beam design

7/17/2019 Beam Design

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BEAM DESIGN

ERT352FARM STRUCTURES

It doesnt matter what the subject is; once youvelearnt how to study, you can do anything you want.


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INTRODUCTION Steel beams of various cross-sectional shapes are

commercially available. Sections of steel beams are indicated ith a

combination of letters and a number. !"# $%& ' (%) U* ) $%& mm by (%) mm

universal beam ei"hin" )& +",m.


3/84

N /SIS 0 O DDISTRI*UTION

oads from slab are normally de1ned in 234+N,m ) . These loads are transferred to supportin" beams

in either 2 4+N,m or 254 +N. oads from reinforced concrete solid slab#

o oads may be distributed to the supportin" beamsdependin" on the ratio of lon" side,short side of the slab6 y, 78

o Type of distribution loads from slab# One- ay spannin" slab9 hen y, 7 : ).% T o- ay spannin" slab9 hen y, 7 ; ).%


4/84

N /SIS 0 O D DISTRI*UTION6cont.8

oad from precastconcrete slab

o oads may bedistributed to the

supportin" beams inone direction only andnot dependin" on theratio of y, 7.

o This is becauseprecast concrete slabare one- ay spannin"as they are supportedby beams at the endsof the slab only.


5/84

N /SIS 0 O DDISTRI*UTION 6cont.8

(. One- ay spannin" slab 6 y, 7 : ).%8o


6/84

N /SIS 0 O DDISTRI*UTION 6cont.8

). T o- ay spannin" slab 6 y, 7 ; ).%8o


7/84

!7ample (# oad distributionand beam analysis

The follo in" 1"ure sho s s =oor plan of a steel buildin". The =oor consists of precast concrete hollo core slabs.

oad carried by the slab are follo s#o Unfactored dead load from self ei"ht of precast slabs9

self ei"ht of steel beams and 1nishin" > &.% +N,m)

o Unfactored imposed load > ?.% +N,m ) Determine the ma7imum shear force and moment in beam

(, -*.


8/84

Solution# oad distribution#

o for precast slab9 considered as one- ay spannin" and it issupported by beam * and CD.

Desi"n load9 nn > (.$&@+ A (.&B+ > (.$& 6& +N,m ) 8 A(.& 6?.% +N,m ) 8 > (). & +N,m )

Total desi"n load9 > n 7 idth of load transferred to beam (, -*

> (). & +N,m ) 7 6) m8 > )&.& +N,m


9/84

Solution# 6cont.8 Since beam * is simply supported and loads are

symmetry9 then#o a7imum shear force9 E !d > 6 8,)

> 6)&.& +N,m86& m8,)> 63.75 kN

o a7imum moment9 !d > 6 ) 8,F> 6)&.& +N,m86& m8 ) ,F> 79.7kNm


10/84

!7ample )# oad distributionand beam analysis

The follo in" 1"ure sho s the plan of a buildin". The slab is cast in-situ reinforced concrete.Unfactored dead and imposed loads are sho n onthe dra in". In addition9 beam ), -* alo carries

?m hi"h bric+ all of $ +N,m). determine thema7imum shear force and moment of beam ), -*.


11/84

Solution# oad distribution on reinforced concrete solid slab

o Slab (# y, 7 > m,?m > (. & G ).%9 t o ay spannin"slab

o Slab )# y, 7 > m,$m > ).$ : ).%9 one ay spannin"slab


12/84

Solution#6cont.8 oad from slab (# beam ), -*

o Desi"n load9 n ( > (.$&@+ A (.&B+> (.$& 6& +N,m ) 8 A (.&6).%+N,m ) 8> H. &+N,m )

o Total desi"n load9 5( > n( 7 trape odial area of loadfrom slab (

> H. &+N,m) 7 J6$m A m8,) 7 )mK

> H .&+N


13/84

Solution#6cont.8 Since beam is simply supported and loads are

symmetry9 then#o a7imum shear force9 E !d( > 6 8,)

> 6H .& +N,m86& m8,)> 48.75 kN

o )ma > )m , m > %.)FM

o a7imum moment9 !d( > J6$ 0 ? a ) 8 ,)? 6( - a8K 5 > J6$ 0 ? 6%.)FM ) 8 ,)? 6( - %.)FM8K H .&+N

