Pismeniispit, 16.09.2015., grupeA+Bimksus.grf.bg.ac.rs/nastava/beton/PROJEKTOVANJE I... ·...

1Pismeni ispit, 16.09.2015., grupe A+B

B

7.20

POS 1

6.006.00

2.40

1 2 3

A

2.40

POS 2 POS 2

A

POS

1d p

=20

cm

5.00

Pres

ek A

-A

POS

S125

/25

cmPOS

225

/60

cmPO

S 2

POS

325

/60

cm

A

POS S1

POS

S3

POS S2

POS 2 POS 2

POS S3

POS S325/50 cm

POS S225/25 cmPOS S1

25/25 cm

POS

3

25/2

5 cm

25/25 cm 25/25 cm25/50 cm

(25/

60 c

m)

POS

4(5

0/60

cm

)

POS

3(2

5/60

cm

)

0.500.25 0.25

0.25

0.25

W x = ±160 kN

Wy = ±160 kN

2

6.00

7.20

6.00

B

A

31 2

2.40

2.40

B

A

B

A

POS

3

POS 2 POS 2

POS 2 POS 2

POS

3

POS

4

g = 0.20×25 = 5.0 kN/m2

p = 10 kN/m2

konzolna ploča (deo “A”)Rg = 5.0×2.4 = 12.0 kN/mRp = 10.0×2.4 = 24.0 kN/mqu = 1.6×5.0 + 1.8×10.0 = 26.0 kN/m2

Mu = 26.0×2.42/2 = 74.88 kNm/m

krstasta ploča (deo “B”)Kako se povremeno opterećenje može naći u proizvoljnom položaju na konstrukciji, potrebno ga je razdvojiti na simetrični deo p/2 (deluje po čitavoj površini ploče) i antimetrični deo ±p/2(razlikuje se u zavisnosti od razmatranog uticaja).

3

g = 0.20×25 = 5.0 kN/m2

p = 10 kN/m2

Ly/Lx = 7.2/6.0 = 1.2G = 5.0×6.0×7.2 = 216 kN = P/2

Qu1 = 1.6×G + 1.8×P/2 = 734.4 kN

Mxu1 = 0.032×734.4 = 23.5 kNm/mMyu1 = 0.023×734.4 = 16.9 kNm/m-Xu1 = 0.071×734.4 = 52.1 kNm/m-Yu1 = 0.062×734.4 = 45.5 kNm/m

-X

-Y

Lx = 6.00

L y =

7.2

0

Mx

My

Simetrični deo (g, p/2)

4Antimetrični deo (±p/2)

B3

-X-X

-Y-Y

B2B1

AA A

A

B3B2

YY

AA

Mx

My

B1

+p2 +p

2

–p2

–p2

+p2 –p

2

+p2

–p2

+p2 –p

2

–p2

+p2

P/2 = 5.0×6.0×7.2 = 216 kNQu2 = ±1.8×P/2 = ±1.8×216 = ±388.8 kN

Šeme povremenog opterećenja koje daju najveće momente nad osloncima i u poljima su prikazane na skici. Pri tome se srednja šema (maksimalni momenti –Y) neće ni razmatrati, s obzirom da se na spoju konzolne i krstaste ploče uvek usvaja moment sa konzole.

5Maksimalni oslonački moment -X

918044415

1061526552

052084

17kab

.%.

‰/./.

..

=ζ=µ=εε

⇒==

mcm4613

400521744415A

2

a ... =××=

-X

Lx = 6.00

L y =

7.2

0

Mx

My

-Xu2 = kB1×Qu2 = 0.082×Qu2

-Xu2 = 0.082×388.8 = 31.9 kNm/m-Xu = (-X)u1 + (-X)u2 = 52.1 + 31.9 = 84.0 kNm/mpretp. a1x = 3.0 cmhx = d - a1 = 20 - 3 = 17 cm

⇒ usv.: RØ16/15 (13.40 cm2/m)

mcm692461320A

2

ap ... =×= ⇒ usv.: RØ10/25 (3.14 cm2/m)

6Maksimalni oslonački moment -Y-Yu = 74.88 kNm/m (moment sa konzolne ploče)a1y = 2.0 + 1.6 + 1.6/2 = 4.4 cmhy = 20 – 4.4 = 15.6 cm

⇒ usv.: RØ16/15 (13.40 cm2/m)

mcm632131320A

2

ap ... =×= ⇒ usv.: RØ10/25 (3.14 cm2/m)

914042716

1076325812

0528874615k

ab

.%.

