Elaborat Iz Fundiranja

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Grañevinski fakultet univerziteta u Beogradu - Fundiranje 1.

ZADATAK

20

dz = 0,2 m MB 30

Df

H

50

V M

Hg = 70 kN/m Vg = 800 kN/m

e d

Mg = 150 kNm/m σdoz = 200 kN/m2 γ = 18 kN/m3

1.1. Centrisanje temelja

∑M = 0 e=

⇒

V ⋅e − M − H ⋅d = 0

d = 0,70 m

Pr etpostavljamo :

M + H ⋅ d 150 + 70 ⋅ 0,7 = = 0,249 m V 800

1.2. Odreñivanje dimenzija kontaktne površine

D f = t + d = 0,5 + 0,7 = 1,2 m F pot = B ⋅ 1 =

σ doz

V 800 = = 4,585 m 2 − 0,85 ⋅ γ b ⋅ D f 200 − 0,85 ⋅ 25 ⋅ 1,2

Usvojeno : B = 4,6 m

0,50

1.3. Odreñivanje potrebne visine trakastog temelja

V 800 = = 173,91 kN / m B 4,6 1 1 M = ⋅ q n ⋅ l 2 = ⋅ 173,91 ⋅ 2,45 2 = 521,95 kNm / m 2 2 T = q n ⋅ l = 173,91 ⋅ 2,45 = 426,08 kN / m qn =

0,70

e= 0 ,2 5

1 ,9 5

0 ,2 0

2 ,45

qn

Matić Slavko 63/03

Trial version of ABC Amber PDF Merger, http://www.processtext.com/abcpdfmg.html

1


Grañevinski fakultet univerziteta u Beogradu - Fundiranje Dimenzionisanje

-

MB 30 ⇒ f b = 20,5 MPa = 2,05 kN / cm 2 γ = 1,60

oo

k = 2,311

Tu 1,6 ⋅ 426,08 = 68,86 cm = z b ⋅ τ b 0,9 ⋅ 100 ⋅ 0,11 h = 70 cm

Usvojeno :

Pr etpostavlja se : hstat = h − 5 − k=

o

Mu 1,6 ⋅ 521,95 ⋅ 100 = 2,311 ⋅ = 46,64 cm fb ⋅ b 2,05 ⋅ 100

hM = k ⋅ hT =

ε b / ε a = 3,50 / 10

hstat Mu fb ⋅ b

Aa , pot = µ ⋅

φ 2

=

φ 22 (a a(1) = 3,80 cm 2 )

= 70 − 5 −

2,2 = 63,9 cm 2

63,9

= 3,166

1,6 ⋅ 521,95 ⋅ 100 2,05 ⋅ 100

⇒ ε b / ε a = 1,925 / 10

o

oo

µ = 10,552

o

o

100 ⋅ 63,9 2,05 b ⋅ h fb ⋅ = 10,552 ⋅ ⋅ = 34,56 cm 2 100 σ v 100 40

A pod = 0,2 ⋅ Aa , pot = 0,2 = 0,2 ⋅ 34,56 = 6,91 cm 2

Usvojeno:

Glavna : Podeona :

(

R φ 22 / 10 38 cm 2

(

)

R φ 14 / 20 7,70 cm 2

)

1.4. Analiza opterećenja - Korisno opterećenje: - Težina temelja: - Težina nadsloja:

800 kN/m B ⋅ d ⋅ γ b = 4,6 ⋅ 0,7 ⋅ 25 = 80,50 kN / m (4,6 − 0,20) ⋅ 0,50 ⋅ 18 = 39,60 kN / m

∑V = 920,10 kN / m σ rač =

∑V B

=

920,10 = 200,022 ≈ σ doz = 200 kN / m 2 4,6

Matić Slavko 63/03


2



3



4



ZADATAK

Stub 50 x 50 cm

50

MB 30 Hg = 80 kN/m

50

V M

Vg = 800 kN

Vp = 450kN

Mg = 160 kNm/m Mp = ±120 kN/m

H

γ = 18 kN/m3

2.1.

c = 20 kN/m3

φ = 21°

Fφ = 1.50

Fc = 2.50

Centrisanje temelja za stalno opterećenje

∑M = 0 e=

Mg + Hg ⋅d Vg

Vg ⋅ e − M g − H g ⋅ d = 0

⇒ =

Pr etpostavljamo :

d = 0,60 m

160 + 80 ⋅ 0.6 = 0,26 m 800

2.2. Odreñivanje sila u težištu temeljne spojnice V = Vg + V p = 800 + 450 = 1250 kN ←

←

→

→

M = M p + V p ⋅ e = 120 + 450 ⋅ 0.26 = 237 kNm M = M p − V p ⋅ e = 120 − 450 ⋅ 0.26 = 3.0 kN / m 2.3. Odreñivanje dimenzija kontaktne površine Usvojeno:

k=

L = 1 .5 B

pp. σ doz = 300 kN / m 2

D f = 0.5 + 0.6 = 1.1 m

V M + ≤ σ doz − 0.85 ⋅ γ b ⋅ D f F W 1250 237 ⋅ 6 + ≤ 300 − 0.85 ⋅ 25 ⋅ 1.1 k ⋅ B2 k 2 ⋅ B3

Matić Slavko 63/03


5


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 1250 1422 + 2 3 ≤ 276.63 2 1 .5 ⋅ B 1 .5 ⋅ B 833.33 ⋅ B + 632 − 276.63 ⋅ B 3 ≤ 0 276.63 ⋅ B 3 − 833.33 ⋅ B − 632 ≥ 0 ⇒ B = 2.0435 Usvojeno : B = 2 m

L=3m

2.4. Odreñivanje dozvoljenog napona

D f = 1.1m tgϕ r = -

B=2m

ϕ = 21°

tgϕ tg 21 = = 0.256 ⇒ ϕ r = 14.35° Fϕ 1.50

Faktori nosivosti

ϕ  14.35  π ⋅tg14.35   N q = tg 2  45 + r  ⋅ e π ⋅tgϕr = tg 2  45 + = 3.704 ⋅e 2  2    N γ = 1.8 ⋅ (N q − 1) ⋅ tgϕ r = 1.8 ⋅ (3.704 − 1) ⋅ tg14.35 = 1.245 N c = ( N q − 1) ⋅ ctgϕ r = (3.704 − 1) ⋅ ctg14.35 = 10.570

-

Faktori nosivosti

B 2 = 1 + 0.2 ⋅ = 1.13 L 3 S q = S c = 1.13 S c = 1 + 0 .2 ⋅

S γ = 1 − 0 .4 ⋅ -

B 2 = 1 − 0.4 ⋅ = 0.73 L 3

Faktori dubine Df

1.1 = 1.193 B 2 d −1 1.193 − 1 dq = dc − c = 1.193 − = 1.141 Nq 3.704

d c = 1 + 0.35 ⋅

= 1 + 0.35 ⋅

d γ = 1.0 tgδ = 0 ⇒ ic = iq = iγ = 1

σ dop = γ ⋅ D f + N q ⋅ d q ⋅ S q + 0.5 ⋅ γ ⋅ B ⋅ N γ ⋅ S γ ⋅ d γ + c ⋅ N c ⋅ S c ⋅ d c = = 18 ⋅ 1.1 + 3.704 ⋅1.141 ⋅1.13 + 0.5 ⋅18 ⋅ 2 ⋅1.245 ⋅ 0.73 ⋅ 1 + 20 ⋅10 .57 ⋅1.13 ⋅1.193 = 325 .92 kN / m 2 ∆=

325 .92 − 300 ⋅100 = 7.95 o o 325 .92

Usvojeno : B = 2 m

L =3m

Matić Slavko 63/03


6


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 2.5. Odreñivanje visine temelja II

2

I

0 .99

0 .5

1.51

3

I

-

II

Stalno optertećrnje V g=800 kN

σ= -

800 = 133.33 kN / m 2 2⋅3

Korisno opterećenje ( I ) V p=450 kN

M p=3 kN

450 3 ⋅ 6 ± = 75 ± 1 2 ⋅ 3 2 ⋅ 32 σ 1 = 76 kN / m 2

σ 1, 2 =

74 7 4 .6 6

7 4 .9 9

76

σ 2 = 74 kN / m 2 -

Korisno opterećenje ( II ) M p=237 kN

1 0 1 .7 8 154

V p=450 kN

4

450 237 ⋅ 6 ± = 75 ± 79 kN / m 2 2 2⋅3 2⋅3 σ 1 = 154 kN / m 2

σ 1, 2 =

7 5 .4 2

σ 2 = −4 kN / m 2

Matić Slavko 63/03


7


Grañevinski fakultet univerziteta u Beogradu - Fundiranje   2  0.99 2 0.99 2 154 − 101.78 M U , I − I = 1.6 ⋅ 133.33 ⋅ + 1.8 ⋅ 101.78 ⋅ + ⋅ 0.99 2 ⋅   ⋅ 2 = 450.06 kNm 2 2 2 3     154 − 101.78   TU , I − I = 1.6 ⋅ 133.33 ⋅ 0.99 + 1.8 ⋅ 101.78 ⋅ 0.99 + ⋅ 0.99   ⋅ 2 = 878.19 kN 2    M U , II − II

  1.512 1.512 76 − 74.99 2   = 1.6 ⋅ 133.33 ⋅ + 1.8 ⋅  74.99 ⋅ + ⋅ 1.512 ⋅   ⋅ 2 = 796.94 kNm 2 2 2 3   

 76 − 74.99   TU , II − II = 1.6 ⋅ 133.33 ⋅ 1.51 + 1.8 74.99 ⋅ 1.51 + ⋅ 1.51  ⋅ 2 = 1054.64 kN 2    pp.

