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kN m⋅ DETERMINAREA SECTIUNILOR ACTIVE ALE DIAFRAGMELOR TRANSVERSALE

1.DETERMINAREA SECTIUNILOR ACTIVE PENTRU MONTANTI



DIAFRAGMA DT1

Aria activa la compresiune excentrica:

Am1=3. 6875 m2

Aria activa la forta taietoare

Av1=2.5625 m2

Pozitia centrului de greutate a diafragmei

X=0.3919 m

Y=5.1250 m Momentul de inertie a diafragmei

Im1=41.1911 m4

DIAFRAGMA DT

M!"TA"T#$ 1




Am1=2. 3875 m2


Av1=1.35 m2

Pozitia centrului de greutate a montantului

X=0.1873 mY=2.9551 m Momentul de inertie a montantului

Im1=9.7004 m4

M!"TA"T#$


Am2=1. 575 m2


Av2=0.9125 m2


X=1.5444 m

Y=1.1099 m Momentul de inertie a montantului

Im2=2.4231 m4

INTREAGA DIAFRAGMA




X=1.6305 m

Y=5.1439 m

Momentul de inertie a diafragmei

I0=54.3645 m4

DIAFRAGMA DT%

M!"TA"T#$ 1


Am1 =1.5258 m2


Av1=0.9258 m2


X=0.2118m


Im1=2.1462 m4

M!"TA"T#$




Am2=0.725 m2


Av2=0.2 m2

Pozitia centrului de greutate a montantuluiX=1.2603 m


Im2=0.0157 m4

MONTANTUL 3


Am3=2.7 m2


Av3=1.6625 m2


X=0.2935 m


Im3=11.0729 m4

INTREAGA DIAFRAGMA




X=0.1903 m

Y=6.6127 m


I0=93.6736 m4

DIAFRAGMA DT4

MONTANTUL 1


Am1=1.6125 m2


Av1=0.8875 m2


X=0.1494 m


Im1=2.0222 m4

MONTANTUL 2




Am1=0.95 m2


Av1=0.2 m2

Pozitia centrului de greutate a montantuluiX=1.625 m


Im1=0.0265 m4

INTREAGA DIAFRAGMA


X=1.6404 m

Y=2.7213 m


I0=11.7767 m4

DIAFRAGMA DT5



MONTANTUL 1


Am1=0.9375 m2


Av1=0.2125 m2


X=0.083 mY=0.657m

Momentul de inertie a montantului

Im1=0.0314m4

MONTANTUL 2


Am2=1.6 m2


Av2=0.875 m2


X=1.6004 m


Im2=1.944 m4

INTREAGA DIAFRAGMA




X=1.594 m

Y=2.5075 m


I0=11.6395 m4

DIAFRAGMA DT6

MONTANTUL 1


Am1=2.4381 m2


Av1=1.3125 m2


X=1. 4904 m


I m1=7.7873 m4

MONTANTUL 2’




Am2’=0.925 m2


Av2’=0.2 m2


X=0.1676 m


Im2’=0.0263 m4

MONTANTUL 2’’


Am2’’=1.6 m

2


Av2’’=0.875 m2


X=1.5496 m


Im2’’=1.944 m4

MONTANTUL 2




X=1.5562 m

Y=2.498 m


I2=11.5933 m4

DIAFRAGMA DT7

MONTANTUL 1


Am1=1.4562 m2


Av1=0.8562 m2


X=0.2615 mY=2.3666 m Momentul de inertie a montantului

Im1=1.7292 m4



MONTANTUL 2


Am2=2.5812 m2


Av2=1.4063 m2


X=1.5022 m


Im2=10.9905 m4

INTREAGA DIAFRAGMA


X=1.5957 m


I0=55.5066 m4



DIAFRAGMA DT8

Aria activa la compresiune

excentrica:

Am1=3.6875 m2


Av1=2.5625 m2

Pozitia centrului de greutate a

diafragmei

X=1.3581 m


Im1=41.1911 m4

1.DETERMINAREA SECTIUNILOR

ACTIVE PENTRU RIGLE

RIGLA 1 (l01.20 !"

