50091086 calculul grinzii stalpului si planseului en

7/29/2019 50091086 Calculul Grinzii Stalpului Si Planseului En

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GIRDER CALCULATION

The girders take the loads from the plates, the closing and the dividing walls andtransmit them to the columns.

A construction element is a girder if the following condition is respected:L

5max(b,h)

where:

L girder lenght;b girder cross-section width;h girder cross-section height.

As the girders are constructions main elements, they are calculated in the elasticstage.

For the column and girder calculation I considered a rigid node in the structure

In order to impose the energy disipation structural mechanism, that will fulfill the request:

for the framed storeyed structures, the plastic deformations must appear first in thesections at the edge of the girders and, after that, in the sections at the base of thecolumns. So there must be respected the following condition (according to P100-2006):

Rc Rd RbM M where:


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3/35

M[daNm]

12803.24

164.54

12776.68

(3)

M[daNm]

9756.99

1964.44

7286.45

(4)


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3.Girders reinforcement

The girders minimum reinforcement percentage according to STAS 10107/0-90 is:

p = 0.45 (frame girders that take part at anti-seismic structures in the AE

calculation zones, for the reinforcement that take the negative moments in the bearings).p = 0.15 frame girders that take part at anti-seismic structures in the AEcalculation zones, for the other tensed reinforcement).

3.1. Calculation of reinforcement in the bearings:

As the concrete compressed area is in the girder, the calculation cross-section is

considered a rectangular double reinforced section.

Grirder 1:

According to Stas 10107/0-90, the concrete coverage for monolith reinforcedconcrete is ab = 3 cm.


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= =

b

2

0

a 0 b

20 c

lim

lim

h = h-a -d/2 = 50-2.5-1.8/2 = 46.1 cm

h = h -a -d/2 = 46.1 - 3 -1.8/2 = 42.2 cm

M 1306540B = 0.1025

b h R 40 46.1 150

B = 0.22

B

cap 0

Aa Ra 10.05 3000x= 5.025

b Rc 40 150

M =b x Rc (h -0.5 x)=40 5.025 150 (46.1-0.5 5.025)=13141.63 daNm

10.05100 100 0.5 0.45

50 40t

cm

Atp

Ab

3.2.Calculation of the reinforcement in the field:

Depending on the compressed concrete area we wil have two design cases.

The establishment of the design case:


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Plates capable moment when x=hp has the formula:

cap.pl p p c 0 pM = b h R (h -0.5h )

cap.pl ext p p

cap.pl ext p

M >M xM 4347.95daNm x


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= =

= =

= =

=

b

2

0

2p 0 c

20

2ef

h = h-a -d/2 = 50-2.5-1.8/2 = 46.1 cm

M 434795B = 0.0341

b h R 40 46.1 150

= 1- 1-2B 1 1 2 0.0341 0.0347

150Aa = 0.0347 40 46.1 3.2

3000

reinforcement 3 14 Aa =4.62 cm

100

Cp

a

Rb h cm

R

Atp

Ab

= = >

= =

= = = ==

= =

= = = >

2 2

2 2

2 2

4.62100 0.23 0.15

50 402 :

:

9.72 5 16 . 10.05

' 0.4 0.4 10.05 4.02 3 14 ' 4.62

13141.63

:

0.704 3 14 4.62

4.62100 0.23 0.1550 40

tt

b

Grider

Bearing

Aa cm Aaef cm

Aa Aa cm Aa ef cm

Mcap daNm

Field

Aa cm Aaef cm

Ap A


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2 2

2 2

2 2

3:

:

9.78 5 16 10.05

' 4.02 3 14 ' 4.62

13141.6310.05

100 100 0.5 0.4540 50

:

0.119 3 14 4.62

4.62100 100 0.23 0.15

40 50

Girder

Bearing

Aa cm Aaef cm

Aa cm Aa ef cm

Mcap daNmAt

pAb

Field

Aa cm Aaef cm

Atp

Ab

Girde

= =

= =

== = = >

= =

= = = >

2 2

2 2

2

2

2

4 :

:

7.35 4 16 8.04

' 3.216 3 14 ' 4.62

1063450.8

8.04100 100 0.4 0.45

40 50

0.45 20009 5 16

100 100

10.05

:

1.43 3 14 4.62

r

Bearing

Aa cm Aaef cm

Aa cm Aa ef cm

Mcap daNm

Atp

Ab

p AbAt cm

Aaef cm

Field

Aa cm Aaef

= == =

=

= = = <

= = =

=

= = 2

4.62100 100 0.23 0.15

40 50

cm

Atp

Ab= = = >


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COLUMN CALCULATION

Because I chose for a static spacial calculation for the structure, I will choose to

dimension and reinforce a column (from the first to the top level) having the biggestefforts.

We consider the buckling lenght equal to the storeys height (h):

3,00

6 100,50

fl

h= =

The flexibilitys influence can be neglected for all levels 1= , where:= eccentricity increasing coefficient for the calculation eoc by which the second orderefforts are introduced.

The column has the following cross-section on the whole height of the building:

50cm

50 cm

x

y

Calculation of the longitudinal reinforcement:

The calculation for the columns longitudinal reinforcement is made according to

STAS 10107/0-90 and normative P100-92.


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The columns are calculated in straight eccentrical compression.

The eccentricities evaluation is made as follows:

oc o ae e e= + , where:eoc calculation eccentricity

eo initial eccentricity - o MeN

=

M exterior moment that acts on the considered cross-sectionN axial force that actson the considered cross-section

ea additional eccentricity

501,667

max 2,0030 30

2a

hcm

e cm

cm

= = = =

Establishment of the eccentricity case:

- limo

x

h = - eccentric compression of big eccentricity (case I)

- limo

x

h = - eccentric compression of big eccentricity (case II)

where: - relative height of the compressed zone- lim maximum relative height of the compressed zone

The reinforcement quantities are calculated as follows:

- case I:

( )'

'

0

2 ca a

a

NN e h

b RA A

R h a

+ = =

;

- case II:( )

2' lim 0

'

0

ca a

a

N e B b h RA A

R h a

= =

Note: For the framed structurs, in practise, it is a must to use symmetrical reinforcementagainst non-symmetrical reinforcement, beceause, at least from the seismic load, wehave alternating loads on the elements. Also, it is eliminated the possibility to reverse thereinforcement on the working site.

The element is checked at oblique eccentrical compression with the relation:


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0 0

1yx

x y

MM

M M

+

p

where: Mx,My calculation bending moments on the two directions

Mx0 cross-sections capable moment on the X direction for a given axial force,when My=0My0 cross-sections capable moment on the Y direction for a given axial force,

cnd Mx=0

coefficient depending on the cross-sections reinforcing way and on n:

0 c

Nn

b h R=

1,3 2,5 =

According to NP 007/97 Cod de proiectare pentru structuri din beton armatparagraf 6.1.4. point b, the bending moment in the columns external cross-sections,

corresponding to the (SG) loads special group, is determined with the relation:

,M s cap gr s

S

gr

k M M MM

M

=

where: Ms columns bending moment in the (SG) loads special group,considering theseismic load on the perpendicular directions of the columns cross-sections;

|Mcap,gr| - capable momentss sum in the cross-sections where plastic hingesappear at the considered level;

MgrS corresponding bending moments algaebric sum, obtained in the (SG)

loads special group;

KM coefficient with the values:1,4 for constructions located in the AC seismic zones1,2 for constructions located in the DF seismic zones1,0 for restrained cross-sections and the ones at the columns last level

Also, it must:

1,2

st

cap

s gr

cap

MKM

=

It is observed that the bending moments corresponding to the columns, calculated inthe (SG) loads special group, will be increased KSKM times

1, 2 1, 2 1, 44S MK K = =


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LONGITUDINAL REINFORCEMENT CALCULATION AT THE BASE-FLOOR LEVEL

A. On the X direction

We consider the concrete coverage ab=3.0 cm for the reinforced concrete elementsfrom the first category according to STAS 10107/0-90

We initially propose reinforcement with the diameter d=20mm

'