‰/./.

...

=ζ=µ=εε

⇒==

mcm1313

4005261542716A

2

a .... =××=

7Maksimalni momenti u polju Mx , My

Lx = 6.00

L y =

7.2

0

Mx

My

Mxu2 = kB3×Qu2 = 0.049×Qu2

Mxu2 = 0.049×388.8 = 19.1 kNm/m

Mxu = Mxu1 + Mxu2 = 23.5 + 19.1 = 42.6 kNm/m

Myu2 = kB3×Qu2 = 0.038×Qu2

Myu2 = 0.038×388.8 = 14.8 kNm/m

Myu = Myu1 + Myu2 = 16.9 + 14.8 = 31.7 kNm/m

a1x = 2.0 + 1.2/2 = 2.6 cm

hx = 20 – 2.6 = 17.4 cm

mcm796

404179010642A

22

ax ...

.=

×××

≈ ⇒ usv.: RØ12/15 (7.54 cm2/m)

8Maksimalni momenti u polju Mx , My

Myu = Myu1 + Myu2 = 16.9 + 14.8 = 31.7 kNm/m

a1y = 2.0 + 1.2 + 1.0/2 = 3.7 cm ⇒ hy = 20 – 3.7 = 16.3 cm

⇒ usv.: RØ10/15 (5.24 cm2/m)m

cm4054031690

10731A22

ay ...

.=

×××

≈

0824k2686052316100

40245fhb

A

B

va .%...

.=⇒=

×××

=××σ×

=µ

yu

2

u Mm

kNm731m

kNm68320520824

316M =>=×

= ...

..

Za usvojenu armaturu moguće je dokazati moment nosivosti:

9Analiza opterećenja za grede

L y =

7.2

0PO

S 4

POS 2

POS

3

POS 2

Q1

Q2

Q3

Q4

Lx = 6.00

opterećenje sa konzolnih prepusta:Rg = Rp/2 = 5.0×2.4 = 12.0 kN/msopstvena težina greda:gg2 = gg3 = 0.25×0.60×25 = 3.75 kN/mgg4 = 0.50×0.60×25 = 7.50 kN/m

10Analiza opterećenja za grede – simetrični deo

6.78

9.93

6.70

9.25

6.78

9.25

9.93

6.70

B

A

B

12.0

6.00

7.20

6.00

B

A

31 2

2.40

2.40 A

12.0

11Proračun POS 2 – prvo polje

6.00

7.20

6.00

B

A

31 2

2.40

2.40

B2

A

B3

A

qB22 = 10.73 qB3

2 = -8.21

12.0

kN4642162980Q 2B2 .. =×=

mkN7310

06464q 2B

2 ...

==

kN2492162280Q 3B2 .. =×−=

mkN218

06249q 3B

2 ...

−=−=

kN2162

0102706Q =××=...

g1 = 3.75 + 12.0 + 9.25 = 25.0 kN/mg2 = 3.75 + 6.70 = 10.45 kN/mp1 = 12.0 + 9.25 + 12.0 + 10.73 = 43.98 kN/mp2 = 6.70 + (-8.21) = -1.51 kN/mqu1 = 1.6×25.0 + 1.8×43.98 = 119.17 kN/mqu2 = 1.6×10.45 + 1.8×(-1.51) = 13.99 kN/m