ε b / ε a = 3.5 / 10 o oo

⇒ k = 2.311

 1.94 ⋅ 796.64 ⋅ 100 = 44.87 cm 200 ⋅ 2.05  ⇒ Usvojeno :  TU 1053.64 hT = = = 53.21 cm  0.9 ⋅ b ⋅ τ r 0.9 ⋅ 200 ⋅ 0.11  hM = 2.311 ⋅

h = 55 cm

2.6. Armiranje -

Pravac “L”

M U = 796.94 kNm

(

pp. φ16 a1a = 2.01 cn 2

)

1 .6 = 49.2 cm 2 49.2 k= = 3.528 ⇒ ε b / ε a = 1.65 / 10 796.94 ⋅100 200 ⋅ 2.05 49.2 ⋅ 200 2.05 Aa = 8.471 ⋅ ⋅ = 42.72 cm 2 100 40 42.72 n= ≈ 22 kom 2.01 h = 55 − 5 −

I: II : III : IV :

−

o

µ = 8.471 o oo

oo

3 7 .5

I

II

III

IV

0.06 ⋅ 22 = 1.32 2 Rφ16 150 0.08 ⋅ 22 = 1.76 2 Rφ16 0.13 ⋅ 22 = 2.86 3 Rφ16 0.23 ⋅ 22 = 5.06 5 Rφ16 Ukupno : 2 x12 = 24 komada (usvojena je po jos jedna sipka sa strane ) Matić Slavko 63/03


8


Grañevinski fakultet univerziteta u Beogradu - Fundiranje -

Pravac “B” V g=800 kN

σn =

800 = 133.34 kN / m 2 6

σn =

450 = 75 kN / m 2 6

V p=450 kN

0.75 2 ⋅ 3 = 293 .92 kNm 2 TU = (1.6 ⋅ 133 .34 + 1.8 ⋅ 75 ) ⋅ 0.75 ⋅ 3 = 783 .71 kN M U = (1.6 ⋅ 133 .34 + 1.8 ⋅ 75 ) ⋅

pp.

(

Rφ12 a 1a = 1.13 cm 2

h = 55 − 5 − 1.6 − k=

n=

1 .2 = 47 .8 cm 2

47.8 293 .92 ⋅ 100 300 ⋅ 2.05

Aa = 2.154 ⋅

)

= 6.914

⇒

47.8 ⋅ 300 2.05 ⋅ = 15.83 cm 2 100 40

o

oo

µ = 2.154

25

15.83 ≈ 14 1.54 0.06 ⋅ 14 = 0.84 0.08 ⋅ 14 = 1.12

2 Rφ14 1Rφ14

III : 0.13 ⋅ 14 = 1.82 IV : 0.23 ⋅ 14 = 3.22

2 Rφ14 3Rφ14

I: II :

−

ε b / ε a = 0.725 / 10

I

II

III

IV

100

Ukupno : 2 x8 = 16 komada (usvojena je po jos jedna sipka sa strane )

Matić Slavko 63/03


9


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 2.7. Kontrola proboja 1 ⋅ (47.7 + 49.2 ) = 48.45 cm 2 2 Fb = (0.50 + 2 ⋅ 0.4845) = 2.16 cm 2

hsr =

Pr = V − σ n ⋅ Fb = 1250 −

τp =

1250 ⋅ 2.16 = 800 kN 6

Pr 800 = = 419.29 kN / m 2 < τ r = 1100 kN / m 2 F p 4 ⋅ (0.5 + 0.4845) ⋅ 0.4845

2.8. Analiza opterećenja - Stalno opterećenje: - Težina tla iznad temelja: - Težina temelja:

σ rač

1250 kN 3 ⋅ 2 − 0.5 2 ⋅ 0.5 ⋅ 18 = 51.75 kN 3 ⋅ 2 ⋅ 0.55 ⋅ 25 = 82.5 kN

(

)

Σ V = 1384.25 kN ∑V ± M = 1384.25 ± 237 ⋅ 6 = 230.71 ± 79 = F W 6 2 ⋅ 32

σ 1 = 300.71 kN / m 2 

2  < σ doz = 325.92 kN / m σ 2 = 151.71 kN / m  2

Matić Slavko 63/03


10



11



12



ZADATAK

1600

1400

1600

a 0 xb0 = 0,50 x0,50

1400

σ doz = 160 kN / m 2 t=0.50

5,7

3.1.

6,5

γ = 18 kN / m 3

5,7

Odreñivanje dužine prepusta 1600

1400 5,7

a1

R

1600

1400

6,5

5,7

x

a2

x' L

R = ∑ Pi = 1400 + 1600 + 1600 + 1400 = 6000 kN 1600 ⋅ 5.7 + 1600 ⋅ 12.2 + 1400 ⋅ 17.9 = 8.95 m 6000 x' = 17.9 − 8.95 = 8.95 x=

a1 = 0.25 ⋅ l max = 0.25 ⋅ 6.50 = 1.63 m a 2 = 1,63 m L = 5.70 + 6.50 + 5.70 + 2 ⋅ 1.63 = 21.16 m q=

3.2.

R 6000 = = 283.55 kN / m L 21.16

Statički uticaji

3.2.1. Statički odreñen sistem 1600

1400

1600

1400 q=283.55 kN /m

6,5

5,7

1,63

M

362.58

362.58

376.68

5,7 376.68

1,63

Matić Slavko 63/03


13


Grañevinski fakultet univerziteta u Beogradu - Fundiranje Na osnovu dijagrama momenata savijanja po dužini nosača može se zaključiti da je potrebno usvojiti promenljivu širinu nosača kako bi se dobile racionalne dimenzije temeljnog nosača. Pretpostavljamo dubinu fundiranja Df=1.5 m

5,7

6,5 21,16

1400 5,7

1,63

B2

1,63

1600

B1

1600

1400

2.13

σ doz ‘

16.9

2.13

R = 4.26 ⋅ B1 + 16.9 ⋅ B2 − 0.85 ⋅ γ b ⋅ D f

6000 = 4.26 ⋅ B1 + 16.9 ⋅ B2 160 − 0.85 ⋅ 25 ⋅ 1.5 46.829 = 4.26 ⋅ B1 + 16.9 ⋅ B2 Iz gornje jednačine usvajanjem B2= 2.0 m dobijamo da je B1= 3.0 m. -

Dijagram reaktivnog opterećenja tla:

R 6000 = = 128.811 kN / m 2 F 3 ⋅ 4.26 + 2 ⋅ 16.9 q1 = B1 ⋅ q = 3.0 ⋅ 128.811 = 386.43 kN / m q=

q 2 = B2 ⋅ q = 2.0 ⋅ 128.811 = 257.622 kN / m

Matić Slavko 63/03


14


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 3.2.2. Merodavni momenti savijanja za dimenzionisanje temeljnog nosača u statički odreñenom sistemu 1600

1400

1600

1400

q = 2 5 7 .6 2 2 k N /m q = 3 8 6 .4 3 k N /m 5 ,2

7 7 0 .1 2

2 ,1 3

659.54

8 3 7 .2 7

M

466.30

692.99

466.30

659.54

6 ,5

513.31

5 ,2 513.31

2 ,1 3

6 2 9 .8 8

7 6 2 .7 3

T 6 2 9 .8 8

7 6 2 .7 3

8 3 7 .2 7

7 7 0 .1 2

3.2.3. Merodavmi momenti savijanja za dimenzionisanje temeljnog nosača u statički neodreñenom sistemu

q = 2 5 7 .6 2 2 k N /m q = 3 8 6 .4 3 k N /m

8 3 7 .2 7

2 ,1 3

372.47

485.97

372.47 7 3 3 .8 4

5 ,2

513.31

6 ,5 866.54

513.31

5 ,2

866.54

2 ,1 3

M

6 2 9 .8 9

7 9 9 .0 1

T 6 2 9 .8 8

7 9 9 .0 1

8 3 7 .2 7

7 3 3 .8 4

Matić Slavko 63/03


15


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 3.3. Dimenzionisanje Merodavni uticaji: M = 866.54 kNm T = 837.27 kN

3.3.1. Visina temeljnog nosača MB30 ⇒ f b = 2.05 kN / cm 2 RA400 / 500 ⇒ σ v = 40 kN / cm 2

τ r = 0.11 kn / cm 2 −

τ = 2 ⋅ τ r = 2 ⋅ 0.11 = 0.22 kN / cm 2 γ u = 1.6 ⋅ 0.7 + 1.8 ⋅ 0.3 = 1.66 Prema momentima savijanja

ε b / ε a = 3.5 / 10 hm = 2.311 ⋅ -

o

oo

⇒ k = 2.311

1.66 ⋅ 866.54 ⋅ 100 = 73.16 cm 70 ⋅ 2.05

Prema T-silama

ht =

837.27 ⋅ 1.66 = 99.279 0.9 ⋅ 70 ⋅ 0.22

Usvojeno: d = 100 cm

a

3.3.2. Visina temeljne ploče M a − a = 0.5 ⋅ 1.15 2 ⋅ 128.81 = 85.176 kNm / m

115

70

115

70

-

1.66 ⋅ 85.176 ⋅ 100 = 19.19 cm 2.05 ⋅ 100 148.13 ⋅ 1.66 hT = = 12.41cm 0.9 ⋅ 100 ⋅ 0.22 hM = 2.311 ⋅

30

Ta − a = 1.15 ⋅ 128.81 = 148.13 kN

q = 1 2 8 .8 1 3 0 0 a

Usvaja se debljina ploče 30 cm celom dužinom temeljnog nosača.

Matić Slavko 63/03


16


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 3.4. Analiza opterećenja - Od stubova 6000 kN - Težina temelja (3.0 ⋅ 0.30 ⋅ 4.26 + 2.0 ⋅ 0.3 ⋅ 16.9 + 0.7 ⋅ 0.7 ⋅ 21.16) ⋅ 25 = 608.56 kN - Tlo (3.0 ⋅ 0.5 ⋅ 4.26 + 2.0 ⋅ 0.5 ⋅ 16.9 + 2 ⋅ 0.65 ⋅ 0.7 ⋅ 16.9 + 2 ⋅ 1.15 ⋅ 0.7 ) ⋅ 18 = 819.497 kN

∑V = 7428.056 kN σ rač =

7428.06 = 159.48 kN / cm 2 < σ doz = 160 kN / cm 2 3.0 ⋅ 4.26 + 2.0 ⋅ 16.9

3.5. Armatura u podužnom pravcu -

Presek ispod srednjih stubova ( Presek 1-1 )

M u = 1.66 ⋅ 866.54 = 1386.46 kNm hstat = 100 − 7.5 = 92.5 cm k=

ε b / ε a = 2.125 / 10 o oo = 2.976 ⇒  − 1386.46 ⋅ 100 µ = 12.027 % 70 ⋅ 2.05 92.5

Aa = 12.027 ⋅ -

70 ⋅ 92.52 2.05 ⋅ = 39.91 cm 2 ⇒ Usvojeno : 8 Rφ 25 39.28 cm 2 100 40

(

)

Presek ispod spoljašnjih stubova stubova ( Presek 2-2 )

M u = 1.66 ⋅ 513.36 = 821.38 kNm hstat = 100 − 7.5 = 92.5 cm k=

ε b / ε a = 1.45 / 10 o oo = 3.866 ⇒  − 821.38 ⋅ 100 µ = 6.962 % 70 ⋅ 2.05 92.5

Aa = 6.962 ⋅

70 ⋅ 92.52 2.05 ⋅ = 23.10 cm 2 ⇒ Usvojeno : 5 Rφ 25 24.55 cm 2 100 40

(

)

Matić Slavko 63/03


17



Presek u srednjem polju ( Presek 3-3 )

Zategnuta gornja zona ⇒ Pretpostavljamo da se neutralna linija nalazi u ploči tj. da je oblik pritisnute zona pravougaoni širine B.

