how to desing rc structures to ec2
TRANSCRIPT
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A J Bond
O Brooker
A S Fraser
A J Harris
T Harrison
A E K Jones
R M Moss
R S Narayanan
R Webster
cementconcrete
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Foreword
Acknowledgements
Published by The Concrete CentreTel:Fax:www.concretecentre.com
Tel:email:
www.concrete bookshop.com
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1. Introduction to Eurocodes
2. Getting started
3. Slabs
4. Beams
5. Columns
6. Foundations
7. Flat slabs
8. Deflection calculations
9. Retaining walls
10. Detailing
11. BS 8500 for building structures
12. Structural fire design
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The design process
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Design life
4
Actions on structures 5
6
3
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Load arrangements
Load set 1. Alternate or adjacent spans loaded
g
g
Load set 2. All or alternate spans loaded
g
Load set 3. Simplified arrangements for slabs
2
2
Figure 2Adjacent spans loaded
Figure 3All spans loaded
Figure 1Alternate spans loaded
Table 1Indicative design working life (from UK National Annex to Eurocode)
Design life (years) Examples
Table 2Selected bulk density of materials (from Eurocode 1, Part 11)
Material Bulk density (kN/m3)
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Combination of actions
Material propertiesConcrete
7
Reinforcing steel
8
9
Table 4Selected concrete properties based on Table 3.1 of Eurocode 2, Part 11
Table 3Selected imposed loads for buildings (from draft UK National Annex to Eurocode 1, Part 11)
For members supporting one variable action the ULS combination
1.25 Gk + 1.5 Qk (derived from Exp. (6.10b), Eurocode)
can be used provided the permanent actions are not greater than
4.5 times the variable actions (except for storage loads).
Symbol Description Properties
Key
a
b
Category Example use qk (kN/m2) Qk (kN)
Reinforcing steel
8
9
11
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Structural analysis
Minimum concrete cover
D
D
Table 6Bending moment and shear co-efficients for beams
Table 7Exposure classes
Class Description
No risk of corrosion or attack
Corrosion induced by carbonation
Corrosion induced by chlorides other than from seawater
Corrosion induced by chlorides from seawater
Freeze/thaw with or without de-icing agents
Chemical attack (ACEC classes)
Table 5Characteristic tensile properties of reinforcement
Moment Shear
Keya
Notes1 2 3 4
D
Class (BS 4449) and designation (BS 8666) A B C
e
Notes1 2
3
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Minimum cover for bond
Minimum cover for durability
12
13
Design for fire resistance 14
f f
R
E I
Table 9Minimum column dimensions and axis distances for columns withrectangular or circular section method A
Standard fire Minimum dimensions (mm)resistance Column width (bmin)/axis distance (a) of the main bars
Column exposed on more Exposed on one sidethan one side (m f i = 0.7) (m f i = 0.7)
Notes
1
2 m m
*
m
Standard Minimum dimensions (mm)fire One-way Two-way spanning slab Flat slab Ribs in a two-way spanning ribbed slabresistance spanning slab ly/ lx 1.5 1.5 < ly/lx 2 (bmin is the width of the rib)
Notes
1 2 3
Figure 4Sections through structural members, showing nominal axis distance,a
Table 10Minimum dimensions and axis distances for reinforced concrete slabs
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Table 8Selected recommendations for normal-weight reinforced concrete quality for combined exposure classes and cover to reinforcement for at least a 50-year intended working life and 20 mm maximum aggregate size
Exposure conditions Cement/combinationdesignationsb
Nominal cover to reinforcementdTypical example Primary Secondary
15 + D cdev 20 +D cdev 25 +D cdev 30 +D cdev 35 +D cdev 40 +D cdev 45 +D cdev 50 +D cdev
Strength classc, maximum w/c ratio, minimum cement or combinationcontent (kg/m3), and equivalent designated concrete (where applicable)
X0
XC1
XC2
XC3XC4
XD1
XD3
XS1
AC-1
XF1
XF3
XF2
XF4
XF4
XF3 orXF4
XF1
XF3
Key
a
b
c
d D
e
f
g
h
j
-
14
Cement/combinationdesignationsb
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a) Bracing system b) Floor diaphragm c) Roof diaphragm
Figure 5Examples of the effect of geometric imperfections
Stability and imperfections
y
y a a
a R
a 0.5
y
y
y
15
Steelstress(ss)MPa
OR OR
Table 11Maximum bar size or spacing to limit crack width
wmax = 0.4 mm wmax = 0.3 mm
Maximum Maximum Maximum Maximumbar bar bar barsize (mm) spacing (mm) size (mm) spacing (mm)
Note
s
g d
g
d
Crack control
Figure 6Determination of steel stress for crack width control
To determine stress in the reinforcement (ss), calculate the ratio Gk/Qk,read up the graph to the appropriate curve and read across to determine ssu.
ss can be calculated from the expression: ss = ssu As,req
As,prov1d( () )
15
ApproximatesteelstressatSLSforA
s,req,
ssu
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References1
2
3
4 5
6
7
8
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f f f f
K
d K
K z/d
K z/d
d d
d
f f
z/d
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7
s s
s s s
d
s
s
s
20
,ssu
A
ApproximatesteelstressatSLSforA
s,req,
ssu
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8
9
b 2
b 2
b b
3
v
r dAbd
k
r I
Rr I
r I
f
r
r r
r r
r
r
r
r
r r
r
r
r
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4
ll
l l
d d d
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5
l
a g
d d
a
d
g g
g
r R
r
r
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1
2
3
Fire resistance
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ecnadiugrehtruFksaTpetS
dradnatSChapter in this publication
Note
Table 1Beam design procedure
a a
a b a b
a
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Outside scope of thispublication
d d
d
f f
Table 4Values for K
% redistribution d (redistribution ratio) K
Key
a
K z/d
K z/d
Key
a
for singly reinforced rectangular sections
Table 6Minimum percentage of required reinforcement
fck fctm Minimum percentage (0.26fctm /fyka)
Key
a
Table 5z/d
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Determine vEd wherevEd = design shear stress vEd = VEd/(bwz) = VEd/(0 .9 bwd)][
Yes (cot y= 2.5)
No No
START
Determine the concrete strut capacity vRd,max cot y= 2.5from Table 7
Redesign
section
Determine yfrom:
Calculate area of shear reinforcement:
Check maximum spacing for vertical shear reinforcement:s l,max = 0.75 d
=s
Asw
y= 0.5 sin-10.20fck (1 fck /250)
vEd
fywd cot y
vEd bw
IsvEd < vRd,max cot
y= 2.5?
IsvEd < vRd,max coty= 1.0?
