National codes in relation with the CEP-FIP model code

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DELFT UNIVERSITY OF TECHNOLOGY

Faculty of Civil Engineering

Department of Structural Concrete

Report 5-80-D4

NATIONAL CODES IN RELATION

WITH THE CEB-FIP MODEL CODE

Ing. J.P. Straman

^ | ^ ^ ^ ^ # March 1981

^ A ^ ^ ^ ^ ^ ^ r RESEARCH REPORT

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Technische Hogeschool Afdeling: Civiele Techniek

Stevinweg 1 postbus 5048 2600 GA Delft

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National Codes in relation with

the CEB-FIP Model Code.

. a

I n g . J . P . Straman March 1981

Technische Hogeschool Delft Afdeling der Civiele Techniek

u i t g a v e

1

-Contents

1 Introduction 2 Symbols

3 Load-prescriptions 3.1 Characteristic imposed loads 3.2 Wind loads

4 Comparison of corresponding articles U.l Grades of concrete

4.2 Partial factors of safety

4.3 Modulus of elasticity 4.4 Shrinkage

4.5 Creep

4.6 Stress-strain diagrams 4.7 Relation moment-normal force

4.8 Bending

4.9 Shear

4.10 Punching shear 4.11 Staggering rule

4.12 Effective width of T-beams 4.13 Deflection

4.14 Minimum cover

4.15 Minimum area of reinforcement 5 Comparison calculation

5.1 Design methods of plane, unbraced frames

5.2 Office building 5.3 General data

5.4 Ratio total moments- first order moments 5.5 Moments in beams

5.6 Theoretical quantities of reinforcement

5.7 Horizontal displacement

5.8 Ratio failure load- working load 6 Conclusions.

- 2 - . . •

Introduction

In the treaty of Rome an article has been included to abolish commercial hindrances between the different countries of the E.C. This means among other things, to smooth out the differences between national concrete codes, so that contractors can build in

every country of the E.C. and dealing in building products will be possible from one country to another without problems of national

kind.

To get information about these differences, a comparison has been

made of some West-European codes for structural concrete and of the CEB code, all as they were available in 1978.

The codes investigated are:

- CEB-Bulletin 124/125 . ,, - CP 110 England

- CCBA 68 • France

- DIN 1045 Germany - VB 1974 Netherlands.

To compare every subject of these codes is a tremendous work, so that this comparison has been limited to the design of the bearing structure of an office building, according to the above mentioned codes.

The required corresponding articles for the calculations have been investigated and some of them have been included in this report.

Because it was the intention to compare concrete codes - and not load prescriptions - equal loads have been taken for the calculation of the building. However to get information about the characteristic loads, prescribed in the different countries, these values have been investigated too.

So the report can be devided into:

- Load-prescriptions

- Comparison of corresponding articles of the different concrete codes

- Comparison of detailed analysis and design of reinforcement of the unbraced bearing structure of an office building.

- Conclusions

The following students have contributed to this report: J. Laurens

R.B.G.H. Markslag J. Vlagsma

3 -Symbols A area of concrete c A area of reinforcement s

E modulus of elasticity of concrete em

E modulus of elasticity of steel s

b width

d effective height of the section e eccentricity of first order

o

e eccentricity due to creep c

e„ eccentricity of second order e total eccentricity

tot

h total height of the section i minimum radius of gyration 1 effective buckling length

s standard deviation z lever arm

f nominal compressive strength of concrete f , characteristic strength of concrete

CK

f . mean value compressive strength of concrete

cm '^ °

f concrete design strength cd

f , steel design strength

f characteristic strength of steel yk

e strain value of concrete c

E strain value of steel s

X slenderness

Y partial factor of safety of concrete Y partial factor of safety of steel

Y partial safety-coefficient of permanent actions

Y " " " variable actions Y " " " wind actions.

4

-3 Load-prescriptions

2 3.1 Characteristic imposed loads (kN/m )

1

Use

Houses Offices Stores

1

Warehouses Code/Country CEB England France Germany Netherlands CEB England France Germany Netherlands CEB England France Germany Netherlands CEB England France Germany Netherlands floors 1,5 1,5 1,75 2,0 1,5 2,5 2,5 2,5 2,0 2,0 depending 4,0 4,0 2,0 3,0 depending stairs/corridors balconies

1 1

1,5 4,0 2,5 3,5 2,0 2,5 4,0 4,0 3,5 2,5 on loaded area 4,0 5,0 3,0 on loaded area > 2,4 metre storage height 5,0 5,0 4,0 5,0 4,0 1,5 1,5 3,5 3,5 2,5 2,5 2,5 3,5 2,5

and type of good

4,0

3,5

3,0

and type of gooc

3,5

5

-If the loads according to the CEB are assumed to be 100%, the following values can be found:

houses offices floor loads (%) CEB 100 100 England 100 100 France 117 100 Germany 133 80 Netherlands 100 80 houses offices loads stairs/corridors (%) CEB 100 100 England 267 160 France 167 150 Germany 234 140 Netherlands 133 100 1

3.2 WINDLOADS 5

-

+-CEB

pressure \ suction J.O. • • »

England France Germany Netherlands

AL

M^

_M-^

*T~*1 I *1 I n I — T ~ * l I *l~*l I 'T"*! » - r ——fc • • > — • • —• • —« « _ • h » *-~fc • —» • —t • —• — — » » * — • • —• — — « - * « —• k • V — • 1 — • # —* •• —» k • » — • • — • 9-4 4*—r l — — » » ' »» t J—J * J—J I 4—»

^^Mè^

.60 E 35 30 25 20 15 10 « ^ / • i ^ - — r _{I I I I} UOO 1600 1B00 1900 > total wind load

I N/m2) CCBA '68: the wind load is equally devided over the

height and depends on the height of the building.

