Flowcharts and Design Equations - Eurocode 2

Slab Design Flowchart

2

for f <= 90 N/mm

_ck

BENDING

Carry out check to

L/d 7.4.2 (2) See separate flowchart Calculate to 7.4.3 No cracking check required

CRACKING

Use Table 7.3 7.3.2 (2) for bar spacing Check A , min tos See separate flowchart As 0.13% 26%f /fctm yk

but not 1.5As & n

Dist 0.12% 0.2Asprov

>= >= > >= >= 9.2.1.1 9.3.1.1(1) 9.3.1.1(2) Primary s 3h 400 Secondary s 3.5h 450

(smaller near point loads)

<= <= <= <= 9.3.1.1 (3) &(4)

Concrete Stress Block

Design stress = hfcd where f = a f /gcd cc ck c and h = 1 - (f -50)/200 <= 1 ck l = 0.8 - (f -50)/400 <= 0.8ck

3.1.6 (1) (3.21) & (3.22) (3.19) & (3.20)

(

)

d x_max = d -0.4 from (5.10)

(

)

[

K K

]

d d z 1 1 2 _c min , ' 0.95 2 + - £ = g

(

'

)

0 '=bd2f K-K ³ M _ck

Design Formulae

ck f bd M K = ₂ ÷ ø ö ç è æ -= 2 ' max 2 max _d x d x K c l g l

(

'

)

' ' d d f M As sc -= yd sc yd f f As z f M M As= - '+ '

Steel Design Stress

f = f /g yd yk s Fig 3.8 s yk sc f x d x f g £ ÷ ø ö ç è æ -=700 '

SHEAR

Construction overload or striking < 7 days? Is h > 200 mm?

DEFLECTION

No No Yes Yes If VRd,ct V no links,Ed otherwise see beams.

>= Where

(

f

)

f b d k d b V _{l ck ctd w} c w ct Rd 100 0.4 18 . 0 3 1 , = r ³ g (6.2) 2 200 1+ £ = d k = £0.02 d b As w l l r and (3.16) c ctm ct ctd f f g a 7 . 0 =

Concrete Stress Block

As for Slabs s ,L = 0.75d max s ,T 0.75d max 600 (9.6) (9.8) = <=

For point load near support, use 6.2.2.5or 6.2.3 (8)

Tee & Ell Beam Flowchart

2

for f <= 90 N/mm

_ck

BENDING

Steel Design Stress

As for Slabs

Treat as rectangular section, substituting b for b.f Is M > MORf ? No Yes

(

)

f f ORf f b bw b M M =

-(

)

÷ ø ö ç è æ -+ ÷÷ ø ö çç è æ -= 2 2 x d M M M h d M M z f f f l 0 '=M -M_OR ³ M

(

'

)

' ' d d f M As sc -= yd sc yd f f As z f M M As= - '+ ' As 0.13% 26%f /fctm yk Ascrack

Part of top steel at supports must be spread across the width of the flange

>= >= >= As 9.2.1.1 7.3.2 (2) 9.2.1.2 (2)

SHEAR

Compression & tension flange widths to 5.3.2.1

At d from support face

max cot min s A zf V s A s A _sw ywk s Ed sw sw _{£ = £} q g (6.7) yk ck w sw f f b s A 0.08 min = where (9.4) & (9.5) y w k s w c d s w f b f s A 2 m a x = n g and (6.9) Process as Slabs See separate flowchart

CRACKING

DEFLECTION

Use Table 7.3 for bar spacing See separate flowchart Check main steel tensile

force due to shear) at zcotq/2 from support

T (d

Or use Shift Rule 6.2.3 (7) Fig 9.2 z may be taken as 0.9d Flange MOR ÷÷ ø ö çç è æ -= = 2 f f f cd ORf h d h b f M h

(

)

max 2 6 10 2 1 1 x d f b M M d x cd w f £ ú ú û ù ê ê ë é _- -= h l

(

)

[

f f w w

]

cd OR f zh b b xb M =h - +l At support face Ed cd w V f z b n q q w =cot +tan = (6.8) ÷ ø ö ç è æ -= 2 5 0 1 6 . 0 fc k n (6.5) where 5 . 2 2 4 cot 1 2 £ -+ = £ q w w 6.2.3 (2) 2 200 1+ £ = d k = £0.02 d b As w l l r where and (3.14)