12

kN7610616

45100257Ag ....=×

−×=

kN0180616

02545107Cg ....=×

−×=

( ) kN9132068

45100255Bg ....=×

+×=

L1 = 6.00

qu1 = 119.17 kNm qu2 = 13.99 kN

m

Ag = 61.7 Bg = 132.9 Cg = 18.0Ap = 116.0 Bp = 159.3 Cp = -20.5

p1 = 43.98 kNm p2 = -1.51 kN

m

L2 = 6.00

g1 = 25.0 kNm g2 = 10.45 kN

m

( ) kN01160616

51198437Ap ....=×

−−×=

( ) kN5200616

98435117Cp ....−=×

−−×=

Proračun POS 2 – prvo polje

13Proračun POS 2 – prvo poljequ1 = 119.17 kN

m

Au = 307.6307.6

xmax = 2.58

396.9

L0 = 5.16 Mu

Tu

Au = 1.6×61.7 + 1.8×116.0 = 307.6 kN

m165x2Lm582171196307x 0 ..

..

maxmax ==⇒==

kNm9396171192

6307q2

AM2

u

2u

u ..

.max, =

×=

×=

cm154425202025

1544

51625B =

=×+

=+= min

%...‰/./

.

.. 6404

cm45x1010s101281

7284

052154109396

53kab

2

=µ=⇒=

=εε⇒=

××

=

2a cm4219

40052

100531546404A ... =×

××=

usvojeno 6RØ22 (22.81 cm2)

14

UØ10/25

25

2Ø12

2Ø22

3Ø22

3Ø22

5.2

2040

Proračun POS 2 – prvo polje

τ<τ>

=××

=τr

r2n 3cm

kN2580539025

6307 ...

m48125801101582 .

... =

−×=λ

( ) 2ARu cm

kN2220110258023 ... =−×=τ

cm31140222025

78502eu ..

.=×

××

=

usvojeno URØ10/10 (m=2)

( ) 2a cm8439045

4026307A .cotcot.

=°−°××

=∆


15Proračun POS 2 – drugo polje

6.00

7.20

6.00

B

A

31 2

2.40

2.40

B2

A

B3

A

qB22 = -10.73 qB3

2 = 8.21

-12.0

p1 = 12.0 + 9.25 + (-12.0) + (-10.73) = -1.48 kN/mp2 = 6.70 + 8.21 = 14.91 kN/mqu1 = 1.6×25.0 + 1.8×(-1.48) = 37.35 kN/mqu2 = 1.6×10.45 + 1.8×14.91 = 43.54 kN/m

( ) kN590616

91144817Ap ....min, −=×

−−×=

( ) kN7390616

48191147Cp ....max, =×

−−×=

Cu = 1.6×18.0 + 1.8×39.7 = 100.3 kN

kNm511554432

3100M2

u ..

.max, =

×=

m6043022Lm30254433100x 0 ...

..

max =×=⇒==

16Proračun POS 2 – drugo poljequ2 = 43.54 kN

m

Mu

Tu 2.30

100.3

115.5

L0 = 4.61

Cu = 100.3

cm6318520825

631246025B =

=×+

=+= min

%...‰/./

.

..

.

9682cm44x0800s

1086908875

05263105115

555kab

2

=µ=⇒=

=εε⇒=

××

=


2a cm355

40052

100555639682A .... =×

××=

r2n cmkN0540

5390253100

τ<=××

=τ ...

usvojeno URØ8/30 (15)

17Proračun POS 2 – oslonackN216

20102706Q =××=

...

p1 = 12.0 + 9.25 + 12.0 + 9.25 = 42.50 kN/mp2 = 6.70 + 7.05 = 13.75 kN/mqu1 = 1.6×25.0 + 1.8×42.50 = 116.51 kN/mqu2 = 1.6×10.45 + 1.8×13.75 = 41.47 kN/m

6.00

7.20

6.00

B

A

31 2

2.40

2.40

B

A

B1

A

qB3 = 9.25 qB1

3 = 7.05

12.0

mkN259

062162570qB

3 ..

