Reinforced Concrete Design to Eurocode 2
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Reinforced concrete design to Eurocode 2
..
r-1
t
Other titles of interest to civil engineers: Civil Engineering Materials, fifth edition
EDITED BY N. JACKSON & R. K. DHIR
Civil Engineering Quantities, sixth edition 1. H. SEELEY Design of Structural Elements
w. M . c. McKENZIE
Design of Structural Timber to EC5, second edition w. M. c. McKENZIE & B. ZHANG
w. M. c. McKENZ IE Design of Structural Steelwork w. M. c. McKE N ZIE Design of Structural Masonry
Engineering Hydrology, fourth edition
E. M. WIL SO N
Highway Traffic Analysis and Design, third edition
R.J . SALTER
& H. B. HOUNSEll
Soil Mechanics, second edition
G. E. BARNES
Structural Mechanics, second edition
R. HULSE & J. A. CAIN
J. UR EN &: w. F. PRI CE DES CH AND J. M . DINWOOD IE
Surveying for Engineers, fourth edition Timber, seventh edition
H. E.
Understanding Hydraulics, second edition Understanding Structures, third edition
LES HAM ILL
DEREK SEWA RD
m ~jjj t·:
92 'Design of Concrete Structures' - commonly known as Eurocodc 2 (or EC2) recommends values for use in such cases. The precise shape of the stress-strain curve is very dependent on the length of time the load is applied. a factor which wi ll be further discussed in section 1.4 on creep. Figure 1.2 is typical for n short-term loading. Concrete generally increases its strength with age. This characteristic is illustrated by the graph in figure 1.3 which shows how the increase is rapid at first. becoming more gradual later. The precise relationship will depend upon the type of cement used. That shown is for the typical variation of nn adequately cured concrete made with commonly used class 42.5 Portland Cement. Some codes of practice allow the concrete strength
Strain
0.0035
Figure 1.2
Stress-strain CUIVe for concrete in compression
40 Figure 1.3
'E
E
£01 related to the mix, but a general relation ship is considered to exist between the modulu~ or da~ticity and the compressive strength.
Properties of reinforced concrete
~5
Figure 1.4 Moduli of elasticity of concrete
I I I .;
tangent or dynamic modulus
_ - - \ • secant or static modulus
I Strain
Typical vulues of Ecr11 for various concrete classes using gravel aggregates which are suitable for design arc ~hown in table 1.1. For limestone aggregates these vulues should be reduced by a !'actor of 0.9, or for basalt increased by a facLOr of 1.2. Thu magnitude of the modu lus of el asticity is required when investigating the de!lection and cracking or a structure. When considering short-tenn effects. member stjffness will be based on the static modulus Ecm defined above. If long-term effects are being considered, it can be shown that t·he effect of creep can be represented by modifying the value or Ec 111 to an effective value Ec,eff· and this is discussed in section 6.3.2. The elastic modulus at an age other than 28 days may be estimated from this table by using t·he anticipated strength value at that age. lf a typical value of Poisson's ratio is needed, this should be tuken as 0.2 for regions which arc not subject to tension cracking.
1.2.2
Steel
Figure 1.5 shows typical stress-strain curves for (a) hot rolled high yield steel. and (b) cold-worked high yield steel. Mi ld steel behaves as an clastic material , with the stmin proportionul 1o the stress up to the yield, ut whic:h point there is a sudden increase in strain with no change in stress. After the yield point, !his becomes a plastic material and the strain increases rapid ly up to the ultimate value. Iligh yield steel. which is most Table 1.1
Short-term modulus of elasticity of normal-weight gravel concrete
28 day characterlst:ic strength (N/mm 2) fck!fck. cube (cylinder/cube)
Static (secant) modulus £, 111 (kN/mmP) Mean
20/25 25/30 30/37 35/45 40/50 45/55 50/60 60/75 70/85 80/95 90/105
30 31 33 34 35 36 37 39 41
42 44
~l
6 :·. Reinforced concrete design figure 1.5 Stress- strain curves for high yield reinforcing steel
0.2% proof stress
Strain
0.002
(a) Hot rolled steel
(b) Cold worked steel
commonly used for reinforcement. may behave in a similar m;mner or may, on the other hand, not have such a definite y ield point but may show a more gradual l:hange from elastil: to pla:-:tic behaviour and reduced ductility depending on the manufacturing prol:css. A ll mater.ials have a simi lar slope of the clastic region with elastic modul us Es 200 kN/mm 2 approximately. The speciricd strength used in design is based on either the y ield stress or a speci ried proof' stress. A 0.2 per cent proof stress is defined in 1'\gurc 1.5 by the broken line drawn parallel to the linear part of the stress-strain curve. Removal of the load within the plastic range would result in I he stress- strain diagram following a line approximately parallel to the loading pori ion - sec line BC in fi gure 1.6. The steel will be left wi th a permanent strain AC. which is known as ' t-.lip'. If the steel is again loaded. lhe su·ess-strai n diagram will follow the unlonding curve unti l it almost reaches the original stress at B and then it will curve i n the direction of the first loading. Thus, the proportional limit for the second loading is higher I han for the in itial loading. This action is rel'tm-ed tO as ·strain hardening' or 'work hardening'. The load deformntion of the steel is also dependent on the length of' time the load is applied. Under a constant stress the strains will gradually increase - this phenomenon is known as ·creep' or 'relaxation'. The mnount of creep that takes place over a period of ·time depends on the grade of steel and the magnitude of the stress. Creep of the steel is of little significance in normal reinforced concrete work, hut i1 is an important factor in prestres:-:ed concrete where the prestressing steel is very highl y stressed.
=
ll
A
C
Figure 1.6 Strain hardening
1.3
Shrinkage and thermal movement
As wncrete hardens there is a reduction in volu!lle. T his shrinlwge is liable to l:Huse cracking of the concrete, but it also has the benefic ial c!Tccl of strengthening the bond between the concrete and the steel reinforcement. Shrinkage begins to take place as soon as the concrete is mixed, and is cause(.] initially by the absorption of the water by lhe l:Oncrcte and the aggregate. Furrher shrinkage is caused by evaporation of the water which ri ses to the concrete smfnce. During the setting process the hyi
'd
tl
II"
2.3.2 Partial factors of safety for actions (l't) Errors anti inaccuracies may be due to a number of causes:
1. design assumpt ions nnd inaccuracy of calculation: 2. possible unuf'ual increases in the magnirude of the actions; 3. unforeseen stress redistributions: 4. constructional inaccuracies. These cannot he ignore,
11 0
""'J 121
151
Figure 3.8
~
Bending-moment and
.-::-----&'---====---->,~-----,., kN.m
\ ·-. ----/,i
11
~
" \: ;::.- - -- :
158
-:l
158
171
~
110
~
(b)
shearing-force envelopes
~
124
~
--- ~
kN
~~ 110 124 171
Continuous beams with approximately equal spans and uniform loading
or
The ultimate bending moments and shearing forces in continuous beams three or more approximately equal spans without canLilevers can be obtained using relevant coefficients provided that the spans c.liff'er by no more than 15 per cent of the longest span, that the loading is uniform, and that the characteristic variable action does not exceed the characteristic permanent action. The values of these coefficients are shown in diagrammatic form in figure 3.9 f'or beams (equi valent si mplified values for slabs are given in chapter R). End Span 0.11FL
_.-/1
(a) Bending Moments
~
Interior Spa11
0.10FL
./1
. ~ if
., a
in I I. its ;oiS
lmd •\'0
When considering the critical loading arrangements for a column. it is sometimes necessary to include the case of maximum moment and minimum possible ax ial load, in order to investigate the possibility of tension fa ilure caused by the bending.
( EXA MPL E 3.3
Analysis of a substitute frame The substitute rrnme shown in figure 3.12 is part of the complete frame in fi gure 3.10. The characteristic actions carried by the beams are permanent actions (including selfweight) G~ = 25 kN/m, and variable action, Qk = 10 kN/m, uniformly distributed along the beam . The analysis of the sub frame will be carried out by moment distributiou: thus the member stiffnesses and their relevant distribution factors are fi rs! required.
!he pan
be
K
Figure 3.12
M
L
E ,.,.;
"' A
B
c
D
E
Beam 60 X 300
ues
C! ....
:mg t-
be
"lll
=ial
'3so
6.0m
~ 300 Typical column
section
6.0m
~ I
I
Substitute fram e
~ 39
~~~
40 · Reinforced concrete design Stiffnesses, k
Beam I =
OJ x o.63 = 54 .
kAB
= kco =
12 Spans AS and CD
X
lo- 3 m4
5.4x iO J
6
.0
= 0.. 133
(3) 92
n: lfl
(4)
133 147
147
Figure 3.15 Bending-moment and shearing-force envelopes
kNm
118
118
:2 Mcol table 3.2 gives
Thus. for the first loading arrangement and raking Column moment MAJ - 68.8 x
0.31 _ 0 58
= 37 kN m
0.27
= 32 kN
MAt:. = 68.8 X _ 0 58
0.31
m
MoK
= 40 x -0.58 = 21 kN
m
MoF
0.27 9 = 40 x - = I kN 0.58
m
This loading arrangement gives the maximum column moments, as plotted in figure 3.16.
44
~~
WReinforced concrete design
Figure 3.16 Column bending moments (kN m)
(
EXAMPLE 3.4
Analysis of a substitute frame for a column n1e substitute frame for this example, shown in fi gure 3.17, is taken from the building frame in figure 3. 10. The loading to cause maximum column moments is shown in the fi gure for Gk = 251cN/m and Qk = IOkN/rn. ~/,
Figure 3.17
___-..-
Substitute frame
E
~ l. lSG, tant. The ultimate design :.tress is gl\·cn hy 0.85f.l 1.5
- 0.567l~l rc the factor ol 0.1!5 allow~ for the dillcrence bet\\ecn the bcnd111g \trength and the 1 fer cru1-.h ing stn.:ngth of the concrete. and;'< = 1.5 ;, the w.uul partial \lllety factor 'le strength of concrete. The ultimate strain of feu~ = 0.0035 ~~typical for cla~'c' of CS0/60. Concrete cla:.sc:. < C50/60 will, un less otherwise stated. be
Figure 4.1 Parabolic-recta ngular stress -strain diagram for C011crete m compression
~
E
-
.z
Porabolic
0.851,.
e
,, 0.0020
0.0035 Strain
59
60
Reinforced concrete design
cono;idcred throughout this book a~ these are the classes mo~t commonly used in reinforced concrete construction. Also for concrete clas~cs higher than CS0/60 the delining propertie such a-, the ultimate \train . ,111 vary for each of the hjgher classes. Oel>ign equations for the higher cla\),CS of concrete can in general be obtained using similar procedures to those shO\\n in the tC\t with the relative properties and coefficients obtained from the Eurocodes.
4.1.2
Reinforcing steel
The representative !.hort-term design !>tre~s strain curve for reinforcement is given in tigure 4.2. The behaviour of the steel i~ identical in ten,ion and comprc!>sion, being linear in the clastic range up to the de1-ign yteld ~tress of f>kh, where jyk b the characteristic yield ~>tress nnd ), is the partial tnctQr of safely. Figure 4.2 Short-term design wess-stmin CUIVe for relnforcEC>ment
1,.
T. Tens1on and
compression
eE z
200kN/mm1
Stra1n
\\ 1thm the cla:-.tic range. the Stres\
rclauon~hip hct\\~Cn
elaqlc mouuiU\ ,.. stralll
= £, 'iO
that the uel>ign yield E)
Mraut
the Mrel>' and strain is (4.1 )
i1>
= (l~k);r.· ~' ~~,
ulthe ultim:tte limit for.f),~ - 500N/mm 2 ~y = 500/( 1.15
X
200
X
101)
= 0.00217 It !.houlu be noted that EC:! permits lhc u~c of an ulternutive design strc!>s-strain curve to that shown in figure 4.2 with an inclined top branch and the maximum strain limited to a value which is dependent on the clth:-. of reinforcement. However the more commonly used curve shown in figure 4.2 \\ill be U!.Cd 111 th1s chapter and throughout the text.
4.2
Distribution of strains and stresses across a section in bending
The theol) of bending for reinforced concrete assumes that the concrete will crack in the regions of tcnllilc . When moment redistribution is uppl ied, reference ~hou l d be mudc to 1011 4. 7 which uc,cribe~ how to modify the design equmions.
uld be I and . .,f the mum eo~ure
'o to rpth of
4.4
Singly reinforced rectangular section in bending at the ultimate limit state
4 4.1 Design equations for bending
lfDhcd lkd in I arc
a.:
te of
Bendang of the \ection will mduce a resultant tcn,ilc force F, 1 m thc reinforcmg \lecl, 31ld a resultant compre"t''e force Ill the concrete /\, which act' thmugh the centrotd of the- effective arca nf concrete in compres~ion. a!> 1-hnwn in figure -1...1. f-ur eqtullbnum. the ultimate destgn moment. M. mu't he balanced by the moment of ,t,tance of the 'ection so thai
lerl'\1011
1-.ion remforcement (t\,) required given the charactcnMic 11aterial ~uocngths are fu = 500 :-..tmm1 and f., = 25 N/mm'
"
M
A,
• •
hd'J~ ~
185 :!6()
10"
4401
X
25
00147 < 00 167
tncrefore comprc~~ion :.teel is not required.
Figure 4.6 Design example - singly reinforced section
6
66
Reinforced concrete design
Lever arm:
~ = d{ 0.5
T
/
(
0.25-
).~14)}
=-wo{o.5~ /(o.:?5 O.l·H)} 1.134 = 373mm cor allcmalively. the value ol
~
= f.,d
he obtained from the lever-arm diagram,
figure 4.5.)
M
A,== 0.87}~~ :: 185
X
106
= 0.87 X 500 X 373 I 140 mrn 2
Analysis equations for a singly reinforced section 'I he following equations may be used to calculate the moment of resistance of a given
:-.cction \\ ith a kno'' n area of \lee! retnforcement. For equilibrium of the compres-.tve force 111 the concrete and the tensile force in the \tcl'l tn figure 4.4: or
= 0.87J~vl Therefore depth or ~lfC).S block. ·~ 0.567J~~b
~
0.87/yk/\,
(4. 11 )
0. 567~~b
and x = s/0.80 Therefore the moment
or resi\lalll:e of the section
i~
P,1 X::
M
-
0.8~~vl,(d -
.
=- o.87}yiA
(
.1/ 2)
O.!l7{1 ~t\,) d - u3.iJ~~..b
(4.12)
The~e
equations assume the tension reinforcement ha'> >ielded. which will be the case if ca~c. the problem would require solving by trying -.uccessh·e value' of..\ unttl
.\ < 0.617d. If thi., i-. not the
with the Meel Mrains and hence '>tresses being determtncd from equations 4.2 and 4.1. to be u\cd in equation 4.12 instead of 0.87f~k·
Analysis of the section EXA MPLE 4.2
Analysis of a singly reinforced rectangular section in bending Determine the uJlimatc moment of resistance of the cm'>s-~ecuon ~ho\\n in fiuure 4.7 ~I\ en that the chamctcri!>tic strengths are i)l = 500 Nlmm1 for the rcinforcen;ent and :!5 N/mm2 for the concrete. 0.567f,
agram,
Figure 4.7
- ,0
- ---
:;:;
•••
_____)
Analysis example - singly reinforced sect1on neutral axi~
A, .. t470 mm1
For equilibrium of the compressive and
tcn~ite
forces on the
~cct ion
Frc - /·,1 t
a gi\'en
tree m the
0.567/..lb.l' O.X7/;kA 1 0.567 < :!5 3()() X I= 0.87
X
5(X)
X
1470
therefore 150mm .1110 .1
t4.ll)
.1/ 0.8 15CJ; O.X lHH mrn
fhi~ value of .1 is l c~s than the value of 0.6l7d derived froml>Cction 4.2. and therefore the steel has yielded and /,1 = O.X7/y~ a~ a~"umed. Moment ol' re1>istance of the section is
0.~7/yki\,(d
,1'/2)
O.H7 x 500 x 1470{520- 150/2) x lO
6
284 kN m
-l.J2)
case if trying
.ru -l.l, to
4.5
Rectangular section in bending with compression reinforcement at the ultimate limit state
(a) Derivation of basic equations It -.houhJ be noted that the equauons in thb !-.Cc.:tion ha\'e been deri\ed for the case ot zero moment rcdiMrihution. When this is not the ca"e, reference should he made to 'ection 4.7 '' hich deal' with the effect of moment rcdbtrihut1on
68
Reinforced concrete design
Figure 4.8 Section w1Lh compress1on remforcemenl
~
0.567(,,
0 0035
b
..... 1
.--- ---,_j_d
1 X~ 0 45d
• A,' •
neutral ax1s
d
.
lbll
A, .
Section
Strains
Stress Block
from the ~ection deal ing with the analysis uf u 1>ingly reinforced section nnd for concrete class not grcall:r than CS0/60 when M
> 0.167fckbd~
the design ultimate moment exceeds the moment of' re~il>turu.:c of' the concrete(Mbal) and therefore compression reinforcement i~ required. For 1his condition the depth of neutral axis, 1 .... 0.45c/. the maximum value allowed hy the code in order to en&ure a tension failure with a ductile ~cc1ion. Therdorc
::t-111 = d- !it-aJ/'2 = d O.Rlh:ll/2
= d - O.t{ = 0.8:!d
0.45cl,2
For equilibrium of the section in ligurc 4.K
~o
that Wtth the reinforcement at yteld O.R?f)~l\~
= 0.567j~~bs
O. K~f~~~~:
or with
0.8
S
X
0.!:\7}~1.A,
0.45d
0.36d
(4. 13)
0.204fdbd 1- 0.8~/ykA~
and taking moments about the centroid of the tension l.lecl.
M - F.c x ;:""' I F\ (-U6)*
Analysis of the section .. areas of compression steel. A~. and tension l>tecl, A,, can he calculated from 5 and 4.16. 1 1g Arut 0.167 and K = M I bd7cl into these equations would convert
(4.17 )* (4.18)* a }sts it hru, been assumed that the compressiOn steel ha' yielded so that the 'J ... = 0.87}yl.· From the proportions of the strain distribution diagram: 0.0035
and tor
(4. 19)
r ) and neutr.d
a ren ...ion 0.0035
=
\\ith };1. 500 N/mm~. the steel strain l l the compresston o,teel () 002 17 olvcd to give \aluc~ for A j btl and M ' /}{/' \o that a ~ct of dc.,1gn charts such as the one 'hm' n in figure 4.9 ma> be plotted. Before 1hc equation' can be ~olved. 1he steel stresse~ and./~ f11U'1,l be calculmed for each value of 1/ d. Thi~ i!> achieved by first determining the rclc\ ant strain:- from 1he strain dwgram (or by applyu1g cquauons 4.2 and 4.3) and 1hen by e\aluating the stresse:-, from the Mres., ·Slram curve of figure 4.2. Values of t:/d below 0.45 will apply when mome111s are rcdi~tnbllled. ltl>hould be noted that EC2 does not give design charts for bending. Hence although 1t 1~ po'~1ble to derive cham; a~A~ (d ·0.567 x 25 x 280 x 195(510
= 319 -
124
as~umcd.
d')
195/ 2)+0.!!7
>
'i00
:>sive or tensile yielded. it would have been nccc1.sary to try successive vnluc:, of .r unlil
~teel
had not
F" = Fe,· + F.., balances with the &teel 4.1. The btcel
~;tresses
~trains
and
).Lrcs~es
being c.:ulculaled from C4m11ions 4.2, 4J and
at halancc would !hen be used to calculate lhe moment of
resistance.
4.6
Flanged section in bending at the ultimate limit state
T-sections and L-~ections which have their flanges in comprc~~•on can botl1 be designed or analy:-ed in a similar manner. and the equations \\ hich arc derived can be applied to either type or cross-section. As the Range~ gene rail} pro\ ide a large compressiYe area. it j.., usually unnecessary to cons1der the ca~c where compressiOn steel i:-, required; if it .should be required. the design would be ba~cd on the principles derived in ..,ection 4.6.3
Analysis of the section
73
lbc '1ngly reinforced section it is necessary to consider two conditions: --e'~
block lies within the compression tlonge, and .:'' block extends beiO\\ the flange.
Flanged section - the depth of the stress block lies within the :ge s hr (figure 4.12) acpth of stress block. the heam can be considered as an equivalent rectangular 1 llreadth bt equal to the flange width. Thi~ is because the non-rectangular belm the neutral axb i' in tcns10n and b. therefore, con'>idered to be cracked :U\c. rhus K = M /btd~f..~ can be calculated and the lever arm determ1ncd from a m curve of figure 4.5 or equation 4.H. The relation between the lever arm. ;:. \.of the neutral axis is given by d
I
2
- d- ::) O..S671,,
"1 •
, - --n-eu-tr_a_la-xi-5-~ X
Figure 4. 12
1r-1 -~ ·8rx - "F.,
T-~cctlon,
-r· r --S
stress block with1n the rldnge, s h1
s/2
l
, F.,
10 Section
Stress Block
e" than the flange thtd..nes-; Chr). the we~s block ....._.._..___...... 1nd the area of retnforccment is given hy
doc~
lie wtthtn the nange as
M ll S~fvk:.
de,ign of aT-section beam is described further in section 7.2.3 with a worked e.
It
E.
MP LE 4.5
~lysis
of a flanged section
n nc the ultimate moment of resistance of the T-section ~hown in figure 4.l3. dl.tractcri
F" x :.:
;::: 0.567}~kbrs:.: 0.567
X
25
X
8()()
X
56
X
392
X
10
1 '
249 kN m If in the analysio; it had been found that s > h1 • then the procedure would have been :.unilar to that in example 4.7.
4.6.2 Flanged section -the depth of the stress block extends below the flange, s > h, For the design of a nanged section, the procedure dcscribt.:d in section 4.6.1 wil l check if the depth of the stress block extends below the nangc. An alternative procedure is to calculate the moment of resi tance. M1, of the section with s = lt1, the depth of the
Analysis of the section
75
~e ~h
(see equation 4.22 of example 4.6 following). Hence if the de~ign moment. Md. is that
\/d > Mr '1
the l :'iOO N/mm 2 and 2 25 N/mm • Calcu late the area of reinforcement required.
~1 ment
1
Figure 4.14 Design example of a T·sewon
0.5671,,
'the
j
I
h,. 100
'
X
axo}
A
• •
·r-
s I
l
· -~
'""
j
lo
ll
F,.
l~b,. 20'L.
Section
Stress Block
In figure 4. 14 Fe~
is the l'orcc developed in the flange
I c"' is the force developed in the area of web in comprcsl.ion
been
_)
below
. eck if -e
IS
to
f the
\lomcnt of
re~istance.
Mr. of the flange is
Ml
F.~ X ;:1
Jfr
0.567}ckbl ltr(d
ur
0.567
25
= 170 I..N m
similar to equation 4.8 but with Krut replacing K.
(4.3-t
Analysis of the section
81
Table 4.2 Moment redistribution design factors x001/d
{o
l=-ed to ~JO )
t5. but mem
Zt>al/d
Kbal
d'/d
0.821 0.853 0.869 0.885 0.900 0.917
0.167 0.142 0.129 0.116 0.101 0.087
0.171 0.140 0.125 0.109 0.094 0.079
According to EC2, UK Annex, k1 - 0.4 and k; - 1.0 0 1.0 0.45 0.82 10 0.9 0.45 0.82 0.82 15 0.85 0.45 0.40 0.84 20" 0.8 0.75 0.35 0.86 25 30b 0.70 0.30 0.88
0.167 0.167 0.167 0.152 0.137 0.120
0.171 0.171 0.171 0.152 0.133 0.114
According to EC2, k, - 0.44 and k2 = 1 25 0 1.0 0.448 10 0.368 0.9 15 0.85 0.328 20. 0.8 0.288 0.248 25 0.75 30b 0.70 0.206
• Maximum perm1tted redistribution for class A normal ductility ste~l b Max1murn perrnltte?d redistribution lor class Band C h1gher ductility ~teel, see section 1.6.2
/EX AMPLE 4 . 9
Design of a section with moment redistribution applied and to
112)*
I 3.3)*
"t n e of IJVC
~
the ons
and
08
The -;cellon 'hown 1n figure 4.17 i::. subject to an ultimate design moment of 230 k.. m niter a 20'/r reduction due to momem redistribution. The characten,tit matcnal Mrcngths Jre ~~~ 'iOO t\/mm 2 and /.;~ = 25 N/mm 2• Determmc the area' of reinforcement required Ul>ing the con,tants ~~ ond ~1 from (a) the EC2 and (b) the UK unncx to EC2.
b e 260
d'• l
so
• • A,'
II
(a) Using EC2
•
(1) From first principles (b - k,)d/k~
Limiting neutral axis depth,xhJI I rum EC2 clau!>e 5.5 therefore StI'C!.S blocI,. depth Lever arm
~1 .thul
~hal :h•l
0.44 and k~ - 1.25, (0.8 - 0.44)490/1.25 = 14 1mm = ox~hJI 11 3 mm = d - .lbal/2 490 l l3/2 = 434mm
'-'1omcnt of rc~btance of the concrete MhJI
F,.,
:t>al
from
0.567 f~~hl'bal
X Zbal
0.567
260
X
25
X
X
113
X
-l34
X
10
II
181kNm 'uhject ton uniform compressive stres of 0.567./~... fn thi ~ concrete pmvide 4.35 and 4.36 should be used, taking the neutral ax.t!'l depth equal the overall section depth. h.
88
Reinforced concrete design
Such M-N interaction diagram~ can be constructed for any o;hape of cross-section vv hich has an axis of symmetry h) applying the ba!>ic equilibrium and strain compatibility equations vv ith the :-.tn.:~!>-~train rclauon\, a\ demonstrated in the foUowmg examples. These diagram., can be very u~cful for design purposes.
(
'\
EXA MPLE 4 .1 0
M-N interactio n diagram fo r a no n-symme trical section
Construct the interaction diagram for the ~cction :.hown m figure 4.21 wirh 1:~ = 25 Nlmm 2 and J, . . = 500N/mm2 The bending causes maximum compression on the face udjm:cnt to the steel area tl~. For a for calcul..ning points on the interaction diagram varying depth~ of neutral a>.b nre: (I)
'"th
Compatibility of stram~ (used in table 4 3. columns 2 and 3): E"
£,
= 0 0035 (" ~·
d') 0.0035 (d ~ ')
(4.41
Analysis of the section ---ection .. strain • n the
Table 4.3
M-N interaction values for example 4.10 (7)
(2)
(3)
X
tK
.Es
d' 60 263d 158 0617d-241 XbJI d 390 h- 450 X·
. The
1 .JCrDS~
, 1ero
(4)
(5)
'~
fs
(N/m~) (N/mm2)
(mm)
Qcrete Th1s is • cally.
89
0 >0.00217 0 0.00217 >0.00217 0.87 fyt. >0.00217 0.00217 0.87 fyk >0.00217 0 0 87 fy~ >0 00217 0.00047 0 87 fyk 0.00217 0.00217 0.87 fyk
0 87 fy 0.87 fyl 0 .87 fy.. 0 93.3 0 87 fyk
(6) N (kN)
(1) M (kNm)
189 899 1229 2248 2580 3361
121 275 292 192 146 0
or when the neutral axis depth extends below the bottom f the section _
., 7{.t
d')
.
,.
c,, - 0.00. (?.I·_ 311) ,tnd ~' t ii 1 Stres~ \train relation~
~ ~ "Y
= 0.00217
=
7(x
0.002 (7.l _
d) 311
(x > II):
)
for the ~tee! (table 4.3. columns 4 and 5):
J - 0.87/yk f = l:: xc
(4.42 )
(Iii) Equil1hrium (t:lble 4.3. column~ 6 and 7):
0.! in Utblc ·U lor a range of key \ in tahlc 4.3 us a 1-erie~ of straight line!>. or courr.e. N and '"' could have been calculated for imcrmediulc val ue!> x to provide u more accurate curve.
or
N(kN)
Figure 4.22
(0, 336 1)
M-N interaction diagram for
3000
\\ ith
a non-symmelrical sec11on
2000 1000
0
-lAL )
M (kNm)
90
Reinforced concrete design
(
EXAMPLE 4 . 11
M-N interaction diagram for a non-re ctangular section Con~truct
the interaction diagram for the equilateral triangular column section in figure -l.23 ''ith fck = ~5 N/mm 2 and /)k 500 Nlmm 2. The bending is about an axis parallel to lhe side AA ami cau~e' max1mum compres!-tion on the corner adjacent 10 the tee I an:a
,- -.
Figure 4.23 Non-rect~ngular 1nter~ction
i\:.
.•
GO
,..,II
M N
exam pip
"'.......
0
'
.,II
,..
2~3s
3
II
""
31125 bars A
~
•
A,
--
400
•
A
I or thi~ triangular 'ection. the plastic centroid i~ at the ~ame location a~ the geometric centrOid, 'ince the momcnt off"' equals the moment of r, ubout thi~ axi1. when aU the bur h 0.!\.l
hlocl- applies to ela,tic condition' during the ),ervtceahiiH) limit ,t,Hc. In practice it i' nllt generally used in design calcultHion), cxcept for lrqllldetaming :-.tructure1>. or ror the calculations of cruel- width' :tnd dcflectiorh a' dc1-.crihed n chupter 6. With the triangular strc1-.s h loc~. the cro~s·M!Ct ton can he con,idcrcd as (t ) cradcd in the tcn,ion 1.onc, or (i t)
uncrackcd with the concrete re~isting a .~mn ll tlll\Ount or tension.
4.10.1
Cracked section
\ crad.cu \Cction i~ \hown in figure 4.27 with a stre)>s resultant /·,1 at'ting through the of the )>\eel and /·_. acting through the centrotd of the triangular \II\!\~ block. r:or cquilihrium of the ),ection
~..:cntro id A A:
=2wf 3 wf
3 5w
0
8
Dz
4
w
/:~
or
F,.
0 5/J.\f... = AJ~1
(4.46)
and the moment of 1'1.!\1\tance M
or
Fe, >
< ~
M - 0.5bxj~c(d- .\f3)
= AJ,M - 1( J )
(4.-+7)*
5
94
Reinforced concrete design
Figure 4.27 Triangular stress block cracked sectton
--- b
-, d
_!~tral
_
-
axos
h
_,_
A.
• • Section
1
' '
Strams
Stress
(I) Analysis of a specified section The t.lcpth of the ncutrnl axis, x. can he determ ined by converJ ing the section into nn ·equivalcnl' area concrclc as shown in fi gure ~.2X. where O'c £,/ Ec. the modulur
or
rnlio. b
Figure 4.28 f.qUJv~len t trc~mlorml'd
sect1on with the concr~tl' cr.1cked
d
h
Transformed • C,A, • rr..A steel arta f.
(,tl..ing the area momcnh about the upper edge: ,I
}JAI )
2.:::\
Then.: fore
or
Sofvtng thi~ quaurallc equation gives
(4A8)* f-.quation 4.48 may be \ohed U\tng a chart 'uch n~ the one ~hown in figure ~.29. Equattons ~.46 to ~.48 can be u~cd to analy).c a ~pectficd reinforced concrete section.
Analysis of the section a.A.'Ibd
Figure 4.29 Neutral-axis depths tor cracked rectangular sections elastic behaviour
0 30
-+
~
~
0 20
0.10
mto an mndular 0
0.2
0.4
0.6
x/d
(ii) Design of steel area, As, with stresses fn and fcc specified 1 e depth of the neutral axi~ can abo be expres~cu in tcnm. ollh~.: ~tr~uns and sucsses ol tl ~ concrete anu \!eel. I rom the ltnear Mrain distribution of figure 4.27: ~--·
\
/.;./£,
I = ~ - .t;.f '~" -1 f," £, J1,~. refore
1 ; -J.., --
"
oJ,.,.
Ec uations 4.47 and 4.49 may be used to design the area of tcn:-.ton \ tccl required, at a cified ~'>Ires::.. in order to resist a given moment.
, EX AMPLE 4. 12
Analysis of a cracked section using a triangular stress block
or the 1\ection ~how n in figure 4J O, delenninc the concrete and srcel slrc\scs caused hy
4.48)*
momcnl of 120 kN m, asl\uming a cracked section. TaJ.. c J:: .j t., 15 X 147() 0.16 'OO > 460 l '111g Ihe chart of figure 4.29 or equation .! 4X gi\'C\ ' - 197 mm. hom equation 4.47
.! 29.
e ...ection.
\(
~b,\ fc, ( d
9
nt
I..
15.
-
b- 300
Figure 4.30 Analysis example with triangular stress block 0
N .,..
•
j)
.c:
3H2S
'
1470 mmz
•••
L----l -
'
96
Reinforced concrete design
therefore 1:!0
10"
X
= 21 X 3000 X
197
X
fcc ( 46() 197) 3
therefore 10..1 N/mm 2
fcc
From equation 4.46
J~1A, = ~II.\ .t~~ therefore 300
10.3
I
197 '( -:;- x 1470
>
Jic stre% resultant Fc1 acts through the centroid of the triangular stress hloek 111 the tension Lone as ~hO\\ n in figure 4.31 Fnr equil ihrium of the section
F"
(4.50)
F,1 -+ F,t
= 0.5bx J,~ F" = 0.5b(ll r =A, f.,
I ,,
''here and
rl.(t
I
Taking moment!. about F._, the moment of resistance nf the section is given by
,
~ (It
M
(4.51 }
The depth ol the neutral axi~. '· can he determined by taking area moments about the upper edge /\A of the equivalent concrete section ~h(m n tn figure 4.32, :-uch that L(A1)
LA . n,. = 1:., .1s tcrme'I l he moduIar ratiO E,
Figure 4.31 Triangular stress block uncwcked section
b
,dl h'
..J
l"
1,.
L.
t 2x/3
'
·rA,
• • Sectton
Strains
Stress
2(h- x)/3
Analysis of the section b
Figure 4.32 Equivalent transformed section witil the concrete uncracked
d -
Transformed= fA steel area E.
=M
Therefore v · -
u
X
11/ 2 + OcA,
X
d
bll ~ o. A, h +2ncn/
(4.52) "'
2 I 2ncr A,/bit
r n
bh
!-rom the linear
proportion~
of the strain diagram in figure 4J I:
.\'
fl,
= 1-1 -
r
d
\'
"
C',,
£,.
j,,
J~
(4.53 )•
·XC',,
\
a~
Thtrcfore
X C'ct
'tress-scction in order to .:tt•rminc the moment ol rc~i~t:HlCC or the uncrad.cd 1-ection.
( EX AMPLE 4 .13
Analysis of an uncracked section lw the section shown in figure 4.30, calculate Lhc serviceability momcm of rc~istancc \\ ith nn cracking or the concrete, given ,k, - 3 Nlmm ~ . He - 30 I.Nimm' and l:.'s 200 I.Nimm ~ .
,.
A, bh 1470
300
"·
520
£, !:.",;
200
- 30
6.67
0 .()()94
97
98
Reinforced concrete design
,l
11 + '2o..,rd 2 + 2n.,r = 520 + 1 X 6.67 =-:-~-
2+2
f, = G~
X
X 0.009-f X 460 6.67 < 0.0094
= '2?2 mm
=:) uJ..,
= (460- 272)6.67 x 3 (420- 272 )
. N/mm1 15 2
M = il~!-, (d-~)-~h(h-Jlfc,c .l ~(II 1470 X 15.2 (460X
G X
272
i
272 3
~ (520
= 8.3 + 38.7 = 47 kN m
) J(}
t))
·I>+ ~~ X 30()(520 -
272)) 10 ,,
272)
X
3
CHAPTER
5
Shear, bond and torsion CHAPTER INTRODUCTION This chapter deals with the theory and derivation of the design equations for 5hear, bond and torsion. Some of the more practic lm\ andthc bending Mrc~~c., arc dominant. the direction of the !>tresses rends to be parallel 10 the beam axi'>. Ncar the support.'>. where the shearing forces are greater. the principal :.tres!>es become inclined and the greater the -;hear force the greater the angle ot mcilnation. The tensile stresses due to l)henr are liable to cau-;e diagonal cracktng of the concrete near to the support so that 1>hear reinforcement must be provided. this reinforcement i~ either in the form of (I) 'ltirrups. or (2) inclined har-. (used in conjunction with Min·ups) as shown in ligure~ 5.4 ant.l 5.5. The :.tcel stirrup~ urc al~o often referred to as link~. Figure 5.1 Pnncipal \ltl'I\C'I in d lJ(ldm
Load
Comprtmlon
D1agonal tension cracks
The concrete uc;elf can rest\1 'hear hy a combinatton of the un-cracked concrete tnthe wmpresston zone. the dowelltng action of the bendtng remforcement and aggregate mterlocJ... across ten,ion cracks but. becau-;e concrete ts weal tn ten'>ton. the shenr remforcement is destgned to restM all the tetl'ille stn~"'e' cau mu..,t be provtded 111 order to form a cage supporting the longttUlltnal reinforcement and to resiM any ten~ilc ~>tresses due to factors ~uch own rna} have l>Ufficient ,hear capacity ectiono; where Vtd ::::; VRd, then no calculated .,hear reinforcement is requ1red. The shear capacity of the concrete. VRtt c. in such situations i:. given by an empirical expres~ion :
(5.1) with a m1nlrnum value of: VRd c =
[0.035k 3 1~f~~ 1/~] hwd
(5.2)
where:
I'Rd,
the dc.,ign shear rc!tistancc of the section without
(I +\ ~10)
/..
