slab and beam ihfds

8/13/2019 slab and beam IHFDS

1/21

CHAPTER I

BEAM AND GIRDER FLOOR

1.1. Data and specifications

A concrete floor slab type beam and girder is considered as a part of a building

(Fig. 1.1), with two transversal spans L= 9. m and longitudinal spans l= !. m.

"he resistant structure is reali#ed of columns and girders, that are part of the floor.

Fig. 1.1

"he floor is reali#ed of o monolith slab that support on beams that transmit the loadto girders (Fig. 1.1).

"he ratio =a

l is considered bigger than $, the slab being reinforced on one

direction% $!&.$$'.$

.!>== .

Concrete uality is chosen function the eposure class of the building in

environment conditions. "he design building is a tetile production hall and there is notecessive humidity. According to "able *.1 and Fig. *.$a, Appendi *, the construction is

included in class +1 without corrosion ris- or chemical attac-. For this class the

minimum concrete grade is $$'.

"he floor will have a minimum resistance to fire of ! minutes (/!).

For concrete C20/25 the design strength in compression, using relation $.11 and"able $.1%

00.10'.1

$1 ===

c

ck

cccd

ff

mm$


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"ensile strength is obtained with the relation%

1'.1

'.11'. ===

c

ctkctctd

ff

mm$

$.$=ctmf mm$

Steel reinforcementtype 2' and welded mesh type 2"3 is used for the slaband beams.

For 2' '=ykf mm$

40'1'.1

'==ydf mm

$

For 2"3 (for d5 &.1 mm) 0601'.1

44==

ydf mm$

"he finish is reali#ed of mosaic (0 mm depth) on a layer of cement plaster of 0

mm depth.

"he service load is%

n

up

= ' m

$

.

1.2. Desin of !"a#

1.2.1. $%e"i&ina%' si(in of t)e s"a#

Fleural design will ordinarily consist in selecting a slab depth which will permit touse of an economical low steel ratio and which will not allow unsightly or damaging

deflections.

For slabs elastically fied on perimeter, the rigidity condition is%

hp70'

1a, where a is the opening parallel with the short side, a= $.$' m.

hp= $$'0'

1 = !4.0 mm

And the minimum depth from technological condition to monolith slab, is ' mm,

the depth from the above conditions will be%

hp= & mm

From the fire condition the minimum depth is hp= 6 mm.

1.2.2. *idt)s fo% #ea&s and i%de%s


3/21

Fig. 1.$

From technological conditions the width of beams (bb) can be between 1' mm and$' mm, and the width of girders (bg) and edge beams (beb) between $ mm and 0'

mm. From fire resistance condition b>1$ mm.

8ne chooses%

bg = 0 mm

beb = 0 mm bb = $ mm

1.2.+. C"ea% spans

Fig. 1.0


4/21

$

$.0.$'.$

$1

=

=

bbal

bebc = $. m

lc$= a bb= $.$' .$ = $.' m

1.2.,. Loads (according to 2"A2 1118A:&& ;0< and 1111:&6 ;4ermanent load

: from own:weight of slab (hp= & mm)nslabg = .& 1. 1. $' = 1&' m

$

: from plaster weight (hplaster= 0 mm)nplasterg = .0 1. 1. $1 = !0 m

$

: from mosaic weight (hmosaic= 0 mm)nmosaicg = 1 m

$(according to 2"A2 ...)

"otal%gn= 006 m$

"emporary load

: service load given in pro?ect data%nup = ' m

$

"otal load%nu

nnpgq += = 006 @ ' = 606 m$

3. esign Boad

"he design loads are determined by multiplying the unfactored loads by loading

coefficients, ni, tht are given in 2"A2 ... function the load type.

>ermanent load

: from own weight of slab 1&' 1.1 = 19$' m$

: from plaster weight !0 1.0 = 6$ m$

: from mosaic weight 1 1.0 = 10 m$

"otal% g = 4' m$

"emporary load

pu= ' 1.0 = !' m$

"otal load% q=g@pu= 4' @ !' = 1'' m$

Boads on linear meter (for a strip of slab of 1 m width, Fig. 1.$)%

qn= 606 m$1. m = 606 m

q= 1'' m$1. m = 1'' m

1.2.5. !tatica" Co&p-tation


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For determining the bending moment it considers a strip of 1. m width, that has as

support the beams (points A, 3, , , C, F, D, E, *).>lastic analysis may be used to determine the bending moments because the

permanent loads are important ;'


