1

1Svođenje opterećenja na grede

C

B

A

3 41 2

POS

104

POS 103

POS

105

POS

106

POS

107

POS 102

POS 101

7.205.40 5.40

6.00

6.00

2.40 A

C D

F G

A

C

E

1.80

B

2.40

2

2

3

4

3

5

6

5.407.20

6.00

6.00

5.40

2.40

1.80

2.40

g [kN/m]

7.03

7.06

10.1

4

9.86

6.48

9.32

9.14

9.32

9.55

10.5

8

9.55

10.5

8

10.04

10.04

7.56

10.80

6.48

9.32

9.14

9.32

9.23

6.55

9.23

8.93

10.8

0

14.40

A

C D

F G

A

C

E B

14.40C

B

A

3 41 2

Stalno opterećenje (bez s.t. greda POS 101-POS 107)

4

7

5.407.20

6.00

6.00

5.40

2.40

1.80

2.40

p [kN/m]

11.7

2

11.76

16.9

0

16.44

10.8

0

15.54

15.2

3

15.54

15.9

117

.64

15.9

117

.64

16.74

16.74

12.60

18.00

10.8

0

15.54

15.2

3

15.54

15.3

9

10.92

15.3

9

14.88

18.0

0

24.00

A

C D

F G

A

C

E B

24.00C

B

A

3 41 2

Povremeno opterećenje

8Procena sila u stubovima

C

B

A

3 41 2

POS

104

POS 103

POS

105

POS

106

POS

107

POS 102

POS 101

7.205.40 5.40

6.00

6.00

2.40

A2

1.80

2.40

3.60 2.70

3.00

3.00

3.00

2.40

5.40

6.00

3.00

6.30

A1A1

5

9

A1 = 6.3×6.0 = 37.8 m2

G+P ≈ (g+p)×A1 = (6.0 + 10.0)×37.8 = 604.8 kNOvu silu treba povećati barem 25% zbog povećanog

naprezanja prve linije unutrašnjih stubova, kao i zbog zanemarenja sopstvene težine greda POS 102 i POS 106

(G+P)max ≈ 1.25×604.8 = 756 kNZa presek napregnut ovolikom EKSPLOATACIONOM

silom potrebna površina poprečnog preseka je približno 756 cm2, pa se pretpostavlja dimenzija stubova 30×30 cm.

Širina greda je jednaka širini stubova, dakle usvojeno b = 30 cm.

10

0.375

0.625

0.375

0.40.5

0.6

0.375 L 0.4 L 0.5 L0.625 L

L L

p

L

p

TT

(x pL)

(x pL) (x pL)

A = 38 B = 54 A = 25 B = 1110A = 38

1.25 L

0.6 L

1.1 L0.625

L2

L

p

L L

0.3940.528 0.492 0.503

0.394 L 0.528 L 0.492 L 0.503 L

0.4970.5080.4720.606

L

T (x pL)

A = 153388 B = 110

97 C = 187194 D = 98

97 E = 193194

(x pL)

(x pL)

1.134 L 0.964 L 1.01 L

6

1111POS 103 (osa C) – analiza opterećenjapretp. b=30 cm ; d = L2/12 = 720/12 = 60 cmsopstvena težina grede:

g0 = 0.3×0.6×25 = 4.5 kN/m

stalno opterećenjepolje 1-2: g1 = 4.5 + 7.06 = 11.56 kN/mpolje 2-3: g2 = 4.5 + 10.04 + 14.4 = 28.94 kN/mpolje 3-4: g3 = 4.5 + 9.32 + 14.4 = 28.22 kN/m

povremeno opterećenjepolje 1-2: p1 = 11.76 kN/mpolje 2-3: p2 = 16.74 + 24.0 = 40.74 kN/mpolje 3-4: p3 = 15.54 + 24.0 = 39.54 kN/m

1212POS 103 – uticaji od stalnog opterećenja

L1 L2 L3 = L1

M1

M2

g1 = 11.56 kNm

0 1 2 3

2221

21

21311

1g1 L3LL8L4LL

2LgM

×+××+×+

××

−=,

kNm71927327458454

27452

455611M 22

3

1g1 .....

...., −=

×+××+×+

××

−=

( ) ( ) kNm6527452

27719LL2

LMM21

21g11g2 .

....,, =

+××=

+××−=

7

1313POS 103 – uticaji od stalnog opterećenja

L1 L2 L3 = L1

0 1 2 3

M1

g2 = 28.94 kNm

M2

( ) 2g221

322

2g1 ML3L22

12

LgM ,, =×+××

××

−=

( ) 2g2

3

2g1 MkNm4832734522

12

279428M ,, ...

