10 - Primer Koriscenja Dijagrama Interakcije

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


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


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


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


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

16

10 - Primer Koriscenja Dijagrama Interakcije

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