/-;;-
1
A ~,
form a 11! due w ICU it tO
IC\tigndopCU. :> tho:-c tndard The use • shear e at the to the :nding a the 'hear force. \'tt· at the face of the beam·~ 'upport~ \O that A~
pre\iou,ly noted EC2 hmitr. B to a 'aluc het\\Cen 22 and 45
(i) With 0
degree~.
22 degrees (this is the usual case for uniformly distributed loads)
1-rom equation 5.4: VRd , ma\(21J -
0. l24b~~od( l - fck/'250)fck
(5.6) *
It v~" mu\r!l , < v, r then a larger value of the angle() mu~t be u~ed ~o thm the diagonal concrete Mrut ha5 a larger vertical component to balance Vcd·
(ii) With
e
45 degrees (the maximum value of 0 as allowed by EC2)
From equation 5.4: ressive
• latlure •a mate enicnl ct ng as
1.5.
area
VRJ
mnx !4.~1 - 0. IBb,.d( I - ./~k/250lt~~
{5.7)*
which is the upper limit on the compres~ive strength of the concrete diagonal member in the analogous tnt).~. When \1~;1 > VRd. 013 ~t~ 5 ,, from equation 5.7 the diagonal ~trut wi ll be over ~tre~scd and the beam's dimensions must be mcrensed or a higher clas~ of concrete be u~ed. (iii) With 9 between 22 degrees and 45 degrees
The requtred value for 0 can be obtained by equating Vc.t to VMcl•• ,, and solvang for(} in equation 5.4 a~ follows: \' "' -
"" -
VRd
0.36b..,.d( I - fck / 250)/.;k (Cot () T tan fJ)
- _ _.::...._..:....,-___:..::.::..:...-::-:--=:.;.:
' ma' -
104
Reinforced concrete design
and I I (cot 0 l tan 8)
= sine X cos e - 0.5 sin 20 (sec proof in the Appendix)
therefore b} substitution
B=O.Ssin-'{
V1-.J
o.l8b"d( 1 -
fc~o./
..,
_so}f.;l
} S 4S
(5.8a)*
which alternatively can be expressed as: 0 = 0.5 sin-' { V VEr
}
< 45'
(5.8b)
Rd mu,l-l'il
where Vr:1 is the shear force at the face of the ~upport und the culculutcd vulue or the angle(} can then he used to uctcrmim: col 0 and cak:ulate the !hear reinforcement A~w/s from equation 5.9 hclow (\\hen 12 < 0 reinforcement provide~ hen.imum ~hear force governed by diagonal compression failure is a function ot the inver~e of (cot 0 +tan 0) Cequmion 5.4) and the additional longitudinal tensile force. j.f 1cJ varies with cot 0 (equation 5.1 J). Figure 5.3 ~how~ the variation of the~c funl:tion~ from which it can be ~ecn that a~(}~~ reduced le'' ~henr reinforcement IS required, but thi.., is compenion bars as described in section 7.9. Example., illustrating the design of shear reinforcement for a beam arc given in Chapter 7.
( EXAMPLE 5. 1
Shear resistance of a beam
:tJon of
I
The beam in figure 5.4 spans 8.0mcLres on 300 mm wide :-.uppons. It is requin:d to ~uppon a uniformly distributed ultimate load, ll'u of 200 kN/m. The c hara~.:tcrist i c material strengths arc hk 30 N/mm 2 for the eontrcte and ./y~ 500 N/mm • for the steel. Check if the shear reinfrcement in the fonn of the vert ical links shown can supron, in shear, the given ultimate load.
= 2.5 •
H~ups
b: 350
at 17S spcg ...
[dII [1 11111 1 fO ~: 1
p (3).
' lrom
Hl2
Sectron 2H2S. A,
982mmz
MJid be Total ultimate load on beam Support renetion Sheur. V1 1 ut face of ~upport
= 200 "
8.0 1600 kN = 1600/ 2 HOO kN
- ROO - 200 X 0.3/ 2 770 kl"\ Shear. VChear force (I 1.a).
where A,.. i' the cross-.,ectional area of the bent-up bar. ror a multaplc 'Y'tem of bent-up bar~. a' in part (bJ of ligurc '\.5. the shear rc~i,tum:c " ancrca,cd proporttonately to the ~pacing, ~. Hence: 1
of
Va d
·~d
rom
. . ().!l7fvkA,w Sill 0
.
0. 9d(cmn
X --
cot 0)
s
or
-A,w = 0. 7Rrl/yk (col Vt-.a --:-:--:-·'' n + col 0}sin n urc nd
_)
(5.13)
Thi~ equation ic; ana logou~ to equation (5.9) for the shear rc!.istanc:e of shcua· lin i-s. In a !.imi lur way it can be ~hown that, bused on cru~hing of the concrete in the compressive ~truls, the analogous equation to (5.4) is given by:
(cot 0 + cotn ) VRdma~ < 0. 36h,.d ( l - fcl./250l.fck ;< ( I ' (J) ·
I cot·
-
(5.14)
und the additional tensile force to be provided by the provi!.ion of additional tension \lee I j.., gaven by a modtficd 'crsion of equation 5.12:
•11., ~ered
""by
~Frd
= 0.5va.J(cotfJ
cotn )
(5.15 )
EC2 abo require., that the max1mum longitudinal spacmg of bent-up ban, as hmJted to 0.6d( I + cot o ) and '>pecifies that at least 50 per cent or the rcquared shear reinforcement should be in the form of shear link'>.
109
11 0
Reinforced concrete design
5.1.4
Shear between the web and flange of a flanged section
The provision of shear links to resist vertical ~hear in a nangcd beam i., identical to that previously de cribed for a rectangular section. on the as!>umption that the web carries aJJ of the vertical shear and that the web Width. b..,. is used as the minimum width of the \Cction in the relevant calculations. Longitudinal complemental) shear stresses also occur in a nangcd !>ection along the interface between the weh and flange as shown in tigurc 5.6. Thi~ i~ allowed for b) providing transverse reinforcement over the width of 1he flange on the a:.sumption that thi~ reinforcement actl> a!> tic!> combined \\ ith compre.,s1ve stnlls in the concrete. It i., necesl>ary to check the posJ.ibility or fai lure by cxce. .sive compressive stresses in the ~truts and to provide sufficient steel area to prevent tensile fai lure in the ties. The variable strut inclination method is used in a similar munncr to that for the design to resist vertical shear in a beam described in 5CClion 5. 1.2. The design is divided into the following ~tngcs: 1. Caleul::~te lhe longitudintll design shear stres1.cs, ,.,~ ut the web-flange interface.
The longitudinal she:lf stresses arc ut a maximum in the region~ of the maximum changes in bending stresses that, in turn. occur ut the stccrest parttand of the flange b.., the breadth of the weh h1
and
6M
the
thicknc),~
b.,. )/2
(b1
of the flange
the change in moment over 1he distance
~~
Therefore 6M
---...,.X (d hr/2)
(br - b-.)/2 hr
Figure 5.6 Shear between flange and web
•h
Shear, bond and torsion
that all the the r by
that i~
I
the The l.to
The longitudinal
V[d h t
1r -
O.R?f~k cot Or
which is derived by considering the tensile force 4. The reqUirements of tran~vcr~c ~teel.
(5. 18) 111
each tic.
EC2 require' that the area of transverse steel should he the greater of (a) that given by equation 5. 18 or (b) half that given by equation 5.1 R plw. the area of steel required by transverlle bending of the flange. The minimum amount of tranwerse steel required in the flange is 2 1\, nun 0.26bdt.ifc~m/J).-.. (> 0.0013bdr) mm /m. where h - 1000 mm (),Ce table 6.8). Example 7.5 (p. 184) illustrates the approach to calculating reinforcement in flanged beam~.
tran~ver~e
shear
111
11 2
Reinforced concrete design
5.2
Anchorage bond
The reinforcing bar subject to direct ten'>ion 'ihown in figure 5.7 mu'>t be firmly ancll· • if it is not to be pulled out of the concrete. Bar.. suhject to forces induced by flexure rr. be ~imilarly anchored to develop their de),ign \tresses. The anchorage dependtreso.,
= m:rf'
-4 '
anchorage force
= comact area
.tnchorage hond '>trcll.,
- (lt> ryd7r 600 mm
I
Good bond condition) for all bars
11
Good bond conditions in unhatchPd zone Poor bond conditions 1n hatched zone
Good hontl cond iti on~ are t:onsidered to he when (a) bors are inclined at an angle of hctwccn 45 and 90 to the hori.amlal or (h) zero to 45 provided thnt in thi:-. second case additional requirements arc mel. These additional condition~ nre that bar~ Ute
1. either placed in members whofte depth in the vcrsc rcinfon:cment along the design anchorage length l:A,~.mon A~
= the
cross-sectional nrea of the minimum tnmsvcr~c rcinforccmcm ( = 0.25A, for beams nnd Lero for ~labs)
= the area of a single anchored bar with maximum har diameter
Thi!-. minimum design length must not be for tension bars:
le!>~
than:
0.3/b.rqd
for compression bars:
0.6/tuq.J
In both ca~es the mmimum value mu~t also exceed both lO bnr diameters and 100 mm. Anchorage~ may also be prm iUch a~ shcar links will he atlcyuate. Otherwi!>c transvcr~e reinforccmcm mu~t be prm 1ded. a:-, 1\hown in figure 5.13, having a total area of not less than the urea of one -.pliccc.l har. The arrangement of lapped bar~ must al~o confom1 to figure 5. 14. 'I he clear ~pace bet11 ecn lapped ha~ 'hould not he greater than ole;) or 50mm other11 i!>c an additional lap length e4ual to the clear ion reinforcement in the form of closed lin~ mU'.t hi! pro.. ided to rest\t the lullwn.ional moment. The cquauon!- for tor.;iorlJI destgn arc developed from a !>tructural model where it b :Mumed that the concrete beam m tor~ron beha1c-. in a tonal moment. TCd· lienee A,J~ tl
1.15
(5.26)
The required cross-secrional area of torsional link, can be determ1ned b) considering one tace of the hox section m• .-.hown in figure 5.16c. If ir is assumed that the urea of one
120
Reinforced concrete design leg of a link (Asw) is acting at it)> design yield strength([.,~ / 1.15) the force in one link is given by
AsJ;i../ 1.15 = q
X
h
H1mc'"er if the linh arc spaced at a distance proportionately and i~ given by
A,.J,l I s · =qxtx--
h cot 0
1. 15
~
apart the force
10
each link b reduced
ql cOL ()
---
T&t-"
(5.27)*
2A• cot B
Equations 5.26 and 5.27 can be used w de~ign a section to res1st torsion and an example of their u~e i~ given in chapter 7. The calculated amount oJ reinforcement must he provided in addition to the full bending and ),hear reinforcement requirements for the ultimate load comhinalions corresponding to the tor:.ionul lofld cnse considered. Where longitudinal bending reinforcement is required the ndditional torsional steel nrea ma) either be provitll:d by increasing the size of the bars. or by additionul bnrs. Torsional l ink~ must consist of tully anchored clo~>cd links spaced longitudinally no more than 11~ /H apart. The longitudinal steel must con~i'>t of at lea'>! nne har in each corner of the :-.cction with other bar5 di\trihutcd around the 111ncr periphery of the links nt no mor~ than 350 mm centres. Where the reinforcement I!) known equation~ 5.26 and 5.27 can lx rearranged for :malysis purpo).es to gi\c TEd and 0 a-; follows:
rh, Cction an equivalent thickne~ Urtl 1-; used. detinet! a~ equul to the towl area of the cro!\)o-l'>CCtion dtvided hy the oute circumfen:ncc of the ~ection. In the case of un aduul hollow section the cro~:.-~cctio aren woulcl include uny inner hollow a rea~ and the culcu latcd thicknel.s should not be taken u~ grcnter than the actual wollthickncs!'l. In no ca~c :-.hould the thickness be take'" as lc:-.s than twice the cover 10 the longitudinal hm~. When analysing or dc)oigning a ~ectinn it is abo necess~1ry to check that excess"" cnmpresl>ive stresses do nol occur in the diagonnl compre~sivc strut!>, leading possibly t compre~sJve failure of the concrete. With reference to figun: 5. 16c and taking th.. lim1ting torsional moment for strut comprc~sive l"adure as '/~J 11111, : rOI'CI!
in ~trut
= (q X h )/~i n(;J
An•a of strut = r~ 1 x (II co~ 9)
Stri!H in
~trut =
rorcc; Area
= --. CB/ J,~.; l5 let Sill C0\11 II -
"here f.:~o. IS the characteristic comprc-.si\ e stre~ in the concrete. As q then the above equauon can be e\pre~scd a-.
~ma.•/(2AI.)
k cot B)
('f',. 1 u~ / 2Ad cot 0/( 0.87{> 1d
Further infonnation on the pmctieal details of design fo1 tor~ion and a de11ign example arc given in charter 7.
>e taken ce~sivc
'1hly to • ng the
5.4.2
Torsion in complex shapes
A ~et:tion consi~ting of a T. Lor I :,hapc should be divided into component rectangles and each component h. then de~igncd separately to CIOnal moment (/ 1,1) ~o.ati•.f) the interaction formula ~tres.,cs
FigureS 17
Combined shear and tonton
(5. ~' ''here TRd ,..,
the dc\ign tor-.JOnttl
I'Rd ma~
the design shear
resi~tancc
resi~tancc
(equation 5.31)
(equation 5.8)
I r thi!o. imcraction equation i~ Mlti ~lied. the de~ign of the ~hear andLOr!.ionnllinks can b.. carried ()Lit separately providing th:.H the assumed angle of the compressive su·ll!s (0) ' the same for both tor:-.ionul and lihe;lr design. However for a solid rectangular sectiou, ~>uhj cctlo rcluti vcly small torsional and shea s trcl>se~. neither shear nor torsional reinlon:cmcnl i.., necessary if (5.3~
where VRd ~ is the ~hear capacity nf the concrete a:. given by equation 5. 1. TR.J.c i~ th~ torsional crad.ing momem \\hich can he calculated from equation 5.23 for a shear me, equal to the dc!>ign tens1le strc..,..,, /tonal moment Hence no calculations tor tor~iOll OfC generally llCC.:CS~ttry for the tlltitnti!C limit Stale Of bending of reinl'orced concret~o: unless tor~ion has been iucludccl in lht.: original analysis or is required for equi lihriurn. When combined n~.:xure nnd lor~ion il> COll\tdered the longttuclinal ~lecl tor both C:l'\C:. can be determined :-.cparately. In the flexura l tension tone the longitudinal .~tet: l required for both cases can be added. llowever in the flexural compressive ,.:one no additional tor'lonul longitudinal 'lee! i'> ncces \\ hil:h mu\t be tuJ..en into account at the member \t.!ing and remforcement dctathng \tagc. In \OillC ca~e' tabulated value' arc pronded for I) p1cal common ca~c)>. "h1ch are ba,cd on more complc\ formulae gi\en 111 the code of practtce. Reinforcement detathng may al~o he aflcch.:d hy \tahi lity ctm~tderattOnll as tlc!>crihed in sectton 6.7. a~ well a~ rull!~ concerning anchorage and lapptng of bars which have been dbcu,~>cd in section' 5 2 and 5.3.
6. 1.1
Minimum concrete mix and cover (exposure conditions)
rhese requirement~ arc interrelated and. although not rully dcwiled in
EC'2, EN 206 Pt't.fnmtance, Prod11clirm. Plal'ing and Complirmce Criteria and the comrlememory Briti~>h SIUndard BS 8500 give more detailed gu idance on minimum Concre/1!
e in ~oped
s 1kely
e sure iO..Il its ~u of
and
........
comhinattons of thickness nf covel' anu mix chan.tcteris tics for variou~ clas~>cs of cxpthurc. It ~hould be noted that the UK national Annex to F.C2 (and B~ 8500) induuc 'igni licant modification~ to EC2 itself. The mixes arc CXJWC~'eu in tcnm of minimum cemen t content. maximum free water/cement ratio and corre~ponding lnwe~t concrete ~ln:ngth cl:l\S. bxpo~urc clal>sification~ are given intahlc 6.1 whtch then define~ the mix und cmcr reqUirements and so on "hich must be complied \\ith. Con~r to re111forcement 1s specified and ~hown on drawing~ a~ a nominal \:tlue. This ~~ obtaincu from i'""m -
ign life change, then adjuMmcnts will be nece.~sary. Design ~ hou l d he hu:-.cd on the most severe exposure clu~sifkation if more than one nre combined. Minimum com:rcle mix requirement:. f'or ca~>es where freeze/thaw apply are summnriseu in luble ().3, wh ilst exposure to chemical attack (class XA exposure) may place fun her limit), nn mix dctalls nnd mny nl))o require additional protective mea:-.urc),. Reference should he mode to the appropriate documentation in !>uch cases (e.g. BS R.'i00).
and
Jtlon
I or ~t
Table 6.2 Cover to reinforcement ~50-year des1gn life, Portland cement concrete with 20mm maximur aggregate size) [Based on UK Nationa Annex]
Exposure class
Nomma/ Cover (mm)
xo XCl XC2 XC3/4
Not recommended for reinforced concrete
25 35 45
35 40
---~
1
XDI XD2 XD3
45 50 2
35
35
40 1
40 45 1
45'
XS2 XS3 Maximum free water/cement ratio Minimum cement (kg/m 1) Lowest concrete
50
0.70
0.65
0.60
45
1
0.55
35 1
35 40 50 1
35 40 50 40 40
60 1
401 551
so 1
45 1
1
40 2 60 2
45 40 55 1
0.45
0.35
0.35
XSl 2
30
45
0.50
55
240
260
280
300
320
340
360
380
C20/25
C25/30
C28/35
C32/40
C35/45
C40/50
C45/55
CS0/60
Notes. l. C~mt>nt contt>nt should be mneased by 20 kgtm 1 abolll' thE' values shown in the tdble. 2 Cement contem should be Increased by 40 kg/m 1 AND wat4'f 'E'ment rauo reduced by 0.05 compdred woth the value$ shown in I he table.
Ctnerol Notes These v,llues may be reduced by S mm 1f an approved quality control sy~tcm os spcclfoed . Cover ~hould not be less than the bar diameter + 10 mm to ensure adequate bond perform part 1-2 of l:.C2. wh1ch gi-.cs several po~sible methods rang1ng from detailed calculations to 'impltried tabk:-. as presented here. Hre effect\ are con,idered further m ~ection 6.6.2 and dc).1gn 1~ based on ~ati1>fying load heanng (R). 1ntegrit} (E). and msulatmg (II performance as appropnate. !'he approach offers the destgner permts~lhlc comhtnallon~ ol member dimcn~iun and a-.:1' dt,tance a" indicated 111 Tables 6.-t to 6.C1. for a range of \tandard lire resi~Lunce period' (minutes). These '' 111 generally appl) ''hen nonnul deta11tng ntleo;, have been follm' cd und ''hen moment redt,tribuuon doc:-. not C\cecd 15q f-or benms nnd !.labs.
Table 6.4 Minimum dimensions and ax1s distance for RC beams for fire resistance Minimum dimensions (mm)
Standard fire n•slstance
Possible combinations of a and bm1n where a is the overage axis distance and brn 111 is the width of the beam Cotllinuou~
Simply supported
A
R60 R90 R120 R240
bll111l
a bmon a b,n•na bm~n-
120 40 150 55
200 65 280 90
B 160 35 200 45 240 60 350 80
c 200 30 300 40 300
D 300 25 400 35 500
55
50
500
700 70
75
c
r
120 25 150 35 200 45 280
200 12 250
75
G
H
450 35 650 60
500 30 700 50
25
300 35 500 60
Note: The oiXt) do~tdnce o,1 from the sidt' of a beam to the comer bar should be o+ IOmm except whtrl.' bn,., tsgreater lhdn the values in
cofumm C and f
Serviceability, durability and stability requirements
1~
further detailing requiremenb may appl} for higher fire periods. \\htlrature cond•llons as 0 7
2. Mmlmum of 8 bars reqwed
130
Reinforced concrete design
6.1.3
Maximum spacing of reinforcement
Crad..ing of a concrete member can result from the effect of loading or can an~ hccau~e of re~tmint to \hrinkage or thermal mo\l.:ment. In addition to providing minimum area of bonded reinforcement (see sectton 6.1.5). cracking due to loading minimised b} ensuring that the maxtmum clear 'pacing-. hetween longirudm~ reinforcing hars in beams itn.:.,!-> level (j~) can be complicated and an acceptable approximalion is to take l as
.f..
(6.1
I'm orlicc and domestic ~ituations (sec fable 2.4 for other circum&tances), where .r;k j, the characteristic )>trcngth of' the reinl'orccment. h will hove a value of 1.0 unle' mnml.!nt reubtribution hm. been earned out. in whil.'h ca~e t i~ the ratio of the distribute.. moment to the undi~tributcd moment at th~: ~ection at the ultimate Iunit. Table 6.7 Maximum clear bar spacings (mm) for high bond bars 1n tension caused predom~nantly by loading
Steel stress (N/mm 1)
160 200 240 280 320
360
Maximum bar spacing (mm)
-----
300 250 200 150 100 50
Thcl'te ~racing rule~> do not apply to ~ l 11hs with u11 overall thickness of200 mm or lc"' In this cu!.c the srmcing of longitudinal rcinforcemem '>hould be no greater than thrt:o; times the ovcrull ~luh dcpth or . wo mm, whichever is the le!>ser, and secondar: reinforcement three-and-a-half time' the ucpth m 450 mm gt:nerally. In arens o concentrated loadi't or maximum mo111cnh these shou ld he rl!duccd to 2h < 250111111 al1' 311 400 mm respectively.
6.1.4
Minimum spacing of reinforcement
To permit concrete llO\\. around reinforcement during con . . truction. the clear di~tan~.-~ between bar~ should not be less than (t) the maximum bar siLc. (ii) 20mm. or (iii) the maximum aggregate size plu~ 5 mm. \\h1chever ts the greater figure.
Serviceability, durability and stability requirements
6. 1.5
~on
the and an
6. 1
Minimum areas of reinforcement
For most purpo~es. thennal and shrinkage cracking can be controlled \~ithin acceptable mit 20% main reinforcement
Longitudinal reinforcement in columns A '"'" 0 1ON..., 0 87 f>~ 0 002Ac where N!d is the axial compression force Vertical reinforcement in walls A, m11• 0 002-A. Note· b, es the ml'dn wodth of the tenseon zone.
500N/mm 2 ) C40/50
CS0/60
0.0018
0.0021
13
132
Reinforced concrete design
6.1 .6
Maximum areas of reinforcement
The~e
are determined largely from th~ pracllcal need to achic\e adequate compaction of the concrete around the reinforcement. The limns specified arc a.-. follows
(a) For a
~lab
or beam, ten'>ion or compressiOn reinforcement
IOOA, A, < -l per cent other than at lap.., (b) For a column IOOA,/ Ac ::; 4 per cent other than at laps and 8 per cent at laps (c)
For a wall. verticnl reinforcement lOOA, j A, ::; 4 per cent
6.1.7
Maximum bar size
Sl·ction 6.1.3 dc\crihed the.: limitations on har \ ptlcing to cn:-.un.: that crack widths due to loading nrc h.epl within acceptable limi t\, When cnnsidcring lmu..l-induccd cracking bar diameters may be rc~ t rit'tcd a' indicated 111 tahlc 6.9 which is ha~ed on ('30/.n concrete nnd 25mm cover as an altcmativc to limiting -;pac1ng. Jn calculatmg the ~!ee l stre~s. the approximation given in equation 6. 1 may he U\ed. Table 6.9
Maximum bar diameters (0.3 mm crack width)
Steel stress (Nimm')
160 200 240 280 320 360 400 450
Mox1mum bar lize (mm)
32
----
15
16 12 10
8 6 5
When cracldng occurs as u re~ult of re:-.truint to ~ hrinh.nge or thcrmul effects then the hur si;ws nutst be limited us indicated in table 6.9. but the maximum ~pucing lim it~ of tnbk 6.7 do not need to be applied. l'he \Ice! ~tre!>s to he used in table 6.9 can be cnlculated from equation 6.3 where A, P''" is the steel area pmvidcd at the ~cc ti on under con~iderntion and A, nun i5. given in equation 6.2. (6.3 )
6.1.8
Side face and surface reinforcement in beams
In beams over I m deep addiuonal remforcemcnt mu ... t he prO\ 1ded in the 'Ide faces to control crad.ing as indicated 111 figure 6.21al Thts reinforcement ~hould be distnbuted evenly between the main tension steel and the neutral a\.t\ and within the \tirrups. The
Serviceability, durability and stability requirements I
Figure 6.2 Side-lace and surface re1nlorcement
I.
N/A
t(d - x)
< 600mm
Side-face reinforceml'nt (a)
rcmforcemcnt (b)
m111imum area of this reinforcement can be cakuluted from equation ()_2 with k taken as 0.5. In assessing the maximum spacing nnd ~itc of lhi:. reinforcement from tables 6.7 and 6.9 a -.tress -.alue equal to one half of that calculated for the main tenc;tle reinforcement may be used and it mny he a11sumcd that the side face reinforcement is in pure tension. In addauon to the ahove requirement, EC:? requires that surface reinforcement ~~ provided where it is necessary LO control spalling of the concrete due to tire (axis distance> 70 mm) or where hundled hun, or har~ greuter than -10 nun diameter are u~cd U\ main rcanforccment. In the Ul\, however. thi'> " not adopted due to practical dirtkullie~ in providing such relJ1fnrccmenl. !·or high covers il i~ n:conunencled that udditional fire protection i' provided and crack wiuth calculations are recommended with largl! daametcr har~. The sua·face reinforcement, if provided, :.hou ld consist of' welded mesh or small diumeter high bond bars located 0111\iclt• the lin"-' as indicated in lagurc 6.2(b). Cover to thi' reinforcement mu't comply\\ llh the requirements of )o.ectton 6. 1.1 and the mimmum area of longitudiawl ~urfacc reinforcement should be I per ecnt the area ol the cnncrete ouhide the link.!> and in the tension zone below the neutral axb: shO\\ n as the ~l1.1ded nre.t in figure 6.2(b). The ..urface reanfon.:cment bar~ 'hould he ~puccd no further than 150 mm ~tpnrt und if properly nm:hored can he taken into account as longitudinal bending and shear reinforcement.
or
6.2
Span-effective depth ratios
The appearance and function of a reinforced concrete heum or skah may he impaared if the deflectaon under 'e" iccahiltty Joadang ts c.\ce~stve. Deflections can he calculated il~ indicated in ~ection 6.3 but it is more usual to control dellcctions hy placing a limit on the rutio of the spun to the effective depth of the hcam or ,Jah. EC2 specilie' equations to calculnte ha,ic span-effecta\e depth ratio~. h> comrol deflection' to n maxamum of span/250. Some typical values are given in table 6.10 for rcct::tngular sections of cluss CJ0/35 concrete and for grade 500 steel. The ratios can abo he U~>ed for Hanged sections except where the rtltl() of the width of flange to the \\idth of web exceeds 3 when the h(l~ie values shou ld be multiplied by 0.8. For two-way ~>panni ng slabs. the check for the hn~;ic -;pau effective depth ratio ~hould be based on the \horter span whereas for nat 'lab~ calculations \hould be ba~cd on the longer span. The two columns given in table 6. 10 correspond to levels concrete !'>trcs!. under "erviceability conditaons: highly Slrc-,sed when the steel ratio p exceeds 1.5 per cent and
or
133
1 34
Reinforced concrete design Table 6.10
1
Basic span-effective depth ratios (fyk- 500 N/mm , C30/35 concrete)
Basic span-effective depth ratio Factor for structural system K
Concrete highly stressed (p- 1 5%)
Concrete lightly stressed (p- 0.5%)
1.0
14
20
2. End span of continuous beam or one-way continuous slab or two-way slab continuous over one long side
1.3
18
26
3. In tenor span of continuous beam or one-way or two-way spanning slab
1.5
20
30
4. Slab on columns without beams
1.2
17
24
0.4
6
8
Structural system
1. Simply supported beam or one/two-way spanning simply supported slab
(!lilt slab) based on longer span
5. Cantilever
lightly strc:.\ed when p equal:. 0.5 per cent. pi-. gtven hy IOOA, rr.: ho reinforcement rntio and concrete &trcngth it may be more convenient to use the char in rigurc 6.3 \Vhich is for a simply supported '>pan with no compression steel togcthc with a modification factor K (as -;hO\\ n 111 table 6.1 0) accord111g to member type. T1 upprouch is basl.!d on the same ha~ic equations and offer!> greatl.!r flexibility than rchnn~ • placet! on tabulated values.
Figure 6.3
Graph ot basic ~pdn effective depth ratios for different classes or concrete
N 36
v"'
.,
~
b c 0
....vc
32
.
28
~
24
~
:t: -Q 0 Zl
~
-5
20
"' "'2:
16
CL
"Q
~ 'I'
::::
c
"'
Q.
"'
100A, .,1 12 0.40%
0.80%
1.2%
1.6%
2.0%
bd
Serviceability, durability and stability requirements
:-te)
The basic rutios are modillcd in particular cases a:- follows: (a)
For spans longer than 7 m (except tlat slabs) and where it is necc1-sary to limit dellcr.;tions to ensure that finishei>, such as partitions. are not damaged. the basic values -,hould be multiplied by 7/ span.
tb) For llat ,Jabs with span-, in e'he~ or tl of partition., should not exceed span, soo to a-.osd damage to fixture~ and tutm the creep cflect~. Lateral delkction mu't not be ignored. especially on tull ~ lender structures. and limitation:-. in these case~ mu~t be Judged b) the engineer. It i~ smpot1ant to realise th. l there arc many factors which may have \ignificant effects on dcOcclion~. anti me difficult to allow for. Thus any calculated values mu-.t be regarded a' an cMtmatc onh The mo.,t unportunt of these factors arc: 1. suppon re-,traint mu\t he c:-.timatcd on the ham ol si mpllticd as~ umpti ons. whll \\ill have varying degree~> of aceur:tl.); 2. the prl'ci'e loadmg cannot he predicted and crrnrs in permanent loading may have a signsflcalll l'lfect, 3. a cr;K~cd rneml1t:1 will heha-.c differently from one that is uncracked - thi.i' depth. Alternativcl). momenb ol area can be taken to c'tabli~h the neutral-a>;~). depth d1rectly. 'I he l>Ccond moment ol area of thc cracked section can then be determined hy laking econd moments of area about the neutral axi'
(6.6 where a~ is the modultlr nttio equal to the ratio of the cla!>tic modulus of the reinforcement to that of the concrete. For a given moment, M , and from clastic bending theory, the curv:uurc of the cracked section. { I / r).:,·• is therefore given by
M ( 1/rlcr=-E I
(6.7)
~. elf ~r
6.3.2
Creep and shrinkage effects
Creep The effect of creep will be to increase ddlccuons v.ith t1me and thu' 'hould be allowed for in the calculations by using an cffccthe modulus. /.:., ell· u-.ing the equatton t.~ cfl
where tt>
= Ecm/ ( I IS
- 6( ~. ro))
{6.8
a creep coefficient equal to the ratio of creep strain to initial elastic strum
Serviceability, durability and stability requirements Table 6.12
Final creep coefficient of normal weight concrete (Class C25/30)
Age at
Notional size (2fl.cfu) mm
loading (days)
100
200
300
500
Dry atmosphere (inside: 50% RH) 1
5.5
3 7
4.6 3.8 3.0 2.7
28 100+ Note:
139
A-
5.0 4.0 3.5 2.8 2.5
cross-~eclional
area ot concrete,
4.7 3.8 3.2 2.6 2.3 u
100
200
4.3 3.6 2.9 2.3 2.1
3.8 3.1 2.6 2.1 1.9
3.5 2.9 2.4 2.0 1.8
perimeter of that area exposed to drymg.
The value of rJ>. \\ hile oeing affected oy aggregate properties, mix design and curing conditions. is also governed by uge at lir~t loatling. the durm.ion of loatl and the section dunenl>ions. Table 6.12 gives some rypit:al long-term value~ of q'( XJ, to) a)< ~>ugge:-ted by EC2 for a class ('25/30 concrete made with a type N cement. The~e arc valid if the concrete i~ not subjected to a compressive ~Ires~ greater than 0.4~f...~ 1 1 ,, , at age fn (age at time of loading) and wi ll reduce as the concrdtic hcnthng theory which for smnll deflccuon~ 1~ b:1~ed on the cxprC'>'>ion d!,.
HI ~ tl\·-
M,
(6.10
whcre M~ ts the bending moment nt a section distun~.:e 1 rmm the origin 3); shown in ligure 6.6. l·or small deflections the term d'y/tl.t~ upproximalely equals the cu rvuturc which is the reciprocal of the radiu~ of curvalurc. Douhlc integnuion of equation 6.10 will yield un expression for deflection. This may be illu ~ t ratcd by considering the case of a pin-ended beam subjected to con.,lunl M throughout its length. ~o thttt M, - M. d2 1• El ., ·, u\-
(6.11 )
M
therefore
Figure 6.6
y
:::;
Pin-ended beam subji'Cl to a constdnl moment M
i. X
-
/
L
M
-, 8
"
Serviceability, durability and stability requirements ::25 30)
but if the ~lope ~ ~ zero at mid-span where x
L/2, then
C=- ML
2 am.l
d\· ML f1....:....-MI--
dl
2
.
Integrating agarn grves
M.1.:.
Ely
2
ML\
- -2+
D
0. y = 0.
•lit at support A whcn .t
D ncrete
Hence
()
ll\
1r a cia' ngth
M f,J
('"I2 - J..'·) 2
(6. 12)
ut any !>ection
The maxtmum dellcction in tht ~ cu~e will occur
m mid-'>pan, where
t
L/2.
in which
cac ,.ntJ\
but
=
M J2 £1 R
(6.1.3)
'"1c.:c at any uncracked section
H II
r
tne ma\imum dcllcction may be expressed a\ 6.10
Yrn l\
I ,I p, I. r
In general, the bending-moment di~tribution along a member wil l not be con~tant, hut Ill be a function or.\. The ha1.ic form of the result wi ll however he the l\ame. antlthc c:tlcction may be expres~cd as
the ca.~ I = \f
1
. mux .rmum del1cellon
a
= kL' · -I
rh
(o. I·W
6.11
k
n con~t unt , the value of which depends on the di~trihution nl bendrng moment~ in the member
L
the effective span
..!_ ,,, =-
the mid-span curvature for beams. or the support curvature for cantilevers
T)plcal valucll of k arc given in table 6.14 for varioull common shape!> of bendingment diagramll. If the loading is complex. then a value of k muM be estimated for the c :nplete load smce summing deflection" of simpler components will y1eld incorrect 1. ults.
141
142
Reinforced concrete design Table 6. 14
Typical deflection coefficients
Loading
B.M. diagram
(.M M ;;
!~d
w ....'aq-
~(1 -o)
t
~ w
~ wLl/8
Fi
-WaL
~~
~
k 0.125 4a 2 - Ba 1 1 ~a (if a- 0.5 then k 0.104
End deflection (if a - 1 then k
1--oL...
l
jltlw
wo1 L'/2
0.83)
End deflection
(if a
1 then k
at3 - a) 6
0 33)
-a(412 a) 0.25)
Although the derivation hn'> been on the bu-,h of an uncracked -;cction, the final cxprcl>sion is tn a form that will deal with u crad.cd ~ection ~imply by the :-.uh~titution of the appropriate curvature. Since the exrrel>~ ion involve11 the square of the lipan, it is important thut the true el'fccti vc span as dclincd in chapter 7 i~ used. particularly in the cu~c of cnmilevers. Dcllections of canti levers may al~o he im:rcascd by rotation of the ).Upporting member. and tim mu.,t he taken into :u.:count when the ~upporting structure 1s fairly 11cxible.
(
EXAMPLE 6.2 Calculation of deflection ~ - 30~
J ~
8,.... II
"'
II 't>
.• ....• Ll A,
5 No 25mm bars Figure 6.7
Deflection caltulalion example
E~llmatc the lung-term deflection of the he:.un ~h ow n in figun.: 6.7. It spans 9.5 metre~> nnd '·' designed to carry a unirormly di.~trihuted load giving rise to :1 qun~>i - pcrmuncm moment of 200 kNm. It i~ l:Onstructcd with class C25/30 concrete, is made of normal aggregate\ and the construction props are removed at 28 duys.
(a) Calculate curvature due to uncracked section I rom equation 6.5: ( I /r)""
M
where from table 6. Ll. Ecm - 31 1-.N/mm·. From table 6.12, a\suming loadtng at 28 day!. with indoor exposure, the creep cocflkienl ri> ~ 2.R hecau~c 2A~fu
(2 X [700
X
300])/2000
= 210
and hence from equation 6.X the effective modulus Ecctf
31/(l-2.8)
8.15k.N/mm~
i~
given by
Serviceability, durability and stability requirements lienee
(1/r)
2()()
-
uc -
8.15
X 101 X
2.86
X
X
106
(400
X
7003I 12)
10- 6 1 mm
• ote that in the above calculation lu..: has been calculated on the ba\i~ of the gross concrl.!tc \CCtional area ignoring lhe contribution of lhe remforcemcnt. A more accurate calculation could have been performed. as in example 4. 13 in chapter 4, but such accuracy i~> not JUstified and the 111mpier approach imlicatct.l will be ~ufficient l y accurate.
