1)
POS S − AB STUB
0 0 3 0 1 5
0 0 3
0 0 3
0 0 4 1
0 1 5
0 0 5 0 1 5
600
600
600
600
2400
0 1 5
600
− AB ploča d pl=
λ 35
=
510 =14,57 cm=¿ usvajamd pl=16 cm 35
− AB greda
g
d
bg
b g=30 cm;d g=
L 12
=
600 =50 cm 12
−analizaopterećenja meuspratna!onstru!cija
b / d =30 / 50
600
600
600
Blindit d = 2.5 cm Mršavi beton d = 4.0 cm Monolitna MSK d = 16 cm Malter d = 1.5 cm
6 1
− par!et : g p=0,024∗8,0 =0,192 !" / m2 −blindit : gbl= 0,025∗18,0 =0,450 !" / m2 −mr#avibeton : g mb=0,040∗22,0=0,880 !" / m2 − $S% : g ms! = 0,160∗25,0 =4,000 !" / m2 −malter : gmal =0,015∗19,0=0,285 !" / m2
∑ g =5,807 !" / m
2
s
;
ps=1,50 !" / m
2
−!rovna !rovna $S% Cementna *oš+ljica d=#$0 cm )idroizolacija Cementni malter d=1.5 cm Sloj za pad 1.5!" d=#$0 cm %ermoizolacija d=5$0 cm Mr&avi beton d=4$0 cm MSK d=16$0 cm 'la(on d=1.5 cm
−cementna!o#uljica : g c! =0,03∗22,0= 0,660 !" / m2 −cementnimalter : g cm= 0,015∗19,0 =0,285 !" / m2 sloj za pad pad : g szp= 0,03∗22,0 =0,660 !" /m 2 −sloj
−termoizolaci ja : gti =0,05∗1,0=0,050 !" / m2
−mr#avibeton : g mb=0,04∗22,0 =0,880 !" / m2 − $S% : g ms! = 0,16∗25,0 = 4,000 !" / m2 − pla&on : g pl =0,015∗19,0=0,285 !" / m2
∑ g ! =6,820 !" / m s ! = 0,75 !" / m 2
−sopstvenate'inagrede ggr =0,30∗0,50∗25,0 =3,75 !" / m
−u!upno svedeno opterećenje( $S% ) u
gs = g s∗ λ + g gr =5,807∗5,1 + 3,75 =33,37 !" /m
pus = p s∗ λ =1,5∗5,1 =7,65 !" / m
−u!upno svedeno opterećenje( !rovna !onstru!cija− pro(odna ) u
g! = g ! ∗ λ + g gr = 6,82∗5,1 +3,75= 38,53 !" / m p! = p ! ∗ λ + s ! ∗ λ =( 1,5 + 0,75 )∗5,1=11,48 !" / m u
−dimenzionisanjestubova 38.53
11.48
0 0 3
33.37
0 0 3
7.65
0 0 3
33.37
0 0 3
7.65
0 0 3
33.37
0 0 3
7.65
g
0 0 5
600
p
0 0 5
600
2
2
;
p! =1,50 !" / m
" g =( 38,53 + 3∗ 33,37 )∗6,0 =831,34 !" ma)
" p =( 11,48 +3∗7,65 )∗ 6,0=206,58 !" ma)
ma)
ma)
ma)
" * = " g + " p =1038,42 !"
+ 0 0,7∗ & b!