6/21

mmd

mmccc

chd

tolnom

nomp

'.4&$'$&

$11

$

G

min

s

===+=+=

=

"he value cmin is obtaining from "able '.$ function of eposure class (+1) andstructure class. For establishing the structure class the life of the building was considered

' years (class 4) and provisions from "able '.4 that consider a reduced class for

members type flat (4:1=0)

cmin=1mm

According to "able !.! the minimum distance to the centroid of resistant steel is a= $ mm. For a reinforcement with diameter of ' mm, effective distance to the centroid

is%

aeff= $@'$=$$.' mmH$ mm

"he section of tensioned steel will be%

Fi%st span

:/educed moment is%

M1= 060! 10mm

060.1$&.00.10'.4&1

160!.0lim$0

!

$

1=


7/21

1&4060

00.10'.4&11'!. 0 ==slA mm$= 1.&4 cm$

8ne chooses% welded mesh 111DK 19! L'1:L'1 (Asleff= 19! mm$).

Inte%io% spans and inte%io% s-ppo%ts

M$= :MC=M0= :MD= ... = $&&1 m

060.9$.00.10'.4&1

1&&1.$lim$0

!

$

$=


8/21

For ta-ing over the local moments on continuous supports from the short direction

and on edge supports in the long direction, constructive welded meshes are provided type

11$ DK 19! L'1: L'1, which are etended on both sides of support with .$' ln

1.2.. Desinin of ties cent-%a

"he ties are reinforced according to standard provisions because the computation is madeonly for a current floor, not for the entire structure.

0== !c bb mm H $'min, =cb mm

$$min, === cc hmmh mm

!$0, ==effcA mm$H 'min, =cA mm$

Fig. 1.!

46!16.16.$

,

$

===

effcs AA mm$

1! with 4&1min, =sA mm$ (Fig. 1.!)

"he value of minimum cover with concrete depends on the type of eternal finish.

"he value of minimum cover is given function the structure class (4)%

$'min =c mm H Lsl=1 mm

0'1$'min =+=+= cccnom mm

*f the eternal wall is with plaster and lime, the cover layer can be reduced with'

, = adddurC mm.

For transversal stirrups (having Lsw=! mm) the concrete cover is%

$9!0',

===s"slnom#

cc mm


9/21

Finally, the concrete cover will be considered 0'=slc mm for longitudinal steel

and 0=s"c mm for stirrups.

1.2.3. Anc)o%in of %einfo%ce&ent


10/21

1.+. Bea& Desin

1.+.1. $%e"i&ina%' si(in of t)e #ea&

For secondary beam the depth of the cross:section can be considered with the

following value%

"he rigidity condition%

1$..1

!

1$..1==

lhb = '..! mm

*t is adopted hb= ' mm

From the condition% 0..$=b

b

b

h it results that%

0..$

'

0..$ == b

b

h

b = $..$' mm

*t is adopted bb= $ mm.

From technological condition% the depth must be a multiple of ' mm

From the condition of fire resistance%

hb>1$ mm

For the beam the following si#es are adopted%

hb4 500 &&and bb4 200 &&

1.+.2. !tatica" &ode"

"he beam can be considered as a continuous beam with eual opening, that

supports on the girders (Fig. 1.&).

Fig. 1.&.

1.+. !pans


11/21

Fig. 1.6.

"he design spans are function of clear spans, (Fig. 1.6) which are determined as itfollows%

l= ! m

$11

bbl

gebc

= lc$= '.& m lc$= l!

lc%= 1 bg lc%= '.& m lc%= l0= l4= l'lc$&lc%&: clear spans

beb width of the edge beam (0 mm)

bg width of the girder (0 mm)

"he design spans are computed ta-ing into account the provisions given in Anne

**.

,!$

4.O

$

0.min$&.'$111 =

+=++= aall ceff m

,!$

4.O

$

0.min$&.'$1$$ =

+=++= aall ceff m

1.+.,. Loads

"he loads from the slab are transmitted to the beam from a strip eual to a as

uniform load, Fig.1.9.


12/21

Fig. 1.9

A. nfacto%ed "oad

>ermanent load (gn)

: transmitted by the slab and finish

a= $.$' m

aggg nplasternmosaic

nslab ++ )( = (1&' @ !0 @ 1) $.$' = &!' m

: self weight of the beam

(hb hp) bb'b= (.' .&) .$ 1. $' = $1' m"otal permanent load%gn= &!' @ $1' = 9&'' m

"emporary load (pn)

pn=puna= ' $.$' = 11$' m

"otal load% qn=gn@pn= 9&'' @ 11$' = $1' m.

