04 ec2 ws_walraven_ulssls

25 October 2011

1

Limit state design and verification

Joost Walraven

25 October 2011 2

Flat slab on beams

To be considered:

beam axis 2

25 October 2011 3

b

b1

b1

b2 b2

bw

bw

beff,1

beff,2

beff

Determination of effective width (5.3.2.1)

beff = S beff,i + bw b

where beff,i = 0,2bi + 0,1l0 0,2l0 and beff,I bi

l3

l1 l

2

0,15(l1 + l2 )l =0

l0 = 0,7 l2 l0 = 0,15 l2 + l3l0 = 0,85 l1

25 October 2011 4

Cross-section of beam with slab

beff,i = 0,2bi + 0,1l0 0,2l0 and beff,I bi

beff,i = 0,22875 +0,1(0,857125) = 1180 mm (<2875mm)

beff = S beff,i + bw = 21180 + 250 = 2610mm

25 October 2011 5

Beam with effective width

Cross-section at mid-span

25 October 2011 6

Beam with effective width

Cross-section at intermediate support

25 October 2011 7

Maximum design bending moments and shear forces

Maximum design moments

Med in kNm (values for

different load cases)

Maximum shear forces Ved in

kN (values for different load

cases)

25 October 2011 8







cases)

25 October 2011 9

Determination of bending reinforcement using method with simplified concrete design stress block (3.1.7)

As

d

fcd

Fs

x

s

x

cu3

Fc Ac

400

)508,0 ck

(ffor 50 < fck 90 MPa

= 0,8 for fck 50 MPa

= 1,0 for fck 50 MPa

= 1,0 – (fck – 50)/200 for 50 < fck 90 MPa

25 October 2011 10

Factors for NA depth (n) and lever arm (=z) for concrete grade 50 MPa

0.00

0.20

0.40

0.60

0.80

1.00

1.20

M/bd 2fck

Fac

tor

n 0.02 0.04 0.07 0.09 0.12 0.14 0.17 0.19 0.22 0.24 0.27 0.30 0.33 0.36 0.39 0.43 0.46

z 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88 0.87 0.86 0.84 0.83 0.82

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17

lever arm

NA depth

Simplified factors for flexure (1)

25 October 2011 11

Factors for NA depth (=n) and lever arm (=z) for concrete grade 70 MPa

0.00

0.20

0.40

0.60

0.80

1.00

1.20

M/bd 2fck

Fa

cto

r

n 0.03 0.05 0.08 0.11 0.14 0.17 0.20 0.23 0.26 0.29 0.33

z 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.91 0.90 0.89 0.88

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17


lever arm

NA depth

25 October 2011 12

Determination of bending reinforcement (span AB)

Example: largest bending moment in span AB: Med = 89,3

kNm

001,0253722610

103,892

6

2

ck

Ed

fbd

M

Read in diagram: lever arm factor = 0,99, so:

26

, 56343537298,0

103,89mm

fz

MA

yd

Edreqsl

25 October 2011 13

Determination of bending reinforcement (span AB)

Example: largest bending moment in span AB: Med = 89,3 kNm

Moreover, from diagram: neutral axis depth factor is 0,02, so xu = 0,02180 = 4 mm. So height of compression zone < flange thickness (180 mm), OK

25 October 2011 14

Determination of bending reinforcement (intermediate support B

Bending moment at support B: Med = 132,9 kNm

154,025372250

100,1322

6

2

ck

Ed

fbd

M

Read: lever arm factor 0,81

26

101443537281,0

109,132mm

fz

MA

yd

Edsl

25 October 2011 15







cases)

Shear may be determined at distance d from support, so Ved 115 kN

25 October 2011 16

Design of beams for shear (6.2.2)

First check (6.2.2): if VEd ≥ VRd,c then shear reinforcement is

required:

where: fck in Mpa

k = with d in mm

l =

with d = 372mm, bw = 250mm, l = 0,61%, fck = 25MPa

so shear reinforcement is required

bdfkV cklccRd

3/1

, )100()/18,0(

0,2200

1 d

02,0db

A

w

sl

kNkNV cRd 1158,4710372250)2561,0(73,1)5,1/18,0( 33/1

,

25 October 2011 17

Expressions for shear capacity at stirrup yielding (VRd,s) and web crushing (VRd,max)