(b) Calculate curvature due to cracked section To calculate the curvmure of the crackct.l section the I value or the tra n~formcd concrete section must he calculated. With reference lO fi gure 6.5 the calcul:llionl> can be ~et out as
below. (i) Calculate the neutral axis position Taking arcu moments about the neutral axis:
b
X .1" X
300 nat
of
\~hich
.r/2
s the width hecomcl> excessive. in which ca~e appearance and durability suffer a' the reinfon:cment " expo~ed to corrO!>JOn. I he actual widths of cracks in a reinforced concrete 'truclllre w11l vary between wide limits and cannot he precisely e!-.timated. thu:- the limitutg rcqui remclll to be sflt i~>licd is that the probability of the maximum width exceeding a !>atisfactory value i' mall. The ma\lmum acceptable "alue 'ugge,ted b) EC2 1\ 0.3 mm for all expo~ure classe'> under thi.! action ol the qua~i-permcmmr comtuna11on of load~. Other codes of pract1cc may rccommend lower values ol crnck wicd to aiiO\\ for creep effect,,
6.4. 1
M echanism of flexural cracking
Thi-; can be illu\trated by COihidcring the hcha' 1our of a mi.!mher subject to a uniform moment. A length or beam a~ shown in ligure 6.1 u Wi ll initia lly behave elastically throughout, as the urplied moment M i~> increased. When the limi ting tensile stra1n t'or the concrete i~ reached. a crnck \\ill form and the adjacem tensile ;one will no Iunger be acted on by direct tens1on force!'.. The curvature of the beam. hO\\.C\'Cr, cnusc~ further direct tcn~ion stres!-.cs to develop at some diswncc from the original eruck to maintain equi librium. This in turn cau~es further cracks to form, and the process continue1> until the distance between cracks does not permit sufficient tcn~ile strl.!sscs to develop and cauf.e further cracking. The-.e tnt11al crach are called 'pnmary crack•;. and lhe average spacing in a region of constant moment is largely independent of reinforcement detai ling. A"~> the applied moment is increased beyond this point. Lhc development of cracks is governed to a large extenL hy the reinforcement. Ten-.ilc stresses in the concrete
1·
148
Reinforced concrete design
Figure 6.10 Bending of a length of beam
Strain
surrounding reinforcing bar::. are caused hy homl a~ the strain rn the rcinforceme • increa:.es. These !:>tresscl. increase with distam:c from the primary cracks and m... eventually cause further cracks to form approximately mid-way between the primar cracks. This action may continue with increal.ing moment unti l the bond betwee concrete and steel i~ incapable of developing 1.uflicierH tension in the concrete to cau"t; further cracking in the length between existing crnc~s. Since the development of th~ tensile stresses is caused directl y by the prc),ence of reinforcing bars. the spacing o cracks wi ll he innuenced by the spacing of the rei nforcement. If bar 0.00067
200
_
X
101
'>
0 6 ...:.:.:._ .
200
X
101
151
152
Reinforced concrete design
(iv) Calculate the maximum crack spacing (s, max) ~,rna\= 3.4c - 0.425klk~O/Pptff
where: c = cover = 1000 - 930 1.: 1 0.8 for ribbed bar;
401 2 = 50 mm to main bar-.
0.5 for flexure o - bar diameter = 40 mm
k1
hence
.I,
- 34 ln.l.l< -
X
•
50
+
0.425
X
296mm (which is
O.R X 0.5 0.0539
le~s
X
40
than S(r I r&/2)
350mm)
(v) Calculate crack width 1\'k
().()Ql
X
296
=0.30 mm which ju..,t satisfies the recommended limit. l_______________________________________ ~J
6.4.4 Control of crack widths II I'- ~1pparent from the expre:-.,ion'> derived ahove that there arc four fundamental \\'8) 111 \\ h1ch .,urfacc crack '' idth' may he reduced: (I) reduce the .. tre~' in the retnlorcement (a ) "h1ch w1ll hence reduce "": tiil rc.!duce the har dtamctcr' (ol \\h1eh \\Ill reduce b:u -;pacmg and ha\c the effect d reducing the crud. spacing (1, 111a,): (iii l tncrease the effective rcint'orccment rnrio (Pp.rtt ); (iv) U\C high bond rather thtm plain bar~. The u:-,c of steel at reduced 1>tn:s~c~ i:. generally uneconomicul and, ulthough ll approach is used in the design of watcr-rctllining ~trucwre~ where crading mu\t ofl~ he avoided altogether, it is generally easier to limit the bar diumcter11, incrcal!c f!p,dl u~c high bond bars in prclcrcncc to plnin bnn•. To increase Pr .r1 the effective ten~ ton area A ~.e l'f' should he made a~ small .:. possible. Thil' i~ hest achieved by placing rhc reinforcement close to the tension fal~ &uch that the depth of tension area {2 ..5(11 d)} i~ mudc as ~;mall as possibrecogni . . ing, nevertheless, that duruhility requirements limit the minimum value c. cover. The calculation of the design crack width~ indicated above only applie~ to region• '' ithin the effective tension zone. Since cracking can also occur in the side face of beam it j., also good practice to con!>idcr the provbion ot longitudinal !>tee! in the siti.. faces of hcam\. The critical po panJcularl) important in the ca.-.c of marine structure). nenr the \\ater surface, and cause strco,~e' to develop if the movements producetl arc re),traincd. It ~~ also JX>ssihle for crystal gro\\ lh to occur from drying out of sea water in cracks and pores. and this may cause further mtcmal stresses. leading 10 cracking. Alternate free7mg anti thawing is another maJOr cau~e of physical damage. particularly 111 road anti runwa) slabs and other ~>ituations \\here water in pores and cracks can frccLc und expand, thus lead111g to spalling. It hu~ been found that the cntrainmem of n .small percentage of air 111 the concrete 111 the form of small dllicrete bubble'> offer::. the mot thi~ form or attack. Although tlus reduces the strength or the com:rcte. it is recommended that between 4.0 and 6.0 per cent by volume of entrained air llhould be included in concrete subjected to rcgulur welting and drying combined with ~cvcre lrost. All the:>e form~> o f attack may be minimised hy the production of a dense, wellcompacled and well cured concrete with low pcnncahility. thus rc1-.tricting damage to the surfucc 1011e of the rncmhcr. Aggregates which are likely to react with the a l~ ali matrix ~hou ld be avoidetl (or the alkah levels of the cement carefully limited). a:- must tho~e which exhibit unu:-.ually high -.Jmnknge charactemtic~. If thi\ i:-. done, then pcrmcahili ty, and hence durability. i~ affected by
1. aggregate type and dcn,lly; 2. water cement ratto: 3. degree of hytlr Coupled with this I' the need fnr n nn-porou~ aggregates \\ hich arc hard enough to re,ist any allntion, and lor thorough compuction. It is e~~entiul that the mix i' designed to have adequate workability J'or the situnlion in which it is to be used. thUll the cement content of the mix must be reasonably high. EN 206 1>pecifics minimum cement content:- for vnriou:- expo~ure contlitions according to cement ty pe~. ns well as minimum strength and maximum wmer t·em~·111 ratio which can ulso be related to minimum cover requirement~ a~ dc~crihcd in ~ce ll on 6. 1.1. The con~e4ucnccs of thermal effects on duruhility must not be ove1lookcd, and very high cement content~ ~hould only be used in conjunction with u detai led cmcking 1 u~~es~rnent. A cement content of 550kg/m i!. oJ'tcn regarded a~ an upper ]unit fnr general u~e. Provided thnt ~uch mcallure~ ure taken. and that adequutc cover ol \Ound concrete ~~ gtven to the remforcement, deterioration of reinforced concrete i~ unlikel> Thus although the ~urfnce concrete mu} be affected. the rcinfor'-111!! \ICC] \\Ill remain protected by nn alkaline concrete matrix which hn~ not hcen cnrbonated h) the atmosphere. Once tht!> CO\ er hrcaks dlm n and '' ater and po!>sihl) chem1cab can reach the Mecl, ru-;ting and con\cquent cxpan,ion lead rapid!) to cracking nnd ~pallmg of the cover concrete and /.evert: dnmage - \ i~uully and sometimes :-.tructurally.
157
158
Reinforced concrete design
6.6.2
Fire resistance
Depending on the type of structure under cono,idcration. it may be neces~ary to constde"" the lire reststance of the individual concrete members. Three conditions must examined: 1. effects on structural strength 2. name penetration re:>Jstance
3. hctntclllre to '' nh,tand cxtrcmcl. ot accidental loading as may be caused by colli'>ion, c\plo.,ion or stmtlar, n i' important that resulting dumagc 'hould not be di~proportionate to the cau-.e. It follm'' therefore that a major a rclathely minor mi,hap which ma) ha\'e a rea~onahl) high probabihry of huppcnmg in the anticipated lifetime of the ~tructure.
Serviceability, durability and stability requirements The po~sihilities of a structure buckling or overturning under the 'design· loads will have been con~idered as part of the ultimate limit ~tate analysis. However. in ~orne instances a ~tructure will not have an adequate lateral -;trength e\en though 11 hal> hcen designed to rc~ist the speci fied combinations of wind load and vertical load. Thi' could be t11e case if there IS an explo~ion or a slight earth tremor. since then the lateral load\ are proportional to the ma~s of the '\tructure. Therefore it is recommended that at any floor level. a ~tructure should alwayl> he capable of rcsio;ting a minimum lateral force al> detailed 111 section :\.4.2. Damage and pOl>sihle in,tability should also he guarded again-.L \\hercvcr possible. for example vulnerable load-bearing member~ '\hould be prmected from collision by protective features iluch as hanJ..s or barrier),.
6.7.1 Ties l.n addition to the&c precautions, the general stabi lity and robustness or a bu ilding structure can be increased by providing reinforcement acti ng a~ tics. These ttes ~ohould act both vertically hetween roor and foundations. ami horizontally around anti m:rw.s each noor (figure 6.15). and all external vertical load-bearing mcmhl!rs should he anchored to the noor~ nnd beam~. If a building i~ divided hy expansion joint~ into ~lructurn ll y independent sections. then each section should have an independent tying ~y~tem.
Internal ties
Column ties
Penpheral lie
Vertical ties
Vertical ties Vertical tics are not generally neces~ary in ~tructurc~ of lc),s than f1vc ~torcy~ hut in higher hui l d 1 ng~ should be provided by reinforcement, effect1vely continuou~ from roof 10 foundation by means of proper laps. running through all vcnical load-hcruing member~. 'I hi~ \ tee! ~hould he capahle of resisting a tcnsih: force equal to the maximum tlc~ign ultimate load carried hy the column or \\all from any one ~torcy or the roof. Although the accidental load ca.\c io, an ultimate limit ~talc, the ultimate load u~ed hould reflect the loads hkely to he acting at the umc and the qut1.1i-pemwnen1 value would normally he taken. The mm i~ to contribute lO n bridging sy:-.tem in the event of loss of the member at a lower Jc,cl. In in silu concrete. thi~o. requtrement i~ almost invarinbly satisfied hy a normal design, but joint detailing may he affected in precaed to their charm:tcril>LiC strength is given m terms of a force F,. where F1 60 I. X or (:!0 + 4 x numhcr of storey~> in structure) k.\1. \\ hichever is less. Thil> expression tal.cs into account the increased risk of an accident in a large building and the seriousncs:-. of the collarsc of u tall structure.
(a) Peripheral ties The rcripheral rie must be provided. hy reinforcement which is effectively continuous. around Lhe perimeter of the building at each lloor and roof level. This reinforcement mu~l lie within 1.2 m of the outer edge and at it~ characterist ic strc!>s be capahle of resisting a force of at least /• 1•
(b) Internal ties lntcmal tie ... should abo h~ provided at each noor in two perpendicular directions anu he anchored at each end. c1thcr to the peripheral tic or to the continuou~ column or \\all tiCS The~e ues mu~t be effect!\ cl} contmuous and they may either be ')pread even I) acros'> a floor. or grouped at beam~ or wall~ a-. cun,en1ent. Where walls are used. the tie reinforcement must be withm 0.5 m of the lop or hottom of the floor -,lab. The resi•ilance required h. related to the :-.pan and loading. lntemal uc:-. mu~t be capable of rc~i-,ting a force ol F1 kN per metre w1dth or lltCIIl + lJk )/7.5 1r/5l.N per metre width. if this 1' greater. Ln th1s expression. lr ~~the greatest homontal di~tance 111 the d1rcction ot the tic between the centre!'> of vertkal load-bearing m~.:mbcr . Tht.. load1ng (Rl - q!..) I.N/m! is the avcrag~.: characteristic load on unit ~rca of the floor con~'>ldered. If the tie:-. are grouped their maximum spacing should he limited to 1.5/r.
(c) Column and wall ties Column and wnll tics must be ahle ll) resh.t n Coree of ut lenst ~ per ~.:cnt of the total verlil:al ultimate load at that level for which the member has been designeu Additionally, the resistance provided must not he li.:sl. than the smaller of 2F1 or F1IJ2 .5 kN where/, is the floor to ceiling he1ght in metre:.. Wullti~.:s ure nssesscd on tht basi' ()['the ahovc forces acting per metre length of the wall. while column tics are ~.:onccntrated within lm of ci th~.:r ~1de of the column cemrc lim.:. Part1cular care ~houla be taken with comer column' to cnr,ure they urc tied 111 two perpendicular directions In eon~idering the structure subjected w accidental loading it i., assumed that no othe force:-. are acting. Thu reinforcement pro\'ided lor an} other purpmes may also act a tics. Indeed, peripheral and intcmalt1es mu} abo be con~1dcred to be acting a:-. columr or ""all tie'>. A'!Ctfied degree of structural damage. tn
'
( EXAMPLE 6 .6 Stability ties
Calculate the \lability tiel> required in an eight-storey buildtng of figure 6.17: Clear storey height under lk:ams
= :! 9 m
rloor to ceiling height ({,)
3.4 m
Characteristic permanent load C.~d
=6 kX/m'
Characteristic variable lo not comply nh the vertical-tie requirement~. or when every precast lioor or roof unit docs not have ,ufficient anchorage to resi~t a force equal to F1 kN per metre width acting in the direction of the 1-pan. The an:1ly:-.i1> mu~t ~>how that each key load-bearing m~.:mhcr. it~ ~:onnections. and the horizontal mcmher:-. '' hich provide lateral support. arc able to 1\ ithstand a sped lied loading from any direction. If this cannot he satt:,.hed. then the maly:,.is must demonstrate that the rcmm al of an} single \ en1cal load-hearing element. Hhcr than key membcrs, at each store) 111 turn \dll not re~uh 111 collapse of a !.igniticant part of the 1>1ructun.:. Thc minimum loading that m;,y m·t from any direction tlll a key member is recommended us 34 J..N/m1. The decision as to what load:. should he considered acttng is left to the engineer, but will generally he on the basis of permanent and reali~tie variable OOmewhat arbnral) value. The 'pressure' method wi ll generally be suitable for npplicatinn to column:- in preea~t tmmed structure~; however, where precast load-bearing panel con~truction is being used an approach incorporating the removnl of individual element.-. muy be more uppropriatc. ln this case. \ c111Cal loading/> -;hould be U'>:;e-..,cd a' described. and the ~tructure mve:-.tigated to dcterr111ne '' hcther it i-. able to remain ~tand1ng b) a d1fferent ,true tum I Jet ion. Tills actton may 1nclutle parts of the damaged :.tructurc bchtl\ 111g a. . a canttlcver or a catenary. and it ma> also be necc~sary to con~idcr the strength of non-load heanng pur1itions or cluoding. Whichever upprouch is adopted, such analyses arc tedious. and the provi~ion of ~o:lfeetive tie forces within the structure ~hould be regurded as the preferred solution hmh from the poun of view of design and performance. Continuity reinforcement and good detailing '' 111 greatly enhance the overall ltre resistance of a \tructure with respect to collapse. A tire-damaged ~tructurc with reduced member ~trength may even be ltkened to u ~tructure subjected to accidental overloud, in the wall. These bars should ha\·e at lca't a tension anchorage on either ~ide of the con,truction JOint. Some typical reinforcing !.lccl detail" are given in I:C!! and a typtcal detail for a coupling beam is shown in figure 6.23. X
coupling beam
---
I
~
Figure 6.23
- ·v
TypiCal reinforcement detail for a coupling beam
~hear wall
diagonal cage of rCinforcement
Section X -X
6.8.4
\
opening X _.
v
Columns
Column' and their connection' to beaml> are critical parts of a 'tructurc. ratlurc of a column in a huilding can be cata~trophic leading to a progrcs-.ivc collapse. and the formation of plastic hinges 111 columns above the ba~c of u building 'hould be avoided. llorimntul hoops of helical reinforcing bars have been found to give a •Monger containment to the longitudinal vertical bars than thot provided by rectangu lar links and at a beom-to-cotumn joint horizontal steel reinforcement hoops not les.~ than 6mm diameter are udvisahlc with in the depth of the beam. At extetnal columns the longitudinal reinforcement of beam:-. should he well anchored within the colu mn. Thi!-. may rcquirc special mea~ures ~u ch as the provision of henm hounches or anchorage plate~ and some typical examples of details ure given in hC8.
6.8.5
Beams and slabs
Beam~ ~hould he ducule so that plastic hinges can form. and thc~e should be dtstributed throughout a \tructure. uvotding 'soft' sroreys. This will provide a gradual type of failure und not a 'udden catastrophic frulure such a:; that U!)suciatcd with shear or brittle compre~stve failure. The formotion of plastic hinges also allows the ma\imum moments to he rcdi,trihutcd to other pans of the statically indelenninatc \lructurc. thu)) provtding more overall Mtfcty. The fir:-t pluMic hinges arc likely to form in the sections of the beam ncar the column \\here the maximum moments are hogging, causing compression on the lo""cr fibres i.O that the section acts cffecuvely as u rectangular section. Plastic hinges which form later
168
Reinforced concrete design
at mid-span will have compression on the upper fibre!. -,o that the '\ection is effective!} .. T-section \\ith the slab acting a!-. the flange and there i' a large area to resist the compress10n. Further discu~sion of the de,ign of ducule \ectJOO'> i~ given in Section~ 4.2. 4.4 and 4.7. The beam section. close!) 'paced to resist the c;hear and to prmide greater compre...sive resi,tancc to the cnclo:.t'd concrete The provision of comprc\sivc steel reinforcement abo cn\ures a more ductile ~ection.
The \lab' in a building act :1!> rigid horitontal diaphragms to stiffen the ~tructure against tor11ion during ~ei:.mic di~turbances and nlso tran~fer the hori7ontal forces into the columnl> and shear walls. The l>labs ~hould be well ned into the columns. the ::.hear walls and the perimeter beams wtth contimuty reinforcement a~ indicated previously. When precast concrete slabs nre used they ~hould have good length' of hearings onto the :-ttpporting beams and sheur walls ~hou l d also he provided with continuity 'ltcel over their supports so that they can net :\)., cont inuous indeterminate members. In this way they can also develop their full ultimate reserve of 'llrcngth hy enabling a tensi le (:!liCJHiry action.
CHAPTER
7
Design of reinforced concrete beams CHAPTER INTRODUCTION Reinforced concrete beam design consists primarily of producing member details which will adequately resist the ultimate bending moments, shear forces and torsional moments. At the same time serviceclbility requirements must be considered to en~ure that the member will behave satisfactorily under working loads. It is difficult lo separate these two critf'ria, hence the design procedure consists or a series of interrelated steps and checks. These steps are shown 1n detail in the flow chart in figure 7.1, but may be condensed into three basic design stages:
2. 3.
preliminary analysis and member sizmg; detailed analysis and design of reinforcement; serviceability calculations.
Much of the material in this chapter depends on the theory and design specification from the previous chapters. The loadIng and calculation of moments and shear forces should be carrit.>d out using the methods descnbed in chapter 3. The equations used for calculating the areas of reinforcement were derived in chapters 4 and 5. Full details of serviCeability requirements and calculations are given in chapter 6, but it IS normal practice to make use of simple rules which are specified
__..
169
-1 70
Reinforced concrete design
EC2 Section
Y4ttable actiOns
2.3.1
Concrete class
D
D
Estimated self-weight
Concrett' cover
D
Mtnimum sectton
D 5
4.4.1
D
Permanent actions
2.3.1
3.1
D
PRELIMINARY ANALYSIS
Durab1hty and fire re~istance
4.3 and 4.4
D Trial b
c
·;cOl 0
Esumate d from
~
c .E"'
~
...
6.1 singly reinforced doubly reinforced
r--
.[~ 6.2 7.4.2
m~ximum allow~ble?
Vw
ChPc~
basic 1pan-crfrcttVe depth ratoos _ __
.;L
Select II
.,
-DETAILED ANALYSIS & DESIGN 8endong moment and shear force envPiclpcs
,J. 6.1
Bend1ng re1nforcement design
84
Anchor~ge
,l.. ..:.'"J..
(
8 and 9.2
~
Bending relnforcPmcnl details r-1
../....?
7 4.2
Check
span-cff~tovc
depth rauo
D 6.2
She~r
rconforcement deso9n
D 7.3
C~lculate
crack wodths (of rtQuored)
D 7.4
Calculate denectlom (if required)
D FINISH
Figure 7.1 Beam design flowchart
c
0'
·~
0
;;
c u:
"0
...c
~ ·~
c; 0
Design of reinforced concrete beams
in the Code of Practice and are quite adequate tor most situations. Typical of these are the span-effect1ve depth ratios to ensure acceptable deflections, and the rules for maximum bar spacings, maximum bar sizes and minimum quantities of reinforcement, which are to limit cracking, as described in chapter 6. Design and detailing of the bending reinforcement must allow for factors such as anchorage bond between the steel and concrete. The area of the tensile bending reinforcement also affects the subsequent design of the shear and torsion reinforcement. Arrangement of reinforcement is constrained both by the requirements of the codes of practice for concrete structures and by practical considerations such as construction tolerances, clearance between bars and available bar sizes and lengths. Many of the requirements for correct detailing are illustrated in Lhe examples which deal with the design of typical beams. All calculations should be based on the effective span of a beam which is given by
lrtt
In · a, ~ a2
where
111 is the clear distance between the faces of the supports; for a cantilever In is its length to the face of the support o1, o2 are the lesser of half the width, 1, of the support, or half the overall depth, h, of the beam, at the respective ends of the span
7.1
Preliminary analysis and member sizing
The luynut anti -.i;c of member\ nrc very often controlled hy archttectural tletail1-. and clearances for machtnery nnd equipment. The engineer mut either chcc'- that the hcam ~i.Ge!> arc adequate to carry the londing. or altcmutively. decide on !lite:-. that arc adequate. The preliminary analy..,i~ need only provide the maximum moment~ and ~heors in ordet to asccrtuin reasonahle t.limensions. Beam dimension~ required arc
1. cover to the n.:inlorccmcnt 2. breadth (b) 3. effective depth (c/) 4. overnll depth (h)/ Adequate concrete cover i-. required to ensure adequate bond and to protect the reinforcement from corrosion and t.lamagc. The necessary CO\'er depend~ on the dass of concrete. the exposure of the beam. and the required fire re!.i~tnnce. Table 6.2 give~ the nominal cover that 'hould be provided to all reinforcement. including lin~s. Thi' cover may need to he increased to meet the fire resistance requtrements of the (ode of Practice. The -.trength of a benm is affected considerably more hy its depth th:ln th breadth The spun-depth rauos usually \'ary between say 14 and 30 hut for large spans the ratio\ can be greater. A \ttitahle breadth may be one-third to one-half of the depth: but it mu) be much les:-. for a deep heam. At other times wide shnllow hcam'i are u~ed to conserve
171
1 72
Reinforced concrete design
headroom. The beam should not be too narrow: if it is much less than 200 mm wide there may be difficult)' in providmg adequate side cover and c;pace for the reinforcing bare;. Suitable dimensions forb and d can be decided by a few trial calculations as follows:
1. Por no compression reinforcement K = M/bd~fd ~
Khut
where
0.167 for fc~:
Kt-..t
< C50
With compression reinforcement 11 can he shm' n that M / bd1fck
< 8/.fhould not be grculcr than VRu m.u = 0.1 ~b"d( I - .f,~/2501fd· To avoid congested shear reinforcement. Vl'u 11 ~;1x should preferably be somewhat clo~cr to half (or less) of the maximum allowed. 3. The span -effective depth ratiO for 'pun' not exceeding 7 m should he within the hNc \'aluc' given in table 6.10 or figure 6.3. For ~pans greater than 7 m the basK ratio' arc multiphcd by 7/..,pan.
b
hold
4. The overnll depth of the beam 1s g1ven II
I , covN I
Figure 7.2 Bcdm dimensions
(
i
h)
d +cover + t
where 1 cslimnled distance from the out~> ide of the link to the centre of the tension bnr~ (:-.ec figure 7.2). For exumple. with nominal sited 12 111111 link!- unJ one layer ot 32 mm tension barli. 1 2~. mm approximately. It will. in fact, be slightly larger than this with deformed har~ a~ they have a larger overall dimension than the nominal bur :-.ite.
EXAMPLE 7.1 Beam sizing
A concrete lintel with an ciTcctive ~pa n of 4.0 rn supporLs 3 230 mm hrick wall as shown in ligurc 7.3. The loads on the lintel arc G~ lOOkN nnd Q~ 40kN. Determine ~uitable dimens1ons for the lintel If class C25/30 concre1e i~ u~cd.
I Figure 7.3 Lintel beam
Assumed load
Design of reinforced concrete beams
m wide rcing llows:
The beam breadth b will match the wall thieknl!ss so that
b;;;;: 230mm Allo" ing. say. 14 I..N for the \\eight of the beam. gives the ultimate loac.l /·
1.35
114
X
1.5
X
40
;;;;: 214k.\
Therefore maximum design shear force VFd
= 107 J..l\
Asc;uming a triangular load dbtribution for the preliminary nnalybis. we have
M _
r x span= ::!14 x 4.0 6
6
143 "'-N m For 1>uch a rclati vely minor beam the case with no compre~sion steel ~ h ould be con:-idered K
t n the le basic
M ,}. I )( / ' Ll
< Khu1
therefore 141 ( 101' 230 d' )< 25
.:11\tOn
er of arger ' n the
0.167
0.167
Rcammging, d .;> 386 mm. A\\umc a com: rete cover of 25 mm to the reinforcing steel. So for I0 mm link~ and . s.t}.
n mm han.
O"crall beam c.lcpth h
d + 25 + 10 =d+ 51
32/2
'll1crelorc make II - 5::!5 mm as an integer numher of bril:J.. courses. So that d
525
Maximum
51
474mm
~hear rc~istance
is
I
VRd.n1u'
own nine
0.1 Rbwd( J - /.;~/250lf~k 0. 18
X
230 x 474 X ( I -"15/250) x 25 x 10
J
4461..N -:. ll.,..t = 1071-N Basic :.pan effective depth
~7~0 = 8.35 < ::::: 20 (for a lightly ~trc!.scd heam in C25
concrete tabll! 6. 10) A heam siLc of 230 mm by 525 mm deep would he 'uitablc. Weight of beam
0.23 x 0.525 x 4.0 x 25 - 12.1 kN
l" hieh i!> 'uffic•entl} dose to the assumed value.
1 74
Reinforced concrete design
7.2
Design for bending of a rectangular section with no moment redistribution
The calculation of main bending reinforcement is performed using the equations and charts derived in chapter 4. In the ca~e of rectangular sections which require onl~ tension steel. the lever-arm curve method i~ probably the simplest approach. Whert compression :,teel is required. either design charts or a manual approach with the simplified design formulae may be u'cd. When design charts arc not applicable or nll available. as in the case of non-rectangular sections. the formulae balled on the equivalent rectangular stress block will !>implify calculations considentbly. The grade and ductility class of reinforcing steel to be u~ed must he decided initial -;ince this. in conjunction wtth the chosen concrete class, will affect the areas requirt.-. and also influence such factor' a:, bond calculation\. In moM ctrcumstance~ one of th.. available types of high-yield hars will be used. Areas of reinforcement are calculated the critical M.:ctions with maximum moment~ . antl1-uitable bar si1.c:-. ~-oc lectecl. (Table~ c bar areas arc given in the appendix.) 'I hi ~> permit~ anchorage calculations to bt performed und dewil\ of har arrangement to be produced, tak.ing into account t ~ gutdance gl\cn b) the C'otb of Practice. An exl:e.,~ tv e amount ot n:111forcemcnt u'ually indicate'> thUl a member is undcr!>iZt and it may also cause tid llculty in lixing the bars and pouring the concrete. Therefor~: the code ... tipulates that lOOt\ 1 \, _ 4Cf· except at
lap~.
On the other hand too ltttlc reinforcement i' abo
I00/\, I b1rI where: A,
i~
.>
unde~1rable
therefore
;. .f~trll 2of. - per cent and nol less than 0.I.,.., per cent y~
the area of concrete
= b x II for a n:ctangular ~cctmn
lJ 1 b the mean width of the beam's tension .wnc ~·, 1rn .ts ,'I1e concrete ' s mean ax .to I tenst'Ie MrengtI1 0.•"1 x.f'-d. :. 13 10r. r j'imilar procedure., but using the ultimate concrete '>tratn' and constant~ for c•• ~ clas~ of concrete from EC2 and it~ attonul Annex.
7.2.1 Singly reinforced rectangular sections, no moment redistribution A beam section need in t11e tenstle tone \\hen K
M u < Kh.•t ld ) ./Ck
= 0.167
The singly reinforced section considered ;., \hown in figure 7.4 and it is subjected t< sagging de\tgn moment M at the ultimate limit state. The design calculntion'> for l longimdtnal \teet can be summarised a-. follows:
Design of reinforced concrete beams
'and onI)
0.0035 ~
b
th no
neu ax
dA,
• •
~f
1C
s~
Figure 7.4 Singly reinforced section witl rectangular stress block
I
-s2 I
0.811
__..L.
__
_!"
z =t.d
F,.
Strains
tr uall) e
0.567fu
1--
E,,
Section
-tUired
f
Stress Block
M
= bd~J(~ al
177
= 0.45d and this is the value used in the design of a section with compression steel.
The de~ign method and equatiom are those derired in Chapter .J for fertions lllbject to bending. The design ot ~m
step~
arc:
1. Calculate K = f. M , ff K > Kt>al
=
. dbd· 0.167 compressiOn reinforcement is required and .I
=
.l~>;aJ
=0.45d.
2. Calcu late the area of compression steel from I
A, :..
(M - Kbalf~.bd~) .f-.(d dl)
(7.3)*
where .f~c i ~> the compressive stress in the steel.
If t/1 /x < 0.3S the compression steel hal> yielded and j~. O.S7jy~ If d1/r > 0.3S then rhc strain e,c in the CC>mpressi ve steel mu!>t be calculated fro m the proponion1) of the ~trai n diagrum and .f..L £~'"' '• 200 x 10 1e-". 3. Calculate the area of tension ~tee ! required from Ktmtf.:l. hd2
A,
Cl.S7f)•'
f~r lhe areas of \tee! required and the areas provided thai
(.\: """ A: rc") ~ (A, pro' - A, rtq}
_)
(7.4)*
= 0.82d.
with le'ver arm ::
4. Check
I f.,< 'O.S7],1.
11\ - -
(7 .5)
Thi' i'> w en~ure rhar the depth of rhe neutral axi~ has nnl exceeded the maximum value of 0.45d by providmg an over-excess of tensile rcinforcemem.
5. Check thm the area of 1>lecl actual!) provided i' within the ltmil'> required by rhe Code of Practice. ble Ill
,edon
the
(EXA MPLE 7.3 Design of tension and compression reinforcement, no moment redistribution
f the
,,
The beam section shnwn 1n figure 7.R has chnracteriM ic matcril1l :-.lrenp,lhs or f.k 25 N/mm2 and .f)k 500 N/mm1. The uliirnare design moment is 105 kN m, caul>ing hogging uf the heam:
1.
\O
2.
165
bd~J:.~
230 " 3302 0.26 ,> K~al
X
• b:: 230, 1
1011
M
X
15 0.167
that comprc,.,ion 'tee! is required. x 0.45d- 0.45 330 = 148 mm
tl' I \ 50/ 14X 0.14 < ().3R therefore the compre~~ion \tccl ha~ yielded and
~I
' Figure 7.8 Seam doubly reinforced to resist a hogging moment
1 78
Reinforced concrete design From equation 7.3
. Compress1on steel
1\
1
=
(M - 0 167f.:•bd2} ) 0.87J;dd d'
(165 X 10~ 0.167 X 25 X 230 = - -0.87 X 500(330 - 50)
X
3302)
= 496mm 2 Pro\ ide two H20 bar'> for A~. area 3. From equation 7.4 Tem.ion steel A,
Q.J61f~kbd2
628 mm 2 • bottom 'iteel.
---::-~- ~ A
0.87fyl;. 0.167 X 25
().1{7 X 5()() X8H + 496
X X
1
' 2J0 X 33Q2 - - - + 496 0.82 X 330 1384 mm 2
Prmide three H25 har~ for A" area
I
I
1470mm". top ~teel.
4. Chccl. equatton 7.5 for the area'> ot '>tccl required and prO\ 1ded for the compre-;sll)n and tcn~ion remtorccment to ensure dw.:tility of the section {/\ ~ pnw- A ~.fCIJ) ~ (1\l.prnv- A,, ,cq )
That i.,
628
496 (= 132)
1470- 1384 (
86)mm1
5. The har areas provided arc within the upper and lower limit~> specified by the codl To rc!.train the comJWO!.sion l.>teel, at least 8 rnm links nt 300 mm centres should ~ provided.
7.3
Design for bending of a rectangular section with moment redi stribution
The redl\trihution of the moments obtained from the ela'>tiC analysis ol ;1 concrt strucrurc Hil-es account of the plastic1ty of the reinforced concrete as it upproachc~ uhimatc limit srate. In ul is set according to the amount of redistrihution 6. Por the EC2 code it ,.bal ::::
0.8(b- 0.44)d for hL < CSO
(7.L
where
h = moment at the section after reclistrihulion moment at the ~ecuon before
rcdi&~ribution
However the UK Annex to the EC2 modifies the limit to xro~ as Xbal
< (h 0.-l}d
(7.6b •
Design of reinforced concrete beams Table 7.1
Moment redistribution factors for concrete classes
• M.vumum permitted rediWibutton for
cld$S
A
norm~ I
~
CS0/60
ductility steel
" Maximum permitted redistribution for cl.m 8 and C hogher ductihty ~teel, ~ee sectoon I 6.2
In this chapter the examples will he ha~ed on the L K Annex·~ equation 7.6h. hut. because many of the de~igns in the UK arc for project~ uver~ea~ which may require the u'c or the 1!('2 1-opcci lic ati on~. example 4.9 part (a) was hascr.l on the usc of the EC2 ~4ll upports. Example J.9 oment rediwihu tion 'hows how the hogging moment may be n:duccd without l!a,mg the maximum sagging moment in the bending moment envelope. Thus there n economy un the amount of steel rcinfon.:ement requ ired and I eel must he calculated from th(! proportions of the )>train diagram nnd .h, £,_.,._- 200 x 10 1 .._..
=
5. Calculate the area of tension \Lccl from KNtfclbd: -I-A'~ O.R7ho.:. '0.87/yk \\here ;: - d
O.RxhJI/2.
6. Ch(!c(.. cquauon 7.5 fnr the
area~
(1\~-ll'"'- !\~.rcc 1 ) > (A,,pto•
ol
~tecl
required and the an.:u ... provided that
A,,rcq)
This i~ to ensure that the depth of the neutral axis hns not exceeded th(! maximum va[U(! of' Xb3 1 by providing Ull tlVCf-eXCC~S Of tensile reinforcement. 7. ChccJ., that the area of steel pr within the maximum and minimum limit' required.
(
EXAMPLE 7.4
Design of tension and compression reinforcement, with 20 per cent moment redistribution, 8 = 0.8 (based on the UK Annex to EC2) The beam section sho\\n in figure 7.9 has characreri~tic mutcrial ~trengths of r.~ 25 Nlmm 2 and f.,k 500N/mm 2 . The ultimate momt.:nl j., 370 "-N m, cauing hogging of the beam.
Design of reinforced concrete beams
181
Figure 7.9 Beam doubly reinforced to res1st a hogging moment
Section
Strain s
1. A~ the moment reduction factor b = 0.80, the limiting depth (6 - 0.4)d
X -
= (0.8 2.
or the neutral ax i!> ill
0.4 )x 540 = 2 16mm
M jlu/2}~k - 370
K
X
10('/(300 X 5402
X
25)
0. 169
3.
0.454(b - 0.4) - 0.1R2 (~- 0.4 ) 2
Knal
K
4.
'> Kt>.J
OA )2
0.4) - 0. 182(0.8
0.454(0.8 () 152
therefore compre~~ion ~teel i~ required.
1
tl / \ I00/ 216 0.46 ~ 0.38 therefore f.._. 0. 87/)~
From the
proporuon~
or the \train diagram
.
Stet:! compre!.train the compre~~ion 1>teel, at teaM 8 mrn links m 300 mm centre!> ~hould hi: provided.
