# 2.5 reinforcement of rc frame

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exemplu calcul cadre din beton armat P+4TRANSCRIPT

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2.5 RC FRAME ELEMENTS DESIGNSR EN 1992-1-1-2004

2.5.1 RC GIRDER DESIGN

2.5.1.1 General rules :RC Girder design computation for Bending Moment and Shear Force

The monolith girders cross-section have commonly T shape but also can haverectangular, trapezoidal shape or other shapes.

The minimum height of girders 1/15 of the span.The ratio between the height and width of the transversal cross-section will be less

than 4, but it is recommended to have values h/b = 1.5 3 for the rectangular cross-sectiongirders and h/b = 2 3 for the T shape cross-sections.

The width of the transversal cross-section b must have the minimum value 200 mm forthe safety in case of fire. For monolith girders the dimensions will be multipliers of 50 mm.

Monolith girders commonly have a constant cross-section on span. Function of theratio between the span of the beam (l) and the height of the cross-section (h), the followingclassification is made :

- Long beams (slender), if 8hl ;

- Short beams, if 8hl2 ;

- Deep beams (walls) if 2hl

Constructive aspects regarding longitudinal reinforcement :The minimum and maximum reinforcing percentages will be chosen in function of thefollowing aspects :

- The minimum cross-section of the reinforcement will be determined with thefollowing formula:

dbff0,5A tykctm

smin in seismic zones, no matter the ductility class;

b t - medium width of the beam

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- The effective surface of the reinforcement cross-section must be greater than thenecessary surface in order to resist cracking;

- The concrete cross-sections with reinforcement areas less than the area obtained bythe formula mentioned above will be considered unreinforced cross-sections;

- Maximum cross-section for the tensioned or compressed reinforcement, out of theoverlaying zones, corresponds to the maximum height of the compressed zone

dx where:o 0,25 for girders plastic zones;o lim , for other cases.

But it will be less than 4 % of the concrete cross-section areacsmax A0,04A

The medium reinforcing percent will be between 0.8 and 2 % .

The reinforcement mounting :The cost of the steel reinforcement is influenced by the trimming and mounting

processes:- A very low variation of diameters used;- The use of straight bars;- The adequate jointing systems;- Use of mechanized tools for trimming and assembling.The longitudinal resistance reinforcement is usually composed of flexible bars

associated with constructive and transversal reinforcements. Bounded by spatial enclosures(welded or connected by wires).

The stirrups will be mounted at a maximum 200 mm distance.The materials used:

PC 52 : 30015,1345

ff

s

ykyd N/mm;

Concrete C 20/25 : 33,135,120

ff

b

ckcd N/mm;

2,2fctm N/mm.fctm the medium axial strength of concrete;fyd yielding design limit strength of reinforcement for RC;fyk characteristic yielding limit strength of reinforcement for RC;

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fck characteristic compression strength of the concrete, determined on cylinders at 28 days;

The maximum efforts evaluation was made with Robot, with automatic combinationsaccording SR EN 1990 : 2004 / NA 2006.

The girder from most loaded frame was studied. (see part 2.3 Static and modalanalysis)

My diagram

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Vz diagram

Design Efforts max values Support SpanMy,Ed 176 kNm 94 kNmVz,Ed 119 kN 119 kN

The cross-section will have rectangular shape ( see part 2. Computation Summary)

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2.5.1.2 Longitudinal reinforcement computation in the support (superior part):

hf = hs = 130 mmbw = bg = 300 mmhw = hg = 500 mmbeff = bg = 300 mm

a = 50 mmd = hw a = 450 mm

fcd = 13.33 N/mm2fyk = 345 N/mm2fyd = fyk/s = 300 N/mm2s = 1.15fctm = 2.2 N/mm2 = 0.3188 %Asmin = 430.435 mm2

MEd = 176 kNm

mm

mm2

= 0.3188 %

Asmin = mm2As2 = 1488 mm2

dl = = 21.675 mm2dl = 22 mm

As2r = dl2 = 1521 mm2The longitudinal reinforcement in the support in the superior part is composed of :

4 bars of 22

MEd 106

bw d2 fcd0.217

x d 1 1 2 ( ) 111.622As1 x bw fcdfyd 1.488 10

3

0.5 fctmfyk 3.188 103

bw d 430.435

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2.5.1.3 Longitudinal reinforcement computation in span (inferior part):

mm

mm2

Asmin = 430.435 mm2As1 = 742.202 mm2

dl = = 17.748 mm2dl = 18 mm

mm2

The longitudinal reinforcement in the span in the inferior part will be composed of :3 bars of 18

hf = hs = 130 mmbw = bg = 300 mmhw = hg = 500 mmbeff = bg = 300 mm

a = 60 mmd = hw a = 440 mm

fcd = 13.33 N/mm2fyk = 345 N/mm2fyd = fyk/s = 300 N/mm2s = 1.15fctm = 2.2 N/mm2

MEd = 94 kNm

MEd 106

bw d2 fcd0.126

x d 1 1 2 ( ) 55.665

As1 x bw fcdfyd 742.202

As1r 3 dl2

4 763.407

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2.5.1.4 Transversal reinforcement dimensioning :

maxcw w 1 cdRd,max Ed b zv fV = V (kN)(ctg+tg)acw= 1 coefficient considering the effort in the compressed area

z = 0.9d = 405 mmv1 = 0,6 cracked concrete strength reduction coefficient at shear force

max(ctg1; ctg2) 1 if the condition is not fulfilled the cross-section dimensions mustbe modified

ctg 2.5 if it results a higher value in the computation will be consideredctg = 2.5

s = 100 mm distance between stirrupsfywk = 255 N/mm2 the yielding strength of OB37

fywd = 0.8 fywk = 204 N/mm2, = 0.08 = 0.001

1 300 405 0.6 13.33+ 1 119 10+1/ 122.45= .