M u = 1.66 ⋅ 692.99 = 1108.78 kNm hstat = 100 − 7.5 = 92.5 cm  b + 20 ⋅ d pl = 70 + 20 ⋅ 30 = 670 cm B = min  b + 0.25 ⋅ l 0 = 70 + 0.25 ⋅ 465 = 186.25 cm B = 186.25 cm ε / ε = 0.95 / 10 o oo  b a 92.5 k= = 5.428 ⇒ s = 0.087 1108.78 ⋅ 100 − 186.25 ⋅ 2.05 µ = 3.469 % x = s ⋅ h = 0.087 ⋅ 92.5 = 8.048 < d pl = 30 cm ⇒ Pretpostavka o položaju neutralne linije je dobra. Aa = 3.469 ⋅ -

186.25 ⋅ 92.52 2.05 ⋅ = 30.63 cm 2 ⇒ Usvojeno : 7 Rφ 25 34.37 cm 2 100 40

(

)

Preseci u krajnjim poljima ( Presek 4-4 )

Zategnuta gornja zona ⇒ Pretpostavljamo da se neutralna linija nalazi u ploči tj. da je oblik pritisnute zona pravougaoni širine B.

M u = 1.66 ⋅ 466.39 = 746.22 kNm hstat = 100 − 7.5 = 92.5 cm  b + 20 ⋅ d pl = 70 + 20 ⋅ 30 = 670 cm B = min  b + 0.25 ⋅ l 0 = 70 + 0.25 ⋅ 387 = 166.75 cm B = 166.75 cm ε / ε = 0.80 / 10 o oo  b a 92.5 k= = 6.261 ⇒ s = 0.074 746.22 ⋅ 100 − µ = 2.568 % 166.75 ⋅ 2.05 x = s ⋅ h = 0.074 ⋅ 92.5 = 6.84 < d pl = 30 cm ⇒ Pretpostavka o položaju neutralne linijeje dobra. Aa = 2.568 ⋅

166.75 ⋅ 92.52 2.05 ⋅ = 20.3 cm 2 ⇒ Usvojeno : 5 Rφ 25 24.55 cm 2 100 40

(

)

Matić Slavko 63/03


18


Grañevinski fakultet univerziteta u Beogradu - Fundiranje Armatura ploče

-

(

pp. Rφ16 au1 = 2.01 cm 2

)

hstst = 30 − 5 − 1.6 / 2 = 24.2 cm ε b / ε a = 2.15 / 10 = 2.968 ⇒  − 1.66 ⋅ 85.176 ⋅ 100 µ = 12.209 % 100 ⋅ 2.05 24.2

k=

Aa = 12.209 ⋅

o

oo

100 ⋅ 24.2 2.05 ⋅ = 15.14 cm 2 / m' ⇒ Usvojeno : Rφ16 / 12 5 16.08 cm 2 / m' 100 40

(

)

Za podužnu armaturu usvaja se: Rφ10 / 20

3.6. Osiguranje od glavnih napona zatezanja

τ r = 0.11 kN / cm 2 Tmu = Tu = 1.66 ⋅ T Tmu Tmu = b ⋅ z 0 .9 ⋅ b ⋅ h 3 = ⋅ (τ n − τ r ) 2

τn = τ ru

1 1 7 4 .1 4

1 3 3 9 .6 3

1 2 7 8 .4 2 1 0 0 7 .8 2

Tu [kN ]

1 0 0 7 .8 2 1 2 7 8 .4 2 0 .2 3

0 .2 0 1

1 3 3 9 .6 3 0 .2 1 9

1 1 7 4 .1 4

0 .1 7 3

τ u [kN / cm 2 ] 0 .1 7 3

L1

L2

0 .1 3 7

0 .2 1 9

L3

L4

0 .1 8 0

0 .2 3

L4

L3

0 .2 0 1

L2

L1

0 .1 6 4 0 .0 9 5

τ ru [kN / cm 2 ]

0 .0 9 5 0 .1 6 4

0 .1 8 0

0 .1 3 7

Matić Slavko 63/03


19


Grañevinski fakultet univerziteta u Beogradu - Fundiranje L2 = 123 cm : m = 2 θ = 45 α = 90 eu =

(

pp URφ12 au1 = 1.13 cm 2

m⋅a 2 ⋅ 1.13 σ v ⋅ (cos α + sin α ⋅ ctgθ ) = ⋅ 40 ⋅ (0 + 1) = 10.43 cm ⇒ Usvojeno URφ12 / 10 b ⋅ τ ru 70 ⋅ 0.137

(

L3 = 148 cm : m = 4 θ = 45 α = 90 pp URφ12 a u1 = 1.13 cm 2 eu =

)

m⋅a 4 ⋅ 1.13 σ v ⋅ (cos α + sin α ⋅ ctgθ ) = ⋅ 40 ⋅ (0 + 1) = 15.74 cm ⇒ Usvojeno URφ12 / 15 b ⋅ τ ru 70 ⋅ 0.164 1 u

(

L4 = 170 cm : m = 4 θ = 45 α = 90 pp URφ12 a u1 = 1.13 cm 2 eu =

)

1 u

)

m ⋅ a u1 4 ⋅ 1.13 σ v ⋅ (cos α + sin α ⋅ ctgθ ) = ⋅ 40 ⋅ (0 + 1) = 14.35 cm ⇒ Usvojeno URφ12 / 12 5 b ⋅ τ ru 70 ⋅ 0.180

Matić Slavko 63/03


20



21



22


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 4. V

VI

V II l3

IV

ZADATAK

III

1700

1800

1700

II

1800

1900

2200

1600

I

1500

1800

1900

1400 l2

5.25

6,2

1600

5,7

l1

4,75

5,7

l1

a 0 xb0 = 0,50 x0,50 t = 0.50 m

σ doz = 220 kN / m 2 γ = 18 kN / m 3 4.1.

Odreñivanje dužine prepusta l1 = 0.25 ⋅ 5.7 = 1.45 m l 2 = 0.25 ⋅ 5.25 = 1.30 m l 3 = 0.25 ⋅ 6.25 = 1.55 m - Podužna traka ima ukupnu dužinu: L1 = 2 ⋅ 1.45 + 5.7 + 5.7 + 4.75 = 19.05 m - Poprečna traka ima ukupnu dužinu: L2 = 1.30 + 5.25 + 6.20 + 1.55 = 14.30 m

4.2.

Potrebna naležuća površina - Pretpostavljamo dubinu fundiranja: Df = 1.60 m

F pot =

∑V

i

σ doz − 0.85 ⋅ γ b ⋅ D f

=

20900 = 112.36 m 2 220 − 0.85 ⋅ 25 ⋅ 1.60

Matić Slavko 63/03


23


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 4.3.

Prosečna širina traka

B 1pr = 2 B pr =

F pot

∑L

1

=

112.36 = 0.983 m 3 ⋅ 19.05 + 4 ⋅ 14.30

=

112.36 = 1.096 m 114.35 − 12 ⋅ 0.983

F pot

∑L

2

112.36 = 1.110 m 114.35 − 12 ⋅ 1.096 112.36 = = 1.112 114.35 − 12 ⋅ 1.110 = 1.11 m

B 3pr = 4 B pr

B pr 4.4.

Odreñivanje širine pojedinih traka

4.4.1. Podužne trake I traka: RI 6600 = 3 ⋅ 1.11 ⋅ = 1.05 m R 20900

B I = 3 ⋅ B pr ⋅ II traka:

R II 7500 = 3 ⋅ 1.11 ⋅ = 1.19 m R 20900

B II = 3 ⋅ B pr ⋅ III traka: B III = 3 ⋅ B pr ⋅

R III 6800 = 3 ⋅ 1.11 ⋅ = 1.08 m R 20900

4.4.2. Poprečne trake IV traka: B IV = 4 ⋅ B pr ⋅

R IV 4900 = 4 ⋅ 1.11 ⋅ = 1.04 m R 20900

V traka:

BV = 4 ⋅ B pr ⋅

RV 5400 = 4 ⋅1.11 ⋅ = 1.15 m R 20900

VI traka: BVI = 4 ⋅ B pr ⋅

RVI 5900 = 4 ⋅1.11 ⋅ = 1.25 m R 20900

Matić Slavko 63/03


24


Grañevinski fakultet univerziteta u Beogradu - Fundiranje VII traka: BVII = 4 ⋅ B pr ⋅

4.5.

