Given- Force=230KN fy=250mpa=250N/mm$^{2}$ fu=410N/mm$^{2}$ sw=4mm E.t.T=0.7$\times$4=2.8mm required- Design of tention member

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

AG=$\frac{Tdg\times \gamma mo}{fy}=\frac{230\times10^{3}\times1.1}{250}$

Ag=1012mm$^{2}$

provide 125$\times$75$\times$6mm

A provided=1166mm$^{2}$

P=(Leff$\times$ E.T.T)$\times \frac{fu}{\sqrt{3}\times \gamma mw}$ Assume work shop

230$\times10^{3}=(Leff)\times(0.7\times4)\times(\frac{410}{\sqrt{3}\times1.25}$

leff=4.33.76mm

provide leff=440mm

$\sum$mpx=0

(p$\times40.5)=(P1 \times$125)

P1=74.52KN

P1+P2=230

P2=155.48KN

P1=$lw_{1}\times0.7\times4\times\frac{410}{\sqrt{3}\times1.25}$

74.52-lw$_{1}$$\times0.7\times4\times\frac{410}{\sqrt{3}\times1.25}$ lw1=140.54mm 144.483=lw$\times$0.7$\times$4$\times\frac{410}{\sqrt{3}\times1.25}$

Lw$_2$=293.29mm

lw$_{2}$=299.46mm

1)Tdg=$\frac{Ag\times fy}{\gamma mo}=\frac{1166\times250}{1.1}$=265KN

2)Tdn=0.9$\times …