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2.5 RC FRAME ELEMENTS DESIGN

SR 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 girder’s 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 lessthan 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 t

yk

ctmsmin 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 girder’s plastic zones;o lim , for other cases.

But it will be less than 4 % of the concrete cross-section area

csmax 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 mountingprocesses:

- 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 barsassociated 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 : 30015,1

345γ

ff

s

ykyd N/mm²;

Concrete C 20/25 : 33,135,1

20γ

ffb

ckcd N/mm²;

2,2fctm N/mm².

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 Span

My,Ed 176 kNm 94 kNm

Vz,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/mm2

fyk = 345 N/mm2

fyd = fyk/γs = 300 N/mm2

γs = 1.15fctm = 2.2 N/mm2

ρ = 0.3188 %Asmin = 430.435 mm2

MEd = 176 kNm

mm

mm2

= 0.3188 %

Asmin = mm2

As2 = 1488 mm2

dl = = 21.675 mm2

dl = 22 mm

As2r = dl2 ∙ π = 1521 mm2

The longitudinal reinforcement in the support in the superior part is composed of :

4 bars of Φ22

MEd 106

bw d2 fcd0.217

x d 1 1 2 ( ) 111.622

As1x bw fcd

fyd1.488 103

0.5 fctm

fyk3.188 10 3

bw d 430.435

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2.5.1.3 Longitudinal reinforcement computation in span (inferior part):

mm

mm2

Asmin = 430.435 mm2

As1 = 742.202 mm2

dl = = 17.748 mm2

dl = 18 mm

mm2

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

3 bars of Φ18

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

a = 60 mmd = hw – a = 440 mm

fcd = 13.33 N/mm2

fyk = 345 N/mm2

fyd = fyk/γs = 300 N/mm2

γs = 1.15fctm = 2.2 N/mm2

MEd = 94 kNm

MEd 106

bw d2 fcd0.126

x d 1 1 2 ( ) 55.665

As1x bw fcd

fyd742.202

As1r3 dl2

4 763.407

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2.5.1.4 Transversal reinforcement dimensioning :

maxcw w 1 cdRd,max Ed

α ×b ×z×v ×fV = V (kN)(ctgθ+tgθ)

acw = 1 coefficient considering the effort in the compressed area

z = 0.9d = 405 mm

v1 = 0,6 cracked concrete strength reduction coefficient at shear force

max(ctgθ1; ctgθ2) ≥ 1 if the condition is not fulfilled the cross-section dimensions mustbe modified

ctgθ ≤ 2.5 if it results a higher value in the computation will be consideredctgθ = 2.5

s = 100 mm distance between stirrups

fywk = 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/mm2

fyk = 345 N/mm2

fyd = fyk/γs = 300 N/mm2

γs = 1.15fctm = 2.2 N/mm2

VEd = 119 kN

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maxSWRd,s ywd EdAV = ×z×f ×ctgθ V (kN)s

Asw = ∙ ∙ = 4∙ ∙ = 201.06 mm2

= ∙ ∙ 100 = 0.00672VRd,s = 201.06/100x405x204x2.5 = 415290 N = 415,29 kN > VEd,max = 119 kN

The stirrups will have a diameter of Φ8 and will be mounted at 100 mm distance

stirrups – OB37 Φ8/10/20cm

The stirrups will be mounted on first quarter of the span of the beam at 100 mm distance andon the rest of the span at 200 mm.

2.5.1.5 Anchorage lengths :

Support :

lbd = α1 α2 α3 α4 α5 lbdr = 733.33 mm

α1 = α2 = α3 = α4 = α5 = 1= 4 ∙ = 733.33= = 300 /= 22= 2.25 ∙ ℎ ∙ ℎ ∙ = 2.25 /h1 = 1 , h2 = 1= , . = 1.0 /, . = 1.5 /lbd = 750 mm

2.5.1.6 Overlapping lengths :

lbd = 750

l0 = α6 lbd = 1.125 m

α6 = 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 material

C20/25.