Vu,2

cc1 = f

= fc

Vu,3

s

z

z cot

Afswyw

Vu,2

c c1 = f

= f c

Vu,3

s

z

z cot

A fsw yw

For yielding shear reinforcement: VRd,s = (Asw/s) z fywd cot with between 450 and 21,80

(1 cot 2,5)

At web crushing: VRd,max = bw z fcd /(cot + tan) with between 450 and 21,80

(1 cot 2,5) = 0.6 (1- fck/250)

25 October 2011 18

Design of beams for shear

Basic equation for determination of shear reinforcement:

VEd,s = (Asw/s) z fywd cot

With Ved,s = 115000 N, fywd = 435 Mpa, z = 0,9d, d = 372 mm and cot = 2,5 it is

found that

Asw/s ≥ 0,32 e.g. stirrups 6mm – 175mm

Check upper value of shear capacity (web crushing criterion)

VRd,max = bw z fcd /(cot + tan) with bw = 250mm, d = 372mm, z = 0,9d, = 0,6(1-fck/250) = 0,54, fcd = 25/1,5 = 13,3 Mpa and cot = 2,5 it is found that VRd,max = 1774 kN which is much larger than the design shear force of 115 kN

25 October 2011 19

Stirrup configuration near to support A

25 October 2011 20

Transverse shear in web-flange interface

25 October 2011 21

Shear between web and flanges of T-sections

25 October 2011 22

Shear between web and flanges of T-sections

Strut angle : 1,0 ≤ cot f ≤ 2,0 for compression flanges (450 f 26,50 1,0 ≤ cot f ≤ 1,25 for tension flanges (450 f 38,60) No transverse tension ties required if shear stress in interface vEd = Fd/(hf·x) ≤ kfctd (recommended k = 0,4)

25 October 2011 23

Check necessity of transverse reinforcement

MPahz

V

b

bv

f

Ed

eff

f

Ed 86,01803729,0

115000

2610

1180

No transverse reinforcement required if vEd 0,4fctd For C25/30 fctd = fctk/c =1,8/1,5 = 1,38 Mpa, so the limit value for interface shear is 0,4fctk = 0,41,38 = 0,55 MPa. Transverse shear reinforcement is required at the end of the beam.

25 October 2011 24







cases)

25 October 2011 25

Areas in beam axis 2 where transverse reinforcement is required

25 October 2011 26

Areas in beam axis 2 where transverse reinforcement is required

25 October 2011 27

Example: transverse reinforcement near to support A

Required transverse reinforcement for Ved = 115 kN

e.g. 8 – 250 (=0,20 mm2/mm)

mmmmzf

V

b

b

s

A

fyd

Ed

eff

fst /18,00,2

1

435335

115000

2610

1180

cot

1 2

25 October 2011 28

Design of slabs supported by beams

25 October 2011 29

Design of slabs supported by beams

Load transmission from slabs to beams

Simplified load transmission model Dead load G1 = 0,1825 = 4,5 kN/m2 Partitions, etc. G2 = 3,0 kN/m2 Variable load Q = 2,0 kN/m2 Ged = 1,3(4,5 + 3,0) = 9,75 kN/m2 Qed = 1,52,0 = 3,0 kN/m2

25 October 2011 30

Load transfer from slabs to beams

Loading cases on arbitrary strip (dashed in left figure)

25 October 2011 31

Longitudinal reinforcement in slabs on beams

Examples of reinforced areas

25 October 2011 32

Floor type 2: flat slab d = 210 mm

From floor on beams to flat slab: replace beams by strips with

the same bearing capacity

25 October 2011 33

From slab on beams to flat slab

hidden strong strip

-Strips with small width and large reinforcement ratio favourable for punching

resistance

- Strips not so small that compression reinforcement is necessary

25 October 2011 34

Methods of analysis: Equivalent Frame

Analysis – Annex I

(Informative)

lx (> ly)

ly

ly/4 ly/4

ly/4

ly/4

= lx - ly/2

= ly/2

= ly/2 A

B

B

A – Column strip

B – Middle strip

Negative moments Positive moments

Column Strip

60 - 80%

50 - 70%

Middle Strip

40 - 20%

50 - 30% Note: Total negative and positive moments to be resisted by the column and

middle strips together should always add up to 100%.