7.4
Flanged beams
sections through a T-benm and an L-hcam \\hich form part ol a and slab tloor with the slab spanning between the beam~ and the area~ the :-.lnb n~ting as the llnngcs of the beams ns f,hown in figure 7.1 1. When the heams .., rc~i!.llng ),agging moments. the slab acts a~ a compn.:ssion llonge and the members rn he de~igned as T- or L heams. With hogging moment' the slab will he in tensiOn a a~o,umed to he cracked. therefore the beam must then be des1gncd a!-. J rectangu \CCII on of \\ itlth b,. and overall depth h. At in tel mediate supports of conunuous hcums where hogging moments occur t 1owl area ten~Jon reinforcement ~hould be spread over the effective width of tr flnngc as shown in ligurc 7.10. Part of the reinforcement mny be conccntrnted over the weh \\idth. I he eltccttve flange '' idth bcu i-. specified by the following equation: Figure_ 7.10
~how~
concrete~enm
or
b.,. -
b d1
L b~n
1
\\here 1> within the tlange tlucknc~'· l·or tht!> po~ition of the \trc'' hlod... the \Cction may be designed as an cqutvulent rectangular \Cclion of breadth h1 • Tran~ver11e retnfon:cmcnt .'lhould he placed aero~' the full width of the flange to re.,t\1 the 11hear developed hetween the web and the flange. a., dcscnbed 111 l>Cctton 5.1.4. Qutte often thi\ reinforcement is adequately provided for hy the top !lteel of the hcnding reinforcemen t tn the \lah supported by the beam.
Design procedure for a flanged beam subject to a sagging moment
1. Cu leulatc
~
hrd :fck equm ion 7.I lever arm ~
183
and determine lu !'rom the lever-ann cttrvc of figure 7.5 or from
/Jd and the depth of the stres~ block ~
= 2(d
- ~)
rr ~ the
procedure~
anti equations set out in
(i) Calculate the design longitudinal shear
vEd
~>CCtion
5. 1.4
at the web-flange interface
For a 'lagging moment the longitudinal ~)hear stre.~ses arc the greatest over a distance of ~x mcu.~urcd from the point of :t.ero moment and .6..\ i:. tal...cn tl'> half the diswnce to the maximum moment at mid-:,pan. or~'= 0.5 x L/ '2. = L/ 4 = 1500mm.
Design of reinforced concrete beams TI1en.:forl.! the change in moment :::C.M over distance 6x =L/ 4 from the tero moment i~ ll 'u
XL 2
L
-->(-
4
ll'u X I.
L
3wuL2
4
8
32
3
- - X -= - -
X
44
X 6~
32
= 1491-.:--!m
The change in longttudtnal force .J.F at the Hange- web interface ''
~Fd
=
.J.Af
(d - hr/2)
>< bto
b,
where b1., is the breadth of Hange outstanding from the web. Thus
.J.Fcl
~M (b1 b" )/2 . (t - ,, / "_) x - --. a:, gtven on page 11 0 1 1 1,,
14() X~ X (600 ~ 250)/ 2 (530 150/2) I 600
=_
96
kN
The longitudinal shear l>tre),s ~~~ i nducud is
D.Fd 96 X 10' (It, ) U.l )- 150 )< 1500
0.43 N/mm~
(ii) Check the strength of the concrete strut From equation 5.17. to prevent cru~hing of the compres~ive ~trut in the !lange 0 6( I j~lf''!.50 )J.~ - I 5(cot 01 tan fl1)
l' t d
ho'' n in figure 7.12.
185
186
Reinforced concrete design
7.5
One-span beams
The following example sections. The :,hear rcinfon.:cmcnt for this beam is de:;igned later in
example 7. 7.
(
EXAMPLE 7.6 Design of a beam - bending reinforcement
The beam shO\\n in tigure 7. 13 :-.upports the foliO\\ing uniform!) pennllncnt ;oad Kl ()0 1-N/m. including self-weight v:1r1ahl ~ load
The /~1
di~tributed
= I RkN/m
lfk
eharact~-istic
\lrengrhs of the concrete and :.tcel arc ./:k 30 N/mrn2 ami 500 N/mm". Effective tkpth. d 540 rnm anti breadth, /1 ~()0 mm.
-
480
Figure 7.13 One·span beam-bendmg
2Ht6 2H32 + lHlS
6.0m
(a) Ultimate loading and maximum moment
Ultunatc load
Hu
= (1.35g, 1.5q,) kN/m = (I Vi 60 -+ I 5 I H) X
I 08 k.N/tn
thcrdure max1mum
tlc~.,1gn
moment M
108 ' 6.01 8
8
~86 kN 111
(b) Bending reinforcement K
486 X 1011 lui'f.:k = 300 x 5-J.O"W M
0.185 ,
KbJ.l
= 0.167
1 herdore enmprc~-.ion re10forccment. A~ 1~ reqUJred.
ct 1d
loads
50 5-tO
0.09::!
Cornpn:~~ion ~teel
0.171 I
A,
=
111
(A
table 7.1. therefore/..:= 0.87/>l At.llklbd~ d')
f ... (t! (O.IK5
=-
0.167) X 3U X 300 >( 5402 0.!:!7 X 500(540- 50)
Provide two H 16 bar:., A~ = ~02 mm 2 2
Tcn~ion ..tccl.
A,
=0.167[.du/ 0.87{~1~
- A~
Design of reinforced concrete beams 50%
lin; rom r m
100%
Figure 7.14 Simplified rules for curtailment of bars in beams
50%
0.08l_
Simply supported
where. from the lever arm curve of figure 7.5 /3
-
0.82. Thus
0.167 X '\() X 30() ~ 5~0~ + 2')'l __ 0.87 X 500 X (0.82 ~ 54())
A~ =--
2275
+ 222
2497 mm 2
Provide two 1132 har~ and two H25 bar:.. area
2592 mm 2 , IOOA,/bd
1.6 > 0. 15.
(c) Curtailment at support
The rension reinf'on:cment shnuh.l he anchored over the supportll with a hend as shown in figure 7. 14 which i1. ba~cd on past UK practice. Two b:m, may he curtai led ncnr to the su pport~>.
(d) Span- effective depth ra tio
100. 1, ......41 /bd
f1
( 100 x 2~97)/(300
From tuhle 6 10 or figure 6.3
ba~ic
540)
U4 per cent
span-effective depth ratio
l..f
Modification fur !'ltccl area prov1ded: Modified ratio
1-Ul x
2592 2497
= 14 5
Span cffcctin: depth rauo provided
= 6000 = 11 .1 540
which Ill lcl-\ than the allowable upper limit. thu). deflection rcqturement-, arc likely to he sati.,licd.
7.6
Design for shear
Th~; theory and design requirements for :.hc(ll' were covered in chapter 5 and the relevam design equation!> were derived bused on the requirements of EC2 u~ ing the Variable Strut Inclination MetJ10u. The ~ heur reinforcement wi ll usually take the form of vertical lin~~ nr a combination of l i n~ s and bent-up bars. Shear reinforcement may not he required in very min.. d( l - / ck/ 250)/ck
111 ,
(7.9 •
and if VRJ , m11 , 2: Vr'd then go to Mcp (3) with fJ 22 nnd cot 0 - 2.5 but if VKJ mnx < V1" then () > 22 and 8 mu~t be calculated from equation 7.10 .. fJ
V&~ } < 45 O. !Mb.. tlj,d l - .f, k/ '250) -
() 5 ,in 1 { -
(7.10 •
3. The shear links required can be Chear = '''u x effective span/ 2 - 108 x 6.0/ 2 = 324 k.N Design shear at face of ~upport \'Ed= 324 - 108 x 0.15 = 3081-.N Crushing ~trergth
VRd rna'
of diagonal strut, assuming angle
(J
= 22 . COl B= 2.5 is
OJ?4b"'d{1 - f..l /250}(r~
VRd. max
= 0 124 x 300 ) 5-tO(I - 30/ 250)
530 kl'\ ( > \'F.t
X
30 x 10--'
308 kN)
Therefore angle B = 22' and cot B = 2.5 as n'sumcd. (b) Shear links At distance d from face of support the de~ign ~hear i~ \'F..s 308 - 108 x 0.54 =- 250 kN
308 - '''ud =
ll~,;o
A,w
0.7Rd(,~
.1
cot() 250 10 1 540 x 500
0.7!\
0.475
2.5
Using table !\.4 tn thl! Appendix Provide 8 mm links at 200 mm l:Cntres. A "/ 1 = 0.503. (c) Minimum links
0.0~(1t 5 hw
- ,,k0 ()!{
X
3()0
hear forces it may be found that the usc of links to carry the full force wil l eupport moment). can be reduced and the de-.ign -.pan moments possibly increased.
19
192
Reinforced concrete design
,---
Design of the beam follows the procedures and rule~ set out in the previous section, factors v. hich have to be con),idered in the detailed design are as follow\:
-o~er
1. At an exterior column the beam reinforcing bars which resist the design moment' must ha\e an anchorage bond length within the column. 2. In monolithic construction where a l-imple ~upport has been lll.!>umed in the structural analysis. partial fix.iry of atlcast 25 per cent of the ~pan moment should be allowed for in the design. 3. Reinforcement in the top of the \lao must pa n typical armngement of the bending rcmforcement torn twocontinuous beam. The rcinl'orcemcm hal> been arranged with reference to the bending-moment envelope and in accordance with the rules for anchorage ru1d
~>pnn
Figure 7.17 Continuous beam arrangement or bending remforcement
B.M Envelop!!
Design of reinforced concrete beams
the uld be
10
cment
curtailment dc....cnbed in section 7.9. The application of these rule!'. establishes the cutoff point~ beyond wblch the bars must extent! at leal>t a curtailment anchorage length. It should be noted that at the external columns the reinforcement ha-. been bent to gi\'e a ful l anchorage hond length. The shear-force envelope and the arrangement of the !.hear reinforcement for the same continuou~ beam are shown is figun.: 7.18. On the shear-force envelope the resistance of the minimum stirrups has heen marked and this show~ the length~ of the beam which need shear reinforcement. When de~igning the shear reinforcement. reference should be made l arc 8 4.0 m centre' \\Jlh a slah th1ckness. lrr = 180 mm. a!> :.hown in ligures 7.19 and 7.20. The ~uppm1~> have a width of 300 mm.
=
The unifonnl} distributed ulttmutc dl.!~ig.n load. 11" = 190 kN/m. The ultirnatl.! design and ),hcurs near mid-~ pan und the support~ nrc shown in 1igurc 7.19. The characterbtic 'trcngth~ of the concrete and ~ted arc ./,k = 'O '-'/mm' and
mome nt~>
fy~
500 N/mm,.
I
Shear V(kN)
428 427
428
333
570 522
8
At
0
523
523
0
Moment M (kNm)
1. s.om
t
522
S.Om
F• 1.3SG,
~
570
C l
427
1
ol 5.0m
L
I
1 SOQ,
Total ultimate loud em each ),pnn b
r
950kN
190 x5.0
Design for bendmg
(a) Mid-span of 1sl and 3rd end spans - design as a T-seclion Moment .[!8 k.N m sngging Lffectivc width of Hange ben
hw + 2t0.2b' + 0. 1 x 0.851.1
= 300 bw
( :5 ""' + 2[0.2 x O.X5Lj) (see fi gure 7. 11 )
:! 1(0.2 > (2000- 300, 2)) t (0.085 x 5000)J - 1890 mm
+ 2[0.2 X
Therefore br - btu
0.85/.l
= 300 + 2[0.2 X 0.85
= 1890 mm.
X
5000] = 2{)()0 mm
Figure 7.19 Continuous beam wtth ultimate de~ign bendtng moments and shear rorces shown
194
Reinforced concrete design 428
X
1011
1890 X 600~
X
30 -
(}.():!2
From the le' er-ann cur"e of figure 7.5. 1. ~
=0.95. therefore
= 0.95 x 600 = 570mm
and
d
600 - 570 = 30 ( < ,, ( 1)
;:
so that the ~tress block mu~t lie within the 180 mm thick flange and the eel ion i designed as a rectangular section with b = b1• M
A,
= 0.87}~k;: 6
X J0 = 0.!!7428 X 50() X 570
,
l?Jomm·
Provide lhn:c TI25 bars and two Hl 6 hnr~. area = IH72mm2 (hotlom ~teel). (b) Interior supports - design as a rectangular section M
523 ~ m hogging A/ btf]}, k
523 X (()(I 3()() !\ 5802 X 30
0.173 > 0.167
Therefore. compresICe(). The arrangement or the reinforcement is shown in figure 7.20. At end support A two 1125 har~ have been provided a' top continuity ~ted to meet the requirement ol item (2) in section 7.7. (c) Mid-span of interior 2nd span BC - design as aT-section
M
= 333 kN m. sagging From figure 7 . II. cffccti\'c flange \\ idlh bttr
=- b.., +2[0.:!b' + 0.1 x 0.70L]( < bv. + 2[0 2 v 0.70/, ) =- 300-:! (0.2 x (2000 300/ 2)) (0.07 x 5000)1 1740 mm h.., I :![0.2 x 0.7L] = 300 + 2[0.:! > 0.7 ) 5000] = 1700mm
Design of reinforced concrete beams !.--
H8 @ 200
H8 @ 300
H10@ 200 Figure 7.20
2-H25
4-H25 >
II
2· H20
End-span remforcement details
II
...
3-H25 2-H16
ls5.0m_ __
2525 25
..
252525
1~ 300 Sections
rn1dspan
near the interior support
Calculating M /( brcPf. 600)
M
A,=--.0 87})~ ~
0.95
13-B mm'
Provtde three 112'5 har.... area = 1470 mm2 (hottom ~tccl ).
Design for shear (a) Check for crushmg of the concrete strut at the maximum shear force N'laxtmum !>hear i\ 111 spans AB and CD at supports D and (', At the face ot the support~ V~.J
570 570
ll'u X
support Width/ 2
190 x 0.15
= 542 kN
Cru ~hi 11g ~trenglh of diagonal ~t1111 i~
IlK~. "~"~ - 0. 124bwd( I
.kk/ 250Kk assuming angle {) 22' , cot() = 0.124 x 300 x 600( I - 30/ 250) x 30 = 589 ~N ( > Vm
nt
been
Therefore angle ()
22 and cot fJ
2.5
542 kN)
::.?..5 for all the ~hear calculation ~.
(b) Design of shear links
(i) Shear li11k.l in end rpan.s at
~ upporrs
A and D
Shear distanced from face of suppon b Vl:d A,,. ~
= 427 -
190 x (0.1 5 + 0.6)
VF..s ...,.:.:..:~~ 0.78df,l cotO > 10' = 0.78 X 2R5 - -,-=0.49 600 X 5()() X -·5
Provide 118 linb at 200 mm centres. Aw. j s
0.50 (Table A4 in Appendix)
285 kN
19~
196
Reinforced concrete design
Additional longitudinaJ tensile force is ~/·,d
=0.5VEdcot B - 0.5 X 285 X 2.5
=3561-.N This additional longttudinal tcn~ile force is prO\ ided for by extending the curtailment point of the longitudinal reinforcement, as discus~ed in !.CCtion 7.9 (ii) Shear lin/.. f in end 1pans at 1'11pports 8 and C Shear Vhu distance d fmm face of support is
570
VEd
190(0.15 - 0.58)
= 431 kN Therefore: I\"'
Vtu
0. 78t(fv~ col 0 43 1 X [() -' 0.78 < 580 ')()() 0.762
-- - - - -2.5 I(
Provide 1110 link~ a1 200 mm ccmrc~. A .,. /s =0 762 (Table A4 in Appendix) Additional longitudinal ten~1lc force 1s ~~ ~~~
0.5 \'1.1 cot U 0.5 ~ 431 "2.5
539 ~N I hb additional longiludinal tcn~de force is provided for by extending the curnulment point of the longitudinal reinforcement. a~ di~cus\ed in section 7.9.
(iii) Shear fink.1 in middle span IJC til mpport.\ B and C
Shear d"tance d from the face of ~uppurt = 522- 190(0. 15 I 0.6) 380kN. The calculatiOnS for the shear links \o\OUid be ~imilar 10 thol>c for the other supports in sections (i) and (ii) giving IOmm link'> at 225mm centres. fhe :~ddition:~l longitudinal ten,de force. F1.t 0.5 3RO 2 5 475 kK (i1•) Minimum 1/tear lin/.. v
O.O~f:~ ~ h"" --=-y;:-
A,.,., mm
o.ox 30°5 ---
300
500 = 0.:!63
Provitle HR Shear
Vmon
link~
m 300 mm spadng. A,w Is of links prm ided
= 0.335 (Table A4 in Appendix).
resi~tancc
Asw
=-X
s
0.78d/~1 . ' cotB
=0.335 X 0.78 X 600 X 500 X 2.5 X l96k~
)()
"\\ I
Design of reinforced concrete beams
197
( 1') £trem of vhear links
Links to re:.i't ~hear are required over a dististed by Vn1m 196 k~. a.., provided by the mimmum linl.s. For the face of the end supports A and D the distance r 1 i!>
=
VF.1-V
~ ~-0.15 ll'u
=
427
196
190
For the interior 570 .r2
-0.15=l.07m
~upports B
196
190
and C of the I st and 3rd spans
0. 15 = I.S2m
For rhe links nL ~upporti> 8 and C in the middle spun .\~
522
196 190
- 0.15- 1.57m
Based on these dimcm.ions the link!. are arranged
as shown in figure 7.20.
l ------------------------------------------~) --
7.8
Cantilever beams and corbels
The effective .,pan of a cantilever JS e1ther Ca) the length to the face of the -;upport plm, half the O..:am's overall depth, h or (b) the distance to the centre of the ~uppon 1f the beam " contlnuou.... The moment-., 'hear:. and deflections for a cantilever beam arc l>Uh'ltanually grcatc1 than tho~c for a beam that 1s l>Upponed at both ends \\.ith an equivalent load. Aho the moment:-. in a canlllcvcr can ne\'er be red1'ilributed to other part\ of the 'ltructurc the beam mullt alway!. he capable of resisting the full static moment. Occausc of these factor~. and the pmhlcms thut often occur with increased detlections due to creep, the design and detailing of a cuntilcvcr Ileum !.hould be done with cnre. Particular aucntion ~hnu ld he paid to the anchorage into the support of the top tcn~ion reinforcement. The steel should he anchored ut rhe support by. nt the very leust, a full maximum anchorage length hcyond the end of its effective spnn. Some de~ooign orli cosO x -.inO
.: kiJ X (t/ a' tan O)b,. Stn 20
Re:urangtng I· F..t
(7. 16)
f~udbv.
Thi!- equation cannot be solved directly for f) but table 7.2 (overleaf), which has been developed directly !'rom equation 7.16, can be used.
(c) Main tension steel, As,maln Re~o lving
F,d
horiwntally at B. the force F,J in the steel tie is given hy
I· c1t cos 0
Fc.J ~;os ()/ ~in 0
FFJ cot B
The towf force F~ in the steel tie. including the effect of the hori1ontal force of0.2Ft.d• io; given by
F;d
FEdcot8 t 0.2fht = FEd(COl 0 t 0.2)
(7. 17)
The area of main tension steel, A, m.un• i~ given by
A,. maon
199
= F;d /O.R~f)l
(7.18)
200
Reinforced concrete design Table 7.2 Values of 0 to satisfy equation 7.16 H
f[d
(degs)
f,ddbw d d= 1
22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
(
0.429 0.400 0.371 0.343 0.315 0.288 0.262 0.236 0.211 0.187 0.164 0.143 0.122 0.103 0.085 0.068 0.053 0.040 0.028 O.Q18 0 010 0.005 0.001 0.000
d
d~ . 9
0.458 0.429 0.401 0.373 0.345 0.318 0.292 0.266 0.240 0.216 0.192 0.169 0.147 0.126 0.107 0.089 0.072 0.056 0.043 0.030 0.020 0.011 0.005 0.000
d d- .8
d id - 7
d d - .6
d d- 5
d ld- .4
0.487 0.459 0.431 0.403 0.375 0.348 0.321 0.295 0.269 0.244 0.219 0.195 0.172 0.150 0.129 0.109 0.091 0.073 0.057 0.042 0.029 0.018 0.008 0.000
0.516 0.488 0.461 0.433 0.405 0.378 0.351 0.324 0.298 0.272 0.247 0.222 0.198 0.174 0.152 0.130 0.110 0.090 0.072 0 054 0.039 0 024 O.Q11 0.000
0.545 0 .518 0.490 0.463 0.435 0.408 0.381 0.354 0.327 0.300 0.274 0.248 0.223 0.198 0.174 0.151 0.129 0.107 0.086 0.067 0.048 0 031 0 015 0.000
0.574 0.547 0.520 0.493 0.466 0.438 0.411 0.383 0.356 0.328 0.301 0.275 0 248 0.222 0.197 0.172 0.147 0 124 0.101 0.079 0.057 0.037 0.018 0.000
0.603 0.577 0.550 0.523 0.496 0.468 0.440 0.412 0.385 0.357 0.329 0.301 0.274 0.246 0.219 0.193 0.166 0.141 0.115 0.091 0.067 0.044 0.021 0.000
EXAMPLE 7.9 Design of a corbel l)esign the reinforcement for the corbel ~hown in figure 7.23. The cor·hcl ha~ u breadth h 350 mm and supports an ultimate load of Vhl 400 1--N at a tli!.lancc ac - 200 mm from the face of the column. The bearing is ncxihlc and at u di~>I< to' 0.87 500 Pro\ltlc two II 12 hal".. area
+ 0.2) = 400( 1.40 + 0.2) = 6-tO kN
1471 mm 2
1610 mm 2•
Horizontal/inks A,
hnb
•
0 5A, """"
735 mm,
Pro\itlc four 1116 linb. A, 11 11 ~, 804mm~. Figure 7.2-1 :-,how' the detailing of the reinforcement in the corbel. Figure 7.24
Reinforcement in corbel
2H32
201
202
Reinforced concrete design
7.9
Curtailment and anchorage of reinforcing bars
As the magnitude of the bendtng moment on a ~am decreases along its length so rna~ the area of bending reinforcement be reduced by curtailing the bars since they are no longer required. as shown in figure 7.25. It should he recognised though that because oi the approximations and assumptionpacing minimum bar numbers and curtailment beyond the critical sections for shear. Each curtailed bar should extend a full anchorage length beyond the point at which 1 is no longer needed. The equations for an anchorage length were derived in !>ection 5.2 The equation for the design anchorage length, /htrer: and the bar size and whether the bar is in the top or hot tom of the beam. The honds better wnh the compacted concrete in the hottom of the beam. the design bending moment from the moment envelope
z i~ the lever arm !:J.F1d i~ the additional tensile force obtuined from the design for ~hear 1ot:
Figure 7.25
full anchorage length
CurtallmenLof rclnforcemenl
envelope of tensile forces
c""'"m'"'
'·'~ s~m
'"'"'~'' .-J . tM., "'"'""'' ' ' "''"m (hogglog "''")
~ 5
4
2
71 \~· I
-
4
\
0~~' M,.J z
6~
1
2
y
M,.J l envl'lope dtagram (saggtng region)
1~1 Curtatlment anchorage
Design of reinforced concrete beams
203
~F1d is a maximum where the shear force is a maximum at sections of zero moment, and ~Ftd b zero at the maximum moment near to mid-span and the interior ~upp011. For members where shear reinforcement is not required the tenc;ile force envelope may be estimated by snnply 'shifting· the bending moment envelope diagram honzontally by a dt!>tance a1 (=d) as shown in figure 7.15. To determme the cunailmenr positions of e~ch reinforcing bar the ten\ile force em·elope i" di-.ided into secuons as shown. in proponion to the area or each har. In figure 7.25 the three haro; provided for the sagging envelope and the four for the hogging cm·clopc arc ran'> arc approxi'lnntely equal. Figure 7.28 show~ the rule'> in diagrammatil.: form. llowcvcr 11 -.houlcl be noted that these rule:) do not appear in EC2 and arc based on previou& eMablished UK practice. Figure 7.26 Anchorage of bottom remforcement at end supports
( 1) Beam supported
on wall or column
(2) Beam mtcrsecting
another supporting beam
Figure 7.27 Anchorage ill intermediate supports
204
Reinforced concrete design
!'~ ;I
Figure 7.28 Simplified rules for curtailment of bars in beams
',
0.08L
100%
I
~008L
'
Simply supported
c .. 0.25~ c = 0 1St
60, ;;
20% 100%
30%
Continuous beam
7.10
Design for torsion
The theory and design requirement~ for torsion were covered in section 5.4. The design procedure consiw. of calculations to determme addit1onal arcus of lin~s and longiiUdinal reinforcement to re&ist the torsional moment. u~ing an equivalent hollow box section U),uully 1t is not necessary to design for tor!-.ion m \latically indetcnninate sLructure' \\here the tor From example 7.7: \'hi = 308 kN and Vku. 111 a~ 530 kN The angle of inclinntion of the concrclc strut i.-. 0 = 2:! with cot 0 1.5 a!ld tanH 0 4 For the 'hear
link~
A"
0.475 (required)
I
The additional longitudinal ten.-.de force
~F1u
3!-!5 J..N
2. Con\crt the rectangular M!Ction to an equivalent hollow hox !.ection figure 7.29h) ·n,i cknes~
A
of box section
/(
600
X
30()
= 2(600 + 300)
lOOmm
Area w1th111 centreline
Perimeter of centreline
A~
u~
(I>
t)(h
I)
200
500
100
10' mm~
2(h I h -1t ) 1400 mm
Figure 7.29 Torsion example
1- . - - - - - - ,
H16
H16 H8 lit 12S
••••
H16
H16 TI
Hl6
Hl6
... b = 300 ...J (a) Cross·section
(b) Equivalent box section
(c) Reinforcement details
1~--e
Design of reinforced concrete beams 3. Check if concrete section
i~
adequate
Tt:.u vlid - - + - - < 1.0 \'Rd. m;a~
TRd."'"'
-
where 1.331'dc~lcrAl
. [ Rd.m•~
= (cot 0 ~ tan B)
(~ee
equation 5.31)
tor~ion .
(Note that A__.
with
0.6( 1 - f A... Spacing v
125mm ( .. u~ /8
.1
=0.805 (!>ee Appendix table A4)
175mm)
6. ('olculate the area A, 1 of' the udditional longitudinal reinforcement required for torsion
JJ;Late. Failures of stru~ ... at the ULS are fortunately quite rare hut can get a lot of publicity. \\hereas failure-. to durabiliry and 'crviccability an; much more common during the life of a structure they can quite ea-,ily lead eventually to a :-.tructural failure or be one of the pn ~ causes of a fa.ilurc. Also poor detnal1ng and con\tructaon can be the cause of ' prob l em~ as lcaldng roofs nnd basements and dasfigurement of the structure 1\ const:qucnt high maintenance col!ts and reduced working life. Adequate concrete cover to nil the reinforcement bar.\ is all-important to pre ingress of moisture and corrosion of the steel hnr:- with resu ltant staining and spallin: the concrete. Cover of the concrete is u l ~o required for fire rcsi~>t 10 J
The maximum permi~~iblc 11hcar force ba~ed on the lace of the lnadcd arl'a i~ gf\ en b) the maximum !>hear re' t ~tancc \ ' Rd ""''
0.5ud[o.6(1 0.5
car force 'ble shear "!leter of
C25/30 m) and mtne the
0.94 >< 0.56 246lN
{;~,))f~
2(_1()() ..,.. 400)
X
145
X
[0.6 (I
25 )] 25 250 T5
) 10
1
914kN
v.hkh clearly exceeds the value VRd r based on the first critical perimeter. I fence the lmaximum load thar the ~l ub can carry is 246 kN. ;
8.1.2
Punching shear - reinforcement design
If reinforcement i~ required to resist shear around the contrnl perimeter indicatcd in Figure 8.1, il ~hould he placed between not more thnn 0.5d from the loaded area and a distance 1.5d inside the outer control perimeter m \Vhich shear reinforcement i~ no longer required. The length of this is given hy u0111 tr = II,.J/( ' 'Rd. eel) from wh1ch the nece~sary dio;tance from the loaded area can he calculated. If thi~ i' Je~, than .1d from the face of the loaded area. then reinforcement should be placed in the zone hetween 0.3d and l .5d from this face. Vertical link\ wtll nonnally be used and provtdcd around m lcao;t two perimeterlab. and for vertical links can be e\prcssed a\: l'l(cJ _.
0.751'Rd.c (.--)
/\.-, ~-l.5 ./y"u d .I rill
where the requircu
l'lhl,c'
would be given hy
v~..,.
llj (
A ched, mu"t al:.o be made requirement that:
th~
where ·'l I)> the spncing of link'i around the perimeter and A,11 mrn i~> the area of m indil'idt/(/1 fin/... h•g. Similar procedure:- must be applied to the regums of Om 'ilab' which arc close t Vut =- 650 kN>
(ii) Check control perimeter 2d from loaded face Perimeter
11 1 -
2(a f b) t 4rrd
2(300 -1- 300) + 4r. x '215 an.:: I for
hence for concrete without ~hc~u· reinforcement the ~hear capacity is given hy:
Vnct,c
orced
= 3902 mm
l'fttt ,r X
3902
X
'215
= 838 930vRd .c
bending stc~·l ratio
A, -btl = I 000905:.. .,~1.5 = 0.0042
( ( > 1.40 per cent}
hence from tahlc 8.2. I'Rd , 0.5n fnr eta~~ C30 concrete and. Irom tahlc !U, modification factor for da-.s C25 concrete O.lJ-1 then \ 1Rd .,·
!OX 930
= 442 kN 1murn
0.56
X
l>atil>ficd by a 6 mm diameter har (28.3mm'> Hence the as~umed Hmm lin"-' ''ill be adequate.
215
216
Reinforced concrete design
The area oJ steel required/perimeter is thus given by: A ~~~
> I'Rd ,, -
0.751'Rd.•
-
1.5
(f
/y..,d.tl
)
Sr X IIJ
where. for the outer perimeter 650
VF,t
\'Rd"
= -lltl1 - 3,.,)_ •v.,
~
= 0.94 x 0.56
l'l{d
!.v.d d = 250 1- 0.25 UIIU
Vr -
X
103 ~ _l)
,
= 0.77511.1/mm-
0.526 N/mm2 (u~> above) x 215
= 103 N/mm'
( ~ 500)
160111111
thu~
(0.775 - 0.75
X
0.526)
X
160
X
J9()2
A" ' > - -- -1.5 X303- - -
= 523mm2 (v) Number of links The nre:1 of one leg of nn Xmm link i~ 50.3 mm 2 . lienee the number of II nk-lcgs required 523/50.3 11 on the outer pcrimeler. Thl: same number of link::. cnn conveniently be pro\ tded around each of the 3 proposed perimeter' as 'ummarised in the table hclo\\ The table tndicate' the numhct of ~tng l c-leg g mm diameter link~'. (urea = 50.3 mm~ J proposed for each or lhc three n:inforcemcnt perimeter' taking account of the maximum requm:d spacing and practical h\ing cnn,idcrauon,. Bendtng retnforccment i' 'paced at I25 mm centres in hoth directions: hence link spacing is ~ct at multiple!> or thb value.
--Dl5tance from load face (mm)
Length of perimeter (mm)
85
1734 2739 3713
245 400
Reqtlired ltnk spacing (mm)
Proposed link spacing (mm)
Propo5ed number of finks
158
125 250 250
11
249 323•
14 15
• muxir'num allowed
8.2
Span-effective depth ratios
Excc~sivc arc set on the -.pan-depth ratio. These limit~ arc exactly the ~arne ns those for hcarns a~ described in section 6.2. As n slab i'> usually a !.lender member. the restrictiom; on the ~pan-deplh ratio hecomc more important and this can often comrol the depth of slah required. ln tenm of the 'pan- effective depth nlliO. the depth Of slah IS gtVCI1 by . span mtnimum effective depth = -.--. -.--.- baste rauo x correcuon Iactor:-
Design of reinforced concrete slabs
The correction factors account for slab type and ~upport conditions a~ well as cases of span!> greater than 7 metre~ and for flat slabl> greater than R.5 metres. The ba:-.ic ratio may also be corrected to account for grades of sreel other than grade 500 and for" hen more reinforcement 1\ provided than that requlfed for de~ign at the ult1mate limit -.tate. Initial values of ba~ic ratio may be obtained from tables (e.g. table 6. 10) but thc-.e arc concrete strength dcj)l.!ndcnt. It may normally be a:.~umed that in usUlg such table-., ,Jab., are hghtl) Mres\ed although a more exact determination can he made from ligurc 6.3 when the percentage of tcnllion reinforcement is known. It can he seen 1hat the ba,ic ratio can he mcrcased hy reducing the stress condition in the concrete. The concrete \trel>' may be reduced by providing an area of tension reinforcement grcall.:r than that required to rc1>i1-.t the design moment up (() a maximum of 1.5 x lhm required. In the case of two-way spanning slabs, the check on the span effective depth ratio r.hould be hased on the sltorler span length. This doc!' not apply to llat slabs where the longer span should be checked.
8.3
Reinforcement details
To resist cracking of the concrete. codes of practice ~pccify dct:ul~ ~uch as the minimum area oJ reinforcement required in a ~ection and limit~ to the maxnnum anu min1mum spacmg oJ bar\. Some of thc\e rules are a~ follow~.
re'luircd ~tly be : bcJO\\. .3 mm1 ) imum I!P3CCd at \Ulue.
(a)
M111imum area\ of rdnforcemenr minimum area
0.26f.:1mb1d/f)1.
">
0 0013b1d
in hmh direcuons. where b is the mean '' idth of the terNie tone or scct1on. The m1nimum remforcement provision for crack control. a' spec1hed 111 'cction 6.1.5 may also huve to t>e cons1dered ''here the ,Jut> depth exceeds 200 mm. Secondary tran!>vcrsc re1nforcement should not be less than 20 per cent olthe minimum mam rcinfon.:cment requirement in one way verse reinforcement muximum urea - 0.04A,
bsed
' of links
__
)
. or other :e'e limittu,ually :.1 1ant and 1 e depth
where A, il' the gross cross-sectional area. This limit applic1-. to sections away from arcus of bnr lapping. (c) Maximum spacing of bar!i For slabs not exct:eding 200 mm thicknc~s. hur spacing shou ld not exceed three timet- the overall depth of slab or -100 mm whichever i:-. the l es~er for main reinforcement, and 3.511 or 450 mm for secondary rellllmceml'nt. In areas of concentrated load or maximum moment, thc1-c value~ arc reduced to 211 < 250 mm and 111 e provided. anchored h) top and bottom transverse bars.
2
218
Reinforced concrete design
8.4
Solid slabs spanning in one direction
The -.labs are de-.igned as if they consist of a serie-. of beams of I m breadth. The ma1 ~teel is in the direction of the span and secondary or distribution steel i~ required in tht transverse direction. The main steel should fonn the oUier layer of reinforcement to giH it the maximum lever ann. The cakulations for bending reinforcement foliO\\ a similar procedure to that used ir hcum design. The lever arm curve ol fi gure 4.5 is used to determine the lever arm (· and the area of tension reinforcement i~ then gi\cn by A, -
M -0 87f.,L:
~·or solid slabs spanning one-way the simplified rules for cwtuiling hars as shown 1 hgure 8.2 may be used pro\ 1ded that the load' are uniformly distributed. With continuous sluh it i-. also necessary that tht: spans arc approximately equal. The,, simplihed rule\ arc not gi,en in EC2 but arc recommended on the basis of prO\t satisfactory performance established in previous codes of practice.
8.4.1 Single-span solid slabs The ha-;1c span effective depth ratio lor thl\ type of slah i' 20: I on the bal>l'i that it ' lightly stressed' and that grade 500 steel is u~ctl in the design. For a start-point de~ign a \alue abo\'e thi~ can U\uall} he c),timatcd (uniCl>S the slah ic, known w bhcavily loaded) and subsequently checked once the main tcn~ion n.:mforccmcnt h been designed. fhe effective 1>pan ol the ~lah may he taf..en ~~s the clear d1stance hctween the face l lhc supports plus a distance at both ends taf..en a), the le~scr of (a) the di ~t:Jncc from tl, face of the support 10 11\ centreline and (b) nne-half of the menlll depth of the l!lab.
rlgure 8.2 Slmpliht-d rules for curtailment or ba~ ir1 slab }panning In one direCtion
100% < SO%
~
~
O.JJ. Stmpty Supported 25% or
' • 0.3L
m1d-span SLc!!l
c- 0.20L
'" 0.1SL ~
'-
100%
50%
ay. 5 mm tl" half the har diameter of the rcinlorctng har: O\erall uepth of \lab 170 f 25 f 5 200 mm ').7 X c.f.
Slab loading ~tel f weight ol slah
200 x 25
x 10- 1 = 5.0 kN/m 2
1.0 1 5.0 - 6.0 kN/m2
tolul pcrmancttl load Fm a I tn width of slab:
ultimate load
( 1.35Rk I 1.5qk)4.5
= (J.35 X 6.0 + 1.5 X 3.0)4.5 M
56.7 x 4.5/R
56.H N
31.9 kNm
Bending reinforcement ,\.f
bd2}~L
31.9 X ~= 4 I()(X) X 170 2 x 25 0 .()4
From the lever-arm curve of ligure -t5. I~ = 0.96. Therefore adopt upper limtt of 0.95 and lever-arm ;: -lad- 0.95 x 170 = 161 mm: 6
A, =- -M -.- = 31.9 X 10 O.l:!7})kZ 0.87 X 500 X 161
= ..t55 mm '!m
Provide HI 0 bar.. at 150 rnm centres. As = 523 mm 2/m (al> shown in tahle A.3 in the Appendtx}.