(see Table 7)
Yes
f v y v y
y
yy
2
y y
Deflection
Flanged beams
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14
16
18
20
22
24
26
28
30
32
34
36
Spantodepthratio(
Percentage of tension reinforcement (As,reqd /bd)
l/d
)
0.40% 0.60% 0.80% 1.00% 1.20% 1.40% 1.60% 1.80% 2.00%
fck
ck
ck
ck
ck
ck
ck
ck
ck
= 50
f = 45
f = 40
f = 35
f = 32
f = 30
f = 28f = 25
f = 20
fck
ck
ck
ck
ck
ck
ck
ck
ck
= 50
f = 45
f = 40
f = 35
f = 32
f = 30
f = 28f = 25
f = 20
Figure 7Basic span-to-effective-depth ratios
Notes
1
2 r
3
r r
r r
r
r
r
r
r r
r
r
r
s
s s
s s
s s s
d
s
cg
cg
cg
cg
cg
cg
cg
29
For flanged sections
Where the beam span exceeds 7 m and it supports
To determine stress in the provided reinforcement (ss), calculate the ratio
Percentage of tension reinforcement (As,req/bd)
ApproximatesteelstressatSLSforAs
,req,ssu
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l
d d d
D
D
D
y
30
fyd(d 0.5 hf) fydzz
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Figure 14Procedure for determining longitudinal shear capacity of flanged beams
Yes
No No
No
Calculate the longitudinal shear stressfrom: vEd = D Fd/(hf Dx)
(see Figure 13)
Determine the concrete strut capacityfromTable 8 or from:
vRd = 0.160 fck (1 fck/250)
Calculate area of transverse reinforcement from:
Yes (cot yf = 2.0) Yes (cot yf= 1.25)
Is vRD > vEd? Is vRD > vEd?
Is length of
flange under considerationin tension?
Determine yf from:
Determine the concrete
strut capacity from Table 8or from:vRd = 0.195 fck (1 fck/250)
=s
Asf
yf =0.5sin-1
0.2fck (1 fck/250)
vEd
fyd cot yf
vEd hf
Table 8Concrete strut capacity for longitudinal shear in flanged beams
fck vRd,max
Flange in compression Flange in tension
Minimum area of shear reinforcement
r
4 r
Longitudinal shear
Rules for spacing andquantity of reinforcementMinimum area of longitudinal reinforcement
Maximum area of longitudinal reinforcement
Minimum spacing of reinforcement
Table 9Values for rw,min
fck 20 25 28 30 32 35 40 45 50
r
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References1
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3
4
5
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r
d d
a
d
g gg
r R
r
a g
S eulaVnoitinifeDlobmy Sy eulaVnoitinifeDlobm
Selected symbols
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Designing to Eurocode 2
Design procedure
Fire resistance
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Table 1Column design procedure
Step Task Further guidance
Chapter in the publication Standard
Note
Table 2Minimum column dimensions and axis distances for fire resistance
Standard fireresistance
Minimum dimensions (mm)
Column width bmin/axis distance,a, of the mainbars
Column exposed on more thanone side
Columnexposed on
one side(fi = 0.7)fi = 0.5 fi = 0.7
Note
1
2
3
4
Key
a
b
Table 3Minimum reinforced concrete wall dimensions and axis distances forload-bearing for fire resistance
Standardfireresistance
Minimum dimensions (mm)
Wall thickness/axis distance,a, of the main bars
Wall exposed on one side(fi = 0.7)
Wall exposed on twosides (fi = 0.7)
Notes
1
2
Key
a
Figure 1Section through structural member, showing nominal axis distance a
hb
a
b
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Column design
Structural analysis
Design moments
Figure 2Flow chart for braced column design
Use column chart (see Figure 9) to find As required for NEdand MEd.Alternatively, solve by iteration or by using RC
Spreadsheet TCC53 fromSpreadsheets for concrete design toBS 8110and Eurocode 26
START
Initial column size may be determined using quick designmethods or through iteration.
Determine the actions on the column
using an appropriate analysis method.The ultimate axial load is NEd and the ultimate momentsare Mtop and Mbottom (Moments from analysis)
Determine the effective length, l0, using either:1. Figure 52. Table 43. Expression (5.15) from BS EN 199211
Determine first order moments (see Figure 4)M01 = Min {|Mtop|, |Mbottom|} + eiNEdM02 = Max {|Mtop|, |Mbottom|} + eiNEd
Where ei = Max {l0/400, h/30, 20} (units to be in millimetres).M01 and M02 should have the same sign
if they give tension on the same side.
Determine slenderness, l, from either:l = l0/i where i = radius of gyration or
l = 3.46 l0/h for rectangular sections (h = overall depth) orl = 4.0 l0/d for circular sections (d= column diameter)
Determine slenderness limit, lim, from:l
lim =15.4C
n(See Slenderness section on page 5 for explanation.)
Column is slender(refer to Figure 3).
Yes
Column is not slender. MEd = M02
No
Check rules for spacing and quantity of reinforcement(see page 7)
Is ll lim?
ff
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Figure 3Flow chart for slender columns (nominal curvature method)
Effective length
k = EIc
2EIb0.1
lc
From Figure 2
Determine Kr from Figure 9 or fromKr = (nu - n) / (nu - nbal) 1
wheren = NEd / (Acfcd), relative axial force
NEd = the design value of axial forcenu = 1 + w
nbal = 0.4w =As,estfyd / (Acfcd)
As,est = the estimated total area of steelAc = the area of concrete
Revise valueofAs,est
No
Yes
Check detailing requirements
Calculate Kh from Kh= 1 +bhef 1where
hef= the effective creep ratiob= 0.35 + fck/200 /150l = the slenderness ratio.
See section on creep (page 6)
Calculate e2 from
e2 = 0.1 Kr Khfydl02 0.45d Es
whereEs = elastic modulus of reinforcing steel
(200 GPa)
M0e = 0.6 M02 + 0.4 M01 0.4 M02M2 = NEde2
MEd = Max {M02, M0e + M2, M01 + 0.5 M2}
Use column chart to find As,reqd for NEd and MEdAlternatively, solve by iteration or by using
RC Spreadsheet6
a) l0 = l b) l0 = 2l c) l0 = 0.7l d) l0 = l /2 e) l0 = l f) l/2 < l0 < l g) l0 > 2l
y
ly
y
M
Figure 5Effective lengths for isolated members
First ordermoments forstocky columns
M02
M01
M02
0.5 M2 M M01 2+ 0.5
2 Ed 2
M M0e 2+M0e
Additional secondorder moments forslender columns
Total momentdiagram forslender columns
+ =
M e Ni Ed
M N= e
Figure 4Design bending moments
IsAs reqd&As, est?