25 October 2011 35

Flat slab with “hidden strong strips”

25 October 2011 36

Punching shear control column B2

25 October 2011 37

Punching column B2

Junction column to slab

Vertical load from slab to

column Ved = 705 kN

Simplified assumptions for eccentricity factor according to EN 1992-1-1 Cl. 6.4.3

= 1,4

= 1,5

= 1,15

C

B A

25 October 2011 38

How to take account of eccentricity (simplified case)

du

Vv

i

EdEd Or, how to determine in equation

= 1,4

= 1,5

= 1,15

C

B A

Only for structures where

lateral stability does not

depend on frame action and

where adjacent spans do

not differ by more than 25%

the approximate values for

shown left may be used:

25 October 2011 39

Upper limit value for design punching shear stress in design

cdRdEd

Ed fvdu

Vv

4,0max,

0

At the perimeter of the loaded area the maximum punching shear stress should satisfy the following criterion:

where: u0 = perimeter of loaded area = 0,6[1 – fck/250]

25 October 2011 40

Punching shear column B2

1. Check of upper limit value of punching shear capacity

Further data: dy = 210 – 30 – 16/2 = 172mm

dz = 210 – 30 – 16 – 16/2 = 156 mm

Mean effective depth 0,5(172 + 156) = 164mm

= 0,6(1 + fck/250) = 0,54

vRd,max = 0,4fcd = 0,40,54(25/1,5) = 3,60 Mpa

vEd = Ved/(u0d) = 1,15705000/(4500164)

= 2,47 Mpa < 3,60 Mpa

25 October 2011 41

Definition of control perimeters

The basic control perimeter u1 is taken at a distance 2,0d from the loaded area and should be constructed as to minimise its length

Length of control perimeter of column 500x500mm: u = 4500 + 22164 = 4060 mm

25 October 2011 42

Punching shear capacity column B2

Punching shear stress at perimeter:

No punching shear reinforcement required if:

MPadu

Vv Ed

Ed 22,11644060

70500015,1

1

cRdEd vv ,

25 October 2011 43

Limit values for design punching shear stress in design

cRdEd vv ,

The following limit values for the punching shear stress are used in design: If no punching shear reinforcement required

)()100( 1min1

3/1

,, cpcpcklcRdcRd kvkfkCv

where:

where: k1 = 0,10 (advisory value)

25 October 2011 44

Punching shear capacity of column B2

No punching shear reinforcement required if vEd < vRd,c

With CRd,c = 0,12

k = 1 + (200/d) = 1 + (200/164) < 2, so k = 2,0

= (xy) = (0,860,87) = 0,865%

fck = 25 Mpa

It is found that vRd,c = 0,67 Mpa

Since vEd = 1,22 MPa> 0,67 MPa punching

shear reinforcement should be applied.

3/1

,, )100( cklcRdcRd fkCv

25 October 2011 45

Punching shear reinforcement

Capacity with punching shear reinforcement

Vu = 0,75VRd,c + VS

Shear reinforcement within 1,5d from column is accounted for with fy,red = 250 + 0,25d(mm) fywd

25 October 2011 46

kd

Outer control

perimeter

Outer perimeter of shear

reinforcement

1.5d (2d if > 2d from

column)

0.75d

0.5dA A

Section A - A

0.75d

0.5d

Outer control

perimeter

kd

Punching shear reinforcement

The outer control perimeter at which

shear reinforcement is not required,

should be calculated from:

uout,ef = VEd / (vRd,c d)

The outermost perimeter of shear

reinforcement should be placed at a

distance not greater than kd (k =

1.5) within the outer control

perimeter.

47

Design of punching shear reinforcement

The necessary punching shear reinforcement per perimeter is found from:

1 ,

,

( 0,75 )

1,5

r Ed Rd c

sw

ywd ef

u s v vA

f

with:

vEd = 1,22 N/mm2

vRd,c= 0,67 N/mm2

u1 = 4060 mm

fyd,ef = 250 + 0,25 164 = 291 N/mm2

sr = 0,75 164 = 123 mm 120 mm

It is found that: Asw = 800 mm2 per

reinforcement perimeter

25 October 2011 48

Design of column B2 for punching shear

Determination of the outer perimeter for which vEd = vRd,c The distance from this perimeter to the edge of the column follows from: The outer punching shera reinforcement should be at a distance of not more than 1,5d from the outer perimeter. This is at a distance 5,22d – 1,5d = 3,72d = 610 mm. The distance between the punching shear reinforcement perimeters should not be larger than 0,75d = 0,75164 = 123mm.