21
220
Reinforced concrete design Check span-effective depth ratio I
I
= IOOA, n:q = I()() X 455 bd 1000 X 170
0.268''~ (>0. 13'* minimum requirement)
!-rom figure 6.3. thi~ corre~ponds to a baste span-eftectt\'e depth mtio of 32. The actual ratio= 45001 170 = 26.5: hence the chosen effecuve depth ts acceptable. Shear
At the face of the !>uppon Shear V~::d
55.5 (2.25 - 0.5 X O.J) . =T 2 25 I(){) X
..,
k
=- 59 . N
523
PI = IOOO x t70=0.J I VR~ . . - t'Rd,,bd where vR11., from table 8.2 adjustment since fiJ < 0.4%). Thus:
VKu , - 0.55
X
I 000
X
170
=0.55
(note: no concrete strength
93.5 kN
as VL:d b less than VRu, then no
~hear
reinforcement
i~
required.
End anchorage (figure 7.26)
From the tahle of anchorage lengths in the Appendix the tenswn anchorage length
= 40c•,
40 x 10 = 400 mm.
Distribution steel
Pm\ldc minimum
0.00 JJbd - O.O A,= = . Q.87f}k: 0.87 ) 5()0 X J32
510mm2/m
Provide H 10 han. at 150 mm centre!>. A,
523 mm~/m.
lever-arm ;:
l.,d
=
Design of reinforced concrete slabs Check span-effective depth ratio IOOA, l00 x 510 p bd 1000 )( 140 = 0 '364%
=
f-rom figure 6.3 this correspond~ to a ba~ic span-effective depth rat1o of 24.0. Acrual
Span
Eff. depth
45 000 = 32.1 140
Thi!> is inadequate but can he overcome by increasing the Mccl area. Span . . A,. p01 , Ef ' d th = b at I 00 mm centres, A~ prov £Ience:
;th
A,.rrnv 1\,, ,'C~
Upper limit
785 5 10 = 1.54
w torrection
lienee al lowuhle Therefore d
n..,
= 785 mm 2/m.
factor (UK National Annex)
span . . efleetiVC depth
= 24 x 1.5 -
36 which
I 5. i~> greater than that provided.
140 mm is adequate.
8.4.2 Continuous solid slab spanning in one direction f-or a conllnuous slab, bottom reinforcement j., required \\lthlll the 'pan and top reinforcement over the support~. The effective span i-. the diStance hct\\ecn the centreline of the ..upport and the ba.pan-cffettivc depth ratio of an 1ntcnnr 1-pun i' 10.0 for 'lightly ~tressed' where grade 500 steel and clai-~ C'30/35 concrete arc U\Cd. The cnrre~>ptmding limit for an end span is 26.0. If the conditions given on page 209 are met. the bending moment and ~heur Ioree coefficient~ given in table R.l mt1y he used.
(EXAMPLE 8. 4
Design of a continuo us solid slab The four-spun slab shown in figure H.4 ~urports a variable load of 3.0 ~N/m 2 plus lloor fini\hes and a ceiling load of 1.0 kN/m 1. TI1c characteristic material ~trcngths arc .f.l 25 N/mm2 and f>k 500 N/mm2.
Estimate of slab depth
=
A~ the end 'pan i-. more critical th3 mm2/m.
Check span-effective depth ratio
IOOrl,,,.,~ IJd
= 100
X 349 = 0.2-19 I 000 x I 40
l.,d
Design of reinforced concrete slabs Hl0-400 Hl0-250
. s · ; , lfr.· HlO- 400
HlO
•
jk
200
• >
•
Figure 8.5 Reinlorcement in a continuous
slab
9
I
Hl0-200
Hl0 - 400
Hl0 - 250
Hl0-250
f-rom figure 6.3 thb corre1>pond~ to a basic span-effective depth ratio m excess of 12 x 1.3 (for an end ~pan) 41. The acrual ratio 4500/ 140 = 32.1; hence the chosen effective depth is acceptable. Similar calculations for the supports and the intelior ~pan give the steel area~ l.hown in figure 8.5. At the end ~upports there ill a monolithic connection between rhe slnb nnd the beam, therefore rop steel should be provided to resist t~ny negative moment. The momcnl w he de~igned fori~ rt minimum of 25 per cent of rhe span moment. I haL is 5.1 kN m. In fact, to rrovide a m1nimum of 0. 13 per cent of steel, I-l l0 ban m 400 mm centres have b~.:cn ~pecitied. The layout or the reinfmcemen\ in 11gure 8.5 i.., accmding to the ..,implit1ed rub for curtailment of bar~ in slabs as illuwared in llgure 8.2. fran~versc
remforcemcm = 0.00!3bd 0.0013 x 1000 " I~0 182mm ~/m
Provide Ill 0 m 400 mm centres top and bollom. where,·er there
t\
matn rcinforcem~.:nt
( 190 mm'lm).
lit at
=
8.5
223
Solid slabs spanning in two directions
When a .~olah is ~upportcd on t~ll four of it~ :.ides it effectively spun..; in both dir~.:c:tion~o. and it is sometimes more cconomic~1l to design the slab on this ha~is. Th~.: amount of bending in cuch direction will depend on the nttin of the two 11pan:-. and the conditions of remain! at each support. If the slab i~ square and the rcr.traints are similar along ti\C l'our sidus then the load will s.pon c4ually in both directions. If the slah i~ rccrungu lar I hen more than t)ne-half of the load will be c:urried in the stiffer, shorter direction and less in I hi! longer direction. If one span i~o much longer than the other. a large proportion nf the load will he carried in the ~hort direction and the slab may as well be designed as spanning in only one tlircction. Moment!> in eac:h direction of span arc generally calculated using tabulated coefficient
= the length of the longer side
I, - the length of the shorter side
a,, and a,> arc the moment coefficients from mhlc 8.4. The area of reinforcement in directions/, and /\ respectively arc 1\,~
M,x . = - 'f, . per metre w1dth 0.87 y~:
nnd
ll,y
M,y =-0.87/ykZ
per metre width
111c s!ah should be reinforced uniformly m:ro:-.s 1he full width, in each direction. The effective depth d used in calculating/\,~ should he le'' than that for A,, because Of the different depth~ of the tWO layer~ Of reinforcement. Table 8.4 Bending-moment coefficients for slabs spanning in two directions at right angles, simply supported on four sides 1.0
0.062 0.062
17
0.074 0.061
12 0.084 0.059
1.3
I 4
1.5
1.15
2.0
0.093 0.055
0.099 0.051
0.104 0.046
0.113 0.037
0.118 0.029
Design of reinforced concrete slabs F.stabli hed practice '>uggcsts that at least 40 per cent of the mid-span reinforcement 3
From the lever-ann curve. figurc 4.5. lever-arm
~
I~
-= 0.95. TI1crefore
- 0.95 x 185 - 176 mm
and O.R7
X
500
176
X
2
588 rnm /m
n. ·au~e
A,= 646rnm2/rn.
Provide 11 12 at 175 mm centres,
Span-effective depth ratio ~.right
1
= IOOAueq
It
bd
=
100 X 588 1000 > 185
= 0.318
From figure 6.3. tht'> correl>pondl> to a ba'iic 'pan-effective depth ratio of 28.0:
20 118 ~
actual
029 Thu'i d
-;pan cffecti vc depth
4500 185
1R5 mm b adequate.
?4 3 - ..
225
226
Reinforced concrete design Hl0-200
Figure 8.7 Simply supported slab
I
spanning in two directions
'
H12-17S
4.Sm
Bending - long span
,
a,~nl;
M,) -
= 0.051 X 22.43 X 4.5! = 23.16kN m
Since the rei nforcement for this span will have a reduced effective depth. take
;: = 176 A -
' -
12 = 164mrn. Therefore M,y
0.87.J;~::-
:n. l6 x
11Y·
0.87 )( 500
164
325 mm~lm
Prm ide H10 at 200 mm centres. A, = 393 mm '/m lOOA,
100
.193
-;;;/ =- (()()() v
164
= 0.24 "h1ch i' greater than 0.13, the mm1mum for tranwer'e 'teel, wuh cia'" C25/30 concrete. rhc arrangement of the reinforcement ''shown 10 hgurc 8.7.
l~--------------------------------------~) 8.5.2
Restrained slab spanning in two directions
When the slabs have fixity at the supports and reinl'orccment is added to resist torsion and to prevent the corm:rs of the l>lab from lifting then the maximum moments per unit width are given hy
,
d,,111~
in direc1inn of span I,
= 3,>111~
in direction ol span ly
M,, =
and M,)
where J~\ and d,y are the moment coefficients given in table 8.5. hased on previou' experience. for the specified end condition~. and n - ( 1.35gl l.Sqk ), the total ultimate load per unit area. The slab is di\ided into middk and edge !.trips as \.hown in figure 8.8 and reinforcement is required in the mtddle smp-. lO re'>ist ,H,x and M,y. The arrangement this remforcement should tal..c i-. illustrated in figure 8.2. In the edge strip~ only nominal reinforcement i:-. ncccssaJ). such that A,/ bd = 0.26f.:1m/f}l > 0.0013 for high yield steel.
Design of reinforced concrete slabs Table 8.5
227
Bending moment coefficients for two-way spanning rectangular slabs supported by beams
Short span coefficients for values of ~ 1 1~ Type of panel and moments considered
1.0
7.25
1.5
1.75
2.0
Long-spon coeffidents for all values of ly , lx
0.053 0.040
0.059 0.044
0.063 0.048
0.032 0.024
0 063 0.047
0.067 0.050
0.037 0.028
0.083 0.062
0.089 0.067
0.037 0.028
0.093 0.070
0.045 0.034
Interior ponels Negative moment at continuous edge Positive moment at midspan
0.031 0.024
0.044 0.034
One short edge discontinuous Negative moment at continuous edge Positive moment at midspan
0.039 0.029
0.050 O.Q38
0.058 0.043
One long edge discontinuous Negative moment at continuous edge Positive moment at midspan
0.039 0.030
0.059 0.045
0.073 0.055
Two adjacent edges discontinuous Negative moment at continuous edge Positive moment at midspan
0.047 0.036
0.066 0.049
0.078 0.059
0.087 0.065
- - ---
..
... I Q.
51 .,... .
/,
"
;;
:;; "'
.g'l ....
tte
_)
'2
:I
I i
31.. 4
1..
(cl)
ion unit
1- _ _
I
...i ~
ror span I,
I,
-'
Edge slnp
Mtddle strip
8
3/,
T
I;§'
- · - Edge Slrlp · -
l
I,
...
~
•
l I, 8
8 (b) For span ~
In audition, torsion reinforcement is provided ut c.lbwnrinuous corner~ nntl it ~hould:
1. consist of top and bottom mats, each hoving bar~ in both directions uf ~-opan; 2. extend from the edges n minimum diE.tunce 1, / 5: 3. ut a corner where the slab is discontinuou!l in both directions hnve an area of steel in each of the four layers equal to tJ1ree-quartcr~ of the urea required I'm the muximum mid-span momem; OU\ trihutctl lontl, n ( 1.35gk + l.Sqk) I 0 kN/m 2• The moment coefficient"\ nre taken from table 8.5. ~
I,
6.2) 5.0
I ::!5
Positive momenls at mid-span M,,
,J""'~
:;: ; ().().l5 X
5~
10
II :!5 kl\ m tn dtrc~.:uon I,
M,> - d,)111:
0.028
10
"s'
7.0kN min direction ly
Negative moments Support ad. M, SupportCction or it may be llared to form a column hend or capital. These vanoul> forms of con~truction arc illu11trated in figure R.l 0.
Design of reinforced concrete slabs
rr Floor without drop panel or column head
nnl~
Figure 8.10 Drop panels and column
heads
v
Floor with column head but no drop panel
Floor wllh drop panel and column head
The drop panels ore effective in reducing the shearing stre:-.sc:-. where the column i11 liable to punch through the slab. and they also provide an increase. This should be spread aero~!. the respective strip hut. in additional requirements on the amount and di-;tnbuuon ol re111forcement around column head\ to en,urc that full punching 'hear capacuy io; de\'eloped. The U'iUlll basic \pan effective depth ratios may be u...cd hut where the greater '>PliO exceed' R.5 m the basic rntio should be multiplied by 8.5/span. For flm ~labs the ~pun effective depth calculation should be based on the longer span. Reference ~hould he made to codes of practice for further detatled informntion describing the requirements for the analysis and design of Oat ~labs. including the u~>c ol bent-up har~ to provide punching shear reststance.
or
(EXA MPLE 8. 7
Design of a flat slab The co l umn~ arc at 6.5 m centres in each direcUon and the slab support~ a variable lond 2 of 5 kN/m • I he characteristic material strenglhs are ~k 25 N/mm2 for the concrete, and /y~ 500 N/mm 2 for the reinforcement. 1t i~ decided to use a floor ...tah us ~hown in figure 8.13 wi1h 250 mm overall depth of slab. and drop panels 2.5 m square hy I 00 mm deep. The column heads arc to he made 1.2 m dwmetcr.
=
Permanent load Weight of slab = 0.25 x 25 x 6.51 = 264.1 kl\ Weight of drop - O.l x 25 \( 2.51
=
15.6 kl\
Total= 279.71itivc moment
=( I -
0.55) x 247
llllli m.
Design of reinforced concrete slabs
(a) For the middle strip
_!!_ = bd2fck
I 36 X 106 4000 X 205~ X 25
= 0.032
From the lever-arm curve. figure 4.5. Ia = 0.97. thcrcfon: M 136 X let A - - - - ~=---=-=-=---::-:::-:---::-::-= ' - O.K7i)l~ - 0.87 X 500 X 0.95 X 205
1605 mrn 2 bottom steel Thu\ provide ,\ixteen HI 2 bar~> (A, 1809 mm1 ) each way in the ~pan. di~tributed evenly acros~ the 4 m width of the middle ~trip (spacing 250 mrn maximum allowuble for a slab in an area of maximum moment). (h) The column ~trip morm:nts will require 1310mm2 houom ~tcel which can be provided a~-. twelve H 12 bar~ (/\, 1356 mm 2) in the :.pan diMrihuted evenly across the 2.5 m width of the column strip (spacing approx 2 10 mm).
2. Interior &upport Ncg:uivc moment
= - 0.063FI 0.063
and
thi~
X
695
X
5.65
= 247 kN m
can also be divided into ()....') 5 X - 4 6.512
mrddle
tile or '' ith concrete containing a lightweight aggregate. If the blocks arc suitably manufactured and have adequate strength they can be considered to contribute to the strength of the slab in the design calculations. but in many design~ no such allowance is made. The principal advantage of the~c floors is the reduction in weight achieved by removing part or the concrete below the neutral ax i), and, in the ca'e of the hollow block lloor. replacing il with a lighter form of construction. Ribbed and hollow block lloors are economical for buildings where there are long spans. over about 5 m, and light or moderate live load,, such apital wards or apartment buildings. They would not he suitable for Slrut:turcs having a heavy loading, such as warehouses and garages.
Figure 8. 15 Sections through ribbed and hollow block floor~. and waffle slab (a) Sectton through a rtbbed lloor
II I
=s==-1! I
Ill E 21 I ·f il l ~IiI "' II
I· I
t=J;
Supporting beam
Solid end se and concentrated loads. During construction the hollow tiles should be well soaked in water prior to placing .: concrete. otherwise shrinkage cracking of the top concrete flange 1~ liable to occur. The thickness of tJ1e concrete flange should not be less than:
1. 40 mm or one-tenth of the clear distance between ribs. whichever is the greater. fnr slabs with permanent blocks: 2. 50 mm or one-tenth of the clear distance between nbs. whichever ll> the greater. for slabs without pcnnanent blocks. these requirements are not met. than n check of longitudinal shear between web und i ...ngc should be mude to see if additional tran~verl'e steel il> needed The rib width will be governed hy cover. bar-spacing ami lire resi~tunce (!.ection 6.1 ). The rihs !)hould be ~raced no further apart than 1.5 m and their depth below the flange 'hould not be greater than four times their width. Transver~c rihs should he provided at 'lacings no greater than ten tune~ the overall slab depth Provided that the above dimen"onal rcqu1remt!nt~ arc met. nbbcd ~lab~ can he treated or annly!>it-. as solid ~lah1-. and the clcl>ign rcquiremt!nts can be based on tho~e of a solid ,Jab. Calculaticms of reinforcement will require evaJualion of effective llangc breadths ;.Nng the procedurel> de1-crihcd for T-bcam-. in Chapter 7. Ribbed :-.labs will he del>igned for ~hear u-;ing the approach described previously With ?, taken us the breauth of the rib. Although no &pccific guiuance ~~ given in EC2, previou!> practice suggests that. where hollow block!-i urc used, the rib width may be mcrea'\ed by the wall thickncs~-o of the block on one sitlc of the rib. Span effective depth ratioo; \\Ill be based on the 'horter span \\ ith the ba:-.ic values 21ven 111 ligurc 6.3 multiplied by 0.8 where the ratio of the Oange width to the rih width exceeds 3. Again, no specifi~.: guidance i~> given in the Code but previous practice 'uggcsts that the thidness of the rih width may include the thickness of the two adjacent bloc!. walls. At least 50 per cent of the tensile reinforecmcm tn the ~pun ~hould continue to the supports and be anchored. In some instance~ the slabs are i>llpported by steel henrm and an! de:-.igncd as s1mply supported even though the topping i\ cominuou.:;. Reinforcement -.hould he provtded over the suppOrt!> to prevent cracl.1ng 111 these ca~e~. This top !>tee! if there are concentrated or moving loads. or 1f cracking due to ~hrinkage or thermal movement tl = 400 N/mm 2 . An effective section. as shO\\n in figure lU6. which satisfie, requirements for .. 60 minute fire resistance {see table 6.5) is to he tried. The characteri.,tic permanent load including self-weight and finishe~ 111 4.5 I.. N/m~ and the charactcriMic variable load i~ 2.5 1..N/m 2• The cuJculaLions are for an end ~pan {which wi ll be most critical) for which the moments and shears can be determined from the ~.:oefficient::. in tnble 8.1. Considering a 0.4 m width of floor ns supportctl hy each rib: Ultimate loud = 0.4( 1.35g~ + l.5qd
= 0.4( 1.35 X 4.5
I 1.5
X
2.5)
3.93 kN/m Ultimate load on the span. F = 3.93
>
ign a waffle lre~>s in the ~tee l wi lt be reduced, thil\ wi ll kutl to a higher spuneffective ch.:pth ratio thus ensuring that the spon- effet.:tive depth ratio of the slab is kept wi thin acccptubll: limit~. 2. At the support: design as a rectangular ~>Cction for the \Oiid slab. Negative moment at support m,,
0.01 I x I I.R5 x b~
!1"111;
= 13.22 kN m/m Moment carried by each 0.4 m width
5.29
M
btJ2fcl.
4()()
X
X
101>
160~
X
25
0.87.1;~,::
5.29 0.87
y
500
5.29 kN m
2 O.Q l
hom the lever-arm curve, figure 4.5 1. M A, = - - - -
0.4 < I 3.22
X
= 0.95. Thu.,
1011 0.95 ) 160
80mm 2
Provtde two HI 0 ha~ in each OA m v. idth of 'lab . ./\,
157 mm 2 .
Design of reinforced concrete slabs
1
241
3. At the l>cttion where the ribs terminate: the maximum hogging moment of rc'istance of the concrete ribs is 13.36 kJ.'l m. as in the prcviou!> example. Thi~ b greater than the moment at this section. therefore compression :.tecl is not required.
Span-effective depth ratio
IOOA ~ req be/
100 x 62 = 0.096~
_
4()()
160
X
hence from tigure 6.3, limiting basic span depth ratio J2 )( 1.5 (for interior 1-~pan) X 0.8 (for nange > .1 X web thickness) when p < O.JCJr.
Thw. allowable ratio
= 32 x 1.5 x 0.8 = 38.4 = 6000 = 37.5
actunl span effective depth
160
=
Thus d 160 mm is just adequate. It ha:-. not been necesl-tilry here to allow for the increased ~pun/effective tlcrth resulting from providing an incn.:ascd steel aren. t h u ~ con~iclenu ion could be given to reducing the rib reinforcement to two II X bars ( I 0 I mm 2) which :-till sat i!>lie!. nominal rcquircml.!nts.
Shear
From Tahle 3.6 in the Desifollll!rs Guide (ref. 20) the shear force cocflicicnt for a wntinuous edge suppol1 is 0.33. I lence, for one rib, the shear at the 'upport 1',,
~"11/,
Ma~1mum \;Ld
v /)-
11 .85
)I
6
X
X
0.4- 9.38k:--J
\hear 111 the rib 0.6m from the centre-line''
9.18
0.6
Atthl\ po~IIIOil.
>
0.94x 125x 160 x 10
1
0.94
12.8 J..N
Therefore the unrcinl orccd ~>Cellon i~ adcqume 111 shear, and no linJ..~ arc rctJuircd provided thut the bars in the ribs arc 1->Ccurcly located during const ruction.
Reinforcement In the topping flange
Light reinforcing Area required
mc~h
should he provided in the tOp of the nungc.
0.13 x b x il/ 100
Providc D98 mesh (sec tahlc A.5), A,
0. 13 x 1000 x 60/100 = 7Hmm 2/m 98 mm 2/m.
l~·-------------------------------------------) 8.8
Stair slabs
The U\ual form of stam. can be classified into two type:-.: ( I ) tho'c :-.pannmg horiLontall) in the tran~vertair -;lah may span into landings which b~bcd on the plan area. Loads common on two spans which intersect at right angles and surround an open \\CII may he assumed to be dtvided equally between lhe !.pan&. The effective span (/) i'> mea~urcd hori7ontally between the centres of Lhe :.upports and lhe thickness of lhc wai'>t (h) i~ taken a-; the slnb thickness.
Design of reinforced concrete slabs
243
Figure 8.19 Stairs spanning into landings
Stair ~lab!. which arc continuous and constmcted monolithit:ally \\'ith their supporting being left in the )Upporting beam~ to retch c the tatr.... and with no apprectablc end re~traint the dc~ign moment l>hould he 1'1/R. ah~ or beam~ cun
oc
( EXAMPLE 8 . 10
Design of a stair slab The 'tair' arc of the type ~ho~n in figure 8.20 11panning longitudinally and :.et into pudet' in the two ~upporting beam~. The effective lipan i., 3m and the ri~e of the Main.. '' l .S m with 260 mm treads and I 50 mm ri,ers. The variable load is lO k N/m ' and the chnrncteri&lic material ~trcnglhs arc .fc~ 30 N/mm~ and .h·~ 500 N/mml. Try a 140 mm thick waist, cffc ~>lab:-. the mtcn11cdiate suppon moment 'hould al'o lie between hall' 1J t\.\ icc the mugllttudc of the ~pun moments. It 1:-. a'Mnnl.!d that a pallern of yield line~ can be ~uperimpo,cd on the !-.lah. which w1ll IU'c a collupsc mcchani~m. and that the.: regions between yield line' remain rigid anu 'lCrtld.ed. Pigurc lU I ~>how:-. the yield line mechnnii>m which wi ll occur lor the ~implc 1'e of n lixeu endl.!u ,,lah f>panning in one direction with n uniform load. Rotation aloug •e yield lines wi II occur at a constnnt moment equal t.o the ull imarc.: moment 111 Yield line) Figure 8.21
Development of yield lmes FIXI'd
supports
Plastic hinges
246
Reinforced concrete design
resistance of the section. and will absorb energy. This can be equated to the energy expended by the applied load undergoing a compatible displacement and is known as the virtual work method. Considerable care must be taken over the selection of likely yield line patterns. since the method wi ll give an ·upper hound· 1.olmion. that is. either a correct or unsafe solution. Yield lines will form at right c, while for more complex cases difTerenlial calcu l u~ may be used. The danger of mis~ing the critical layout of yield lines, and thus obtaining an incorrect 'olution. mean" that the method can only be used wi1h confidence b) c\petienced dc.,igners. A number of typical panern~ urc 'hown in figure 8.22. Simple supports -
Figure 8.22
Exdmples of yield line patterns
~
!
Fixed support Positive yield line
- -_ ~.
......
....
Axes or rotation
Negattve yteld line Axe) ol rotation
'
Column rr~e
edge
A yield line caw;ed hy a sagging moment is generally referred to a-. a ·positive' yield line and i~ represented hy n full line, while a hogging momcm causing cracking on lhc top surface of the ~lab cuuses a 'negative' yield line shown by :.1 broken line. The basic approach of the method i:; illustrated for the ~ imple case of a one-way spanning slab in example B. II
Design of reinforced concrete slabs
I
( EXAMPLE 8 . 11
Simply supported, one-way spanning rectangular slab
The ~lab ~>h0\\11 in figure 8.23 is ~ubjected to a uniformJy distributed load'' per unit area. Longitudinal reinforcement is provided as indicated giving a uniform ultimate moment of resi~tance m per unit \\idth. Lonttudinal retn orc;ement
~
'
I
I~
m --
line
~
Fig ure 8.23 One-way spanntng slab
~
'
Yield
el ~
~
Pldn Hinge
Collapse mechanism
The ma"pan and a poltlll\c } ield line can thus be 'upenmpo:.ed a~ ~>hUstain is given by ..,- ( 6 1 J1) x _ J?( - llltO ~ -1112 _,..J1) -mta -nh ,- , , ai o(3 - 2 3)1.o·L-(311- 2 1·) ltts clear that the rc!>ull will var) according to the \alue of J. The maximum value of may be obtained by trial and error using se,·eral values of 3. or alternatively. b) di ftcrentiation. let m2 = Jllllt. then 11·
II'
12m d n 2
11,12 ) n~L2(3;J- 2J2)
and tl{ml /w) d,8
=0
will give the aitical value of I'J
hence JrtrJ2 + 4o 2 ;~- 3n2 =o
and
~] ,\ m:gah\e value
1~
1mpossihlc. hence the critical \aluc of ,j for usc in the analyc;il> ''
gi' en b} the posll1ve mot.
8.9.2
Hilleborg strip method
Thi!-t is based on the 'lower hound' concept of plu-,tic theory which suggests that it J ~trcil~ c.Ji~tribuuon throughout a Mructurc cnn be found "'hich sati'>fies all eqUJiibnum t'onditions without violating yield critcriu, then the structure is safe for the corrc~pond1ng system of external loath. Although safe. the '>Lructure will not necessaril) be ~erv1ceahle or economic. hence considerable sktll is required on the part of the engineer in selecting a suitable distribution of bending moment~ on which the ucsign cnn be ba~ed. Detailed analysi~ or a slab designed on thi~ basi~ is not necessary. but the cle~igner\ structural sense and 'feel' for tile way loads are transmitted to the support~ are of prime imporwncc. Although this method for design of slabs wa~ proposed hy ll il lcborg in the 1950s. developments by Wood and Armer in the 1960s havc pmduced it~ currently used form. The method can be applied to slab~ of any shape, and a~sume:-. that at fnilure the load will he carried by bending in either the x or y direction separately with no twisting action. llcncc the title of ·strip method'. Considering a rectangular ~lah ~1mply ~upported on four ~ides and carrying a uniformly di!.tributed load. the load may be expected to he di'itributed to the supports in the manner shown in figure R.26. Judgement will be required to detcnnine the nngle c\. but 11 can be !'teen that if n 90 the slab w11l be a'>sumed to be one-way spanning and. although safe. is unlikely to be servtceable because of cracking near the upports along the r axis. Hillcborg '>Uggcsts that for such a !>lab. n should be 45 . The load dtagrams causing bending moments along typical and the analy!>is of a 'tructuml frame have been described with example~ in chapter 3. Tn the anal}sis it was necessal) to clas~ify the ~tructure mto one of the following type~:
1. hraced - \\here the lateral loads are resio;ted b) shear wall;., or other form~ of bractng capahle of transmitting all horizontal loading to the foundations. nnd
2. unbraced - where horizontal loads arc resisted h) the frame nction of rigid!) connected column~>. beam1. and slabs.
I!
5
If
With a braced structure the axial forces and moments in the column-; are cau~ed by the vcnicul permanent and variable actions only. whereas with an unhraced ~>tructurc the loading arrangements which include the effects of the lateral load~ must also be considered. Both braced and unhraccd strw.:turcs can be further classilied th swuy or non-swny. ln a sway structure sideswoy is likely to significantly increase the magnitude of the bending moments in the columns whereas in a non-sway ~>tructun: thil'l effect i~> let.s significant. This increase of moments due ro sway. known ns n \econd order' effect. is not considered to be significant if there is less than n 10 per cent im:rcasl! in the normal ('fir~t order') design moments as n result of the sidesway dbplacement ~ ol' the ~tructurc Suh,tunt1ally hraced ~tructures can normally be considered to be non-:-.wny. EC'2 gives further guidance concerning the classification of unhraeed structure'>. In thi' chapte1 only the de.,1gn of hraccd non-~wa} 'lructures will he considered. For a braced \trueturc the critical arrangement of the uhnnatc loud 1s u\uall> that which cau ... e~ the largc... t moment in the column. together" 1th a large axial load. \ in the beam. Hence. for a typical column in a symmetrical frame with -;pans of approximate!} equal length, as shown in figure 9.2. k1 and J..~ can be calculated a!l
y/ /)cCllumn -
2:: beam str ffncss "£ 2(/ / I) t.:.m
(/ / llcolumn 2 x 2(/ / /) btam
I (/ / l)cnlumn 4 (//I) t-eam
Column
non-falling column
-
I End 1
End 2
9.1
non-fatling column
Note: the effective contnbution of the non-failing column to the JOint sttftness may be •gnored
Table 9.1
Column effective lengths
1 (f / l(olvrnn) 4 (Tj/btmnl
0 0.0625 0.125 (fixed end)
k
lo braced (equation 9.2) {x/} lo unbraced (equation 9.3(a) and 9.3(b)). Use greater value {x/}
0.25
0.50
1.0
/.5
2.0
0.5
0.56
0.61
0.68
0.76
0.84
0.88
0.91
1.0
1.14
1.27
1.50
1.87
2.45
2.92
3.32
1.0
1.12
1.13
1.44
1.78
2.25
2.56
2.78
Thus. for thic, '>ituat1on typical values of column efleeti\C kngth' c~tn he tahulated using equation~ 9.2 and 9.3 a\ ~hown in table 9.1.
(3) Limiting slenderness ratio - short or slender columns
113.a
EC2 place 1.7.
rm (c)
For braced members in which the first order momentt
= 4()() X 300'/J:!
90()" 10~ Innl~ 1
4
'"""'" - 300 ,.. 500' 1 12 - 3125 x I0 ' mm fcotfl~ol 900 >< 1011 / 3.0 101 k, =h ==-':..::..:._....:.;.;-~ - "iJ11tx-.llll f ltx:...,) 2(2 3125) HY'/4.0 X 101 ) 0.096 From table 9.1 and b) interpolation; effective column height/o
0.59 >. 3.0
= l.77 m.
Column design
Figure 9.3 Column end support detail
_ 3oo _
length= 4.0m each sid~ \
2
'
Beam
~ 400 -
z
Beam Note: the beams are continuous In both directions
I /·. I \(
Slenderness ratio ,\
"
Rudius of gymtion. i Slendernc's ratio \
3.46 111 /i
= 1.77 x l0'/86.6
86.6 mm
20.4
ror a bmced column the mmimum limiting value of \ will be given hy \h.,
26.2t v Nt;J /(AJ..,J)
''here: N~:t~ /(AJ~'Cl)- 1280 ( 10 /(400 1
X
300
25 / 1.5)
0.6-t
thu\ \1uu -
26.2/ V0.64
32.7
(> 20.4)
llcncc, compurl!d with the mwinwm limiting value of,\ the column
l order moment effect!> would not have to be wf...cn into uccount.
i~
shon and
~ccond
Load N
:r The mg
500
Moment M
(4) Failure modes
Short eolumns usually fail by crushing but a slender column b liable to foil by buckling. The end moments on a slender column cause it to deflect ~ideways and thuN bring into play an additional moment Ne114d as illustrnted in figure 9.4. TI1e moment Nendd cou~e:-. u further lmeral deflection and if the axial load (N) exceed~ a critical value thi~ deli eel ton, and the additional moment become self-propagating until the column hud.les. Euler denved the critical load for a pin-ended strut a~ M
The CfU\hing load Nu.t or a trul) axially loaded column may he taken a-.
Nu.J
= 0 567j~k Ac + 0.87AJ;..
where Ac ill the area of the concrete and A, is the area or the longitudinal
~~teet.
Figure 9.4 Slender column with lateral deflelender column~ and is liable to he preceded by cxce,~ive deflections.
9.3
Reinforcement details
The rule., govern ing the mintmum and maximum amount~ of reinforcement in a load hcarmg column ure as follOW\.
Longitudinal steel 1. A mini mum of four bars i'i required in a rectangular column (one bar in each corner) and six bar., in a circul11r column. Bar diameter should not be less than 12 mm. 2.
fhe minimum area of steel is given by
A,
= O.IONEd > OOO"k 0.87/l~ - . - '
Column design
25
3. The rnaJtimum area of steel, at laps is given by 1\,
ma~ < 0.08
A,
\\.here/\, is the total area of longitudinal steel and Ac is the cro ..s-sectional area of the column. . .m regtons . As.nu\ Oh t erwtse. away f rom Iaps: -
A,
< 0"' . u-t .
Links 1. Minimum si7e = ~ x size of the compre11~ion bar hut not less than 6 mm. 2. Maximum spacing should not exceed the lesser of 20 x size of the smallest compression bar or the least lateral dimension of the column or 400 mm. Thi& ~racing shou l(l be reduced by a factor of 0.60. (a)
for a distance equal to the larger lateral dimension or the column above and below n beam or slab, and
(b)
at lapped joint!. of longiwdinal burs> 14 mm diameter.
~p1C3.1
Wtlh
o \m .
Ilu ''
uch
~ load
at lure
3. Where the direction of the longitudinal reinforcement changes, the ~pncing of' the links should be calculated, while taking account of the lateral force~ Involved. If' the change 111 direction b lesl. than or equal to I in 12 no calculation is nece~sary . 4. Every longitudinal bar placed in a corner ~h oulc.J he held by trans\ersc remforcemcnt.
5. No compression bar should be further than 150 mm from a rcstra111ed bar.
1
hon
I
lh 30
Although links are popular in lhe United Kingdom, helical remtorcemcnt '' popular Mlmc part cranked Ml that it will lit within the :-.muller column above. The cr~,.,~ in the rciuforcemcnt11hould, if po:-siblc, commence uhove the soffit of a beam so that the moment of re~i~tance of the column i~ not reduced. For the same reason, the bnrs in the upper column ~ohou l d be the in
Figure 9.6 Detdils of splices in column reinforcement
load
omen ::m. Beam Soffot
(a)
(b)
(c)
260
Reinforced concrete design one., cranked \\hen both columns are of the same si;e., a:; in figure 9.6b. Links should be provided at the pointign cham are u~ually used for columns ha\ ing a rectangular or circular cross1>cction nnd a :.ymmetrical arrangement of reinforcement but internctton d i agrom~ can be constructed for any arrangement of cross-section us illustrated in examples 4. 10 and 4.11. The bu,ic equations or the approximate method can be u~ed when an unsymmetrical arrangement of rcinforccmenl i it e www .eumcode2.in ro.
6
(EXAMPLE 9.2
Column design using design charts Figure 9.9 show~ a frame of a heavily loaded industrial structure for whid1 the centre along lme PQ are to be designed in t11is example. TI1e frames at 4 m centres, arc braced again~t lateral forces. and l and second floors arc shown in figure 9.10(c). Membc1 stil'l'nc~scs arc I 2
kAB
- X
2
the
.. \\0
0.3 X I 0. 73 12 X 6
0. 71
X
I0 3
:1
2 '"1
12/.Afl
I 2
-- X
I 0.3 X 0.7' O O .,x., - = 1.7xl · 4
AR(
k
blr'
-
!_X
- 0 ·3 x OA' 12
X
3.0
= o53 .•
X
I
263
o-3
therefore
2) = (0.7 1 + 1.07 + 2 X 0.53) 10- 3 = 2.8-l X 10
l
264
Reinforced concrete design and
. "buuon . f acLOr 10r r the co Iumn d1qn
kcign method where the moments are relatively large . .\ would generally be lesll than h). 2. Determine the ~tccl "trains f-;.; and ~, from the ~ train distribution. 3. Determine the steel stresses .he andj~ from the equations relating to the stress-strain curve for the reinforcing bars (see section 4. 1.2). 4. Taking moments about the centroid of As
Nl'cJ(e ..- ~
d2)
-: 0.56~/~~bs(d - .v/2) +.t~~A~(d
(9.7}
d' )
where s = 0 K1. This equation can he solved to give a value for A ~
of
5. A, is then determined from the equilibrium of the axial forces , that
N1 of reinforcement required if the chttracteri~tic material strengths are f.,l = 500 N/mrn~ and }~l = 25 N/mm1.
Figure 9.12 Column with a non-symmetrical arrangeme of reinforcement
266
Reinforced concrete design
0
8...
-
0.0035
.I
300
Figure 9.13 Unsymmetrical column design example
neutral axis
~ -- r----
As
• • • '
I
d,=60
'
Section
Strains
axi~ .
1. Select n depth of neutral 2. From the strain diagram . steel stnun
r = I90 mm.
0.0035 =- (x- d') X
E:,c
0 0035 =- '190 ( 190 - 80)
0 .00"-l)3.
and .
steel stram !,
0.0035 =- (d X
r)
0.0035 (340 190 '
190)
= 0.00276
3. hom the M rc,~-Mrain curve and the relevant equations of section 4. 1.2 yield strain. 0.00217 for grade 500 steel t,
> 0.002 17: therefore /....
