afraz structural

8/10/2019 Afraz Structural

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Acknowledgement

I am most humbly grateful to our lecturer Mr. Mihiran Galagedara at the I.C.B.T, for his kind

guidance and valuable instructions given to me in order to carry out this assignment successfully.

Last but not least, with great love I remember my arents for continuous encouragement incomleting this assignment. !nd also I get this oortunity to thank for my friends, who suort

me to comlete this assignment.

"


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Content

Acknowledgement ##### ##############.... $"

Content ###....###################### $% &$%

'uestion $" #######################.... $( & $)

'uestion $% #....###################### $* &$*

'uestion $( ######################## "$ &"+

'uestion $+ ######################## " & "-

'uestion $ #######################.... ") &%"

'uestion $ #....###################### %% & %+

'uestion $- ######################## % & %-

/eference ######################## #%)&%)

%


3/28

Design 1

012IG3 M4M13T,M

Loading

Dead

2elf weight of beam 5 $.( 5 +.% 63m&"

Total dead load 7gk8 5 %$ 9 +.% 5 %+.% 63m&"

Imposed

Total imosed load 7qk8 5 %" 63m&"

Ultimate load

Total ultimate load 7W8 5 7".+gk 9 ".qk8san

5 7".+ : %+.% 9 ". : %"85 (+$.(+ k3

Ma;imum design moment 7M8 5


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!ssume diameter of main bars 7 8 5 % mm

!ssume diameter of links 7 8 5 ) mm>rom Table 3.6, cover for e;osure class ?C" 5 " 9 c 5 % mm.

Effective depth

d 5 h @ c &

A&

= %

d5 $ @ % @ ) @ "%.

5 $+. mm

K = M/bd

6 5%"%.-" ; "$(=7($ ; $+.%8

65 %.+

ence from Table 3.10Interellation%.$%.+(.$

".%".$+

+


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? 5 ".% &7%&%.+8 7".% &"8

7 % &( 8

Correction factor7?8 5 "."+

Ultimatemoment

Mu5 $."fcu: bd%

5 $." : ($ : ($ : $+.%

5 +".$) : "$3mm5 +".$) 63m

M!" M#inceM! " M no comp$ession $einfo$cement is $e%!i$ed&

MAIN STEEL, As

Coefficient

6 5 M=7fcu.b.d%8

6 5 7%"%.-" ; "$8=7($ ; ($ ;$+.%86 5 $.$-*3mm&(

Lever arm

D 5 d7$. 9 7$.% @ k=$.*8"=%8D

D 5 $+.7$. 9 7$.% @ $.$-*=$.*8"=%8D 5 +.+mm

Check the E value


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D F $.*d

+.+ F +-*.%-

' val!e is ok

Main steel As ($e%)

!s 7re8 5 M=$.)->HD

!s 7re8 57 %"%.-" ; "$8=7$.)- ;$$ ; +.++8

5"$-(.*" mm%

ence from Table 3.10, rovide +%$ 7As 5 "%$ mm%

8.3umber of beams 5 +

Bar siEe 5 %$mm

Shear stress,V

Vx=(/!9 /B8=% 5 +"=%63

= 208KN

V = Vx

V =

V= 1.178

V ermissible 5 $ .) J 5 +.()


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Designconcrete shear stress,Kc

Kc=

Kc=

=0.71

Diamete$ and spacing of links


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ermissible san=effective deth ratio 5

5

5 %%.)

*.*" %%.)

!ctual san/effective deth ratio ermissible san/effective deth ratio

Acco$ding to bs *11+ the beam will not fail d!e to deflection

= ,,&-.

resent a suitable techniue to minimiEe the material cost of this beam with roer calculations.


9/28

Design

N 5 $.(fcuAc9 $.-fc Asc

7

"+&$ &.+% 9"$."%

!ssuming that the column is suare,

b

h 5 J

h 5(%*."mm

8

ence a ($ mm suare column constructed of concretefcu 5 ( 3mm%would be suitable.

