2 section analysis
TRANSCRIPT
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2.0 Structural Behaviour
The objective of section design is to get a safe section dimension b x dand safe reinforcement area to support
the ultimate design moment by utilizing the full strengths of both materials. The section analysis is based onthe following assumptions:
i) The section remains plain before and after bending.
ii) The deformation is small.
iii) The strain-stress within the elastic range obeys the Hookes law.
iv) The concrete tensile strength is negligible.
beam section which is in e!uilibrium under an e"ternal bending moment M will have its faces in
compression and tension on either sides of neutral a"is. The depth of neutral a"is depends on the sectiongeometry and material composition or the strengths of both sides of the section as shown in #igure $.% below.
The compression and tension sides will yield simultaneously if the strength of both sides are e!ual. & of a
symmetrical strain& cand tensile strain& t of symmetrical and homogeneous sections will reach the yield
strains or fail simultaneously& while for unsymmetrical or inhomogeneus sections the strains distributions aredepending on internal resistances of each part. #igure $.% below shows the strain distributions of various
renforce concrete sections of a reinforced concrete beams subjected to bending moment&M. The concrete on
the tension side is assumed ineffective due to its low tensile strentgh and is drawn in broken lines.
Therefore& from the above point of view& a section may fail in any one of the following modes depending onthe strengths of the compression and tension sides.
i. Tension failure - tension side reaches the yield strength before the compression side.
ii. 'alanced failure - tension and compression sides reach the yield strength simultaneously.
iii. (ompression failure - compression side reaches the yield strength before the tension side.
%)
2 SECTION DESIGN
Fc x
c= d
-
*
-
*
-
*
b+ ,ymmetrical section c+ nsymmetrical section
Figure 2.1 Strain distriution.
a) Bent!u" ea#
M
xc= d
.
b
dx
c= dFc Fc
Fs
Fs
Fs
b b
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#igure $.$ shows the types of failure or failure modes of reinforced concrete sections. The member ismodeled by springs and stiff plate as shown in /a+.
The depths of the neutral a"esxcandxtare depending on the properties of the springs. That is
xd
x
s
c
=
c
ssc
c
1
ddx
+
=+
= or d.x =
/$.%+
0!n. /$.%+ shows that the depth or strain of the compression side can be controlled by adjusting the strength
of the tension side. 1f sapproaches 0 thanxapproaches d. This means the section is over crowded with
reinforcement or large amount of As has to be used. 2n the other hand if the ratio s3c increases than x
reduces indicating the section will undergoes tension type failure. 1t can be shown as an e"ample that by
substituting the typical values for cu= 0,0035and yd= 0,00217for forfyk= 500 N/mm2steel& the theoretical
depth of the compression isx4 1,67d (i.e. 4 0,167).
However& 0($ has limited the ma"imum strength of the compression (concrete)side corresponding to a depth
x = 0,45dand the stress block tos = 0,8x. This is to avoid overcrowding the tension side with reinforcementand to allow tension type failure. The types of failure& hence the types of sections according to 0($ are
shown in figure $.5 below.
The ma"imum strength of a section corresponding to the depth of neutral a"is x = 0,45dis known as the
ultimate moment of resistance of the section&Mu.
%%
Figure 2.$ T%"es o& section.
a) nder-reinforced section b)'alanced section c)2ver-reinforced section
b b b cu
syAs> As bal
d
x = 0,45dx < xbal
< cu
syAs< As bal
d
(M < Mu) (M =Mu) (M > Mu)
A!xbal
= 0,45d
cu
syAs bal
d
!s
c
s
x
d " x
fyd fyd fyd
fcd fcd fcd
xc < d xc = dxc > d
a+ 6odel b+ ,ection c+ Tension failure d+ balance failure e+ (ompression failure
Figure 2.2 Failure #odes.
*
M
-F
c= f
cdA
c
Fs= f
ydA
s
Fc1 Fc$ Fc5
Fs 4fydAs1 Fs 4fydAs2 Fs 4fydAs3
b
d
As
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2.1 'lti#ate (o#ent o& )esistance
(onsider a balanced rectangular reinforced concrete section b" dof a beam subjected to an e"ternal momentMas shown in figure $.7 below. The strain and e!uvalent rectangular stress distributions at failure according
to 0($ (0 %88$-%-%:$))7 clause 5.%.9) are shown in (c)and (e).