435 ~/mm 2
500/1.15
and E..:.< 0.00217;
therefore
!...:
£,.,,
200 x 10 1 x 0.00203
4061'!/mm2• compres~>ion. 4. In equation 9.7
Nt::d
(e + ~- d2) = 0.56~(~~/n(d
e
MEd Nt;d
~
230 X 106
11 00 x 103
= 0.8.r = 0.8 x 190
= 209
r f...:A~(d d' )
.1'/2)
mm
= 152 mm
To allow for the area of concrete displaced
I-.e become!> 406 0.567fck = 406 - 0 567 x 25
= 392~/mm
2
and from equarion 9.7 1100 X 103 (209 J40) 0.567 X 25 > 300 X 152(14()- 152/2) 1 A,=------~~----~~~~~~~----~------~~ 392(340
= 2093mm~
RO)
Column design 5. !·rom equati on 9.8
0.567/c.. bJ - /scA: ...L /.,As
Nr..d
(0.567
25
X
X
300
X
152 )- (392
A,
X
2093)- ( JJOO
X
101 )
435
= 843mm2 Thu'
A: +A,
2936 mm 2 for
x = 190 mrn
6. Values of II~ f-A, calculated for other depths of neutral axis.
A. arc ploncd in figure 9. 14. h om this tigure the minimum area of reinforcement required occurs with x ~ 2 10 mm. Using thi s depth of neutral axis. step~> 2 to 5 arc repeated giving
C:sc
fr~hm
Ji., so
0.00217. e,
= 0.00217
435 N/rnm2 and f., = 435 N/nun 2 tcn~ion
that
A:
= 1837 mm~ and /\,
R9 1 mm 2
(Alternati vely separate values of A~ and A, as calcu lated for each value of x could have al~o have been plotted against x and their values remJ from the graph at r 2 10mrn.) This area would be provided \\ith
A:
three
H25 plus two lJ20 ban.
2098mm' and
A,
one 1125 plu' l\\ o 1120 bars - 1119mm 2
With n symmetrical arrangement of reinforcement the area from the design chart of ligure 9.X would he A~ A, ~ 3 120 mm 2 or 14 per cent grl!ater than the area wi th an unsymmetrical arrangement, and including no allowance for the area of concrete displaced by the ~ooteel.
r
Figure 9.14 Design chart tor unsymmetrical column example
2900 2800 2700 180
190
200
2t0
220
230
Depth of neutral axis, x
These types of iterative ealculalionJ> arc readily programmed for solution by computer or using spread~heets that could find the oplimum l>tecl areas without the necel>~ity of ploning a graph.
267
268
Reinforced concrete design
9.4.3 Simplified design method As an alternatiYe ro rhe previous rigoroul> method of design an approximate method ma:r be used when the eccenuiciry of loading. e ;., not lese; than {h/ 2- th). Figure 9.1S Simplified design method
M. = M + N(h/2 - d1 )
M
? A', A,
A', A,
\:-
\/
The moment MEd and the axial force when.:
NEd
arc replaced hy an Increased moment Ma
(9.9 l plus a compressive force NFA acting through the tcn~ilc 'teel A a~ shown tn figure 9.15. Hence the destgn of the reinforcement t'> carried out 111 two pal1!-..
1. The member is designed as a douhly rcmforced :-.ectton to rc,i:-.t Mu acting hy
it~elf.
The equation., for calculating the area' of rctnforcement to reCctinn 4.5 a~: /VI~
O.l67fc~bd~ I 0.87J;v'\:(d
d' )
O.R~/~~11, - 0.2CJ4(dbd I 0.8~()~/\:
2. The area of A, calculated in the
11r~t pun is reduced by the amount
(9.10 (9.11
Nf!.d /O.'d7f)·k·
This preliminary design method i~> prohahly most usc l\ 11 for non-rccttlllgulur column sections as shown in example 9.5, hut the procedure i& hrc;t clemonstn.1ted with a rectangu lar cross-section in the following example.
(
EXAMPLE 9 .4
Column design by the simplified method Calculate the area of steel requtrcd in 1hc 300 x .tOO column of figure 9.1.l = II 00 k . MF.d = 230 k m. /.k 251\/mm, and />l 500 Nlmm 2•
=
NF~
. .
23()
X )()6
I:.cccnlncll) e = ---...,.1
1100 X 10
= 209 mm >
G-(/2)
Column design 1. lncrea ed moment
=MrA+NwG-d1)
Ma
= 230 + 1100(200 - 60) 10-3 = 384k.J"l m The area of sive reinforcement. that •., 0.167fckblP
MJ
+ 0.87J;.k11~(d
d' )
and 0.8~(,lA,
= 0.204/-~bd + 0.87/y~A:
therefore 38-+ x 1o6
0.167 x 25 x 3oo x 3402 ~ o.s7 x 500A~ ( 340 - 80}
-
so lhat A~ = 2115mm~
and ().!{7 \,
99
J(
500
331 I
J about A, so that with reference to figure 9. 16:
NEd ( e-'-
~- d2) 0.567}~kAl., (d
x) I
f~cA: (d -
d')
Solve th1 ~ equation to give A~
5. For no resultant force on the sec11on Nw - 0.567..f..kAcc
+ f
1 to
lind a minimum (A:+ A, ).
(..J ) and (5)
A., b the mea of concrete 111 comprcssmn shown \haded
.\ i:-. the di\tance from the centroid of '1,< to the extreme lihrc m compression j~ l'i the Mre~c; in reinforcement
A" negative 11' ten~ile.
The calculation for a particular cro~s-~cction would be very similur to that described 111 example 9.3 except when using the design equat1ons it would he necessary to determine A,c and x for cuch position of a neutral ax1~.
9.5.2
Simplified preliminary design method
The procedure is similar LO that de~cribed for a column with n rectangular section as described in section 9.4.3 and figure 9. 15. The column is designed to resist a moment M0 only. where
M,
MI\U I
NF.d
G-
(9. 12)
d2)
The slecl area required to resist this moment ~:w1 he calculalcd from
M,.
0.567/.kAcc(d- i } ~ 0.87/.;~A~ (d
d')
(9.13)
and
(9.14) where 11,. i'> the area of concrete in compre~~ion \\ ith 1 0.4Sd for concrete grades CSO and below and x ill the distance from the cenLroid of A•• to the extreme fibre in compression.
Column design
The area of tension reinforcement. A,, as given by equation 9.14 is then reduced by an amount equal to N~-.d/0.87/>1.. Thi~ method should not be used if the eccentricity, e, is less than (h/ 2 - d2) .
9.5.3 M- N interaction diagram These diagram~ can be con!>tructcd using lhc method described in sec11on 4.8 \\ ith example5 4.10 and 4.11. They are particularly u~eful for a column in a mult1-storey building where the moments and associated axial forces change at each ~torey. The diagrams can be constructed after carrying out the approx1mate design procedure in section 9.5.2 to obtain suitable arrangements of reinforcing bars.
(EXAMPLE 9.5
Design of a non-rectangular column section De~ign
the reinforcemeut for the non-rectangular section shown in figure 9. 17 given NEI.J = 1200 kN at the ultimate limit state and the characteristic material ~trenglhs are .fck = 25 N/mm 2 and ./)·• = 500 N/mm 2 •
= 320 ~N m.
MFd
11
t'
Ml.:d
Nl:.d
-
320 X 10 1200 X lQl
'?67
-
- -
lm:rca~cd moment M.
mm >
Mw I Nw
(II2 -d)-'
G
320 t 1200{200
d2)
80) I0
1
=464k..\lm With , 0.45d 144 mm. limit of the Mrcss hlocl.
b,
300
200{400
~
0.8.\
115 mm and the width (11 1) of the 'ection at the
115)
400
443mm A.,~
x(l> -1- "') 2 11 5{500 1443} 2 54 223 rnm 2
L l d'
.3
80
'
0
N
'
M
"
Figure 9.17 Non-rectangular section example
271
272
Reinforced concrele design
The depth of the centroid of the trapezium is given by _
s(b ...L 2bJ)
X=-'---,.;3(b T bJ)
= 115
(500 ..L 2 X 443) _ ) =56.3mm ( 3 500 443
Therefore ubstituting in equation 9.13 4(i.l x 106
0.567
25
X
X
54223(320
56.3) f 0.H7 x 500A:(320- 80)
hence
A: = 2503 mrn~ 2XI2 rnm 2•
Prnviue three H32 plus two H 16 bars, area From equation 9.14 lUl~/ykA,
0.567
25
X
X
54 223 I 0.87
X
500
X
2503
thcr~rorc
= 4269 mm2
A,
R~ducing
i\
'
A, by Ncd/0.87f,k gives
4'~-69
120()
0.87
X X
103 500
1510mm~
l
~.
Provtde one HI 6 plus two H32 hat·~. area IHII mm 2 • The total area of n:inforccmcnt prm ided 4623 mm 2 "hich i~ le~s than the 8 per cent allowed. An M-N mtcraction dtagram could nm' he con-;tructcd for tht'> ~teel arrangement. a~ 1n \CCtton 4.8. to provide a more rigurou~ dc~ign
___________________________________________) 9.6
Biaxial bending of short columns
For mol>t columns, biaxial bending wi ll not govern the design. The loadi ng pattern~ necessary to cause biaxial bending in a building's inLernal and edge columns will not usually cause large moments in borh directions. Corner eolumnf. muy have LO resist signilicant bending about bmh axes, hut the axial loads are u with a rectangular cros'>-,ecrion. \Cparate check M, and M) in the direction of the ZL and YY axe!> rc~pccti\cly (sec figure 9.18) may be dc:-.igncd for a -;inglc axis bending but with an 111crca,ed moment and ~ubjcct to the follm' ing condition~:
. M, tl ,,
(3)
M, h'
then the tncreased smgle axi desrgn moment rs
h'
M, ~
j b' X M)
if M, ..._ AI/~ ,, ll
(h)
then the increased single oxis design moment is M;
My +
I/
lj;,, X M,
The dimension.., II' anull are defined in figure 9. I8 and the coefficient ;:J is ~pecihed in table 9.3. The coefficient~ in table 9.3 ore obtained from the equation d=l
Table 9.3
Ned bhfc• j
Nht
bhf,;k Values of coefficient 1 for biaxial bending
0
0.1
0.2
0.3
0.4
0.5
0.6
1.0
0.9
0.8
0.7
0.6
0.5
0.4
07
0.3
274
Reinforced concrete design
(
EXAMPLE 9 . 6 Design of a column for biaxial bending
The column section shown in figure 9.19 is to be dc~igncd to res1st an ultimate axial load of 1200 k · plus moments of M, = 75 k'J m and M> - 80 kN m. The chamcteristic material strengths are fck = 25 , /mm~ and f.,k = 500 l\/mm2. Figure 9.19 Biaxial bending example
t 0 .....
t
M,
e
I
z
~
Z -·- ·-·-·-
75kNm
l
""
60 I
y
106
M2
75
My
80 x I if
X
c•,
= -NF.d = 1200 X 1()3
C'y
= NFd = 1200 X lO'
62.5 mm 66.7 mm
lhU\
~/('> "
"
= 62.5/66.7 350
0 o.x > '2
300
and l'y/f!( - 66.7/62.5 . - -- - 1.24 /1 It 300 350
> 0.2
lienee the column must he designed for binxiul bending. Mz _ 75 _ O 168 '" (350 - 70) - ·Mv 80 0 333 b' (300 - 60) • M, My II' < b'
therefore the increased M;
M)
Nwfblifcl
~ing lc
axis design moment b
b'
.,..JWx M7 1200 x IO't(300
350 x 25)
h om tahle 9.3, 3 = 0.54 M'
~
X()..,.. 0.54
240
X -
X
280 .
75
= 114.7 kt\ m
0.46
Column design thus
Mtd bh2_{ck
-- -
I 14.7 X 106 2 350 X (300) X 25
=
0 .15
From the de!>ign chart of figure 9.8 AJ)k
blifck
= 0.47
Therefore required A,
2467 mm 2•
Su provide four 1-1 32 bars.
9.7
Design of slender columns
As specified in section 9.2. a column is da:-.silicd as slender if the slcndemess rutio uhout either ax is exceeds the value of AJim· lf A::; AJim then the column may be cla~l>ilicd a~ shmt and the slenderness effect may be neglected. A slender column with A > Alim must be designed for an additional moment cau~ed hy its curvature at ultimate conditions. EC2 identi!ie~ four different apprnachc~ to dc!tigning slender column~: 1. A general method based on a non-linear analys1~ of the 't1111:ture and allm~ing for ~econd-order effect~ that necc~~itates the u'c of computer analy"'·
2. A second-order analy1>i~ based on nominal st1ffness values of the beam' and column' that. agam, require~ computer analysi" using a proces~ of Iterative analy!.i!-1.
3. The ' moment magnification· method \\>here the design momenh arc obtained by factonng the fiN-order moments.
4. The 'nominal curvature' method where second-order momentl- arc dctermmed from an estimation of the column curvature. These second-order momentl- are added to the first-order moment:- to give the tmal column des1gn moment. Only the fourth method. as given above. will be detailed here ll). this method is not greatly dissimilur to the approach in the previous Briti~h Standard for concrete design, BS RI I 0. Fur1hcr information on the other methods can be found in specialist literature. T he expressions given in EC2 for lhc additional moments were derived hy ~ t udying the moment/curvuture behaviour for a member subject to bending plus uxiul loud. The equa t ion~ for calr..:ulating the design moments arc only applicable w colu mn~. of a rectangular or circular section with symmetricul reinfnrcemcnt. A slcnuer column ~hou ld be designed for an ultimate axial load (NCd) plu1. on im:reascd moment given by Mr
Nhll'rur
where
=
l'tor eo + eJ + e~ eo IS an equivalent fif).t-ordcr cccenlrieity
e0
•~ an accidental eccentricity which accounts for geometric Jmperfecuonl- in the column
e: is the
~econd-order
eccentricity.
275
276
Reinforced concrete design The equtvalent eccentricity eo is given by the greater of 0.6eo2 - 0.4eoJ or
0.4eo~
where e01 and eo:>. are the first-order eccentricities at the two ends of the column a.' described above, and le02l is greater than Jl'ot . The accidental eccentriciry is given by the equation In
ea-- ~· 2
where /0 is the effective column height about the axis considered and I
l
100/i
200
1'=-->where 1 is the height of the column in metres. A conservative estimate of e0 can be given hy: /o 1 /o lo 11-=-X-=-
2
200
2
400
The !.econd-order eccentricity e2 is an estimate ol the deflection of the column at failure and i::. given by the equation c~
15 ( -C'v 36.47
S.O
0.4 X 3A6 = 61.55 > 36.47 5
,
Therefore the column b ~lender. and ,.\ i critical. Equivalent eccentricity = 0.6eo2 + O.·kot > 0At·0~
0.6eo2
OAeo1 = 0.6 x 41.2
0.4eu:
+ 0.4 x ( 5.9)
22.15 mm
= 0.4 x 41.2 = 16.47 mm
Thcrl!fore the equivalent eccentricity ec 22.35 mm. Taking ,. a~ l / 200 the accidental eccentricity i!. ea
fry
=
I'
=
2
I
200
6750 x T - l6.88mm
The second-order eccentricity is
Kt K!lij{1 ~
('"t=
11. 1
-·-
7!"1 X
-
I03 500tl
= 1-
(oJs 1 200~~
I~O)Oet = I + ((U5
1
096 ( X 6750' ;;· " 103 500
I >- I ••• t'~
:r~
I 03 :'iOOd
>(
25
:wo
77 !15) 150
0.87
I)
5(Xl 240
92.92 111111 \\ tth K' 1.0 for the initial value. l-or the lir:.t iteration the towl eccentricity
+ 16.88 + 92.92
22.35
i~
132. 15 mm
and the total moment is M,
NL!det(lt
Nt:u 1111/..k
1700
X
1().1
= 45() X :100 X 25
M1 =
1Jh1fck
= 1700 x 11
225 x ~() .f5() X 300· X
'"/~ _;,
132. 15 x 10 ' - 225 kN rn
0.504
= 0. 222
From the dc~ign chart of figure 9.&
Ah = 0.80
blrJ..~
and K' -
= 0.78
This new value of K1 iign o will often rellcct sound insulation and fire resi~lllm:c requirement)>. Nominal
The :nm
lhc
reinfon.:ement will he used to control cracl.ing in such t:a,\e).. More commonly, reinforced cont:rctc wall~ will form part of a structural frame and wil l he designed rnr vertical and horitontal forces and moments obwincd hy normal unulysis methods. In thi ~ situ:ltion a wall is dl!fi ned as being a vertical l oad- hc~1ring member whose length is not less than four timc11 its lhickncss. Where several walls arc connected monolitlucally so that they behave a~ a llllll, they urc de~cnhed as a wall system. Sometimes horilontal f'orl:cs on a strucwrc are rest~ted hy more than one wall or \)Stem of wall~. in \vhich case the dt ... tribution of forces hctwcen the walls or sy the \~all thickncs,, Homontal bar~ !.hould haYe a d1ametcr of not bs than one-quarter of the \'ert ,.. bar , and w1th a total area of not les' than 25'1 of the vertical ba~ or 0.00 lAc whiche a " greater. The horizomal har arc required for each It mit state u-;ing the
partial factor:. of ~afety a~ appropriate from tahlc~ 10. I and 10.2 2. The 'Indirect Method' which allows for a simultaneou.; hlending of ullimatc limit slate and rlwtlthe traditional UK approach to the sinng ol foundation~ t!\ effectively retained ~uch thnt a suitable base ~tLC may he determined based on the :.crviceability limit slate values for action~ nnd an assumed ullowuble ~ufc bearing pressure (see tnble IOJ). In !his way settlement will he controlled. with the exception that for foundations on soft clay full sculemcnt calculutions mw.t he t:arried out.
Table 10.2
Partial safely factors applied to geotechnical material properties Angle of shearing resistance '\
Combination 1 Combination 2
1.0 1.25
Effective cohesion 1
SUre aero~~ the ba~o.e of the rooting 1\ :t'iMtllled :11, shown in figure 10.2(a). This assumption must be ba~ed on the soil acting as an elt1~tlc material and the foming having infinite rigidity. In fact. not on ly do most soi ls exhibit some pln~tic bchmiour and all footings have a finite stiffne~:-. but also the t.hstributton of 'ioil pre!>surc varies with time. 'I he aclltal diwibution of hearing pres~ure many moment may take the form shown in ligure 10.2(b) or (c), depending on the type of ~od and the sriffne'~ of the ha'ic and the structure. But as the behaviour ol foundation\ mvolves many uncenamue~ regarding the actton ol the ground and the loadmg. 11 " u~ually unrealbtic to con\ldcr an analys1Cctions through the ha.,e for chcckmg '>henr. punchmg )~hear and bending arc shown in figure 10.5. The shcan ng force and hcndmg moment~ an.: cmt~ed hy the ultimute load!'~ from the column 1111d the wetght of 1he base ~hould not he included in thc:-c calculations. The thtcknes' of the base i~ oflcn governed h) the requirement' tor 'hear re~•stancc. Following the Prescriptiw! Ml!tlwd the princqxll 'tep~ in the dc:.tgn calculation' arc a!> follows:
1. Calculate 1he plan size of the footing using the permi&sible hcming pressure amlthe critical loading arrangement for the serviceability limit state.
2.
Calculate the heanng prcs'>urc., a. 8. Where upplicablc, both foundations and the given by:
169 I kN ( .-. Vud
= 626 kN)
7. Maximum Shear Force - sec figure 10.7!b)
AI the critical section for shear. I.Od from the column face: De~ign
shear """
A' before. .', \'KJ
r
I'Kd.,
=
239 x 2.8 x 0.68 = 455kN
= OAON/mm~
I'Rd. cbd
= 0.40 X 2800 "
520
X
10- 3
-
582 kN ( ~
Therefore no shear reinforcement b required.
\'hi
455 kf':)
Foundations and retaining walls
291
Instead of as~uming a footing weight of 150kN at the stan of this example it i~ possible to allow for the weight of the footing by using a net safe bearing prc:-.surc flnct where P.wt
= 200 = 200
II x unit \\eight of concrete 0.6 >< 25
= I85.0kN/m2
Therefore Required base area
l~
1.0 x column load
1000 + 350
r>net
IR5.0
7 . 30m~
It should be noted that the self-weight of the footing or ih effect nw~t be included in the calculations at serviceability for determining the area of the base hut at the ultimate limit stme the self-weight should not he included.
_______________________________________)
Example 10.1 shows how w design a pad footing with a centrally located set of actions. If the action~ arc eccentric.: to the c.:entroidal axis or the base then in the chccking of punching shcor the maximum shear stress. ' '&J. i!- multirlied by an enhancement factor ,1i ( l ). This factor nccount~ for the non-linear tli:-.lrihution of strc'" urnund the critical perimeter due to the eccentricity of loading. Refcrcncc should he mudc to EC! Clause 6.4.3 for the details of this dcs1gn approach.
10.2
Combined footings
Where two column-; are clo'e together it ·~ sometime!> necessnl) or convement to combine thetr footing1> to form a continuous base. The uimen,ions of the l'oottng ~hould he chosen so that the resultant loud pa,~cs through the centroid of the ha'e area. This may be a~!.lllned to gne a uniform bearing pres:o.urc under the foming and help to prevent differential settlement. For most ~tructures the ratios of permanent wtd variable loads carried hy each column arc 11imilar so that if the t·c~uhnnt pa~~es through the centroid lor the serviceability limit ~Late then this will also he true or very neurly at the ultimate limit ~tate. and hence in these cu:-.es u uniform prc:-.sure dt.,tribution may be considered for both limit ~tates. The ~hape of the footing may be rectangular or trapezoidal u~ ~hown in figure I O.R. The trape:wi(lol base has the tlisadvant:~ge of tlt:ttrengthen the hase and economi~c on concrete a beam is incorporated bctweeu the two columns :-;o that the bose i~ designed as an inverted T-scction. Centroid of base and resultant load coincide
Figure 10.8 Combined bases
-"
--m
+'
o-.-
Rectangular
1--
Trapezoidal
292
Reinforced concrete design
The proportions of lhe footing depend on many factors. u· it i!> too long. there will be large longitudinal moments on the length~ projecting beyond the column!>, 'Whereas a ~hort base ''ill ha-.e a larger span moment between the columnver::.c reinforcement should be Jllm:ed a1 clo~er centre~ under the columns to allow for greater moment~ in those regions. For the purro~e~ of crack control. the ma"imum bar site or maximum bar 11pacing ~hould al~o be checked as in example I 0. I.
8. Shear Punching shear cannot be checked. since the critical peri meier 2.0d from the column lace lies outside the base area. The critical sec1ion for shear is tn~c n I.Od from the column face. Therefore with d 780 mm. Design ~hear Vr:d
= 1420 -
377 x 2.3(0.7K 1 0.2)
570 kN
The ~hear resistance of the concrete without shear reinforcement can be obtained from tabll: 8.2 where p 1 can be talcn as the average of thl: steel ratio' in hmh direction~
"""" As [ 2830 I340 ] =0.5 Lbd= 0·5 2300 x 790 + 1000 < 770
°·0016 (
0 · 16t;f< 2%)
=
hence from table 8.2 ' 'Rd , 0.36 N/mm2• Therefore 1he shear resistance ot the concrete. VRd 1s g1\Cn by: VR \1~~~
570 kN)
l~------------------------------------~)
Foundations and retaining walls
295
Strap footings
10.3
Strap footings, as shown in figure I0.11. are used where the base for an exterior column must nor project beyond the property line. A strap beam is constructed between the exterior footing ami the adjacent interior footing - the purpose of the strap is to restrain the overturning force due to the eccentric load on the exterior footing. The base areas of the footings are proportioned ~o thai the bearing pressures are uniform and equal under both ba1.cs. Thus it is necessury that the resultant of the loads -::>n the two footing!'! should pass through the cemrotd of the areas of the two bases. The 'trap beam between the footings should not bear agatnst the soil, hence the ground directly under the beam \hould be IOO!.cncd and lefl uncompacted. As well as the loadings indicated in figure 10.11 EC2 recommend& that. where the action of compaction machinery could affect the lie beam. the beam should he designed for a minimum downward load uhjcctcd to the ground hcanng pre~sure~. With a thick rigid footing and 3 hrm -.oil. J linear dtstributmn of bearing pre~o.sure is considered. If the column' are equally '>paced and equally loaded the prcso;ure IS untforrn.ly distributed but if the loading t)> not ')mmctrical then the base i' !'.Uhjectcd to Wl eccentric loau and the bearing pres.,ure \'ancs a' shown 10 figure IO.ll The bearing pre"urcs will not be linear when the footing i' not very ngid and the soil " soft and compressible. In these cases the bending-moment diagram would be quite unlike that tor a continuous beam with lirmly held support' and the moment~ cuuld be qutte large. particularly if the loading is unl.ymmetrical. For a large foundation it may be necessary to have 11 more dctai lctl inveqigution of the soil pressures under the base in order to determine the bending moments and ~hearing forces. Reinforcement is required in the b01tom of the base to resi~t the transverse hencling moments in addition to the reinforcement required for the longitudinal bending. Footings which support heavily loaded columns often require !>lirrups and bent-up bars to resist the shearing forcet-. ~tcppcd
(EXAMPLE 10.3 Design of a strip footing
Design a 'tnp fooung to carry 400 mm square column-. equally ~raced at '3.5 m centre.... On each column the characteristic load~ are I000 k:-.1 permanent and J50 kN \Uriable. The safe beanng preso;ure is 200 k;\/m2 and the characteristic material \trength~ aref.L JON/mm~ and};,= 500 •tmm2. Bal>c the dcl-tgn on the Prt•lcriptil'l' Method.
Figure 10.13 linear pressure distribution under a rigid strip footing
298
Reinforced concrete design
1. Try a thickness of footing - 800 with d reinforcement. Net bearing pressure. p,.,1 = 200 - 2511
200
740mm for the longitudinal 25 x 0.8
1
- 180.0 k '/m . rcqui!C . d= \v1'dlh of ..tooung
Provide a
~trip
1000 -l 350 X 3.4
180.0 footing 2.2 m wide.
2.14m
At the ultimate limit state column load, Nb.J 1.35 x 1000 . 1875 bcanng pressure= . x . 22 35 = 244k.N/m 2
1.5 x 350
1875 kN
2. Punching .\'hear at the C()lumn face Maximum shear resistance.
VKd,m.l\
= o.swt [o.6 ( 1- ;;~>) J {::~ 0.5(4
X
400) "740
X
30 )1301 '\ ( 10 \ [o.6( I 250
6251 k:'-l 8} in-.pection, the normal 'hear on a ~ection at the column face will be ~ignificantly le~~ \C\'crc than thi' value.
3. umxiTtulmal r~infurceme/11 L -;ing the moment and shear coeflicients for an equal-1.pan continuous beam (figure 3.9). for an interior pan
moment at the column1-. M~o.c~
244 x 2.2 x 1.5] x 0. 10
= 605 k
Ill
therefore
A1
665
X
106
= 0.87 X 5()() X 0.95 X 740
2175mm 2
From table 6.8
O. ;~~d = 0.0015
A, m'"
- 2442mm
X
2200
X
740
1
Provide eight H20 bar~ at 300 mm centre~. area
2510 mm,. boltom steel.
In the CCtton for ~hear tS taken I J)d from the column face. Therefore wrth d 740mm Oe~tgn ~hear
244 > 2.2(3.5 x 0.55
Vrtt
0.74 - 0.2}
= 529kN (The cnefftcient of 0.55 is from figure 3.9.) The shear resistance of the concrete without shear reinforcement can be obtained from tahle 8.2 where p 1 can he taken as the average ot the ~teet ratio., in hmh directions
~A, = 0'5 ~bel
s[
25 10 _ ~] _ O.. 2200 X 740 I 1000 X 720 - 00 ' 0165 (
() 165 ~ . centre to l:Cilli'C, ~hould not be les~ lhan (I) the pile perimeter for friction pile~. 01 (2) tw1ce the lca!>t \\idth of the pilc for end hearing pilel>. Bored piles 3rc ~mnetime~ enlarged at the btl\e ~o that they have a larger bearing area or u greater resistance to uplift. A pile is designed as a ~ho1t column unle% it i~ ~>lender and the :-urrounding :-oil is too weak to provide rcstra1nt. Prcca\t pileigncd to resi~t the bending moments caued hy lifting and ~tacking, nnd the hcud of the pile must be reinforced to withstand the impact of the driving hammer. Jt i~ very difficult, if not impOSSible. to determine the true distribution of lond of ll pile group. Therefore. in general. it is more reali'>IIC to u-.c method-. that are Mmplc but log1cal. A \erucal load on a group of vcnical p1b wnh an ax1' of '>ymmetry is considered to he distributed according to the following equation, which is similar in form to that for an eccentric load on a pad foundation: ~oil
or
Pn
= -N ± Ne" Yn ± Ne» 11
'"
\n
In
where
Pn is the axial load on an individual p1le N IS the vertical load on the pile group n i~> the number of piles e.. and e,y arc the eccentricities of the load N about the centroitlul axes XX and YY of the pile group fu and In arc the second moment\ of area of the pile group about axes XX anti YY r 11 and .l'n arc the distances of the inclividu:.~l pile [rom axes YY und XX, re~>pectively.
301
302
Reinforced concrete design
(
EXAMPLE 10.4
Loads in a pile group Determine the distribution between the individual pile~ of a 1000 kN vertical load acting at the position ~hown of the group of vertical piles shown in figure 10.18. To determine the centroid of the pile group take moments about line T- T.
- LY
' ' =f/- =
.
2.0 - 2.0 .L 3.0 + 3.0 l 67 = . m 6
where 11 IS the number of piles. Therefore the eccentricitic~ of the load about the XX and YY centro1dal axis are e._ - 2.0 - 1.67
= 0.33 m
nnd ~'Y>'
-
1,~
= LY~
0.2m
with respect to the centroida l axis XX
= 2 X 1.672 + 2 X 0.332 + 2 X 1.33 ~ 9.33 Similarly 1,\
= I>~ = 3 X
1.0! + 3
X
1.0'
6.0
therefore N -
Nl'n I Yn
II
"
1000
= -6- J:
=Ne,, I .
tn
t)
1000 X 0.33
JOQO
9.33 _In± 166.7 ± 35.4yn ± 33.3.\n
0.2
6.0
111
y
Figure 10.18 Pile loading example
... l .Om +
l.Om _
I T &
@1
2®
I I
E
T ~
.....
10 II
X- · -
:t
E 0 ...;
1000kN
1>-
-@3 \ -·-
-
---
4~ ,
E
..... ..... 0
..= II
E
e-s -
O.Sm
·- X
60 -e,., ...
y
0.2m
~
Foundations and retaining walls Therefore.
166.7 - 35.4
P1
P2
X
for Yn and
Xn
1.67 t- 33.3
X
= 166.7-35.4 X 1.67-33.3 X
1.0 = 140.9!w'l 1.0 = 74Jk'
X
1.0 = 211.7 k.'l
= 166.7-35.4 X 0.33-33.3 X
1.0 = 145.1 k:\l
166.7
P1
p~
~ub~tituting
35.4
P5 - 166.7- 35.4
Pt.
166.7
X
X
0.33 r33.3 1.33- 33.3
35.4 x 1.33
X
1.0 = 247.1 k.\f
33.3 x 1.0 = 180.5 kN Total = 999.6::::: IOOOkN
When a pile group is unsymmetrical about both co-ordinate axes it is nece&sary to consider the theory of bending about the principal axes which is dealt with in mo11t tcxtboo"-~ on strength of materials. Ln thjs case the formulae for the pile l oad~ are f>u
N II
.L 1\•\'n .L IJ ru
where
N (e\~ 2::.>~- ive members of the I russ
and steel reinforcement providing the tensile tie force as 'hown in the two-pile cap in figure I0.20(al. The upper node of the ln1ss is locatctl ut the centre of the loaded are.. and the lower nodes at the inter~ection of the ten:.ile reinforcement with the centreline' of the pile~. Where the piles arc .,paced at a distance greater than three times the p1le diameter only the reinforcement within u d~'>t cxcechown in figure I0.22. The shear force at the column face should be checked LO ensure that it is less than 0.51'l/cdlui = 0.5,· 1(Jc~/ l.5}ud where u is the perimeter of the column and the strength reduction factor. 1· 1 = 0.6( I - /c~/250).
10.7.4 Reinforcement detailing As for al l member~. normal dct:Jiling requirements must be checked. These include maximum and minimum steel areas. har spucings, cover to reinforcement and anchorage lengths of the tension steel. The main tension reinforcement should eontimre past each pi le and ~>hould be bent up vertically to provide u full anchorage length beyond the centreline of each pi le. In orthogonal directions in the top and bottom faces of the pile cap a minimum steel area of 0.26(f..11n/f)dhd ( > 0.0013/}(/) ~hould be provided. It I' normal to provide fully lapped horiL.ontal links of si1.c not le'>~ than 12 mm and a ~pacings of no greater than 250 mm. a~ sho" n in figure I0.2:\(b). The piles ),hould be cu otl ~o that they do not extend Ulto the pile cap beyond th~.: lower mat of reinforcing bar. otherwise the punching shear strength may be reduced.
10.7.5 Sizing of the pile cap In determrntng a suitable depth of pile cap table I0.5 may be used as a guide ''hen thert are up to six piles in the pile group. Table 10.5
Depth of pile cap
Pile size (mm) Cap depth (mm)
(
300 700
350 800
400 900
450 1000
500 1100
550 1200
600 1400
750 1800
EXAMPLE 10. 5
Design of a pile cap A group of four piles supports a 500 mm 'paced at three time~ the pile diameter thi' retnforcement rna) he dt given by: \ ' lld
c
l'Rcl
.hd
- 0.35 "2 100 x 875 X 10)- 643kN (> V~.cd
414k:-.i)
307
308
Reinforced concrete design (d) Check for punching shear A~ the pile spacing is at three times the pile diameter no punching shear check necessary. The shear at the column face should he checked:
i~
Maximum shear resistance. VRd,mn'
= 0.5ud[o.6(1- {~~1) Jf:~ = 0.5(4 X 500) X 875 " [o.6(1 = 9240 kN ( > N'Ell 5000 kN) 10.8
3() ) ] 30
250
15
X
lO
~
Retaining walls
Su~.:h wal l ~ ore usually required to re~ist u combination of earth and hydrostatic loadings. The fundamental requirement is that the wall is capable of holding the retained material in plnce without undue movement arising from deflection, overturning or ).liding.
10.8.1 Types of retaining wall Conc.;rctc retaining walb ma) be constdered in terms of three ba~tc categories: (I) gruvity. (2) counterfort, and (3) cantilever. Within these group~ many common variations extsl. for C\ample cantilever wall~o may have additional surporting tiel> into the rctnined material. The su·uctural action or each type 11-. fundamentally different, but the techniques used in analysis, design and detailing nrc those normally u~ed for concrew structures.
(i) Gravity walls The~;e
arc uwall) con,tructcd of nw. ~ concrete. with reinforcement mcluded tn the face~ to rc-.trict thermal and ..,hrinktUblhty requirements. both in respect of overturning and \lldmg. It 111 generally taken a~ ultnnt of the ~ell'-weight and overturning forces mu~t lie within the middle third at the interface of the busc and soil. This en,ures lhnt uplift is avoided Ht this inte1fac:c, us described in !>Cction I0.1. Friction effects which resi~t 'liding are thus maintained across the entire base.
Figure 10.24 Gravity wall
Ioree
Foundations and retaining walls
Bending, ~hear. and deflection~ of such walls are u~ually insignificant in view of the large effective depth of the section. Distribution steel to conrrol them1al cracking i:. necessary. however, and great care mullt be taken to reduce hydration temperatures by mix destgn. construction procedures and curing technique~. (ii) Counterlort walls
This type of cot1\lructinn will probably be u:.ed where the overall height of the wall is too large to be con~lructed economically either in mass concrete or as a cantilevct. The basis of design of counterfort walls is that the earth pref.~urcs act on a thin wall which :-.pan~> hmizontally between the massive counterforts (tlgure 10.25). These mu~>t be ~ u fficient l y large to provide the necessary permanent load for !>Lability requirements. pussibly with the aid of the weight of backfill on an enlarged hase. The countcrfort~> muM he designed with rcinforl'cment to act as cantilevers to resist the considerable bending moments tiH\1 arc concentrated at these points. Countcrfort Figure I 0.25 Countcrfort wall
JllURl l
Cro~~-sewon
Span Plan
The i>paclllg of countcrfort'i \\til be go\cmcd by the abO\c factor:-.. coupled with the need to rnatntatn a l>atisfactory span-depth ratio on the wull "lab. \\ hich mu'it he del>tgned for bending us a continuou~ slab. The athalllage ol Lhtll form of con~truction i'i thut the volume of concrete involved is con.,idcrahly reduced. thereby removmg many of the problems of large pours. and reducing the quuntJfles of cxcavatton. Bulanc.:d aguin'it thi.s mu)t be con11idcrcd the generally increased -;htHtering complication anti the probable need for increased reinforcement.
