Week 3 Challenge
1.Effective length factor determination problem.
Given: The column is of height 10 m and it is restrained at mid.
Hence, the actual length of the column is 5 m.
Solution:
1.The effective length factor of the 1st cases:
For braced frame and shear wall the effective lenght is
2L = 2*5= 10 m
Result: L = 10 m.
2.The effective length factor of the 2nd cases:
For unbraced frame with a pinned base
2L = 2*5= 10 m
Result: L = 10 m.
3.The effective length factor of the 3rd cases:
For unbraced frame with a fixed base effective length is
1.2L = 1.2*5 = 6m
Result: L = 6m.
2. Design Axial strength determination of a column with its end conditions
Given : Column having fabricated by ISMB 300 with flange cover plate 200x12 mm, fy=250N/mm^2, Size of the column is 3m.
Solution:
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.
(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>
Hence, it belongs to buckling class a
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
Restlt:
- Pd=5169.73Kn.
- fcd=> 91.89 N/mm2
- fi=> 1.65
- lamda=>1.43
- fcr=> 122.85N/mm2
- alpha = 0.21
3. Design circular tube strut member for certain load.
Given : L= 2 m, Axial load = 400Kn, fy= 210Mpa, K=1, KL= 2000 mm, td= 80N/mm2
Solution:
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
So, 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.
Result:
- fcd=> 174.73 N/mm2
- fi=> 0.67
- lamda=>0.52
- fcr=> 788.78N/mm2
- Pd=873.5Kn
4. Design of laced column.
Given : h=10 m F= 1000Kn, Assume fcd = 200N/mm2
Solution:
Area= 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
For 2 ISMC 200 and area provided will be 5642 mm2
From Sp 6.1 we get r as 80.3mm
Actual slenderness ratio is KL/r= 1*10000/80.3=124.53
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
Result:
- A = 5642 mm2
- fcd=40.98N/mm2
- d= 110.89 mm
- length of lacing = 216.94 mm
- width of the lace = 60 mm
5. Design of battened column.
Given : H= 5m, F=2000Kn, Assume fcd = 200N/mm2.
Solution:
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 xss=removed>
C= 1415 mm
C/28.3<0 xss=removed> 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.
Result:
- Cross section area A =10730 mm2
- Fcd = 191.57 N/mm2
- C= 1415 mm
- d= 220 mm
- t = 4 mm