0

0.5 3 0.5 2,0 4

50 4 46

ba a a d cm

h h a cm

= = + = + == = =

50 cm

50cm

hAa'

Aa

x

Aa'Ra

bxRc

AaRa

N

eoc

xx

Y

Y

Maximum moment from the (SP) special group:

max 16399,15

90077,53coresp

M daNm

N daN

==

'

max max 1.44 16399,15 23614,776M SM K K M daNm= = =


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Steel

Heavy

52PC

concrete

lim

lim

0,55

0,4B

= =

Conform STAS 10107/0-90

'

max0

0 0

23614,7760,262 26,2

90077,53

26,2 2 28, 2c a

Me m cm

N

e e e cm

= = = =

= + = + =

0 90077,53 28,2 2540186,346c cM N e daNcm= = =

90077,53

12,0150 150c

NX cm

b R= = =

lim 0

2 2 3,5 7

0,55 46 25,3

x a cm

x h cm

> = =< = =

lim0

12,010,26 0,55

46

x

h = = = < = (for PC 52 according to STAS 10107/0-90, table 12)

The element is subjected to eccentic compression of big eccentricity.

50

28,2 4 49, 22 2

oc

he e a cm= + = + =

( )'

'

0

2 ca a

a

NN e h

b RA A

R h a

+ = =

( )' 2

90077,5390077, 53 49, 2 46

2 50 1509,44

3000 46 4a aA A cm

+ = = =

29,44aA cm= we choose 418 cu Aa,ef= 10.18 cm2

20

min

50 46' 0, 20% 4, 6

100 100a a

b hA A p cm

= = =

According to STAS 10107/0-90 (pct. 6.4.3.1) that reccomends that for PC52 i PC60steel bars, the minimum diameter must be 12 mm, I choose for each side on X direction4 bars 18 mm resulting an effective area of10.18 cm2.


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B. On Y direction

50cm

50 cm

h

Aa'

Aa

x

XX

Y

Y

Aa'Ra

bxRc

AaRa

N

eoc

We consider the concrete coverage ab=3 cm for the reinforced cocnrete elemets fromthe first category according to STAS 10107/0-90

Initially we propose reinforcement with the diameter d=20mm

'

0

0.5 3 0.5 2,0 4

50 4 46

ba a a d cm

h h a cm

= = + = + == = =

Maximum moment from the special group (SG):


15/35

max 15811,81

90077,53coresp

M daNm

N daN

==

'


'

max0

0 0

22769,00640,2527 25,27

90077,53

25,27 2 27,27c a

Me m cm

N

e e e cm

= = = =

= + = + =

0 90077,53 27,27 2456414,243c cM N e daNcm= = =

90077,53

12,0150 150c

NX cm

b R= = =

lim 0

2 2 3,5 7

0,55 46 25,3

x a cm

x h cm

> = =< = =

lim0

12,010,26 0,55

46

x

h = = = < = (for PC 52 according to STAS 10107/0-90, table 12)

The element is subjected to eccentic compression of big eccentricity.

50

27,27 4 48,272 2

oc

he e a cm= + = + =

( )'

'

0

2 ca a

a

NN e hb R

A AR h a

+ = =

( )' 2

90077,5390077, 53 48, 27 46

2 50 1509,75

3000 46 4a aA A cm

+ = = =

29,75aA cm= alegem 418 cu Aa,ef= 10.18 cm2

20min

50 46' 0, 20% 4, 6100 100

a ab hA A p cm = = =



16/35

Shear force check

For the shar force check I choose from the software the maximum value for Fy or Fz ithe most unfavourable case (exceptional group) .

Q = 9819.82daN

2

9819,820,357 0,5

50 50 11t

Q daN

daNb h Rcm cm

cm

= = <

According to STAS 10107/0-90, it is not necessary a calculation for the shar forceaction.

There are considered potentail plastic hinges the zones from the columns edges, for

each level.

The lenght of the potential plastic hinge lp is measured from the superio, inferiorespectivelly, edge of the girder and is considered like so:

30050

6 6

50

60

sp

p

p

H cml cm

l h cm

l cm

= = =

=

the lenght of the plastic hinge = 60 cm.