6 m8 > 106.5 kNm


14/84

Solution#6cont.8 oad from slab )# beam ), -*

o Desi"n load9 n ) > (.$&@+ A (.&B+> (.$& 6& +N,m ) 8 A (.&6$.%+N,m ) 8> ((.)& +N,m )

o Total desi"n load9 5) > n) 7 idth of slab )> ((.)& +N,m ) 7 (.& m> (M.H +N,m


15/84

Solution#6cont.8 oad from beam self ei"ht and bric+ all

o Self ei"ht of beam > ( +N,mo 5ei"ht of bric+ all per meter len"th > ei"ht of bric+ all 7 hei"ht

of all

> $ +N,m)

7 ? m > () +N,mo Total dead load9 @+ > self ei"ht of beam A ei"ht of bric+ all

> ( +N,m A () +N,m> ($ +N,m

o Desi"n load9 $ > (.$&@+ > (.$&6($+N,m8 > ( .&& +N,m

o Total desi"n load for slab )9 bric+ all and self ei"ht of beam9? > ) A $ > (M.H +N,m A ( .&& +N,m

> $?.& +N,m


16/84

Solution#6cont.8 Since beam * is simply supported and loads are

symmetry9 then#o a7imum shear force9 E !d) > 6 8,)

> 6$?.& +N,m86 m8,)> 120.75 kN

o a7imum moment9 !d) > 6 ) 8,F> 6$?.& +N,m86 m8 ) ,F> 211.3 kNm


17/84

Solution#6cont.8 oad combination from slab ( and slab )

o The ma7imum shear force and moment from beam ), -*is obtained by combinin" the loads from slab ( and slab) to"ether as follo #

o a7imum shear force9 E !d > E !d( A E !d)> ?F. & A ()%. &> 169.5 kN

o a7imum moment9 !d > !d( A !d)> (%M.& A )((.$> 317.8 kNm


18/84

*eam Desi"n Section Classi1cation

o ny steel beam sections that are sub ect to compression due tobendin" or an a7ial force should be classi1ed.

o

The purpose of the classi1cation is to determine hether localbuc+lin" in=uences the capacity of the beam.

o The occurrence of local buc+lin" of compressed elements of across-section prevents the development of full section capacity.

o The classi1cation of a section is carried out by comparin" theidth-to-thic+ness ration of the elements9 i.e. c/t ! t"# $%&'#

#(#m#&t and c/t ! t"# )#* #(#m#&t a"ainst the limit of c,t ofthe =an"e and c,t of the eb respectively "iven by T%*(# 5.2EC3


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l


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!7ample# Sectionclassi1cation

Determine the classi1cation of a ?%M 7 (?% 7 ?MU* in "rade S) &.

Solution:o Refer Table $.( pa"e )M#

Steel "rade9 S) & t G ?% mm fy > ) & N,mm )

o Refer Table &.) pa"e ?)# > P 6)$&,fy8 > P 6)$&,) &8 >%.H)

o &$.% G ) > )6%.H)8 > MM.) eb is Class(

R i d d


23/84

Restrained andUnrestrained *eams

Steel beams may be desi"n as either restrainedor unrestrained.

If a beam full lateral restraint to its compression=an"e alon" the span9 the beam is consideredfully restrained.

Cases here beams can be desi"ned as fullyrestrained alon" the spans are#

o *eams carryin" in-situ reinforced concrete slabo *eams ith steel dec+in" =oorin" systems9 ith orithout shear studs. Shear studs function as a simple

concrete anchor.


24/84

Restrained and Unrestrained *eams6cont.8

De=ection of restrained beamo The restrained beam is fully prevented from movin"

side ays.o The de=ection ill only ta+e place vertically about the

ma or a7is ithout any lateral de=ection.

Types of restrainto There are t o conditions of restraint#


25/84

Desi"n of Restrained*eam

The desi"n process for fully restrained beam isfollo s#

o naly e the beam and determine the reaction 6R8 Lma7imum shear force 6E !d 8L ma7imum moment 6 !d 8 and

ma7imum de=ection.o Select suitable steel U* section.o Classify the sectiono Chec+ the bendin" moment resistanceo Chec+ shear resistanceo Chec+ shear buc+lin" resistance of ebo Chec+ combination of shear and momento Chec+ de=ection

* di " 9


26/84

*endin" oment9Clause M.).& !C$

The desi"n resistance for bendin" for classes (and ) cros sections


27/84

Shear9 Clause M.).M!C$


28/84

Shear area for secondary beam9 v

o

5here > "ross sectional area of the section

Shear *uc+lin" Resistance of 5eb9 Clause M.).M 6M8!C$

o If 9 shear buc+lin" resistance no need tobe carried out.

o 5here9 depth of eb9

)2(9.0 f v bt A A =

72

0,

)1(

m

y pl Rd c

f W M

=2

,

12 =

Rd pl

Ed

V V


30/84

!7ample# *eam * ith span &.% m is simply supported at

and *. )?%

+Nm. Usin" "rade S) &9 Desi"n beam *.