The column design is made taking in account the skew bending checking and the relative

level 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 kNm

VEd = 94 kN

NEd = 1796 kN

hc = 600 mma = 60 mm

d = hc – a = 540 mmea = 20 mm

fcd = 13.33 N/mm2

fyd = fyk/γs = 300 N/mm2

= + = 131.91= ℎ2 + − = 381.91= ℎ ∙ = 224= ℎ ∙ ∙ ∙ ( − 0.5 ∙ ) − ∙∙ ( − ) = 658.59= 0.32%= ∙ ℎ100 = 1152= = 19.15= ∙ = 1256.63

The longitudinal reinforcement in the column will be composed of :4 bars of Φ20

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

Ed

Rd

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

α ×b ×z×v ×fV = V (kN)(ctgθ+tgθ) ctgθ

αcw = 1 coefficient considering the effort in the compressed fiber

z = 0.9d

v1=0,6 cracked concrete strength reduction coefficient to shear force

max(ctgθ1; ctgθ2) ≥ 1 if the condition is not fulfilled the cross-section dimensions must bemodified

ctgθ ≤ 2.5 if it results a higher value in the computation is considered ctgθ = 2.5

SWRd,s ywd EdAV = ×z×f ×ctgθ V (kN)s

bwd (8,10)mm

2bw

SW rπ×dA =n ×4

nr=4, s=100 mm , ctgθ = 2.5, db = 10 mm, z = 495 mm

fywd=0,8·fywk=0,8·255=204 (N/mm2) OB37

pemin=0.005 for the base level

SWe emin

c

Ap = ps×h

= 314.16pe = 0.0052

pe > pemin= 793.1 > = 942.5.2.5 Relative level displacement checking:

Checking in the ultimate limit state ULS:

displacement amplification coefficient:

1

c

Tc=3-2.5×T

SLUrad =0.025×H (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 SLSr rad =υ×q×d <d (mm)

reduction factor considering the shortestresilience period of the seismic action:

ν = 0.50SLSrad =0.005×H (mm)

q = 6.75ν = 0.50= 15= 4.5=15.18 mm

2.5.2.6 Horizontal design shear force in the nodes:

central nodes:

jhd Rd s1r s2r yd EdV =γ ×(A +A )×f -V (kN)

γRd = 1.2 overstrength factor

As1r, As2r – real areas of reinforcements from the superior and inferior part of the girders

VEd – design shear force from the column under the node

Vjhd = 720.68 kN

edge nodes:

jhd Rd s1r yd EdV =γ ×A ×f -V (kN)

Vjhd = 180.68 kN

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2.5.2.7 Horizontal shear force checking:

Central nodes:

djhd j c cd

υV η 1- b×h ×f (kN)η

ckfη=0.6× 1-250

Edd 2

c cd

Nυ =h ×f

j c w cb =min(h ;b +0.5×h ) (mm)fck = 20 N/mm2 concrete compressive

characteristic strengthνd – normalized axial force in the column

above

bj = 600 mm

νd = 0.37

η = 0.552

Vjchd = 720.68 kN < 1521 kN

Edge nodes:

djhd j c cd

υV 0.8×η 1- b ×h ×f (kN)η

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×ν )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 girdersνd – normalized axial force of the inferior

column

fywd = 204 N/mm2

As1r = 1521 mm2

As2r = 763 mm2

Ash = 1884.96 mm2

384.53 kN ≈ 385.90 kN

Edge nodes:

sh ywd s1r yd dA ×f >0.8×A ×f ×(1-0.8×ν ) 384.53 kN > 256.98 kN

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2.5.2.9 Longitudinal reinforcement checking in the node

jcsv sh

jw

2hA A ×

3h

Asv – vertical longitudinal reinforcementpassing through the node, including the

longitudinal reinforcement of the columnAsh – total area of horizontal stirrups in the

nodehjc = hc - 2a – interaxial distance between the

edge reinforcements of the columnshjw = hw - 2a – interaxial distance betweenthe reinforcement from the superior and

inferior part of the girders

Asw = Ash + As1r + As2r = 4168 mm2

As1r = 1521 mm2

As2r = 763 mm2

Ash = 1884 mm2

hjc = 480 mmhjw = 380 mm

4168 mm2 > 1570 mm2

In order to see the formwork plan and reinforcement plans for column and girder checkdrawings R3, R4, R5.