lb
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Table 4Effective length factor, F, for braced columns
k2 k1
0.10 0.20 0.30 0.40 0.50 0.70 1.00 2.00 5.00 9.00 Pinned
0.10 0.59 0.62 0.64 0.66 0.67 0.69 0.71 0.73 0.75 0.76 0.77
0.20 0.62 0.65 0.68 0.69 0.71 0.73 0.74 0.77 0.79 0.80 0.81
0.30 0.64 0.68 0.70 0.72 0.73 0.75 0.77 0.80 0.82 0.83 0.84
0.40 0.66 0.69 0.72 0.74 0.75 0.77 0.79 0.82 0.84 0.85 0.86
0.50 0.67 0.71 0.73 0.75 0.76 0.78 0.80 0.83 0.86 0.86 0.87
0.70 0.69 0.73 0.75 0.77 0.78 0.80 0.82 0.85 0.88 0.89 0.90
1.00 0.71 0.74 0.77 0.79 0.80 0.82 0.84 0.88 0.90 0.91 0.92
2.00 0.73 0.77 0.80 0.82 0.83 0.85 0.88 0.91 0.93 0.94 0.95
5.00 0.75 0.79 0.82 0.84 0.86 0.88 0.90 0.93 0.96 0.97 0.98
9.00 0.76 0.80 0.83 0.85 0.86 0.89 0.91 0.94 0.97 0.98 0.99
Pinned 0.77 0.81 0.84 0.86 0.87 0.90 0.92 0.95 0.98 0.99 1.00
105 kNm
105 kNm 105 kNm
105 kNm 105 kNm
rm = 1.0 rm = -1.0rm = 0
a) C= 1.7 - 1 = 0.7 b) C= 1.7 - 0 = 1.7 c) C= 1.7 + 1.0 = 2.7
0
Figure 6Calculating factor C
Slenderness
ll
l
l lim =20ABC
15.4C
n n
hh
ll
Column design resistance
outside
within
0.00175
hx
x x
hingepoint
hingepoint
0.00175 / ( /2)x x h
h d
h/2
0.0035 max 0.0035 max
0.00175 min
d) General relationshipc) When x < hb) When x > ha) Pure compression
0.00175
exex
Figure 8Strain diagrams for columns
a) Strain diagram b) Stress diagram
h
dc
sscecu2
esc
eyd2
sst
n. axis
d2
x
fcd
As2
As
Figure 7Stress block diagram for columns
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0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.3
1.2
1.1
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45
0
0.1
0.2
0.30.4
0.5
0.6
0.7
0.8
0.9
1.0
A f bhf s yk ck/
M/bh f2
ck
N
bhf
/
c
k
Kr
= 0.2
d h2/ = 0.05
Figure 9a
Column design chart for rectangular columns d2 /h = 0.05
l
l
Creep
h
Biaxial bending
MEdz
a+
MEdy a1.0
MRdz MRdy
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0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
0.1
0.2
0.30.4
0.5
0.7
0.9
1.0
d h2/ = 0.15
1.3
1.2
1.1
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40
A f bhf s yk ck/
N
bhf
/
ck
M bh f/2
ck
Kr =1
0.6
0.8
Figure 9c
Column design chart for rectangular columns d2 /h = 0.15
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.3
1.2
1.1
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45
M bh f/2
ck
d h2/ = 0.10
Kr =1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
N
bhf
/
ck
A f bhf s yk/ ck
Figure 9bColumn design chart for rectangular columns d2 /h = 0.10
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A f bhf s yk ck/
d h2/ = 0.25
M/bh f2
ck
N
bhf
/
ck
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.3
1.2
1.1
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.05 0.10 0.15 0.20 0.25 0.30
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Kr = 1
Figure 9e
Column design chart for rectangular columns d2 /h = 0.25
0 0.05 0.10 0.15 0.20 0.25 0.30 0.350
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.2
1.1
1.30.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
A f bhf s yk ck/
d h2/ = 0.20
M/bh f2
ck
N
bhf
/
ck
Kr = 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Figure 9dColumn design chart for rectangular columns d2 /h = 0.20
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Table 5Value of a for rectangular sections
NEd /NRd 0.1 0.7 1.0
Note
Unbraced columns
Walls
Rules for spacing andquantity of reinforcementMaximum areas of reinforcement
Minimum reinforcement requirements
Symbol Definition Value
h
h
h
l
l
h h
h
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References1
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Spacing requirements for columns
Particular requirements for walls
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Eurocode 7: Geotechnical designScope
Limit states
EQU
STR
GEO UPL
HYD
Geotechnical Categories
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Table 1Geotechnical categories of structures
Category Description Risk of geotechnical failure Examples from Eurocode 7
Table 2
Design values of actions derived for UK design, STR/GEO ultimate limit state persistent and transient design situations
CombinationExpression referencefrom BS EN 1990
Permanent actions Leading variableaction
Accompanying variable actions
Unfavourable Favourable Main (if any) Others
Combination 1 (Application of combination 1 (BS EN 1997) to set B (BS EN 1990))
c
c c
c
Combination 2 (Application of combination 2 (BS EN 1997) to set C (BS EN 1990))
c
Keya b gc cd j
Table 3Partial factors for geotechnical material properties
Angle of shearing resistance(apply to tan h)
Effective cohesion Undrained shearstrength
Unconfinedstrength
Bulk density
Symbol gh gc gcu gqu gg
Methods of design and combinations
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45
Geotechnical design report
Spread foundations
Table 4Design values of actions derived for UK design, EQU ultimate limitstate persistent and transient design situations
CombinationExpressionreference
Permanent actions Leadingvariableaction
Accompanying variableactions
Unfavourable Favourable Main(if any)
Others
c
Key
a
b g
c c
Table 5Presumed allowable bearing values under static loading (from BS 8004)
Category Type of soil Presumed allowable bearing value (kN/m2) Remarks
Note
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Reinforced concrete pads
Figure 1Procedures for depth of spread foundations
Design foundation (structural design) using the worst ofCombinations 1 and 2 (ULS) for actions and geotechnical
material properties.
START
Design usingdirect method?
Obtain soil parameters from Ground Investigation report
Size foundation(geotechnical design) usingthe worst of Combinations
1 or 2 (ULS) for actionsand geotechnical materialproperties. Combination 2
will usually govern.
Use prescriptive method.Size foundation
(geotechnical design)
using SLS for actionsand presumed
bearing resistance
Is there anoverturning moment?
Check overturning using EQUlimit state for actions and
GEO Combination 2for material properties.
Yes No
Yes
No
MM M
P
PPe
eee = /M P
P
P
L
L
L
L1 +
e6
6e
1
2P
1.5 3L e
L = width of base
SLS pressure distributions ULS pressure distribution
or
P
2L e
P P P
Figure 2Pressure distributions for pad foundations
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START
Determine value of factor (=1.0 when applied moment is zero; refer to Expressions
(6.38) to (6.42) from BS EN 199211 for other cases)
Determine value of vEd,max(design shear stress at face of column) from:
vEd,max = (VEd DVEd) (from Exp. (6.38))
(u0deff)
where u0 is perimeter of column(see Clause 6.4.5 for columns at base edges)
deff = (dy + dz)/2 where dy and dzare the effective depths in orthogonal directions
Determine value of vRd,max (refer to Table 7)
Determine concrete punching shear capacity vRd (withoutshear reinforcement) from 2dvRd,c/a (Refer to Table 6 for vRd,c)
Yes
Either increase mainsteel, or providepunching shear
reinforcement required.(Not recommended
for foundations.)
No
No shear reinforcement required. Check complete.
Figure 4Procedure for determining punching shear capacity for pad foundations
Yes
Redesign foundationIs vEd,max < vRd,max?No
Determine value of vEd, (design shear stress) from:vEd = (VEd DVEd)
(u1deff)where u1 is length of control perimeter (refer to Figure 5). For
eccentrically loaded bases, refer to Exp. (6.51).