31/84

Solution# Trial section si e based on moment resistance9 5pl

So9 try section ?%M 7 (?% 7 ?M U* in "rade S) & 65pl >FFFcm $8

3

33

2

6

2

7.872

102.872/275

10240/275

240

cm

mm xmm N Nmm x

mm N kNm

f M

W y

Ed pl

=

=

==

=


32/84

Solution# 6cont.8 Section properties ?%M 7 (?% 7 ?M U*

h > ?%$.) mm d > $M%.?mmb > (?).) mm > &F.M cm )

t > M.F mm t f > ((.) mmIy > (& %% cm ? I > &$F cm ?

5 pl9y > FFF cm $ 5 el9 y > F cm $

d,t > &$ c,t f > &9($! > )(%%%% N,mm ) @ > F(%%% N,mm )


33/84

Solution# 6cont.8 Desi"n stren"th

o Table $.(# Steel "rade S) &o TG ?% mm f y > ) & N,mm ) 9 f u > ?$% N,mm )

Section classi1cationo Refer Table &.) pa"e ?)# > P 6)$&,fy8 > P 6)$&,) &8 >

%.H)o &$.% G ) > )6%.H)8 > MM.) eb is Class

(

o Since =an"e and eb are class (9 the section is classi1edas Class (.


34/84

Solution# 6cont.8 Shear Resistance of Section 6Clause M.).M8

o a7imum e7ternal desi"n shear force9 E!d > (H) +No Shear resistance of section9 E c9 Rd > E pl9 Rd

o


35/84

Solution# 6cont.8 Shear area for main beam9 v

Therefore9 use the bi""est v > )H H.$ mm )

2,

2 4.25893.2979

6.3043.31855860

2.11)]2.10(28.6[)2.11)(2.142(25860

)2(2

mm Amm

t r t bt A A

webv

f w f v

=>=+=

++=

++=


36/84

Solution# 6cont.8

Desi"n chec+

section is satisfactory.

kN x

f AV V

m

yv Rd pl Rd c

473101

)]3/275(3.2979[

)3/(

3

0,,

=

=

==

0.141.0473192

,

== Rd c

Ed

V V


37/84

Solution# 6cont.8 *endin" oment Resistance of Section 6Clause

M.).&8o a7imum e7ternal desi"n moment9 !d > )?% +No oment resistance for class ( cross-section#

kNm

x x

f W M

m

y pl Rd c

2.244

101)275)(10888(

6

3

0,

=

=

=

kNm M M Rd pl Rd c 2.244,, ==

0.198.02.244

240

,

== Rd c

Ed

M M


38/84

Solution# 6cont.8 Shear *uc+lin" Resistance of 5eb 6Clause

M.).M6M88o If 9 shear buc+lin" resistance no need to be

carried out.

o 5eb depth 9

o So9

local eb buc+lin" is li+ely to occur and hence shearbuc+lin" chec+ no need to be carried out.

72 )?% +NL

Shear force at ma7imum moment is E > % +N. %.& Ec9Rd > %.& 7 ? $ > )$M.& +N

Since E > % +N G %.& Ec9Rd9 it does not a ect the momentresistance9 c9Rd.

o This beam section is able to carry the most criticalcombination of bendin" and shear.

0.198.02.244

240

,

== Rd c

Ed

M M


40/84

Unrestrained *eam The possibility of lateral-torsional buc+lin" must

be ta+en into consideration hen thecompression =an"e of the beam is not fullyrestrained alon" the span.