+ 0 0,7∗35
, 0,35
=0,35=¿ + 0=8575,0 !" / m2
ma)
" * 1038,42 + 0 = =¿ 8575,0= =¿ A b= 0,121 m2 A b A b
−usvajam !vadratni prese! stuba =¿ 0,121= a2=¿ a =34,79 cm=¿ US-A.A$ a =35 cm
33
34
21 0 0 3
16
0 0 4 1
0 0 3
7 21 7
6
0 0 5
1
1
22
2
15
9
10
23
24
8
9
3
600
28 14
8
2
15
27 13
20
14
26 12
6
32 19
13
25
20
31 18
12
25
19
30 17
11
24
18
29
11
36
23
17 16
0 0 3
35
22
10
4
4
3 600
5
600
5 600
2400
Štapovi od 1 do 20 - A st!bovi
Štapovi od 21 do 36 - A g"#d#
m c 0 , 5 3 = s b
bs=35,0cm
m c 0 , 0 5 = G d
bG=30,0cm
X
−greda :
−stub :
A ) =0,3∗0,5=0,15 m
/ z=
0,3∗0,5 12
2
A ) =0,35∗0,35=0,1225 m2
3
=0,003125 m
4
ggr =0,30∗0,50∗25,0 =3,75 !" / m
0,35∗0,35 / z= 12
3
=0,001251 m4
ggr =0,35∗0,35∗25,0 =3,0625 !" / m
−#ema opterećenja
-'ta(%o opt#"#c#%&# 21 0 0 3
16 16
0 0 3
0 0 4 1
33
0 0 3
29
0 0 5
17
25
12
7
35
33.37 18 18
30
13
26
33.37
31
36
19
27
14
32
9
20 15
28
9 23
25 20
14
8 8
22
24 19
33.37 13
7 21
23
34
12
6 6
22 17
11 11
38.53
15 10
24
10
3.0625
3.0625
3.0625
3.0625
3.0625
1
2
3
4
5
1
2 600
4
3 600
600
5 600
2400
-$ov"#m#%o opt#"#c#%&# 21 0 0 3
16 16
0 0 3
0 0 4 1
33
29
0 0 3
0 0 5
7.65
7.65
25
12 7
21
1
7
7.65 22
2
1
19
14
23
9
25 20
32
20 15
28
9
15 10
24
4
10 5
4 600
2400
−Seizmič!emase
27
3 600
36
14
3
2 600
31
8 8
24 19
13 13
26
35
18 18
30
12
6 6
17
23
34
17
11 11
22
11.48
5 600
m=100 ∗ g + 50 ∗ p
m1=( 33,37∗24,0 + 5∗ 4,0∗3,0625 ) + ( 0,5∗7,65∗24,0 ) =953,93 !"
m2=( 33,37∗24,0 + 5∗ 3,0∗3,0625 ) + ( 0,5∗7,65∗24,0 )=938,61 !"
m3=( 33,37∗24,0 + 5∗3,0∗3,0625 ) + ( 0,5∗7,65∗24,0 )= 938,61 !" m4= (38,53∗24,0 + 5∗1,5∗3,0625 )+ ( 0,5∗11,48∗24,0 )=1085,45 !"
)
m4 21 0 0 3
16
0 0 4 1
24
35
18 17
29
11
25
36
19
30
12
20
18
19
31
13
20
32
14
15
m2 12
25
6
26
7
13
14
27
8
15
28
9
10
m1 6
0 0 5
23 m3
11 0 0 3
34
17 16
0 0 3
22
33
7
21
1
22
2
1
9
23
3
2
600
8
4
3
600 2400
5
4
600
10
24
5
600
X
SASS /0LA0"/ 1A.L STRUCTURE POS AB RAM - SEIZMICKE MASE TYPE PLANE FRAME NUMB OF JOIN 25 NUMB OF MEMB 36 NUMB OF SUPP 5 NUMB OF LOAD 4 TABU ALL JOIN COOR 1 0. 0. SUPP UNTIL 5 STEP 1 6. 0. 6 0. 5. UNTIL 10 STEP 1 6. 0. 11 0. 8. UNTIL 15 STEP 1 6. 0. 16 0. 11. UNTIL 20 STEP 1 6. 0. 21 0. 14. UNTIL 25 STEP 1 6. 0. MEMB INCI 1 1 6 UNTIL 5 6 6 11 UNTIL 10 11 11 16 UNTIL 15 16 16 21 UNTIL 20 21 6 UNTIL 24 25 11 12 UNTIL 28 2! 