K12 :=

st kN 4 kN st := 4 kN h := 0.40m G := 500 kN = 511 kN Stalp 0.40m m := 6 = 10 m := 2 s = m 25 25 := 6 E E + n = 1.611 1.611 10 kN m 52 52 := 6 6 E E b2 = 1.611 1.611 10 10 kN s := 1 UU := 1 2 12y M := 16148.02 kN m M := 15574.55 kN m M := 20142.59 M kN m := 20277.24 kN m 21x 2 2 47b2 74b2 58b2 85b2 II.Determinarea h matricilor de rigiditate laterala ale cadrului h spatial pe cele doua directii h1 + M25a2 + M52a2 + M36a2 + M63a2 M47a2 h +1 M74a2 + M58a2 + M + M96a2 + M69a2 M41a2 + M14a2 1 kN 85a2 k21 U12y k U K12 := 1 EM + := 29000000 12 21x Cadrul B := 16094.50 M kNm := 0.91 UU := 0.237 6-3 2 kNm h1 = = 69b2 96b2 := 15512.68 22y 4 h2 4 Bara 6-3 11x Bara m k m k11 2_1x m1 2 4.593 10 2.672 10 22 2_2y 4 Grinda b := 0.40 m h := 0.60 m KLA =Istgb kN Matricea de rigiditate cadrului A gb K 2.811 10 n kN h1 :==4.25 m +I 0.05 m = laterala 4.8 m st 4 +a 4 12 = 1.611 10 M +M +M M85b2 + Mm +:= M6 4 4 m 36 := 6 E = 1.611 10 kN 63 E kN 47b2 74b2 58b2 96b2 69b2 4 m 36 := 6 E = 1.611 10 kN m 63 := 6 E = 1.611 10 kN k21 U12y + k22 m2 U U22y Verificare 2.672 10 2.324 10 m k2 U m == 00 N N b Verificare 2 2_2y K := = 2.524 12 21x + k11 2_1x 1 11x Stalp 0.40 m10 kN hs := 0.40 22 h h s :=:= h2 M := 4 m := + 15181.87 0.01 n m kN = m 4.11 m M := 15182.42 kN m M 20489.06 M85a1 kN:= m 19992.44 kN m h h h 1 1 47a1 74a1 58a1 2

Prima forma de vibratie pe directia x A B Cadru A Calculul matricei de a cadrului M47c1 := 18564.27 kN m rigiditate M := 15946.61 kN m M58c1 := 20831.69 kN M m := 20501.20 kNm 2998.75 kN m 1499.37 kN m 12693.56 M 3 74c1 85c1 2 14b23 41b2 25b2 52b2 := 1175.29 = b 3.244 10 de a := m m kN Calculul momentelor incastrare perfecta: Calculul momentelor de incastrare perfecta: 1 :=2 2_1y 10 c1 T := 2 = 0.108i n 3.411 := 11 Lucrarea nr.2 kN= M := 16529.78 m M := 16320.45 kN m M := 16393.41 kN M m kN m M := 18727.13 kN m M := 16272.59 kN m 2 a 41a1 14a1 25a1 1512.27 kN m 3024.54 kN m M := 16148.02 kN M52a1 m := := 16675.72 15574.55 kN m 7 69c1 96c1 2_1y Derformata a 36b2 63b2 47b2 74b2 Derformata Pentru structura figura sakN se2efectueze calculul seismic conform normativului b := k11 m2 +a k22 m1 din = 1.176 10 M := 16345.76 kN m + urmatoarele M := 16532.94 kN m+ M M M kN m := 15182.42 kN m A doua forma vibratie pe + directia 3 36a1 63a1 47a1 74a1 + de M M +xM P100-2006 parcurgand M := 20142.59 kN m M := kN m M := 15181.87 16094.50 kN M m kN m 3 M 9M20277.24 := 47c1 74c1 58c1 85c1 2 etape: 96c1 