Distance bewtween stirrups: ae

d = minimum diameter for the longitudinal bars

For the potential plastic hinges

8 8 18 176 17,60

5010

5 5

10

e

e

e

a d mm mm cm

cmha cm

a cm

= = =

= =

ae = 100mm=1cm

For current zones

cma

cmmmmmda

e

e

20

00,27270181515

===

ae = 200mm=20cm

Transversal reinforcement diameter (d)

d > d4

10,25 x 18 mm = 4,50 mm;

6 mm; d = 8mm


17/35

8 mm for group A For the potential plastic hinges (lp=65cm) we choose =8 mm (Ae

ef = 0,503cm2)having ae=12cm.

For the current zones we choose = 8 mm (Aeef= 0,503cm2) having ae=20cm

LONGITUDINAL REINFORCEMENT CALCULATION FOR THE FIRST LEVEL

A. On the X direction We consider the concrete coverage ab=3 cm for the reinforced cocnrete elemets fromthe first category according to STAS 10107/0-90

initially we propose reinforcement with the diameter d=20mm

'

0

0.5 3 0.5 2,0 450 4 46

ba a a d cmh h a cm

= = + = + == = =

Maximum moment from the special group (SG):

max 13003,3

71051,44coresp

M daNm

N daN

==

'


'

max0

0

18724,7520,26 26

71051, 44

26 2 28oc a

Me m cm

N

e e e cm

= = = =

= + = + =

0 71051,44 28 1989440,32c cM N e daNcm= = =

71051,44

9,4750 150c

NX cm

b R= = =

lim 0

2 2 3,5 70,55 46 25,3

x a cmx h cm

> = =< = =

lim

0

9,470,21 0,55

46

x

h = = = < = (for PC 52 according to STAS 10107/0-90, table 12).

The element i subjected to eccentric compression of big eccentricity.


18/35

50

28 4 492 2

oc

he e a cm= + = + =

( )

'

'

0

2 ca a

a

NN e h

b RA A

R h a

+ = =

( )' 2

71051,4471051, 44 49 46

2 50 1504,36

3000 46 4a aA A cm

+ = = =

20

min

50 46' 0, 20% 4, 6

100 100a a

b hA A p cm

= = = the reinforcing is made with the minimium

reinforcement percentage.


B. On the Y direction

For the Y disrection the reinforcing is made also from theminimum reinforcingpercentageloke on the X direction. For this it results 4 bars 18 mm resulting aneffective area of10,18 cm2.

For the columns on II,III, i IV levels rezult it results the same reinforcing areas as withthe number of bars on the to directions:4 bare 18 mm.

CALCULATION OF THE PLATE


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The plates are horizontal constructing elements that divide vertically the buildingsvolume and close it at the superior end. Regarding their role in the resistance structure,the plates have the role to take the gravitational loads and to transmit them to otherstructural elements (girders, walls, columns ), as well as to ensure the work of other

vertical elements (columns, diaphragms) at the horizontal loads action.The plates, being secondary construction elements, are calculated in the plasticstage.

Generally, the plates are dimensioned for vertical loads and checked forhorizontal loads.

Vertical loads in plates:

plate's dead load (plate, grout, floor, plaster); dividing walls weight utile (peolpe, furniture);

temporary ( snow only for the roof plate ).

Conditions for resting and unloading: It is reffered to the plates resting and co-working with the other construction elements on which it rests: girders, bearing walls.

The plates elements rest is considered like so:

simply rested, when the rotation on the rests are possible and themoments that the rest can take are neglectable.

constrained, when the rotations in the rest are very small, the constriningmoment can be fully taken and transmitted and the rests rigidity isconsidereable.

partially constrained, when there are fulfilled conditions intermediar to thesimple rest and constrain.

The efforts to which the floor plates are subjected to are determined statically inthe elastic domain, using one or more loading schemes.

The reinforcing of the reinforced concrete plates can be done:

On one direction, when : 2L

l ;

On two directions, when : 0.5 2Ll

.