16 1 UNTIL 32 33 21 22 UNTIL 36 MEMB PROP PRIS 1 T"RU 20 A# 0.1225 IZ 0.001251 21 T"RU 36 A# 0.1500 IZ 0.003125 CONS E 32500000. ALL LOAD 1 $ SEIZMICKE MASE$ JOIN LOAD 8 FORC # !53.!3 13 FORC # !38.61 18 FORC # !38.61 23 FORC # 1085.45 TRACE SOL%E PROBLEM CORRECTLY SPECIFIED& E#ECUTION TO PROCEED 1 PA'E
1
(((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( STRUCTURE POS AB RAM - SEIZMICKE MASE (((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( REZULTATI PRORACUNA SEIZMICKI" SILA PREMA TEORIJI DINAMIKE KONSTRUKCIJA BROJ ETAZA
N( 4
BROJ RAMO%A M( 1
MATRICA FLEKSIBILNOSTI) .5!208E-04
.634!E-04
.6388E-04
.63!46E-04
.634!E-04
.85405E-04
.88320E-04
.88668E-04
.6388E-04
.88320E-04
.10880E-03
.1112E-03
.63!46E-04
.88668E-04
.1112E-03
.13252E-03
.848E-03
.38516E-03
SOPST%ENE %REDNOSTI .34366E-01
.2301E-02
SOPST%ENI %EKTORI TON OSCILO%ANJA) 1 .35544
.403
.5486
.5!248
TON OSCILO%ANJA) 2 -.6660
-.4516
.0!81!
.58366
TON OSCILO%ANJA) 3 .5808
-.38424
-.51
.410
TON OSCILO%ANJA) 4 -.333!
.64
-.62150
KOEFICIJENAT OBJEKTA
KO(
KATE'ORIJA TLA - PODLO'E
(
KOEFICIJENAT DUKTILITETA
KP(
KOEFICIJENAT SEIZMICNOSTI KC(
.20822 1.00000 2 1.00000 .050
SEIZMICKE SILE - TON OSCILO%ANJA) KOEFICIJENAT DINAMICNOSTI KD( PERIODA OSCILO%ANJA T(
1.1648
SPRAT
SPRATNE
AMPLITUDE
OSCILO%ANJA
1
.600!
C%OR
TEZINE
SEIZMICKE
SILE
1
.5!!!
!54.
*
8+
20.580
2
.!45
!3!.
* 13+
26.81
3
.!261
!3!.
* 18+
31.25
4
1.0000
1085.
* 23+
3!.034
SEIZMICKE SILE - TON OSCILO%ANJA) KOEFICIJENAT DINAMICNOSTI KD( PERIODA OSCILO%ANJA T(
SPRAT
AMPLITUDE
OSCILO%ANJA
2
1.00000
.3283
SPRATNE
C%OR
TEZINE
SEIZMICKE
SILE
1
-1.1438
!54.
*
8+
-64.31!
2
-.3!
!3!.
* 13+
-42.816
3
.1682
!3!.
* 18+
!.308
4
1.0000
1085.
* 23+
63.!84
SEIZMICKE SILE - TON OSCILO%ANJA) KOEFICIJENAT DINAMICNOSTI KD( PERIODA OSCILO%ANJA T(
SPRAT
AMPLITUDE
OSCILO%ANJA
3
1.00000
.153
SPRATNE
C%OR
TEZINE
SEIZMICKE
SILE
1
1.406
!54.
*
8+
56.561
2
-.!213
!3!.
* 13+
-36.424
3
-1.3851
!3!.
* 18+
-54.64
4
1.0000
1085.
* 23+
45.22
SEIZMICKE SILE - TON OSCILO%ANJA) KOEFICIJENAT DINAMICNOSTI KD( PERIODA OSCILO%ANJA T(
SPRAT
AMPLITUDE
OSCILO%ANJA
4
1.00000
.1233
SPRATNE
C%OR
TEZINE
SEIZMICKE
SILE
1
-1.6030
!54.