69c1 4 := 15512.68 58b2 bgc h 85b2 69b2 96b2 b h gc 3 4 c := k gb k gb k k = 3.279 10 kN K21 := 11 1 ga 22 ga 3 rigiditate 4 = 2.697 10 kN b += matricei 12 21 I := m 10 m 1.Determinarea de laterala 7.2 Ig := := = 6.3 4.853 10 M m h2 := = 329.963 T individuala := M 269a1 := 18200.6 = 0.346i M 20489.06 kN := 19992.44 kN m kN M96a1 m := 15137.68 kN m g 2_2y 58a1 85a1 12 12 2 a matricei +M M + M + M + M M + M + M + M + M 2.Determinarea de rigiditate laterala a structurii pentru cele doua directii 13 2_2y M41b2 2 14b2 + 4 63b2 47b2 74b2 58b2 85b2 96b2 + M69b2 M := b 43 a c =25b2 9.565 52b2 10 M kN 36b2 := 15441.50 kN m := 14726.64 kN m M := 19752.24 M kN m := 19894.71 kNm 1 principale + 3 47c2 74c2 58c2 85c2 b h b h s s 3 M + M + M + M + M + M M + M + M + M + M Determinarea vectorului propriu pentru prima forma de vibratie pe directia X h h 4 s s 3 41a1 14a1 25a1 36a1 63a1 47a1 74a1 58a1 85a1 96a1 + M69a1 110 52a1 2 4 I := = 2.133 m Ist := forma = 2.133 10 proprii m K11 := Prima + 3.Stabilirea modurilor de vibratii de vibratie pe directia x st M := 15711.86 4 kN m M := 15045.24 kN m 12 U := 1 69c2 96c2 h2 21y K12 = 12 3.001 10 kN h1 4.Determinarea seismice b incarcarilor fortelor 3 k12 2_1x1-4 := =21y 3.319 10 Bara 4 U Bara 1-4 T := 2 M 0.109i 5.Stabilirea suplimentare din M +M + := M85c2 + M96c2 M M := 18935.54 kN seismice m + M= M0.238 16707.92 kN m+torsiune := = 21026.28 M85b1 kN:= m 20778.08 kNm 2 a U11y K11 = 4.593 fortei 10 kN := 47c2 74c2 58c2 69c1 58b1 4 47b1 74b1 k m I I 2_1x K := = 2.52 10 kN Ist Ist 11 2_1y 1 de 4 4 22 6.Trasarea diagramelor eforturi finale din actiunea st de 4 st seismica 4 h A doua forma vibratie pe directia x 2 := 1803.2 m 14 := 6 E kN =m 1.611 10 kN m 41 := 6 E := = 1.611 10 kN := 1810.19kN m M := 18929.45 kN m M 16700.13 kN m := 3606.4 kN m M 10745.83 kN M m 69b1 96b1 2 2 41a2 14a2 25a2 52a2 7.Determinarea deplasarilor relative 2 2 h h b +h1 U21y + ( k11 2_1y m4 Verificare h1 4 12= 1) U11y = 0 N K 4 4 1k 1 K M Date := 304.471 K :=personale: 2.815 10 kN K21 := 2.815 10 kN 2_2x 4.754 10 2.815 11 12A M := 1816.03 kN m M := 3632.06 kN m M kN m := 13896.85 kN m10 Cadrul 12 T := 2 := 14776.86 = 0.36i 36a2 63a2 47a2 74a2 2 a M + M + M + M + M + M + := 0.35 Bara 2-5 K =m kN 47b1 85b1 96b1 4 h := 0.55m Bara 2-5 G n kN = 74b1 611 kN 58b1 Grinda bga LC := 69b1 2_2x 1 := ga4 K21 :=600 1 kN = 2.751 10 kN K21 K 4 22 I de M M kN m M := 19211.69 kN m forma M := 14719.49 kN pe m := 13826.94 Determinarea vectorului propriu pentru a doua forma vibratie directia XkN h2 2.815 10 2.52 m 10 Determinarea propriu pentru prima vibratie pe 58a2 := 19069.40 85a2 69a2 96a2 Ist vectorului Ist de I 4 4 directia X

