Challenge

Week 3 Challenge

1) Height of column =10m

Restrained in mid height.

10m long column is restrained at ends in both Y-Y and Z-Z direction.

Hence the actual length=5m

a) Braced frame and shear wall

The effective length factor for bottom column is fixed and top of column is free.

=2xL

=2X5

=10m

b) Unbraced frame with a pinned base

=2xL

=2x5

=10m

c) Unbraced frame with fixed base

=1.2xL

=1.2X5

=6m

2) For rolled steel section

Fy=250N/mm2

Fu=410N/mm2

E=2X105N/mm2

For top roller and bottom fixed

KXL=1.2X3=3.6m

For ISMB300

h=300, bf=200, Lf=12mm

h/bf=300/200=1.5

h/bf>1.2, tf<=40mm

It belongs to building class 'a'

from sp-6 steel table.

γmin=γyy=25.4mm

design compressive strength

Pd=AcxFcd

Fcd=(Fy/γm0)/Φ+[Φr-γr]0.5

Φ=0.5[1+∝(λ-0.2)+λ2]

λ=√(fy/fce)

fce=π²E/(KL/r)²

fce=π²x2x105/(3600/28.4)²

=122.84N/mm²

λ=√(250/122.840)=1.426

for building class 'a' '∝' imperfection factorfrom IS-800:2007 Table-7

∝=0.21

Φ=.5[1+0.21(1.426-0.2)+(1.426)²]

=1.645

fcd=(250/1.1)/1.645+[(1.645)²-(1.42)²]0.5

=92.19N/mm²

Pd=AcxFcd

=56.26x92.19

=5186.80KN

3)L=2m, P=400KN, Fy=210MPa, K=1

KL=2000mm

Let Fcd=80N/mm²

Areq=400x10³/80=5000mm²

πD²/4=Areq

D=√(5000X4/π)=80mm

For circular hollow section (semi compact)

D/t<=88ε²

e=√(250/fy)=1.09

Let t=8mm

80/8<=88x1.09²

D=80mm t=8mm

λ=KL/r

r=80/2=40mm

λ=2000/40=50

Circular sections are under buckling class a

using table 9a

KL/r fcd@fy=210

Pd=Acxfcd=5000x1.76=880,000KN

4) Assuming fcd=135N/mm

Area required=1000x1000/135=7407.40mm²

Try 2 ISMC250 @ 298.2N/mm² (SP-6-1)

Area provided =2x3867=7734mm²

r22=99.4mm

distance will be maintained so as to get

ryy>r22

Actual slenderness ratio(KL/r)=1x10000/99.4=100.60

Since it is a leced column

(KL/r)=1.05x100.60=105.63

From table 9C of IS 800 of buckling class C table-10

fcd=107-(105.63-100/110-100)x(107-94.6)

fcd=100.018N/mm²

Load carrying capacity=7734x100.015

=7735.39>100KN

Hence OK

Spacing between channels

let it be a clear distance'd'

Now Ixx=2x3816.8x104

=7633.6x104mm4

Iyy=2[219.1x104+3867(d/2+23)²]

equating Iyy to Ixx we get

[(d/2+23)²]=4368.5

d/2+23=66.09

d=86.18mm

provide 90mm spacing

Horizontal spacing=90+23+23=136mm

vertical spacing=136xsin 45ºx2

=192.304mm

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