RVII 4700 = 4 ⋅1.11 ⋅ = 1.00 m R 20900

Centrisanje temelja 6800 ⋅11.45 + 7500 ⋅ 5.25 = 5.61 m 20900 4700 ⋅16.15 + 5900 ⋅10.45 + 5400 ⋅ 5.70 XR = = 8.06 m 20900

YR =

-

Težište naležuće površine −

(1.05 ⋅1.30 + 1.19 ⋅ 6.55 + 1.08 ⋅12.75) ⋅ (19.05 − (1.04 + 1.15 + 1.25 + 1.0 )) + 4.44 ⋅14.30 ⋅ 7.15 = 7.04 m (19.05 − 4.44)⋅ (1.05 + 1.19 + 1.08) + 4.44 ⋅14.30 (1.04 ⋅1.45 + 1.15 ⋅ 7.15 + 1.25 ⋅11.90 + 1 ⋅17.60) ⋅ (14.30 − (1.05 + 1.19 + 1.08)) + 3.32 ⋅19.05 ⋅ 9.53 = 9.52 m = (14.30 − 3.32) ⋅ (1.04 + 1.15 + 1.25 + 1.0) + 3.32 ⋅19.05

YT = −

XT

−

YT = YT − 1.30 = 7.04 − 1.30 = 5.74 m −

X T = X T − 1.45 = 9.52 − 1.45 = 8.07 m Usvajamo nove širine traka:

-

B I = 1.15 m

B IV = 1.05 m

B II = 1.15 m

B V = 1.15 m

B III = 1.0 m

B VI = 1.25 m B VII = 1.0 m

−

(1.15 ⋅1.30 + 1.15 ⋅ 6.55 + 1.0 ⋅12.75) ⋅ (19.05 − (1.05 + 1.15 + 1.25 + 1.0)) + 4.45 ⋅ 14.30 ⋅ 7.15 = 6.91 m (19.05 − 4.45) ⋅ (1.15 + 1.15 + 1.0) + 4.45 ⋅ 14.30 ( 1.05 ⋅ 1.45 + 1.15 ⋅ 7.15 + 1.25 ⋅ 11.90 + 1.0 ⋅ 17.60) ⋅ (14.30 − (1.15 + 1.15 + 1.0)) + 3.30 ⋅ 19.05 ⋅ 9.53 = = 9.51 m (14.30 − 3.30) ⋅ (1.05 + 1.15 + 1.25 + 1.0) + 3.30 ⋅ 19.05

YT = −

XT

−

YT = YT − 1.30 = 6.91 − 1.30 = 5.61 m −

X T = X T − 1.45 = 9.51 − 1.45 = 8.06 m F = 19.05 ⋅ (1.15 + 1.15 + 1.0 ) + (14.3 − 3.30 ) ⋅ (1.05 + 1.15 + 1.25 + 1.0 ) = 112.82 m 2 p=

∑V F

=

20900 = 186 .91 kN / m 2 112.82

Matić Slavko 63/03


25



Proračun statičkih uticaja u trakama roštilja

4.6.1. Podužne trake-SON

4900

5400

5900

4700

q

P

P

IV

1 ,4 5

P

V

5 ,7

P

VI

4 ,7 5

V II

5 ,7

1 ,4 5

q = (1.15 + 1.15 + 1.0 ) ⋅ 186.91 = 616.80 kN / m PIV = 1.05 ⋅ (14.3 − (1.15 + 1.15 + 1.0 )) ⋅ 186.91 = 2158.81 kN PV = 1.15 ⋅ 11 ⋅ 186.91 = 2364.41 kN PVI = 1.25 ⋅ 11 ⋅ 186.91 = 2570.01 kN

1814.49

1283.80

2112.42

PVII = 1.0 ⋅ 11 ⋅ 186.91 = 2056.01 kN

1846.83

1365.94

648.41

695.44

225.38

648.41

M [kN m ]

1766.13 894.36

T [kN ] 894.36 1668.93

1563.86

1749.63

Matić Slavko 63/03


26


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 4.6.1.1. Raspored momenata savijanja po trakama srazmerno njihovim krutostima -

Traka I i II

30

80

60

115 F = F1 + F2 = 0.3 ⋅ 1.15 + 0.8 ⋅ 0.6 = 0.825 m 2 0.345 ⋅ 0.15 + 0.48 ⋅ 0.7 = 0.470 m 0.825 1 1 2 2 I 1 = I 2 = ⋅ 1.15 ⋅ 0.3 3 + 0.345 ⋅ (0.47 − 0.15) + ⋅ 0.6 ⋅ 0.8 3 + 0.48 ⋅ (0.7 − 0.470 ) = 0.089 m 4 12 12

YT =

-

Traka III

30

80

60

100 F = F1 + F2 = 0.3 ⋅ 1.0 + 0.8 ⋅ 0.6 = 0.780 m 2 0.30 ⋅ 0.15 + 0.48 ⋅ 0.7 = 0.488 m 0.780 1 1 2 2 I 1 = I 2 = ⋅ 1.0 ⋅ 0.33 + 0.30 ⋅ (0.488 − 0.15) + ⋅ 0.6 ⋅ 0.8 3 + 0.48 ⋅ (0.7 − 0.488) = 0.084 m 4 12 12

YT =

∑ I = 0.089 ⋅ 2 + 0.084 = 0.262 m M I = M II = M ⋅ M III = M ⋅

II

∑I

=M⋅

4

0.089 = 0.340 ⋅ M 0.262

I III 0.084 =M⋅ = 0.321 ⋅ M 0.262 ∑I

Matić Slavko 63/03


27


617.09

436.66

718.38


220.46 582.60

236.28 412.26

678.24

220.46

76.46

M I M II [kNm]

208.14

587.33

544.14

609.82

T [kN] 311.67

476.85

637.86

615.37

Podužne trake-SNN

1 4 6 4 .9 0

1 .4 5 648.41

5 ,7

1397.37

104.98

1397.37 1 5 8 8 .6 3

1613.31

q = 6 1 6 .8 0 k N /m 4 ,7 5

5 ,7 648.41

1 ,4 5

1613.11

4.6.2.

223.08

311.67

72.19

208.14

M III [kNm]

M

[k N m ]

1 9 2 7 .1 3 8 9 4 .3 6

T

[k N ]

8 9 4 .3 6 1 9 2 7 .1 3

1 4 6 4 .9 0

1 5 8 8 .6 3

Matić Slavko 63/03


28


475.10

35.69

475.10


220.46 448.55

548.46 33.70

448.55

548.46

220.46

M I M II [kNm]

602.25

311.67

208.14 605.32

561.50

T

[kN]

311.67

529.43

620.19

4.6.3.

517.81

517.81

208.14

M III [kNm]

610.77

Poprečne trake-SON

6600

7500

6800

q

P 1 .3 0

P

I

P

II

5 ,2 5

6 ,2

I II

1 ,5 5

q = (1.05 + 1.15 + 1.25 + 1.0 ) ⋅ 186.91 = 831.75 kN / m PI = 1.15 ⋅ (19.05 − (1.05 + 1.15 + 1.25 + 1.0 )) ⋅ 186.91 = 3138.22 kN PII = 1.15 ⋅ 14.6 ⋅ 186.91 = 3138.22 kN PIII = 1.0 ⋅ 14.6 ⋅ 186.91 = 2728.89 kN

Matić Slavko 63/03


29


999.14 3528.18

115.32

2580.36

702.83


M

[k N m ]

2 2 7 4 .9 5

2 2 8 0 .5 0

1 1 8 9 .2 1

T

[k N ]

9 8 1 .2 8 1 8 8 6 .1 8 2 6 8 1 .9 0 4.6.3.1. Raspored momenata savijanja po trakama srazmerno njihovim krutostima - Traka IV F = F1 + F2 = 0.3 ⋅ 1.05 + 0.8 ⋅ 0.6 = 0.795 m 2

60

0.315 ⋅ 0.15 + 0.48 ⋅ 0.7 = 0.482 m 0.795 1 2 I IV = ⋅ 1.05 ⋅ 0.3 3 + 0.315 ⋅ (0.482 − 0.15) 12 1 2 + ⋅ 0.6 ⋅ 0.8 3 + 0.48 ⋅ (0.7 − 0.482) = 0.085 m 4 12

30

80

YT =

105

- Traka V F = F1 + F2 = 0.3 ⋅ 1.15 + 0.8 ⋅ 0.6 = 0.825 m 2

30

80

60

115

0.345 ⋅ 0.15 + 0.48 ⋅ 0.7 = 0.470 m 0.825 1 2 I V = ⋅ 1.15 ⋅ 0.33 + 0.345 ⋅ (0.470 − 0.15) 12 1 2 + ⋅ 0.6 ⋅ 0.8 3 + 0.48 ⋅ (0.7 − 0.470) = 0.089 m 4 12

YT =

Matić Slavko 63/03


30


Grañevinski fakultet univerziteta u Beogradu - Fundiranje - Traka VI 60

30

80

F = F1 + F2 = 0.3 ⋅ 1.25 + 0.8 ⋅ 0.6 = 0.855 m 2

125

0.375 ⋅ 0.15 + 0.48 ⋅ 0.7 = 0.459 m 0.855 1 2 I VI = ⋅ 1.15 ⋅ 0.3 3 + 0.375 ⋅ (0.459 − 0.15) 12 1 2 + ⋅ 0.6 ⋅ 0.8 3 + 0.48 ⋅ (0.7 − 0.459) = 0.092 m 4 12

YT =

- Traka VII 60

YT = 0.488 m I VII = 0.084 m 4

30

80

F = F1 + F2 = 0.78 m 2

100

∑ I = 0.085 + 0.089 + 0.092 + 0.084 = 0.35 m M IV = M ⋅

I IV 0.085 =M⋅ = 0.243 ⋅ M 0.35 ∑I

MV = M ⋅

IV

=M⋅

0.089 = 0.254 ⋅ M 0.35

I VI

=M⋅

0.092 = 0.263 ⋅ M 0.35

∑I

M VI = M ⋅ M VII = M ⋅

∑I

4

I VII 0.084 =M⋅ = 0.240 ⋅ M 0.35 ∑I

Matić Slavko 63/03


31


M IV [k N m ]

857.35

M V [k N m ]

896.16

M V I [k N m ]

927.01

M V II [k N m ]

846.76

27.68

619.29

168.68

293.79

30.33

678.64

184.84

262.77

29.29

655.41

178.52

253.78

28.02

627.03

170.79

242.79


7 9 3 .2 5

64 6 .89

3 62 .1 4

T [k N ] 3 0 3 .7 3 6 7 0.8 4

7 6 0 .2 2

Poprečne trake – SNN

5 ,2 5

q = 8 3 1 .7 5 k N /m

6 ,2

1 ,5 5

M [k N m ]

3044.38

1111.56

2034.22

702.83

1 .3 0

999.14

4.6.4.