WINDMOMENT 60 r 55 ^ 5 0 J C 40 35 30 25 20 15 10 5 500 1000 1500 2000 2500 3000 ^ ^ windmoment (kNnYm) wind moment DibliothooU

afd. Civiele Techniek T.H. Stevinweg-1 - Delft

4 Comparison of corresponding articles

4.1 Grades of Concrete

The description of the quality of concrete often exists of a letter followed by a number, similar to the required characteristic compressive strength of the concrete of a certain age.

A comparison of the grades of concrete according to different codes is hardly possible because the way of testing is not the same.

Some differences are:

- the shape of test-pieces;

- the dimensions of the test-pieces;

- the way of storage; - the speed of the tests.

According to CCBA 68 the conception "characteristic strength" is

unknown. The concrete grade is represented by means of the formula:

f = f - 0,Bs c cm

f = nominal compressive strength. ==• f = mean compressive strength

s = standard deviation.

To make a comparison with regard to the concrete grades used in France, it is necessary to determine a frequency distribution curve and a standard deviation.

o ^J V >^ l|v^< - ^ tvV. ^; _t>t VÊ.

V&

ft ^7,r I r ' I t = SH

9

-Correspondence between Concrete Grades assuming that compliance criteria are the same.

Charact. cylinder strength *ck ' 10 12 14 16 18 20 24 26 28 30 32 34 36 38 40 42 44 46 48 50 CEB cylinder ^ClO 012 - 1/- ' 016 018 020 025 030 -035 C40 045 050 CP 110 cube- grade strength 15 gr.l5 — '" 20 gr.20 25 gr.25 30 gr.30 • -40 gr.-40 50 gr.50 60 gr.60 CCBA 68 t , an = a 28 s = 3 a'= 12,52 " s = 3 a'= 16,52 s = 4 a'= 21,36 s = 5 0 = 28,2. s = 5 a'= 34,2 s = 5 . .9\:L^O^2 s = 5 ' a'= 46,2 s = 5 a'= 52,2 DIN 1045 cube- grade strength 14,7 B15 -• — r -23,9 /v B25 34,8 B35 45 B45 — — — -54,5 B55 VB 19 74 cube- grade strength 12,5 B12,5 17,5 B17,5 — ... 1 22,5 B22,5 30 B30 37,5 B37,5l - - - — 45 B45 52,5 B52,5

j

60 B60 1 . . ---- '\

10 -4.2 P a r t i a l factors o f s a f e t y Loads permanent p r e s t r e s s variable M a t e r i a l c o n c r e t e s t e e l ^f ^g ^P ^q ^m ^c CEB 1,35 0,9-1,2 1,5 1,5 1,15 CP 110 1,2-1,4^^ 1,0 1,2-1,6^^ 1,5 1,15^) CCBA 68 1,0 1,0 1,0-1,2^^ 2) 1,4-2,8 -^ 1,5 DIN 1045 1,75-2,1 1,0 1,75-2,1 1,0 1,0 VB 1974

1,7

1

1,0 1,7 2 1,0-1,33 1,0

1) depending on load combination 2) depending on kind of loads

3) for pressure f , - 2~ Y + —^

^s 2000

Global safety indications

1

^

hf^c-Kf^s

1 a CEB •a—w- a.r- T __ fcyl \ | 0,85 2,38-2,64 1,55-1,73 OP 110 0,84 2,14-2,86 1,38-1,84 CCBA 68 -1,4-3,36 1,5-1,8 DIN 1045 0,76-0,86 2,03-2,75 1,75 VB 1974 0,75-1 0 1,7-2,27 1,7 \

11

-Moduli of elasticity of concrete

OEB Modulus of longitudinal deformation 1/3

cm

9,5 (f^^ t 8)- f , = characteristic compressive CK

Strength for the age of 28 days. f , (N/mm^) 1 ck

1

12

16 20 25 30 35 40 45 50 E (kN/mm^) cm J

26

1

27,5 29,0 30,5 32 33,5 35 36 37 CP 110 static modulus E = (1,25 E c cq cq 19) kN/m

dynamic modulus of elasticity.

cube strength N/mm2 20 25 30 40 50 60 modulus of elasticity kN/mm2 25 26 28 31 34 36 CCBA 68 Short-term: E. = 21000 1 a. (bars) (< 24 hours)

a- denotes compression strength for the age of j days.

Long term: E 7000 y a- (bars). If only a „. is known: E. = 23000 \/ a' (bars) 1 V 28 E - 7670 V 28 (bars)

- 12 - . • •

DIN 1045 Short-term elastic modulus, relevant to the serviceability

limit states: Cube strength (N/mm2)

1 ^°

15 25 35 45 55 E, (kN/m^)

22

1

26 30 34 37 39

VB 1974 Short-term elastic modulus: . .

E' . = (C,p '^'^ - C^f' .)\/lO f' . E ; = 0,9 E ' .. ob] 1\ m 2 cm] cm] b ob j

f . = mean value cube strength for an age of j days cm]

3/2

0,0 - 1800 (normal weight concrete) 1\ m

0 = 4 (normal weight concrete)

Cube strength (N/mm2)

1 12,5

17,5 22,5 30 37,5 45 52,5 60 E^ (kN/m^)| 21,5 25,0 28,0 30,5 32,5 34,0 35,5 37,0

The figure on page 13 shows the moduli of elasticity of concrete

in relation with the concrete grade:

a. without influence of creep

b. with creep influence (t =<=o) in relation with a relative humidity (R.H.) of 40 a 50%.

4 Shrinkage

CEB, DIN 1045 and VB 1974 prescribe extensive formulae to

determine the influence of shrinkage. CP 110 refers to CEB-FIP recommendations

CCBA prescribes an equally divided stress E x e depending on

the location of the building.

The differences are very small as shown on page 15. Only

according to VB 1974 shrinkage depends on the concrete grade.

Also according to that code the influence of the reinforcement can be taken into account.

5 Creep

CEB, DIN 1045 and VB 1974 prescribe extensive formulae to

determine the influence of creep.

CP 110 refers to CEB-FIP recommendations.

