)10 =
12 −2 11 +10 −3.2642 −2∗−1.4128 −0.7634 2 = = -1.202 q 2 2
(σy)11 =
4 ′−2 12 +11 −6 −2∗−3.2642 −1.4128 2 (σy)12 = = = -0.884 q 2 2 14 −2 13 +14 −1.0497 −2∗−0.3296 −1.0497 2 = = -1.440 q 2 2
(σy)13 =
15 −2 14 +13 −3.1126 −2∗−1.0497 −0.3296 2 = = -1.343 q 2 2
(σy)14 =
9
4 ′−2 15 +14 −6 −2∗−3.1126 −1.0497 2 (σy)15 = = = -0.825 2 2 14 ′−213 ′ +14 ′ −0.75 −2∗0−0.75 2 = = -1.500 q 2 2
(σy)13’ =
15 ′−214 ′ +13 ′ −3 −2 ∗−0.75−0 2 = = -1.500 q 2 2
(σy)14’ =
16 ′−215 ′ +14 ′ −6 −2 ∗−3−0.75 2 = = -0.75 q 2 2
(σy)15’ =
Tensiuni tangentiale : τxy (τxy)k =
− 1 + +1 −( +1 + −1 ) , unde: F j-1, F j+1, Fl-1, Fl+1 sunt valorile functiilor 4 ∗∗
necunoscute in jurul punctului k : =
=a
(τxy)1’ =
2 +2 −(2 +2 ) =0 q 4 ∗ 2
(τxy)2’ =
3 +1 −(1 +3 ) −3.4893 −1.4523 −(−1.4523 −3.4893) = =0 q 4 ∗ 2 4∗ 2
(τxy)3’ =
4 +2 −(2 +4 ′ ) −6−1.9645 −(−1.9645 −6) = =0 q 4 ∗ 2 4 ∗ 2
(τxy)1 =
2 ′ +5 −(2 ′ +5 ) =0 q 4 ∗ 2
(τxy)2 =
3 ′ +4 −(1 ′ +6 ) −3.5 − 1.3272 −(−1.5 − 3.4519 ) = =0.031 q 4 ∗ 2 4 ∗ 2
(τxy)3 =
4 ′ +5 −(2 ′ +4 ′ ) −6 −1.8668 −(−2 − 6) = =0.033 q 4∗ 2 4∗ 2
(τxy)4 =
2 +8 −(2 +8 ) =0 q 4∗ 2
(τxy)5 =
3 +7 −(1 +9 ) −3.4893 −1.1054 −(−1.4523 −3.3802 ) = =0.060 q 4∗ 2 4∗ 2
(τxy)6 =
4 ′ +8 −(2 +4 ′ ) −6−1.6902 −(−1.9645 −6) = =0.069 q 4 ∗ 2 4 ∗ 2
10
(τxy)7 =
5 +11 −(5 +11 ) =0 q 4∗ 2
(τxy)8 =
6 +10 −(4 +12 ) − 3.4519 −0.7634 −(− 1.3272 −3.2642 ) = =0.094 q 4∗ 2 4 ∗ 2
(τxy)9 =
4 ′ +11 −(5 +4 ′ ) −6−1.4128 −(−1.8668 −6) = =0.114 q 4∗ 2 4 ∗ 2
(τxy)10 =
8 +14 −(8 +14 ) =0 q 4∗ 2
(τxy)11 =
9 +13 −(7 +15 ) −3.3802 −0.3296 −(−1.1054 −3.1126 ) = =0.127 q 4∗ 2 4 ∗ 2
(τxy)12 =
4 ′ +14 −(8 +4 ′ ) −6−1.0497 −(−1.6902 −6) = =0.160 q 4∗ 2 4 ∗ 2
(τxy)13 =
11 +14 ′−(11 +14 ′ ) =0 q 4∗ 2
(τxy)14 =
12 +13 ′−(10 +15 ′ ) −3.2642 +0−(−0.7634 −3) = =0.125 q 4∗ 2 4 ∗ 2
(τxy)15 =
4 ′ +14 ′−(11 +16 ′ ) −6−0.75−(−1.4128 −6) = =0.166 q 4∗ 2 4 ∗ 2
(τxy)13’ =
14 +14 −(14 +14 ) =0 q 4∗ 2
(τxy)14’ =
15 +13 −(13 +15 ) −3.1126 −0.3296 −(−0.3296 −3.1126 ) = =0 q 4∗ 2 4∗ 2
(τxy)15’ =
4 ′ +14 −(14 +16 ) −6−1.0497−(−1.0497 −6) = =0 q 4 ∗ 2 4 ∗ 2
11
12
13
14