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1. Establishment of loads

a) Current floor cold floor

b) Dividing wall

1 2 1

Nr.crt Name of material d

( )m

( )3/daN mkG

( 2/daN m )

1 Ceiling plaster M50T

0.01 1900 19

2 R.C. plate 0.15 2500 375

3 Leveling concreteM100

0.02 2100 42

4 Ceramic + adhesive 0.015 2100 46.2

Total dead loads 491.2


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Nr.crt Name of material d

( )m

3

/daN m

nq

2/daN m

1-3 Interior plaster M100 0.015 2100 31.5

2 Brick masonry 0.25 1450 362.5

TOTAL 425.5

The utile load is: qu = 200 daN/m2.

2. Calculation of moments in field and bearings

2.1. Calculation of moments in field

For a simple calculation of the maximum moments in the field, the real system willbe equivalent with two systems thatsdeformation modes are known: system A+systemB.


22/35

. / 2

. / 2

;

.

sist a p cp u

sist b u

q q q q

q q

= + +

= p cp+ u/2q = q +q q =825.94 daN/m

2.1.1. Determination of 1q and 2q loads for each plate in sistem a:

P1 P2 P3 P4 P5 P6 P7 P8 P9

P10

P11

P12

P13

6 5 3,6 3,6 4,8 3,6 3,6 5 6

3

5,5

3,5

3

direction 1

direction

2

F1

F2 qp=545.94daN/m

qcp=150daN/m

qp=200daN/m

1 1syst.a. 1syst.b

2 2syst.a. 2syst.b

q = q +q

q = q +q

q

a) Plate 1=Plate 9 type 4


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q1

q2

3,5

5

P1 type4

41

42

1 41

2 42

0.79355001.43 2

0.2065350

0.7935 825.94 655.38 /

0.2065 825.94 169.81 /

q q daN m

q q daN m

== = < == = == = =

b) Plate 2 = Plate 8 type 5


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q2

q1

3,5

6

P2 type5

51

52

1 51

2 52

0.20583500.58 2

0.7942600

0.2058 825.94 169.98 /

0.7942 825.94 655.96 /

q q daN m

q q daN m

== = < == = == = =

c) Plate 3=4=6=7=11 type 6


25/35

q2

q1

P3 type6

3,6

3,5

61

62

1 51

2 52

0.53500.97 2

0.5360

0.5 825.94 412.97 /

0.5 825.94 412.97 /

q q daN m

q q daN m

== = < == = == = =

d) Plate 5 type 5


26/35

q2

q1

P5 type5

4,8

3,5

51

51

1 51

2 52

0.32443500.73 2

0.6756480

0.3244 825.94 267.93 /

0.6756 825.94 558.01 /

q q daN m

q q daN m

== = < == = == = =

e) Plate 10 type 4


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P10 type4 q2

q13,6

3

41

42

1 41

2 42

0.29063000.83 2

0.7094360

0.2906 825.94 240.02 /

0.7094 825.94 585.92 /

q q daN m

q q daN m

== = < == = == = =

f) Plate 12 type 6


28/35

q2

q1

3,6

2

P12 type6

61

62

1 61

2 62

0.05882000.53 2

0.9412360

0.0588 825.94 48.57 /

0.9412 825.94 777.37 /

q q daN m

q q daN m

== = < == = == = =

g) Plate 13 type 4


29/35

P13 type4q2

q1

3,6

3

61

62

1 61

2 62

0.0588200 0.53 20.9412360

0.0588 825.94 48.57 /

0.9412 825.94 777.37 /

q q daN m

q q daN m

== = < == = == = =


30/35

Statical schemes for system a

Strip 1:

Strip 2:

2.1.2. Determination of 1q

and 2q

loads for each plate in system b:

1.3 130 daN/m2

uqq = = .