CCBA prescribes that the long term elastic modulus is 1/3 of

the short elastic modulus, so c (t«>,) = 2. The figure on page 16 shows small differences.

15 -LU SHRINKAOE in X • l/l u 30 20 10 CCBA6'8^ V B 1974 ' • • • R H = 40 a 50 % R.H.= 90 a 95 % CIO C14 C18 C24 C30 — » concrete grades CEB = DIN 1045 =CP 110

4.5 CREEP

I

18 -CEB DIN 1045.1 R H , ^ ^ ^ ' " f 4 0 a 5 0 % V B .1974 J VB 1974

+ + + + -•--»-•

— CCBA 68 + DIN 1045. _{RH =} 90 a 9 5 % — CEB.CPnO CIO C14 C18 C24 C30 — • concrete grades

« O n l y according to VB 1974; creep depends on concrete grade - CCBA 68 ' l o n g term elastic modulus i s

17

-4.6 S t r e s s - s t r a i n diagrams 4.6.1 Steel

The stress-strain diagrams of steel grade S 400 are shown

on the next page.

According to CEB and OP 110 the yield stress has to be reduced

by the material-factor y .

4.6.2 Concrete

The stress-strain curves of concrete grade 020 are shown on page 19.

VB 1974 is the only code which describes the possibility of

18

-STRESS STRAIN DIAGRAM REINFORCEMENT Design diagram S 400 ^(NAnm*!) I 3 4 7 .

— «^s ru)

tension

U3 ip 3^78 — i^i (•/..)

no diagram

W —•^,(%.)

I 4X

— i , (•/•.)

CEB S 400 yk 400 N/mm

tension and compression

_ Jik _ 400

yd Y, 1,15 347 N/mm

CP 110

Hot rolled high yield f

yk 410 N/mm tension: f yd yk _ 410 _ -,„ ^,. 2 --^^— = -—rr = 357 N/mm Y^ 1,15 0. compression: CCBA 68 FeE 40 ^ = 0,8.357 = 285 N/mm ^^f f ^ = ^^ ^ — = 303 N/mm^ ^s 2000 (j) < 20 mm yield stress 412 N/mm 2 (j) > 20 mm yield stress 392 N/mm Permitted: 2/3 of the yield stress E = 200 kN/mm^

DIN 1045

BSt 42/50 f , = 420 N/mm^ yk

tension and compression: f , = f , = 4 2 0 N/mm^

yd yk

VB 1974

FeB 400 f , = 400 N/mm^

yk

tension and compression; 2 yd f , = 400 N/mm yk

- 19

STRESS STRAIN DIAGRAM CONCRETE

Design diagram 0 20

f (N/mm*)

1 n.3 .>

^(N/mm)

t •« I

15 ^ ^ r / . . )

15 —^glV-)

CEB 0 20 (cylinder-strength) f ^ = 0,85 - ^ = 0,85.|^ cd Y 1,5 11,3 N/mm OP 110

Grade 25 (cube strength) 0,67 f 0,67.25 CU c d Y 1,5 1 1 , 2 N/mm

no diagram

CCBA 68 a = o - o = 2 3 , 4 N/mm n 2o Permitted: 0,6 x 23,4 = 14,0 N/mm^

f^(N/mm)

f )7,5..

ID

35 — ^C/..)

DIN 1045 B- 25 (cube-strength) f , = 0,7 f = 0,7 X 25 = 17,5 N/mm^ cd cu

Z5 a s

— ^CA.)

VB 1974 B 25 (cube-strength) f J = 0,6 f = 0,6 X 25 = J-5 N/mm^ cd cu

20

-4.7 Relation moment-normalforce

The design charts on page 21 are based on the characteristic

values of M en N.

To determine this values the following load-factors have been used: 1,35 + 1,50 1,4 + 1,6 CEB OP 110 DIN 1045 VB 1974 N

According to OP 110 the maximum value of -x— corresponds to a

M V "^ N J

minimum design eccentricity given by — I'D,05 h so that -7—3e. 20

= 1 , 4 2 5 = 1,5 = 1 , 7 5 -= 1,7 2 , 1 M

DIN 1045: If the value N increases, the loadfactor increases from 1,75 to 2,1.

22

-RELATION ^ t o t " ^ °^ un'iaxialy loaded column IGQOn E E ^

900- 800- 700-600 500- 400- 300- 200- 100-CEB t t > »>>*CP110 CCBA'68 i -^-^-•—^-^ Dl N 1045 !\ VB 1974

i\

i •\

/ / \

N'

I I . ^

P-1^ depends on ratio normalforce to long l e r m - t o t a l normalforce eo=h eo=0,5h

«0=0.25 h

J I I I I I I L J 100 150 200 250 300 350 400 450 500 — • h (mm) 139 92 59 55 46 40 35 31 28 — * X

23 -8 Bending Example fNI 5,00 m ,300, Materials : 018 S400 Actions : deadweight g „ = 14,4 kN/m live load q „ = 14,4 kN/m' Codes load-coefficient Yr M sd (kNm) Stress distri-bution comp. zone Design strength Concrete f ,(N/mm ) cd Steel :y^(N/mra ) f mm) (mm) N c (kN) ((mm ) 2. OEB CP 110 CCBA 68 0,85f ck 1,35 1,5 128,25 1,40 1,0 DIN 1045 1,75 1,60 1,2 135,0 99,0 1,75 157,5 VB 1974 1,7 1,7 153,0

°t^=10,2

1,5 400 1,15 = 34f 126 410 313,3 900 0,45f cube

-f^H-0,45x22,5=10,1 12,82 400 1,15 348 2x400 =266,7 120 193 407 li 398 0,7f cube 0,7x21,7 =15,2 400 0,8f J. c cube 0,8x22,5 = 18 400 102 332 954 272,3 934 421 374 934 104 425 361 903

24

-4.9 Shear

• Calculations of shear stresses and shear reinforcement in beams are based on comparing the average shear stress on a section with a nominal value of ultimate shear stress.

