Week 3 Challenge
1. Effective length factor determination problem
- 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
Ans
The column is of height 10 m and it is restrained at mid. Hence the actual length of the column is 5 m.
The effective length factor of the following cases are given below.
For braced frame and shear wall the effective lenght is
2L = 2*5= 10 m
for unbraced frame with a pinned base
2L = 2*5= 10 m
For unbraced frame with a fixed base effective length is
1.2L = 1.2*5= 6m
2. Design Axial strength determination of a column with its end conditions
Calculate the Axial compressive capacity of Column made up of Hot rolled ISMB 300 with flange cover plate of 200x12 mm on each flange having length of beam as 3 m. Assume the bottom of the column is Fixed and Top as Roller (Rotation Fixed, translation free). Yield strength of material fy= 250 N/mm2
Ans:
Data given :
Column having fabricated by ISMB 300 with flange cover plate 200x12 mm
fy=250N/mm^2
Size of the column is 3m
The top of the column is having roller suport and the base is fixed.
For the column the KL is 1.2*L= 1.2*3= 3.6 m.
Let's calculate wcich buckling class it falls under as per IS800 2007 table 10
h/bf = 300/200= 1.5 and it is greater than 1.2.
and tf < 40 mm.
Hence it belongs to buckling class a
Now let's design it by following formula
Pd= Ae*fcd => 56.26*91.89= 5169.73Kn.
fcd= [(0.909*fy)/{fi+(fi^2-lamda^2)^0.5}] => [(0.909*250)/{1.65+(1.65^2-1.43^2)^0.5}]=> 91.89 N/mm2
here fi=0.5[1+alpha (lamda-0.2)+lamda^2]=> 0.5[1+0.21(1.43-0.2)+1.43^2]=> 1.65
lamda=sqrt ( fy/fcr)=> sqrt(250/122.85)=>1.43
fcr= Pi^2*E/(KL/r)^2=> (3.14^2*2*10^5)/(3600/28.4)^2=> 122.85N/mm2
From IS 800 2007 alpha = 0.21
3. Design circular tube strut member for certain load
Design 2 m long tubular strut member to carry factored axial load of 400 kN. Assume the ends of the member are simply supported and the yield strength fy=210 Mpa. Also find Equivalent Square hollow sections.
Ans
For tubular strut
Data given L= 2 m, Axial load = 400Kn, fy= 210Mpa.
K=1
KL= 2000 mm
td= 80N/mm2
Areq=F/td= (400*1000)/80= 5000 mm2
Pi*d^2/4=5000
d= sqrt((5000*4)/3.14)= 80 mm
For hollow section
Data given L= 2 m, Axial load = 400Kn, fy= 210Mpa.
K=1
KL= 2000 mm
td= 80N/mm2
Areq=F/td= (400*1000)/80= 5000 mm2
Pi*d^2/4=5000
d= sqrt((5000*4)/3.14)= 80 mm
For hollow circular section
D/t should be less than equal to 88E^2
E= sqrt(250/fy)= Sqrt(250/210)= 1.09
let t be 8 mm
hence: 80/8=10 and 88*1.09^2=104.55 Hence Ok.
fcd= [(0.909*fy)/{fi+(fi^2-lamda^2)^0.5}] => [(0.909*210)/{0.67+(0.67^2-0.52^2)^0.5}]=> 174.73 N/mm2
here fi=0.5[1+alpha (lamda-0.2)+lamda^2]=> 0.5[1+0.21(0.52-0.2)+0.52^2]=> 0.67
lamda=sqrt ( fy/fcr)=> sqrt(210/788.78)=>0.52
fcr= Pi^2*E/(KL/r)^2=> (3.14^2*2*10^5)/(2000/40)^2=> 788.78N/mm2
r=80/2
Pd= Ae * fcd= 5000 * 174.7= 873.5Kn hence safe.
4. Design of laced column
Design a single laced column made up of two channels back to back of 10 m height to carry factored axial load of 1000kN. The lacings are bolted. Design of bolted connection not required.