#rom (e)
(ompressive force bsf567,0bsf
F c#c
c#ccc ==
/$.$+
,ubstituting s = 0,8x bxf454,0F c#c =
/$.$a+
Tensile force sy#ss
y#s Af87,0A
fF ==
/$.5+
The moment of resistance of the compression and tension sides are obtained by taking the moment
e!uilibrium about the centroids of the tension and compression sides.
rt. (ompression side bx$f454,0$FM c#cc ==
/$.7+
2r 2c#c bd#fM = /$.7a+
where
=
2
x8,01454,0#
rt. (ompression side $Af87,0$FM sy#ss == /$.;+
#or balanced sections& the ultimate moment of resistance& Mu 4 Muc 4 Mus& since both sections fail
simultaneously. #or under-reinforced and over-reinforced section& the ultimate moment of resistance& Muis
always the resistance of full compressed compression side at (x = 0,45d). Therefore the ultimate moment of
resistance can be defined as the resistance of balanced section& i.eMu4Mbal.
%$
x = 0,45d
yd
As
b
Figure 2.* Balanced section stress!strain distriutions+
, 1.0
, 0+-
d
a)S"ring #odel )Section c)Strain d)Stress e)Euivalent stress.
*
M
-
fcd
fydFs
Fcs =0,8x
M
$= d " s/2
ccfc#/c = 0,567fc#
Fc
Fsfy# /s= 0,87fy#
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,ubstitutingxbal4 0.45dinto 0!n. ($.7) and($.;)andMuc= Mus= M& we get
0!n. ($.7): ltimate 6oment of resistance wrt. concrete:
2c#
2c#bal bdf%#bdf167,0M ==
/$.etermine also the depth of the neutral a"is&xbal.
Solution
(heck the moment of resistance wrt. concrete
2c#bdf
M#= 4 166,0
380x250x25
10x150
2
6
=
%# 'ala(c)d s)c*+(
( ) mm6,31138082,02
d45,0x8,0d$bal ===
baly#bals $f87,0
MA = 4
6,311x500x87,0
10x150 6
= 1106,6 mm2
%5
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-l .
.2.1.1(1)
-l .
.2.1.1(3)
4,128380x250500
6,226,0db
f
f26,0A *
y#
c*mm+(s
===
0,003b*d = 285 mm2
2cxams
mm3800380x250x04,0A04,0A =
)e&. Calculations Out"ut
-l .
.2.1.1(1)
-l .
+d) 420 As= 1257 mm2
) d)* f *) ()u*al ax+s xbal=0,45d = 0,45 x 380
= 171 mm
ote: 1n practice& a given section is hardly designed as a
balanced section due to the given dimensions band d
is unlikely match the balanced section criteria.
E/a#"le 2.2.1(2)
rectangular beam section has to support an e"ternalmomentMof %)) km. >etermine the section dimensions b
" d, b = 0,6d) and reinforcement area. ssume fc# 4 $;3mm$andfy#4 ;)) 3mm
$. ,ketch the section detail.
Solution
#or balanced section2c#bdf
M# = 4
167.0d6,0x25
10x100
3
6
=
mm8,341d =
b = 0,6" 341,84 205,1 mm
ay b "d = 205 "340 mm
mm8,278d82,0
2
d45,0x8,0d$bal ===
baly#bals $f87,0
MA = 4
8,278x500x87,0
10x100 6
=825
mm2
*y#
c*mm+(s
m2,4340x205500
6,226,0db
f
f26,0A ===
b "d = 205 "340
%7
$;)
5=)
4H$)
%$;9 mm
$
$);
57)
3H$)
875 mm$
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.2.1.1(1) 0,0013b*d 0,6 mm2
2cxams
mm2788340x205x04,0A04,0A =
+d) 320 As= 43 mm2
2.2.2 'nder!rein&orced Section
#or the under-reinforced section where the depth of the compression zone is less than that of the balancedsection& we have
M
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2
134,1
#411
a2
ca4bb
d
$ 2 =
=
=
134,1
#25,05,0d$ 0,5d9
/$.8+
The lever arm can also be obtained from the e!uation below after having solved the depth of the neutral a"is
x.
