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fe 40 and bolt of grade 4.6 case I) Design the column with two channel placed back to back case II) Design the column with two channel place face to face case II) Design the lacing system with side welding connection from channel back to back
Pu=1080 ×103N
l=10m=10000 mm
leff=kl
=1.0× 10000
leff=10000mm
fe=410
fy=250mpa
Required -design single lacing built up column
step 1 Design of column section
Pd=Ae×fcd
1080×103=Area×150
Area=7200mm2
for 1 channel=72002=3600mm2
provide 215 mc 300@ 351.2 N/m
A=4564mm2
bf=90mm
tf=13.6mm
γxxx=118.1mm
γyy=26.1mm
cyy=23.6mm
Ixx=63626×104mm4
Pd=Ae×fcd
To find fcd
kLγmin=10000118.1=84.67
λ=1.05×84.67=88.9
80 136
88.9 fcd
90 121
90−8090−88.9=121−136121−fcd
fcd=122.65
fd=2(4564)×122.6
=1119.5×103>1080×103N safe
Step II Provide spacing
spacing between channel
Ixx=Iyy
2[Ixx1+Ah2=[Ixx2+Ah2]
[6362.6×104+10]=[310.8×104+4564×(52+cyy)2]
S=183.10mm
provide 184mm
step III Design of lacing
1) Angle of inclination
θ=45∘ clause 7.6.4
2) width of lacing =3×16 (Assume d=16mm)
=48mm
width of lacing =50mm
3) leff
Sinθ=s+2gleff
leff=S+2gsinθ=(184+2×50)sin 45=401mm
provide leff=402mm
tanθx(5+29)
x=5+29(tanθ)tanθ
x=(184+2x50)(tan45∘)
x=284mm
distance between lacing point
=2x284
=568mm
aγy≤50
568261<50
aγy<0.7×λc
56826.1<0.7×88.9
21.76<62.23
v) f=140×eff length
140×402
f=10.06
provide 12mm thickness
vi) Radius of gyration
γmin=t√12
12√12
γ min=3.46mm
Step IV Check for compression
V=(2.5%)1080×1032
V=2.5100×1080×1032
v=13.5kN
V=Fsinθ
F=13.5
F=19.09KN
PDAex×fcd
=(50×12)×fcd
λ=klγmin=4023.46=116.18
110 94.6
116.18 fcd
120 83.7
(120−11)(120−116.18)=(83.7−94.6)(83.7−fcd)
fcd=87.86
Pd=(50×12)× 87.86
Pd=52.71 kN=F(19.09) Hence safe Assumption
should be correct
Step V
Check for tention
1) Tdg=Ag×fyγmo
=(50×12)×2501.1
Tda=136.36KN
2) Tdn=0.9×An×fuγml
=0.9×[(50−18)×12]×4101.25
Tdn=113.35kN
Tdn=113.35 19.09 hence safe
step VI bolted connection
vdsh=An.nn.fub√3×γmb
=157×2×400√3×1.25
Vcl sb=58.01
vdpb=2×kb×ϕ×t×fu
=2×0.51×16×12×410
vdpb=46.35kN
Bolt value =46.35>19.09----safe
provide 1 NO.of bolts 16mm ϕ
Step VII Design of Tie member plate
effective depth=(5+2.cyy)
=(184.4+2×33.6)
=231.2mm
Provide 235 mm effective depth
overall depth=235+2× 30=295mm
length of tie plate=s+2× bf
=184+2×90
=364mm
t=150(s+29)
t=150(184+2×30)
t=4.88
t=≈8mm
Case II Channel place Toe to Toe
Step I design built-up column
Au=Area × fcd
1080 ×103=Area×150
Area=7200mm2
Required for each channel= 3600mm2
provide Ismc 300@ 351.2 N/m
γ xx=118.1mm Ixx=6363.6×104mm4
γyy=26.1mm Iyy=310.8×104mm4
A provide =4564mm2
A provide area=2×4564=9128mm2
Pd=Ae×fcd
=9128×122.65
=1119.25×103>1080×103 Analysis is safe
λact=kLγ=10,000γxxx=10000118.1=84.97
λe=1.005×84.64
=88.9
kLγ
fcd
80 6
88.9 fcd
90 12
fcd=122.65 --by interpolation
Pd=Ae×fcd
=9128×122.65
=1119.25×103>1080×103 safe
Step II
IXX=Iyy
2[Ixxx+Ab2=2[Iyy+Aehe2]
[63626×104+0[310×104+456×(52−Cyy)2]
x=2775mm
s=278mm
Step III Design of lacing
L=251.73
W=3×(dia of bolt)=8×16=48≈50mm
θ=45
Sinθ=5−29byp
leff=s−29sinθ
=278−2×50sin45
leff=251.7mm
t=140(leff)
=140×251.73
=6.29mm
f=8mm
Spacing between lacing point=2x
tan(90-θ=x6−29
x=(278−2×90)×tan(90−45)
x=178mm
lacing point=2x
=2× 178
=356mm
Step IV Vt=2.5%pu2
=2.5100∗1080×1032
vt=13.5×103N
Fsinθ=vt
F=vtsinθ
F=13.5×103sin45∘
F=19.09KN
qγ1<0.7λe & aγ1<50
35626.1<0.7×88.9
13.64<62.23
hence safe
Pd=Ae× fcd
=(50×8)×fcd
γmin=t√12=8√12=2.31
λe=2522.31=109.09
KL/rmin fcd
100 107
109.09 fcd
110 04.6
fcd=95.88mpa
Pd=(50×8)× 95.88
Pd=38.35KN>19.09 KN Hence safe
Step V Tention capacity of end plate
Tdg=Ag×fyγmn
=(50×8)×2501.1
Tdg=90.9KN