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 $\times 10^{3}$N

l=10m=10000 mm

leff=kl

=1.0$\times$ 10000

leff=10000mm

fe=410

fy=250mpa

Required -design single lacing built up column

step 1 Design of column section

Pd=Ae$\times $fcd

1080$\times10^{3}$=Area$\times$150

Area=7200mm$^{2}$

for 1 channel=$\frac{7200}{2}=3600mm^{2}$

provide 215 mc 300@ 351.2 N/m

A=4564mm$^{2}$

bf=90mm

tf=13.6mm

$\gamma$xxx=118.1mm

$\gamma$yy=26.1mm

cyy=23.6mm

Ixx=63626$\times10^{4}mm^{4}$

Pd=Ae$\times$fcd

To find fcd

$\frac{kL}{\gamma min}=\frac{10000}{118.1}$=84.67

$\lambda=1.05\times 84.67$=88.9

80 $ \ \ \ \ \ $ 136

88.9 $ \ \ \ \ \ \ $fcd

90 $ \ \ \ \ \ \ \ \ \ $ 121

$\frac{90-80}{90-88.9}=\frac{121-136}{121-fcd}$

fcd=122.65

fd=2(4564)$\times$122.6

=1119.5$\times10^{3}\gt 1080\times10^{3}$N safe

Step II Provide spacing

spacing between channel

Ixx=Iyy

2$[Ixx_{1}+Ah^{2}=[Ixx_{2}+Ah^{2}]$

[6362.6$\times10^{4}+10]$=$[310.8\times10^{4}+4564\times(\frac{5}{2}+cyy)^{2}]$

S=183.10mm

provide 184mm

step III Design of lacing

1) Angle of inclination

$\theta=45^{\circ}$ clause 7.6.4

2) width of lacing =3$\times16$ (Assume d=16mm)

=48mm

width of lacing =50mm

3) leff

Sin$\theta=\frac{s+2g}{leff}$

leff=$\frac{S+2g}{sin\theta}=\frac{(184+2\times 50)}{sin \ 45}$=401mm

provide leff=402mm

tan$\theta\frac{x}{(5+29)}$

x=$\frac{5+29(tan\theta)}{tan\theta}$

x=(184+2x50)(tan45$^{\circ})$

x=284mm

distance between lacing point

=2x284

=568mm

$\frac{a}{\gamma y}\leq 50$

$\frac{568}{261} \lt 50$

$\frac{a}{\gamma y}\lt0.7 \times \lambda c$

$\frac{568}{26.1}\lt0.7 \times 88.9$

21.76$\lt$62.23

v) f=$\frac{1}{40}\times eff$ length

$\frac{1}{40}\times$402

f=10.06

provide 12mm thickness

vi) Radius of gyration

$\gamma min=\frac{t}{\sqrt{12}}$

$\frac{12}{\sqrt{12}}$

$\gamma$ min=3.46mm

Step IV Check for compression

V=(2.5%)$\frac{1080\times10^{3}}{2}$

V=$\frac{2.5}{100}\times\frac{1080\times10^{3}}{2}$

v=13.5kN

V=Fsin$\theta$

F=13.5

F=19.09KN

PDAex$\times$fcd

=(50$\times12)\times$fcd

$\lambda=\frac{kl}{\gamma min}=\frac{402}{3.46}$=116.18

110 $ \ \ \ \ \ \ \ \ \ \ $94.6

116.18 $ \ \ \ \ $ fcd

120 $ \ \ \ \ \ \ \ \ $ 83.7

$\frac{(120-11)}{(120-116.18)}=\frac{(83.7-94.6)}{(83.7-fcd)}$

fcd=87.86

Pd=(50$\times12)\times$ 87.86

Pd=52.71 kN=F(19.09) Hence safe Assumption

should be correct

Step V

Check for tention

1) Tdg=Ag$\times\frac{fy}{\gamma mo}$

=(50$\times12)\times\frac{250}{1.1}$

Tda=136.36KN

2) Tdn=0.9$\times An\times\frac{fu}{\gamma ml}$

=0.9$\times[(50-18)\times12]\times\frac{410}{1.25}$

Tdn=113.35kN

Tdn=113.35 19.09 hence safe

step VI bolted connection

vdsh=$\frac{An.nn.fub}{\sqrt{3}\times \gamma mb}$

=$\frac{157\times2\times400}{\sqrt{3}\times1.25}$

Vcl sb=58.01

vdpb=2$\times kb\times \phi \times t\times fu$

=2$\times 0.51\times 16\times 12 \times410$

vdpb=46.35kN

Bolt value =46.35$\gt$19.09----safe

provide 1 NO.of bolts 16mm $\phi$

Step VII Design of Tie member plate

effective depth=(5+2.cyy)