*


10/28

Design -

!

"%63m&" dead load"%63m&" imosed load

/!

"$ m

(63 imosed load

/B

B

4verall loads 5 ". gk9 ".+ k

5 7". : "%8 9 7".+ : "%8

5 ( 63

oint load 5 ( 63 : artial factor of loads

5( 63 : ".

5 63

7from able .&Partial factors for loads (Table 2, BS 5950)

Considering >orces

/!9 /B5 63 9 (63m&":"$m

/!9 /B5 +"63

! /B: "$m 5 63 : m 9 ($63 : m

/B5 %$) 63

/!5 %$) 63

"$


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$ ?

!

%$)

?

T K;M

K;5 %$)63 @ (? 63

T M 5 %$)63 : ?m & ( 63 : ?

M 5 %$)? 63m & ") ?%63m

$ ? "$

63

!

?

%$)

2

K;

M

K;5 %$) 63 &63 @ ( ? 63

2 M 5 %$)63 : ? & ( 63 : ? & : 7?&8m

M 5 "%? 63m & ") ?%63m 9 %)$ 63m

Ma;imum bending oint?5

M 5 "%? 63m & ") ?%63m 9 %)$ 63m

M 5 *$63m

""


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M

*$63

%$)

%)

&%)

&%$)

Bending moment diagram

2hear force diagrams

"%

;

?


13/28

I3ITI!L 21CTI43 21L1CTI43

!ssumingpy5 %- 3=mm%

M5 ma;imum bending moment

5 design strength of steel

2;

2;

2; %."+ "$&( m(

2; %"+ cm(

3e;t using the able 01Dimensions and properties of steel universal beams (structural sections to BS4: Part 1).we can select the suitable beam sections.

2uitable sections areO

". +- : "*" : *) PBO S; 5 %%($ cm(

%. ($ : ($ :"(- PBO S; 5 %($$ cm(

(. (( : %"$ : *% PBO S; 5 %(-$ cm(

Check the beam of deflection

select rofile

Check whether the selected beam section is suitable to cover with a laster or any kind of

brittle. 7using table +.8

If laster O

"(


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2ma;5

5 %-.-- mm

2tranded case and relevant deflection

2actual 5

5 $.$"- m 9 $.$$--)m

5 %+.+)mm

2ma; 2actual

Check for deflection under other beams 7 Table +.8

4ther beams 5

5

5 $ mm

There have many tye of beams. But we ne to think our scoe.


15/28

Design .

Psing table +.% artial factors for load in lecture notes 7Table %, B2 *$8

Tye of load and load combination

0ead loadImosed load

>actored load 7>c8 5 ".+gk9 ". k

>actored load 7>c8 5 %"$$ 9 "$$

5 (-$$ 63

1ffective length 7 8, from table +."

5 $.-L

5 $.- : -

5 +.* m

".+".

"

>actor


16/28

2ection using the table of dimensions and roerties of steel universal beam,

Initial trial., 254:254:107 UC:

y 5 %63m

Q 5

ry 5 .- cm !g 5 "(-$$ cm b=t 5 .( d=t 5 ".+

5 -.)

c 5 !

5 "$$$ : %$)=

5 ("%$

("%$ (-$$, so this is not ok

second trial.,305:305:118 UC:

y 5 %63m ry 5 --. cm !g 5 "$$$ cm

Q 5

5 (.%%

c 5 "*" 3=

c 5 !

5 "$$$ : %%%=

5 ((($

((($ (-$$, so this is not ok

"

b=t 5 ).% d=t 5 %$.-


17/28

Third trial. Try($ : ($ : ")

y 5 % ry 5 -).* cm !g 5 %"$$$ cm

Q 5

5 %

c 5 "*" 3=

c 5 !