When the average shear stress is greater than the nominal stress, shear reinforcement is provided in proportions, calculated under

the assumption that the reinforcement forms the tension members of one or more series of pin-jointed trusses (truss analogy). The figure on page 30 show the required shear reinforcement by using vertical stirrups and the maximum shear stresses.

With regard to punching shear two values has been defined T^ and T (page 35).

The lowest value T indicates the ultimate value above which

shear reinforcement should be provided.

In no case, even with shear reinforcement, shear stress should

25

-Shear

CEB

There are two methods: - Standard method - Accurate method.

Standard method

a) T, = 2,5 T„,b d no shear reinforcement should be provided. 1 Rd w ^ck ^Rd 12 0,18 16 0,22 20 0,26 25 0,30 30 0,34 35 0,38 40 0,42 45 0,46 50 0,50 b) T2 -- 0,3 f^^

where inclined stirrups are used (45 <a < 90 )

T^ = 0,3 f , (ltcotga);if 0,45 f ,. 2 cd cd ^ck ^cd 12 8 16 10,7 20 13,3 25 16,6 30 20,0 35 23,3 40 26,7 45 30 50 33,3

c) Design shear stress: V

sd b d

w sd

shear force due to ultimate limit loads

width of the section effective depth. d) Shear reinforcement A f , V - ^ .0,9 d. - ^ (1 + cotga) sina = r:^ - T , . B , s Y b d 1 1 w

A = cross-sectional area of the two legs of a stirrup

6 = coefficient by which the value x may be increased for members subjected to longitudinal compression (included prestress) 1,0<6^<2,0.

26

-030 S400 vertical stirrups 3. = 1

^w'^yk 1,15 x 6 1,15 x 0,85 ^ ^ .,/ 2 F-b = —0^9 ÖT9 = ^'^ N/'"'"

Accurate method

This method is provided for special cases in particular where both

shear and torsion are involved. It allows a choice for the inclination of web compression (3/5<cot0<5/3). Reduction of the shear reinforcement is only possible in a limited transitional range.

^sw ^ k

. 0,9 d. — ^ (cot 0 + cota) sina = s ' Y b d sin 29 ^^ °'3 f^^) 030 S400 vertical stirrups 0 = 45 o A .f sw' yk 1,15.6 _ „ „ . , , 2 6 ,, ^ ., , 2 - T b = - ^ ^'^ N/"™ " = Tj25 = ^'2 N/mm CP 110

a) T. = V depends on longitudinal tension reinforcement A

100 A s bd 0,25 0,50 1,00 2,00 3,00 concrete 20 0,35 0,45 0,60 0,80 0,85 25 0,35 0,50 0,65 0,85 0,90 grade 30 0,35 0,55 0,70 0,90 0,95 40 ore 0,35 0,55 0,75 0,95 1,00 morel b) X, '2 concrete grade 20 3,35 25 • 3,75 30 4,10 40 ore morel 4,75

c) Design shear stress: V

V = -—T V = shear force due to ultimate loads bd

b = width of the section d = effective depth.

27 -d) Shear reinforcement A b(v-v ) sv c > 3V 0,87 f yv 030 =grade 37,5 f = 400 N/mm & ' yv s b 0,9 0,9 b d V w • w 4,6 _ „ . „, 2 -r-^ = 3,1 N/mm 1,5 10 OA 1,15.4,6 1,15.1,0 _ ,, c M/ 2 /

- ^ ^ T I ^ ^ 7 9 - -

^'^ ^Z'"'" ^"b d

' ' T.T = 3,0) CCBA 68

a) Shear reinforcement must always be provided

b) X2 ^ 3,5 0^ if a^ < a^^

%o

a, = admissible tensile strength of concrete

a' = concrete compression strength in the same section b

a' = admissible compression strength with normal force bo

X = 5 a, if inclined reinforcement is used.

010 0,44 3,76 014 0,51 4,96 018 0,59 6,41 024 0,72 8,46 030 0,83 10,26 036 0,93

c) Design shear stress: T

X = z—~ T = shear force due to serviceability loads b.z

b = width of the section 7 z = lever arm (— d) d) Shear reinforcement A f , .z sw yk . T 030^ a' = 34,2 N/mm2 Steel: a = 2/3 f * n at yk A f ^ \ ^ ^ = |.3,5.0,83 = 4,4 N/mm^ x = 2 x 3,5 x 0,83 = 2,5 N/mm^ S . D ^ O w * A d m i s s i b l e s t e e l s t r e s s a - p .a at a en P = 1 - — ^ P„,.„ = 2/3 9 a, m m

28 DIN 1045 a) X. = -012 if x„,„ ^ X < T„„ it is allowed to reduce x ->- x = — 012 ^ o ^ 02 o x 02 ^0,4x^

X = design shear stress.

^) ^2 = "^03

B15 ^x = 2,0 N/mm^ B55 ^x = 5 , 0 N/mm^

c) Design shear stress Q,

b .z o

Q = shear force due to serviceability loads b = width of the section

o z = lever arm omax

r o i 2

k2

k3

B15 0 , 5 0 1 , 2 0 2 , 0 0 B25 0 , 7 5 1 , 8 0 3 , 0 0 B35 1 , 0 0 2 , 4 0 4 , 0 0 B45 1 , 1 0 2 , 7 0 4 , 5 0 B55 1 , 2 5 3 , 0 0 5 , 0 0 d) Shear reinforcement 2 C30/V B35 f , = 400 N/mm , vertical stirrups -' X x„^ = 1,75 X 4 X 0,9 = 6,3 N/mm^ (z = 0,9d) ^^ 03 1,75 3,6 N/mm A .f sw' yk 1,05.5,3 „ ,, ., , 2 - T X ^ = 0^9' = 7,4 N/mm w