2
sd$ =
The reinforcement area is obtained from
$f87,0
MA
y#
s = /$.%)+
)e&. Calculations Out"ut
E/a#"le 2.2.2(1)
rectangular beam section of $;) mm " 5=) mm /b" d9issubjected to an e"ternal momentMof %)) km. >etermine
the reinforcement area if fc# 4 $; 3mm$ and fy# 4 ;))
3mm$. ,ketch the section detail.
>etermine also the depth of the neutral a"is&x.
Solution
2c#bdf
M#= 4 2
6
380x250x25
10x100#=
= 0,111 < #!
-m)ss+( )+(fc)m)(* +s (* ):u+)d.
+=
134,1
#25,05,0d$ 4
134,1
111,025,05,0d
= 0,8d = 338,2 mm
%etermine the reinforcement
area iffc#4 $; 3mm$andfy#4 ;)) 3mm
$. ,ketch the section detail.
Solution
2c#bdf
M#= 4 2
6
400x250x25
10x250#=
= 0,25 < #!
-m)ss+( )+(fc)m)(* +s ):u+)d.
38,0d45,0
d%
x
d%
bal
== 400x45,0x38,0%d
ay 70 mmd! = 70 mm
%=
9)
$;)
7))
7H$;
%8
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-l. .2.2.1(1)
-l. .2.2.1(3)
( )
( )%ddf87,0
bdf%##A%
y#
2c#
s
=
4( )
( )2
2
mm2,57870400500x87,0
400x250x25167,025,0=
75400x82,0x500x87,0
400x250x25x167,0A%
$f87,0
bdf%#A
2
sbaly#
2c#
s +=+=
4 1748,7 mm2
*y#
c*mm+(s
m2,135400x250500
6,226,0db
f
f26,0A ===
0,003b*d = 300 mm2
2
cxams mm4000A04,0A
=+d) 425 As= 163 mm
2; 220 A!s= 628 mm2
1n constructions& flanged sections may occur in the forms of monolithic beam-slab and T or @
beams as shown in #igure $.< below. The composite action between flange and web resulting in thewhole section bend as one peace.
%8
2.* FNGED SECTION
bf
b b
f f
d d
Figure 2.3 Flanged sections.
/a+ 6onolithic beam-slab /b+ T-beam /c+ @-beam
slab
b)am
bf
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( fffc#f 5,0dbf567,0M = /$.%5+
) s)c*+( +s d)s+()d as a )c*a(ula s)c*+( fbf" d.
C
a
E
/A
>
e
$%
)e&. Calculations Out"ut
-l. .2.1.119
-l. .2.1.139
Solution
) mm)(* f )s+s*a(c) f *) fla()
fffc#f 5,0dbf567,0M =
9100x5,0350100x400x25x567,0 =
Nm10x1,170 6=
Mf> M&d s*)ss blc# +*+( *) fla().
)s+( as a )c*a(ula b)am f bf"d?
2fc#
&d
dbf
M#= 4 2
6
350x400x25
10x120#=
%#0.8,0 >=
+d) 320 As= 42 mm2
mm315$ =
bf= 400
f
= 100
d=
b
= 200
400
100
350
200
5H$) 87$ mm$
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The flange width bfin the design e!uations is the effective flange width& beff. 0($ clause ;.5.$.% specifies beffas follow:
i+ beff4 beff,+; bwb for monolithic cast beam-slab.
where
beff,+4 0,2b+* 0,1l*) m+(+mum f0,2l b+
l+s *) d+s*a(c) b)*))( +(*s f $) mm)(*s.
*,ee figures ($.9)and ($.=)below for notation.
ii+ True width bffor T or @ beam.
)e&. Calculations Out"ut
E/a#"le 2.* 2)8 Flanged Bea#
part of floor plan of a general retail shop building is shown
in the figure below. The beams and slabs are monolithically
cast in-situ. The design data are as given below.
>etermine the effective flange width b)fffor the beams
'(3 % to < .
$$
l1 l
2 l
3
l0= 0,8l
1 =0,15(l
1; l
2) 40,7l
2 l
= 0,15l
2; l
3
Figure 2.9 De&inition o& l0+ &or calculation o& e&&ective &lange 5idth.
b%
b%
b$
b$
b%
b%
beff %
beff $
bw
beff
bw
b
Figure 2.- E&&ective &lange 5idth "ara#eters
$;)";))
5))"Mbal= 187,43 #Nm
-m)ss+( )+(fc)m)(* +s ):u+)d.
a) -m)ss+( )+(fc)m)(*?
