2 Section Analysis

8/11/2019 2 Section Analysis

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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


5/23

-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$


6/23

.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


7/23

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/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


10/23

-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


11/23


12/23

( 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

$%


-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$


13/23

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/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


b) )(s+( )+(fc)m)(*?

sy#

f


19/23

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===


20/23

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.


21/23

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


22/23

;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


23/23

%. 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$

$;)

Documents

2 Section Analysis