2.5 reinforcement of rc frame
Post on 12-Dec-2015
21 Views
Preview:
DESCRIPTION
TRANSCRIPT
56
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
57
- 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;
58
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
59
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)
60
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
61
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
62
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
63
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
64
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
65
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
66
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
67
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
68
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
69
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.
top related