..=−=

×+×××

×−=

1414POS 103 – uticaji od stalnog opterećenja

L1 L2 L3 = L1

0 1 2 3

M2

M1

g3 = 28.22 kNm

2221

21

21313

3g2 L3LL8L4LL

2LgM

×+××+×+

××

−=,

kNm04827327458454

27452

452228M 22

3

3g2 .....

...., −=

×+××+×+

××

−=

( ) ( ) kNm71327452

27048LL2

LMM21

23g23g1 .

....,, =

+××=

+××−=

8

1515M1g = -19.7 + (-83.4) + 13.7 = -89.3 kNmM2g = 5.6 + (-83.4) + (-48.0) = -125.8 kNm

kN952458125

2452228Dg .

....

=−×

=

( )

27

27457148125227942827

245455611

B

2

g .

.......... +×−−×+

+××

=

kN9146Bg .=

kN71445389

2455611Ag .

....

=−×

=

( ) ( )95291467142794284522285611Cg ........ ++−×+×+=

kN8208Cg .=

1616POS 103 - stalno opterećenje (G)

89.3Mg

Tg

L1 = 5.4Ag = 14.7

L2 = 7.2Dg = 52.9Bg = 146.9 Cg = 208.8

L3 = L1 = 5.4

g1 = 11.56 kNm

g2 = 28.94 kNm g3 = 28.22 kN

m

125.8

146.

9 kN

208.

8 kN

52.9

kN

14.7

kN

9

1717POS 103 – uticaji od povremenog opterećenja

L1 L2 L3 = L1

M1

M2

p1 = 11.76 kNm

0 1 2 3

2221

21

21311

1p1 L3LL8L4LL

2LpM

×+××+×+

××

−=,

kNm02027327458454

27452

457611M 22

3

1p1 .....

...., −=

×+××+×+

××

−=

( ) ( ) kNm7527452

27020LL2

LMM21

21p11p2 .

....,, =

+××=

+××−=

1818POS 103 – uticaji od povremenog opterećenja

L1 L2 L3 = L1

0 1 2 3

M1

p2 = 40.74 kNm

M2

( ) 2p221

322

2p1 ML3L22

12

LpM ,, =×+××

××

−=

( ) 2p2

3

2g1 MkNm31172734522

12

277440M ,, ...

..=−=

×+×××

×−=

10

1919POS 103 – uticaji od povremenog opterećenja

L1 L2 L3 = L1

0 1 2 3

M2

M1

p3 = 39.54 kNm

2221

21

21313

3p2 L3LL8L4LL

2LpM

×+××+×+

××

−=,

kNm36727327458454

27452

455439M 22

3

3p2 .....

...., −=

×+××+×+

××

−=

( ) ( ) kNm21927452

27367LL2

LMM21

23p23p1 .

....,, =

+××=

+××−=

2020M1p = -20.0 + (-117.3) + 19.2 = -118.1 kNmM2p = 5.7 + (-117.3) + (-67.3) = -178.9 kNm

kN673459178

2455439Dp .

....

=−×

=

( )

27

2745999178227744027

245457611

B

2

p .

.......... +×−−×+

+××

=

kN9191Bp .=

kN99451118

2457611Ap .

....

=−×

=

( ) ( )6739191992774404554397611Cp ........ ++−×+×+=

kN0295Cp .=

11

2121POS 103 - povremeno opterećenje (P)

p1 = 11.76 kNm

p2 = 40.74 kNm p3 = 39.54 kN

m

Ap = 9.9 Dp = 73.6Bp = 191.9 Cp = 295.0

118.1178.9

L1 = 5.4 L2 = 7.2 L3 = L1 = 5.419

1.9

kN

295.

0 kN

73.6

kN

9.9

kN

Mp

Tp

2222M1u = 1.6×89.3 + 1.8×118.1 = 355.5 kNm (zateže gornju ivicu)M2u = 1.6×125.8 + 1.8×178.9 = 523.2 kNm (zateže gornju ivicu)Au = 1.6×14.7 + 1.8×9.9 = 41.2 kNqu1 = 1.6×11.56 + 1.8×11.76 = 39.66 kN/m

m0820412Lm0416639241x 101 ...

..

,max, =×=⇒==

kNm4212

0416639041241M2

01u .....max, =×

−×=

Du = 1.6×52.9 + 1.8×73.6 = 217.2 kNqu3 = 1.6×28.22 + 1.8×39.54 = 116.33 kN/m

m7338712Lm871331162217x 303 ...