Aria riglei:

Ar =0.1968 m2


Avr =0.15 m2

Pozitia centrului de greutate a riglei



X=0.125 m

Y=0.3559 m Momentul de inertie a riglei

Ir =0.0065 m4

RIGLA 2 (l00.# !"

Aria riglei:

Ar =0.1851 m2


Avr =0.15 m2


X=0.125 m

Y=0.3446 m Momentul de inertie a riglei

Ir =0.0061 m4

RIGLA 3 (l00.#5 !"

Aria riglei:



Ar =00.1871 m2


Avr =0.15 m2


X=0.125 m

Y=0.3465 m

Momentul de inertie a rigleiIr =0.0061 m4

5.CALCULUL CARACTERISTICILOR DE RIGIDITATE PENTRU O

DIAFRAGMA

CU 2 SIRURI DE GOLURI

CADRUL REAL



CADRUL INLOCUITOR



CADRUL REAL$

• RIGIDITATEA MONTANTILOR LA DISTORSIUNEkm1 = (Im1 / He) x Em=(2.1462 x2.8) x Em=6.00936 x Em

km2 = (Im2 / He) x Em = (0.0157x 2.8) x Em = 0.0396 x Em

km3 = (Im3 / He) x Em = (11.0729 x 2.8) x Em = 31.0012x Em

• RIGIDITATEA RIGLEI LA DISTORSIUNE

! = 0.3" x #$ = 0.3" x 0.6 = 0.21 m

% = %1&2 ' 2 x ! =0.9 ' 2 x 0.32 = 1." m

E$ = 2 x Em

1 = 1 / (1 ' (30 x I $ / A x %2)) =

1

1+ 30×0.0061

0.15×1.54❑2

=¿0.66

k$1&2 = ((6 x E$ x I$)/L1&2) x (L1&2/%)2 x =((6 x 2 x Em x 0.0061)/ ") x (" / 1.")3 x 0.66

=0.33 x Em



! = 0.3" x #$ = 0.3" x 0.6 = 0.21 m

% = %2&3' 2 x ! =1.2 ' 2 x 0.32 = 1.8 m

E$ = 2 x Em 2 = 1 / (1 ' (30 x I $ / A x %2)) =0.*2

k$2&3 = ((6 x E$ x I$)/L2&3) x (L2&3/%)2

x =((6 x 2 x Em x 0.006")/ 3) x (3 / 1.8)3

x 0.*2=0.08x Em

CADRUL EC%IVALENT$

RIGIDITATEA MONTANTILOR LA DISTORSIUNE+ ME=(6.00936'0.0396'31.0012)/2 x Em=18."3 x Em

RIGIDITATEA RIGLEI LA DISTORSIUNE+ RE=0.33 x Em'0.08x Em=0.1 x Em

M&'!l) *+ ,+ -+ I0+´ I es

, = k$e / kme = 0.1 x Em / 18."3 x Em = 0.022

- = 1 ' ((Im1 ' Im2' Im3) / L1&"2) x (1/Am1 ' 1 / Am2'1/ Am3) = 1 ' ((2.1462 ' 0.0157+11.0729) /

132) x (1/1.5258

' 1 /0.725+1/2.7) = 1.188

= x √ λ x γ = x √ 0.022 x 1.188 = 0.6* =DIARAGMA U GOLURI MARI

I0 = (- / (-&1)) x (Im1 ' Im2' Im3) = (1.188 / (1.188&1)) x (2.1462 ' 0.0157+11.0729) = 83.632* m

4 = 11.2" 5 e%7$

1 ......... 0.131

0.022 ... x

0.0" .......... 0.1*3

x = 0.198

0.0......... 0.16*



0.022 ...