2 9 0 8 .3 0 1 7 3 7 .3 3

1 2 8 9 .2 1

T [k N ] 1 0 8 1 .2 8 2 2 4 8 .5 5 2 6 2 9 .3 5

Matić Slavko 63/03


32


270.11

494.32


[kNm]

242.79

516.69

739.79

282.34

170.79

M IV

253.78

535.0

773.27

292.34

178.52

M V [kNm]

262.77

488.21

800.67

266.78

184.84

M VI [kNm]

239.79

730.65

168.68

M VII [kNm]

820.73 663.49

362.14

T

[kN]

303.73 712.42

4.7.

Dimenzionisanje MB 30 ⇒ f b = 2.05 kN / cm 2 τ r = 0.11 kN / cm 2 RA 400 / 500 ⇒ σ v = 40 kN / cm 2 −

τ = (2 − 2.5 )τ r = 0.248 kN / cm 2 max

M u = 1.65 ⋅ 927 .91 = 1531 .05 kNm

max

Tu = 1.65 ⋅ 820 .73 = 1354 .21 kN

1531 .05 ⋅100 = 81 .53 cm 2.05 ⋅ 60 1354 .21 hT = = 101 .12 cm 0.9 ⋅ 60 ⋅ 0.248 hM = 2.311 ⋅

Usvojeno: d=110 cm Matić Slavko 63/03


33



Analiza opterećenja - Opterećenje: - Težina temelja:

- Tlo:

20900 kN (19.05 ⋅ 3 ⋅ 0.6 ⋅ 0.8 + 4 ⋅ 0.6 ⋅ 0.8 ⋅ (14.3 − 3 ⋅ 0.6 ) + 19.05 ⋅ 0.3 ⋅ (1.15 + 1.15 + 1.0 ) +

+ 0.3 ⋅ (14.3 − 1.15 − 1.15 − 1.0 ) ⋅ (1.05 + 1.15 + 1.25 + 1.0 )) ⋅ 25 = 2124.41 kN (19.05 ⋅ 0.5 ⋅ (1.15 + 1.15 + 1.0 ) + (19.05 − 4 ⋅ 0.6 ) ⋅ 2 ⋅ 0.8 ⋅ (0.2 + 2 ⋅ 0.275) +

+ (14.3 − 1.15 − 1.15 − 1) ⋅ 0.5 ⋅ (0.225 + 0.275 + 0.325 + 0.2 )) ⋅ 18 = 1690.70 kN

∑V = 24715.11 kN σ rač =

∑V F

=

24715.11 = 219.07 kN / m 2 < σ doz = 220 kN / m 2 112.82

4.9.

Proračun potrebne armature

4.9.1.

Trake I i II -

Oslonac

M U = 1.65 ⋅ 548.46 = 904.96 kNm h = 110 − 7.5 = 102.5 cm ε a / ε b = 1.5 / 10 o oo k= = 3.779 ⇒  − 904.96 ⋅ 100  µ = 7.337 % 2.05 ⋅ 60 60 ⋅ 102.5 2.05 Aa = 7.337 ⋅ ⋅ = 23.12 cm 2 ⇒ Usvojeno : 5 Rφ 25 24.55 cm 2 100 40 102.5

(

)

- Polje Zategnuta gornja zona ⇒ Pretpostavljamo da se neutralna linija nalazi u ploči tj. da je oblik pritisnute zona pravougaoni širine B.

M u = 1.65 ⋅ 718.38 = 1185.33 kNm h = 100 − 7.5 = 92.5 cm  b + 20 ⋅ d pl = 60 + 20 ⋅ 30 = 660 cm B = min  b + 0.25 ⋅ l 0 = 60 + 0.25 ⋅ 520 = 190 cm B = 190 cm k=

ε / ε = 0.875 / 10 o oo  b a 102.5 = 5.876 ⇒ s = 0.080 1185.33 ⋅ 100 − µ = 3.007 % 190 ⋅ 2.05

Matić Slavko 63/03


34


Grañevinski fakultet univerziteta u Beogradu - Fundiranje x = s ⋅ h = 0.080 ⋅ 102.5 = 8.20 < d pl = 30 cm ⇒ Pretpostavka o položaju neutralne linije je dobra. Aa = 3.007 ⋅

4.9.2.

190 ⋅ 102.5 2.05 ⋅ = 30.01 cm 2 ⇒ Usvojeno : 7 Rφ 25 34.37 cm 2 100 40

(

)

Traka III -

Oslonac

M U = 1.65 ⋅ 517.81 = 854.38 kNm h = 110 − 7.5 = 102.5 cm

ε a / ε b = 1.45 / 10 o oo k= = 3.889 ⇒  − 854.38 ⋅ 100  µ = 6.962 % 2.05 ⋅ 60 60 ⋅ 102.5 2.05 Aa = 6.962 ⋅ ⋅ = 21.94 cm 2 ⇒ Usvojeno : 5 Rφ 25 24.55 cm 2 100 40 102.5

(

)

- Polje Zategnuta gornja zona ⇒ Pretpostavljamo da se neutralna linija nalazi u ploči tj. da je oblik pritisnute zona pravougaoni širine B. M u = 1.65 ⋅ 678.24 = 1119.10 kNm h = 100 − 7.5 = 92.5 cm

 b + 20 ⋅ d pl = 60 + 20 ⋅ 30 = 660 cm B = min  b + 0.25 ⋅ l 0 = 60 + 0.25 ⋅ 520 = 190 cm B = 190 cm k=

ε / ε = 0.85 / 10 o oo  b a 102.5 = 6.047 ⇒ s = 0.078 1119.10 ⋅ 100 − µ = 2.858 % 190 ⋅ 2.05

x = s ⋅ h = 0.078 ⋅ 102.5 = 7.99 < d pl = 30 cm ⇒ Pretpostavka o položaju neutralne linije je dobra. Aa = 2.858 ⋅

190 ⋅ 102.5 2.05 ⋅ = 28.53 cm 2 ⇒ Usvojeno : 6 Rφ 25 29.46 cm 2 100 40

(

)

Matić Slavko 63/03


35


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 4.9.3.

Traka IV -

Oslonac

M U = 1.65 ⋅ 739.79 = 1220.65 kNm h = 110 − 7.5 = 102.5 cm


(

)


 b + 20 ⋅ d pl = 60 + 20 ⋅ 30 = 660 cm B = min  b + 0.25 ⋅ l 0 = 60 + 0.25 ⋅ 550 = 197.5 cm B = 197.5 cm k=

ε / ε = 0.93 / 10 o oo  b a 102.5 = 5.484 ⇒ s = 0.087 1414.63 ⋅ 100 − µ = 3.469 % 197.5 ⋅ 2.05


197.5 ⋅ 102.5 2.05 ⋅ = 35.99 cm 2 ⇒ Usvojeno : 8 Rφ 25 39.28 cm 2 100 40

(

)

Matić Slavko 63/03


36



Traka V -

Oslonac

M U = 1.65 ⋅ 773.27 = 1275.90 kNm h = 110 − 7.5 = 102.5 cm


(

)



ε / ε = 0.975 / 10 o oo  b a 102.5 = 5.364 ⇒ s = 0.089 1478.66 ⋅ 100 − µ = 3.627 % 197.5 ⋅ 2.05


197.5 ⋅ 102.5 2.05 ⋅ = 37.63 cm 2 ⇒ Usvojeno : 8 Rφ 25 39.28 cm 2 100 40

(

)

Matić Slavko 63/03


37



Traka VI -

Oslonac

M U = 1.65 ⋅ 800.67 = 1321.11 kNm h = 110 − 7.5 = 102.5 cm


(

)



ε / ε = 0.975 / 10 o oo  b a 102.5 = 5.274 ⇒ s = 0.089 1529.57 ⋅ 100 − µ = 3.627 % 197.5 ⋅ 2.05


197.5 ⋅ 102.5 2.05 ⋅ = 37.63 cm 2 ⇒ Usvojeno : 8 Rφ 25 39.28 cm 2 100 40

(

)

Matić Slavko 63/03


38



Traka VII -

Oslonac

M U = 1.65 ⋅ 730.65 = 1205.57 kNm h = 110 − 7.5 = 102.5 cm


(

)



ε / ε = 0.95 / 10 o oo  b a 102.5 = 5.462 ⇒ s = 0.087 1397.15 ⋅ 100 − µ = 3.469 % 197.5 ⋅ 2.05


197.5 ⋅ 102.5 2.05 ⋅ = 35.99 cm 2 ⇒ Usvojeno : 8 Rφ 25 39.28 cm 2 100 40

(

)

Matić Slavko 63/03


39



40



41


Grañevinski fakultet univerziteta u Beogradu - Fundiranje VEŽBA 5 1400

1600

1400 40 kN/m

0.4/0.6

0.4/0.4

0.4/0.4

0.5

4

0.4/0.4

5

6

MB 30

γ = 18 kN / m 3 σ doz = 160 kN / m 2 E 0 = 8000 kN / m 2

ν = 0.3 5.1. Odreñivanje dužine prepusta

R 1486.73 25.32

1843.41 13.91

1509.86 33.70

17.71 a1

9.09 5 x'

26.80 6 x''

a2

L

R = 1486.73 + 1843.41 + 1509.86 = 4840.0 kN 1843.41 ⋅ 5 + 1509.86 ⋅ 11 + 33.70 − 13.91 − 25.32 x' = = 5.33 m 4840.0 x' ' = 11 − 5.33 = 5.67 m 1 1 a1 = Lmax = ⋅ 6 = 2.0 m 3 3 L = 2 ⋅ (5.33 + 2 ) = 14.65 m a 2 = 14.65 − 11 − 2.0 = 1.65 m R 4840.0 q= = = 330.38 kN / m L 14.65

Matić Slavko 63/03


42


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 5.2. Klasični postupak 5.2.1. Statički uticaji 5.2.1.1. Merodavni momenti savijanja za dimenzionisanje temeljnog nosača u statički odreñenom sistemu 1 4 8 6 .7 3

1 5 0 9 .8 6

1 8 4 3 .4 1

2 5 .3 2

1 3 .9 1

3 3 .7 0

q = 3 3 0 .3 8 k N /m 5

2

1 ,6 5

6

6 3 5 .3 4

991.46

6 6 0 .7 6 6 3 5 .4 4

397.05

M 5 7 4 .4 7

4 1 6 .0 3 4 4 9 .7 3

1 0 1 7 .5 5

8 2 5 .9 7

5 4 5 .1 3

T

6 6 0 .7 6

8 2 5 .9 3

9 6 4 .7 3

5.2.1.2. Merodavmi momenti savijanja za dimenzionisanje temeljnog nosača u statički neodreñenom sistemu q = 3 3 0 .3 8 k N /m 6