*
2
3.2536
!3!.
* 13+
8+
-32.158 64.221
3
-2.!84
!3!.
* 18+
-58.!15
4
1.0000
1085.
* 23+
22.82
1 PA'E
2
STRUCTURE POS AB RAM - SEIZMICKE MASE ((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((( TON OSCILO%ANJA) 1
UTICAJI OD SEIZMICKI" SILA
(((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((
−2ema opterećenja
'1 )
33
34
21 0 0 3
22
16
0 0 4 1
26
11 0 0 3
6
0 0 5
7
1
22
2
1
15
9
20.580
10
23
24
8
9
3
2 600
28 14
8
21
15
27 13
7
20
14
26.817
12
6
32 19
13
25
20
31 18
12
25
19
31.257
17
11
36 24
18 30
16
35 23
17 29
0 0 3
39.034
10
4
3 600
4 600
2400
5
5 600
X
'2 )
33 0 0 3
22
16
17
11
0 0 4 1
6
0 0 5
1
10
23
24 9
10
4
3
2 600
15
9
8
2
1
28 14
20.580
22
15
27
13
7
20
14
8
21 6
32 19
26.817
7
20
31
13 12
25
19
18
26
11 0 0 3
31.257
12 25
36 24
18 30
16
35
23
17 29
0 0 3
39.034
34
21
3 600
4 600
2400
5
5 600
X
− %lasi&i!acija !onstru!cije premaosetljivosti na (orizontalno pomeranje
3
√
∑ 4 , 0,2+ 0,1 n ( n , 3 )=¿ %onstru!cija je sanepomerljivimčvorovima ∑ 5 ¿ / , 0,6 ( n6 4 ) b
b
n −spratnost
3 − visina de&ormabilnog dela !onstru!cije
∑ 4 =( 436,08 +898,54 +872,50 +898,54 + 436,09 ) +( 97,46 +212,69 +205,96+ 212,69+ 97,49 ) ∑ 4 =4368,07 !" 7
5∗ 5 b ¿ / b=5∗3,3∗10 ∗ 0,001251=206415 !" m
2
5b =3,3∗10 !" / m ( za $B 35 ) 7
0,35 / b= 12
2
4
=0,001251 m4
3 =14 m
√
14∗
4368,07 206415
=2,0366 > 0,6 =¿ %onstru!cija je sa pomerljivim čvorovima
STUB 2−7
− /zvijanje stuba (i= (∗!
7 =
5∗ / ∑ ( ()
s
5∗ / ∑ ( l)
8
! =min
(
1 + 0,15∗( 7 a+ 7 b) 2+ 0,3∗7 min
0,3∗0,5 / grede = 12 0,35 / stuba= 12
)
3
=0,003125 m4
4
=0,001251 m4
7 2=0 ( u!lje#tenje )
+ (= ∗ ) +( ∗ ) = + ( ∗ ) +( ∗ ) )= ∗( + = ( + )=¿ = + ∗ = 5 / (
7 7
5 / l
2 −7
5 / (
7 − 12
6− 7
5 / l
7 −8
0,001251 5 0,003125 6
1 0,15 0 0,641 1,096 2 0,3 0 2
! min
0,001251 3 0,003125 6
=0,641
! 1,096
(i=1,096∗500 cm=548,2 cm=5,482 m
−vit!ost stuba λ =
(i
=
i min
imin=
√
5,482 =54,170 =¿ Oblast srednje vit!osti 25 < λ < 75 0,1012
0,001251 0,35
2
=0,1012 m
¿> $etoda dopuns!e e!scentričnosti e =e 1+ e 0+ e9 + e // e 1−e!scentricitet poteoriji/ reda $ * 86,37 e 1= = =0,0401 m " * 2153,87 e 0−e!scentricitet usled netačnosti priizvoenju e 0=
( 300
=
5,0 = 0,01667 m 300
e 9 −dodatnie!scentricitet usled tečenjabetona 1 ¿ λ , 50 ( λ =54,170 > 50 )=¿ nije ispunjenuslov 2¿
e1 d
62
(
0,0401 = 0,1146 < 2 0,35
)=¿
nijeispunjenuslov
3 ¿ " g , 0,2∗ " * ( 898,54 > 0,2∗2153,87 =430,774 )=¿ nijeispunjen uslov
e 9 =( e 1 g
+ e )∗( 0
: 5
2,718
1+ : 5
∗9
−1,0
)
e 1 g=
$ g " g
=
1,50 898,54
=0,0,0017
" g : 5 = " 5 2
" 5 = : 5 = d m=
5b ¿ / b∗ 2
(i
=
3,3∗10
7
∗0,001251∗ 2 5,482
2
= 13557,916 !"