0.0" .......... 0.22

= 0.19032

1.1" ..... 0.198

1.188 ... 4

1.20 ...... 0.19032

4 = 0.180"9"2

´ I es = (4 x I0) / (1'4 x (4 x I0 / Am: x H2)) = (0.180"9"2x 83.632*) / (1'11.2" x

((0.180"9"2x 83.632*/ 2.*883 x (2.8x)2)) = 10.16 m

6.DISTRIUTIA INCARCARII ORI/ONTALE TOTALE LA DIAFRAGMELE

TRANSVERSALE

6.1.1 DETERMINAREA AEI CENTRULUI MASELOR



G;7!<$7>?e:7$!>4:!%@>ex:=0.00"1900'0.120'0.2"2"00'0.01"1900=66" <!N/m2

G;7!<$7>?e:7$!>4:!%@>:=0.01"1900'0.2"2"00'0.01"1900=682 <!N/m2

DT1 : calculul greutatii

G1 10.25 3 0.25⋅−( ) 2.54⋅ 1.50 3⋅ 2.54⋅+[ ] 2.5⋅ 3 0.25⋅ 2.54⋅ 6.65⋅+ 10.25 3 1.50⋅+( ) 0.26⋅ 6.65⋅+[ ] 4⋅ 508.284=:= k

TRAMA 1:

G2 22.56 4.275 1.5 0.3⋅+( )⋅ 3⋅ 4.75 1.625+( ) 2.59⋅ 1.62⋅ 3⋅+ 22.56 8.81 1.2 0.4⋅+( )+ 609.615=:= kN

TRAMA 2:

15.43 4.275 1.5 0.3⋅+( )⋅ 4.68 5.46 1.5 0.3⋅+( )⋅+ 2.43 4.56 1.5 03⋅+( )⋅+[ ] 3⋅ 4.75 1.5+( ) 2.59⋅ 1.62⋅ 3⋅+ 22.56 8.81 1.2 0.4⋅+( )+ 655.998=

kNDT2 :

G4 5.4 2 0.25⋅−( ) 2.54⋅ 1.5 1+ 0.15+ 1.5+( ) 2.54⋅+ 3.65 0.25−( ) 2.54⋅+ 1.5 1.15+( ) 2.54⋅+[ ] 2.67⋅3 0.25⋅ 2.54⋅ 6.82⋅ 10.25 1.5+ 1+ 0.15+ 1.5+ 1.15+ 1.5+( ) 0.26⋅ 6.82⋅+ 1.2 0.34⋅ 6.82⋅++

... 4⋅ 593.652=:= kN

TRAMA 3:

G5 16.62 4.275 1.5 0.3⋅+( )⋅ 3⋅ 16.62 8.81 1.2 0.4⋅+( )⋅+ 389.988=:= k

TRAMA 4:

G6 16.62 4.56 1.5 0.3⋅+( )⋅ 3⋅ 3.5 2.59⋅ 1.62⋅ 3⋅+ 16.62 8.81 1.2 0.4⋅+( )⋅+ 448.254=:= kN

DT3:

G7 3.7 0.25−( ) 2.54⋅ 2.4 2.54⋅+ 0.8 0.25−( ) 2.54⋅+ 2.1 2.54⋅+ 6.65 3 0.25⋅−( ) 2.54⋅+ 4.15 2.54⋅+[ ] 2.67⋅5 0.25⋅ 6.82⋅ 13.25 2.4+ 2.1+ 4.15+( ) 0.26⋅ 6.82⋅+ 2.1 0.34⋅ 6.82⋅++

... 4⋅ 712.1198=:=k

TRAMA 5:

G8 18.93 4.56 1.5 0.3⋅+( )⋅ 3⋅ 4 2.59⋅ 1.62⋅ 3⋅+ 18.93 8.81 1.2 0.4⋅+( )⋅+ 510.727=:= k

TRAMA 6:

G9 13.07 5.19 3 0.3⋅+( )⋅ 3⋅ 13.07 8.81 1.2 0.4⋅+( )⋅+ 360.209=:= k

TRAMA 7:

G10 22.5 4.275 1.5 0.3⋅+( )⋅ 3⋅ 22.5 8.81 1.2 0.4⋅+( )⋅+ 527.963=:= k

DT4:

G11 3.55 0.25−( ) 2.54⋅ 2.9 2.54⋅+ 0.8 0.25−( ) 2.54⋅+ 3 2.54⋅+[ ] 2.67⋅ 3 0.25⋅ 2.54⋅ 6.82⋅+ 5.25 2.9+ 3+( ) 0.26⋅ 6.82⋅+0.9 0.34⋅ 6.82⋅+ ... 4⋅ 403.891=:= k

TRAMA 8:

G12 14.84 4.275 1.5 0.3⋅+( )⋅ 2.84 4.56 1.5 0.3⋅+( )⋅+ 4.87 5.46 1.5 0.3⋅+( )⋅+[ ] 3⋅ 4.75 1.625+( ) 2.59⋅ 1.62⋅ 3⋅+22.56 8.81 1.2 0.4⋅+( )⋅+

... 629.214=:=kN



TRAMA 9

G13 4.95 4.56 3 0.3⋅+( )⋅[ ] 3⋅ 4.95 8.81 1.2 0.4⋅+( )⋅+ 127.066=:= k

TRAMA 10

G14 7.425 5.19 3 0.3⋅+( )⋅ 3⋅ 7.425 8.81 1.2 0.4⋅+( )⋅+ 204.633=:= k

TRAMA 11:

G15 19 4.56 1.5 0.3⋅+( )⋅ 4 2.59⋅ 1.62⋅+[ ] 3⋅ 19 8.81 1.2 0.4⋅+( )⋅+ 512.430=:= kN

DT 5:

G16 0.85 0.25−( ) 2.54⋅ 2.9 2.54⋅+ 3.5 0.25−( ) 2.54⋅+ 2.90 2.54⋅+[ ]2.67

2 0.25⋅ 2.54⋅ 6.82⋅ 5.25 2.9+ 2.9+( ) 0.26⋅ 6.82⋅+ 0.9 0.34⋅ 6.82⋅++... 4⋅ 383.146=:= k

DT 6:

G17 5.25 3 0.25⋅−( ) 2.54⋅ 4.5 2.54⋅+ 0.8 0.25−( ) 2.54⋅+ 3.5 0.25−( ) 2.54⋅+ 2.9 2.54⋅+ 2.9 2.54⋅+[ ] 2.67⋅5 0.25⋅ 2.54⋅ 6.82⋅ 5.25 2⋅ 4.5+ 2.9 2⋅+( ) 0.26⋅ 6.82⋅+ 0.95 0.34⋅ 6.82⋅++

... 4⋅ 747.522=:= kN

TRAMA 12:

G18 22.56 5.19 3 0.3⋅+( ) 3⋅ 22.56 8.81 1.2 0.4⋅+( )⋅+ 621.754=:= kN

TRAMA 13:

G19 G12 629.214=:= kN

DT 7:

G18 3.43 0.25−( ) 2.54⋅ 2.4 2.54⋅+ 5.64 2 0.25⋅−( ) 2.54⋅+ 4.5 2.54⋅+[ ]2.673 0.25⋅ 2.54⋅ 6.82⋅ 10.25 2.4+ 4.5+( ) 0.26⋅ 6.82⋅+ 1.2 0.34⋅ 6.82⋅++

... 4⋅ =:= k

TRAMA 14:

G20 20.78 4.56 1.5 0.3⋅+( )⋅ 4.375 1.875+( ) 2.59⋅ 1.62⋅+[ ] 3⋅ 20.78 8.81 1.2 0.4⋅+( )⋅+ 584.041=:= k



TRAMA 15:

14.21 4.27 1.5 0.3⋅+( ) 2.43 4.56 1.5 03⋅+( )⋅+ 4.125 5.46 1.5 0.3⋅+( )⋅+ 4.37 1.5+( ) 2.59⋅ 1.62⋅+[ ] 3⋅ 20.78 8.81 1.2 0.4⋅+( )⋅+ 607.331==

DT 8:

G22

G1

508.284=:= k

y!164048.5595

11840.756813.855=:=



6.1.3. DISTRIUTIA INCARCARII SEISMICE LA DIARAGME TRANSVERSALE

6.1.2 Determinarea pozitiei centrului de rigiditate

DIARAGMA A)> A)>

DT1 2."62" 0 0

DT2 2.262" " 11.312"

DT3 2.*883 8.*" 2.39*62"

DT 1.08*" 13 1.13*"

DT" 1.0*" 13.*" 1.*812"

DT6 2.38*" 18 2.9*"

DT* 2.262" 23 "2.03*"

DT8 2."62" 2*.62" *0.*89062"

16.9883 230.303*"

DETERMINAREA BOCITIEI ENTRULUI DE RIGIDITATE

"#$ y! y% − 0.291=:=

y% 230.4304375

16.988313.564=:=

"#$ y! y% − 0.291=:=

& 27.625:=

"' 0.05 &⋅ 1.381=:=

"$I "' "#$− 1.091=:=

"$II "' "#$+ 1.672=:= 799.6064:= k !*$I "$I⋅ 872.1705=:= kN⋅



E%eme

:

A>

m2

R

m

R2

m

A> x R

m3

A> x R2

m

; x

Av ,i

Σ Av , i

kN

M:>4 x

Av , i x yRi

Σ( Av ,i x yRi2)

;>DT

kN

I

kN

II

kN

DT1 2."62"

13."6

183.9821 3.*"**" *1."

1

120.61191"3 20.*99

*

&31.8*8

9

88.*302

DT2 2.26

2"

8."6

*3.321 19.3*60"

16".936

"

106.91

9"911."9"

&

1*.**1

2

88.*2

29

DT3 2.*8

83

.81

23.1*6 13.2288

6.61**

3

131.239

882 8.032"

0

&

12.311

1

118.9

88

DT 1.08

*"

0."6 0.318096 0.6133"

0.3"92

9

"1.186"

2013 0.36*0

&

0."62"

"

"0.62

9*

DT" 1.0*

"

0.186 0.03"96 0.1999"

0.03*19

1

"0."981

6933

&

0.1196

"

0.1833

89

"0.*8

"6

DT6 2.38

*"

.36 19.6*81 10."909"

6.981

"

112.3*"

00

&

6.33*8

3

9.*13*

"6

122.0

88

DT* 2.26

2"

9.36 89.0381 21.389"

201.8

*

106.91

9"9

&

12.**"

6

19."80

*3

126.0

22

DT8 2."62"

1.061

19*.*11* 36.03131"06.636

3

120.61191"3

&21."61

8

33.0*

03

1"3.689

1"*."

8

*99.606

&

0.0006

8

0.0011

36

*99.6

*"

6.2. DISTRIUTIA INCARCARII SEISMICE PE VERTICALA

P DT3

1=2F;DT/('1)=2F118.9288/('1)=11.89 +N

2=23.*9 +N

3=3".68 +N



=*."* +N

4=2FF;DT/('1)H=2FF118.9288/("F2.8)=6*.96 +N

7.DETERMINAREA EFORTURILOR AIALE DIN INCARCARI

VERTICALE(NGR"

MONTANT 1 (DT1"

!) DETERMINAREA SUBRAETELOR AERENTES1=2.8203 m2

S=".606 m2

S*=1.12" m2

S10=1.12" m2

S1=".606 m2

S1*=2.8203 m2

;) INARAREA DATA DE BLANSEE

• A?@e$4 :e$!4!