..

,'

max, =×=⇒==

kNm82022

871331168712217M2

23u .....max, =×

−×=

12

2323

m413641194407x2 .

..

max, ==

kNm233853552

413641194134407M2

12u ......max, =−×

−×=

kN9172456639241T lBu ...., −=×−=

kN3580919181914661Bu ..... =×+×=

kN440735809172T dBu ..., =+−=

mkN64119744081942861q 2u ..... =×+×=

m76464119

23388L 20 ..

., =

×=

kN045427641194407T lCu ...., −=×−=

kN041145331162217T dCu ...., =×+−=

2424POS 103 - granično opterećenje (1.6×G+1.8×P)

523.2

355.5

338.2

21.4

202.8

41.2

172.9

407.4

454.0

217.2

411.0

L0 = 2.08

1.04

3.795

L0 = 4.756 L0 = 3.734

1.867

Mu

Tu

qu1 = 39.66 kNm

qu2 = 119.64 kNm qu3 = 116.33 kN

m

Au = 41.2 Du = 217.2Bu = 580.3 Cu = 865.0L1 = 5.4 L2 = 7.2 L3 = L1 = 5.4

13

2525POS 103 – dimenzionisanje

Mu,max = 523.2 kNm (osa 3, gornja zona)MB 30 ⇒ fB = 20.5 MPa = 2.05 kN/cm2

RA 400/500 ⇒ σv = 400 MPa = 40 kN/cm2

pretp. a1 = 7 cm ⇒ b/d/h = 30/60/53 cm

%.‰././

.

.. 51737

0524538171

05230102523

53k ab

2 =µ=εε

⇒=

××

=

2a cm5730

40052

100533051737A ... =×

××=

usvojeno: 2RØ16 + 6RØ25 (33.47 cm2)

2626POS 103 – kontrola glavnih napona zatezanja

Tu,max = 454.0 kN (presek u osi 3 levo)MB 30 ⇒ τr = 1.1 MPa = 0.11 kN/cm2

usv. z = 0.9×h = 0.9×53 = 47.7 cm

=τ<=τ>

=×

=τ 2r

2r2lC

n cmkN3303cmkN110

cmkN317074730

0454/.

/./.

..,

)()(

..

1u

1u

u a58840311030

a2e ×=×××

=

( ) 2Ru cmkN3110110317051 /.... =−×=τ

usvojeno: m=2 ; α = 90° ; θ = 45°

URØ10 + URØ8 ⇒ eu = 8.58×(0.785 + 0.503) = 11.05 cmusvojeno: URØ10/10 + URØ8/10 (m=2)

14

27

L0 = 4.76

c=1.67

L0 = 3.73

2 3 4

x2,max = 3.41

a=1.03 2.38 b=1.42

L2 = 7.20 L3 = 5.40

2.38

x'3,max = 1.87

m0312764413

2L

xa 202 ...,

max, =−=−=

( ) ( ) m42103176427aLLb 202 ...., =+−=+−=

m67173345LLc 303 ..., =−=−=

Procena dužine šipki

28

c = 1.67

3

b = 1.42

2Ø16

2Ø25

2Ø25

2Ø25

2Ø16

+ 6

Ø25

c1 = 1.47b1 = 1.25 v+Lsv+Ls

cm753953750v .. =×=

m25147334529421

25Ø616Ø225Ø6bb1 .

.

.. =×=+

×=

MPa75130MB p .=τ⇒

Ø7531Ø751814

400L 1s ×=××

= ...

Ø647L51L 1s2s ×=×= ..

526477539Lv s ... ×+=+

cm8158Lv s .=+

2a cm4733452902491460122A ..... =+=×+×=

m47147334529671

25Ø616Ø225Ø6cc1 .

.

.. =×=+

×=

Procena dužine šipki u gornjoj zoni

15

29

c = 1.67

3

b = 1.42

2Ø16

2Ø25

2Ø25

2Ø25

2Ø16

+ 6

Ø25

c1 = 1.47b1 = 1.25

0.980.83

0.490.42 v+Lsv+Ls

v+Ls

v+Ls

v+Ls

v+Ls

Ø647L51L 1s2s ×=×= ..

