Week 3 Challenge
1.Height of column is 10 m and it is effectively restrained in mid height in one direction (in Z direction) but free to move in other direction (Y direction). Bottom of column is Fixed and top of column is Free.Calculate effective length factor of columns in following cases
The height of the column is 10m and is restrained in mid height. Since they are restrained in both y-y and z-z direction we take the actul length as 5m.
a. braced frame and shear wall = Here top is free and bottom is fixed ; Eff Length = 2L = 2x5=10m
b. Unbraced frame with pinned base = Eff length = 2L = 2x5 =10m
c. Unbraced frame with a fixed base = Eff. length = 1.2L =1.2 x 5=6m
2.
Fy =250N/mm2
Fu = 410N/mm2
effective length = 1.2 x L =1.2x 3= 3.6m
height of ISMB 300 = 300mm , flange cover= 200 , Tf cover=12mm, tf =12.4mm . breadth of flange bf= 140mm
h/Bf =300/200 = 1.5 -> which should be greater than or equal to 1.2 acc. to IS800 tabl 10.
Hence the class is c, because of built up member.
rmin =ryy =25.4 from SP6-1 steel
From tble 9c , under fy=250
KL/r = 3600/25.4 = 141.7
for 140 => 66.2 & 150 => 59.2
by extrapolating we get 141.7 => 65.01 as design compressive strength.
To calculate the design strngth =
imperfection factor- alpha = 0.49
Pd =Ac Fcd
fcc =122.84N/mm
lamba = root of (250/122.84) = 1.426
alpha =0.49
∅∅= 3.38
Fcd = 92.19N/mm2
Pd = 5186.6KN
3.
L= 2m ,P= 400kN, fy = 210Mpa, K=1, KL =2000mm, let fcd = 80N/mm2
Area = 400x 10^3 / 80 = 5000mm2
π⋅d24=5000/24=5000
d= 80mm
For circular hollow section (semi compact)
D/t <=88 E22
e=2√250fy2502= 1.09
let t =8 mm
80/8 = 88 x1.09^2
λ= (KL)/r
r = 80/2 = 40mm
λ= 2000/40 = 50
Circular sections ae under bukling class a, table 9a
Kl/r = 50 at fcd @fy =210
Pd = Ac x fcd
= 5000x 1.76
=880000 KN
4.
Assuming fcd = 135N/mm2
Area reuired = 1000x1000/ 135 = 7407.40mm2
Assuming 2 ISMC 250 @ 298.2N/mm2
Area provided = 2 x 3867 = 7734 mm2
rzz=99.4m
Distance will be maintained so as to get ryy<rzz
Actual slenderness ratio = KL/r = 1x10000/99 = 100.60
For laced column effective slenderness ratio = 1.05 x actual slenderness rtio = 105.63
From table 9c of "IS 800 = buckling class c, table 10
fcd = 107 -(105.63 -100)/(110-100) x (107.94.6)
fcd = 100.018 N/mm2
Load carrying capacity = 7734 x 100.015
7735.39KN > 100KN hence ok.
Spacing between channels
Let it be a clear distance "d"
Now Ixx= 2 3816.8 x10^4= 7633.6 x 10^4mm4
Iyy =2(219.1x10^4 + 3867(d/2 +23)^2)
d= 86.18mm
Provide 90mm spacing
Horizontal spacing = 90+23+23 =136mm\
Vertical spacing = 136sin45 =192.304mm
ryy =23.8
kl/r of channel between lacing = 192.3/23.8=8.08<50;
Traverse shear to be resisted by lacing system = 2.5/100 * 1000 *10^3 =12500N
Shear to be resisted by each lacing system = 25000/2 = 12500N
Length of lacing: 90+23+23 * 1/cos45 = 192.30
Min thickness of lacing = 1/40 * 192.30 = 480mm
Min width of lacing if 20mm bolts are used = 3*20 =60mm
5.
P = 2000KN, fcd = 135 assumed,
L= 5m
Effective length = 1.1x 5= 5.5m
C/s area required = (2000x10^3)/135 = 14814.8mm2
Provide 2ISMC 350 @ 413N/mm2
A =5365mm2 , rxx =136.6mm , ryy =28.3mm
Ixx =10008x10^4mm4, Iyy = 430.6x10^4mm4
Area provided =2*5365
=10730mm2
l/r =x10^3 /136.6 =36.60
for l/r =36.6 & fy = 250 Mpa
Since it is a battened column KL/r = 1.1x5x10^3 /136.6 =40.26
From table 9c IS 800,
fcd = 211- (40.26-30)/(40-30)x (211-198)
=197.60N/mm2
Load carrying capacity =10.730 x 197.66 =2120.89 KN >2000KN , hence ok.
Spacing of battens,
C/28.3 <50, C<1415
It should also satisfy the condition
c/28.3< 0.7*40.26 ie C<797.55
Let us select C= 1200mm
vt =2.5/100x2000x1000 =50000N
Vb =vtc/ns =50000*120 / 2*250 =12000N
M=Vtc/2N = 50000*1200/ 2*2 =150000n-m