STEEL FORMULA

$f_v=\frac{V}{dt_w}

$F_y=248 MPa$
or
$F_y=345 MPa$

$F_u=400 MPa$

$F_u=428 MPa$

$q=\frac{VQ}{I}$

sample unit
(N/mm)

$f_v=\frac{VQ}{I\ t_w}

$f_b=\frac{Mc}{I}=\frac{M}{S_x}

$s=\frac{RI}{VQ}

$R=F_{vbolts}(A_v)(no.\,of \,bolts)

$S_x=\frac{I}{c}$

$F_{cr}=\frac{\pi^2\ E}{\left(\frac{k\ L_u}{r}\right)^2}

$SR=\frac{k\ L_u}{r}

$r=\sqrt{\frac{I}{A}}

$C_c=\sqrt{\dfrac{2\pi^2\,E}{F_y}}

FACTOR of SAFETY
is equals to
$\dfrac{5}{3}+\dfrac{3}{8}\left(\dfrac{\;\;\dfrac{kL}{r}\;\;}{C_c}\right)-\dfrac{1}{8}\left(\dfrac{\;\;\dfrac{kL}{r}\;\;}{C_c}\right)^3

$SHORT\; COLUMN$
$F_a=\left(1-0.5\left(\dfrac{\;\;\dfrac{kL}{r}\;\;}{C_c}\right)^2\right)\left(\frac{F_y}{FS}\right)

$LONG\,COLUMN\,$
$F_a=\dfrac{12\pi^2\,E}{23\left(\dfrac{kL}{r}\right)^2}

$LONG\,COLUMN\,$
$F_a=F_{cr}=0.877\,F_e

$SHORT\,COLUMN\,$
$F_a=F_{cr}=\left(0.658^{\dfrac{F_y}{F_e}}\right)\,F_y

$P_a=\ 0.6\ F_y\ A_g$

$P_a=\ 0.5\ F_u\ A_e$
$ ;\ A_e=U{\ A}_n$

$P_a=\ 0.3\ F_u\ A_{ev}$

$P_a=\ 0.3\ F_u\ A_{ev\ }+0.5\ F_u\ A_{et\ } $
$\; A_e=U{\ A}_n$

$P_a=\ F_v\ A_{v\ }$

$P_a=\ F_p\ A_{p\ }$

SAMPLE CALC

SAMPLE CALC

$\left(F_{v_{weld}}\right) = 0.3\ F_{u_{weld}}

$A_n=\ A_g-A_{holes}\ +\ \left(t\right)\ \sum\left(\frac{s^2}{4g}\right)$

$J=\ n\ \left(x\sigma n\right)^2+n\ \left(y\sigma n\right)^2$

$I_{\bar{x}}=\ \sum{\ y}^2\ \ \ \ \ \ \ I_{\bar{y}}=\ \sum{\ x}^2$

the reaction due to the force alone is equal and opposite in direction to the
reaction due to moment alone
$F\ =\ \frac{J}{e\ n}

$R\ =\ \left|\frac{P\ \angle\ \theta\ }{n}+\ \frac{T}{J}\left(y\ -x\ i\right)\right|$

$R\ =\ \left|\frac{P\ \angle\ \theta\ }{L_T}+\ \frac{T}{J}\left(y\ -x\ i\right)\right|$

$APPS\ \Longrightarrow\ \bar{x}\ \ \&\ \ \bar{y}$

$J=\ \sum_{1}^{n}\left(\frac{{L_n}^3}{12}\right)\ +\ n\ \left(\left(x\sigma n\right)^2+\ \left(y\sigma n\right)^2\right)$

$\downdownarrows \left[STRENGTH\ CRITERION\right] \downdownarrows$
$\frac{f_a}{F_a}\ +\ \frac{f_{bx}}{F_{bx}}\ +\frac{f_{by}}{F_{by}}\ \le\ 1.0$

_____________________________________
$\downdownarrows\ \left[STABILITY\ CRITERION\right] \downdownarrows$
$\frac{f_a}{F_a}\ +\ {\rm MF}_x\frac{f_{bx}}{F_{bx}}\ +{\rm MF}_y\frac{f_{by}}{F_{by}}\ \le\ 1.0\ $
_____________________________________
$\downdownarrows\ \left[STRENGTH\ CRITERION\right]\downdownarrows$
$\frac{f_a}{{0.6F}_y}\ +\ \frac{f_{bx}}{F_{bx}}\ +\frac{f_{by}}{F_{by}}\ \le\ 1.0\ \ \ \ \ $
_____________________________________

${MF}_x\ or\ {MF}_y=\ \frac{C_m}{\left(1-\ \frac{f_a}{{F'}_e}\right)}$

${AF}_x\ or\ {AF}_y=\ \left(1-\frac{f_a}{{F'}_e}\right)$

${F'}_e=\ \frac{12\ \pi^2\ E\ }{23\ \left(\frac{k\ L_b}{r_b}\right)^2}$

$F_p=\ 0.35\ {f'}_c$

$F_p=\ 0.35\ {f'}_c\ \sqrt{\frac{A_2}{A_1}}\ \le\ 0.70\ {f'}_c$

$t_p=\ 2\ l_{(x\ or\ y)}\ \sqrt{\frac{f_p}{F_y}}$

$R_{max}=\left(0.66 F_{yw}\right) t_w\ \left(N+2.5\ k \right)$
$f_{yw}=\dfrac {R}{t_w \left(N+2.5\ k\right)}$