.=

×=

kN3422161960Q 1B3 .. =×=

mkN057

06342q 1B

3 ...

==

18Proračun POS 2 – oslonackN410606

16751350427Ap ....

=×−×

=

kN2200616

504275137Cp ....=×

−×=

( ) kN0211068

75135425Bp ....max, =×

+×=

kNm43550616

474151116M 2u ....

max, −=×+

−=

%.‰././

.

.. 01729

2646530122

05225104355

53k ab

2 =µ=εε

⇒=

××

=

2a cm7019

40052

100532501729A ... =×

××= ⇒ usvojeno 6RØ22 (22.81 cm2)

5.2

3Ø22

3Ø22

2Ø12

2Ø22

UØ10/12.5UØ10/12.5

+−=

16Lq

16LqM

222

211

u max,

19Proračun POS 2 – oslonac

Au = 1.6×61.7 + 1.8×106.4 = 290.3 kN

11uulB

u LqAT ×−=,

kN840806511163290T lBu ...., −=×−= L1 = 6.00

qu1 = 116.51 kNm

Ag = 61.7 Bg = 132.9

Ap = 106.4 Bp = 211.0

2.49

183.6290.3

408.8

Tur2

lBn 3

cmkN3430

5390258408

τ>=××

=τ ...'

( ) m3823430110149206 .

.... =

−×−=λ

cm3740343025

78502eu ..

.=×

××

= ⇒ usvojeno U2RØ10/12.5

m492511163290x1 .

..

==

20Proračun POS 2 – oslonac

τr

L0,T = 350.8

λ = 238.3

τB,ln =3.43

τ(1)u,u=2.01

3τr

13.188λ1 = 137.2m1310

34303301513x .

... =

−×=

21uu cm

kN20104051225

78502 ..

.)(, =×

××

=τ

( ) ( ) m37113103823302010x

3 r

1uu

1 .....)(

, =−×=−λτ

τ=λ

Jednostruke uzengije URØ10/12.5mogu prihvatiti napon:

i dovoljne su na dužini:

dok je na preostalom delu dužine osiguranja 2.38 – 1.37 = 1.01 m potrebno postaviti dvostruke uzengije (ili, kao varijantno rešenje, vertikalne uzengije i koso povijene profile).

∆Aa = 0 (»špic« momenta)

21Proračun POS 2 – oslonac qu2 = 41.47 kNm

Bg = 132.9 Cg = 18.0

Bp = 211.0 Cp = 20.2

L2 = 6.00

1.57183.6

408.8

65.2

Tu

Cu = 1.6×18.0 + 1.8×20.2 = 65.2 kN

kN6183265064741T dBu ...., =−×=

( ) m2711540110157106 .

.... =

−×−=λ

cm1202025

50302eu .%.

.=

××

= ⇒ usvojeno URØ8/20

u22udB

u CLqT −×=,

2dB

n cmkN1540

5390256183 .

.., =

××=τ

m5714741265x2 .

..

==

( ) 2dB

Ru cmkN06601101540

23 ..., =−×=τ

2Ru2uu cmkN0660

cmkN080040

202550302 ...

max,, =τ>=××

×=τ

∆Aa = 0

22Proračun POS 3

mkN168

27858q 3B

1 ...

==

g = 3.75 + 6.78 = 10.53 kN/mp = 6.78 + 8.16 = 14.94 kN/mqu = 1.6×10.53 + 1.8×14.94 = 43.74 kN/mMu = 43.74×7.22/8 = 283.4 kNm 6.00

7.20

6.00

B

A

31 2

2.40

2.40

B3

A

B3

A

qB3 1 =

8.1

6

cm851272025B =+=

%....

‰/./.

.. 0436

cm20dcm26531170x1170s103261

1564

05285104283

53k p

ab

2=µ

=<=×=⇒==εε

⇒=

××

=

2a cm9513

40052

10053850436A ... =×

××= ⇒ usvojeno 4RØ22 (15.21 cm2)

kN8582162720Q 3B1 .. =×=

23Proračun POS 3Tu = 1.6×37.9 + 1.8×53.8 = 157.5 kN

m60013201101

227

cmkN1320

5390255157

2n .....