526477539Lv s ... ×+=+

cm8158LvLv 2ss .=+=+

cm590159147125159LvcbLvL s11s1 =+++=+++++≈

cm50015931472

31252159Lv

3c2

3b2LvL s

11s2 =+

×+

×+=+++++≈

cm4101593

1473

125159Lv3c

3bLvL s

11s3 =+++=+++++≈

3030POS 103 – dimenzionisanje – ostali preseci

Mu = 355.5 kNm (osa 2, gornja zona)MB 30 ⇒ fB = 20.5 MPa = 2.05 kN/cm2

RA 400/500 ⇒ σv = 400 MPa = 40 kN/cm2

pretp. a1 = 7 cm ⇒ b/d/h = 30/60/53 cm

%.‰././

.

.. 38923

6148532042

05230105355

53k ab

2 =µ=εε

⇒=

××

=

2a cm0619

40052

100533038923A ... =×

××=

usvojeno: 2RØ16 + 4RØ22 (19.23 cm2)

16

31

L0,2 = 4.76

2

x2,max = 3.41

a = 1.03 2.38 2.38

1

L0,1 = 2.08

d = 3.32

L1 = 5.40

m0312764413

2L

xa 202 ...,

max, =−=−=

m32308245LLd 101 ..., =−=−=

32

2

2Ø16

2Ø22

2Ø16

+ 4

Ø22

a1 = 0.81 v+Ls

a = 1.03d = 3.32

2Ø22

d1 = 2.63v+Ls

m63223192115323

22Ø416Ø222Ø4dd1 .

.

.. =×=+

×=

2a cm2319211502480340122A ..... =+=×+×=

m81023192115031

22Ø416Ø222Ø4aa1 .

.

.. =×=+

×=

Procena dužine šipki u gornjoj zoni

17

332

a = 1.03d = 3.32

2Ø16

2Ø22

2Ø22

2Ø16

+ 4

Ø22

d1 = 2.63

1.31

v+Ls

v+Ls 0.41

a1 = 0.81

v+Ls

v+Ls

cm145226477539Lv s =×+=+ ...

cm63563414581263145LvadLvL s11s1 ≈=+++=+++++≈

cm465462145281

2263145Lv

2a

2dLvL s

11s2 ≈=+++=+++++≈

3434Mu = 338.2 kNm (polje 2-3, donja zona)pretp. a1 = 7 cm ⇒ b/d/h = 30/60/53 cm

%..

‰/./.

.. 0804

0950s100451

0355

052149102338

53kab

2

=µ=

=εε⇒=

××

=

2a cm5016

40052

100531490804A ... =×

××=

usvojeno: 5RØ22 (19.01 cm2)

cm14914947625030

5402402600

350162030Bm764L0 =

=×+=+=×+

=⇒=.

.min.

cm16dcm05530950x p =<=×= ..

18

35

0.5×L0,2 = 2.381.03

2

x2,max = 3.41

2Ø22

3Ø22

5Ø22

3Ø22

2Ø22

x2 = 3.68

x1 = 3.00

v+Lsv+Ls

v+Lsv+Ls

L0,2 = 4.76

0.5×L0,2 = 2.38

Procena dužine šipki u donjoj zoni

3636

m0034076422Ø522Ø2L

MMLx 0

101 ... =×=×=×=

m6836076422Ø522Ø3L

MMLx 0

202 ... =×=×=×=

cm61092275317539LvLv 1ss .... =×+=+=+

cm520110300110LvxLvL s1s1 =++=++++≈

cm590588110368110LvxLvL s2s2 ≈=++=++++≈

Procena dužine šipki u donjoj zoni

Šipke su simetrično postavljene u odnosu na mesto maksimalnog momenta u polju (x = 3.41 ≈ 3.40 m od ose 2 ka osi 3)

19

3737Mu = 202.8 kNm (polje 3-4, donja zona)pretp. a1 = 5 cm ⇒ b/d/h = 30/60/55 cm

%..

‰/./.

.. 7232

0760s108270

1426

052123108202

55kab

2

=µ=

=εε⇒=

××

=

2a cm479

40052

100551237232A ... =×

××=

usvojeno: 3RØ22 (11.41 cm2)

cm12312337325030

5402402600

350162030Bm733L0 =

=×+=+=×+

=⇒=.

.min.

cm16dcm24550760x p =<=×= ..

38

0.5×L0,3 = 1.87

x3 = 2.16 v+Ls

L0,3 = 3.73

0.5×L0,3 = 1.87

4

1Ø22

2Ø22

3Ø22

v+Ls

Procena dužine šipki u donjoj zoni

m16231733

22Ø322Ø1L

MMLx 0

303 .. =×=×=×=

cm440436110216110LvxLvL s3s3 ≈=++=++++≈