The buc+lin" capacity of unrestrained beamdepends on the

o Section type.o

Unrestrained len"tho Restraint conditiono Type of load applied

D i" f U t i d


41/84

Desi"n of Unrestrained*eam

The follo in" are the steps for desi"nin"unrestrained beams#

o Divide the beam into se"ments bet een lateralrestraints

o Chec+ the moment buc+lin" resistance for each se"menthere is "iven asL

o Section modulus9 5yL 5y > 5pl9yL lastic modulus for Class ( and Class )

sections 5y > 5el9y9 !lastic modulus for Class $ section

1,

m

y y LT Rd b

f W M

=


42/84

The reduction factor T9 is "iven by

5here9o Non dimensional slenderness9

o *uc+lin" parameter9

22

1

LT LT LT

LT

+=

cr

y y LT M

f W =

])2.0(1[5.02 LT LT LT LT ++=


43/84

The critical elastic buc+lin" moment9

5here#o !9 @ are material propertieso I 9 It9 I are section propertieso cr is the buc+lin" len"th of the membero C( is the factor that depends on the shape of bendin" moment

dia"ram

5.0

2

2

2

2

1 += z

t cr

z

w

cr

z cr

EI

GI L

I

I

L

EI C M


44/84

If the moment alon" the beam is not uniform9then cr is modi1ed by the C ( factor

The modi1cation factor9 C ( is "overned by theratio V hich is the ration of the smaller to lar"ermoments of restrained ends.

The ratio of smaller moment to lar"er moment is"iven as

V > small , lar"e for restrained ends

The C ( factor can be obtained from the follo in"table refer *S&H&% .


45/84


46/84

The C ( factor is never less than (.%.

If C( > (.%9 its means the moment alon" the

beam is uniform.

C( > (.FF 0 (.?% V A %.&) V )

9f d i" i "


47/84

9cr for desi"nin"unrestrained

beams Typical e7ampleof beams ithoutintermediatelateral restraintsand theircorrespondin"buc+lin" len"ths.


48/84

intermediate lateral

restraints Simple supported beams ith intermediate lateralrestraints are described as beams hereo ateral restraints are provided at beam supports and at

intermediate of beams


49/84


50/84

The C ( factor#o V > small , lar"e for restrained endso C( > (.FF 0 (.?% V A %.&) V )

!7ample# Desi"n of


51/84

!7ample# Desi n ofUnrestrained *eam


52/84


53/84

Solution# Desi"n load9 3 > (.$&@+ A (.&B+

> (.$&6$.%+N,m ) 8 A (.&6&.%+N,m ) 8> ((.&&+N,m )

Reaction of beam CD at support C> point load at point C of beam !> 63 7 idth 7 len"th8,)> 6((.&&+N,m) 7 ).& m 7 M m8,)> FM.M +N

Self- ei"ht beam !9 > &$+",m 7 H.F( 7 (%-$> %.&) +N,m


54/84


55/84

Section properties ?%M 7 (?% 7 &$ UW*h > ?%M.M mm d > $M%.?mmb > (?$.$ mm > M .H cm )

t > .H mm t f > ().H mmIy > (F$9%% cm ? I > M$& cm ?

I > %.)?M dm ? It > )H cm ?

5 pl9y > (%$% cm$

5 el9 y > FHH cm$

d,t > ?&.M c,t f > ?.?M

! > )(%%%% N,mm ) @ > F(%%% N,mm )

cr > ).& m r > (%.) mm


56/84

Desi"n stren"tho Table $.(# Steel "rade S) &o TG ?% mm f y > ) & N,mm ) 9 f u > ?$% N,mm )


%.H)o ?&.M G ) > )6%.H)8 > MM.) eb is

Class (



57/84

Solution# 6cont.8

Shear Resistance of Section 6Clause M.).M8o a7imum e7ternal desi"n shear force9 E!d > ??.M +No Shear resistance of section9 E c9 Rd > E pl9 Rd

o 5eb shear area

o So9 shear are of eb only

0,,

)3/(

m

yv Rd pl Rd c

f AV V

==

wwwebv t h A =,

mmmmmm

t hh f w8.380)2.11(22.403

2

===

2

3.3008

)9.7)(8.380(0.1

,

mm

mmmm

t h A wwwebv

=

==


58/84

Solution# 6cont.8 ore accurate calculation of shear area

considerin" the 1llet area

Therefore9 use the bi""est v > $?&F mm )

2,

2 3.30083458

1.3651.369767909.12)]2.10(29.7[)9.12)(3.143(26790

)2(2

mm Amm

t r t bt A A

webv

f w f v

=>=+= ++=

++=


59/84

Solution# 6cont.8

Desi"n chec+

section is satisfactory.