=(184.4+2$\times$33.6)

=231.2mm

Provide 235 mm effective depth

overall depth=235+2$\times$ 30=295mm

length of tie plate=s+2$\times$ bf

=184+2$\times $90

=364mm

t=$\frac{1}{50}(s+29)$

t=$\frac{1}{50}(184+2\times 30)$

t=4.88

t=$\approx$8mm

Case II Channel place Toe to Toe

Step I design built-up column

Au=Area $\times$ fcd

1080 $\times10^{3}$=Area$\times $150

Area=7200mm$^{2}$

Required for each channel= 3600mm$^{2}$

provide Ismc 300@ 351.2 N/m

$\gamma$ xx=118.1mm Ixx=6363.6$\times 10^{4}mm^{4}$

$\gamma yy=26.1mm \ \ \ \ Iyy=310.8\times10^{4}mm^{4}$

A provide =4564mm$^{2}$

A provide area=2$\times 4564=9128 mm^{2}$

Pd=Ae$\times$fcd

=9128$\times$122.65

=1119.25$\times10^{3}\gt 1080\times 10^{3}$ Analysis is safe

$\lambda act=\frac{kL}{\gamma}=\frac{10,000}{\gamma xxx}=\frac{10000}{118.1}=84.97$

$\lambda e=1.005\times 84.64$

=88.9

$\frac{kL}{\gamma}$

$ \ \ \ \ \ \ \ \ $ fcd

80 $ \ \ \ \ \ \ \ \ \ \ $ 6

88.9 $ \ \ \ \ \ $ fcd

90 $ \ \ \ \ \ \ $ 12

fcd=122.65 --by interpolation

Pd=Ae$\times$fcd

=9128$\times$122.65

=1119.25$\times 10^{3}\gt1080\times 10^{3}$ safe

Step II

IXX=Iyy

2[Ixxx+Ab$^{2}$=2[Iyy+Aehe$^{2}]$

[63626$\times10^{4}+0[310\times 10^{4}+456\times(\frac{5}{2}-Cyy)^{2}]$

x=2775mm

s=278mm

Step III Design of lacing

L=251.73

W=3$\times(dia \ of \ bolt)=8\times16=48\approx$50mm

$\theta$=45

Sin$\theta=\frac{5-29}{byp}$

leff=$\frac{s-29}{sin\theta}$

=$\frac{278-2\times 50}{sin45}$

leff=251.7mm

t=$\frac{1}{40}$(leff)

$=\frac{1}{40}\times$251.73

=6.29mm

f=8mm

Spacing between lacing point=2x

tan(90-$\theta=\frac{x}{6-29}$

x=$(278-2\times 90)\times tan(90-45)$

x=178mm

lacing point=2x

=2$\times$ 178

=356mm

Step IV Vt=2.5%$\frac{pu}{2}$

=$\frac{2.5}{100}\ast\frac{1080\times10^{3}}{2}$

vt=13.5$\times 10^{3}$N

Fsin$\theta$=vt

F=$\frac{vt}{sin\theta}$

F=$\frac{13.5\times 10^{3}}{sin45^{\circ}}$

F=19.09KN

$\frac{q}{\gamma 1} \lt0.7\lambda e$ & $\frac{a}{\gamma 1}\lt 50$

$\frac{356}{26.1}\lt 0.7\times$88.9

13.64$\lt$62.23

hence safe

Pd=Ae$\times$ fcd

=(50$\times$8)$\times$fcd

$\gamma min=\frac{t}{\sqrt{12}}=\frac{8}{\sqrt{12}}=2.31$

$\lambda e=\frac{252}{2.31}$=109.09

KL/rmin $ \ \ \ \ \ \ $ fcd

100 $ \ \ \ \ \ \ \ \ \ \ \ $ 107

109.09 $ \ \ \ \ \ \ $ fcd

110 $ \ \ \ \ \ \ \ \ \ \ \ $ 04.6

fcd=95.88mpa

Pd=(50$\times8)\times$ 95.88

Pd=38.35KN$\gt$19.09 KN Hence safe

Step V Tention capacity of end plate

Tdg=Ag$\times\frac{fy}{\gamma mn}$

=(50$\times 8)\times\frac{250}{1.1}$

Tdg=90.9KN