( )
( )
( )
( )
6
y#
lab
s 3,650350500x87,0
1043,187200
%ddf87,0
MM
%A =
=
=
Su##ar% &or &langed section design
The design of flanged sections can be summarized as shown in #igure $.%$ below.
2. Design chart
$9
)e&. Calculations Out"ut
b) )(s+( )+(fc)m)(*?
sy#
f
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1nstead of using lengthy e!uations and formulae& the calculations can be simplified by means of design charts.The charts normally in used are the lever arm $versusM/fc#d
2or #. The construction of chart for rectangular
sections is briefly e"plained in the ne"t paragraph.
6oment of resistance wrt. concrete:
Mc= 0,567fc#bs$
'ut$ = d @ s/2 s =2(d @ $) Mc= 0,567fc#b2(d @ $)$
6oment of resistance wrt. tension reinforcement:
M = 0,87fy#As$ =0,87fy#As2(d @ $)$
1ntroducing $ =lad
M = 0,87fy#As2(d @ lad)lad= 0,87fy#As2d2(1 @ la)la
fc#bd2 ( ) aa22c#
sy#
2c#
ll1d2bdf
Af87,0#
bdf
M
==
( ) aac#
y#ll12
f
f87,0=
#orfc#4 $; 3mm$andfy#4 ;)) 3mm
$& we get
ow a graph of # versuslafor a group of percentages of reinforcement area&can be plotted as shown in
#igure $.%5.
$=
( ) ( ) aaaa2c#
ll18,34ll1225
500x87,0#
bdf
M===
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2.3 Design Flo5 chart
The flow charts for the section design of rectangular and flanged sections are given in #igures $.%7 and $.%;.
$8
Figure 2.1$ Design Chart ! fck, 2 N:## fyk, 00 N:##2.
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5)
Figure 2.1* Design &lo5 chart &or rectangular sections 8 #ain rein&orce#ent.
M > Mbal # > #!
-m. )+(f. +s ):u+)d
M < Mbal # < #!
-m. )+(f.+s (* ):u+)d
>ata
Mbal
= 0,167fc#
bd 2
2r # =M&d/fc#bd2
)s
=
134,1
#25.05,0d$
As=A
s=
d!/xbal0,38B
As=
0nd
$ = d " =0,82d
A!s=
d!
0nd
0nd
As> As m+(E
As< AsE
As> As m+(E
As< AsE
As> As m+(E
As< AsE
M&d, b x dC
fc#, fy#,-(m
M&d= MbalE
D # = #!E
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;role#s
5%
)s, s+(ly )+(fc)d
Figure 2.1 Design &lo5 chart &or &langed sections.
E)c*. s)c*+( bfx d9
>ata
0nd
0nd
Mf> ME
%f= 0.567
(1 " b
/b
f)(1 @
f/2d)
f/d ;01,67b
/b
f
Mbal
= !ff
c#b
fd2
Mbal
> ME
N
As> A
s m+(E
As< A
s maxE
)s
)s
0nd
# = M/fc#
bfd2
$ = d " =0,82d
As=
As> A
s m+(E
As< A
s maxE
)s
N, dubly )+(fc)d
Mf= 0,567f
c#b
f
f(d " 0,5
f)
x = f
A!s=
As=
; A!s
As> A
s m+(E
As< A
s maxE
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%. hat is meant by the followingE
a) tension failure& balanced failure and compression failure.
b) under-reinforced& balanced and over-reinforced sections.
$. rectangular section of a beam is subjected to a bending moment of %7; km is shown in the figurebelow. >etermine the bending reinforcement for the section iffc#4 $; 3mm
$andfc#4 ;)) 3mm$.
5. a) >etermine the main reinforcement for the section in the previous !uestion if the bending moment is
increased to $;) km.G325 As= 1473 mm
2H 216 A!s= 402 mm2I
b) >etermine the suitable effective depth for balanced section.
G510 mmI
7. flanged section is subjected to a bending moment of 5%$&; km. >etermine the main reinforcement iffc#4 $; 3mm
$and
fc#4 ;)) 3mm$.
G325 As= 1473 mm2I
BmmC
;. >etermine the main reinforcement for the section in the previous !uestion if the bending moment is increased to %)%$&; km.
G332 A!s= 2413 mm2 H 632 As= 4825 mm
2I
5$
$;)