..

=

−×=λ⇒=

××=τ

( ) 2Ru cmkN03301101320

23 ... =−×=τ

cm1202025

50302eu .%.

.=

××

= ⇒ usvojeno URØ8/20

2Ru2uu cmkN0330

cmkN080040

202550302 ...

max,, =τ>=××

×=τ

( ) 2a cm9719045

4025157A .cotcot.

=°−°××

=∆ ⇒ usvojeno 2RØ22 (7.60 cm2)

24Proračun POS 4

g = 7.5 + 2×9.93 = 27.36 kN/mp = 2×(9.93 + 10.86) = 41.58 kN/mqu = 1.6×27.36 + 1.8×41.58 = 118.62 kN/mMu = 118.62×7.22/8 = 768.7 kNm

6.00

7.20

6.00

B

A

31 2

2.40

2.40

A

A

qB1

1 =

10.

86

B1 B1

qB1

1 =

10.

86

cm2304

72050B =+=

⇒ usvojeno 10RØ22 (38.01 cm2)

kN2782163620Q 1B1 .. =×=

mkN8610

27278q 1B

1 ...

==

%....

‰/./.

.. 0576

cm20dcm26531170x1170s103271

1514

052230107768

53k p

ab

2=µ

=<=×=⇒==εε

⇒=

××

=

2a cm8437

40052

100532300576A ... =×

××=

25Proračun POS 4Tu = 1.6×98.5 + 1.8×149.7 = 427.0 kN

usvojeno URØ10/10 (m=2)


( ) 2Ru cmkN10401101790

23 ... =−×=τ

cm11240104050

78502eu ...

=××

×=

( ) 2a cm3459045

4020427A .cotcot.

=°−°××

=∆

UØ10/20

50

2Ø12

2Ø22

6Ø22

4Ø22

5.2

2040

m39117901101

227 .

...

=

−×=λ

τ<=τ>

=××

=τr

r2n 3

MPa11cmkN1790

5390500427 .

...

26Vetar u podužnom pravcu - WxH

= 5

.00

m

L = 6.00 m L = 6.00 m

b/d

= 25

/50

b/d

= 25

/25

b/d

= 25

/25

Wx2 = 80

b/d = 25/60

POS

S1

POS

S2

POS

S3

POS 2Wx1 W x2 W x3

Wx prihvataju dva rama jednake krutosti (ose A i B)

kN064WkN08W

kN802

WW10WW8W2x

1xx1x1x1x1x .

.==

⇒===++

1s1s

3

2s3s1s J8J2550JJJ =×

== ;

3Wx1x

1Wx M2

HWM ,, =≈

kNm1602

05064M 2Wx =×

≈..

,

kNm202

0508M 1Wx =×

≈..

,

2HWM 2x

2Wx ≈,

kN7166

8020CA WW .=+

=−=

27

H =

5.0

0 m

L = 6.00 m L = 6.00 m

b/d

= 25

/50

b/d

= 25

/25

b/d

= 25

/25

Wx2 = 80

b/d = 25/60

POS

S1

POS

S2

POS

S3

±80

16020 20

2020

±MWx[kNm]

16.7 16.7

80

160

80

POS 2

Vetar u podužnom pravcu - Wx

28Vetar u poprečnom pravcu - Wy

Vetar Wy prihvataju tri rama, od kojih su ivični ramovi (u osama 1 i 3) širine 25 cm, a srednji ram (osa 2) širine 50 cm, dakle dvaput veće krutosti. Stoga srednji ram prihvata polovinu, a ivični po četvrtinu ukupne sile:

1y

1y

1y

1y

3y

2y

1y W4WW2WWWW =++=++

kN80WWkN40WkN160WW4 2y

3y

1yy

1y ===⇒== ;

Svaki ram formiraju po dva stuba istog preseka, visine i konturnih uslova, pa svaki stub prihvata polovinu odgovarajuće sile od vetra:

kNm100205

280M 2Wy =×≈

., kN827

27100100BA ww .