mmdvVu cRdEdout 737816467,0/()70500015,1()/( ,

dmmhua out 22,5856)2/()50047378(2/)4(

25 October 2011 49

Punching shear design of slab at column B2

Perimeters of shear reinforcement

25 October 2011 50

Design of column B2

25 October 2011 51

General background: Second order effects at axial loading (EC2, 5.8.2, 5.8.3.1 & 5.8.3.3

- Second-order effects may be ignored if they are smaller than 10% of the corresponding 1th order effects

- “Slenderness”: is defined as = l0/i where i = (l/A) so for rectangular cross-section = 3,46 l0/h and for circular cross section = 4l0/h - Second order effects may be ignored if the slenderness is smaller than the limit value lim

- In case of biaxial bending the slenderness should be calculated for any direction; second order effects need only to be considered in the direction(s) in which lim is exceeded.

25 October 2011 52

General background: “Slender” versus “short” columns

Definition of slenderness

)/(

00

AI

l

i

l

l0 effective height of the column i radius of gyration of the uncracked concrete section I moment of inertia around the axis considered A cross-sectional area of column

Basic cases EC2 fig. 5.7

25 October 2011 53

General background: when is a column slender?

Relative flexibilities of rotation-springs at the column ends 1 en 2 k = (/M)(EI/l) where = rotation of restraining members for a bending moment M EI = bending stiffness of compression member l = height of column between rotation-springs

25 October 2011 54

General background: when is a column slender?

)45,0

1)(45,0

1(5,02

2

1

10

k

k

k

kll

)101(21

210

kk

kkll

Determination of effective column height in a frame

For unbraced frames: the largest value of:

and

where k1 and k2 are the relative spring stiffnesses at the ends of the column, and l is the clear height of the column between the end restraints

For braced frames: Failing

column

Non failing

column

End 1

End 2

Non-failing

column

1 2

0

1 2

1 11 1

k kl l

k k

25 October 2011 55

General background: determination of effective column length (1) (5.8, 5.8.3.2)

Failing

column

Non failing

column

End 1

End 2

Non failing

column Simplifying assumption: * The contribution of the adjacent “non failing ” columns to the spring stiffness is ignored (if this contributes in a positive sense to the restraint) * for beams for /M the value l/2EI may be assumed (taking account of loss of beam stiffness due to cracking)

Assuming that the beams are symmetric with regard to the column and that their dimensions are the same for the two stories, the following relations are found: k1 = k2 = [EI/l]column / [SEI/l]beams = [EI/l]column / [22EI/l]beams = 0,25 where: = [EI/l]column / [EI/l]beams

25 October 2011 56

General background: Determination of effective column length (2) (5.8, 5.8.3.2)

or

0 (fixed end)

0.25

0.5

1.0

2.0

(pinned end)

k1 = k2 0 0.0625 0.125 0.25 0.50 1.0

l0 for braced

column

0.5 l

0.56 l

0.61 l

0.68 l

0.76 l

1.0 l

l0 for

unbraced column:

Larger of the values in the

two rows

1.0 l

1.14 l

1.27 l

1.50 l

1.87 l

∞

1.0 l

1.12 l

1.13 l

1.44 l

1.78 l

∞

The effective column length l0 can, for this situation

be read from the table as a function of )/(

00

AI

l

i

l

25 October 2011 57

General background: when is a column slender ?

nCBA /20lim

)2,01/(1 efA

21B

A column is qualified as “slender”, which implies that second order effects should be taken into account, if lim. The limit value is defined as:

where:

mrC 7,1

ef = effective creep factor: if unknown it can be assumed that A = 0,7 = Asfyd/(Acfcd): mech. reinforcement ratio, if unknown B = 1,1 can be adopted n = NEd/(Acfcd);

rm = M01/M02: ratio between end-moments in column, with M02 M01

25 October 2011 58

Design of column B2

Configuration of variable load on slab

B2

25 October 2011 59

Determination of columns slenderness

First step: determination of rotational spring stiffness at end of column:

Column: EI/l = 0,043106 kNm2

Beam: EI/l = 0,052106 kNm2

K1 = [EI/l]col/[SEI]beams = 0,043/(20,052) = 0,41

If the beam would be cracked a value of 1,5 k1 would be more realistic. This would result

in l0 = 0,80l = 3,2m.

llk

k

k

kll 70,0)

02,1

41,01(5,0)

45,01)(

45,01(5,0 2

2

2

1

10

25 October 2011 60

Verification of column slenderness

Actual slenderness of column:

Limit slenderness according to

EC2, Cl. 5.8.3.1:

With the default values A = 0,7 B=1,1 C = 0,7 whereas the value n follows from

n= Ned/(Acfcd) = 438400/(500220) = 0,88, the value of lim becomes:

Because the actual slenderness of the column is larger than the limit slenderness second

order effects have to be taken into account.