kN x

f AV V

m

yv Rd pl Rd c

549101

)]3/275(3458[

)3/(

3

0,,

=

=

==

0.108.0549

6.44

,

== Rd c

Ed

V V


60/84


M.).&8o a7imum e7ternal desi"n moment9 !d > (%H.H +No oment resistance for class ( cross-section#

kNm x

x

f W M

m

y pl Rd c

3.283101

)275)(101030(6

3

0,

=

=

=

kNm M M Rd pl Rd c 3.283,, ==

0.139.03.2839.109

,

== Rd c

Ed

M M


61/84

Solution# 6cont.8 ateral torsional buc+lin" 6 T*8

o a7imum e7ternal desi"n moment9 !d > (%H.H +Nmo oment buc+lin" resistance

o

If the moment alon" the beam is not uniform9 then cris modi1ed by the C( factor.

1,

m

y y LT Rd b

f W M =


62/84

o Determine the modi1cation factor9 C(small > % +Nm9 lar"e > (%H.H +Nm

V > small , lar"e > %,(%H.H > % for restrained ends >

%

C( > (.FF 0 (.?% V A %.&) V )

> (.FF 0 (.?%6%8 A %.&)6%8 )

> (.FFo !lastic critical buc+lin" moment

5.0

2

2

2

2

1 += z

t cr

z

w

cr

z cr EI

GI L I I

L

EI C M


63/84

( dm > (%% mm9 hence ( dm M > (% () mm M

Non-dimensional slenderness for T*

kNm

Nmm x

mm x xmm N

mm x xmm N

xmmmm x

mm xmm

mm x xmm

N

EI GI L

I I

L EI

C M z

t cr

z

w

cr

z cr

3.884

1030.884

)10635()21000(

)1029()81000()2500(1063510246.0

)2500(

)10635()21000(88.1

6

5.0

442

2

44

2

2

44

612

2

44

2

2

5.0

2

2

2

2

1

==

+

=

+=

566.0103.884

)/275(1010306

233

=== Nmm x

mm N mm x M

f W

cr

y y LT


64/84

o Determine the buc+lin" parameter9 X TImperfection factor9 Y T for T* curves

h,b > ?%M.M mm,(?$.$ mm > ).F? :)curve Zb[ \ hence9 Y T > %.$?

o Reduction factor9 ] T for T*

722.0

]566.0)2.0566.0(34.01[5.0])2.0(1[5.0

2

2

=++=

++= LT LT LT LT

855.0566.0722.0722.0

112222

=+

=+

= LT LT LT

LT


65/84

oment buc+lin" resistance

Desi"n chec+ section is satisfactory

kNm x

mm N mm x

f W M

m

y y LT Rd b

18.242100.1

)/275()101030(855.0 62

33

1,

=

=

=

0.145.018.2429.109

,

== Rd c

Ed

M M

!7ample# Desi"n of


66/84

!7ample# Desi n ofrestrained beam


67/84

l


68/84

Solution#

l


69/84

Solution# Section properties &$$ 7 )(% 7()) UW*

h > &??.& mm d > ? M.&mmb > )((.H mm > (&& cm )

t > (). mm t f > )(.$ mmIy > M9%%% cm? I > $$H% cm ?

I > ).$) dm ? It > ( F cm ?

5 pl9y > $)%% cm $ 5 el9 y > ) H% cm $

d,t > $ .& c,t f > ?.%F! > )(%%%% N,mm ) @ > F(%%% N,mm )

cr > ?.% mr > (). mm

S l i 6 8


70/84

Solution# 6cont.8 Desi"n stren"th

o Table $.(# Steel "rade S) &o TG ?% mm f y > ) & N,mm ) 9 f u > ?$% N,mm )


%.H)o $ .& G ) > )6%.H)8 > MM.) eb is

Class (


S l i # 6 8


71/84


M.).&8o a7imum e7ternal desi"n moment9 !d > F)( +Nmo oment resistance for class ( cross-section#

kNm x

x

f W M m

y pl

Rd c

880101

)275)(103200(6

3

0,

=

=

=

kNm M M Rd pl Rd c 880,, ==

0.193.0880821

,

== Rd c

Ed

M M

S l i # 6 8


72/84

Solution# 6cont.8 ateral torsional buc+lin" 6 T*8

o a7imum e7ternal desi"n moment9 !d > F)( +Nmo oment buc+lin" resistance

o

If the moment alon" the beam is not uniform9 then cris modi1ed by the C( factor.