.=

+=−=

3Wy1Wy MkNm50205

240M ,,

.==×≈ kN913

275050BA ww .

.=

+=−=

(S2)

(S1,S3)

29

H =

5.0

0 m

L = 7.20 m

b/d

= 50

/25

b/d

= 50

/25

Wy2 = 80

b/d = 50/60

POS

S2

POS

S2

±80 100

±MWy[kNm]

100 100

100

POS 4

27.827.8

H =

5.0

0 m

L = 7.20 m

b/d

= 25

/25

b/d

= 25

/25

Wy4 = 40

b/d = 25/60

POS

S1

POS

S3

±40 50

±MWy[kNm]

50 50

50

POS 3

13.913.9

OSA 2 OSA 1 (3)

Vetar u poprečnom pravcu - Wy

30Dimenzionisanje stubova S1, S3Maksimalna sila usled p u stubu S1, istovremeno i minimalna sila u stubu S3:

6.00

7.20

6.00

B

A

31 2

2.40

2.40

B2

A

B3

A

qB22 = 10.73 qB3

2 = -8.21

12.0

qB2 1 =

7.3

2

qB3 1 =

-8.1

6

mkN327

272162440

LQq

y

2B12B

1 ..

.=

×==

mkN114327786p 3POS ...max =+=

kN850227114R 3POSp ...max, =×=

kN8166850116RAP 3POSp

2POSp ..max,max,max =+=+=

mkN381168786p 3POS ...min −=−=

kN05227381R 3POSp ...min, −=×−=

kN52505520RCP 3POSp

2POSp ...min,min,min −=−−=+=

31Dimenzionisanje stubova S1, S3Ukupna sila u POS S1 i S3 se dobija sabiranjem reakcija POS 2 i POS 3:stub POS S1: G = Ag2 + Rg3 = 61.7 + 37.9 = 99.6 kNstub POS S3: G = Cg2 + Rg3 = 18.0 + 37.9 = 55.9 kN

32Dimenzionisanje stubova S1, S3

33Dimenzionisanje stubova S2

mkN8610

272163620

LQq

y

1B11B

1 ..

.=

×==

7.20

B

A

31 2

2.40

2.40

B

A

B1

A

qB3 = 9.25 qB1

3 = 7.0512.0

qB 1 =

9.9

3

qB1 1 =

10.

86

6.00

7.20

6.00

B

A

31 22.

402.

40

B

A

B1

A

qB3 = -9.25 qB1

3 = -7.05-12.0

qB 1 =

-9.9

3

qB1 1 =

-10.

86

mkN654086109399392p 4POS ....max =++×=

kN31462276540R 4POSp ...max, =×=

kN33573146211RBP 4POSp

2POSp ..max,max,max =+=+=

mkN93086109399392p 4POS ....min −=−−×=

kN33227930R 4POSp ...min, −=×−=

02591225912p 2POS1 =−−+= ..min,

mkN350057706p 2POS2 ...min, −=−=

( ) kN41068

35005Bp ...min, −=×

−×=

kN743341RBP 4POSp

2POSp ...min,min,min −=−−=+=

G = Bg2 + Rg4 = 132.9 + 98.5 = 231.4 kN



36

UØ10/20

253Ø25

3Ø251 (3)

A (B)

UØ10/20

50

3Ø252

A (B)

3Ø25

3Ø25

3Ø25

Pismeniispit, 16.09.2015., grupeA+Bimksus.grf.bg.ac.rs/nastava/beton/PROJEKTOVANJE I... ·...

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Transcript of Pismeniispit, 16.09.2015., grupeA+Bimksus.grf.bg.ac.rs/nastava/beton/PROJEKTOVANJE I... ·...