5 %"$$$ : "*"=

5 +$""

+$"" (-$$, so this is ok

b=t 5 .%" d=t 5 ".-

($ : ($ : ") section beam used our building, there have lots of beams sections, but we wantdecide to what is best for our scoe. gross cross&sectional area of section is main factor of thissection changed, its mainly connect with width ,length, Inertia and thatAs why thinks, I chooseits over ($$ cross sections, its rise direct change in this value, our >actored load is high, andne;t i think good weight may be over "(. !nd medium every values good for our house. Thisis single story house, we donAt need high sectional beams.

"-


18/28

Design

EFFECTIVE SPANClear spam =3000mm

bearing length= 150mm

effective san 5 ("$mm

2e$missible deflection

7

7 8

longer-span domestic floor joists,

0eflection due to bending

0eflection due to bending

8

")


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2te "

0eflection due to bending ermissible deflection0eflection due to bending *.+: "$&(

I

I

7

7

8

8

"*


20/28

I

using the able ,&* ,select the beam section

- : ($$

"$$ : ($$

this section can be consider as the most suitable

2te %

-:($$ section

%$


21/28

7

5 *."( mm 9 ".") mm

5"$.(%" mm

*.+

"$$:($$ section

7

5 .)) mm 9 $.*$% mm

5-.-)% mm

*.+

8

8

7

7

8

8

Therefore a beam with a section is adequate for bending and deflection. Its very suitable for our scopeand agree with all equations

"$$:($$ for beams our building.

%"


22/28

Design ,

#lende$ness $atio

Q = R

>rom table ."", 1ffective length of comression members

3est$ained at both ends in position b!t not in di$ection = 1&+

Effective length

Le5 ".$ : h

Le5 ".$ : ($$

Le5 ($$mm

he slende$ness $atio sho!ldnot e4ceed 1*+ fo$ comp$essionmembe$s ca$$5ing dead andimposed loads

i 5J

I 5

! 5 bd

%%


23/28

i 5 J

i 5 J

i 5 J

R

i = -1&6.

#lende$ness $atio

Q 5 R

Q 5

Q = 11+&-

GRADE STRESSESAND MODULUS OF ELASTICITY

>rom table .%, Grade G2 whitewood belongs to strength class C"

7omp$ession pa$allel to g$ain (8c&g&ll)

,&*

M9DI:I7AI9; :A793

5

= *&

Modification factor for comression members 7 6"%8

By interolation,

1++ 11+& 1+

*++ +&-61 +&*+

* +&--


25/28

Design 6

L4!0I3G

Pltimate design load,N 5 "+$ 63 m&" 5 "+$ 3 mm&"

Small plan areamodification factor does not apply since horizontal cross-sectional area of wall,

A5 $.%"m (.$m5 $.( m% $.% m%

Hence modified characteristic compressive strength is 1.15fk

#afet5 facto$ fo$ mate$ials (m)Manufacture and construction controls categories are, resectively, UIIA and UnormalA. encefrom Table 5.10, m forcomression 5 (.

7apacit5 $ed!ction facto$ (V)Eccent$icit5

2ince wall is a;ially loaded assume eccentricity of loading, e; $.$t

%


26/28

#lende$ness $atio

#lende$ness $atio5

5 $.- : height 5 $.- : ($$$ 5 %%$ mm

5 actual thickness 7single leaf8 5 %"$ mm

#lende$ness $atio

5"$.-"

ence, from Table 5.11, V 5$.* modified characteristic strength

Design ve$tical load $esistance of wall (;3)

01T1/MI3!TI43 4>

>or structural stability

%


27/28

21L1CTI43 4> B/IC6 !30 M4/T!/ TH1>rom Table 5.!a", any of the following brick=mortar combinations would be aroriateO

!ssuming the wall will be lastered on both sides, the aearance of the bricks is not an issue.

%-


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3efe$ences

Chanakya, !. 7%$$*8#esign of Str$ct$ral %lements, (rd edn, 2on ress.

ibbler, /.C. 7%$$+8 Statics and Mechanics of MaterialsN 2I ednN rentice& all, Inc.

Beer, >. ., 0e&

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