1,225,0

2,346,346,3 0

h

l

n

CBA

20lim

5,1188,0

7,01,17,020lim

25 October 2011 61

General : Method based on nominal curvature

Mt = NEd (e0 + ei + e2) Different first order eccentricities e01 en e02 At the end of the column can be replaced by an equivalent eccentricity e0 defined as:

e0 = 0,6e02 + 0,4e01 0,4e02

e01 and e02 have the same sign if they lead to tension at the same side, otherwise different signs. Moreover e02 e01

25 October 2011 62

General : Method based on the nominal curvature

2

0lvei

Mt = NEd (e0 + ei + e2) The eccentricity ei by imperfection follows from (5.2(7)):

where l0 = effective column height around the axis regarded

200

1

100

1

lv

where l = the height of the column in meters

25 October 2011 63

General: Method based on nominal curvature

1)150200

35,0(1 efckf

K

Mt = NEd (e0 + ei + e2) The second order eccentricity e2 follows from:

where

0,1

balud

Edudr

NN

NNKand

2

02 2 0, 45

yd

r

le K K

d

25 October 2011 64

Calculation of bending moment including second order effects

The bending moment on the column follows from:

e0 = Med/Ned = 42/4384 = 0,010m = 10mm . However, at least the maximum value of

{l0/20, b/20 or 20mm} should be taken. So, e0 =b/20 = 500/20 = 25mm.

ei = i(l0/2) where i = 0hm 0 = 1/200 rad, h =2/l = 1 and

so that ei =(1/200)(4000/2) = 10mm

where and

and finaly

)( 20 eeeMM iEdt

1)1

11(5,0)

11(5,0

mm

2

02 2 0, 45

yd

r

le K K

d

effckf

K

)150200

35,0(1

tEdEqpeff MM ,00 )/( balu

Edur

nn

nnK

25 October 2011 65

Calculation of bending moment including second order effects

where (estimated value = 0,03)

so Kr = 0,62 and finaly:

4,0225,1

23,0

0

0

Ed

Eqp

effM

M

14,14,0)150

9,22

200

3035,0(1)

15020035,0(1 eff

ckfK

balu

Edur

nn

nnK

65,1

20

43503,011

cd

yd

uf

fn

88,0cdc

EdEd

fA

Nn 4,0baln

mmd

lKKe

yd

r 1445425,0

1017,2320062,015,1

45,0

3

2

2

2

2

02

25 October 2011 66

Calculation of bending moment including second order effects and reinforcement

Determination of reinforcement

kNmeeeNM Edtot 21510)141025(4384)( 3

210

58,030500

43840002

ck

Ed

bhf

N

06,030500

21500032

cd

Ed

fbh

M

From diagram: So: (1,4%)

15,0ck

yks

bhf

fA

22

3448435

3050020,0mmAs

25 October 2011 67

Design of shear wall

25 October 2011 68

Design of shear wall

The stability of the building is ensured by two shear walls (one at any end of the

building) and one central core

shear wall 1 core shear wall 2 I = 0,133 m4 I = 0,514 m4 I = 0,133 m4 Contribution of shear wall 1: 0,133/(20,133 + 0,514) = 0,17 (17%)

25 October 2011 69

Second order effects to be regarded?

“If second order effects are smaller than 10% of the first

order moments they can be neglected”.

Moment magnification factor:

]1/

1[0

EdB

EdEdNN

MM

2

2

)12,1( l

EINB

lqN vEd

NB is the buckling load of the system sketched, l = height of building, qv = uniformely distributed load in vertical direction, contributing to 2nd order deformation.

qv

25 October 2011 70


The moment magnification factor is:

where n = NB/NEd

Requiring f < 1,1 and substituting the corresponding

values in the equation above gives the condition:

(Eq.1)

Assuming 30% of the variable load as permanent, the

load per story is 3014,2510,65 = 4553 kN. Since the

storey height is 3m, this corresponds with qv=1553

kN/m’ height.

