Design of Steel Structures

\begin{aligned} \text{At collapse,} \quad M_{p} \theta+M_{p} \phi&=P \Delta\\ \Rightarrow \quad \quad 3 M_{P} \frac{\Delta}{l}+\frac{3 M_{P} \Delta}{2 l} &=P \Delta \\ M_{P} &=\frac{2 P l}{9}\\ \text{Also,} \qquad\qquad \quad M_{P ,\text { static }}&=M_{P,\text{ kinematic}} \end{aligned}

\begin{aligned} F_{1}&=\frac{P}{n}=\frac{60}{6}=10 \mathrm{kN}\\ F_{2}&=\frac{P \cdot e}{\sum r_{i}^{2}} \times r_{R}=\frac{60 \times 100\mathrm{~mm}}{4 \times 50^{2}+2 \times 40^{2}} \times 40\\ F_{2}&=\frac{60 \times 100 \times 40}{10000+3200}=\frac{60 \times 100 \times 40}{13200}=\frac{200}{11} \mathrm{kN}\\ F_{R}&=F_{1}+F_{2}\\ &=10+\frac{200}{11}=\frac{310}{11}=28.2 \mathrm{kN} \end{aligned}

Design capacity of welded connection
P_s=f_b \times l_{eff}\times t_t

\begin{aligned} P&=\frac{410}{\sqrt{3} \times 1.25} \times 520 \times 0.7 \times 6 \\ &= \frac{716352}{\sqrt{3}}=413586N\\ &= 413.586kN \end{aligned}

Ratio of \frac{M_p}{M_y}= Shape factor

q = Shear stress at the level mn in the weld = 100 MPa =\frac{SA\bar{y}}{Ib}
Shear force at the given section
A = Area of the cross-section above the level mn=100 \times 10 mm^2
\bar{y}=C.G. of shaded area above the level mn = 60-5 = 55 m
I=7.73 \times 10^6 mm^4
b = Width of weld at mn (4 welds) = 4 x t = 4 x 7 = 28 mm
t = Throat thickness
\begin{aligned} &=0.7 \times s=0.7 \times 10 \times 4=28 mm \\ 100&= \frac{F \times (100 \times 10) \times 55}{7.73 \times 10^6 \times 28}\\ F&=\frac{100 \times 7.73 \times 10^6 \times 28}{1000\times 55} \\ &=393.527kN \end{aligned}

Web bearing strength
\begin{aligned} &=[b+2.5(t_f+R)]\times t_w \times \frac{f_y}{\gamma _{mo}} \\ &= [75+2.5(20+10)] \times 8 \times \frac{250}{1.1}\\ &= 272.73kN \end{aligned}

\lambda =\sqrt{\frac{f_y}{f_{cr}}}=\sqrt{\frac{250}{1000}}=0.5

Shear flow in horizontal member (flange) is linear with zero at free end and in vertical member (web) it is parabolic.

The given support conditions indicates the following support/ end conditions of column

l_{eff} as per theoretical conditions = 1.0l_{o}
l_{eff} as per IS 800 : 2007 = 1.2 l_{o}
Considering the errors that may occur due to construction of supports on site.

Direct concentrated load (P) = 350 kN
Eccentricity (e) = 100 mm
Depth of weld (d_w) at each face = 500 mm
Size of weld (S) = 10 mm
Throat thickness of weld (t_t) = 0.7S = 7 mm
It is a case of combined shear (P) and Bending moment (M = P.e)
Shear stress acting on weld due to direct shear force P,
\begin{aligned} (q)&=\frac{P}{A_w} \\ &= \frac{P}{2d_wt_t}\\ &=\frac{350 \times 10^3}{2 \times 500 \times 7}=50MPa \end{aligned}
Bending normal stress acting at i^{th} point from Neutral axis [f_i]=\frac{M}{I}y_i
Maximum bending stress at extreme point in weld
\begin{aligned} [f]&=\frac{M}{I}y_{max} \\ &= \frac{Pe}{2t_t \frac{d_w^3}{12}} \times y_{max}\\ &= \frac{350 \times 10^3 \times 100}{2 \times 7 \times \frac{(500)^3}{12}} \times 250\\ &= 60MPa \end{aligned}

As per IS800:2007. Maximum resultant stress an equivalent stress in fllet weld due to combined shear and bending stress
\begin{aligned} (f_{eq})&=\sqrt{f^2+3q^2} \\ &= \sqrt{60^2+3 \times 50^2}\\ &=105.3MPa \end{aligned}