29

VB 1974

a) X. = 0,5 f, f = design tensile strength of concrete

" l "2 B12,5 0,5 2,5 B17,5 0,55 3,5 B22,5 0,65 4,5 B30 0,75 5,0 B37,5 0,90 7,5 B45 1,00 9,0 B52,5 1,10 9,0 B60 1,25 9,0 b) x„ = 0,25 f,, ( ^ 9 N/mm ) f^, = characteristic compression 2 bk bk strength of concrete.

c) Design shear stress: Td

T , = -rr- T , = shear force due to ultimate loads d bh d

b = width of the section

h = effective depth

d) Shear reinforcement

2

O30 = B37,5 f = 4 0 0 N/mm , vertical stirrups

A .f sw s.b sw'^yk _ 7,5 0,9 _ „ „ ... 2 7,3 _ 2 —•—''•— = —^— - —^— = 7,3 N/mm x = -—r - 4,3 N/mm 0,9 0,9 1,7

30

-C E B

r.^J^[ N/mm»)

y^.b^d

CCBA 68

CP 110

DIN 1045

VB 1974

4 4,3

r=.^^(N/mm')

lf,b^d

31

-4.10 Punching shear

OEB a. X depends on: - concrete grade C

percentage longitudinal reinforcement in both directions p and p

X y - effective height d. x ^ = 1,6. Xj^^. K . ( I t 50 p^) K = 1,6 - d (m) -t 1,0 p Ix P^y ^ 0,00^ ^ck ^Rd 12 , 0,18 16 0,22 20 0,26 25 0,30 30 0,34 35 0,38 40 0,42 45 0,46 50 0,50 012 -> T = 0 , 2 9 - 0 , 5 6 N/mm 050 ^ X, 0,E 1,6 N/mm 2 012 C50 1,6 X, 0,47 1,3 0,9 N/mm 2,5 N/mm"^

c. Design shear stress

Td Td.

U

ph W

e denotes the eccentricity of the load with respec"t to the centroid of the critical section.

W denotes the modulus of inertia of the critical section n is a factor depending on the shape of the periphery The critical section should be taken on a perimeter 0,5d

from the boundary of the loaded area.

CP 110 a. X depends on:

C .V s c

- concrete grade (grade)

- percentage of reinforcement in both directions

- overall depth (h) 100 A b.d 0,25 0,50 1,00 2,00 3,00 1 1 grade 20 N/mm2 0,35 0,45 0,60 0.80 0,85 grade 25 N/mm^ 0,35 0,50 0,65 0,85 0,90 grade 30 N/mm2 0,35 0,55 , 0,70 0,90 0,95 grade 40 or more 0,35 0,55 0,75 0,95 1,00 Overall slab depth (mm) 250 or more 22 5 200 175 150 or less 1

^.

1

1,00 1,05 1,10 1,15 1,20

32 g r a d e 20 ->-x, g r a d e 40 o r m o r e - > x ^ 0,35 0,35 1,02 N/mm 1,20 N/mm^

X depends on concrete grade and should not exceed half the appropriate value given in the tabel below

grade 20 N/mm^ 3,35 grade 25 N/mm 3,75 grade 30 N/mm^ 4,10 grade 40 or more | 4,75 1 grade 20 ->x, grade 40 or more-^x,

c. Design shear stress v

1,68 N/mm 2,38 N/mm^

V

crit U .^ = length of critical crit °

parameter There are no data about the determination of eccentricity. The critical section should be taken on a perimeter 1,5 h from the boundary of the loaded area and if h > 200 mm a similar amount on a parallel perimeter at a distance of

0,75 h inside it. CCBA 58 a. X <1,2 a 1,5 P d p .h ^c o P is concentrated load p is perimeter h is overall depth (h) _o

a is tensile strength of concrete

2 0, (N/mm ) b 010 0,44 014 0,51 018 0,59 024 0,72 030 0,83 036 0,93 010 -> T = 0,53 N/mm 036 -»• T^ = 1,11 N/mm^

b. The critical section should be taken on a perimeter 0,5 h

33

-DIN 1045 a. T depends on:

^1 = ^r

oil

concrete grade (B)

way of anchorage of tension reinforcement percentage of reinforcement in both

directions

quality of the steel

X, = 1,3a V U ; 1,5 > U > 0,5

U g

mean percentage of reinforcement

B St 22/34 B St 42/50 B St 50/55 1,0 1,3 1,^ l a * 1 l b 2 ^ 0 1 1 ^ 0 1 1 ^ 0 2 B15 0 , 2 5 0 , 3 5 1 , 2 0 B25 0 , 3 5 0 , 5 0 1 , 8 0 B35 0 , 4 0 0 , 6 0 2 , 4 0 B45 0 , 5 0 0 , 7 0 2 , 7 0 B55 0 , 5 5 0 , 8 0 3 , 0 0

* Values if reinforcement has been anchored in tensile area.

' I . . .

>.serviceability state B55

B15 ^ X = 0,23 - 0,78 N/mm

X = 0,50 - 1,80 N/mm

T depends on: - concrete grade

- percentage of reinforcement - quality of the steel

"^2 " ^2" ^^02 0

,45a y r

B15 B55 -v X X = 0,38 - 0,93 N/mm 0,96 - 2,31 N/mm serviceability state

c. Design shear stress x rmax

u.h u = perimeter

h = effective depth (d)

At edge and corner columns x has to be increased with 40%. The critical section should be taken on a perimeter 0,5 d

from the loaded area. At edge and corner columns the

VB 1974

- 34

a. X depends on the concrete grade

X = f ; f = design tensile strength of concrete

Concrete grade B12,5 B17,5 B22,5 B30 B37,5 B45 B52,5 B60 N/mm2 1,0 1,1 1,3 1,5 1,8 1,9 2,2 2,5 fbk N/mm2 10 14 18 24 30 36 42 48 B12,5 -> X = 1,0 N/mm B50,0 -^ T-, = 1,9 N/mm^

b. X depends on the concrete grade

X = 0,25 f' (< 9 N/mm ) f = characteristic compression strengths of concrete B12,5 ^ T = 2,5 N/mm^ .. -^ 2 B45 or more->• T = 9 , 0 N/mm T T c. Design shear stress x^ = —r— + —r

^ d ph ph 2e h+d (inner column) h = effective depth (d) p = perimeter d = 2 (a + a )

-2L

iL_

The critical section should be taken on a perimeter 0,5 h from the loaded area.