(iii) Cantilever walls These arc tlc.signetl as vertical cantilevers ~panning !'rom a large ngicl ha~o.c which often t'Ci les on the weight of backfill on the hato.c to provide ~labi l ity . Two fo ll11l> of thito. rlgure 10.26
Cant1lever walls
~ -- H,
'
-- Heel beam
c. (a)
(b)
309
31 0
Reinforced concrete design
con.,Lruction arc illustrated tn figure 10.26. In both ca.c,es. stability calculations foliO\\ similar procedures to those for gravity walls to ensure that the re1-oultant force lie within the middle third of the base and that overturning and sliding requirements are met.
10.8.2 Ana lysis and design The design of retaining walls may be split into three fundamental !>tttges: (I) Stability analysis- ult1mtlle limit \tate (EQU and GEO), (2) Bearing pressure analysis - ultimate lunll '>tate (GEO). and (3) Member design and detmling - ultimate limit state (STR) and c,erviceability limit states.
(i) Stability analysis l~ nder the action of the loads corresponding to the ultimate limit state (EQU). a retaining wr~ ll must be stable in term!-- of resistance to Ol'errurning. This is clemon~lrated hy the simple case of a gravity wall as shown in figure 10.27. TI1e critical condition' for overtummg are when a maximum horilOntal force t~ch with a minimum verucal load. To guard again\! fmlure by overturning, it ic, u~ual to upply conservative fr~ctorl-1 of safety to the forces and loads. Tahle IO.I(c) gives the factors that arc relevant to these calculation1.. A panjal factor of safety of ~·c ; = 0 9 i' applied to the permanent load Gl if 11oil mechanics. or to usc increa.~ed factorl> of sufety. L~r.:.:m.
I I - ~
D/2
Beanng pressures
Re;ultant force Hk
D/ 2
Figure 10.28 Forces on a cantilever wall
311
312
Reinforced concrete design
(ii) Bearing pressure analysis The bearing pressures underneath ret:Hntng walls arc a:-.sessed on the basis of the ultimate limit state (GEO) when determining the size of base that is required. The anaJy,is will be 'imilar tc> thai dJscussed in ~ection 10.1 with the foundation being subject to the combined effects of an eccentric verucal load, coupled with an overturning moment. Considering a unit length of the cantilever wall (figure 10.28) the resultant moment about the centroidal axis of the base i'> ( 10.11 )
and the vertical load is ( 10.12)
where in lhi~ ca:-.c ol the STRand GEO ultimute limit state~ the part ial factors of safely are gi\'cn in Tahlc 10.1: l-or load comhinmion I· " 1
-
1.35 and 'irz
For load comhinm ion 2: 1r1
= l 'r:!
")'f'
'in
= 1.0
= l.O
as,uming that. for load comh1nat10n I. the effect of the moment due to the hori7ontal load on the maxtmum hcanng pressure ut the toe ot the wall at A 1s ·unfavourahlc' whi lst the moments of the ~te l f-wcight of the wall and the earth acting on the heel or the wall act in the opposite sense and are thus ·ravourahle'. This assumption may need chcck1ng in mdi\1dual ca~es and the appropriate partwl ft~ctor:-. applied depending on whether the cllccl of the load can be con:,1dcrcd to be f~1vourable or unfavourahlc. The distribution of bearing prcssure11 wi II be os ~hown in tigurc 10.28. provided the cl'fcctive eccentricity lies within the 'middle third' of 1hc base. that i~
M
D
N
6
The ma'\imum hcanng
prc~),ure
1s then £1\en by
N M D
(11
D+ I
"2
Therefore N
D-t
PI
6M 1)2
( 10.13)
and N
"~
=
o
6M D~
( 10.14)
Foundations and retaining walls
(iii) Member design and detailing A with foundations, the design of bending and ~hear remforccment " ba\ed on an analystign the bending reinforcement u~ing high-yield ~ tee!. concrete cia's C30/.l7.
fvk
2. determine the bcanng pressure:. at the ultimate limit
3.
= 500 kN/mm2
and
(1) Stability Horizontal force
=
It b a:>sumed that the coefficient of active pressure K4 0.33. \\hich is Ccllculate lOlSe~
under quasl·permanent lo.td~
D 7.'1
8.10.3 6.1. 5.10.8
6.2
8.10.3 Figure 11 .1
Prestressed concrete destgn flow chart
Check delle( lion~
D D Ultimate moment of reslsta11ce D Untenstoned reinforcement Design end block
D Shear remforcPmPnl des.gn D Check end-block (unbonded) D FINISH
Pr~strtss
lystem
, . . _ Ultimate moment
~
~ -'
, . . _ Ultimate shear lore~
~
~
§"' 5
Prestressed concrete
Principles of prestressing
11.1
In the dei>ign of a retnforced concrete beam subjected to bending it i1- accepted that the concrete Ill the ten~ile .tone !!> cracked, and that all the ten!>ilc rc-.i'>tancc " prov1ded hy the reinforcement. The '>trel>i> that rna) be pcrmiued in the reinforcement i-. limited by the need to keep the cracls in the concrete to acceptable '' idth~ under working condillons. thus there is no advamage to be gained from the u~c of the very high Mrength steels wh1ch are available. The design is therefore uneconomic in two rc)!pccts: (I) dead \\eight includes 'uselesl> · concrete in the tensile 7one. and (2) economic uM: of Mccl resource~ is nm po~'ihle. ·Prestressing' mean:- the artificial creation of Mres.,es in a structure before loading, l>O that the stre~ses which then exist under load arc more favour applied force acting over the length of a beam. The stres!-. & .. tributiun at any ~;ecuon \\Ill equal the 'um of the compression and bcndmg ..,trcsse\ if it i., avmmcd that the concrete behave-. elastically. 'I hu., 11 is poss1hle to determine the applied force 'o that the combmed stresses arc alway!> comprel>SI\·e. By appl) ing the compre11-.ive force eccentrically on the concrete cross-section. a funhe1 Mrc"s distribution. due to the bending effects of the couple thu' created, 1s added lC) lho\e 'hown in figure 11.2. rhi'l effect i:-, illuI·~·1
p
c
c
c
c
! 0![7 c
c
T Bending strain dlstnbution
Prestress
Seclion 8-B
Stress distribution - Section 8-8
T
Bending
Total
Figure 11.2 Elfecls of axial preslre~s
321
322
Reinforced concrete design
Figure 11.3 Effects of eccentric prestress
t t
+
t '
' e 1
t
•
l_s C
C
T
C
Of\· [7 C
Axial
C
T
Bending
Eccentricity
C
or
Total
prestress prestress Stress distribution- Section B B
Early
attempt~ to
achieve this effect were hampcn:d both by the limited steel availahk and by shrJllJ..agc and acep ol the concrete under sustained romprcs•.1on. cnuplco with rela\atmn of the '>tccl. This meant that the steel lo~t a large part nl Its innial prcten~ion ano as a n:~uh re~idual stre~~e~ were so small a~; to be t"cle". It is no\\ posSible. howc\er. to produce stronger conere::tes which have good cn:ep propcrttcs, and \er} htgh stn.:ngth '>teclc. whtch can he stressed up to a lugh pcn.:cntage of their 0.2 per cent prool -.tres'> arc al-;o avatlable. l·or example. hard-drawn ''ires m.l) c:arr) stre:-.seo.; up to ahout three tunes tho-.e fXl''-ihle in grade 500 remforcmg \tccl. This nut onl} result\ in saving-. Oll>tcel quantity. hut al'o the cfli!ct!) ot shnnkage and creep become relamcly ~mailer and rna) t}ptcally amount til the loss of only about 25 per cent of the tmtial applied force. Thu,, modern material-, mean that the pn:strcssing of concrete i' a pracllcal propo~Jtton. with the force' being provided b} 'teel p!retched tn the required ten'>ion and anchored 10 the ends of the moulds for the concrete. l'hc concrete is cast around the Lc!n,toned \lee!. and \\hen it ha' reached o;;uffictent 'trength. the anchors are released and the force in the c;teel ts tmn,ferred 10 the concrete hy hond. In addition to long-term lo,ses due to creep. -,hrinl-uge and relaxation. an 1mmediate drop 111 pre,tress force occurs due to elastic shortening of the concrete. TheM! features are illustrated in figure II A.
Prestressed concrete Beam with pretensioned tendons
Figure 11.4 Tendon stres~es pretensionmg
, Before transfer After transfer and losses
-I
bond length
length
Becau))e or the dependence on bond. Lhe tendon~ for this form of c:on.~truction generally conflisl of small diameter wires or small strands which have good bond characteristics. Anchorage near the ends of these wires is often enhanced by the provision of small indentations in the surface of the wire. The melhml is ideally suited for factory production where large number\ or identical unit~ can he economkal ly made under comrolled conditJon~. a development of thi'> being the 'long line' :-y~tem where several unit\ can be cast at once end to end - and the tendon~ merely cut between each unit after rclea\e of the anchorage~. An adv:mtage of factory production of prestrc~scd units i~ that ~pcciali)..ed curing technique~ ~uch a~ stenm curing cnn he employed to increase the rate of hardening of the concrete nnd to enable earlier 'transfer' or the stress to the concrete. Thi~ is particularly important where re-use of mould!. t\ reqtured. but 11 t~ essential that under no circum:,tancc., rnu.,l cnlcium chloride be U'>ed ac; an ncceleraLOr becnuse of its severe corrosi' c acuon on small diameter ~teel wtres. One maJor hmlfallon of this approach is that tendons mu~t be straight. whtch may cau~e difficulue~ when attempting to produce acceptable final !.lre:,:, level\ throughout the length of a member. It may therefore be necessary 10 reduce either the pre~tress or eccentnctty of force nenr the ends of a member. in which cu~c tcntlon-; muM either be 'dcbondcd' or 'denccted'. 1. Deboncling consist~> of applying a wrapping or coating to the Mcel to prevent bond developing with the surrounding concrete. Treating some of the wires in thi~> way over pun of their length allows the magnitude of effective prc:-tn:s~ force to be varied along the length of a member. 2. Dellecting tendons is a more complex operation and i~ U)..Ulllly restricted to lurge member!'., such as bridge benms. where the individual member:- may he required to form purl of a continuous structure in conjunction with in .1it11 concrete ~labs and ~ill beam!.. A typicnl arrangement for det1ecting tendons is shown in figure I 1.5, but it must be appreciated that substantial ancillary equipment is rc4uircd w provide the necessary reactton~. Deflection supports (cut off after tr.1nsfer)
Concrete
l
~
Tojacks~ ~
l
Prestressed tendons
~
- ~,
Y"" 7 - \;..,
I
.
22 , 7 2 2 ,~Tofacks
Figure 11.5 Tendon deflection
323
324
Reinforced concrete design
11.2.2
Post-tensioning
This method. wtuch is the most suitable for in .1i111 con~truction. involves the stressing again~! the hardened concrete of tendon~ or steel bar\ which are not bonded 10 the concrete. The tendons are passed through a flexible ~heathing. wh1ch is cast into the concrete in the correct position. They arc tcm.ioncd by Jacking against the concrete. and anchored mecharucally by mean\ of \tccl thrust plate:. or anchorage blocks at each end of the member. Alternatively. steel bar!> threaded at their ends may be tensioned against hearing plates by means of tightening nms. It i~ of cour e usually necessary to wait a con~iderable time between casting and stre~sing to permit the concrete to gain sufficient strength under in siw conditions. The use of tendons cons1sting of u number of ~trunds pas:-.ing through tlexible sheathing offer~ considerable advantage~ in that curvet.! tcnt.lon protiles may be obtarned. A post-tensioned strucruml member may be constructed from an n~semb l y of sep:~rme pre-cast units which [Ire constrained to act LOgcther by means of tensioned cnb l e~ which ~u·e often curved as illu~tntted in figure 11 .6. Alternntively, the member may he cast a:-. one uni1 in rJ1e normal way but a lighl cage of untensioned reinforcing steel i~ nl:ccssary to hold the ducts in their correcl position during concreting. Af1er strcs~ing. the remaining ~puce in the ducts may be left empty ( 'unbondcd· con~>truction). or more usually will be filled with grout under high prcs~urc ("bonded" con~truction). Although this grout a~sistc; 111 tmnsm1l1111g forces between the steel and eom:rcte under live loads, and improves the ultrmate ~trength of the member. the principal use rs to protect the highly stres-;ed ~trands from corro~ton. The quality of ~orl..manship of grouting i!> thu!:. criucal to avoid air pocl..cts which may permit corro~ion. The honding of the highly stressed o;tecl \\llh the 'urrount.ling concrete beam al'o greatly a''ist-. demolition. since the hcam may th~.:n \afcly be 'chopped-up· into 'mall length' wrthout rclea!)rng the energy ~torcd 111 th~.: -.tccl. Par;~bohc
Figure 11.6 Po~t·tensloned
tendons
s!'gmental
construction
Precast segments
11.3 Sine~:
Analysis of concrete section under working loads
the object of prestressing is to marntain favourable ~tress conditions 111 a concrete member under load. the ·worJ,.ing load' for the mcmhl:r must be considered 111 terms of hmh maximum and minimum values. rhu' at any . .ection. the stre!>se~ produced by the prcl>tress force mu:-.t he considered in conjunction with the \tres~es cau~ed by maximum and rmnimum values of applied moment. Unlike rernforccd concrete, the primary analysi\ of pre'>tressed concrete is b:t\ed on serv1ce condition'>. and on the U\)..Umption thai stres:-.es in the concrete are limited to values which will correspond to elu~tic behaviour. ln this l>Cction. the following assumption-. arc made in analy~'"·
Prestressed concrete lo= -
I
rr___
y,
b---, -
e-ve
h
- - - -T - f-
t a
f'!C:d tiber
JOg
1. Plane
and
the ) of
mm
section~ remain plane.
2. Stress strain rclation1>hip1. arc linear.
3. Bending occurs ahout a principal axb. 4. The prc1-trcssing force is the value remaining al'tcr all
5. Changes in tendon stress due to applied
lo~l.es have occurred .
load~ on the member have negligible effect
on the behaviour of the member.
l~tf
ded'
t+ve Bottom fibre Prestressing tendon
_,__ L - -- - --' _ _
Compressive stresses +Ve Tensile stresses - ve
nent t1ble be of
_.___
- - axis
Q
I =-;;;: ,.
' Centroidal --
e
)'b
lb
Figure 11.7 Sign convention and notation
Top fibre
y,
6. Section properlies are generally based on the
gro~s wncrete eros~ ~ection.
The ~tre~~ in the ~tecl is unimportant in the analy'>i'> of the concrete ~ection under wor~ing condition~. it being the force provided by the 'tccl that i., con.,tdered 111 the analy'>i~.
The sign convention!> and nmalion~ used for the analy'•' are indicated in ligurc 11.7.
mto
11 .3.1
Member subjected to axial prestress force
If secuon BB ol the member shown in figure II.X i\ subjected to moment' ranging between Mm 3 , and M111111 • the net \lrc~sc:. at the outer fibres of the beam urc given hy
{!
, ---+--P M~, A ;:,
fo
p
Mrnu'
A
~b
p
under M,,;n
{fi ./i.
:Is of
-- I A
Mmiu
at the top
(II. I)
ut the bottom
( 11.2)
at the top
(11.3)
at the bottom
( 11 .4)
:::c
p
Mmln
;\
Zh
where Zh nnd ::1 are the ela&tic section moduli and P b the final pre~tre!>s force. The critical condition for tension in the beam is g1ven by equation J1.2 which for no tens ton, that is fh = 0, becomes
or
p
. . r . d -Mm;,,i\ -= mmtmum prestress •orce requtre
325
326
Reinforced concrete design
Figure 11.8 Stresses in member with axial prestress force
PIA
Mlz,
f,
0![7
PIA Prestress
Mlzb
f.,
Bending
Total
Stress distribution- Section B-B
For thi~ value of pre!>trcs!> force. subl>titution in the other equation~ wi ll yield the ~trcs~e!> in the beam undt:r maximum load and also under minimum load. Similarly the stressc~ immediately after prestressing, before losses have occurred, may be calculated if the value or los~es i!. known. For example. the maximum stress in the top of the member is given by equation I l.J
\\here Mma\ A P--~~
therefore
,t;
p
p ~h
A
;\ tn!~ses Ill the top libre~. TI1is is paniculatl) likely at the iniunl transfer of the prestress force to the unloaded beam.
(e) Calculate stresses at ends In this situation M 0. Hence 4.3
A
5.6 - - 1.3 N/mm 2
and I' 1'1' jj, - A+ 7
'·
11.4
= 4.3 + 5.6 = 9.9N/mm2
Design for the serviceability limit state
The destgn of a pre:.tressed concrete member is based on maintaining the concrete '> tresses within specified limits at all stages of the life of the member. Hence the primary design ,., bac;ed on the serviceabilicy limit state. with the concrete ~tress limits based on the acceptnble degree of flexural cracking. the necessity to pre\ ent e\cc~\ive creep and the need to eno,ure that excessive compressiOn does not resuh in longitudinal and micro cracking.
329
330
Reinforced concrete design
Guidance regarding rhe allowable concrete compressive stress in bending is given in EC2 as limited to: (i) 0.6/c~ under the action of characteristic loads
and (ii) 0.45/ck under the action of the
qua.~i-pen11anem load~ .
The qzwJi-permwzemloads are the permanent and preme~~ing load. Gl + Pm.t· plus a proportion of the characteristic variable impo ed load. This proportion is taken as 0.3 for dwellings. offices and stores. 0.6 for parking areas and 0.0 for snow and wind loading. Jf the tensile stress in the concrete is limited to the values offc1m given in table 6.11 then all stresses can be calculated on the a~sumption that the ~ection is uncracked nnd the gross concrete section is re~isting bending. If th11> is not the case then ca l cu l ation~ may have to be based on a cracked section. Limited cracking is permissible depending on whether the beam is pre- or pol>t· tensioned and the appmpriate exposure class. Generally for prestressed members with bonded tendons crack witlths shoultl be limited to 0.2 mm under the action of the frequent loading combination taken as the permanent characteristic and prestre~sing load. Gk Pm I· plus a rroportion of the characteristic variable imrosed load as given by equation 2.3 and table 2.4. ln some, more aggre!.sive exposure contlitions, the pos~>ibility of decompre~sion under the quasi pemument load conditions may need to be consitlered. At initial tran1.fer of prestrest. to the concrete. the pre,tress force will be considembly higher than the 'long-term· value a' a result of sub~equent losses which are due to a number of causes including elastic shortening, creep and shrinkage of the concrete member. Esumation of losses is dc).cribed in section II .4. 7. Since these lossc' commence unmedtately. the condition at tran.,fcr represent\ a trunsitory stage in the life of a member and further confer i). u critical stage which 'hould be considered carefully. The comprcs\ive stress at transfer \hould be limited to 0.6/,~ wherefc• is based on the Mrcngth on the concrete at transfer. The tcn1.ile stress should be limited to I N/mm 2 for sections designed not to be in tension in service. Where limited Aexural 'tress under service loads h. permillcd, some limited tensile stress is permitted at transfer. The choice of whether to permit cracking 1.0 take place or not wil l depend on a number of factors which include condllions of exposure nnd the nature of loading. Ira member consists of precast segments with monnr joint~. or if it is essential that cracking ~>hould not occur. then it will be designed to he in comprcs~ion under ulllond condition~. However a more efficient use of materials can be made if the tensile strength of the concrete. f~tm· given in table 6.11 is utilised. Provided thet--c ~trcs~c~ arc not exceeded then the section can be designed, ba!-.cd on the gm1>s uncrad.ctl section. Unless the section is designed to be fully in comprcs~ion untlcr the characteristic loads. a minimum amount of bontlcd reinforcement ~hould be provided to conrrol cracking. Thio; j., calculated in an identical manner to the minimum reqULrement for reinforced concrete (sec ~ection 6.1.5) with the allowance that a percentage of the prestressing tendons can be counted tO\\ ards this m111imum area. The de\ign of prestressing requirements ts based on the manipulation of the four basic exprc!.1>ions given in section 11.3.2 descnbing the stress distribution acros~ the
Prestressed concrete m
3 31
concrete section. These are used in conjunction with permissible ~tresses appropriate to the type of memher and covering the following conditions:
1. Initial tran~fer of prestress force with the associated loading (often JUSt the beam's sel f-weight); 2. At ~ervicc. after prestress losses. with minimum and maximum characteri~tic loading; 3. At !>ervice with the quasi-permanent loading.
The loadings mu~t encompass the full range that the member will encounter dunng its life, and the minimum values will thus be governed by the construction techniques u~ed. The partial factors of safety applied to these loadl> will he those for serviceability limit ~o.tat e, that is 1.0 for both permanent and variable load!-.. The qunsi-permancnt loading situation is considercCf\ 1cc that wil l govern. as shown in figure I 1.1 0. From tigure 11.1 0 the govcmi ng equauon~ for a ~i nglc-span beam arc:
At
tran~fer
Po
Pue
A
:.
Po
Po J.'mm
M mon
.......
( 11.9)•
.f~ ~f/na~
( II 10)
~ervice
KPn
KPoe
A
~(
KPo
KPoe
A t
Zh
Mmnx -;·
J,. - mon
(11 .12)*
-r - - ~l
I
Zh
where }~ 11 ~, ;;~in· j;11 u~ nnd };nln arc the appropriate permissible ~ trcssc~ ut tram, fer anti conditions. Po is the prc:-.tressing force at transfer and K is a loss factor that account~ for the prestress losses - for example, K = 0.8 for 20 per cent loss. ~erviccnbi lity
w ,.,·ln
f f f +
'
+ +
Transfer w"'...
f f f f f f
Figure 11 .10
r•, > r;...,.
I'• -~ I'm., f,c;f.,.,
[7 Service
Prestressed beam al lransfer and servoce
332
Reinforced concrete design
11.4.1
Determination of minimum section properties
The two pairs of expres~ions can be combined as follows: 11.9 and 11. I I ( 11.13)
II.!Oand 11.12 (Mmu~ - KMnun ) :::; (Kfm:tx -
Hence. tf (Mmal
.fmm):Oh
(11.1 4)
KMmm ) is written as M,. the moment variation
M,
•,>---.. - ifrru, KJ~,n)
( 11.1 5)
and
M,
( ll. J6)
Zb> - - - . - ( KJ~., - };mn)
In equations 11.15 and 11.16, for~~ and ::11 1t can be assumed with ~uftic.:ient accuracy, for preliminary siting that M10. , will depenu on both the imposed and dead (self-weight) load ami Mnun will depend on the dead (self-weight) load only. ~o that in effect the c.:n lculations for Mv bec.:ome independent of the self-weight of the beam. These minimum \ulues of section moduli muM be ~atistied by the cho,cn section in order that a prestress force and eccentricity exist which will permit the stress limits to be met; but to ensure that practical con~iderntion~ arc met the chosen section must have a margin above the minimum values calculated above. The equations for minimum moduli depend on the difference between maximum and minimum values of moment. The ma\imum moment on the section has not directly been included in these figure ... thu~> it i!. possible that the rc:-.ulttng prestresl> force may not he economic or practicable. However. it is found in the majority of cases t·hat if a section is cho,en which satislles thel>c minimum requirements. coupled wtth any other speci fied requirement!. regarding rhc shape or the section, then a snw;fuctory dc~>ign is usually possible. The ratio of acceptable '>pan to depth for a prcstres~cd beam cannot he categorised on the haiotis of dcnccttons as ea~tly as for reinforced concrete. In the ab~ence of any other criteria, the following formulae may be used as a guide und will generally produce reusonably consen ative de,ignc; for post-tensioned memhcr~. span< 16m
I1=
span > 36m
II
spnn
.,-+ _s 0.lrn
span m 20
:::.--
In the case of :-.hort-span members it may be possible to use very much greater spandepth ratios quite satisfactorily. although the resulting prestress forces may become ver) high. OU1er factors which must he consic.lcred at this stage include the slencleme~>s ratio or beams. where the same criteria apply as for reinforced concrete. and the possibility of web and flange splitting 111 tlanged members.
Prestressed concrete
(EXAMPLE 11 . 2
Selection of cross-section Select a rectangular
Therefore h ' /(2.93 x 106 x 6/200) - 297 mm
The minimum depth ol' beam b therefore 2'->7 mm and 10 allow a mnrgw 111 subsequent det {;:.hfrniu + Mrnax) o_ K(ideratton. Such limitil ~ill include consideration of lhc required minimum cover to the prcsLressing tendons which will depend on the e\po~urc and ~tructural cia~' assumed for the design. The ciTect of this limitation will he moM severe when considcnng the maximum moment' acti11g on the :-.cction. thnt is, the inequalities of equations II . II and 11.1 2. If the !uniting value for maximum eccentricity e""''' depend~ on cover requirements. equation I 1.11 becomes
Mma' $ };nax:.l
KPoC · ema~)
( 11.21 )
nnd equation 11.12 becomes
Mm." $ KPo e~ t
C two expressions, when
KPo (~ + ema,)
fmm:.t>
=fma,:.l KPoC elll3,)
Prestressed concrete 22
21
Figure 11 .1 1 Maximum moment and prestress force relationship
h
Max. moment lnequahttes satisfied In thts zone
'--'+---___;___;_ _ _ __ Po Y'
thus
p _};nax";t +};n.n;.h
()- K (- A of eccentricity further limit& on the prestress force cnn be determined in an iucntknl manner although the ca l culation~ would he tedimt' and repetitive. In uddition. it is po~~ihle to U\sumc values of eccentricity for which there is no '\olution for the prestreso; force a~ the upper und lower limit~ wuld overlap. A much more u,cful approach to design can be developed if the equations arc treated graphicully a~ follow~. Equal ron-. 11.9 to II. I 2 can be rearranged into the following
fonn: K( I 1 A -
t•/':.1)
{ equauon 11.1 I }
( 11.24)
{ equauon I 1.9}
( 11.25)
K ( 1/ A e/':..h) (/;,"' I Mmu\/":.t>)
{equation 11.12}
( 11.26)
I ( l /A +e/:.h) ">-Pn • ({~,.., I M1111n/ /.h)
{equntion 11.10}
( I I .27)
(fmJ\ - Mm." /~.)
Pn Pu
.
t•f:.,)
({~1ln- M111u1 /~l)
!.tble for line I. derived from equation 11.2~. to have either a po~itive or a negative '>lope depending on whether fm.t, is greater or less than Mm., /:.1 •
338
Reinforced concrete design
Figure 11.12 Magnel diagram construction
Q)
Po
--~~-L----LL--~• e
r
z.IA
zJA
· I· · I (b)
The Magnel diagram is a powerful design tool as it cover force which will permit all stress conditions to be :-.ati~fied at the critical section, it is necessary to determine the eccentricity at which this force must be provided. not only at the critical section but also throughout the length of the member. At any section along the member. e is the only unknown term 111 the four equations 11.9 to 11.12 and these will yield rwo upper and two 10\\er limits '' hich mu~t all be simultaneously satisfied. This reqturemenr must be met at all sections throughout the member and \\ill reflect both \'ariations of moment. prc:.trcs~ force and section properties along the member. The design expressions can be rewritten as: At transfer ( 11.28) (11.29) At service
fmJ,:t] -r- NJ_ma\
e>
[x
l' ~
[-~
KPu
( 11.30)
!\Po
fmm::t>] + Mm3.\ KPo
(11.3 1)
KPn
Equauons 11.28-11.31 can be C\'aluated at any 'cct1on to dctcrmmc the range of ccccntncltles "11h111 which the resultant force Po mu't lie. The moment-. Mn 1a~ and Mmm arc those relating to the secuon be111g con\ldcrcd. For a memher of constant croso,-~ection, if minor change., in prc~tress force along the length are neglected. the terms in hrackct~ in the uhovc expre:.s10ns are constant Therefore the LOne within which the centroid mu:-.t lie is governed by the :.hape of the bending moment envelopes. a~> ~ohown in ligure 11.14. In the case of uniform loading the bending moment envelopes arc p [- ~ +/,ninZh] I
1
A
-
II 29)
MnM'
!\Po
KPn
As there are no moment\ due to c.xtcrnalloadmg at the end of a 'unpl} 'upported hcam equation 11 .29 bccnme~
< [- 4.0H X 10~ -~)( 4.08 (350
(!.
200)
X
-58.2!i
7()()
X
10"] t () 10 1
93.25
c. 35 111111
Similarly equation 11.31 become:.
4.08
I'
> [ • (350
106
v
]
200) I O I· ()
>-58 29mm At the ends of the beam whcre the moments are lero. and for : 1 ;:h. the im:quality cxpre~sions can apply with the tendon eccentricities above or below the neutrul axis (e po,itive or negative). So that e must lie within the range ±35 mm. (b) Mid-span Rq ur~lion
(!
11 .2H hecomcs:
4.08 X 1011 (- I )4.0X )' 10~] [ ~ (350 200) -]()() I~
21.lJ
y
+ 700 X
1 0~
JO'
< 64. 1 i 31
< 95. 1mm Equation 11.29 mighl be more critical than cquauon ll.2X checked. From equation 11 .29: 4.0!-l x 106
C'
16 x 4.08 x 106 ] 2J.9 ~ [ - (35() X 2()()) +-700 )I 103 f 700 ~ ~
X
r~nd
-.hould be ubo
1()6
10 1
58.3 + 93.3 + 31
66 mm
Hence equation 11 .29 is critical and the eccentricity mu-;t be less than 66 mm.
J
342
Reinforced concrete design
Equation 11.31 gives e
> - 4.0H X JW -0] - [ (350 X 200)
0.8
X
700
X
1() 1
-58.3 + 106.1 ~ 47.8mm ~
Hence at mid--.pan Lbe resultant of Lbe tendon force mu!>t lie at an eccentricity in the range of 47.H to 66 rnm. Provided that the tendons can be arranged Ml that their resultant force lies within the calculated l imit~ then the dc11ign will be acceptable. If a Magnet diagram for the ~tress conuition at mid-span had been drawn. a:. in example 11.5. then the eccentncity range could have t>een determined directly from the dtagram without further calculation. For tendontratn
1l1c \aluc of ~pecific creep used m thi~ calculation \\'Ill be mflucnced by the facto~ di\CU'>~cd m I'> based on cmptrical figures tor hnnkugc/unit length of concrete (£C>) for particular curing condjuono; and transfer malllrity a... di~CU\ ed 1n chapter 6. Typical values range from 230 x 10 6 for UK outdoor expo~ure (SOtq relative humidil}) to
346
Reinforced concrete design
(
EXAMPLE 11 . 7 Estimation of prestress losses at mid-span
A po~t-tensioned beam shown in figure 11 .15 IS stressed by two tendons\\ ith a parabolic proille and having a total cross-c':
of friction 11 = 0.19 \\Ohhle factor J.. 0.0 1/mc tr~ Cm:lfi ~P(x)
= Po ( 1 - e_,,(OH•)) = Po( 1 _ e - o111 (UI~Sl+U.tll x 15)) 0.0-MPo
= 460k..~ = 4.-t per Cl!nt
(2) Elastic shortening for post-tensioned construction
P'
ra~c
und
p
= ----.,.----- -
l - 0.5nc ~ (I e~ y)
lht: 3\erage eccentriClly for the parabolic tendon a:o. 5/8e, = E-/E,m = 205/32 = 6.41.
Oe
5/8 x 640
= 400 mm
I~
Prestressed concrete
Po
P'
-
1 + 0.5 0.968P0
3 X 10 ( X 1.05 X 1Q6
7.5
X
6.41
,
I
1.05
X
+ ..f()()· 0.36 X
IOIJ )JO I2
= 10 160kN
Los ~J = 10500 lO 160 340kN = 3.2 per cent Total l)hort-term los~es = 460 -'- 340 = 800 k.~ P0
p'
-
short-term losses
= 10500 (3) Creep
800
9700kN
m F.,Ap (I+e2 ~)I P' ' ( 1.05£.m)A
Los!. t:.P
6
205
1
10 7.5 X 10] ( ' 1.05 X IO'') - I , l ( 1.05 x 32) x I 03 x 1.05 X 1()1> I + ..JOO OJ6 Xi(iTi 9700 X
992 kN ( = 9.-1 per cent of Pn) (4) Shrinkage Los~
.:::,p = •c,c',Ar - JJ()
J0
1 ' X
2()5
X
7.5 X 10'
507 kN ( = ..J .8 per cent of Pu) (5) Relaxation
Long-tc.:nn rda\atJOn IO\S !actor = 2.5 for cia be checked since lrihution in terms of an equivalent uniformly distributed load. Fm the beam in figure 11.16 the moment due to prestre-., loading 31 any section is A1, -Pe, but 'Iince e~ t puroholic, the prestre'l-. loading may be IJkened to a umfonnly dt of curvature due to prestre~s along the memher length. There arc three pnncipal 'otages in the life of a prestrcs\Cd member m whtch deflections may he critical and may need to be as\e:o..,ed.
1. At tranllfcr - a chcd. of actual dcnection a1 tran~fcr for compari,on "ith estimated values i-. a useful guide that a pre. tres. ed beam hal> been correctly wnstructcd. 2. Under dead load, before application of lini!>he!-. - dcOecuons muo;t he evaluated to permit :-.ubscqucnt movement and possible damage to he csumated. 3. Long-term under full quasi-permanent action:-. deOect1ons are reqlllrcd. both to determine the sub~equent movement and al!lo to a~sess the arpearuncc of the final structure. Short-term dellections will be based on materials rroperrie!> u!.socimcd with characteristic strengths hm = I ) and wi th actual loading ("}1 I ). Long-term ns~essment however must not only take intn account loss in prestrcs~ force, but al!.o the effects of creep both on the applied louding and the prestress loading component~ of the del~ection. Creep is allowed for hy using an cffectivl! mmJulus of elusticity for the concrete, us discussed in section 6.3.2. Thu~ if f:..'.:r ~ 1 is the in~tantancou~ value. the effective vnlue after creep is given by
where the value of ¢{-x:.r0 ) , the creep coefficient can be obtained from table 6. 12 It can be ~hown in -.omc instances that when net upward deflections occur, the~e often increa.,c bccau~c of creep. thus the most criucal downward deftectton mny well be before creep IO!>~es occur. while the most critical upward dcftecuon may he long-term. This further complicates a procedure which alread) ha:-. many uncertainties as di)o.Cll~\ed in chapter 6: thus deflection!. must always he regarded as e~;ltmute-. only.
350
(
Reinforced concrete design
EXAMPLE 11 . 8
Calculation of deflection
Estimate transfer and long-term deflections for a 200 x 350 mm beam of I0 m span. The pre,tressing tendon has a parabolic profile with mid--.pan eccentricity 75 mm and the end eccentricity = 0 at both ends. The initial prestre:,-. force at traru.fer. Po. i~ 560 kl and there arc 20 per cent losses. The imposed load consbts of 2.0 1-N/m finishes and L.O kN/m 'ariable load. Ecm = 35 kl\/mm~ and the creep factor ~( oo.t0 ) = 2.0. Self-\\ eight= 0.2 x 0.35 x 25 = 1.75 kN/m bh3
1 =-= 12
(a)
2()() X 35()3 12
715x 106 mm4
At transfer 5 ~~'minL4 5 (Poer)L2 DeflccLtOn Yn = 384 Ecml -48 E,.ml .
5 1.75 X JO~ X 10 12 - 384 35 X l 03 X 715 X I 06 9.1 17.5 = -8 mm
5 560 48
X
35
103
X
75 X 102 X 106 7 15 X l 06
X I 01 X
(upward~)
(h) At application of ftni~hes A~sume
that only a small proportion of pn:strcss lo..sc-. have occurred:
Weight of
fini~hes
- 2 0 kN/m
therefore 5 ~ 2.0 : 10~ 10 1 ~ · -.a 384 x 35x 101 x 715 x I(J~ res~
force after losses
Assuming 30 per cent of the variable load contribute), to the acrion:
qua~i-pcrmancnl
In the long term due to the quasi-permanent action plus
Quasi-permanent action
self-weight I finishes I OJ x vuriablc load
= 1.75 + 2.0 + 0.3 x 1.0 Prestres~
forces after losses
4.05 kN/m
= 0.8Po
0.8 x 560 44l'l kN 35 , Ecetf=( ( )) =( 1- ., ) = 11.7kN/mm· l + cJ> oo. ru i -· 0 5 4.05 X 104 X 10 11 5 448 X 10 1 Y 75 X 102 X (()6 ."c = 38411.7 X 103 X 715 X !Qfl 48 11.3 X t01 X 715 ~ ,
Ecm
63.0- 43.3 = 20mm (downwards) < span/ 250 (d)
40 mm
1l1eretore sat.J\factory. \t1ovement after application of finishes )'d
y., -
)'b
= 20- 2 = 18 mm $ spnn/ 500 = 20 mm ('>allsfactory).
Prestressed concrete 11.4.9
End blocks
In pre-tensioned members. the prestress force is transferred to the concrete by bond over a definite length at each end of the member. The transfer of !>tre~s to rhe concrete is thus gradual. In post-tensioned members however, the force is concentrated over a small area at the end faces of the member. 11nd this leads to high-tensile forces at light angles to the direction of the compression force. This effect wi ll extend some distance from the end of the member until the compression has distributed itself aero:.' the full concrete cross~ection. Thi~ region is known as the 'end block' and muo;t be heavily reinforced by steel to resist the bursting tension forces. End block reinforcement will generally wnsist oJ closed links which surround the anchorages. and the quantities provided arc u~uaily obtained from empirical methods. Typical 'How line~· of compressive stre~~ arc sho\\n in figure 11.18. from \\hich it can be seen that ~hmever type of anchorage il> u~ed. the required di:.tribuuon can be expected to have been attained at a distnnce from the loaded face equal to the lateral dimension of the member. Thil> is relatively independent of the anchorage type. In dc:.igning the end block it i~> ncccs~ary to check that the bearing strcsl-. behind the anchorage plate due to the prestres:.ing force doc~ not e'\cecd the limiting ,\Ires\. /ll.du· given by /Rdu
= 0.67f._k(Ad/Aco) 0'5 "' 2.Qf:k
where b the loaded urea of the anchorage plate A.t is the maximum aren. having the ~ame ~hape a~ of 0.567fed to I 120 .:--l/mm~ and 30 per cent ((h~C~ are anticipated. estimate the ultimate moment of n.:l>istance of thl! ~cction it clas~ C35/45 concrete ~~ used. The ~tre~s-!>Lrain curve lor prcl>Lre~sing wirc i~ ~hown in tlgure 11.23. Area of 5mm wire
rr x 52 /4
Stress in 'tccl after los.,cs
)p :>-.