898,54 = 0,0663 13557,916 2
2 A b
2∗0,35 = =0,175 m 4∗0,35 O
9 =3,6 ( iz tablice 11, PBAB 87 )
(
e 9 =( 0,0017 + 0,0401 )∗ 2,718
0,0663 ∗3,6 1+ 0,0663
)
−1,0 =0,0105 m
e // −dodatnie!scentricitet // reda e1 d
=
(
e 0,0401 =0,1146 0 < 1 < 0,3 0,35 d
¿> e // =
d∗ λ −25 100
∗
√
0,1 +
)
√
e 1 0,35∗54,17 −25 0,0401 = ∗ 0,1 + = 0,0473 m 100 0,35 d
e =0,0401 + 0,01667 + 0,0105+ 0,0473=0,1146 m e =0,1146 m
−
nu=
$ u 2
b∗d ∗& b " u
b∗d ∗& b
=
=
246,834
0,35∗0,35
2
∗23∗103
2153,87 3
0,35∗0,35∗23∗10
=0,250
= 0,764
¿> iz dijagrama intera!cije ´== 0,42 ==
= ´ ∗& b
2 −7
+ v
=
0,42∗23 =0,0403 240 2
A a = =∗b∗d = 0,0403∗35∗35= 49,31 cm
STUB 7−12
− /zvijanje stuba (i= (∗!
7 =
5∗ / ∑ ( ()
s
5∗ / ∑ ( l)
8
! =min
(
1 + 0,15∗( 7 a+ 7 b) 2+ 0,3∗7 min
0,3∗0,5 / grede = 12 0,35 / stuba= 12
5 / (
5 / l
( ) = ∗ +( ) ∗ +( ) = ∗ +( )
+
2 −7
5 / l
! min
=0,003125 m4
=0,001251 m4
( ) = ∗ ( ) ∗ ( ) = ∗ ( ) = ( +
7 12
3
4
5∗ / (
7 7
)
6− 7
5∗ / (
7 − 12
5 / l
7 −8
7 −12
5 / (
11−12
5 / l
12 −17
12−13
0,001251 0,001251 + 5 3 0,003125 0,003125 + 6 6
=0,641
0,001251 0,001251 + 3 3 =0.801 0,003125 0,003125 + 6 6
1 0,15∗( 0.641 + 0,801 )=1.216 2 + 0,3∗0.641=2.192
)=¿ =
! 1.216
(i=1,216∗300 cm=364.8 cm= 3.648 m
−vit!ost stuba λ =
(i i min
imin=
√
=
3,648 =36,05 =¿ Oblast srednjevit!osti 25 < λ < 75 0,1012
0,001251 0,35
2
=0,1012 m
¿> $etoda dopuns!e e!scentričnosti e =e 1+ e 0+ e9 + e //
e 1−e!scentricitet poteoriji/ reda $ * 54,06 e 1= = =0,0332 m " * 1630,17 e 0−e!scentricitet usled netačnosti priizvoenju e 0=
( 300
=
3,0 300
= 0,01 m
e 9 −dodatnie!scentricitet usled tečenjabetona 1 ¿ λ , 50 ( λ =36,05 < 50 )=¿ ispunjenuslov
2¿
e1 d
62
3 ¿ " g , 0,2∗ " *
e 9 =0 m e // −dodatnie!scentricitet // reda e1
0,0332
d
0,35
=
(
)
e
=0,0949 0 < 1 < 0,3
¿> e // = d∗ λ − ∗ 25
100
√
d
0,1 +
√