With l = 19m, E = 33.000/1,2 = 27.500 MPA, I = 0,78 m4

1

n

nf

84,0EI

lql vEd

84,070,078,05,27

191518

10

193

25 October 2011 71


However, in the calculation it was assumed that the stabilizing

elements were not cracked. In that case a lower stiffness

should be used.

For the shear wall the following actions apply:

Max My = 66,59 kNm = 0,0666 MNm

Corresp. N = -2392,6 kN = 2,392 MN/m2

So the shear wall remains indeed uncracked and 2nd order

effects may be ignored.

2/78,425,02

2392mMNN

2/99,301667,0

0666,0mMN

W

MM

25 October 2011 72

Alternative check by Eq. 5.18 in EC2

According to Cl. 5.8.3.3 of EC-22nd order effects may be ignored if:

Where

FV,Ed total vertical load (both on braced and unbraced elements)

ns number of storeys

L total height of building above fixed foundation

Ecd design E-modulus of the concrete

Ic moment of intertia of stabilizing elements

The advisory value of the factor k1 is 0,31. If it can be shown that the

stabilizing elements remain uncracked k1 may be taken 0,62

21,6,1 L

IE

n

nkF ccd

s

sEdV

S

25 October 2011 73

Alternative check by Eq. 5.18 in EC2

Verification for the building considered:

Condition:

or: 27.318 29.084

so the condition is indeed fullfilled

21,6,1 L

IE

n

nkF ccd

s

sEdV

S

2

6

19

78,0105,27

6,16

662,045536

25 October 2011 74

Monodirectional slab with embedded lighting elements

25 October 2011 75

Bearing beams in floor with embedded elements

25 October 2011 76

Design for bending of main bearing beam in span 1-2

Med = 177,2 kNm

Effective width:

Midspan: beff = 2695 mm

from diagram z = 0,98d = 365mm

bbbb wieffeff S , 0, 1,02,0 lbb iieff

02,0253722695

102,1722

6

2

ck

Ed

fbd

M

26

1367435365

102,172mm

fz

MA

yd

Edsl

25 October 2011 77

Design for bending of main bearing beam in span 1-2 (intermediate support)

Med = 266 kNm

Effective width:

Internal support: beff = 926 mm

At intermediate support: !?


31,025372250

102662

6

2

ck

Ed

fbd

M

25 October 2011 78

Factors for NA depth (n) and lever arm (=z) for concrete grade 50 MPa

0.00

0.20

0.40

0.60

0.80

1.00

1.20

M/bd 2fck

Fac

tor

n 0.02 0.04 0.07 0.09 0.12 0.14 0.17 0.19 0.22 0.24 0.27 0.30 0.33 0.36 0.39 0.43 0.46

z 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88 0.87 0.86 0.84 0.83 0.82

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17

lever arm

NA depth


25 October 2011 79


Med = 266 kNm

Effective width:


At intermediate support compression reinforcement required:

e.g. 320


222

826)35372(435

37225025)167,031,0(

)'(

)'(mm

ddf

bdfKKA

yd

cksc

25 October 2011 80


Med = 266 kNm

Effective width:


Calculation of tensile reinforcement:

For K = 0,167 z = 0,81372=301 mm

e.g. 720 = 2198 mm2


26

2031435301

10266mm

fz

MA

yd

Edsl

25 October 2011 81

Design of one-way beams with embedded elements

Loads:

G1 = 2,33 kN

G2 = 3,0

Q = 2,0

Qed = 1,3(2,33+3,0) + 1,52,0=9,93 kN/m2

25 October 2011 82

Beams with embedded elements: design for bending at intermediate support

Compression reinforcement required

In any rib 203 mm2

167,0294,025189240

10632

6

2

ck

Ed

fbd

Mk

222

406)35189(435

18924025)167,0294,0(

)'(

)'(mm

ddf

bdfKKA

yd

cksc

25 October 2011 83

Beams with embedded elements: design for bending at intermediate support

Tensile reinforcement: for K = 0,167 z = 0,8189=151 mm

e.g. 12-100 = 1130 mm2 or

26

959435151

1063mm

fz

MA

yd

Edsl

10-75 = 1040 mm2

25 October 2011 84

Beams with embedded elements: design for bending at midspan

From diagram z = 0,95d = 0,95189 = 180 mm

044,0251891000

102,392

6

2

ck

Ed

fbd

MK

26

501435180

102,39mm

fz

MA

yd

Edsl

251 mm2 per rib

25 October 2011 85

Deflection control by slenderness limitation

23

0ck

0ck 12,35,111

ffK

d

l if 0 (7.16.a)