Ist Ist 4 4 4 4 4 U := 1m 2.197 st 4 st 44 K K 12x Grinda l m := U 4.5 1 U+ 0.01 + m n m = U= 4.61 U22x m = 0 b := 0.35 m hgc := 0.60 m 10 4.807 10 10 2.876 2.876 10 2.524 10 m := 6 E 2.197 10 kN m 74 := 1 6 E = kN 11 12 2x 1 11x 12x 2 21x gc m 47 47 := 1 6 E = 2.197 10 kN m 74 := 1 6 E = 2.197 10 kN kN + M + M58a1 + M85a1 + M96a1 M69a1 M47a1 2 2 KLB+ := = 4 2 74a1 2 U h K21 := 1 k21h = 2.535 10 kN h 12x h2 K22 4 0.40m 4 := 2 m 2.876 h l1y := 3.75 n m = 4.85 m Stalp K21 bs 0.40 := U22x := m + 0.12 = 0.939 h 10 2.524 10 s 4 4 2 4.754 10 2.815 10 22 2_2x m2 k Bara 5-8 = kN Matricea de rigiditate laterala a cadrului C m KLCM l2y M := 4 := m 14776.86 + 0.05 n m kN = m 4.55 m M74a2 := 13896.85 kN 19069.40 M85a2 kN:= m 19211.69 kN m 47a2 58a2 := Calculul matricei de rigiditate a cadrului C 4 4 I I I I st 4 st 4 2.815 10 2.52 10 k21st U12x + k22 2_2x m2 U22x = 0 N Verificare 4 kN E st = 2.197 104 kN m 1 = m 85 m 58 58 := := 1 6 6 E E kN = 2.197 2.197 10 10:= kN m 85 := := 1 1 6 6 E 2 = 2.197 10 kN M := 14719.49 m M 13826.94 kN m 2 69a2 96a2 MA := 16563.69 kN m M := 16337.40 kN m M kN M52c1 m := 16574.97 kNm 2 2 41c1 14c1 25c1 := 16343.04 h2 h2 2 2 Verificare derigiditate ortogonalitate intre forma 1spatial si forma Matricea de laterala a cadrului pe 2 directia X M36c1 := 16352.23 kNm+ M M63c1 := 16571.21 kN+ mM M47c1 := 18564.27 kN M m := 15946.61 kNm M + M + M + M Bara 6-9 47a2 74a2 58a2 85a2 96a2 69a2 4 74c1 Bara 6-9 K22 := = 2.324 10 kN 1.399 105 8.22 104 c1 m U11x U +m =2 := 0 20501.20 h I I U22x B M 20831.69 m M85c1 kN m kN M69c1 := 18727.13 kN M96c1 m := 16272.59 kNm 1 12x+ 2 U 21x IkN Ist st 4 4 58c1 KLx := := 2 K K = st 4 st 4 LA LB = 2.197 10 m69 := 1 6 E 4 kN 4 m 96 := 1 6 E = 2.197 10 kN 4 4 2 2 K12 := 2.672 10h kN K21 := 2.672 10 kN 2 8.22 10 7.172 10 h22 h h 2 2 2 + M14c1 + M25c1 + Mmodurilor + M63c1 M47c1 +pentru M74c1 + M58c1 +X M85c1 + M96c1 + M69c1 M41c1 III.a.)Determinarea proprii de 4 vibratii directia 52c1 + M36c1 4 4.593 de K rigiditate laterala a cadrului spatial K11 := Matricea + pe directia Y K11 Derformata b 10 2.672 10 12 b h h2 = kN A :=Derformata 1 KLA 5 4 l1x l2x 5 kN 4 C := 4 K l1x k11C := = 1.426 10 k4 KLy = 8.445 10 l2x kN Bara 4-7 C K22 Bara 4-7 