The control perimeter will have to be found through iteration;it will usually be between dand 2d
Is vEd < vRd atcritical perimeter?
Design for punching shear
Table 6vRd,c resistance of members without shear reinforcement, MPa
rl Effective depth, d (mm)
300 400 500 600 700 800 900 1000a
0.25% 0.47 0.43 0.40 0.38 0.36 0.35 0.35 0.34
0.50% 0.54 0.51 0.48 0.47 0.45 0.44 0.44 0.43
0.75% 0.62 0.58 0.55 0.53 0.52 0.51 0.50 0.49
1.00% 0.68 0.64 0.61 0.59 0.57 0.56 0.55 0.54
1.25% 0.73 0.69 0.66 0.63 0.62 0.60 0.59 0.58
1.50% 0.78 0.73 0.70 0.67 0.65 0.64 0.63 0.62
1.75% 0.82 0.77 0.73 0.71 0.69 0.67 0.66 0.65
2.00% 0.85 0.80 0.77 0.74 0.72 0.70 0.69 0.68
k 1.816 1.707 1.632 1.577 1.535 1.500 1.471 1.447
Key
a
Notes1r
r rr r r
2r
f 25 28 32 35 40 45 50
Factor
Beam shearfaces
2d
d
d
h
Bends mayberequired
Punching shear perimeters,(load within deducted from V )Ed
Figure 3Shear checks for pad foundations
bz
2d2d
by
u1
u1
Figure 5Typical basic control perimeters around loaded areas
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y
y
Raft foundations
Piled foundations
Table 7Values for vRd, max
fck vRd,max
a a
bF
hF
Figure 8Dimensions for plain foundations
Stress zone
45o
As contributing to shear capacity
Figure 7Shear reinforcement for pilecaps
Punching shear 5 2dfrom column face
f/5
f/5
f
Beam shear 5 dfrom column face
Figure 6Critical shear perimeters for piles
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49
Table 8Minimum percentage of reinforcement required
fck fctm Minimum % (0.26fctm/fyka )
Key
a
Plain concrete foundations
a
s
Rules for spacing and
quantity of reinforcementCrack control
Minimum area of principal reinforcement
Maximum area of reinforcement
Minimum spacing of reinforcement
Deep elements
Symbol Definition Value
a g
dd
a
b
d
g
r
r
c c
c
Selected symbols
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50
References1
2
3
4
5
6
7
8
9
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Designing to Eurocode 2
Analysis
Design procedure
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52
Table 1Flat slab design procedure
Step Task Further guidance
Chapter in this publication Standard
Note
Table 2Minimum dimensions and axis distances for reinforced concrete slabs
Standard fire resistance Minimum dimensions (mm)
Slab thickness,hs Axis distance,a
Notes
1
2
3
4
5
Key
a
Fire resistance
ff
Flexure
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53
Table 4Values for K
% redistribution d (redistribution ratio) K
Keya
Table 5z/dfor singly reinforced rectangular sections
K z/d K z/d
Keya
Table 6Minimum percentage of reinforcement required
fck fctm Minimum % (0.26 fctm /fyka )
Keya
Figure 1Procedure for determining flexural reinforcement
Table 3Bending moment coefficients for flat slabs
End support/slab connection Firstinteriorsupport
Interiorspans
Interiorsupports
Pinned Continuous
Endsupport
Endspan
Endsupport
Endspan
Notes
1
2 3
4
Carry out analysis of slab to determine design moments(M) (Where appropriate use coefficients from Table 3).
No compression reinforcement required
Check maximum reinforcement requirements.As,max = 0.04 Ac for tension or compression
reinforcement outside lap locations
Determine K from Table 4 orK = 0.60d 0.18d2 0.21 where d 1.0
START
Yes
Yes
Outside scope ofthis publication
Concrete classC50/60?
No
Compressionreinforcementrequired not
recommended fortypical slabs
Is KK ?No
Determine K from: K=Determine K from: K=bd2fck
M
Obtain lever arm z from Table 5 or
1 + 0.95dz =2d
1 3.53 K[ ]
Calculate tension reinforcement required from
As = fydzM
Check minimum reinforcement requirements (see Table 6)
As,min = where fyk 25fyk
0.26 fctmbt d
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Deflection
Punching shear
bb
Figure 2Simplified rectangular stress block for concrete up to class C50/60from Eurocode 2
Figure 4Basic span-to-effective-depth ratios for flat slabs
Figure 3Procedure for assessing deflection
Is basic l/dx F1 x F2 x F3 Actual l/d?
IncreaseAs,prov
Determine Factor 3 (F3)F3 = 310/ss
Where ss = Stress in reinforcement at serviceabilitylimit state (see Figure 5)
ss may be assumed to be 310 MPa (i.e. F3 = 1.0)
Note:As,prov 1.5 As,reqd (UK National Annex)
Determine basic l/dfrom Figure 4
Check complete
Determine Factor 1 (F1)For ribbed or waffle slabs
F1 = 1 0.1 ((bf/bw) 1) 0.8
(bf is flange breadth and bw is rib breadth)Otherwise F1 = 1.0
Determine Factor 2 (F2)Where the slab span exceeds 8.5 m and it supports
brittle partitions, F2 = 8.5/leff
Otherwise F2 = 1.0
No
Yes
START
The Eurocode is ambiguous regarding linear interpolation. It is understood that thiswas the intention of the drafting committee and is in line with current UK practice.
Notes
1
2 r
3
rr
rr
r rr
r
r
[ ]
r
r r
r[ ( ) ]
fck = 50
fck = 45
fck = 40
fck
= 35
fck = 32
fck = 30
fck = 28
fck = 25
fck = 20
Percentage of tension reinforcement (As,req/bd)
39
37
35
33
31
29
27
25
23
21
19
17
150.40% 0.60% 0.80% 1.20% 1.80%
Span-to-effective-depth
ratio
(l/d)
1.00% 1.40% 1.60% 2.00%
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55
Figure 5Determination of steel stress
Figure 6Procedure for determining punching shear capacity
START
Determine value of factor (refer to Figure 7 or Expressions (6.38) to (6.46)
of the Eurocode)
Determine value of vEd(design shear stress at face of column) from:
vEd = VEd /(uideff)where ui is perimeter of column
deff = (dy + dz)/2 (dy and dzare the effective depths in orthogonal directions)
Determine value of vRd,max from Table 7
Determine value of vEd, (design shear stress) from:
vEd = VEd /(uideff)where u1 is length of control perimeter (see Figure 8)
Yes
Redesign slabIs vEd,vRd?No
Determine concrete punching shear capacity(without shear reinforcement), vRD,c from Table 8
where r l = (r lyr lz)0.5
(r ly, r lz are the reinforcement ratios in two orthogonaldirections for fully bonded tension steel, taken over a
width equal to column width plus 3deach side.)
Is vEd > vRd,c?