449.73

1 ,6 5

763.51

198.58

660.76

1006.96

5

2

M

1 0 8 4 .0 1 7 5 6 .7 1 5 4 5 .1 3

T

6 6 0 .7 6 8 9 5 .1 9

8 9 8 .2 7

Matić Slavko 63/03


43


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 5.2.2. Dimenzionisanje Merodavni uticaji:

M = 1006.96 kNm T = 1084.01 kN 5.2.2.1. Visina temeljnog nosača

MB30 ⇒ f b = 2.05 kN / cm 2 RA400 / 500 ⇒ σ v = 40 kN / cm 2

τ r = 0.11 kN / cm 2 −

τ = (2 − 2.5) ⋅ τ r = 0.275 kN / cm 2 γ u = 1.6 ⋅ 0.7 + 1.8 ⋅ 0.3 = 1.66 -

Prema momentima savijanja

ε b / ε a = 3.5 / 10

oo

⇒ k = 2.311

1.66 ⋅ 1006.96 ⋅ 100 = 85.19 cm 60 ⋅ 2.05

hm = 2.311 ⋅ -

o

Prema T-silama

ht =

1084.01 ⋅ 1.66 = 121.17 cm 0.9 ⋅ 60 ⋅ 0.275


5.2.2.2. Širina temeljnog nosača

σ doz = 160 kN / m 2 D f = d + t = 125 + 50 = 175 cm

B pot =

σ doz F pot L

R 4840.0 = = 39.41 m 2 − 0.85 ⋅ γ b ⋅ D f 160 − 0.85 ⋅ 25 ⋅ 1.75 =

39.41 = 2.69 m ⇒ Usvojeno B = 270 cm 14.65

10 40

5.2.2.3. Visina temeljne ploče

1.66 ⋅ 67.45 ⋅ 100 = 17.08 cm 2.05 ⋅ 100 Usvaja se ploča debljine 35 cm. hM = 2.311 ⋅

a 105

60

105

35

4840.0 = 122.36 kN / m 2 14.65 ⋅ 2.7 M a − a = 0.5 ⋅ 1.05 2 ⋅ 122.36 = 67.45 kNm / m

q* =

10

90

F pot =

q * = 1 2 2 .3 6 2 7 0 a Matić Slavko 63/03


44


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 5.2.3. Analiza opterećenja - Od stubova: - Težina temelja: - Težina tla:

4840.0 kN (2.7 ⋅ 0.35 + 0.9 ⋅ 0.6) ⋅ 25 ⋅ 14.65 = 543.88 kN (0.5 ⋅ (2.7 − 0.4) + 2 ⋅ 1.05 ⋅ 0.9) ⋅ 18 ⋅ 14.65 = 801.65 kN

∑V = 6185.53 kN σ rač = 5.3.

6185.53 = 156.38 kN / cm 2 < σ doz = 160 kN / cm 2 2.7 ⋅ 14.65

Winkler-ov model za tlo

5.3.1. Sleganje s=

1 −ν 2 V 1 − 0.3 2 4840 ⋅ ⋅ B ⋅α = ⋅ ⋅ 2. 7 ⋅ α E0 ⋅ F F 8000 14.65 ⋅ 2.7

L 14.65 = = 5.43 B 2.7 α (L / B = 5) = 2.10   ⇒ α (L / B = 5.43) = 2.137 α (L / B = 2.53) = 2.53 s= -

1 − 0.3 2 4840 ⋅ ⋅ 2.137 ⋅ 2.7 = 0.0803m = 8.03 cm 8000 14.65 ⋅ 2.7

Ukupna krutost celokupnog tla ispod temeljnog nosača: k* =

R 4840.0 = = 60273.97 kN / m s 0.0803

2x1.0=2.0 1

2

3

5x1.0=5 4

5

6

6x1.0=6 7

8

9

10

11 12

2x1.83=1.65 13

14 15 16

Fi F F1 = 0.5 ⋅ 1 ⋅ 2.7 = 1.35 m 2

ki = k * ⋅

F2 = .......... = F13 = 1 ⋅ 2.7 = 2.7 m 2 F14 = 0.5 ⋅ (1 + 0.85) ⋅ 2.7 = 2.46 m 2 F15 = 0.825 ⋅ 2.7 = 2.23 m 2 F16 = 0.5 ⋅ 0.825 ⋅ 2.7 = 1.11 m 2

Matić Slavko 63/03


45


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 1.35 = 2057.13 kN / m 14.65 ⋅ 2.7 2.7 k 2 = ... = k13 = 60273.97 ⋅ = 4114.26 kN / m 14.65 ⋅ 2.7 2.46 k14 = 60273.97 ⋅ = 3748.55 kN / m 14.65 ⋅ 2.7 2.23 k15 = 60273.97 ⋅ = 3398.08 kN / m 14.65 ⋅ 2.7 1.11 k16 = 60273.97 ⋅ = 1691.42 kN / m 14.65 ⋅ 2.7 l Fi = k i ⋅ i Eb

k1 = 60273.97 ⋅

1. 0 = 65.31 ⋅ 10 − 6 m 2 6 31.5 ⋅ 10 1.0 F2 = ... = F13 = 4114.26 ⋅ = 130.61 ⋅ 10 − 6 m 2 31.5 ⋅ 10 6 1. 0 F14 = 3748.55 ⋅ = 119.00 ⋅ 10 −6 m 2 6 31.5 ⋅ 10 1.0 F15 = 3398.08 ⋅ = 107.88 ⋅ 10 −6 m 2 31.5 ⋅ 10 6 1. 0 F16 = 1691.42 ⋅ = 53.69 ⋅ 10 − 6 m 2 31.5 ⋅ 10 6 F1 = 2057.13 ⋅

Matić Slavko 63/03


46


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 5.3.2. Dimenzionisanje Merodavni uticaji: M = 850.08 kNm T = 856.19 kN

5.3.2.1. Visina temeljnog nosača

MB30 ⇒ f b = 2.05 kN / cm 2 RA400 / 500 ⇒ σ v = 40 kN / cm 2

τ r = 0.11 kN / cm 2 −

τ = (2 − 2.5) ⋅ τ r = 0.275 kN / cm 2 γ u = 1.6 ⋅ 0.7 + 1.8 ⋅ 0.3 = 1.66 -

Prema momentima savijanja

ε b / ε a = 3.5 / 10

o

oo

⇒ k = 2.311

1.66 ⋅ 850.08 ⋅ 100 = 78.28 cm 60 ⋅ 2.05 Prema T-silama

hm = 2.311 ⋅ -

ht =

856.19 ⋅ 1.66 = 95.71 cm 0.9 ⋅ 60 ⋅ 0.275


5.3.2.2. Širina temeljnog nosača

σ doz = 160 kN / m 2 D f = d + t = 100 + 50 = 150 cm

B pot =

σ doz F pot L

R 4840.0 = = 37.77 m 2 − 0.85 ⋅ γ b ⋅ D f 160 − 0.85 ⋅ 25 ⋅ 1.50

=

37.77 = 2.58 m ⇒ Usvojeno B = 260 cm 14.65

10 40

a

5.3.2.3. Visina temeljne ploče

hM = 2.311 ⋅

1.66 ⋅ 63.53 ⋅ 100 = 16.58 cm 2.05 ⋅ 100

100

60

100

35

4840.0 = 127.07 kN / m 2 14.65 ⋅ 2.6 M a − a = 0.5 ⋅ 1.00 2 ⋅ 127.07 = 63.53 kNm / m

q* =

10

65

F pot =

q * = 1 2 7 .0 7 2 6 0 a

Usvaja se ploča debljine 35 cm.

Matić Slavko 63/03


47


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 5.3.3. Analiza opterećenja - Od stubova: - Težina temelja: - Težina tla:

4840.0 kN (2.6 ⋅ 0.35 + 0.65 ⋅ 0.6) ⋅ 25 ⋅ 14.65 = 476.13 kN (0.5 ⋅ (2.6 − 0.4) + 2 ⋅ 1.00 ⋅ 0.65) ⋅ 18 ⋅14.65 = 632.88 kN

∑V = 5949.01 kN σ rač =

5949.01 = 156.18 kN / cm 2 < σ doz = 160 kN / cm 2 2.6 ⋅ 14.65

Matić Slavko 63/03


48

Trial version of ABC Amber PDF Merger v3.01, http://www.processtext.com/abcpdfmg.html Grañevinski fakultet univerziteta u Beogradu Katedra za geotehniku

Fundiranje

Vežba 5

Matić Slavko 63/03 Улазни подаци - Конструкција Координате чворова No

X [m] 0.0000 0.0000 1.0000 1.0000 2.0000 2.0000 3.0000 3.0000 4.0000 4.0000 5.0000 5.0000

1 2 3 4 5 6 7 8 9 10 11 12

Y [m] 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

Z [m] -1.0000 0.0000 -1.0000 0.0000 -1.0000 0.0000 -1.0000 0.0000 -1.0000 0.0000 -1.0000 0.0000

No

X [m] 6.0000 2.0000 6.0000 7.0000 7.0000 8.0000 8.0000 9.0000 9.0000 10.000 10.000 11.000

13 14 15 16 17 18 19 20 21 22 23 24

Y [m] 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

Z [m] -1.0000 4.0000 0.0000 -1.0000 0.0000 -1.0000 0.0000 -1.0000 0.0000 -1.0000 0.0000 -1.0000

No

X [m] 7.0000 11.000 12.000 12.000 13.000 13.825 13.000 14.650 13.825 14.650 13.000

25 26 27 28 29 30 31 32 33 34 35

Y [m] 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

Z [m] 4.0000 0.0000 -1.0000 0.0000 -1.0000 -1.0000 0.0000 -1.0000 0.0000 0.0000 4.0000

Табела материјала Назив материјала Beton MB 30 Beton MB 30

No 1 2

E[kN/m2] 3.150e+7 3.150e+7

µ 0.20 0.30

γ[kN/m3]