(

)

Ed cd w Rd V f z b V ³ + = q q n tan cot max , (6.8)

(

f

)

f b d k d b V _{l ck ctd w} c w ct Rd 100 0.4 18 . 0 3 1 , = r ³ g (6.2) c ctm ct ctd f f g a 7 . 0 =

Rectangular Columns

Design Flowchart

Final design moment

M = M + M Ed 0Ed 2 M >= MEd 02

(5.33)

First order moment

M = M + M0Ed 0E imp Imperfections M = q N.l /2imp i 0 where q =(2/3<=2/ l<=1)/200i Ö Joint stiffnesses

At each end, k = relative column stiffness ie. EI/l / S(EI/lcol beams), but assume 50% of beam stiffnesses to allow for cracking

simplification of 5.8.3.2 (3)

Slenderness ratio

l = l /i 0

where i = radius of gyration of section (including reinforcement)

(5.14) Curvature 1/r = K . K .1/rr f 0 where 1/r = f /(0.45d.E )0 yd s = f /103500d andyk K = 1 + bf >= 1 f ef (5.34) (5.37)

Axial load correction factor

K = (n - n)/(n - n ) <= 1 r u u bal where n = N /(A .f )Ed c cd

n = 1 + wu

(5.36)

n may be taken as 0.4bal

Creep correction f = fMef 0Eqp/ M 0Ed (5.19) f from 3.1.3 Is f <= 2, M/N >= h and l <= 75? No check needed Is eyb/exh<= 0.2 or exh/eyb <= 0.2? b = 0.35 + f /200 - l /150ck k = 1f

Second order moment

M = N e2 Ed 2 2 where e = (1/r) l /c2 0

(5.33)

2

c normally p unless constant M0E

Second order moment

M = 02 Effective length (5.16) þ ý ü î í ì ÷÷ ø ö çç è æ + + ÷÷ ø ö çç è æ + + + + = 2 2 1 1 2 1 2 1 0 1 1 . 1 1 ; 10 1 max . k k k k k k k k l l

SLENDERNESS

BUCKLING

MOMENTS

BIAXIAL BENDING Is Column braced? No No No No Yes Yes Yes Yes Is l <= 25(w+0.9)(2 - M /M )?01 02 where w =A f /(A f )>=0.05s yd c cd (from 5.8.3.1) M and M to01 02

include any global second order

effects

Equivalent end moment

M = 0.6M + 0.4M >= 0.4M0E 02 01 02 where M >= M02 01 (5.32)

Determine Rebar and MRd

from N:M interaction charts

ú û ù ê ë é ÷ ø ö ç è æ + ÷ ø ö ç è æ + = 3 1 5 , 2 , 12 11 10 , 1 , 7 . 0 Max n Min n n If a p y d s c d c E d f A f A N n + = . Let , then nd

Repeat all for 2 axis then check 1 £ ÷ ÷ ø ö ç ç è æ + ÷÷ ø ö çç è æ a Rdy Edy a Rdx Edx M M M M (5.39) ÷÷ ø ö çç è æ + + ÷÷ ø ö çç è æ + + = 2 2 1 1 0 45 . 0 1 . 45 . 0 1 5 . 0 k k k k l l Effective length (5.15)

may be taken as for internal columns for edge columns or for corner columns

b

1.15 1.4 1.5 6.4.3 (6)

Enhancement

factor

b

Slab Punching Flowchart

(rectangular columns)

Do adjacent spans differ by <= 25%?

c

₁ = column dim in direction of M.

c

₂ = column width.

c

_x = column dim parallel to edge.

c

_y = column dim normal to edge.

u

₁ = full control perimeter (2d locus from column face, allowing for holes as ).