hf = hs = 130 mmbw = bg = 300 mmhw = hg = 500 mmbeff = bg = 300 mm

a = 60 mmd = hw a = 440mm

fcd = 13.33 N/mm2fyk = 345 N/mm2fyd = fyk/s = 300 N/mm2s = 1.15fctm = 2.2 N/mm2

VEd = 119 kN

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maxSWRd,s ywd EdAV = zf ctg V (kN)sAsw = = 4 = 201.06 mm2= 100 = 0.00672

VRd,s = 201.06/100x405x204x2.5 = 415290 N = 415,29 kN > VEd,max= 119 kNThe stirrups will have a diameter of 8 and will be mounted at 100 mm distance

stirrups OB37 8/10/20cmThe stirrups will be mounted on first quarter of the span of the beam at 100 mm distance and

on the rest of the span at 200 mm.

2.5.1.5 Anchorage lengths :Support :

lbd = 1 2 3 4 5 lbdr= 733.33 mm1 = 2 = 3 = 4 = 5 = 1= 4 = 733.33= = 300 /= 22= 2.25 = 2.25 /

h1 = 1 , h2 = 1= , . = 1.0 /, . = 1.5 /lbd = 750 mm

2.5.1.6 Overlapping lengths :lbd = 750

l0 = 6 lbd = 1.125 m6 = 1.5

l0 = 1.20 m

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2.5.2 RC column design

Efforts evaluation was made with the help of Robot Str. A. (see 2.3 Static and modalanalysis)

For pre-dimensioning (see part 2 Computation Summary 2.1.2)2.5.2.1 Design Efforts Computation:

The most loaded column to bending moment was selected. Column with materialC20/25.

The column design is made taking in account the skew bending checking and the relativelevel displacement checking.

2.5.2.2 Longitudinal reinforcement computation :

C R60x60 designefforts

NEd,max VEd,max,y VEd,max,z MEd,max,x MEd,max,y MEd,max,z

Diagrams- scheme

Values(kN /kNm)

1796 94 68 1.13 168 201

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MEd = 201 kNmVEd = 94 kNNEd = 1796 kN

hc = 600 mma = 60 mm

d = hc a = 540 mmea = 20 mm

fcd = 13.33 N/mm2fyd = fyk/s = 300 N/mm2

= + = 131.91= 2 + = 381.91= = 224= ( 0.5 ) ( ) = 658.59= 0.32%= 100 = 1152= = 19.15= = 1256.63

The longitudinal reinforcement in the column will be composed of :4 bars of 20

Resisting moment of the cross-section will be:= 2 + ( ) = 481.252.5.2.3 Eccentric skew compression checking :

EdRd

M2 1M

2Rd c cdN =h f

NRd = 4798.8 kNNEd/ NRd = 0.37

a = 1.220.48 < 1

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2.5.2.4 Transversal reinforcement dimensioning :

maxcw w 1 cdRd,max Ed b zv fV = V (kN)(ctg+tg) ctgcw= 1 coefficient considering the effort in the compressed fiber

z = 0.9dv1=0,6 cracked concrete strength reduction coefficient to shear force

max(ctg1; ctg2) 1 if the condition is not fulfilled the cross-section dimensions must bemodified

ctg 2.5 if it results a higher value in the computation is considered ctg = 2.5SW

Rd,s ywd EdAV = zf ctg V (kN)s bwd (8,10)mm

2bw

SW rdA =n 4

nr=4, s=100 mm , ctg = 2.5, db = 10 mm, z = 495 mmfywd=0,8fywk=0,8255=204 (N/mm2) OB37

pemin=0.005 for the base levelSW

e eminc

Ap = psh = 314.16pe = 0.0052pe > pemin= 793.1 > = 94

2.5.2.5 Relative level displacement checking:Checking in the ultimate limit state ULS:

displacement amplification coefficient:1c

Tc=3-2.5TSLUrad =0.025H (mm)

T1 = 0.67 s , Tc = 0.7 sq = 6.75c = 0.60= 75= 12=48.60 mm

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Checking SLS:SLS SLS SLS

r rad =qd

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2.5.2.7 Horizontal shear force checking:Central nodes:

djhd j c cd

V 1- bh f (kN)

ckf=0.6 1- 250

Edd 2

c cd

N = h fj c w cb =min(h ;b +0.5h ) (mm)

fck = 20 N/mm2 concrete compressivecharacteristic strength

d normalized axial force in the columnabove

bj = 600 mmd = 0.37 = 0.552

Vjchd = 720.68 kN < 1521 kN

Edge nodes:d

jhd j c cdV 0.8 1- b h f (kN) Vjehd = 180.68 kN < 1216.8 kN

2.5.2.8 Transversal reinforcement checking:Central nodes:

sh ywd s1r s2r yd dA f >0.8(A +A )f (1-0.8 )2

sh e r sbwA =n n A (mm )Asbw = 78.54 mm2

Ash total area of horizontal stirrups in thenode

ne = 6 number of horizontal stirrups in thenode

nr= 4 number of barsAs1r, As2r real areas of reinforcement in the

superior and inferior part of the girdersd normalized axial force of the inferior

column

fywd = 204 N/mm2As1r