αt[1/C] 1.000e-5 1.000e-5

25.00 25.00

Em[kN/m2] 3.150e+7 3.150e+7

µm 0.20 0.20

Сетови греда Сет: 1 Пресек: Правоугаони

60

2

T

Мат. 1

P/Z

A1 2.400e-1

A2 2.000e-1

A3 2.000e-1

I1 7.512e-3

I2 3.200e-3

I3 7.200e-3

Мат. 1

P/Z

A1 1.600e-1

A2 1.333e-1

A3 1.333e-1

I1 3.605e-3

I2 2.133e-3

I3 2.133e-3

Мат. 1

P/Z

A1 4.000e-2

A2 3.333e-2

A3 3.333e-2

I1 2.533e-4

I2 1.333e-4

I3 1.333e-4

Мат. 1

P/Z P+Z

A1 6.531e-5

A2 3.333e-2

A3 3.333e-2

I1 2.533e-4

I2 1.000e-6

I3 1.000e-6

Мат. 1

P/Z P+Z

A1 1.306e-4

A2 3.333e-2

A3 3.333e-2

I1 2.533e-4

I2 1.000e-6

I3 1.000e-6

Мат. 2

P/Z P+Z

A1 1.190e-4

A2 3.333e-2

A3 3.333e-2

I1 2.533e-4

I2 1.000e-6

I3 1.000e-6

Мат. 2

P/Z P+Z

A1 1.079e-4

A2 3.333e-2

A3 3.333e-2

I1 2.533e-4

I2 1.000e-6

I3 1.000e-6

Мат. 2

P/Z P+Z

A1 5.369e-5

A2 3.333e-2

A3 3.333e-2

I1 2.533e-4

I2 1.000e-6

I3 1.000e-6

3

40 [cm]

Сет: 2 Пресек: Правоугаони

40

2

T

3

40 [cm]

Сет: 3 Пресек: T-пресек

125

90

2 60

T

3

270 [cm]

Сет: 4 Пресек: Произвољни





Сетови тачкастих ослонаца Сет

K,R1 K,M1

1

K,R2 K,M2

K,R3 K,M3 1.000e+10

x 1.000000

Оса 1 y 0.000000

z 0.000000

x 0.000000

Оса 2 y 1.000000

z 0.000000

Tower - 3D Model Builder 5.4 DEMO


Radimpex - www.radimpex.co.yu

49


Fundiranje

Vežba 5

Matić Slavko 63/03 Контуре греда No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

Чвор I

Ослобађање утицаја Чвор I Чвор J M1 M2 M3 N1 T2 T3 M1 M2 M3 N1 T2 T3

Чвор J

2 4 14 6 8 10 12 15 25 17 19 21 23 26 28 35 14 25 31 33 2 4 6 8 10 12 15 17 19 21 23 26 28 31 33 34

4 6 6 8 10 12 15 17 17 19 21 23 26 28 31 31 25 35 33 34 1 3 5 7 9 11 13 16 18 20 22 24 27 29 30 32

Оса 1

Оса 2

Оса 3

0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 0.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000

0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

1.000000 1.000000 0.000000 1.000000 1.000000 1.000000 1.000000 1.000000 0.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 0.000000 1.000000 1.000000 1.000000 1.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

Мимоилажење

Контуре тачкастих ослонаца Чворови 1, 3, 5, 7, 9, 11, 13, 16, 18, 20, 22, 24, 27,

Сет 1

Чворови

Сет

29, 30, 32

Tower - 3D Model Builder 5.4 DEMO



50


Fundiranje

Vežba 5

Matić Slavko 63/03 Улазни подаци - Оптерећење, Статички прорачун Опт. 1:

14

P=1400.00

P=1400.00

P=1600.00

p=40.00

25

35

2

4

6

8

10

12

15

17

19

21

23

26

28

31

33

34

1

3

5

7

9

11

13

16

18

20

22

24

27

29

30

32

Диспозиција греда Tower - 3D Model Builder 5.4 DEMO



51


Fundiranje

Vežba 5

Matić Slavko 63/03

-28.42

-14.37

-146.36

Опт. 1:

-28.42

-295.34

-850.08

97.43

54.75

14.37

6.15

739.39

660.57

475.52

-1.78

100.34 -770.06

-439.66

-672.08

-341.69

-115.89

139.80

205.27 -14.80

-495.98

-164.58

-3.70 420.06

-0.34

312.00

4.04

528.00

642.12

856.19

-139.66

-74.56

125.44

Утицаји у греди: max М3= 739.39 / min М3= -850.08 kNm Опт. 1:

Утицаји у греди: max Т2= 856.19 / min Т2= -770.06 kN Tower - 3D Model Builder 5.4 DEMO



52


Fundiranje

Vežba 5

Matić Slavko 63/03

-164.58 -331.40

-336.45

-330.13

-326.80

-326.89

-330.39

-336.82

-328.19

-322.73

-321.16

-323.77

-330.40

3.70

3.70

3.70

3.70

3.70

3.70

4.04

-1500.34

4.04

4.04

4.04

-1865.10

4.04

-1474.56

-3.70

-4.04

Опт. 1:

-310.23 -280.26

-139.80

Утицаји у греди: max N1= 4.04 / min N1= -1865.10 kN Опт. 1:

-0.43-0.43-0.42-0.42-0.42-0.41 -0.41 -0.41

-82.66

-81.78

-84.99

-0.45

-82.76

-0.45

-0.45-0.45-0.45-0.45-0.44-0.44-0.44

-81.87

-83.54

0.44

Утицаји у греди: max u2= 0.44 / min u2= -84.99 m / 1000 Tower - 3D Model Builder 5.4 DEMO



53


Grañevinski fakultet univerziteta u Beogradu - Fundiranje VEŽBA 6 Vp Vg 0,5

Mg

Opterećenje:

Humus Hg

H g = 480 kN

3

V g = 4900 kN V p = 1900 kN

Muljevita glina

M g = 420 kN

NPV

4

Pesak

2d

Šljunak

- Karakteristike slojeva: Humus:

γ = 18 kN / m

3

Muljevita glina:

Pesak:

γ = 17.5 kN / m

γ = 18 kN / m

3

Šljunak:

γ = 18.5 kN / m 3

3

ϕ = 22 o

ϕ = 29 o

c = 10 kN / m 2

c=0

E = 5000 kN / m 2

E = 17000 kN / m

ϕ = 30 o c=0 2

ν = 0.30

ν = 0.35

E = 30000 kN / m 2

ν = 0.25

- Temeljna stopa se izvodi na “FRANKI” šipovima φ 520 d = 0.60 m

d baze = 1.5 ⋅ d = 0.90 m

S max = 1100 kN

- Pretpostavljamo da je visina naglavne grede h=1.0 m

6.1. Potreban broj šipova:

Vmax = V g + V p = 4900 + 1900 = 6800 kN n=

1.1 ⋅ Vmax 1.1 ⋅ 6800 ⋅η = ⋅ 1.1 = 7.48 S doz 1100

⇒ Usvojeno : 7φ 520 Matić Slavko 63/03


54


Grañevinski fakultet univerziteta u Beogradu - Fundiranje 6.2. Centrisanje šipova za stalno opterećenje:

∑M

=0 ⇒ e=

g

∑M ∑V

=

g

g

420 + 480 ⋅ 1.0 = 0.18 m 4900

0,6

1,8

4,8

1,8

0,6

6.3. Raspored šipova:

0 ,6 0 ,9

1 ,8 4 ,8

0 ,9 0 ,6

6.4. Analiza opterećenja: - Korisno opterećenje: - Težina naglavne grede: - Težina tla:

6800 kN (4.8 ⋅ 4.8 − 4 ⋅ 0.6 / 2)⋅ 25 ⋅ 1 = 558 kN 4.8 ⋅ 4.8 − 4 ⋅ 0.6 2 / 2 ⋅ 18 ⋅ 0.5 = 200.88 kN 2

(

)

‘

S max = S max =

∑V + V

p

n

⋅ e − H ⋅ 1.0

∑y

2 i

∑V = 7558.88 kN

⋅ yi

7558.88 1900 ⋅ 0.18 − 480 ⋅ 1.0 + ⋅ 1.8 = 1054.28 kN < S doz = 1100 kN 7 0.9 2 ⋅ 4 + 1.8 2 ⋅ 2

6.5. Odreñivanje dužine šipa: - Dozvoljena sila u šipu:

S = Sb + S0 6.5.1. Nosivost baze šipa:

S b = Fb ⋅ σ doz

σ doz = k s ⋅ (∑ γ i ⋅ hi ) ⋅ N q

∑γ

i

⋅ hi = 0.5 ⋅ 18 + 3 ⋅ 17.5 + 4 ⋅ 10.5 + h ⋅ 11 = 103.5 + 11 ⋅ h

ϕ = 30 o

tgφ r =

tgϕ tg 30 = ⇒ φ r = 21o ⇒ N q = 15 FS 1.5

k s = 1 − sin 30 = 0.50 Matić Slavko 63/03


55


Grañevinski fakultet univerziteta u Beogradu - Fundiranje σ doz = 0.5 ⋅ (103.5 + 11h ) ⋅ 15 = 776.25 + 82.5h 0. 9 2 ⋅ π Fb = = 0.64 m 2 4 S b = Fb ⋅ σ doz = 0.64 ⋅ (776.25 + 82.5h ) = 496.8 + 52.8 ⋅ h

6.5.2. Nosivost omotača šipa

S 0 = ∑ t i ⋅ F0i ci tgϕ + q i ⋅ (1 − sin ϕ ) ⋅ 2.5 1.5 10 tg 22 t1 = + (0.5 ⋅ 18 + 1.5 ⋅ 17.5) ⋅ (1 − sin 22 ) ⋅ = 9.94 kN / m 2 2.5 1.5 tg 29 t 2 = (0.5 ⋅ 18 + 3 ⋅ 17.5 + 2 ⋅ 10.5) ⋅ (1 − sin 29 ) ⋅ = 15.71 kN / m 2 1.5 h tg 30   t 3 =  0.5 ⋅ 18 + 3 ⋅ 17.5 + 4 ⋅ 10.5 + ⋅ 11 ⋅ (1 − sin 30 ) ⋅ = 19.92 + 1.06h 2 1.5   ti =