e 1 0,35∗36,05− 25 0,0332 = ∗ 0,1+ =0,0171 m 100 0,35 d
e =0,0332 + 0,01 + 0 + 0,0171=0,0603 m
e =0,0603 m
−
nu=
$ u 2
b∗d ∗& b " u
b∗d ∗& b
=
=
== = min=1,000
98,299
0,35∗0,35
2
∗23∗103
1630,17 3
0,35∗0,35∗23∗10
=0,100
= 0,579
7 −12
A a
=
= 100
∗b∗d =
1 ∗35∗35=12,25 cm2 100
STUB 12−17
− /zvijanje stuba (i= (∗!
7 =
5∗ / ∑ ( ()
s
5∗ / ∑ ( l)
8
! =min
(
1 + 0,15∗( 7 a+ 7 b) 2+ 0,3∗7 min
0,3∗0,5 / grede = 12
/ stuba=
7 12
7 17
(= ( ( = (
0,35 12
3
=0,003125 m4
4
=0,001251 m4
) ∗ ) ∗ ) ∗ )
5∗ / (
7 − 12
5 / l
1 − 12
5 / ( 5 / l
! =min
)
( ) ∗ +( ) ∗ +( ) ∗ +( ) +
5 ∗ / (
0,001251 0,001251 + 3 3 12 −17 = =0.801 0,003125 0,003125 5 / + 6 6 l 12 −13
12 −17
5 / (
17 −22
16 −17
5 / l
17 −18
=
0,001251 0,001251 + 3 3 = 0.801 0,003125 0,003125 + 6 6
(+
1 0,15∗( 0.801 + 0,801 )=1.240 2+ 0,3∗ 0.801=2.240
)=¿ =
! 1.240
(i=1,240∗300 cm=372 cm=3.720 m
−vit!ost stuba λ =
(i i min
=
3,720 0,1012
=36,76 =¿ Oblast srednje vit!osti 25 < λ < 75
imin=
√
0,001251 0,35
2
=0,1012 m
¿> $etoda dopuns!e e!scentričnosti e =e 1+ e 0+ e9 + e // e 1−e!scentricitet poteoriji/ reda $ * 36,45 e 1= = = 0,0323 m " * 1127,85 e 0−e!scentricitet usled netačnosti priizvoenju e 0=
( 300
=
3,0 300
= 0,01 m
e 9 −dodatnie!scentricitet usled tečenjabetona 1 ¿ λ , 50 ( λ =36,76 < 50 )=¿ ispunjenuslov 2¿
e1 d
62
3 ¿ " g , 0,2∗ " *
e 9 =0 m e // −dodatni e!scentricitet // reda e1 d
=
(
e 0,0323 =0,0923 0 < 1 < 0,3 0,35 d
¿> e // =
d∗ λ −25 100
∗
√
0,1 +
)
e1
0,35∗36,76 − 25
d
100
=
e =0,0323 + 0,01 + 0 + 0,0180= 0,0603 m e =0,0603 m
−
√
∗
0,1 +
0,0323 0,35
=0,0180 m
$ u=e∗ " u =0,0603∗1127,85=68,01 !"m m u=
nu=
$ u 2
b∗d ∗& b " u
b∗d ∗& b
=
=
68,01
0,35∗0,35
2
∗23∗103
1127,85 3
0,35∗0,35∗23∗10
=0,069
= 0,400
== = min=1,000 12 −17
A a
=
=
100
∗b∗d =
1 ∗35∗35=12,25 cm2 100
STUB 17−22
− /zvijanje stuba (i= (∗!