Yes
Punching shearreinforcement not
required
No
Determine area of punching shear reinforcement per perimeterfrom:
Asw = (vEd 0.75vRd,c)sru1/(1.5fywd,ef)where
sr is the radial spacing of shear reinforcement (see Figure 9)fywd,ef = 250 + 0.25 defffywd (see Table 9)
Determine the length of the outer perimeter where shearreinforcement not required from:
uout,ef = bVEd/(vRd,cd)
Determine layout of punching shear reinforcement(see Spacing of punching shear reinforcement
Section and Figure 9).
To determine stress in the provided reinforcement (ss), calculate the ratioGk/Qk , read up the graph to the appropriate curve and read across todetermine ssu.
ss can be calculated from the expression: ss = ssuAs,req
As,prov
1
d( ) )
Ratio Gk/Qk
Approximatesteelstressat
SLSforA
s,req
180
200
220
240
260
280
300
320
1.0 2.0 3.0 4.0
c2
= 0.8, gG
= 1.35
c2= 0.6, gG= 1.35
c2= 0.3, gG= 1.35
c2= 0.2, gG= 1.35
c2= 0.6, g
G= 1.25
c2= 0.3, gG= 1.25
c2= 0.2, gG= 1.25
(c
2is the factor for quasi-permanent value of a variable action. For furtherexplanation refer to How to design concrete structures using Eurocode 2:Introduction to Eurocodes3.
AppromximatesteelstressatSLSforA
s,req,ssu
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Rules for spacing andquantity of reinforcement
Minimum area of reinforcement
Maximum area of reinforcement
Minimum spacing of reinforcement
Maximum spacing of main reinforcement
b= 1.5
b= 1.4 b= 1.15
Internal column
Corner column
Edge column
Figure 7Recommended standard values for b
Table 8vRd,c resistance of members without shear reinforcement, MPa
rI Effective depth, d(mm)
200 225 250 275 300 350 400 450 500 600 750
0.25%
0.50%
0.75%
1.00%
1.25%
1.50%
1.75%
2.00%
k
Notes
1r I
RrIRr r r r
2 r
I
f 25 28 32 35 40 45 50
Factor
Figure 8Typical basic control perimeters around loaded areas
bz
2d2d
by
u1
u1
Table 7Values for vRd,max
Table 9Values for fywd,ef
fck vRd, max deff fywd,ef
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Table 10Factor, F, for determining Asw, min
fck Factor, F
Note
Spacing of punching shear reinforcement
Symbol Definition Value
a g
dd
a
d
g g g
r R
r
r
Selected symbols
Figure 9Punching shear layout
sr
Outer controlperimeter
s 0.75r d
0.5 d
kd
Section A - A
st
Outer perimeter of shearreinforcement
Outer controlperimeteruout
0.5 d
0.75d
1.5
(2 if > 2from column)
d
d d
A A1.5d
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58
References1
2
3 4
5
6
7
8
9
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Methods for checking deflection
Overview
What affects deflection?There are numerous factors
that affect deflection. These
factors are also often time-
related and interdependent,
which makes the prediction
of deflection difficult.
The main factors are:
Concrete tensile strength
Creep
Elastic modulus
Other factors include:
Degree of restraint
Magnitude of loading
Time of loading
Duration of loading
Cracking of the concrete
Shrinkage
Ambient conditions
Secondary load-paths
Stiffening by other elements
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60
Factors affecting deflection
Tensile strength
Creep
h
Elastic modulus
Figure 1Typical floor layouts
a) Favourable layout of restraining walls (low restraint)
b) Unfavourable layout of restraining walls (high restraint)
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61
Cracking
K f Wctm= 0.5^ h
z
h
h h
Loading sequence
Figure 2Loading history for a slab an example
0
2
4
6
8
10
12
14
0 50 100 150 200 250 300
Duration (days)
Load(kN/m) a
b
c
de
f
gh
Loading sequence
a Slab struck
b 1st slab above cast
c 2nd slab above cast
d 3rd slab above cast
e Floor finishes applied
f
g Quasi-permanent variable actions
h Frequent variable actions
Partitions erected
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z
Shrinkage curvature
Methods forcalculating deflections
Rigorous method
Panel 1Determining long term elastic modulus of elasticity
EE
W
E
W
E
W
E
W
E
W
eff,1
1
eff, 2
2
eff, 3
3
eff, 4
4
eff, 5
5LT = + + + +RW c m
h
h
Figure 3Outline of rigorous method for calculating deflection
Collate input data
Determine the curvature of the slab
Assess whether the element has flexural cracking
h aa
zz z
z
] g11r E I Mfl e c E Ie uQPMQP
= +g g
Repeat
at1/20pointsforallthreeloadingsta
ges
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63
Table 1Concrete properties
fck MPa 320 325 328 330 332 335 340 350
e e e e e e e
Key
a
Panel 2Useful Expressions for a rectangular section
2 1x
bh A d d Ase s2 2
2
u =
+ - +a] ]g1bh A Ae s s2- +a] ]g g
g+
bh+ + +x- 1I 12bh
2h
A d x x du
3
u u
2
s Ae 2u s2= - - -aa ] ] ]k g g 22 g6 @
1 2 A b A d d bs 2s2x A s2As2Ac e2
s e
0.5= + +- +a As eaea a] 1-ea]^ ^ ^g 1- - +ea]h g h g h7 A# -
g g g3 1I bx A d xc c3
e s As2 d22 2
c xc= + - - -+a ea^ ^ ^1r I
S
cscs e
u
u= + -1g gf a cs ef a^ h IScc
a
100 300 500 700 900 1100 1300
ho (mm)h ?( , )t0
C20/25C25/30
C30/37C35/45
C40/50C45/55
C50/60
100 500 900 1300h0 (mm)h
100 100
N NR RS S
t0 t0
1 1
2 2
3 35 5
10 10
20 20
50 50
30 30
7.0 7.06.0 6.05.0 5.04.0 4.03.0 3.02.0 2.01.0 1.00 0( ,? t )0
A
E B
D
C
a) Inside conditions - RH = 50% b) Outside conditions - RH = 80%
1100700300
Notes1
2
3 Intersection point between lines D & E can also be above point A4 For t0 > 100 it is sufficiently accurate to assume t = 100
t0 = age of concrete at time of loading
h0 = 2Ac/u
KeyHow to use Nonogram
Figure 4Method for determining creep coefficient h(,t0)
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64
8
Loading Bending moment diagram
M M
M
M
al W
l
0.125
M Wa a l= (1- )
0.0625
al alW/2 W/2
0.104
0.102
M =Wal
2
q
ql
q
q
q
0.125
3 4a48 (1-a)
If a = , K =
MA
MA
MC
MC
MC MC
MB
MB
WalWal
al
al al
a a(4 )
if = = 0.25a l , K
K = 0.083 (1 )
b=
1
80Wl 2
24(3 4 )a
2
qa l2 2
2
K = 0. 104 ( 1 )
b=M MA + B
2
15.6
ql 2
K
a6
112
12
2
2
End deflection(3 )a a6
load at end = 0.333K
12
b
4
M + MA B
(5 4 )a
3 4a
2 2
=
b
10
Simplified method
Figure 5Simplified method for calculating deflection
Figure 6Values for Kfor various bending moment diagrams
Calculate the moment, MQP, due to quasi-permanent actions atthe critical section (i.e. mid-span or at support for cantilever)
Calculate creep coefficient, h(,t0), using either Figure 4or Annex B (in which case look-up fcm in Table 1)
START
Calculate flexural curvature 1r E I
M1
n ef f c E Ie ff u
QP MQP= + g g^ h
Obtain concrete properties,fctm, and Ec28 from Table 1
1 Calculate long term elastic modulus, Eeff from: Eeff= Ec28/[1+h (,t0)]2 Calculate effective modulus ratio, ae from ae = Es/Eeff, where Es is
elastic modulus for reinforcement (200 GPa)3 Calculate depth to neutral axis for uncracked condition,xu4 Calculate second moment of area for uncracked condition, Iu
Calculate depth to neutral axis for crackedcondition,xc and calculate second moment of area
for cracked condition, Ic
Yes
FinishNo
Calculate the deflection that will occur at the time of application ofthe load due to partitions and/or cladding.