B. Desin "oad

>ermanent load (g)

: transmitted by the slab and finish

(gslab@gmosaic@gplaster) a= (190 @ 6$ @ 10) $.$' = 911$.' m

: self weight of the beam

(hb hp) bb'b1.1 = (.' .&) .$ $' 1.1 = $0!' m

"otal permanent load%g= 911$.' @ $0!' = 114&&.' m

"emporary load

p=pua= !' $.$' = 14!$' m

"otal load% q=g@p= 114&&.' @ 14!$' = $!$$.' m.

1.+.5. !t%ess Co&p-tation


13/21

1.+.5.1. Bendin &o&ents fo% se%6ice stae

For computing to limit state of crac-ing and to limit state of deflection is necessary

to determine the values of bending moments for second stage of wor-ing (service loads).

3ecause of that the bending moments are computed in elastic domain, as for a continuousbeam with constant moment of inertia.

Fig. 1.1.

For determining the bending moments one can use the values given in Appendi 6.

"he values and the coefficients for the continuous beam with five spans, and themaimum and minimum bending moments are given in "able. 1.1.

"able 1.1.

!ec

tio

n

Coefficient of inf"-ence!pan

Teff

&

$e%&anent

"oad

g

7N/&

8e&po%a%'

"oad

p

7N/&

M&a9: M&a9;

a # cag:bpll ag:cpll

1 ,&$ ,99 :,$! !. 11,4&6 14,!$' &0,691& 14,49!

3 :,1' ,14 :,1$ !. 11,4&6 14,!$' :0$,'40 1&,6!00

$ ,00 ,&9 :,4! !. 11,4&6 14,!$' 49,644' :9,''10

:,&9 ,0$ :,111 !. 11,4&6 14,!$' :14,$''4 :6$,$4$

0 ,4! ,6' :,09 !. 11,4&6 14,!$' '&,'40' :1,0&&$

For the limit state of crac-ing%


14/21

Fig. 1.11.

1.+.5.2. Bendin &o&ents fo% desin "oads

"hese moments are used for designing the ultimate limit state in normal section.

"he bending moments are computed in plastic domain, with the following relations%According to 2"A2 11&$:9$, current floors of slab and beams reali#ed of

reinforced concrete and Pprestressed concreteQ when the ratio between long and

temporary loads and total is smaller than .&' is sufficiently the computation of stressesin plastic domain.

*n the case of beams the ratio is'.$!1$

14!$'= .'!

: First span%

11

!'$!1$ $

11

$1

1

lqeff

M =

= = 6'4$!.4 m

: First support%

14

.!'.$!1$ $

14

$1 =

=

lqeff

MB= !&1$.&m

: 8ther interior spans and supports%

1!

.!'.$!1$ $

1!

$$

0$

=

====

lqeff

MMMM DC='6&0 m

1.+.5.+. !)ea% fo%ces

2hear forces are used in computation at ultimate limit state in inclined sections. "he

values of shear forces are determined in plastic domain with the following relations (Fig.

1.11)%: in ad?acent sections of edge support%

(A= .4' qleff$= .4' $!1$.' !. (A= &4&&

: in first span, ad?acent to first interior support%

(Bleft= .!' qleff$= .!' $!1$.' !. (B

left= 11&99

: other sections ad?acent to interior supports%

(Bright

= .''q

leff%= .''

$!1$.'

!. JB

right

= 6!106


15/21

Fig. 1.11.

1.+.. Co&p-tation at t)e -"ti&ate "i&it state in no%&a" sectionharacteristics of materials%

fyd= 40' mm$ for 2'

fcd= 10.00 mm$ for $$'

"he depth of the cross:section is%

mmd

mmccc

chd

tolnom

nomg

4!$$0'

01$

$

G

min

s

===+=+=

=

1.+.. !tee" Desin

"he beam has flanged section in spans and rectangular section in support.

Fi%st span

Cffective width of the slab must be established from the conditions given in theAppendi **.


16/21

Fig. 1.1$.

mm14$'

1$'$

$

$

$$'

$$bwhere

mm$$'

mm14$'$'.!1$$

min

$1

$1

,

=

====

=++

=+=+

= eff

b

b

bieff

eff b

mmba

b

bbb

bb

b

Rhere%

mm1$!6'.$.mm'.!1$

!6'.1.$

1$'$.1.$. $,1$,1

=


17/21

Fig.

