Steel Staircase Design

Design of steel staircase Design of step Design of Flat INPUT Density of steel Try Flat of size

Width Thickness

Tread of step !mber of Flats pro"ided per tread #rade of steel  %llo&able bendin' stress  %llo&able shear stress

b t T f y

 = 0))f y

σ bt

τ "m

 = 04f y

k1m3 mm mm mm mm

= = = = =

78 30 4 220 8

= = =

2$0 1mm2 ()$ 1mm2 (00 1mm2

*+,- 200

.oads (/ .i"e .oad = .i"e load on each flat = 2/ Dead load of each flat Tryin' a flat of crossection 3054 mm2/

=

$ k1m2 0($7(43 k1m 0003) k k 1m

-ombination Dead load  .i"e load/ on each flat

=

0())$03 k1m

=

00208(3 km

,a5im!m 6endin' ,oment on each & flal2t 1 8

,

=

 = * 1y = bd 1)

+ection ,od!l!s



=

)00 mm3

b y d 5

5

Flat section  %ct!al ma5 6endin' +tress in each flat

=

,1

et ffecti"e %res of crossection

% τ

 %ct!al shear stress

"

= =

34)88( 1mm2 O.K (20 mm2 0)37)2 1mm2 O.K

 %ct!al 6endin' stress and shear stress is .ess than %llo&able 6endin' stress and shear stress respecti"ely hence the tried section is ok

Design of Double Angle section Tryin' a section 2 *+% 4054054/ .en'th of 9!al an'le section

=

022 m

Load .i"e .oad Dead load of Do!ble an'le section Total Dead  .i"e load of Flats Dead load  .i"e load transfer to each do!ble an'le

= = = =

$ k1m2 0047 k1m (332023 : 302732$ k1m

=

302732$ :1m

,

=

00(83($ km

 *55

=

3(2$ mm3

Load Combination  Total Dead .oad  .i"e .oad/

,a5im!m 6endin' ,oment

& l21 8

+ection ,od!l!s

  = 2 ; *55 1 y

y y

4$ cm4 288 cm From teel Table

=

$8)( 1mm2

2 *+% 4054054

5

 %ct!al 6endin' +tress

= =

5

,1

O.K Net !ffecti"e Area of cross#section as per I$ %&&claus '.(.) For !ntacked do!ble an'le section % a"ailable = 2;  % of sin'le an'le / = 2; %(  : %2/  %( = l( > t12 > nd?/5t

%(

=

8 mm2

 %2 = l2 >t12/5t

%2

=

($2 mm2

: %

= =

0)$(3 3)34) mm2

τ

=

: = 3%(13%(%2/

 %ct!al shear stress

"

084 1mm2 O.K

 %ct!al 6endin' stress and shear stress is .ess than %llo&able 6endin' stress and shear stress respecti"ely hence the tried section is ok

Design of *olted connection input Thickness of &eb of *+,- 200 as per +teel Table Thickness of %n'le section  *+% 4054054 /  %ss!min' ominal Dia of 6olt 6olt in clearance holes

t& t d

= = =

)( mm 4 mm (2 mm


τ "f 

=

80 1mm2


σ pf 

=

2$0 1mm2

2 *+% 4054054 6ack to 6ack

t& *+,- 200

e= *+,- 200

t

20 mm 20 mm 2 *+% 4054054

*n calc!latin' the shear and bearin' stress of bolt the effecti"e diameter of a bolt is !sed that is its

nominal diameter as per *+  800 cla!se 8(

#ross dia of 6olt +tren'th of 6olt in 6earin'

d? d tmin σpf 

+tren'th of 6olt in sin'le shear

d 2; τ"f 1 4

π

6olt "al!e Total .oad &hich is to be transfered o of 6olt re9!ired

<

= = = = =

047787  048 k

=

048 k

=

0)))0(( k

=

<  6olt "al!e

(3$ mm (2000  (2 k

0073)(

~

( 6olt

,in d'e distance as per *+ 800 Table 82

emin r 

=

(7 mm


epro"ide

=

20 mm

b

= = =

4 mm 022 m )24 1m


Design of side plate$ Try a plate of cross>section 200 5 4/ .en'th of plate Dead .oad of
=

3(3)72$ :1m

,a5im!m 6endin' ,oment

,

=

00(877 km



=

2)))))7 mm3

=

07(()44 1mm2 O.K 800 mm2

& l21 8

+ection ,od!l!s

2

  = *55 1y = bd 1) b y d = 200 5

5

 %ct!al 6endin' +tress et ffecti"e %res of crossection

mm

,1 % τ

=

08)2$ 1mm2 O.K  %ct!al 6endin' stress and shear stress is .ess than %llo&able 6endin' stress and shear stress  %ct!al shear stress

"

=

respecti"ely hence the tried section is ok