1 Calculate the creep coefficient h(t,t0), where t is the age whenpartition/cladding loads are applied and t0 is the age of striking.h(t,t0) h(,t0) bc(t,t0). For bc(t,t0) refer to Figure 7, alternativelyrefer to Annex B of Eurocode 2.
2 Calculate the moment due to self-weight, partitions/cladding and anyother loads which have been applied prior to the installation of thecladding/partition,Mpar and use in place of MQP
3 Recalculate the section properties, curvature and hence deflection,dpar, using h(t,t0) or equivalent instead of h(,t0)
4 The approximate deflection affecting cladding and partitions isd = dQP dpar
Calculate cracking moment, Mcr from:0.9
Mh
f Icr
xu
ctm u=
(Note the factor 0.9 has been introduced into this method
because the loading sequence is not considered)
Yes No
Section is uncrackedz= 0
Is Mcr > MQP?
Section is crackedz= 1 0.5(Mcr/MQP)2
Calculate total shrinkage strain ecs from ecs = ecd + eca where:ecd = khecd,0 = Drying shrinkage strainkh = Coefficient based on notional size, see Table 2ecd,0 = Nominal unrestrained drying shrinkage, see Table 1eca = bas(t) eca() = eca() for long-term deflection, see Table 1
Calculate curvature due to shrinkage strain 1/rcs (see Panel 2)
Calculate total curvature = +1
rt, csnQP
1 1r r
Calculate quasi-permanent deflection from1
KLQP2=d rt,QP
where Kcan be obtained from Figure 6 and L is the span.
Do you needto calculate deflectiondue to cladding and
partitions?
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65
100 300 500 700 9000.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
Coefficient,
bc
(t,
t0)
h0 (mm)
t = 90, t0 = 7
t = 60, t0 = 7
t = 28, t0 = 7
t = 90, t0 = 3
t = 60, t0 = 3
t = 28, t0 = 3
Precamber
Flat slabs
Accuracy
Figure 7Coefficient for development of creep with time after loading
Table 2Values for Kh
h0 kh
Notes
Notes
Figure 8Precambering of slabs
Deflection affecting partitions
Just before installationof partitionsPrecamber
Deflection due to
frequent combinationDeflection due toquasi-permanentcombination
Figure 9Recommended acceptance criteria for flat slabs
a
X
Notes
d d
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References1
2
3
4
5
6
7
8
9
10
Cladding tolerances
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Introduction
Geotechnical Categories
Limit states
A J BondO BrookerA J Harris
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682
gc
g
c c cc
g
g
hhhhgh
Calculation models forstrength limit states
favourable
unfavourable
Figure 1Ultimate limit states for reinforced concrete retaining walls
a) Overall stability
b) Sliding c) Toppling
d) Bearing e) Structural failure
Table 1Partial factors to be used for retaining wall design according todesign approach 1 (UK National Annex)
Combin-ation
Partial factors onactions
Partial factors on materialproperties of soil
gGa gG,fav gQ g
b gc gcu g
Keyag
bghh
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693
Calculation model A
c = bc hh
Figure 2Calculation model A
h
a
Figure 3Calculation model B
h
cchh
Calculation model B
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704
inclined
yd
y
d
h
Figure 4Overall design procedure
Design procedure
both combinations.
Overall stability of the site
Initial sizing
8
Figure 5Symbols for initial sizing
70
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715
Material properties
g
h
Figure 6Procedure for determining material properties, geometry and actions
From Figure 4
Determine characteristic material propertiesgh
g
h
g
Determine initial geometry and actions (see Panel 1)1
(y2 3
Carry out separately for both Combinations 1 and 2
Determine design material properties and earthpressures (see Panel 2)1
gggggg h
hg g
2 hb
To Figure 4
Panel 1General expressions for geometry and actions
g
g
For calculation model A:
b
g
&
For calculation model B:
&g
y
b
Wb
Wy
Panel 2General expressions for material properties and earth pressures
For calculation model A:
bRh
b
bRh
bb
g
gg
For calculation model B:
h
b
b
h
h
h
h
bhyh
b
g
bby
g
gg
g
71
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726
g
Design against sliding
Figure 7Procedure for design against sliding
From Figure 4
gS
To Figure 4
Panel 3Expressions for drained sliding resistance
Undrained sliding resistance:
Drained sliding resistance:
dh
d
g
/
h
&g
/
h
R D
df
dfff
f
df
&R2D2
Design against toppling
Figure 9Procedure for design against toppling
From Figure 4
W
W
To Figure 4
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73
Figure 8Effects of shear key
0
Panel 4Expressions for bearing resistance
g/W W
W
Undrained bearing resistance:
p
R W
Drained bearing resistance:
g
g
gg
W
g
/
W
g
g
Figure 11Bearing capacity factors, N, from ground properties
Angle
gc
,
Bearing
capacity
factor,
N
significant
Design against bearing failure
Figure 10Procedure for design against bearing failure
From Figure 4
To Figure 4
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748
Structural design
10
Compaction earth pressures
s
Figure 12Effective base width, B
At-rest earth pressures
hRbb
Figure 13Design procedure for structural design
From Figure 4
11
To Figure 4
Figure 14Compaction earth pressures for structural design of cantileverretaining walls
zJ
zK
zD
74
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759
Figure 15Pressure diagram for design of reinforced concrete base
Panel 5General expressions for designing walls for compaction pressures
Forces for stem design (see Figure 14)
Forces for heel design (see Figure 15b)
&
&
Forces for toe design (see Figure 15a)
g
R
R
sg
sR
s
g
pg
pg
g
p
bRh
b
bR
h
b b
bRh
b
bRh
bb
Table 2Typical compaction equipment12
Description of compactionequipment
Mass (kg) Centrifugalvibratorforce (kN)
Designforce (kN)
Panel 6Expressions for the structural design of basement walls for at-restpressures
h
Rb
s
g
ss
s
g
s
s
ss
Forces for stem design
75
Soil pressure abovetoe ignored
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7610
s
s
Detailing
Control of cracking
13
14
3
restraint
Figure 16Earth and pore water pressures for structural design of retaining
walls subject to 'at-rest' conditions
z
W
zD
W
loading
s
1
Table 3Maximum bar size or spacing to lmit crack width (mm)
Steelstress(ss)MPa
Wmax = 0.3 Wmax = 0.2
Maximumbar size(mm)
Maximumbar spacing(mm)
Maximumbar size(mm)
Maximumbar spacing(mm)
Note
s
g d
g
d
76
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7711
s
Large radius bends
f
ff
f
f
15
Rules for spacing and quantity ofreinforcement
Vertical reinforcement
Figure 17Typical drainage layout for a retaining wall
300 mm widegranular backfill
Weep hole
Drainage pi pe
Large radius bendif required
f
f
Horizontal reinforcement
Practical issuesDesign for movement
16
2
3
1
Drainage
77
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7878
Construction
References
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
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80
Table 1
Notation for steel reinforcement
Type of steel reinforcement Notation
Note
Figure 1Description of bond conditions
a) 45 < a< 90
Key
b) h < 250 mm d) h > 600 mm
c)h > 250 mm
a
Cover
D
D
D
Anchorage and lap lengthsaa
Anchorage of bars and links
Table 2Anchorage and lap lengths for concrete class C25/30 (mm)
Bondcondition,(see Figure 1)
Reinforcement in tension, bar diameter, f(mm) Reinforcementincompression8 10 12 16 20 25 32 40
f
f
f
f
a
f
f
a
f
f
Notes
1 a
2 aaaa
3 ss
4
5
6
Concrete class C20/25 C28/35 C30/37 C32/40 C35/45 C40/50 C45/55 C50/60
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81
Bars in compression
f
Figure 2Design anchorage length lbd, for any shape measured along thecentreline
flbd
Figure 3Anchorage of links
a) Bend angle > 150 b) Bend angle 150
f f
f
f
Figure 4Percentage of lapped bars in one lapped section
a
Figure 5Arranging adjacent lapping bars
f
f
f
Arrangement of laps
1.
