gusset is connected to 100mm leg b) gusset is connected to 75mm leg. Assume edge distnace 30mm ^ tield stress & ultimate stress is angle 250N/$mm^{2}$ & 410N/mm$^{2}$ gauge distance is 100mm leg=60mm gauge distance in 75mm leg=40mm

a) gusset plate connected 100mm leg

i) Strength due to yielding

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

=$\frac{1650 \times 250}{1.1}$

Tdg=375KN

Ag=1650mm$^{2}$-steel table

or

$(100-\frac{10}{2})\times10$=950

$(75-\frac{10}{2})\times10$=700=1650$mm^{2}$

ii)Strength due to rupture

Tdn=0.9Anc $\frac{fu}{\gamma ml}+B Ago\frac{fy}{\gamma mo}$

Anc=$(100-\frac{10}{2}\times-(18)t$

$=(100-\frac{10}{2}-18)\times 10$

Anc=770mm$^{2}$

Agn=$(75-\frac{10}{2})+$

Agn=700mm$^{2}$

1. Strength of due to rupture

Tdn=0.9Anc$\frac{fu}{\gamma ml}+B Aga \frac{fu}{\gamma ma}$

B=1.4-0.075$(\frac{w}{t})\times(\frac{fy}{fu})\times (\frac{bs}{le})$

w=75

t=10

bs=75+60-10=125mm

Lc=$(5\times50$)=250mm=centre to centre distance between two hole

B=1.4-0.076$(\frac{75}{10})(\frac{250}{410}(\frac{125}{250})$

B=1.23 $\leq 1.4$

Tdn=0.9$\times 770\times\frac{410}{1.25}+1.23\times700\times\frac{250}{1.1}$

Tdn=422.98KN

3) Strength due to block shear

Tdb1$\frac{Avg fy}{\sqrt{3}\gamma mo}+\frac{0.9 Afn.fu}{\gamma m1}$

Tdb2=$(\frac{0.9 Avn.fu}{\sqrt{3}.\gamma m1})+\frac{Afg \ fy}{\gamma mo}$

Avg (length $\times$thickness)

(Lckt 30)$\times$10

=280$\times$10

Avg=2800mm$^{2}$

Avn =[(280$\times10)-(5.5\times18)10]$

Avn=$1810mm^{2}$

Afg=(40$\times10)$

Afg=400mm$^{2}$

A+n=[40-$\frac{1}{2}\times18]10$

A+n=310mm$^{2}$

Tdb1=$\frac{2800\times250}{\sqrt{3}\times1.1}+\frac{0.9\times310\times250}{1.25}$

Tdb1=470.7ekN

Tdb2=$\frac{0.9\times1810\times410}{\sqrt{3}\times1.25}+\frac{400\times250}{1.1}$

Tdb$_{2}$=399.39$\times10^{3}$N

select 399.39 for block shear

Design tensile strength =375KN

(least of 375 KN,422.98KN,399.39KN]

b)when 75mm leg connect to gusset plate

i) Tdg

I) Tdg=$\frac{Ag\times fy}{\gamma ma}$

Tdg=$\frac{1650\times250}{1.1}$

Tdg=375KN

ii)Anc=$[75=\frac{10}{2}\times10-[18]10$

Anc=520mm$^{2}$

Ago=$[100-\frac{10}{2}]t$

Ago=950mm$^{2}$

B=1.4-0.076$(\frac{w}{t})(\frac{fy}{fu})(\frac{bs}{Ls})$

w=100

t=10

bs=100+40-10=130

Lc=5$\times50$

B=1.4-0.076$(\frac{100}{10})(\frac{250}{410})(\frac{130}{250})$

0.7$\geq\beta=1.159\leq1.4$

Tdn=$0.9\times 520\times\frac{410}{1.25}+1.159\times 950\times\frac{250}{1.1}$

Tdn=403.74KN

Tdb$_{1}=\frac{Avg.fy}{\sqrt{3} \gamma mo}+0.9\times Afn\times\frac{fu}{\gamma m1}$

$Tdb_{2}=\frac{0.9\times Avn\times fy}{\sqrt{3} \gamma m1}+Afg\times\frac{fy}{\gamma \ mo}$

Avg=280$\times10=2800mm^{2}$

Avn=[(280$\times10)-(5.5\times18)]10=1810mm^{2}$

Afg=$35\times10=350mm^{2}$

Afn=$(35-\frac{1}{2}\times18)10=260mm^{2}$

Tdb$_{1}=\frac{2800\times250}{\sqrt{3}\times1.1}+0.9\times360\times\frac{410}{1.25}$

Tdb$_1$=444.20KN

Tdb$_2=\frac{0.9\times1810\times250}{\sqrt{3}\times1.25}+\frac{350\times250}{1.1}$

Tdb$_2$=388.030KN

design tensile strength of given angle 375 KN

[least of 378, 403.74,388.030]