Given

L=5.9m = 5900 mm

Pu=2000$\times 10^{3}$N

required two channel back to back with battern & site welding $\gamma$ mw=1.5

1. Pu=Aea$\times$ fcd

2000$\times 10^{3}$=Area$\times$ 150

Area=13333.33mm$^{2}$

Area of each section=$\frac{13333.33}{2}=6666.67mm^{2}$

provide fcd=190mpa

Pu=Area$\times$ fcd

2000$\times 10^{3}=Area\times$190

Area=1056.31mm$^{2}$

Area of each section=$\frac{10526.31}{2}$=5263.15mm$^{2}$

2) Provide ISMC 350

Aprov=5366mm$^{2}$

$\gamma$ xx=136.6mm $\ \ \ \ \ \ \ $ Ixx=10008$\times 10^{4} mm^{4}$

$\gamma$yy=28.3mm $\ \ \ \ \ \ \ \ \ \ $ Iyy=430.6$\times10^{4}mm^{4}$

Cyy=24.4

$\frac{kL}{\gamma}=\frac{5900}{136.6}=$43.19

$\lambda e=1.1\times$43.19

$\lambda$e=47.51

40 $\ \ \ \ \ \ \ \ $ 198

47.51 $\ \ \ \ $ fcd

50 $\ \ \ \ \ \ \ \ \\ $183

fcd=186.74mpa

Pd=Ae$\times$ fcd

=2$\times5366)\times$186.74

Pd=2004.09kN$\gt$ 2000 KN s safe

Step II

Ixx=Iyy

2[Ixx$_{1}+Ab^{2}]$=2[Ixx+Ah$^{2}]$

[10008$\times 10^{4}+0]=[430.6\times 10^{4}+5366(\frac{5}{2}+cyy)^{2}]$

S=218.99mm

S=220mm

Step III Sapcing between batten

$\frac{c}{\gamma yy}\lt50$ & $\frac{c}{\gamma yy}\lt 0.7 \lambda e$

c$\lt$50$\times \gamma yy$

c$\lt 50\times$28.33

c$\lt$1415mm

c$\lt0.7\times \lambda \times \gamma yy$

$\lt0.7\times97.81\times28.3$

$\lt$941.17

c=940mm

Step IV Size of batten

end batten

over all depth=eff depth d$^{'}$

d$^{'}$=s+2cyy

=220+2(24.4)

=268.8

D=270mm

thickness=$\frac{1}{50}\times s$

=$\frac{1}{50}\times$220

=4.4mm

provide 8mm

overlao=$4\times$ 8 =32mm

length provide 60mm

length=s+2$\ast$ overlap

=220+2$\ast$ 60

=340mm

Intermedient

d=$\frac{3}{4}d^{\prime}$

=$\frac{3}{4}\times$270

D=d=205mm

t=8mm

Step V check for shear stress

$\frac{V}{dt}\lt\frac{fy}{\sqrt{3}\ast \gamma mo}$

Vb=$\frac{Vt\times c}{2N}$

Vt=2.5$\ast$ P4

$\frac{2.5}{100}\times 2000\times 10^{3}$

$\frac{Vt=5000 N}{Vt=50kN}$

shear stress=$\frac{Vb}{d\times t}$

$\frac{69.12\times 10^{3}}{250\times 8}$

=42.15$\times131.26$

safe

Vb=$\frac{50\times10^{3}\times 940}{2\times (s+2 overlap)}$=$\frac{50\times10 ^{3}\times 940}{2(5+260)}$

Vb=69.12$\times10^{3}$N

Shera Stress=$\frac{69.12\times 10^{3}}{270\times 8}$=32KN$\lt \frac{fy}{\sqrt{3}.\gamma m}$=131.26

Step VI Check for bending stress=$\frac{6m}{td^{2}}$

end batten

m==$\frac{vt\times c}{2N}=\frac{50\times 10^{3}\times 940}{2(2)}$

m=11.75$\times 10^{6}$ N.mm

bending stress=$\frac{6\times 11.75\times10^{6}}{8\times 270^{2}}$

=120.88mpa

$\lt\frac{fy}{\gamma mo}$=227.27

Intermedient

b. stress=$\frac{6\times 11.75\times 10^{6}}{8\times 205^{2}}$

=209.70mpa

$\lt$ 227.27

Step VII

Connection

(rn)$^{2}=30^{2}+(\frac{205}{2})^{2}$

direct shera stress=(f1)=$\frac{Vb}{Lw\times t}$

=$\frac{69.12\times 10^{3}}{530\times 1}$

f1=130.41N/mm

Torrtn stress(F2)=$\frac{m.\gamma n}{IP}$

IP=IXX+IYY

Ixx=$2[\frac{60\times t^{3}}{12}+Ah^{2}]+2[\frac{t\times 205^{3}}{12}+0]$

Ixx=2.69$\times10^{6}mm^{4}$

Iyy=2[$\frac{1\times60^{3}}{12}$+0]+2[$\frac{25\times t 3}{12}+(205\times )\times 30^{2}]$

Iyy=405$\times10^{3}mm^{4}$

Ip=Ixx+Iyy

=2.69$\times10^{6}+405\times 10^{3}$ $\ \ \ \ \ \ $ rn$^{2}=(30)^{2}+(\frac{205}{2})^{2}$

=3.0595$\times10^{6}mm^{4}$ $\ \ \ \ \ \\ $ $\gamma$n=106.800

bending stress=$\frac{m.rn}{2D}$

=$\frac{11.75\times10^{6}\times106.8}{3.095\times10^{6}}$

cos$\theta=\frac{30}{106.8}$

f2=405.46N/mm

$\theta=0.28^{\circ}$

Resultant stress=$\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F{2}cos\theta}$

$=\sqrt{130.47^{2}+405.46^{2}+2\times130.42\times 405.46\times \gamma cos\theta}$

R=459.37N/m--------eq A

force in p=Lw$\times FT.T\times \frac{fu}{\sqrt{3}\gamma mw}$

459.37=1$\times(0.7\times sw)\frac{410}{\sqrt{3}\times 1.5}$

459.37=1$\times(0.7sw)\frac{410}{\sqrt{3}\times 1.5}$

sw=4.16mm

sw=5mm