f
2.
3.
f
Transverse reinforcementBars in tension
f
f
f
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82
BeamsCurtailment
y
y
Figure 6Transverse reinforcement for lapped splices
a) Bars in tension
b) Bars in compression
f f
Table 3Bar sizes for transverse reinforcement
Lap length(mm), fortransversebars at 150mm centresa
Numberof barsat eachlap
Bar size (mm)
20 25 32 40
As = 314 As = 491 As = 804 As= 1260
Key
a
Figure 7Illustration of curtailment of longitudinal reinforcement
D
D
Figure 8Simplified detailing rules for beams
a) Continuous member, top reinforcement
l al
l a
b) Continuous member, bottom reinforcement
l a
l
c) Simple support, bottom reinforcement
l
Notes
1
2
3
4
5 6
7
8
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83
Reinforcement in end supports
s
Flanged beams
Minimum area of longitudinal reinforcement
Maximum area of longitudinal reinforcement
Figure 9
Placing of tension reinforcement in flanged cross section
beff
beff1 beff2
As
bw
hf
Minimum spacing of reinforcement
Shear reinforcement
r r
SlabsCurtailment
Reinforcement in end supports
Table 4
Minimum percentage of reinforcement required
fck fctm Minimim percentage(0.26 fctm / fyka)
rw, min
x 10-3
Keya
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84
Minimum spacing requirements
Maximum spacing of reinforcement
ss
s
Figure 10Simplified detailing rules for slabs
a) Continuous member, top reinforcement
l l
l
b) Continuous member, bottom reinforcement
l
c) Simple support, bottom reinforcement
l
Notes
1
2
3
4
5 6
7
Minimum areas of reinforcement
Maximum area of longitudinal reinforcement
Edge reinforcement
Flat slabs
Figure 11Edge reinforcement for slab
2h
h
Table 5Values for rw,min
Steelstress
(es) MPa
wmax = 0.4 mm wmax = 0.3 mm
Maximumbar size(mm)
OR
Maximumbar spacing(mm)
Maximumbar size(mm)
OR
Maximumbar spacing(mm)
Table 6Apportionment of bending moments in flat slabs equivalent frame method
Location Negative moments Positive moments
Notes
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85
Figure 12Division of panels in flat slabs
lx> ly
ly/4 ly/4
ly
ly/4
ly/4
Middle strip = lx ly /2
Middle strip = ly /2
Column strip = ly/2
Punching shear reinforcement
Figure 13
Effective width,be of a flat slab
a) Edge column b) Corner columnNote
Figure 14Punching shear layout
s d
d
kd
u
d
d
d
d
dd
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86
Columns and wallsMaximum areas of reinforcement
Minimum reinforcement requirements
Particular requirements for walls
Table 7
Factor, F, for determining Asw, min
fck
Factor, F
Note
Table 8Requirements for column reinforcement
Bar dia. (mm) 12 16 20 25 32 40
Keya
b
c
Lapping fabric
f
f
f
f
Tolerances
Table 9Minimum area of vertical reinforcement in walls (half in each face)
Wall thickness (mm) As,min/m length of wall (mm2)
Table 10Tolerance
Cutting and bending processes Tolerance (mm)
Bending:
Table 11Deductions to bar dimensions to allow for deviations between twoconcrete faces
Distance betweenconcrete faces (mm)
Type of bar Total deduction(mm)
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88
Table 13Anchorage and lap lengths
Bond condition(see Figure 1)
Reinforcement in tension, bar diameter, f(mm) Reinforcementin compression
8 10 12 16 20 25 32 40
Concrete class C20/25
l
f
f
f
f
l
a
f
f
a
f
f
Concrete class C25/30
l
f
f
f
f
l
a
f
f
a
f
f
Concrete class C28/35
l
f
f
f
f
l
a
f
f
a
f
f
Concrete class C30/37
l
f
f
f
f
l
a
f
f
a
f
f
Concrete class C32/40
l
f
f
f
f
l
a
f
f
a
f
f
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9090
Table 14Sectional areas of groups of bars (mm2)
Bar size(mm)
Number of bars
1 2 3 4 5 6 7 8 9 10
Table 15Sectional areas per metre width for various spacings of bars (mm2)
Bar size(mm) Spacing of bars (mm)75 100 125 150 175 200 225 250 275 300
Table 16Mass of groups of bars (kg per metre run)
Bar size(mm)
Number of bars
1 2 3 4 5 6 7 8 9 10
Table 17Mass in kg per square metre for various spacings of bars (kg per m 2)
Bar size(mm)
Spacing of bars (mm)
75 100 125 150 175 200 225 250 275 300
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92
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93
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94
94
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95
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96
96
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97
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98
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99
Introduction
A S FraserA E K Jones
Start
Finish
Can thetabular method conditions
be met?
Is the element abraced column?
Is the element aslab or beam?
Is there anacceptable solution?
Is there anacceptable solution?