:e$!4!=( S1' S' S*' S10' S1' S1*)F(8.81'1.2F0.)=1*8.106 +N• Ne% ?7$e:

me<>TR1=me<'@<'0.3Fk

me<=(1.2"'.8*").2*"'2.606F."6"/(1.2"'.8*"'2.606)=.3102 +N@<=0.8F0.*969=0.63*"2 +Nme<>TR1=.3102'0.63*"2'0.3 F1."=".39**2 +N

me<>TR2=me<'@<'0.3Fk

me<=1".*"F.2*"'2.3*"F."6"'."F".62"/(1".*"'2.3*"'.")=.96 +N@<=0.8F0.*813=0.62"0 +Nme<>TR2=.96'0.632"0'0.3 F1."="."* +N



(S1'S'S*) me<>TR1'(S10'S1'S1*) me<>TR2=10".13"" +N?) GREUTATEA BROBRIE BE UN NIJEL A UNUI MONTANT

G="08.28/=12*.0*1 +N<) RECULTANTA EORTURILOR AKIALE LA NIJELUL 0

NDT1=:e$!4!'3@%'G=1*8.106'3F10".13""'F12*.0*1=1001.*96" +N

MONTANT 2 (DT2"

!) DETERMINAREA SUBRAETELOR AERENTE

S2=2.8203 m2

S3=1."312 m2

S"=".606 m2

S6=".2" m2

S8=0.3301 m2

S9=1.12" m2

S11=0.3301 m2

S20=0.2813 m2

S12=0.2813 m2

S13=1.12" m2


• A?@e$4 :e$!4!:e$!4!=( S2' S3' S"' S6' S8' S9' S11' S*' S20' S12'

S13)F(8.81'1.2F0.)=18.*19F(8.81'1.2F0.)=1*3.861 +N

• Ne% ?7$e:me<>TR3=me<'@<'0.3Fk

me<=16."F.2*"/16."=.2*" +Nme<>TR3=.2*"'0.3 F1."=.*2" +N

me<>TR=me<'@<'0.3Fk

me<=(".2"'10.9)F."6"'/(".2"'10.9)=."6" +N



@<=0.8F0.3*"=0.3" +Nme<>TR=."6"'0.3"'0.3 F1."=".36" +N

(S2'S"'S8) me<>TR1'(S11'S20) me<>TR2'(S3'S6'S9) me<>TR3'(S12'S13) me<>TR=

.6816F".39**2 '0.611F"."*'*.9062F.*2"'1.063F".36"=*3."**1 +N

?) GREUTATEA BROBRIE BE UN NIJEL A UNUI MONTANTG=((".0&

2F0.2")2."'.1"F2.")2.6*'2F0.2"F2."F6.82'(".'2.*"'1.9)F0.26F6.82=

8*.8"*3 +N<) RECULTANTA EORTURILOR AKIALE LA NIJELUL 0

NDT2=:e$!4!'3@%'G=1*3.861'3F*"."**1'F8*.8"*3=*"2.0219 +N

MONTANT 3 (DT2"

!) DETERMINAREA SUBRAETELOR AERENTES1"= ".3"9 m2

S16=.9688 m2

S18=2.8203 m2

S19=1."312 m2


• A?@e$4 :e$!4!

:e$!4!=( S1"' S16' S18' S19)F(8.81'1.2F0.)=1.6*9* F(8.81'1.2F0.)=136.3*

+N

• Ne% ?7$e:

(S1"'S18) me<>TR2'(S16'S19) me<>TR=8.1*9*F"."*'6."F".36"=80.33 +N?) GREUTATEA BROBRIE BE UN NIJEL A UNUI MONTANT



G=((3.6"&

0.2")2."'2.6"F2.")F2.6*'0.2"F2."F6.82'(3.6"'2.6")F0.26F6.82="6."318

+N<) RECULTANTA EORTURILOR AKIALE LA NIJELUL 0

NDT3=:e$!4!'3@%'G=136.3*'3F80.33'F"6."318=603.8018 +N

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