7 =
5∗ / ∑ ( ()
s
5∗ / ∑ ( l)
8
! =min
(
1 + 0,15∗( 7 a+ 7 b) 2+ 0,3∗7 min
0,3∗0,5 / grede = 12 0,35 / stuba= 12
7 22
7 17
5 / l
5 / ( 5 / l
(
3
=0,003125 m4
4
=0,001251 m4
( = ∗ ( ) ∗ ( ) = ∗ ( )
! =min
)
) ∗ +( ) ∗ +( ) ∗ +( )
5∗ / (
17 −22
21 −22
5 / l
12 −17
16 −17
=
0,001251 3 =0.400 0,003125 0,003125 + 6 6
=
0,001251 0,001251 + 3 3 = 0.801 0,003125 0,003125 + 6 6
22 − 23
5 / (
17 −22
5 / l
17 −18
1 + 0,15∗( 0,400 + 0,801 ) =1,180 2+ 0,3∗ 0.400=2.120
(i=1,180∗300 cm=354 cm=3.540 m
)=¿ =
! 1,180
−vit!ost stuba λ =
(i
=
i min
imin=
√
3,540 =34,98 =¿ Oblast srednje vit!osti 25 < λ < 75 0,1012
0,001251 0,35
2
=0,1012 m
¿> $etoda dopuns!e e!scentričnosti e =e 1+ e 0+ e9 + e // e 1−e!scentricitet poteoriji/ reda $ * 30,30 e 1= = =0,0481 m " * 630,52 e 0−e!scentricitet usled netačnosti priizvoenju e 0=
( 300
=
3,0 300
= 0,01 m
e 9 −dodatnie!scentricitet usled tečenjabetona 1 ¿ λ , 50 ( λ =34,98 < 50 )=¿ ispunjen uslov 2¿
e1 d
62
3 ¿ " g , 0,2∗ " *
e 9 =0 m e // −dodatnie!scentricitet // reda e1 d
=
(
e 0,0481 =0,137 0 < 1 < 0,3 0,35 d
¿> e // =
d∗ λ −25 100
∗
√
0,1 +
)
e =0,0481 + 0,01 + 0 + 0,0170=0,0751 m
e =0,0751 m
√
e 1 0,35∗34,98− 25 0,0481 = ∗ 0,1+ =0,0170 m 100 0,35 d
−
nu=
$ u 2
b∗d ∗& b " u
b∗d ∗& b
=
=
47,352
0,35∗0,35
2
∗23∗103
630,52 3
0,35∗0,35∗23∗10
=0,048
= 0,224
== = min=1,000 17 − 22
A a
=
= 100
∗b∗d =
A a =49,31 cm
1 100
∗35∗35=12,25 cm2
2
US-A.A$ 8 > 32 ( A a =64,34 cm stv
uzengije : > u=
> ma) 3
2
)
= 32 =10.66 =¿ US-A.A$ U> 12 / 7 .5 / 15 cm 3
−Speci&i!acijaarmature
redni
ϕ
Obli! i dimenzije
broj
[ mm ]
n
[ !om]
ln
[ m]
g , ln &m-
¿ ¿ */
8
[ !g ]
m 5 # 1
1
20
#2
6
2.66
15.6
6.#1#
100.6
#2
6
.1
4.46
6.#1#
2.61
#2
6
.20
55.20
6.#1#
#4.4
#2
6
6.5
41.5
6.#1#
262.4
#2
6
5.0
#5.40
6.#1#
22#.4
#2
6
2.0
1.40
6.#1#
10.5
12
155
1.6
22.0
0.
242.25
660
2
111 20
20
3
65
4
50
5
20
6
#032
1#
1#
7
#0
#0
#0
,
156.2