31/35

a) Plate 1= Plate 9

1

2

103.2 /

26.8 /

q daN m

q daN m

==

b) Plate 2= Plate 8

1

2

26.8 /

103.2 /

q daN m

q daN m

==

c) Plate 3= Plate 4= Plate 6= Plate 7= Plate 11

1

2

65 /

65 /

q daN m

q daN m

==

d) Plate 5

1

2

42.2 /

87.8 /

q daN m

q daN m

==

e) Plate 10

1

2

q = 37.8 daN/m

q = 92.2daN/m

f) Plate 12

1

2

7.6 /

122.4 /

q daN m

q daN m

=

=

g) Plate 131

2

q =87.7 daN/m

q =42.3 daN/m


32/35

Statical scheme for system b:

Strip 1:

Stryp 2:


33/35

Centralization of maximum moments:

6 5 3,6 3,6 4,8 3,6 3,6 5 6

directia 1

directia

2

F1

F2

qp=545.94daN/m

qcp=150daN/m

qp=200daN/m

1681.3+70.4

1751.7

106.7+1648.41755.1

354.77+92

446.7

211 +

308.4519.4

778.5+332.41110.9

1486 +1621.83107.8

273.4+1913.82187.2

565.8+26

591.8

560.8+1079.6

1640.4

560.8+1079.61640.4

354.77+92

446.7

211 +308.4519.4

106.7+1648.41755.1

1486 +1621.83107.8

273.4+

1913.82187.2

565.8+26

591.8

427.5+

161.7

589.2

427.5+

161.7

589.2

308.2+

298.2

606.4

427.5+

161.7

589.2

346.8+

291.7

635.5

653.6+

459.8

1113.4

124.6+

470.9

559.5

356.1+

234.9

591

-207.0+

67.7

139.3

1681.3+70.4

1751.7

From the calculation there resulted the following maximum moments in field andbearings:

field

max

bearing

max

Strip 1 - M = 1755.1 daN m

M = 3107.8 daN m

bearing

max

field

max

Strip 2 - M =1113.4 daN m

M = 595.5 daN m


34/35

Plate reinforcement

A. Bearing

a) direction 1:

=

= = =

= = =

= =

= = == =

=

max,1

0

20

20

ef 2reazem 1a

3107.8

0.8- - 15 -1.5 - 13.1

2 2

3107800.12073;

100 13.1 150

1- 1- 2 0.12905;

150

0.12905 100 13.1 8.45 ;3000

8 12 / 9.05 ;

p b

c

ca

a

M daN m

dh h a cm

MB

b h R

B

R

A b h cmR

A m cm

p = =

9.05100 0.6%.

100 15

aef

bs

A

A

b) direction 2:

=

= = =

= = =

= =

= = =

= =

max, 2

0

20

20

ef 2reazem dir 2a

1113.4

0.8- - 15 -1.5 - 13.1

2 2

1113400.04325;

100 13.1 150

1- 1- 2 0.04423;

1500.04423 100 13.1 2.89 ;

3000

6 8 / 3.02

p b

c

ca

a

M daN m

dh h a cm

MB

b h R

B

RA b h cm

R

A m cm

= = =

;

3.02100 0.2%.

100 15

aef

bs

Ap

A

B. Field


35/35

a) direction 1

=

= = =

= = =

= =

= = =

= =

=

max,1

0

20

20

ef 21a

1755.1

0.8- - 15 -1.5 - 13.1

2 2175510

0.11;100 13.1 150

1- 1- 2 0.070679;

1500.070679 100 13.1 4.63 ;

3000

6 10 / 4.71 ;

p b

c

ca

a

camp dir

M daN m

dh h a cm

MB

b h R

B

RA b h cm

R

A m cm

Ap = =

4.71100 0.314%

100 15

aef

bsA

b) direction 2:

=

= = =

= = =

= =

= = =

= =

=

max, 2

0

20

20

ef 22a

595.5

0.8- - 15 -1.5 - 13.1

2 2

595500.02313;

100 13.1 150

1- 1- 2 0.02341;

1500.02341 100 13.1 1.53 ;3000

6 8 / 3.02 ;

p b

c

c

a

a

camp dir

aef

M daN m

dh h a cm

MB

b h R

B

RA b h cmR

A m cm

Ap = =

3.02100 0.2%.

100 15bsA

CONCLUSION:

: 6 10 /inf

: 8 12 /

field mre orcement

bearing m

50091086 calculul grinzii stalpului si planseului en

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