1,

m

y y LT Rd b

f W M =


73/84

o Determine the modi1cation factor9 C(small > H? +Nm9 lar"e > H? +Nm

V > small , lar"e > H?, H? > ( for restrained ends

C( > (.FF 0 (.?% V A %.&) V )

> (.FF 0 (.?%6(8 A %.&)6(8 )

> (.%o !lastic critical buc+lin" moment

5.0

2

2

2

2

1 += z

t cr

z

w

cr

z cr EI

GI L I I

L

EI C M


74/84

( dm > (%% mm9 hence ( dm M > (% () mm M

Non-dimensional slenderness for T*

kNm

Nmm x

mm x xmm N

mm x xmm

N

xmmmm xmm x

mm

mm x xmm

N

EI GI L

I I

L EI

C M z

t cr

z

w

cr

z cr

46.1397

1046.1397

)103390()21000(

)10178()81000()4000(103390

10232)4000(

)103390()21000(0.1

6

5.0

442

2

44

2

2

44

612

2

44

2

2

5.0

2

2

2

2

1

==

+

=

+=

794.01046.1397

)/275(1032006

233

=== Nmm x

mm N mm x M

f W

cr

y y LT


75/84

o Determine the buc+lin" parameter9 X TImperfection factor9 Y T for T* curves

h,b > &??.& mm,)((.H mm > ).M :)curve Zb[ \ hence9 Y T > %.$?

o Reduction factor9 ] T for T*

916.0

]794.0)2.0794.0(34.01[5.0

])2.0(1[5.02

2

=++=

++= LT LT LT LT

728.0794.0916.0916.0

112222

=+

=+

= LT LT LT

LT


76/84

oment buc+lin" resistance

Desi"n chec+ section is not satisfactory9

hence9 lar"er section should be used.

kNm x mm N mm x

f W M

m

y y LT Rd b

4.640100.1 )/275()103200(728.0 6

2

33

1,

=

=

=

0.128.14.640

821

,

>== Rd c

Ed

M M

D i


77/84

De=ection beam may not fail due to e7cessive de=ection9

ho ever9 it is necessary to ensure that de=ections arenot e7cessive under unfactored imposed loadin" toprevent#

o Dama"e to various architectural features such as interior alls9

partitions9 ceilin"s and e7terior claddin"o Severe crac+in" in brittle 1nishes such as bric+ all ith plaster

1nishedo Dama"e to non-structural element such as "lass fa^ade9 ceilin"s9

partitions and other fra"ile elements.

Eertical de=ection limito The follo in" table "ives su""ested limits for calculated vertical

de=ections of certain members under the characteristic loadcombination due to variable loads and should not includepermanent loads.


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Desi"nsituation

Eertical De=ectionlimit


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a7imum Eertical De=ection due to e7ternal loado The follo in" table sho s the formula to calculate

vertical de=ection at mid-span of a simply supportedbeam ith di erent loadin" types.

o The formula can be used to calculate de=ection of arestrained or unrestrained beam


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!7 l #


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!7ample# Chec+ de=ection of beam sho n in 1"ure belo

S l ti #


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Solution# De=ection is chec+ed under serviceability. Therfore9 only unfactored imposed load 6(.%B+8

are considered in calculatin" the de=ection. Deadloads are not included

*eam ?%M 7 ( F 7 M% U* Clause $.).M !C$#

o odulus of !lasticity9 ! > )(% %%% N,mm )

)( M%% cm ?

S l ti #6 t8


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Solution#6con.t8 a7imum de=ection due to unfactored imposed loads

If the beam is carryin" all ith plaster 1nish or anyother brittle 1nish9 de=ection limit9

choose lar"er beam section

mm

mmmmmm xmm N

x

mm xmm N

mmmm N EI

wL EI

wLc

2.28

8.134.14)1021600)(/210000(48

)10000(1030

)1021600)(/210000(384

)10000)(/5(548384

5

442

33

442

4

34

=+=

+=

+=

mmmm L

8.27360

10000360lim

===

mmmmc 8.272.28 lim =>=


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To acquire knowledge, one must study; but to acquire wisdom, onemust observe .

TQ NW /OU

beam design

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