u

₁* = reduced control perimeter. d = (dx + dy) 6.4.2 (3) (6.33) ½ Internal Column

b

= 1 + k MEd

u

₁ /(VEd W ) ₁ where, if

c

₁/

c <

₂ = 1, k = 0.45 + 0.3(

c

₁/

c

₂ - 0.5) >= 0.45 or if

c

₁/

c

₂ > 1, k = 0.6 + 0.1(

c

₁/

c

₂ - 1) <= 0.8 and to (6.40) Table 6.1 (6.42) 1 2 2 2 1 2 1 1 4 16 2 2 cc c d d dc c W = + + + + p Edge Column

where k is as internal column, but replace c /c with c /2c1 2 1 2

and to

1

(6.46) p a r e W u k u u 1 1 1 1 * + = b 2 2 1 2 1 2 1 1 4 8 4 cc cd d dc c W = + + + +p

(

d c

)

d c u₁*= ₂ +min3 , ₁ +2p 02 . 0 . £ = _{lx ly} l r r r No No No Or Yes Yes Yes

Shear Stress

Links

At Column Face vEd = bVEd/(u .d)₀ (6.54) Is vEd > 0.3fcd[1-fck/250]? Is vEd > vRd,c ? Increase h

Links not required

At Control Perimeter vEd = bVEd/(u .d) 1 (6.39)

(

l ck

)

ctd c c Rd k f f v _, =0.18 100r 1/3 ³0.4 g (6.48)

(

)

e f y w d c R d E d r s w f u v v S A , 1 , 5 . 1 7 5 . 0 -= y w d e f y w d f d f = + £ 4 2 5 0 , where

Asw = total link area on 1st

perimeter (at >=0.3d &<=0.5d) Outer control perimeter

u = V /(v d)out Ed Rd,c

(6.55)

Last link perimeter <= 1.5d from u (polygonal shape) out

st St,max <= 1.5d on 1 and <= 2d on last perimeter y k t r s w f S f c k S A . 5 . 1 8 . 0 m i n , ₌ (9.11) Internal

u

₀ = 2(c1+c2) Edge

u

_{0 =} min(cx+3d,cx+2cy) Corner

u

_{0 =} min(3d,cx+cy) 6.4.5 (2) Corner Column (6.47) where * 1 1 u u = b d c d c d u ÷+p ø ö ç è æ + ÷ ø ö ç è æ = 2 , 5 . 1 min 2 , 5 . 1 min * 1 2 1 If e is towards outside, treat as internal column Min h = 200

Flexural Crack Width Calculation

Flowchart

(rectangular sections)

M = full SLS moment t = age at cracking in days

fs = maximum tension bar diameter S = maximum tension bar spacing s = cement type coefficient

0.20 for rapid hardening high strength 0.25 for normal & rapid hardening 0.38 for slow hardening

MATERIALS

Concrete modulus 0.3 Ecm = 22[(fck+8)/10] Modular ratio ae = Es /Ecm Time factor,

Mean concrete strength at cracking fcm(t) = bcc(t).fcm

Average concrete tensile strength If f >50, fck ct,eff = 2.12ln(1+fcm,t /10) 2/3 otherwise, fct,eff = cm,t 0.3(f - 8) Table 3.1 (3.2) ú ú û ù ê ê ë é ÷ ÷ ø ö ç ç è æ -= t s t cc 2 8 1 exp ) ( b

STRESSES

CRACKING

Neutral axis depth

Concrete stress

Stress in tension steel

(

)

(

)

ú ú û ù ê ê ë é ÷ ø ö ç è æ + + -+ -= d d d x _{ae r r' ae}2 _{r r'}2 _{2ae r r'} 2

(

)(

)

(

)

_ú û ù ê ë é -+ ÷ ø ö ç è æ -= x d x d d As x d bx M e c 2 2 2 1 3 2 a s ÷ ø ö ç è æ -= x x d s s_ca_e s Crack width W = s (e - e )k max sm cm (7.8)

Effective tension area

and 7.3.4 (2)

(

)

(

)

_ú û ù ê ë é -= 2 , 3 , 5 . 2 min , h x h d h A_ceff eff c eff p A As , , = r

Average strain for crack width calculation

(7.9)

(

)

s s s eff p e eff p eff ct s cm sm E E f s r a r s e e 0.6 1 4 . 0 _, , , ³ + -=

-Max final crack spacing, if spacing <=5c+2.5f

otherwise, sr,max = 1.3(h-x)

where k = 0.8 for high bond, otherwise 1.61 and c = cover (7.12) e f f p r k c s , 1 m a x , 2 1 2 5 . 0 4 . 3 r f + = (7.11)

d x cu       + + ≥ ε δ 0.4 0.6 0.0014 Equation (5.10)       + − = cu d x ε δ 0014 . 0 6 . 0 4 . 0 max