Punching shear reinforcement

CEB

OP 110

CCBA 58

DIN 1045

VB 1974

The vertical component of the force resisted by this reinforcement should be not less than 75% of the load. Only if the slab is at least 200 mm thick it is allowed to apply shear reinforcement.

No articles about punching shear reinforcement

like OEB

PUNCHING SHEAR serviceability state 35 -?^[ rVmm^) 2 . 1 • 10 20 30 40 50 ^^ck ( N / n m ^ j ?:iNmm^) 5 . 4 • 3 • 2 . 1 . — • • • • CEB CP no CCBA 68 DIN 1045 VB 1974 ' f 1,425 1,5 1,1 1,75 1,7 10 20 30 40 50 * ^ck(N/mm2)

36

-4.11 Staggering rule

The envelope line of the tensile forces is obtained by a horizontal M

displacement a of the envelope line F = 1 t z

jiii i

TTTTT

la

^ •

h

envelope of tensile forces

OEB CCBA 68 : DIN 1045; V sd.s 1 2 A .f , . , sw ywd.smcj)

d cotga (if shear reinforcement is used)

2 400 2 stirrups ((> 6 s = 100 A = 57 mm f , = ^ j — — = 348 N/mm ^ sw ywd 1,15 V ^ = 162 kN sd ^ - 2 "^ ^ a = 90 0,9 d ^ a = 410 mm ^ a = 200 mm a = 0,25 d inclined stirrups

a = 0,50 d bent-up bars + vert, stirrups

a = 0,75 d vertical stirrups a = 340 mm

By reducing the shear stress the values a have to be increased by 0,25 d.

If no shear reinforcement is used a, = 0,75 d,

VB 1974 a^ = d 450 mm.

37

-4.12 EFFECTIVE WIDTH FLANGED BEAMS

one-span beam

ht

in

2.0

continuous beam

0.1

li 02 l7

r

hf/h-0.1 02 03 Ql Q2 03 CEB » « » CP110 CCBA 68 - * - - ^ - * - DIN 1045 VB 1974

38 4.13 DEFLECTION

Coses for which checking of the deflections may be omitted

I . s p a n d . effective height (7^ = steel stress CEB CP 110 CCBA 68 DIN 1045 VB 1974

Km)

7d

t

+-l / h ^ 44 42 33 31 25 23 18 10 <143 20 •>• Km)

39 -beams

-I

4-

1

1

. 5EP 46,7 l^ * - 4 l 6 373 26 +••+--^^• + ••-+ + + 4- i- 4-H--H--»"* q-'• i143

^\ m

^ • • — • — • — • — • — • — • — S I , . ^ * - ^ • • ' - .

UmL

10 = 235 «US « 240 20 > • > • • ^^mm «.I - ^ - - ^ . CEB CP no CCBA 68 DIN 1045 VB 1974 ^ 17,5 146 7 •h -•• 4--»--»- -t-H- •+4--l--«--f4-4-4-= 145 240 i 143 = 235

UmL

10 20

•^e»-dO

slobs

spanning in two directions simply supported

35

2 ^

20 - ^ - , - 1 ^ » . ^ t " » ^ ' ^ ^ ^ ^ ^ ' * " » ^ ^ ^ ^ ^ ^ ; ^ i 4 3 = 235 *4 -r '9' _{• • ^ .}

Ivlm)

10 . , ^ ' - ^ =145 * ^ * . = 240 20 restrained or continuous

50

46^7 3a7 374 2^7 24 ^ - f 4 - 4 - 4 - 4 4 - 4 - « - 4 - * - l - H - 4 =s » z r < » ' _ r w . £ ; ~ r ^ • ^ , ' - - - . . * • * . i:143 = 235 = 145 ,=240

Uml

10 20

41 -4.14 Minimum Cover (mm) Code CEB CP 110 CCBA 58 • DIN 1045 VB 1974 Grade of concrete 012, 016, C20 general case slabs, shells 025, C30, 035 general case slabs, shells 040, 045, 050 general case slabs, shells grade 20 grade 25 grade 30 grade 40 grade> 50 all grades B15 general flat structures > B25 general flat structures all grades slabs walls beams columns Conditions of exposure Slightly aggressive. 20 15 15 15 15 15 dry 25 20 15 15 15 10 20 15 15 10 10 15 20 25

1

Moderately agressive 30 25 25 20 20 15 humid -40 30 25 20 20 25-30 20-25 20-25 15-20 15 20 25 30 Highly aggressive 40 35 35 30 30 25 agressive -50 40 30-60 25-50 40 40 35 35 30 \ 25 30 35 40 Additions

minimum cover shall not be less than - 15 mm

- maximum seize aggregate plus

5 mm.

subject to salt used for de-icing

. -50 40 25 minimum cover: diameter of bar minimum cover: diameter of bar if inspection is impossible: to increase values with 5 mm.

- 42

4.15 Minimum area of reinforcement

CEB OP 110 Slabs S220 0,0025 0,0025 depends 0,0014-0,003 -0,002 S400 0,0015 0,0015 Beams S220 , 0,0025 0,0025 S400 0,0015 0,0015 on concrete grades 0,0008- 0,0014-0,0016 -0,0015 0,003 -0,002 0,0008-0,0016 -0,0015 Wall S220 0,004A c 0,0025 -0,005 0,002 3 S400 0,004A 0,0025 -0,005 0,0015 Columns ) ] S220 0,008A c 0,01 or A =

°'^5 Id

fyd S400 0,008A 0,01 depends on the location of the column 0,008 0,004 0,008 0,003 CCBA 68 DIN 1045 VB 1974

The minimum areas of reinforcement with regard to slabs, beams and walls

do not differ much. '•

The required minimum reinforcement in the columns however differs much, which finds expression in the example on page 47..