Ly 12 1.426 10 8.445 10 21 K 1 , 1 4 1, 2 10 2.672 10 2.324 KLy := 3 KLC kN A=2 buc = 10 kN 11 = 4.754 Ist Ist 4 4 4 4 4 4 st 4 st =10 4 kN 8.445 10 7.561 10k22 m k 47 := := K 6 E = = 2.197 10 kN kN := m K 74 := = 6 7.561 E 2.197 kN 10 m 47 := 6 E = 2.197 10 kN m 74 := 6 E = 2.197 10 kN 21 Ly Ly Calculul matricei de rigiditate a cadrului B 2 2, 1 2 2 , 2 2kNm 2 M41c2 := 3321.34 M14c2 := 1660.67 kNm M25c2 h := 12148.33 kN M m kNm h2 2 52c2 := 1409.84 2 2 G = 611 kN G = 511 kN 1 2 M kNm modurilor M14b1 :=proprii 16342.75 kN m M25b1 := 16310.99 kN M52b1 X m := 16510.86 kNm III.a.)Determinarea de vibratii pentru directia 41b1 := 16574.40 Bara M36c25-8 := 1597.27 kNm M63c2 := 3194.55 kNm M47c2 := 15441.50 kN M74c2 m := 14726.64 kNm Bara 5-8 G1 G 2 M36b1 := 16349.91 kNm kN5 M63b1 := 16579.86 kN m := 18935.54 M74b1 m := 16707.92 kNm m1 := = 62.283 m2 := M47b1 = 52.09 kN 4 kN I I I= Ist st 4 4 k := := K 10 kN k12 := K =6 8.22 10 kNkN st 1.399 4 kN:= 19894.71 st 4 := M := 19752.24 kN m M kN m M := 15711.86 M m 15045.24 kNm 9.81 9.81 11 Lx Lx m 58 6 E = 2.197 10 m 85 := E = 2.197 10 kN 58c2 85c2 69c2 96c2 m58 := 61 E 2 = 2.197 10 kN m85 ,1 1 , 2:= 6 E 2 = 2.197 10 kN M58b1 := 21026.28 M := 20778.08 kN m M69b1 h := 2 18929.45 kN M96b1 m := 16700.13 kNm 2 kNm h 3 4 2 85b1 h2 h2 4 a := m m = 3.244 10 kN k21 := KLx 8.22+ 10 kN k22 := KLx = 7.172 10 kN + M 1 2 += +M M + M74c2 +M M41c2 14c2 25c2 M52c2 + M36c2 + M 63c2 M 47c2 58c2 85c2 + M96c2 + M69c2 2, 1 2, 2 Bara 6-9 K12 := 1 M + Bara 6-9 + M14b1 + M25b1 + M52b1 + M36b1 + M63b1 M47b1 + M74b1 + M58b1 + M85b1 + M96b1 + M69b1 41b1 7 h2 2 611 kN G2h = 511 kN 1 1.214 K11 := + G b1 := =k11 m + k m = 10 kN I I l1y l2y l1x l2x 2 Ist 22 1 h Ist 4 h24 st 4 kN st = 2.197 10 4 kN 1 m 69 := 6 E = 2.197 10 m 96 := 6 E G2 m69 := G 61 E = 2.197 10 kN m96 := 6 E = 2.197 10 kN B= 1buc C=3 buc 2 4 kN 2 kN 9 2 := = 62.283 m := = 52.09 h h Km = 2.933 10 kN 4 13 h h 1 2 2 2 := k11 k22 k12 k21 = 3.652 10 kN 2 4 12 2 Kc 10 kN 9.81 9.81 := b 4 2 a c = 9.995 10 kN 11 = 4.807

4.807 104 2.876 104 Verificare de n ortogonalitate intre forma kN 1 si forma 2 Bara 4-7 Bara 4-7 l1x M := 4 := m 18200.6 + 0.05vectorului kN 4 m m = 4.55 m M m forma Cadrul A Determinarea propriu pentru de vibratie pe directia X 4 a doua kN 69a1 96a1 := 15137.68 KLB = Matricea de rigiditate laterala a cadrului K12 := 2.876 10 10 kN B 21 := 2.876 I kN K I

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