$P_{max}=\left(0.66 F_{yw}\right) t_w\ \left(N+5\ k \right)$
$f_{yw}=\dfrac {P}{t_w \left(N+5\ k\right)}$

$P\ \le\ 177.2\ \left(t_w\right)^2\left[1+3\ \left(\frac{N}{d}\right)\left(\frac{t_w}{t_f}\right)^{1.5}\right]\sqrt{F_{yw}\left(\frac{t_f}{t_w}\right)}$

$R\ \le\ 89.3\ \left(t_w\right)^2\left[1+3\ \left(\frac{N}{d}\right)\left(\frac{t_w}{t_f}\right)^{1.5}\right]\sqrt{F_{yw}\left(\frac{t_f}{t_w}\right)}$

$\frac{f_{bx}}{F_{bx}}\ +\frac{f_{by}}{F_{by}}\ \le\ 1.0$

$\frac{f_t}{F_t}+\frac{f_{b_x}}{F_{b_x}}+\frac{f_{b_y}}{F_{b_y}}\ \le1.0$

$L_1=\frac{200\ b_f}{\sqrt{F_y}}\;\;\;\;\;\;\;\;\ L_2=\frac{137,\ 900}{\left(\frac{F_y\ d}{A_f}\right)}$

COMPACT BEAM in
STRONG AXIS BENDING
$$
therefore CHECK if FLANGE and
WEB is COMPACT

$\Downarrow\ \ \ FLANGE\ COMPACTNESS\ \ \ \Downarrow$
$\frac{b_f\ }{2{\ t}_f}\ \sqrt{F_y}<\ 170\ \ \left[\begin{matrix}if\ satisfy\\\therefore COMPACT\\\end{matrix}\right]$

$\Downarrow\ \ \ WEB\ COMPACTNESS\ \ \ \Downarrow$
$\frac{d\ }{{\ t}_w}\ \sqrt{F_y}<\ 1680\ \ \left[\begin{matrix}if\ satisfy\\\therefore COMPACT\\\end{matrix}\right]$

$F_b=\ 0.66\ F_y$

$170<\frac{b_f\ }{2{\ t}_f}\ \sqrt{F_y}<\ 250\ $
$\ \left[\begin{matrix}if\ satisfy\\\therefore PARTIALLY\\COMPACT\\\end{matrix}\right]$

$F_b=\ F_y\ \left(0.79-0.000762\ \frac{b_f\ }{2{\ t}_f}\ \sqrt{F_y}\ \right)$

$F_b=\ 0.60\ F_y$

NON-COMPACT BEAM in
STRONG AXIS BENDING

therefore
$F_b=\ 0.60\ F_y$

SLENDER BEAM in
STRONG AXIS BENDING

therefore
CHECK $\frac{L_b\ }{{\ r}_T}$ with $\lambda_1\ and\ \lambda_2$

ALWAYS USE RESISTANCE
in STONG AXIS BENDING

$e.g.\ \ \frac{M_{max}}{S_x}$
$S_x=\ Section\ Modulus\ in\ Strong\ Axis$

NSCP 2001
ASD
$$
DOES NOT USE FACTOR LOADS

NO LATERAL TORSIONAL BUCKLING
$$
NO LATERAL TORSIONAL BUCKLING

$\Downarrow\ \ \ FLANGE\ COMPACTNESS\ \ \ \Downarrow$
_____________________________________
$\frac{b_f\ }{2{\ t}_f}\ \sqrt{F_y}\leq\ 170\ \ \left[\begin{matrix}if\ satisfy\\\therefore COMPACT\\\end{matrix}\right]$
_____________________________________
$170<\frac{b_f\ }{2{\ t}_f}\ \sqrt{F_y}\leq\ 250\ \ \ \left[\begin{matrix}if\ satisfy\\\therefore PARTIALLY\\COMPACT\\\end{matrix}\right]$
_____________________________________
$\frac{b_f}{2\ t_f}\ \sqrt{F_y}>\ 250\ \left[\begin{matrix}if\ satisfy\\\therefore NON-COMPACT\\\end{matrix}\right] $

$F_b=\ 0.75\ F_y$

$F_b=\ F_y\ \left[1.075\ -\ 0.0019\frac{b_f}{2\ t_f}\ \sqrt{F_y}\right]$

$F_b=\ 0.60\ F_y$

$\# \ of \ STUDS= \dfrac{V_h}{Capacity\ of\ Each\ Connector}$

$S.F. \ = \dfrac{Z_x}{S_x}$
$S.F. \ = \dfrac{M_p}{M_y}$

$ M_p= Z_x \bullet F_y$

$F_v=0.4\ F_y=\ \frac{V}{d\ t_w}\ $

$\frac{k\ L_u}{r}\ <\ 200$

$0.6\ F_y=\frac{V_n}{d\ t_w}$