43 -5 Comparison calculation

5.1 Design methods of plane, unbraced frames

It is mentioned in and permitted by all investigated codes (except the French code) to analyse this type of structure by non-linear

elastic methods (normally by means of a computer).

Most codes permit the use of simplified methods for solution by hand calculations.

OEB First order moments by linear elastic design

Second order moments are replaced by moments produced by

equivalent horizontal forces acting at the top of each storey (Iterative method F-A).

CP 110 First order moments by linear elastic design

Second order moments by means of additional eccentricity dependent on effective column-height.

CCBA 68 like CP 110

DIN 1045 First order moments by linear elastic design

Second order moments by means of additional excentricity of substitute, imperfectly restrained columns

VB 1974 First order moments by linear elastic design

Second order moments by means of additional excentricity of substitute cross-shaped parts of the frame.

CEB

A<25

1

load distribution first order linear elastic theory slope 1/200

fno second 1 order effects slenderness 25<A<140 ^ iterative method F - A 1 determination of real stiffness of elements iterative method F - ^ second order eccentricity e 50<A<140 additional eccentricity e ' determination of the sections ^ o t = ^o + «2 + ^c A<140 non-linear elastic design] CP 110

load distribution first order linear elastic theory

slender columns -no • slenderness A<40 • y e s . A<70 no no second order effects in beams second order eccentricity e„ determination of the sections e^ . = e -I- e„ tot o 2 short columns no second order effects 4=

CCBA 68

load distribution first order linear elastic theory

determination 1 and A o A .^35 35< A <50 no second order effect: additional eccentricity 62 (1) determination of reinforcement of columns e tot %^^2 determination of the reinforcment of the beams e^ ^ = e tot o 50< A «150 additional eccentricity e^ (2) DIN 1045

load distribution first order linear elastic theory

• no no A>45 additional eccentricity . ^2 1 determination 1 and A _ D ' A >20 A <:20 determination e /d for decisive section A^70

1

e /d »35 yes A>a'0 yes no

1

' determination reinforcement no second order effects .e^/d,3,5;^. non-linear elastic design 1 •fr 1 ye s

1

1 no second order effects additional eccentricity e -' c determination of reinforcement

VB 1974

Load distribution first order linear elastic theory

determination points of contraflexure

(only wind load)

1 •

1 assume hinges at the points of contraflexure and divide the construction into cross-shaped elements r not possible \ ; ; : : - ; i . i . '•'', '' •'J'--non-linear elastic design |

1 determination of e^ (first orderl) determination e (second order)

1 assess reinforcement in beams to determinate the influence 1 of the beam on the total ecc.

1 calculation of the reinforcement 1 of the beam 1 1 calculated reinforcement > 1 assessed reinforcement? no yes determination of reinforcement in columns

47

-2 Office building

As an example an unbraced bearing structure of an office building has been analysed.

The structure has been divided into a series of plane frames and has been simplified for solution by hand calculations.

The dimensions, materials and loads are determined in advance.

The differences between the first order moments, shear forces etc.,

based on linear elastic design, are small and mainly caused by the differences of the bending resistance o.f the beams.

After that the amounts of reinforcement of the different members have been determined.

With these theoretical quantities of reinforcement the ratio failure load-working load has been determined.

48

-3 General data

Geometrical properties of the structure are illustrated on page 51.

Materials:

2 - concrete: : characteristic cylinder strength 18 N/mm - reinforcement : S 400

Actions:

- Permanent Loads:

- dead weight of reinforced concrete - finish + pipes + separations

- front wall load (inclusive beam) : - side wall load (inclusive beam) :

24 kN/m ; 1,5 kN/m^; 10 kN/m'; 16 kN/m'; - Live loads: 2 2 - live load: 4 kN/m on floors, 2 kN/m on roof;

2

- wind load: 1 kN/m equally divided over height and only perpendicular to the respective walls,

suction included.. Determination of actions

~ Characteristic values of actions Permanent loads on floors:

r

weight of floor 0,125 x 24 = 3 kN/m^

finish, pipes and separations = 1,5 " 4,5 kN/m

- Loads on beams of central bays:

- permanent loads 3,60 x 4,50 - weight of beam 0,3 x 0,375 x 24

distributed permanent loads

- live loads floor beam 3,60 x 4,0 q - live loads roof beam 3,60 x 2,0 g fk fk 16,20 kN/m 2,70 " 18,90 kN/m' 14,40 kN/m' 7,20 kN/m'

49

-Weight of front wall and columns:

- weight of columns 3,60 x 0,5 x 0,5 x 24 = 21,6 kN - weight of frontwall and beam 3,5 x 10 = 36,0 "

F, ., = 57,6 kN K 1 - weight of columns F^^ = 21,6 kN Wind load: - floors: 3,6 x 3,6 x 1 - roof : 2 ^ 3 , S ^ 3 , 6 5 < 1

The following values have been used.

CEB CP 110 CCBA '68 DIN 1045 VB 19 74 ^g 1,35 1,20 1,00 1,75 1,70 ^q 1,50 1,20 1,20/1,00. 1,75 1,70 \ 1,5 1,20 1,00 1,75 1,70 W = 12,96 kN K 1 k2 6,50 kN

I

GEOMETRICAL PROPERTIES 39600 50 Tlxl36Q0

n r " ^ t f i n^^n ' * ~ir^f~'

f

4-Mc '4

•^—w— J —f f~'^'""P-'-i f • ' p

8

ro X «0 14400 6000 .24001 6000

^^"""1

IT) 'M LT) ro 300/500 500/500

i

o o to in •~-r i n r-ro O o 150 500

m.

