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PRIMER TESTIMONIO GUATEMALA
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ADMINISTRACIÓN FINANCIERA
Go d išn ji zad atak –L is t 2: PRIMER 6
Ø Za stub pravougaonog popre nog nog preseka za koji su date t ri stalnog i r i k o m b i n ac a c i je j e u t ic i c aj a j a usled povremenog optere# enja, enja, odrediti potrebnu površinu armature. Uticaj izvijanja zanemariti. a . N g = 1 3 8 5 . 8 k N b . N g = 3 6 8 2 . 9 k N
M p = ±82 ±826.3 6.3 k Nm M p = ±63 ±637.7 7.7 k Nm
c.
M p = ±23 ±238.9 8.9 k Nm
N g = 4 8 2 0 . 8 k N
b = 40 cm d = 85 cm
⇒ f B = = 25.5 MPa M B 4 0 0 = 400 MPa RA 400/500 400/500 ⇒ σv =
1
Stari go dišnji za zadatak datak – Lis t 4: PRIMER 6
Ø ko m bin acija uticaja a . Ø pretp. εa 1 > 3‰ 3‰ (zat.) zat.) ⇒ γ u,g = 1.6 ; γ u,p = 1.8 M u = 1 .8 ×826.3 = 1487.3 1487.3 k Nm N u = 1 .6 ×1385. 1385.8 8 = 2217. 2217.3 3 kN INA: R A UNA U NA NJE B EZDIMENZI EZDIMENZIONA ONA L NIH NIH VELI I NA:
nu =
N u 2217 .3 = = 0 .256 b × d × f B 40 × 85 × 2 .55
M u 1487 .3 × 10 2 mu = = = 0 .202 2 2 b × d × f B 40 × 85 × 2 .55
Ø pr etp. a = 6.5 6.5 cm
⇒
a /d /d = = 6.5/85 = 0.076 # 0.075
2
3
4
0.255
0.202
0.256
Stari go dišnji za zadatak datak – Lis t 4: PRIMER 6
Ø ko m bin acija uticaja a . Ø p o v o l j n o dejstvo stalnog $ $enja e nja stal nog optere Ø
M u = 1 .8 ×826.3 = 1487.3 1487.3 k Nm N u = 1 .0 ×1385.8 = 1385.8 1385.8 k N
INA: R A UNA U NA NJE B EZDIMENZI EZDIMENZIONA ONA L NIH NIH VELI I NA:
nu =
N u 1385 .8 = = 0 .160 b × d × f B 40 × 85 × 2 .55
M u 1487 .3 × 10 2 mu = = = 0 .202 2 2 b × d × f B 40 × 85 × 2 .55
5
6
0.320
0.255
0.202
0.16
0.256
Stari go dišnji za zadatak datak – Lis t 4: PRIMER 6
Ø ko m bin acija uticaja b . Ø pretp. εa 1 < 0‰ ( p r it i t .) .) ⇒ γ u,g = 1.9 ; γ u,p = 2.1 M u = 2 .1 ×637.7 637.7 = 1339.2 kN m N u = 1 .9 ×3682.9 = 6997.5 6997.5 k N INA: R A UNA U NA NJE B EZDIMENZI EZDIMENZIONA ONA L NIH NIH VELI I NA:
N u 6997.5 nu = = = 0 .807 b × d × f B 40 × 85 × 2 .55 M u 1339.2 × 10 2 mu = = = 0 .182 2 2 b × d × f B 40 × 85 × 2 .55
7
8
0.320
0.255
0.202 0.182
0.315
0.16
0.256
0.807
9 εa = 2‰
εa = 0.5‰ εa = 0‰
0.182
0.315
0.807
10
Stari go dišnji za zadatak datak – Lis t 4: PRIMER 6 ZATEZANJE 2
a
Aa2
PRITISAK 2
εb2 εa2
3.5 d 7 3
b 2
d
a h
a
a'
c
d
e f g
Aa1 1
a
εa1 εb1
10‰
3
ε
v
0
C h
2
d 7 4
3.5
εa1 = 0.38‰
b
γ u,i 1.6 (1.8)
γ u ,g = 1.9 −
1.9 − 1.6 × 0 .38 = 1.862 3 − 0
1.862 (2.062)
γ u p .1 − , = 2
1.9 (2.1)
2 .1 − 1.8 × 0 .38 = 2 .062 3 − 0
Stari go dišnji za zadatak datak – Lis t 4: PRIMER 6
Ø ko m bin acija uticaja b . ×637.7 = 1314.9 M u = 2.062 1314.9 k Nm ×3682.9 N u = 1.862 3682.9 = 6857.6 k N INA: R A UNA U NA NJE B EZDIMENZI EZDIMENZIONA ONA L NIH NIH VELI I NA:
N u 6857.6 = = 0 .791 nu = b × d × f B 40 × 85 × 2 .55 M u 1314.9 × 10 2 mu = = = 0 .178 2 2 b × d × f B 40 × 85 × 2 .55
11
12
0.320
0.255
0.202 0.182 0.178
0.315
0.290
0.16
0.256
0.791
0.807
Stari go dišnji za zadatak datak – Lis t 4: PRIMER 6
Ø ko m bin acija uticaja c . Ø pretp. εa 1 < 0‰ ( p r it i t .) .) ⇒ γ u,g = 1.9 ; γ u,p = 2.1 M u = 2 .1 ×238.9 = 501.7 501.7 kN m N u = 1 .9 ×4820.8 4820.8 = 9159.5 k N INA: R A UNA U NA NJE B EZDIMENZI EZDIMENZIONA ONA L NIH NIH VELI I NA:
N u 9159.5 nu = = = 1.056 b × d × f B 40 × 85 × 2 .55 M u 501.7 × 10 2 mu = = = 0 .068 2 2 b × d × f B 40 × 85 × 2 .55
13
14
0.320
0.255
0.202 0.182 0.178
0.315
0.290 0.068
0.23 0.16
0.256
0.791
0.807
1.056
15
Stari go dišnji za zadatak datak – Lis t 4: PRIMER 6 Rezime:
Ø Ø Ø
kombinacija a : :
≈ 0.32
kombinacija b :
≈ 0.29
kombinacija c :
≈ 0.23
Aa = µ × b × d ×
f B σv
2 .55 2 Aa = 0 .32 × 40 × 85 × = 69 .36 cm 40 69 .36 2 Aa1 = Aa 2 = = 34 .68 cm 2
usvojeno:
( ( ±34.37 ±7RØ25
cm2 )
Stari go dišnji za zadatak datak – Lis t 4: PRIMER 6 5RØ25 5 . 4
5 . 5
2RØ25 0 2
2RØ12 UØ8/25
5 8
a II = 10 cm a 1 = (5× (5 ×4 .5 + 2 ×1 0 )/7 a 1 = 6.07 6.07 cm = a 2
5 2
2RØ12
0 2
5 . 4
a I = 4.5 c m
2RØ25 5 . 5
5RØ25 4.5 8
15 40
8 4.5
h = 85 - 6.07 6.07 = 78.93 78.93 cm a /d .0 7 /85 /8 5 = 0 .07 .0 7 1 1 = 6 .07