(

)

      + − = ∴ cu d x ε δ 0014 . 0 6 . 0 4 . 0 max Where

(

)

.0035 40 90 0009 . 0026 . 4 ≤     − − = ck cu f

ε Code equation is wrongTable 3.1& Fig 3.5

(

)

8 . 0 400 50 8 . 0 − − ≤ = fck λ Equations (3.19) & (3.20)

(

)

0 . 1 200 50 0 . 1 − − ≤ = fck η Equations (3.21) & (3.22) c ck cc cd f f γ α = where 85α_cc =0. Equation (3.15) c ck cd c f b x f bx F γ ηλ λ η = _max = ∴ when 2 max x d z= −λ Figure 3.5 ck c ck cc c conc K bd f x d bx f z F M max max _' 2 2 _=     − = = ∴ λ γ ηλ α       − = ∴ 2 ' max 2 max _d x d x K c cc λ γ ηλ α

Design Equations (Rectangular section)

ck f bd M K = ₂

(

min , '

)

2 2_{f K K} bd x d bx f z F M _ck c ck cc c conc =      − = = λ γ ηλ α

(

)

2 ' , min 2 2 _x dx K K d cc c λ ηλα γ − = →

(

min , '

)

0 2 2 2 = + − cc c K K d dx x ηλα γ λ

(

)

(

)

_      _{− −} = − ± = ∴ 1 1 2 min , ' ' , min 2 1 K K d K K d d x cc c cc c ηα γ λ λ ηα γ or

(

)

                − − − = − = 1 0.5 1 1 2 min , ' 2 d K K x d z cc c ηα γ λ

(

)

        − + = ∴ 1 1 2 min , ' 2 K K d z cc c ηα γ but limit to z ≤0.95d Residual _{M =bd}2_f

(

_{K −K}'

)

_≥0 ck       − = x d x cu sc ' 3 ε ε and       − = x x d cu st ε 3 ε s yk sc sc f f γ ε ≤ = 200000 and s yk st st f f γ ε ≤ = 200000

(

'

)

' d d f M As sc res − = and st sc st res f f As z f M M As= − + '

(

)

      + − = ∴ cu d x ε δ 0014 . 0 6 . 0 4 . 0 max Flange MOR _      − = = 2 f f f cd ORf h d h b f M η If M ≤M_ORf

treat as rectangular section, substituting bf for b If _{M f}M_ORf take

(

)

f f ORf f b bw b M M = − and 2 f f h d z = −

For web section, 

     − = − 2 x d x b f M M _f η _{cd w} λ λ

(

)

max 2 6 10 2 1 1 x d f b M M d x cd w f _≤         _{− −} − = η λ and 2 x d z_w = −λ

Composite concrete lever arm,

(

)



     − − +       − = 2 2 x d M M M h d M M z f f f λ

Remaining equations as rectangular section

Equations for Shear

Ed cd w V f z b ν θ θ

ω =cot +tan = (6.8) where VEd is at support face, and

(

)

250 1 6 . 0 − fck = ν (6.5) 5 . 2 2 4 cot 1≤ θ =ω+ ω2− ≤ 6.2.3 (2)

(

f

)

f b d k d b V _{l ck ctd w} c w ct Rd 100 0.4 18 . 0 1₃ , = _γ ρ ≥ (6.2) where =1+ 200 ≤2 d k = ≤0.02 d b As w l l ρ c ctm ct ctd f f γ α 7 . 0 = (3.16)

(

w cd

)

Ed Rd V f z b V ≥ + = θ θ ν tan cot max

, at support face (6.8) (z may be taken as 0.9d)

max cot min A _s zf V s A s A _sw ywk s Ed sw sw ≤ = ≤ θ γ

(6.7) where VEd is at d from support face

where yk ck w sw f f b s A _min=0.08 _{(9.4) & (9.5)} and ywk s w cd sw f b f s A 2 max=ν γ (6.9)