_{Q — 4 .}

I

4 £ K M 4

50

sa

• g p jOQ section a-o

SCHEME OF LOADS - 51

w,

_l<2 Wkl-•"kl

I

L

L

''k2 f'k2

— 1

-—i

W

I

Dibliotliool(

afd. Civiele Techniek T.H. Stevinweg 1 - Delft.

- _ ^ _

5.4 RATIO TOTAL MOMENTS - FIRST ORDER MOMENTS

f^ total

O - i .

left fafode colunnn

- i

t ^ f [

n

If IV / > i » central columns

I

ri'ght fafode column

CEB CP 110 DIN 1045 CCBA 68 VB 1974

5.5 MOMENTS IN BEAMS

53

-first order moment

beam 7^^ f l o o r

L

second order moment

beam 2 " ^ floor

second order moment

CEB

DIN 1045 VB 1974

According to CCBA and CP no second order effects in beams ( i n this case)

- 54

S6 THEORETICAL QUANTITIES OF REINFORCEMENT added per section for 8 floors (mm )

- r ^

II

iTIi

J l

ITT^Ï

U

PLACE 1 2

1 ^

1 ^

1 ^

_total beams facade column central 1 column total columns TOTAL CEB 10.500 4000 10600 4800 Z200 32400 12550 13325 25875 57975 CP 110 8000 4100 8300 4000 1400 26200 7320 7320 14460 40660 CCBA 68 8400 4300 8700 4000 1400 26800 22800 7300 30100 56900 DIN 1045 16300 5500 9 200 4500 2100 37600 8400 7200 15600 53200 VB 1974 1

13400 1

4600 1

12000 1

5600 1

3000 1

38000 1

6000

6000 1

12000

50000 1

- 55 5.7 HORIZONTAL DISPLACEMENT 8 4 20 30 40 - ^ displacement (mm) CEB CP 110 CCBA 68 DIN 1045 VB 1974

5 6

5.8 Ratio failure load-working load

Each structure, with the amounts of reinforcement as shown in page 5 4 , has been submitted to increasing loads (live load, dead load and wind load) and calculated with the help of a computer.

Depending on the amount and the situation of the reinforcement, plastic hinges develop.

The situation of these hinges with the accesary load-factors are shown on page 58.

If at one section the ultimate strain of the concrete or of the steel has been attained, the failure load is supposed to be reached. Material properties: ^ ( N/mm 2) - ( 7 o o ) ^ c 2,5 3,5 ^ CONCRETE

RATIO FAILURE LOAD-WORKING LOAD 59 1.A 1.5

1

^-^

1.6 1.6t 1.61.6 r n ' n. 1.4 1.6 1.6 1.6 1.6 K T 1.8 1.8,

21-1

1.6 —•— 1.6 1.6 1.6 1.6 — • -1.7 1.6 /f7 1.8 1.8 1.7 1.5 1.8 —•— 1.6 — • -1.6 1.8 1.7 1.71.7 1.7 1.8 1.8 1.8 f^ fr 1.5 13 L«1.9 1.9 1.6

_J4

1.8 1.5 •--» 1.8 1 6 19 1.8 17 13 17 I91.G 1.9 1.7 1.8 — • -ffr 1.7 IBIBJ 1.8 17 1.9 1.8 1 _Si.g 1.8 1.9 CEB VB 1974 DIN 1045 ifr 1.4 1.2 1.0 1.2 >7r 7fr 1.A — • 1.4 1.A 1.4

U

—•

1.4 — • 1.4 ffr 1.3 —•— 1.3 —•— 1.3 1.3 m-1.2 • — 1.1 1.2 /fr ffr 1.3 1.3 1.3 1.3t ——• 1.3 1.3

- f plastic hinge and accesary load-factor T failure and

accesary load-factor

58

-Conclusions

The differences between the corresponding articles are rather divergent

and can not be discussed in generality.

With regard to the comparison calculations the following can be noticed: a. Loaddistribution

- The differences between the first-order moments, shear forces etc.,

based on linear elastic design, are small and mainly caused by

the differences of the bending resistance of the beams (effective width T beams).

- The second-order moments, calculated according to the schemes on the pages

44,4 5 and 4 6 , differ much. These moments are shown on the pages 52 and 53. According to CCBA 68 and CP 110 no second-order moments have to

be taken into account at the beams.

b. Reinforcement

The theoretical quantities of reinforcement are shown on page 54. The differences are mainly caused by:

- the calculated second-order moments - the loadfactors

- the prescribed minimum reinforcement at the columns (except for the fa(jade columns of the French structure all columns have been provided .of minimum reinforcement).

c. Relation failure load-working load

The bearing structures have been checked by means of a computerprogram STANIL. This program is especially suitable to non-linear elastic

design of plane frames.

By increasing the loads each time with 10%, the ratio failure load-working load has been determined.

If at one section the ultimate strain of the concrete or of the steel has been attained, the failure load is supposed to be reached.

By cracking of the beams at the span, the moments near the supports increase. Because of the higher percentage of reinforcement near the

supports calculated according to CEB, DIN 1045 and VB 1974, the plastic hinges develop especially in the span, as supposed by the English and French structures.

All structures have attained the load factor required.

A reason for the low value (1,3) according to CP 110 will be: - the loadfactors are low (1,2)

2

- t h e w m d l o a d s m England a r e c o n s i d e r a b l e h i g h e r t h a n 1 kN/m . • d. Deflection

The horizontal displacements at the top of the buildings (y = 1,0) amount to a maximum of -^^ of the total height of the building.

59

-The effect of the amounts of reinforcement in the beams is evident.

d. Calculations

According to all codes simplified methods can be used for analysis by hand, except according to CEB.

The P-A method is labourious and by using this method a computer is

indispensable.

e. Main differences

- The methods and formulae for determination of second-order effects. - The fact that according to some codes it is not necessary to introduce

these effects in the beams.

- The required minimum areas of reinforcement.

Note. The difficulty of the comparison calculations is that it is

not possible to determine the relation between the adopted live loads