0

ck0

ck

'

12

1

'5,111

ffK

d

lif > 0 (7.16.b)

l/d is the limit span/depth

K is the factor to take into account the different structural systems

0 is the reference reinforcement ratio = fck 10-3

is the required tension reinforcement ratio at mid-span to resist the moment

due to the design loads (at support for cantilevers)

’ is the required compression reinforcement ratio at mid-span to resist the

moment due to design loads (at support for cantilevers)

For span-depth ratios below the following limits no further checks is needed

Deflection control by slenderness limitation

)(

500

310

,

,

provs

reqsyk

s

A

Af

The expressions given before (Eq. 7.6.a/b) are derived based on many different

assumptions (age of loading, time of removal of formwork, temperature and humidity

effects) and represent a conservative approach.

The coefficient K follows from the static system:

The expressions have been derived for an assumed stress of 310 Mpa under the quasi

permanent load. If another stress level applies, or if more reinforcement than required

is provided, the values obtained by Eq. 7.16a/b can be multiplied with the factor

where s is the stress in the reinforcing steel at mid-span

Rules for large spans

For beams and slabs (no flat slabs) with spans larger than 7m, which support partitions liable to damage by excessive deflections, the values l/d given by Eq. (7.16) should be multiplied by 7/leff (leff in meters).

For flat slabs where the greater span exceeds 8,5m, and which support partitions to be damaged by excessive deflections, the values l/d given by expression (7.16) should be multiplied by 8,5/ leff.

25 October 2011 88

0

10

20

30

40

50

60

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Reinforcement percentage (As/bd)

lim

itin

g s

pan

/dep

th r

ati

o

fck =30 40 50 60 70 80 90

Eq. 7.16 as a graphical representation, assuming K = 1 and s = 310 MPa

25 October 2011 89

Tabulated values for l/d calculated from Eq. 7.16a/b The table below gives the values of K (Eq.7.16), corresponding to the structural system. The table furthermore gives limit l/d values

for a relatively high (=1,5%) and low (=0,5%) longitudinal

reinforcement ratio. These values are calculated for concrete C30 and s = 310 MPa and satisfy the deflection limits given in 7.4.1 (4) and (5).

Structural system K = 1,5% = 0,5%

Simply supported slab/beam

End span

Interior span

Flat slab

Cantilever

1,0

1,3

1,5

1,2

0,4

l/d=14

l/d=18

l/d=20

l/d=17

l/d= 6

l/d=20

l/d=26

l/d=30

l/d=24

l/d=8

25 October 2011 90

Beams with embedded elements: design for bending at midspan

From diagram z = 0,95d = 0,95189 = 180 mm

044,0251891000

102,392

6

2

ck

Ed

fbd

MK

26

501435180

102,39mm

fz

MA

yd

Edsl

251 mm2 per rib (e.g. 214 = 308 mm2)

25 October 2011 91

Control of deflection slab with embedded elements

Reinforcement ratio at midspan = Asl/bed = 501/(1000189) = 0,265%

According to Cl. 7.4.2(2) no detailed calculation is necessary if the l/d ratio of

the slab is smaller than the limit value:

So:

23

0ck

0ck 12,35,111

ffK

d

l

49)1265,0

5,0(52,3

256,0

5,0255,111[3,1 2/3

d

l

25 October 2011 92

Control of deflection slab with embedded elements

Moreover correction for real steel stress versus 310 N/mm2 as default value:

Quasi permanent load: Qqp=2,33 + 3,0 + 0,32,0 = 5,93

Ultimate design load: Qed = 9,93

Steel stress under quasi permanent load 2 = (5,93/9,93)435 = 260 Mpa

Corrected value of l/d is:

Actual value is l/d = 7,125/189 = 38 so OK

4,5849260

310)(

310

,

d

l

d

l

qps

25 October 2011 93

Theory of crack width control

sr

se

steel stress

concrete stress

ctmf

t t

w

The crack width is the difference between the steel deformation and the concrete deformation over the length 2lt, where lt is the “transmission length”, necessary to build-up the concrete stength from 0 to the tensile strength fctm. Then the maximum distance between two cracks is 2lt (otherwise a new crack could occur in-between). It can be found that the transmission length is equal to:

bm

ctmt

fl

4

1

25 October 2011 94

EC-formula’s for crack width control

For the calculation of the maximum (or characteristic) crack width, the difference between steel and concrete deformation has to be calculated for the largest crack distance, which is sr,max = 2lt. So ( )

cm sm k w

r s

max ,

where sr,max is the maximum crack distance and (sm - cm) is the difference in deformation between steel and concrete over the maximum crack distance. Accurate formulations for sr,max and (sm - cm) will be given

sr

se

steel stress

concrete stress

ctmf

t t

w

Eq. (7.8)