Ans
Data given : h=10 m F= 1000Kn.
Assume fcd = 200N/mm2
Areq= 1000*1000/200=5000 mm2
From IS sp 6.1 steel table we can take 2 number of ISMC 200
The area of each ISMC 200 is 2821 mm2 hence for 2 ISMC 200 will be 5642 mm2
The fcd for ISMC 200 will be (1000*1000)/2821= 354.49N/mm2
hence for 2 ISMC 200 and area provided will be 5642 mm2
From Sp 6.1 we get r as 80.3mm
Hence actual slenderness ratio is KL/r= 1*10000/80.3=124.53
and since it is a laced column the efective slenderness ratio for laced column is 1.5*124.53=186.80
from table 9C of IS 800: 2007 taking fy= 250 Mpa
fcd=40.98N/mm2
Load carrying capacity = 5642*40.98= 2312.09KN > 1000Kn.
Spacing between channels
Let it be clear distance d
Ixx= 2 * 1819.3x10^4 mm4= 3638.6x10^4 mm4
Iyy= 2(140.4x10^4+2821(d/2+21.7)^2)
Equatin Ixx=Iyy
3638.6x10^4=2(140.4x10^4+2821(d/2+21.7)^2)
d= 110.89 mm
Provide 110 mm spacing
Horizontal spacing = 110+21.7+21.7=153.4 mm
Vertical spacing = 153.4sin45 * 2 = 216.94 mm
ryy= 22.3
KL/r= 216.94/22.3=9.73 < 50
Transverse shear to be resisted in lacing system is
1000Kn*2.5%= 25000N.
Shear to be resisted by each lacing = 25000/2= 12500N
length of lacing = (110+21.7+21.7)*(1/cos45)=216.94 mm
Minimum thickness for single lacing
Le/40=216.94/40 = 5.42mm
Provide 6 mm thick flat
considering 20 mm bolt hole
Hence width of the lace will be 3*20 = 60 mm
5. Design of battened column
Design a battened column made up of two channels back to back of 5 m height to carry factored axial load of 2000kN. The Battens are welded. Design of weld not required. Use Hot rolled channel sections.
Ans
Data given : H= 5 m F=2000Kn Assume fcd = 200N/mm2
Areq= 2000*1000/200=10000 mm2
from steel table sp 6.1 Provide 2 ISMC 350
Area = 5365mm2, rxx=136.6mm ryy=28.3mm, Ixx = 10008x10^4 mm4, Iyy= 430.6x10^4 mm4
Area provied = 2*5365= 10730 mm2
Kl/r= 5000/136.6=40.26
effective slenderness ratio = 1.1*KL/r= 1.1*40.26=44.29
From table 9c of IS 800 2007
Fcd = 191.57 N/mm2
Load carrying capacity = 10730*191.57= 2055.55Kn > 2000Kn hence safe.
Spacing of batten = C/ryy<50 =
C/28.3 = 50
C= 1415 mm
C/28.3<0.7* effective slenderness ration => 0.7*44.29*28.3 =877.38mm
Let's provide C = 1200 mm
Transverse shear (V)= 2.5% * 2000*1000= 50000N
Moment = VC/2N = 50000*1200/2*2=15Knm
Longitudinal shear = VC/NS = 50000*1200/2*320 =93.75 Kn.
Let it be clear distance d
Ixx= 2 * 10008x10^4 mm4= 20016x10^4 mm4
Iyy= 2(430.6x10^4+5365(d/2+24.4)^2)
Equatin Ixx=Iyy
20016x10^4=2(430.6x10^4+5365(d/2+24.4)^2)
d= 218.42 mm Provide d= 220 mm
Hence S = 220+100=320 mm
Thickness of batten t> Lb/50
Taking Lb = 220-50= 170 mm
Henc t = 170/50 = 3.4 mm
Provide batten thickness t = 4 mm