F0,1 = d ⋅ π ⋅ h1 = 0.6 ⋅ π ⋅ 2 = 3.77 m 2 F0, 2 = d ⋅ π ⋅ h2 = 0.6 ⋅ π ⋅ 4 = 7.54 m 2 F0,1 = d ⋅ π ⋅ h3 = 0.6 ⋅ π ⋅ h = 1.88h S 0 = ∑ t i ⋅ F0i = 9.94 ⋅ 3.77 + 15.71 ⋅ 7.54 + (19.92 + 1.06h ) ⋅ 1.88h

S = S b + S 0 = 496.8 + 52.8h + 155.93 + 33.45h + 1.99h 2 S = 1.99h 2 + 90.25h + 652.73 ⇒ h = 6.34 m L = 2 + 4 + 6.34 = 12.34 m 6.6. Metoda deformacija -

Prosečni modul elastičnosti za slojeve date deformabilnosti:

ES = -

1 n 1 Ei ⋅ hi = ⋅ (5000 ⋅ 2 + 17000 ⋅ 4 + 30000 ⋅ 6.34 ) = 21734.20 kN / m 2 ∑ L i =1 12.34

Koeficijent relativne krutosti k=

Eb 30 ⋅ 10 6 = = 1380.31 kN / m 2 E S 21734.20

Matić Slavko 63/03


56



Potrebni koeficijenti prema dijagramima parametarske analize:

d b / d = 0.9 / 0.6 = 1.5

  ⇒ I 0 = 0.085 L / d = 12.34 / 0.6 = 20.57 

k = 1380.31 kN / m 2   ⇒ Rk = 1.080 L / d = 20.57  L/h = 0   ⇒ R h = 1. 0 L / d = 20.57 

νs =

1 1 ν i ⋅ hi = ⋅ (2 ⋅ 0.35 + 4 ⋅ 0.30 + 6.34 ⋅ 0.25) = 0.28 ∑ L 12.34

k = 1380.31kN / m 2   ⇒ Rv = 0.94 ν s = 0.28   k = 1380.31  Eb 30 ⋅ 0.60  = = 1.38 ⇒ Rb = 0.96 E s 21734.20   L / d = 20.57 I = I 0 ⋅ Rk ⋅ Rh ⋅ Rv ⋅ Rb = 0.085 ⋅ 1.080 ⋅ 1.0 ⋅ 0.94 ⋅ 0.96 = 0.083 -

Sleganje usled jedinične sile P=1.0 F22 =

-

P 1. 0 ⋅I = ⋅ 0.083 = 6.36 ⋅ 10 −6 Es ⋅ d 21734.20 ⋅ 0.6

Odreñivanje koeficijenata horizontalne reakcije prema Vesiću:

ks =

E 0.65 E s ⋅ d 4 ⋅ 12 ⋅ s d Eb ⋅ I b 1 − ν s

ks =

0.65 12 21734.20 ⋅ 0.6 4 21734.20 0.65 12 ⋅ = ⋅ 0.01476 ⋅ 23583.12 = 17979.89 kN / m 3 ⋅ 6 0.6 0.6 30 ⋅ 10 ⋅ 0.00636 1 − 0.28

λ=4

ks ⋅ d 17979.89 ⋅ 0.6 =4 = 0.345 4 ⋅ Eb ⋅ I b 4 ⋅ 190800

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-

Koeficijenti fleksibilnosti za uticaje jedinične horizontalne sile ( Ho=1.0 , Z=0 ) F11 =

2 ⋅ H 0 ⋅ λ − λz 2⋅λ 2 ⋅ 0.345 ⋅ e ⋅ cos λz = = = 63.96 ⋅ 10 −6 ks ⋅ d k s ⋅ d 17979.89 ⋅ 0.6

F31 =

2 ⋅ H 0 ⋅ λ 2 − λz 2 ⋅ λ2 2 ⋅ 0.345 2 ⋅ e (cos λz + sin λz ) = = = 22.066 ⋅ 10 − 6 ks ⋅ d k s ⋅ d 17979.89 ⋅ 0.6

Koeficijenti fleksibilnosti za uticaj jediničnog momenta savijanja ( Mo=1.0 , Z=0 )

2 ⋅ M 0 ⋅ λ 2 − λz 2 ⋅ λ2 2 ⋅ 0.345 2 F13 = ⋅ e ⋅ (cos λZ − sin λZ ) = = = 22.066 ⋅ 10 −6 ks ⋅ d k s ⋅ d 17979.89 ⋅ 0.6 F33 = -

4 ⋅ M 0 ⋅ λ3 − λ z 4 ⋅ λ3 4 ⋅ 0.345 3 ⋅ e cos λZ = = = 15.23 ⋅ 10 −6 ks ⋅ d k s ⋅ d 17979.89 ⋅ 0.6

Matrica fleksibilnosti

0 22.066  63.96  F= 0 6.36 0  ⋅ 10 −6  22.066 0 15.23  -

Matrica krutosti

K=F

−1

0 − 0.4529  0.3126   ⋅ 10 5 = 0 1.5723 0   − 0.4529 0 1.3128  1

Za šipove 1:

0  1 0  I1 = 0 1 − 1.8 0 0 1 

2

3

4

5

Za šipove 2:

0  1 0  I 2 = 0 1 − 0.9  0 0 1 

Za šipove 4:

Za šipove 5:

1 0 0  I 4 = 0 1 0.9  0 0 1 

1 0 0  I 5 = 0 1 1.8 0 0 1 

Za šipove 3:

1 0 0  I 3 = 0 1 0  0 0 1 

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Grañevinski fakultet univerziteta u Beogradu - Fundiranje 0 − 0.4529  0.3126  K1 = I ⋅ K ⋅ I 1 =  0 1.5723 − 2.8302 ⋅ 10 5 − 0.4529 − 2.8302 6.4071  −

T 1

0 − 0.4529  0.3126  K2 = I ⋅ K ⋅ I2 =  0 1.5723 − 1.4151  ⋅ 10 5  − 0.4529 − 1.4151 2.5864  −

T 2

0 − 0.4529  0.3126   ⋅ 10 5 K3 = I ⋅ K ⋅ I3 =  0 1.5723 0   − 0.4529 0 1.3128  −

T 3

0 − 0.4529  0.3126  K4 = I ⋅ K ⋅ I4 =  0 1.5723 1.4151  ⋅ 10 5 − 0.4529 1.4151 2.5864  −

T 4

0 − 0.4529  0.3126  K5 = I ⋅ K ⋅ I5 =  0 1.5723 2.8302  ⋅ 10 5 − 0.4529 2.8302 6.4071  −

-

T 5

Matrica krutosti sistema −

−

−

−

−

K 0 = 1 ⋅ K1 + 2 ⋅ K 2 + 1 ⋅ K 3 + 2 ⋅ K 4 + 1 ⋅ K 5

0 − 0.3170  0.2188   ⋅ 10 6 K0 =  0 1.1006 0  0 2.4473  − 0.3170 -

Nepoznata generalisana pomeranja:

H0  u 0   V  = K ⋅ ϑ  0  0  0   M 0  ϕ 0 

H 0 = 480 kN V0 = 7558 .88 kN M 0 = 480 ⋅1 + 420 = 900 kNm

u 0  H0  0.5626 ϑ  = K −1 ⋅  V  = 10 5  0 0  0  0   ϕ 0   M 0  0.0729

0 0.0909 0

0.0729   0.48  0.0034     3 0  ⋅ 7.5588  ⋅ 10 = 0.0069  0.0008  0.0503   0.9 

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Grañevinski fakultet univerziteta u Beogradu - Fundiranje  u1  0.0034   q1 = ϑ1  = I 1 ⋅ q = 0.0054 ϕ1  0.0008

 H1   68.57   V  = K ⋅ q =  852.70  1  1    M 1  − 46.65

u 2  0.0034   q 2 = ϑ2  = I 2 ⋅ q =  0.0061 ϕ 2  0.0008

H2   68.57   V  = K ⋅ q =  966.27  2  2    M 2   − 46.65

u 3  0.0034   q 3 = ϑ3  = I 3 ⋅ q = 0.0069 ϕ 3  0.0008

H3   68.57   V  = K ⋅ q =  1079.8  3  3    M 3   − 46.65

u 4  0.0034   q 4 = ϑ4  = I 4 ⋅ q = 0.0076 ϕ 4  0.0008

H4   68.57   V  = K ⋅ q =  1193.4  4  4    M 4   − 46.65

u 5  0.0034   q 5 = ϑ5  = I 5 ⋅ q = 0.0083 ϕ 5  0.0008

H5   68.57   V  = K ⋅ q =  1307  5  5    M 5  − 46.65

∑V = 852.70 + 2 ⋅ 966.27 + 1079.8 + 2 ⋅ 1193.4 + 1307 = 7558.88 kN

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Grañevinski fakultet univerziteta u Beogradu - Fundiranje 6.7. Dimenzionisanje naglavne grede 6.7.1. Podužne trake

1,72

2,08

1 I

I

0,6

0,9

0,9

0,9

0,9

0,6

M I − I = 1307 ⋅ 1.48 + 2 ⋅ 1193.40 ⋅ 0.58 = 3318.7 kNm M u = 1.65 ⋅ 3318.7 = 5475.86 kNm b = 60 cm h = 100 − 10 = 90cm k=

90 5475.86 ⋅ 100 3 ⋅ 60 ⋅ 2.05

Aa = 20.546 ⋅

Usvojeno:

= 2.336

−

µ = 20.546%

⇒

60 ⋅ 90 2.05 ⋅ = 56.86 cm 2 100 40

(

12 Rφ 25 58.92 cm 2

)

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Grañevinski fakultet univerziteta u Beogradu - Fundiranje 6.7.2. Poprečne trake

2,15

0,5

2,15 a

a

0,6

1,8

1,8

0,6

M a − a = 966.27 ⋅ 1.8 + 1193.40 ⋅ 1.8 = 3887.41 kNm M u = 1.65 ⋅ 3887.41 = 6414.22 kNm b = 60 cm h = 100 − 10 = 90cm k=

90 6414.22 ⋅ 100 5 ⋅ 60 ⋅ 2.05

Aa = 13.987 ⋅ Usvojeno:

= 2.787

⇒

−

µ = 13.987%

60 ⋅ 90 2.05 ⋅ = 38.69 cm 2 100 40

(

8 Rφ 25 38.69 cm 2

)

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64

Elaborat Iz Fundiranja

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