Use simplified methods
Use tabular method
Use 500oc isotherm methodor zone method
Use Annex C of Part 12:Buckling of columns under fire
Use Annex E of Part 12:Simplified calculation method
for beams and slabs
NoNo
No
No
Yes
Yes
Yes
No
Yes
Yes
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100
Basic concepts
Fire types
Level of protection
R
E
I
Material factors
g
y
yy
Combinations of actions
nn
nc
Coefficientkc(y(fck) of concrete
1.0
0.8
0.6
0.4
0.2
00 200 400 600 800 1000 1200
Temperature, ( C)o
Coefficient,
(
)
kc
y
Calcareousaggregates
Siliceousaggregates
y
Coefficientks(yfck)of tension and compression reinforcement (class N)
1. 0
0.8
0.6
0.4
0.2
00 200 400 600 800 1000 1200
Hot-rolled tensionreinforcement, 2%s,fi
Cold-worked tensionreinforcement,
Compression
reinforcement andtension reinforcement,where < 2%
Temperature, ( C)o
y
Coefficient,
(
)
ks
y
e
2%s,fie
s,fie
Coefficientkp(y(bfpk
1.0
0.8
0.6
0.4
0.2
00 200 400 600 800 1000 1200
Quenched and temperedprestressing steel (bars)
Cold-worked prestressingsteel (wires and strands)Class A
Cold-workedprestressingsteel (wires andstrands) Class B
Temperature, ( C)o
y
Coefficient,
(
)
kp
y
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102
y
SS
Columns
Method A
m
m
mn
aa
m
a
0.5 0.6 0.7 0.8 0.9 1.020
18
16
14
12
10
8
6
4
2
0
fi = 0.2
fi = 0.5
fi = 0.7
Ratio, /s,req
Reductioninaxisdistance
(mm)
a
D
n
n
n
AA
Standardfireresistance
Minimum dimensions (mm)Column width bmina, of themain bars
on one side
mfi = 0.2 mfi mfi mfi
a
a a
a
a a a
a a -
Keya
Note
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105
Tensile members
Slabs
Simply supported slabs
Continuous slabs
0.3 eff 0.3 eff0.4 eff
BM from Exp. (5.11)
BM in fire location
Design BMaccording toBS EN 199211
BM when
= 0t
Two-way slabs
a
prestressed solid slabs
Standard fireresistance
Minimum dimensions (mm)
One-wayslab
a Flat slab
ly/l ly/l 2 dc dc
b b b b b
b b b
b
Key
a
b
cd
Notes
2
b b b
bw
(a) Constant width (b) Variable width (c) -SectionI
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106
Table 6
supported ribbed slabs in reinforced or prestressed concrete
Standard fireresistance
Minimum dimensions (mm)
Possible combinations of width ofribs bmina
Slab thicknesshsdistance a inSimply supported
restrained
a a a
a a a a
a a
Key
a
Notes
2
Simplified calculation method for beamsand slabs
n n
Simply supported members
ggy
Flat slabs
Ribbed slabs
Simplified calculationmethods
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107
g
g
y
y
y
Continuous members
gg
Minimum width of cross-section as function of fire resistance
Fire resistance
Minimum width ofcross-section (mm)
Flow chart for simplified calculation method for beams and slabs
Finish
Is the element asimply supported?
Is MEd, fi. MRd, fi?
Are the support
moments exceeded?
Calculate the support designmoment of resistance,
MRd, fi, support
Fit the free bending
moment so that MEd,fi = MRd,fi
Redesign section or usealternative methods
Start
Calculate MRd, fi
Calculate MEd, fi.
Determine ks(y) from Figure 12
Determiney
, using temperature profiles in Annex A of Part 1-2.
Yes
No No
Yes
Yes
No
1.0
0.8
0.6
0.4
0.2
00 200 400 600 800 1000 12
Coefficient,
ks
(yc
r)or
kp
(yc
r)
Reinforcing steel
Prestressing steel (bars)
Prestressing steel(wires and strands)
Temperature, ( C)o
y
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108
Reduced cross-section of reinforced concrete beam and column
c) Fire exposure on four sides (beam or column)
500 Co
500 Co500 Co
hhfi
bfi
bfifi
b
bb
b
a) Fire exposure on three sideswith tension zone exposed
d dfi =dfi
b) Fire exposure on three sides withthe compression zone exposed
Compression Tension
Tension Compression
900
100
200
300
400
500
600
700
800
240
220
200
180
160
140
120
100
80
60
40
20
00 20 40 60 80 100 120 140
Distance from bottom left corner of element (mm)
Distancefrom
bottom
leftcornerofelement(mm)
cross-section with compression reinforcement.
As1
As
z d1
b1
x
Mu1zz
F A fs )s1 scd,fi m
Mu2
A fs1 sd,fi( )m
fcd, 1(20)
xb f n cd, 1(20)xl
l
n
y
y= (
gy
y
y
y
Note:yy
ln
y
y
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109
Finish
Is MEd,fiMRd,fi?Redesign section or use
alternative methods
Start
Calculate MEd, fi (see simplified calculation method for beams and slabs)
Check the minimum dimensions exceed the values in Table 7
Determine reduced section size (bfidfi) using Figure 13 andtemperature profiles in Annex A of Part 12
No
Yes
Determine y, using temperature profiles in Annex A of Part 12
Determine ks (y), from Figure 3 or Figure 4
Calculate Mu, using stress distribution shown in Figure 15. Mu = Mu1 + Mu2
Finish
Does the sectionalso resist torsion?
Calculate the referencetemperature at points P alongthe line AA see Figure 20
Calculate the torsion resistanceand interaction with shear usingsection 6.3 of Eurocode 2, Part 11
Start
Determine the reduced cross-section using either 500Cisotherm or zone methods
Calculate the compressive and tensile concrete strengths:
For isotherm method,fcd,fi = fcd,fi(20) =fck andfctd,fi =fctd,fi(20) = fctkFor zone method, fcd,fi = kc(ym)fcd,fi(20) andfctd,fi = kct(ym)fctd,fi(20),
where kc(ym) and kct(ym) may be taken as kc(y) and can be determined
from Figure 2
Determine position P, the point at which the reference temperature,
yp is calculated. P is located along section AA, which isdetermined from hc,ef (see Figure 19)
Yes
No
Determine yp using temperature profiles in Annex A of Part 12
Calculate the reduced design strength of the shear reinforcement,fsd,fi,from:fsd,fi = ks(y) fsd(20) = ks(y)fywd
where ks(y) can be determined from Figure 3 or Figure 4
Calculate the shear resistance using the methods given for ambienttemperature design, see Chapter 4 Beams11
Coefficient kc,t(y fck,t) ofconcrete at elevated temperatures
1.0
0.8
0.6
0.4
0.2
00 100 200 300 400 500 600
Temperature, ( C)o
Coe
fficie
nt
,
(
)
kc
,t
y
y
Calculation methods for shear and torsion
y
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110
Unbraced structures
References
2
6
9
temperature yp at point P
A
AA
c,eff
c,ef = MIN {2.5 ( ); (
hd
x
hc,ef
2 = 0
1
e
e
The reference temperature ypthe calculation of torsion resistance
A
AA
A
A
AA
A
p in linksy
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How to Design Concrete Structures using Eurocode 2
Dr Andrew Bond
Owen Brooker
Dr Andrew Fraser
Andrew Harris
This publication brings together in one place How to...
guidance for the design, specification and detailing of a broad
range of concrete elements.