19.6 mm 2 1120 x 0.7
=0.9 x
thcrdore . .111 'tcel alter . S • rra111
lossc~
= ·£(,., -
., -705
- 05 X I 01
wh1ch is les' than _ , the yield l>train.
= ()·{)urable o;ection of the web. For non-grouted duct~. grouted plastic duct~ and unbonded tendons the web thickness shou ld be reduced hy 1.2 times the !-Um of the duel diameters. u· the de:.ign ~hear force exceeds VRd m3~ then it will be necessary to increase the !tiLe of the ~cct ion. (2) The vertical shear reinforcement
As in reinforced concrete. ~hear reinforcement musl he provided to resist the shear force if it can not be sustained by the concrete '>ection including tbe enhanced shear resistance
364
Reinforced concrete design 4. The shear links required can be calculated from equation 11.47
A"'
VEd
s
0.78df~L cotB
where A~" is the cross-sectional area of the legs of the links (2 x 1r9~ / 4 for 1>ingle stirrups). For a predominate!) unifonnly distributed load the maximum shear Vw can be calculated at a distance d from the face of the ~upport and the shear reinforcement should continue to the face of the support. The shear resistance for the links actually specified is Vmin
Asw
-
s
x 0.78d/yL cot f)
and this value will be used together with the shear force envelope to determine the cunailment position of each set of designed links. 5. Calculate lhe minimum links required by EC2 Ji·om A,w
ndn _
0.08fc~ ~bw
.r-~ 6. Calculate the additional longitudinal tenile force
s
ct~uscd
hy the shear
:::.f td = 0 5 \ '~"A~ cot 0 The al.1o'e procedure ~hould be repeated at different allustrated in the following example
(
~cctions
along the beam. as
EXAMPLE 11 . 12 Design of shear reinforcement
l hc beam cross-section l>hO\\-n in figure 11 .27 1s constant over a 10m :-.imply supported l)pan with a parabolic tendon prolile and an eccentricity varying between 300 mm at the ends and 750 mm at mid-span, measured hclt>w the ncutrulllxis in both cases. The beam 1.upports an ultimate uniformly distributed loud of 40 kN/rn and j~~ 35 N/mm 2•
Figure 11.27 Shear reinforcement example
1000
0
0\ "'
II 0
"'""+ 0
~ I
8
___._,"'" location of tendons at the supports
Prestressed concrete Given data: Prc...,tres~
force after losses
2590 kN
= 145 106 x A = 500 X I03 mm 2 Ar = 3450mm2 /yL = 500 1':/mm~ for the shear links fctk = 2.2 N/mm~ 106 mm4
I
The calculation~ will he presented for a section at the ~upport nnd then repeated and tnbulntccl at 3m intervals along the span. (1) Calculate shear force at the section
Although th e m:~ximum shear force can be taken at the fac;c or the ~upport , in thi~ example we wi ll, for illustrative purpo~e:., luke the 'ection nt the middle of the support itself. llcncc: 40 x 30/2
Vrd
600 kN
(2) Check if shear reinforcement is reqwred
rrom equation I 1.40 the concrete ~hear \trength i., gi"cn hy: I'Rd, = [o. I2k( lOOpJ.~)
1
'+ O.ISa,"• b,.d
\\here:
cl :: 1.5
(1 -1
1.
e
0.85
~2~))
= "'fq,KPn/11
= 0 95 m at the ~uppurt
~ f:o) = 1.46
(1
3450 150 X 950
rr,p
0.85 + 0.3
1.5
(::; 2.0)
( > 0.02) ... PI - O.D2
0.0242
- 0.9 x 2590 x 10 1/(500 x 101 )
4.66 N/mm 2
{5
0.133.f.·L = 0.133
X
35
4.66 OK}
lienee:
VRLI c
[o. I2k( IOOpt/~k) l/l + 0. I5crrp] b,..d [0.12
X
1.46(100
X
0.02
X
35)
111
-j
0.15
X
4.66] 150
X
950
X
JO-J
202kN Noll!: a check on equation 11.41 will show that the minimum value of VKd, as given by
equation I 1.41 1., not cruical in this case. As thb is u \imply supponed beam equation 11.42 should also he used to check the l>hear capac it) of the concrete ~ection. From equation 11.42: b~J
~
VMt~c= AJ vU~id+nlacJc•d
)
36:
366
Reinforced concrete design
where:
= axial stress in section due to prc:-.tress = 4.66 Nlmm2• as before
rr,p
= the design
tensile strength of the concrete = 2.2/ 1.5 o 1 = I for post-tensioned tendonl>.
.f.ld
= 1.47 N/mm2
Hence by reference to the dimensions shown in figure 11.25: bwf . j ,
VRd .c
= 1\y V (fc~d +
(I
I rTq/tld)
150 X 145 106 X 106 f. , = ~;{-IO_OO_x_l-75-x-56-2-.5-)_t_( 150 x 475 x 237.5 )) V ( l.47- + I x 4·66 x 1.4?
-3 X
lO .
S66kN This is considerably greater than the figure of 202 k~ calculated from equation II .40. We will take the lower value of202 kN as representing the shear capacity of the concrete section The effective resistance of the section is the sum of the shear resi~tnncc of the concrete. VRd , . plus that of the ,·crlical 'hear rclobtance of the inclined tendons. Shear strength Including the shear resistance of the inclined tendons
The vertical component of the prestress force is P sin 11 where r3 tendon slope. The tendon profile" y - c,.:! with the origin of the cable prohlc taken at mid-lipan; hence at .1 15000. ~ 750 - 3CXl 450 and
c x 15 ooo2
,150
c
2.0 . . . 10 (, 2.0 x 10· 11 rl and tendon :.lope
Therefore the tendon profile '' .'
dy/dt
2C\ .
At end
dyj dx
= 2 X 2,() X
10·
X
15 000
= ().()60 -
tan i
lienee, ,J
3.43
und 'in ;3::::: tan ii
= 0.06
Therefore vertind component, V1• of prc~trcs~ force 111 259!1:-in i = 2590 x 0.06 1551..1'\
at
the suppo11s
i~:
and the total shear capacit) i-,: VRd. c + V1
202 I / p X 155 202 + 0.9 142 kN at the support!-..
X
155
At the end ot the beam the design ~hear rorce is (40 >< 10/ 2) = 6001.:-1 and hence the shear capacity uf the concrete &ection i~ inudequatc and ~hear reinforcement mu~t be provided.
(3) Check the crushing strength
VRd max
of the concrete diagonal strut
A check must be made to cnwrc that the !-hear force doc:-. not cause excessive cnmpression to develop in the diagonal strut:. of the assumed Lrus~. From equation 11 .4-l (COle - 2.5):
VRu mn.t1 2~
= U
X
!.:k
0.133- 1.200
m"' 22) = oC\\O. l24b.. d(l- /ck / 250lfck - 1.200
X
0.124 X 150
950(1
X
35) 250
X
35
X
I0- ·l
= 638 kN
As the ~hear force at the end of the beam is 600 kN then the upper li mir to the shear force i~ not exceeded. (4) Calculate the area and spacing of links
Where the shear force exceeds the capacity of the concrete section, allowing for the enhancement from the inclined tendon force. shear reinforcement must he provide to resist the net shear force taking into account the beneficial effect of the inclined tendons. From equation I I A~ thi~> is given by:
A,w S
~ = (600 - 0.9 X 155) X 103 1.95dfyk 1.95 X 950 X 500
= 0.497
(5) Calculate the minimum link requirement
2
0.08fc~ b" = 0.08 X 35 1 ~ J..... 500
A,"•'"'" -
}
150 = O.l 4
X
Therefore provide I 0 mm links at 300 mm centres rcsi,tance of the linb actually ~pecified is: A,. Vnun -
I
(I\ ,..
h
0.523) \Uch that the 'hear
X 0.18dfy~ COL 0
0.523
X
0.7!!
X
950
X
500
X
2.5
X
10
J
4X4 kN
(6) Calculate the additional longitudinal force
The additional longitudinal tensile force is:
AF1d
0.5Vt:ucot()
0.5 x (600
0.9 x 155) x 2.5
575 kN
Hence:
575 X 101 0.87 X 500
- - - - = 1322 mm
1
Thi!. additional longitudinal steel can be provided for by four untensioned II25 burs (1960 mm 2) located at the bottom of the beam's cross-seclion and fully anchored pa~t the point required using hooks and bends as nece~sary. Umen ioned longitudinal reinforcement mu~t be provided at every cross-~ection to resist the longitudinal ten~ile force due to ~hear and the above calculation must be repeated at each :.ection to determine the longitudinal ~teel requirement. All of the above calculations can be repeated ar other cross-section)> and are tabulated in table 11 .2 from wh1ch it can be seen that, from mid-span to n liection approximately 9 111 from mid-1.pan, nominal shear reinforcement is rcquin.:d and in the outer 6 m of the !-.pan fully designed shear reinforcement is required. This can he provided a!.
367
368
Reinforced concrete design
Table 11 .2
Shear calculations at 3m intervals
Prestress
Mid-span
0 3 6
9
12 End-span 15
v,
(1)
(3)
(2)
(4) A,wfs
(kN) 1
(kN)
(kN)
(kN)
""'~ (kNi
1400 1382 1328
281 278 270
0 28 56
0 120 240
281 306 325
941 928 892
Minimum reinforcement only
1238 1112 950
255 234 202
84 112 140
360 480 600
339 346 342
832 747 638
Reinforcement 0.229 carries all the 0.339 shear force 0.497
D (m) (mm) X
'Yp
VRd.c
VEd
VRd c
~Pvt
VRd
(S) Asw/S m1111
(6) ~Ftd
(kN)
- -0 0.14 0.14 0.14
115 230 345 460 575
1 Equation 11 .40. 2 Equation 11.44.
Figure 11.28 Shear resistance diagram
z
.. 2 .."'
Ultimate sheM force
600
~
...c
~ ~
v,,.
Concrete plus tendon ~hear resi5Lance
t:
~
400
..c
:::?
V'l
200
0
Q.
:2
Q.
::>
V'l
' 6
9
12
15
Distance alonq IPMI (m)
(figure 11.28) 10 mm link::. at 300 mm centres in the outer 3 metres (A,..., / .1 0.523) changing to I 0 mm link' Jt 450 mm centres (A "/1 - 0 ~49) between 3 and 6 m from the end of the heam and then 8 mm at -150 centres (A ,,J 1 0 223) throughout the rc~t of the ::.pan.
CHAPTER
12
Composite construction CHAPTER INTRODUCTION
)
Many buildings are constructed with a steel framework composed of steel beams and steel columns but mostly with a concrete floor slab. A much sliHcr and stronger structure can be ach1eved by ensuring that the steel beams and concrete slabs act together as composite and so, effectively, monolithic units. This composite behaviour is obtained by providing shear connections at the interface between the steel beilm and the concrete slab as shown in figure 12.1. These shear connect1ons reSISt the horizontal shear at the interface and prevent slippage between the beam and the slab. The shear connectors are usually in the form of steel studs welded to the top flange of the bec~m and embedded in the concrete slab. The steel beam will usually be a universal 1-be.:~m. Other .:~ lternl of symbols for the three codes extends to 21 pages. Tt is intended in this chapter to try and simplify many of the complications and enable the reader to gain a grasp of the basic principles of the design or composite heams.
12.1.1
Effective width of the concrete flang e (EC4, cl 5.4.1.2)
An early step in the design of the composite benm section is to determine the effective breatlth bcrr or the concrete flange. For builtling structures at mid-span or an internal support
bru
= L:b.,
where be~ i!> the effective width of the concrete flange on each side of the steel web and i'> taken as 4 /8. but not greater than half the dist:lnce to the centre of the adjacent beam. The length 1-t is the approximate di!-.tance ~tween points of cero bending moment "hich can be taken as L/ 2 for the mid-.,pan of a continuou:. hcam, or L for a one-span simply supported beam. The length L is the \pun of the hcam being conl>idered. For example. for a continuous hcam with a span oft 16m and the adjacent beams being at 5 m centre to centre the effective hrcatlth. hd1 , of the concrete flange is
bell
2 >. Lc/8 = 2 X 0.5 x 16/ 8
If the beam be 4.0 m.
wa.~
a nne-span
~imply
2.0 m supported beam the effective breadth. bttt. would
12.1.2 The principal stages in the design These
stage~
arc listed with brief description~
u~ fo ll ow~>:
(1) Preliminary sizing
The depth of a universal steel beam mny be taken as approximately the ~>pan/20 for a simply l'lupported span and the span/24 for a continuous heum. The yield strength. j~. and the section classification of the steel heam ~houltl he determined. (2) During construction (for unpropped construction only)
The loading i:-. taken ac; the self-we•ght of the steel beam with any 'buttering or ~lecl deckmg. the \\eight of the wet concrete and an unposed con~truction load of at lea1.t 0.75 kN/m~. The following design checkc; are reqUired: (a) At the ultimate limit state
Check the strength of the steel !>ection in bending untl 'hear.
Composite construction
(b) At the serviceability limit state Check the deflection of the ~tccl beam.
(3) Bending and shear of the composite section at the ultimate limit state Check the ultimate moment of rcsio;tanee of the composite section and compare 1t v. ith the ultimate design moment. Check the shear strength of the ~tccl beam.
(4) Design of the shear connectors and the transverse steel at the ultimate limit state The shear connecters are required to re~>iM the horizontal shear at the interface of the ~>teel and the concrete so !hut the steel beam and the concrete llange act as u cnmposih.: unit. TI1e shear connectors can be either a full shear connection or a partial f>hcur connection depending on the design and dewiling requirements. Tmnwerse reinforcement is required to re~>i:-.t the longitudinal 'hear in the concrete flange and to prevent cracking of the concrete in the reg1on of the ~hear connectors.
(5) Bending and deflection at the serviceability limit state for the composite beam The deflection of the heam i~> chcc"-ed to ensure it is not exces!.IVC and so cau:-.ing crac"-ing of the architectural hni\he~.
12.2
Design of the steel beam for conditions during construction (for unpropped beams only)
The steel beam muM be des1gned to support a dead load of its e~timated self-\\ eight. the weight of wet concrete ond the weight of the proflled ~teel dec"-ing or the formwork. plus a construction li ve loud or til leu:-.t 0.75 k.N/m 2 covering the lloor urt:a. A preliminary depth for the siling of the Meel beam can be raken m. the '-pan/ 20 for a one-::.pun simply ~.upported heam.
(a) At the ultimate limit state (i) Bending The plastic section modulu~ Wpt y· for the steel beam may he calculated from
IVpt.y
= MrcJ r Jy
( 12 I)
where MP.t i~ the ultimate de~rgn moment
J;
i~ the dc,ign strength of the steel as obtained from EC3. tahlc
3.1
1l1is as.,ume!. that the compre1.sion flange of the steel beam is adequatel> n!strallled against bud.. ling by the steel decktng for the ~lab and the !.tccl sect1on used can be classified a~ a plastic or compact section a~ defined in EC3. sections 5.5 and 5.6.
· 373
374
Reinforced concrete design
(ii) Shear The hear iJ. considered to he carried by the steel beam alone at the con~Lruction stage and also for the final composite beam. The ultimate shear strength of a rolled 1-beam is based on the following shear area. A,. of the section (122)
where Aa is the cros!.-scctional area of the ~tecl hcam and h.,. is the overall depth of the web. IJ can he taken as 1.0. The other dimensions of the cross-section arc detined 1n figure 12.5. Figure 12.5 Dimensions for
1-sectlon beam
1-I ~
an
b"----'
-
h.,
fw
d
'
-
/'
'
r''
r " radius ol roof f1llet
For class I and cla-.r. 2 1-beams with a predominately uniformly di~trihutcd load the design shear ~tre~se:-. arc seldom ex~o:c),~ivc and the shear area. A, may be con~c rvatively tal.cn a~. t.., = 9 0 mm and tJte flange tluckne~~. t1 I t4 mm. and both arc h!\l> than 40 mm. Therefore from EC3. l>CCtion 3.2. table 3.1 the yield -.trength. f> 355 N/mm 2 . From EC3, sccuon 5.6, table 5.2 E
(235
O.R I
Vh t!
407.6
lv.
l) .0
45.3
< 72
X f -
58.3
therefore the steel ~cction i:. class !. (b) Loading at construction
and rib!>= 90 + 50/ 2 = 115 mm 0.115 x 25 x 3 8.62 J..N/m
Average depth of concrete Weight of concrete
~l ab
Steel deck
= 0.15 x 3
Stee I beam
= 74 X 9.8 1 X
Tmal dead load
= 0.-15 ( ()-)
= 0.7] = 9.8kN/m
Jmpo~ed con~truction
Ultimate load
load = 0.75 x 3 = 2.25 kN/m U5Gk I 1.5QL = (1.35 x 9.8 + 1.5 x 2.25) =
16.6 k '/merre
376
Reinforced concrete design (c) Bending
= (16.6 x 9~) /8 = 168 kN m Moment of re~istance of Meel section = Wpl J, 1653 x 355 x 10 = 587kNm > 168kNm OK Maxunum bending moment = wL~ /8
3
(d) 5heor
Maximum
~>hear
force V = lrL/2- 16.6 x 9/1
Shear Cl!sbtancc of M:clion -
vpl Rd -
74.7 kl'\
AJ)!'> l\10 v 3
For the steel ~>eclion tile web depth. d = 407.6 mm and the weh thic\..ness U),ing the conservative value of dt.,.,
1\,
Shenr
407.6 x 9.0
rc&i~tnnce
of section
1
= 9 mm.
3.67 x 101 mm 1
= Vpl. Rd
1\ v.f~
"/Mo/3
3.67
X
101
X
355
_,
- - --;:::-- " I 0 ·'
1.0
7'i2 I..N
X
/3
74.7 kl\1
hom the calculation!- for bending and shear il can he seen lhe loading on the beam during l'OINrul·tion is rdati\'cly IO\\ wmpan:d to the '-lrength of the beam. Abo, U1c 'reel th:ckmg "uh the COJTuga!lon~ Jl right angk' to I he 'pan g1vcs lateral and torsional re,lr:Jilll to the 'teeI heam. For the~e n!n,on' it i' con,idcn:d unnecc,sary to carry out tile Jmnht.>d calculatinn' for lateral and tor,ional stabllit} which arl! descnhcd in EC3. De,1gn ol ()reel Structure' I or the calculation of the deftccuon of the ~reel be;1m during construcllon at the M!I'\ICCahllll) hmll state see Example 12.4.
l~·----------------------------------------------------------------) 12.3
The composite section at the ultimate limit state
At the ultimate limit state it h. necc11sllry lll check the compo~ite section tor its moment cupacily and its shear strength, and compurc them aguin~t the maximum design ullimate moment and shear.
12.3.1 Th~:
Moment capacity with full shear connection
moment capacity M, of the composite sed ion i ~ derived in terms of the tensile or of the variou~ clement~ of the section as follows:
comprcl>~ive ~trl!ngths
Rcststancc Resistance Re,i'>tance ResiStance Resi,tancc Resi,tance Re'>l\tance Rest\tance
of of of of of of of of
the concrete flange the steel sccti(m the steel flange overall web depth clear weh depth the concrete above the neutral :ms the ~teeI flange above the neutral axts the \\eh O\'er dl'>tance X•
R" R, R,1 R"
0.56~f,lhcrr(hf~A·
J; btr R\
2R,r
R, J.,dt., R., 0.567/.kbcnx R x -/>b.\ 1 R.,.., =f-,tv." are defined in ligures 12.7 and 12.8.
h11 )
n Composite construction b.t~
,-h
I
j_ h
'hp
37i
Figure 12.7 Composite section dimension
J -
-, c,
Note: dis the distance between the fillets of the steel section
ft is important to note in the figures Lhm the SlrC!-!> hlock for the concrete extends to the depth of the neutrnJ nxis as specified in EC4 for compo~>ite design. There ttre three pos~ible locations or the neutral axis a~ shown in figure 12.8. These are:
(a) The neutrul axis in the concrete flange: (b) The neutral axis in the steel t1ange: (c)
The nculrttl uxis in the steel web.
Figure 12.8 Stress blocks at the ultimate limtt state
A
(a) Neutral axis In the concrete flange X< h: Rc~> R,
b,,
c, '
,, •
!!I
.... b ...1
(b) Neutral axis In the steel flange II < x < h
1
t, and R, > Rtt > Rw
(,
d
It
(c) Neutral axis In the steel web x > h + t,: R.:t < R.,.
378
Reinforced concrete design
The location of the neutral axis is determined from the equilibrium equation of the forces R at the section.
rel>i~tance
'E,R = 0
i.e.
The moment of resistance at the section is then obtained by taking moments about a convenient axis such as the centreline of the ~teel section. 1>0 that where : is the lever arm about a chosen a xi'\ for the resistance R. For Calle~ (bland (c) the analysis i' facilitated by considering an equivalent system of the re~istaoce forces a shown in the relevant diagram~. (a) Neutral axis in the concrete flange, x
< h Figure
12.8(a)
This condition occur" when Rcr > R,. Then the depth x of the neutral axi1. is given by Rex = 0. 567]~kVettX- Rs . R, R$(11 - hp) Therefore x ::.. ," b 1?d 0 ·567J~~ eff The moment of resistance is
= R,:.
M,
where the lever arm (11.12
~
i\
+"- ,,2)
Therefore ( 12.6) (b) Neutral axis in the steel flange, h < x
This condition occurs when Rs > Rd
R..,.
l·or the equilibrium of the rcsiMancc forces Ret I 2/?,,
2R,x
i.e. and
Xt
where b i!. the breadth and
ft
=
1?,
= 2Jyi)),1
1?,
(R,- R, t )
I? cf
(!?,
~{yb
l?d)lt
21?,,
is the thickness of the
~tee t
nange.
The moment of resistance is given by M,
R.r:.t
+ 2 Rs~Z~
l?d::.t
+ (R\- R,r):.1
,.. here :. 1 and :.2 are the tc.. cr arms as shown :1 ;:z
(h4-" + hp)/2 :rt)/ 2
= (ha
tn
figure 12.R(b) and
Composite construction ;he
Therefore substituting for z1• z~ and
x1 and
37
rearranging
( 12.7)
ow a (c) Neutral axis in the steel web, x
of
> h + t1 ( Figure 72.8(c))
Thi\ condition occurs when Rcr < R-.. and is mostly associated \\tth hutlt-up beam !>Cctions wtth small top flanges and larger bottom flanges and also with ~tiffcncd weh~ to avoid wch buckling. For equilibrium of the equivalent arrangement of the rcsi~tance force~
2R-..x
= 2/y lwX2 = Rcr
Therefore
where x2 is the distance between the neutral axil> and the centreline of the steel section . The moment of re~ismnce of the composite ~ection is the moment of the two couples produc.:cd hy R, and Rc~ with 2Rwx so that
Me
\4 I
'
-f
"
R.r(h1
hp
X2)
2
or
A{
I
A' -~o
Rc (h.
+ h + 1111 ) R~rd 2
( 12.8)
4R,
':!6
(EXAMPLE 12.2 Moment of resistance of a composite section Determine the moment or resistance or the tompo).itc scctinn shown in Jigure 12.9. The universal 457 x 19 1 x 74kg/m steel beam has a cross-sectional area ()fAn 94.6em 2 and is grade S355 t'. tcel with .fy 355 N/mm2• Usc concrete clnss C25/30 with characteristic cylinder strength J:k 25 Nlmm 2.
=
btl1 :
0.5671,,
3000
Figure 12.9
Moment of resistance examp
h,- 50
h.= 457
py
457 x 191 x 74 UB
Section
Stress Blocks
380
Reinforced concrete design (a) From first principles Resi~tancc
0.567/..~berr(h
of concrete flange Ret -
hp )
= 0.567 X 25 X 3000 X ( 140- 50) X 10 J = 3827 kN Resistance of Meel beam
R~
/yAu
355
X
9460 x 10- 3 = 3358 kN
As R, < R., the neutral axis is within the concrete flange. Determine the depth of neutral axis: 0.567AkbeffX = R, Therefore
3358 1...1\
3358 X 101 x= 0.567 X 25 X 3()()()
79.0 mm
Moment of rcsi~Lance: Lever mm z to the centre of the steel section is (/lu /2 I h ~/2) = 457/ 2 -f 140 - 79.0/2 M,
R,z
1
3358 x 329 x 10
-
329 mm
1105 kN m
(b) Alternatively using the design equat1ons derived
From pari (a) the neutral axi'> i~ \\ Hhlll the concrete flange therefore. from equation the moment ot resi~tance of the o;ect1on 11- g1vcn by M,
1~ .6.
1?,{';" .. h - :c't (h ~ hr) } BSR{457 140 _ 3358 ( 140 50)} IO 3 = ( 5 k 3827 2 I l J Nm ... 2 t
12.3.2 The shear strength
VRd
of the composite section
f·or the compoc;ite section. as for the con~tmction ~tage. secunn 12.2a(u). the shear is rclliMcd by the shear area A, of the ~teel beam and the shear resistuncc VR.J ic; given by A..(y
VRll
Vpi , Rd
= ----r,; '}MO V 3
(12.9)
and the !\hear (!rca A,. is given hy ( 12.10)
where A1 ll. the cross-sectional area of the steel beam and the othe1 dimension:\ of the cro!ls secuon are defined in figure 12.5. For class I and cia.,~ 2 I-beams with a predominately uniformly dt!>tributed load the des1gn shear stress is seldom exces.,ive and the !.hear area, A,. may be safely taken conservattvely as the area of the web. 1\. - d
X fw
For heams where high shear force:- and moments occur at the same ~eel ion such Uu11 > 0.5VRd it is necessary LO usc a reduced moment capacity for the composite section by reducing the bending stress in the steel weh a~ de~cribed io EC4, c;ection 6.2.2.4. VFAI
Composite construction
12.4
381
Design of shear connectors
The shear connectors are required to prevent slippage between the concrete nange and the Mcel beam thus enabling the concrete and steel to act a' a composne unil. Stud 'hear connector~; welded to the steel nange are the most corrunon t} pc u'ed. The head on the '' tthout 'hear connector.... The "li ppage is a maximum at the ~up ported end or the beam ''here the shear I' and the rate of change or moment dM / d1 are a maximum. The 11lippage reduce:. to tcro at mtd-span where the moment is a maximum and shear V = 0 for a uniformly tlistrihutcd load The connectors restrain the slippage by resisting the horizontal shear at the interface of the concrete nnd the ~tccl. The design is carried out for the conditions at the ultimate limit state. The design shear resistance, PHu· of a hended stud automatically welded i~ given hy I ~C4 as the lesser value of the following two equation;,: plld :
0.&/~ 1Td~ /4
(12.11 a)
or:
0.29nd~ ../J..J.Ecm
n-1
)
for ll..,j d
( 12. 11h)
>~
where -,, d
j;,
t~o.
the ptU1tnl safety factor
1.25
the diameter of the shun~ of the stud, between 16 mm and 25 mm 1s the ullimutc tensile ~o.trcngth of the stud 500 t'oilmm 2 or 450 N/mm, when equation 12.12b npplie~ 11o.
fck is the cylinder c:haraclcrislic compressive strength of the concrete £,111 is the
~cctmt
modu lus of eluMiciry of the concrete. sec table 1.1
A further reduction factor. k1 or k, is w he applied to PKtl ~~~ ~rcctticd in EC'4, section 6.6.4 and its value depend), on whether the ribs of the profiled ~heetulg arc parallel or tranwer~e to the supporting beam.
Horizontal shear
-.
/.
(a) Slippage
(b) Horizontal shear at beam-slab interface
figure 12.10 Slippage . The ~traight interaction line (b) represen ts a lincnr relation between the moment capacity anti 11 which provide:- n simpler and safer hut Jess economic ~o lutinn.
12.4.3
Shear connection for concentrated loads
When the beam !-.Lipp011~ concentrated load~ the \lope dM jdx of the bending moment i!> greater and the "hear is more inten. Fully composite sectoon M,
Figure 12. 12
(a) Stress Block Method
Interaction diagram for partial shear connection
(b) Lonear Interaction Method
Steelloection M,
0
1.0 0.4 Degree or shear connection 11
Composite construction N I -1 NzLI -L - 1
---
, ~- ' ]
'
w.
N, (L
--- -L 1 W1
2-
L) 1
t
Figure 12.13 Distribution of shear connectors with concentr; loads
6a
~ Mz
B.M Oiagrdm (all loads including thE' pomt load~ shown)
M.,,.,
6h
lhe 1:!)
:md
t, a
12.5
Transverse reinforcement in the concrete flange
Transverse reinforcement is required to re1.iM the longitudinal sheur in the concrete Range. Thb shear ucts on vertical planes either side of the shear connectors as shown in figure 12.14.
er
fransversc reinforcement
b
Potential failure planes
be The
.nt
Figure 12.14 Transverse reinforcement l the concrete flanges
b
Ribs parallel to beam
Ribs perpendicular to beam
Tile analy'>i\ uno de!>ign for the trallS\'Cr!>C rcinron:cment to rt.!\i\1 the longiludma) shear in a flanged hcum follo\\s the variable strut inclination method all required 111 PC:! and dellcrihcd in thi' hooJ.. in ~ections 5.1.4 anti 7.4. in COilJI.IIlt:lion with C\Hmple 7.5
part (2). nt
F...C2 ~peci lie' a mini mum of tranwerse ~teet area equal to (0.1311 1 > l 000/ I 00) mm 1 per metre Width. The method of de~igning the tranwer~e steel for u composite beam is ~>hown in example 12.3 part (b) Tra11sverse reinforcellll!l/1.
(EXAMPL E 1 2.3 Shear coonectors and transverse reinforcement The compo~>itc beam of example 12.2 and ligurc 12.9 srans 9.0 m..:1rcs ant! ill provided with 80 ~hear stutl connectors in pair~ at 225 mm ccntn:s. The stud' arc 19 mm tliumcter and of IOOmm hc1ght. The plal>lic -.cction of modulus of the steel ~ection is IV111 ) 1653 N/mm2 ant! the 1 tlestgn 'trc\~ of the \teet. J., - 355 "lmm • The characteristic material strengths are /..-L 251'\/mm, for the con~rctc and frk 500 Nlmm 1 for the rem forcing bal'\.
=
=
(a)
Calculate the degree of shear rcsiswncc and the mom..:nt of rcsi,tance ot the compo,ite heam based upon the l>hear conncctorll proVIded
(b)
Dcllign the transverse reinforcement required to rcsi~t the tranwcrsc llhear in the concrete tlange.
386
Reinforced concrete design (a) Degree of shear connection and moment of resistance
The design :.hear resio;tance. PRd· of each shear 'tud i~ the le~ser value obtained from equations 12. ll a and 12.llb wilh/0 • the ulttmate ten~lle \trcngth of the steel. equal to 450N/mm~. Us~ng
these equations it is found from equauon 12.1I a th S 2 .md 5.3
'I (src Ldble 52). Abo 1ee the nddltlono~l requirement' for minln,um lap lengths ~f'd
Maximum and minimum areas of reinforcement Table A.7
Maximum areas of reinforcement
(a) For a slab or beam, tem1on or compression reinforcement
·1OOA, I Ac
4 per cent other than at laps
(b) ror a column 1OOA,. A.:
_4
per cent other than at laps and 8 per cent at laps
(c) For a wall, vertical reinforcement
1OOA,/A1
c.
4 per cent
Appendix Table A.8
Minimum areas of reinforcement Concrete closs (fy. ~ 500N/ mm 1 )
Tension reinforcement in beams and slabs
\~"'
>
397
0.26
(
'~=
0 001 3)
C25/ 30
C30/ 35
C40/ 50
CS0/60
0.0013
0.0015
0.0018
0.0021
Secondary reinforcement > 20% main reinforcement Longitudinal reinforcement in columns A, min .;. 0 1ONKJ / 0.87 fyk ., 0.002Ac where NKl is the axial compression force Vertical reinforcement In walls A,,rnln "> 0.002At Note: t>. os the ml'an width
or the tension zone.
300
32
E' E'
250
g'
200
2
150
.§,
25
.§,
~
E
?;
bar spacmg
20
ti
sE E
16
E i(
Maximum bar size and spacing for crack control
'2
::>
"' E
£
Figure A.1
100
12
so
10
M
0
-~
2 E ::> E
sE
6
100
150
200
250
300
350
400
Stress In reinforcement under quasi-permanent load (see secuons 6.1.3 and 6.1.7)
(N/mm~)
Span-effective depth ratios 36
32
L
~ - ---+---1---
1I
- i- - t - - t - -
Figure A.2
---i--11--t--
Graph of basic span-effective depth ratios for different classes of concrete
---;---+-+
4
K • 1 0 for a som ply supported span
100A,,... 12 ~------~----~--~--~~~-----
0.40%
080%
1.2%
1 6%
2.0%
bd
398
Appendix
Summary of basic design equations for the design of reinforced concrete (a) Design for bending (see chapters 4 and 7) For a swgl) reinforced section: A,=
M
. -
0.87})~;:
;:=d{0.5+(0.25-K/ 1.134) 1
~}
K - M /bd 2}~k
For a doubly rcinrorccd section I
A,
(K
> KbJI)
- ~cc
ligurc A.3:
1
(K - KbJilf.:khd - -0.8~/"ydd - d') Kh.ltf.,hd 2 I ---'-A
0.87J;.k:.hal
'
When moment rcdi),tribution modified - sec 1ablc -U. Table A.9
ha~
been applted lhl'n
th~.: above equation~
Limiting constant values
Concrete class
--------------------------------xn.)ljd Limit1ng Maximum z1111, Khlol limiting K Um1tmg d'td Maximum percentage steel area 100Ab,,1/bd
0.95
Compression rcmtorccment requ1red (al M1.,,)
II
~
0.90
0.85 082 0
0.05
CS0/60
0 45 0.167 0.171 23 Afckl fyk
m3x1mum valuf.' of 1/d according lo Lh~ Concise Codf.' and previous UK prt~clicc
~
k A,w
.
0.7'6c/}yk COl
\ln1in - - X .I
~Ftd
()
= 0.511~-;.,cotO
(c) Design for torsion (see chapters 5 and 7) t=
for
rt
Area ol the section Perimeter of the section
II u
rectangular section h x h bll Ilk -
2(hhJ
2(b
"
:!t)
' llr..s_ < 1.0
~1M m.l\ -
1 . 33lj~kA~t
cot 0 I tan (}
.,,..,
2Al0.8~/y1.. COl(}
Tt:..tllk cot
n
211 ~.,0.8~/ylk
(d) Design for punching shear in slabs (see chapter 8) VRtl , mJx 11 1
2(a
.r..')].fck1.5 0.6 (I 250
0.5ud [ /J)
~
47Td
0.053 J.Tc.(.l, ..l·, )
r.,.. . I'Rd c'
A," > -
-
0.751'Rt.f c
15/;"t.l d lr
llf
-
3
400
Appendix
Proof of equation conversions used in Chapter 5 section 5.1.2 To prove that !>in 0 cos() =
1 : () tan -cot 8
Con1>ider a right-angled triangle with sides length a, band ft. where h b the length of the hypotenuse and () is the angle between :.ides of length a and b, and use the theorem of Pythagoras where ~~~ = a 2 + b~. l>in ()COS()
-
abab h X II = h1
ab
=a1 +
I b2 = a2
!J2
tan 8 +COl (J
tlb -r- ab
A reason for using this type of conver~ion in the equations for the analysis for shear is that it facilitates the selling up of quadratic equations which can he more readily solved.
Typical design chart for rectangular columns Figure A.4 Rectangular columns (d'·h 0 20)
14
13
e
12
A,
e
2
1.1
• 2 • l. A,
"' 0 .5 l--.:::....t-O-f''cc t ion .vee T-he:nm Flat ~lab 228-35 Floms l'ee S l ab~ Footing~
allowable 'oil prc~~ure1> 284 combined 291 -l factors of tair 241-4 ~trip method 245. 250-1 yield lines 245-50 Slender columns 255-8. 275-9 Spacing of remforcemem 130. 2 17. 259. 397 Span- block 93 Ultimate limit '-lute factors of safety 19, 282 loading arrangements 21 -5. 310 II. 313 pn!,trc,,cd concrete 353-(,8 'tahility 310 II Uncr3cked section 96. 137. 360-1 Untcns10ned 'teet 111 pre,trc,M:d concrete 357 9 Wall\ 279 -80 Weighh or matenal\ 393 Wind loading :25. 30, 45 Yield line' 245 -50 Yield Mrains 6, 60 Young\ modu lus we Ela,tic modulus
ISISN : 9780230500716 - __ TI'TLE : REINFORCED CONCRETE tCAT : T7S - CONCRETE