Week 3 Challenge
1) Height of column = 10m
Restrained in mid height , therefore actual length = 5m
Effective length for the mentioned cases are as follows :
a. Braced frame and Shear wall = 2L = 2 * 5 = 10m
b.Unbraced frame with pinned base = 2*L = 2*5=10m
c.Unbraced frame with a fixed base = 1.2L = 1.2*5 =6m
2) fy=250N/mm2 , Fu = 410N/mm2
Effective length = 1.2 x L = 1.2 x 3 = 3.6m
Properties of ISMB 300 - Height (H) = 300mm, flange cover plate = 200mm Tf cover = 12mm, tf=12.4mm , Breadth bf=140mm
h/bf= 300/200 = 1.5 > 1.2 (IS 800 Table 10)
since it is built up member, the class is c.
rmin = ryy= 25.4 92
from table 9c, fy = 250
kL/r = 3600/25.4 = 141.7, for 140 = 66.2 , 150=59.2
interpolating 141.7 we get 65.01 as design compressive strength.
Design strength = Ac x Fcd
`alpha`=0.49 ; fcc = 122.8N/mm
`lambda`= `root`(fy/fcc) =`root`250/122.8 = 1.426
`phi`=3.38 , fcd=92.19N/mm2
Pd = A* fcd = 5186.6kN
3) L=2m , P=400kN , fy = 210N/mm2 Assume fcd= 80N/mm2
Area = 400 x 10^3/80 = 5000mm2
`5000 = (pi *d^2)/4` ; d=80mm
For circular hollow section D/t < = 88 x e^2
e = squareroot(fy/250) = 1.09
let t=8mm
r= 80/2 = 40mm
`lambda`= 2000/40 = 50
Pd = Ac x fcd = 5000 x 1.76 = 880000kN
4) Assuming fcd = 135N/mm2
area required = 1000 * 1000/135 = 7407.4mm2
Assuming 2 ISMC 250 @ 298.2N/mm2
Area Provided = 2 x 3867 = 7734mm2
rzz = 99.4m
Actual slenderness ratio = KL/r =10000/99.4 = 100.6
since its a laced column
KL/r = 1.05 * 100.6 = 105.63
fcd = 107 - ((105.63-100)/(110-100))*(107-94.6) = 100.018 N/mm2
Load carrying Capacity = 7734 x 100.015 > 1000kN Hence Safe
Ixx = 2 * 3816.8 x 10^4 = 7633.6 x 10^4 mm^4
Iyy = 2 [ 219.1 x 10^4 + 3867[d/2 +23]^2]
Equating Ixx & Iyy , [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 = 136 Sin45 x 2 = 192.304
ryy = 23.8
KL/r of channel between lacing = 192.3/23.8 = 8.08<50
Transverse shear to be resisted by lacing system = 2.5/100 * 1000 * 10^3 = 25000N
shear to be resisted by each lacing system 25000/2 = 12500N
Length of lacing = (90+23+23)/ cos 45 = 192.3
Minimum thickness of lacing = 192.30/40 =4.8mm
Use 100mm flats , minimum width of lacing : if 20mm bolts are used = 3 *20 = 60mm.
5) Battened column design , P=2000kN fcd=135(Assume) L=5m
Effective length = 1.1 x 5 = 5.5m
C/s area required = 2000 * 10^3/135 =14814.8mm2
Provide 2 ISMC 350@413N/m
A= 5365 mm2 , rxx= 136.6mm ryy=28.3mm Ixx = 100008*10^4 mm^4, Iyy = 430.6*10^4 mm^4
Area Provided = 2 * 5365 = 10730 mm2
l/r = 5*10^3/136.6 = 36.6
kl/r = 1.1 * 5 * 10^3/136.6 = 40.26
from table 9(c) of IS 800
fcd = 211-((40.26-30)/(40-30))*(211-198) = 197.6N/mm2
Load carrying capacity = 10.73*197.66 = 2120.89kN >2000kN
Spacing of batten
C/28.3 < 50 ; C < 1415
It should also satisfy the condition
C/28.3 < 0.7*40.26 i.e C < 797.55
Assume C=1200mm
Vt = 2.5/100 *2000*10^3 = 50000N
Vb = Vt x C/Ns = 50000 x 1200/ 2 x 250 = 12000N
M = Vt C/ 2N = 50000 x 1200 /2*2 = 1500000Nmm