25 October 2011 95

EC-2 formula’s for crack width control

where: s is the stress in the steel assuming a cracked section e is the ratio Es/Ecm p,eff = (As + Ap)/Ac,eff (effective reinforcement ratio including eventual prestressing steel Ap is bond factor for prestressing strands or wires kt is a factor depending on the duration of loading (0,6 for short and 0,4 for long term loading)

(Eq. 7.9)

s

s

s

effpe

effp

effct

ts

cmsmEE

fk

6,0

)1( ,

,

,

25 October 2011 96

EC-2 formulae for crack width control

For the crack spacing sr,max a modified expression has been derived, including the concrete cover. This is inspired by the experimental observation that the crack at the outer concrete surface is wider than at the reinforcing steel. Moreover, cracks are always measured at the outside of the structure (!)

25 October 2011 97


Maximum final crack spacing sr,max

eff p r k k c s

, 2 1 max , 425 . 0 4 . 3

(Eq. 7.11)

where c is the concrete cover is the bar diameter k1 bond factor (0,8 for high bond bars, 1,6 for bars with an effectively plain surface (e.g. prestressing tendons) k2 strain distribution coefficient (1,0 for tension and 0,5 for bending: intermediate values van be used)

25 October 2011 98


In order to be able to apply the crack width formulae, basically valid for a concrete tensile bar, to a structure loaded in bending, a definition of the “effective tensile bar height” is

necessary. The effective

height hc,ef is the minimum of: 2,5 (h-d) (h-x)/3 h/2

d h

gravity lineof steel

2.5

(h

-d)

<h

-xe

3

eff. cross-section

beam

slab

element loaded in tension

ct

smallest value of

2.5 . (c + /2) of t/2

c

smallest value of

2.5 . (c + /2)

of(h - x )/3

e

a

b

c

25 October 2011 99

EC-2 requirements for crack width control (recommended values)

Exposure class RC or unbonded PSC members

Prestressed members with bonded tendons

Quasi-permanent load

Frequent load

X0,XC1 0.3 0.2

XC2,XC3,XC4 0.3

XD1,XD2,XS1,XS2,XS3

Decompression

25 October 2011 100

Crack width control at intermediate support of slabs with embedded elements

Assumption: concentric tension of upper slab of 50 mm.

Steel stress s,qp under quasi permanent load:

Reinforcement ratio: s,eff = Asl/bd = 959/(100050) = 1,92%

Crack distance:

MPafA

A

Q

Qyd

provs

reqs

Ed

qp

qps 22043585,0597,0,

,

,

mmkkkckeffs

s 2770192,0

12425,00,18,0194,3

,

4213max,

25 October 2011 101

Crack width control at intermediate support of slabs with embedded elements

Average strain:

Characteristic crack width:

so, OK

s

s

s

effpe

effp

effct

ts

cmsmEE

fk

6,0

)1( ,

,

,

31079,0000.200

)0192,071(0192,0

6,24,0220

cmsm

mmmmsw cmsmrk 30,018,01079,0227}{ 3

max,

25 October 2011 102

Crack width at mid-span beams with embedded elements

Cross-section of tensile bar

Height of tensile bar: smallest value of 2,5(h-d), (h-x)/3 or h/2.

Critical value 2,5(h-d) = 2,529 = 72 mm.

s,eff = Asl/bheff = 308/(12072) = 3,56%

MPaf

A

A

Q

Qyd

provs

reqs

Ed

qp

qps 21043581,0597,0,

,

,

25 October 2011 103

Crack width at mid-span beams with embedded elements

Cross-section of tensile bar

mmkkkckeffs

s 1560356,0

12425,05,08,0294,3

,

4213max,

3

,

,

,

1087,0000.200

)0356,071(0356,0

6,24,0210)1(

s

effpe

effp

effct

ts

cmsmE

fk

mmsw cmsmrk 14,01087,0156)( 3

max, OK

25 October 2011 104

Different cultures: different floors

04 ec2 ws_walraven_ulssls

Documents