$1$.00.104!14$'

14.6'4$!$

0

$

1 ==

=

cdeff fdb

M

S=.$1$

4$'40'

00.104!14$'$1$. ===

yd

cd

effslf

fdbA mm$

8ne chooses 0G14 Aaeff= 4!$ mm$

Inte%io% !pans

"he width of the flange is%

mm64!&.$.mm'.'$$!&.1.$

1$'

$.$,1

===

b

eff

b

b it results that neutral ais passes through the

flange and the section is considered as rectangular cross section having the width eual tobeff

1!&.

00.104!1$4'

1'6&0$

0

$

$ =

=

=

cdeff fdb

M from "able ......it results%

S=.1&'

0&40'

00.104!1$4'1&'. ===

yd

cd

effslf

fdbA mm$

8ne chooses $G14 Aaeff= 06 mm$


18/21

!tee" desin in fi%st s-ppo%t B

0&$.119.00.104!$

1&.!&1$lim$

0

$ ==+

=

1.+.3. Desin fo% s)ea%


19/21

"he minimum coefficient of transversal reinforcing is determined with the

relation%

6$.40'

$6.6.

min, ===

yk

ck

"f

f

"he maimum uantity of transversal reinforcement is%

!'.140'

00.10$'4.1'.'. 1

ma

===

y"d

cd

c"

s"

f

fb#

s

A

Rhere% .c"=1 for reinforced concrete

v1=.! '4.$

$1!.

$1 =

=

ckf

"he maimum distance among stirrups on longitudinal and transversal direction

is%

( )

=>===

==+

=

$04'4!&'.&'.

0

04'4!&'.1&'.

,ma),

ma),

efftt

l

/mmd/

mm

mmctgd/

!-ppo%t B "eft

"he design shear force is%

0dq(( leftBleft

redB 69&9$4!'.$!1$11&99, ===

"he capable shear force of the member without specifically shear reinforcement is

computed with relation%

( ) dbkfC( cpckc1dc1d += 101

11,, 1

Rhere =cp aial force is neglected

1$.'.1

16.16., ===

c

c1dC

.11 = for ordinary concrete

$404.14!

$1

$1


20/21

Rhere% $!&.$40.10'.0'. $0

$0

min === ckfk#

0( c1d $4'!44!$$!&.min, == =$4.'!4 T

20(20( 1dcc1d '!4.$4!1.0$ min,, =>=

3ecause the design shear is bigger than the capacity of the beam without reinforcementfor shear%

20(20( c1dleft

redB !1.0$&9$.69 ,, =>= "he shear reinforcement is

necessary.

"he capacity of compressed struts (1d& ma+is determined for the maimum value of

ctg3=%4

0tgctg

f#!b( cdc"1d0

1ma, 1'.$'

'.$

1'.$

100.10'4.414$1

1=

+=

+=

where #=.9d

*t can observe that J3redleft is between the two values but closer to the inferior

limit, so in this case the shear force is reduced to medium, the minimum uantity of

transversal steel can be obtained for high value of ctgU.

*t can adopt for ctgU=1.&'

"he distance among stirrups for a diameter of mm! and #=.9d=.9V4!=414

is given by%$

,!.'!0.$6$ mmA "s == (two legs)

mm(

ctgf!Asleft

Bred

y"ds"$9.196

69&9$&'.140'414!&.'! ===

"he capacity of compressed struts for ctgU=$ is%

0tgctg

f#!b( cdc"1d0

1ma, 14.$06

.$

1.$

100.10'4.414$1

1 =+

=+

=

left

Bred1d (( >ma),

"he transversal steel percentage is%

6$.141'.$$

!.'!min, =>=

=

= bs

As"

eff"

"he basic anchoring length lb&rqd for the resisting steel L1! used at superior fiber

is obtained from "able+J*.'.

lb,rd=1$& mm


21/21

"he design anchoring length is%

min,,'40$1 brdqbbd lll >=

Rhere coefficients W are obtained from "able '.9

W1 =1 for straight bars

( ) 6&.1!

1!01'.11'.1$ =

== dc

Rhere cd=min(a$, c1, c)=min('4,0)

a=$:$V0:$V1!=16 mm

W0= W4=1.

1' = without transversal pressure

mmlbd 11'1$&1116&.1 ==

bdb

brdq

bd lmml

mm

mm

mml

l

slab and beam ihfds

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