RCD-FORMULA

$f_s=\frac{600(d_t-c)}{c}

$c_b=\frac{600 \ d_t}{f_y+600}

$0.85f'_c\,ab=A_sf_y

$M_n=0.85f'_c\,ab\left(d-\frac{a}{2}\right)$
$M_n=A_sf_y\left(d-\frac{a}{2}\right)

$\rho_b=\frac{0.85{f'}_c\,\beta_1}{f_y}\left(\frac{600}{600+f_y}\right) \left(\frac{d_t}{d}\right)

$\rho_{max}^{TC}= \dfrac{3}{8}\left(\frac{600+f_y}{600}\right) \rho_{bal}

$R_n=\frac{M_u}{\phi\,bd^2}

$\beta_1=0.85-\frac{{0.05(f'}_c-28)}{7}

$M_u=\phi\rho\,f_ybd^2\left(1-\frac{\rho\;f_y}{1.7f_c^\prime}\right)

$M_{umax\ T}\ =0.9\bullet\ \frac{3\ }{8}\left(0.85f_c'\ \beta_1\ bd^2\right)\left(1-\frac{3\ }{16}\beta_1\right)$

$M_{umax\ }\ =\frac{49\ }{60}\bullet\ \frac{3\ }{7}\left(0.85f_c'\ \beta_1\ bd^2\right)\left(1-\frac{3\ }{14}\beta_1\right)$

$COMPRESSION\;CONTROLLED$
$\phi=0.65

$\rho_{min}=\frac{\sqrt{f'c}}{{4f}_y}$
$\rho_{min}=\frac{1.4}{f_y}\ $

$TENSION\;CONTROLLED$
$\varphi=0.90

$TRANSITION\;REGION$
$\phi=0.65+0.25\left(\frac{f_s-f_y}{1000-f_y}\right)

$0.85f'_c\,ab+A'_s(f_s\;or\;f_y)=A_sf_y

$f_s'=\frac{600(c-d')}{c}

$M_n=0.85f'_c\,ab\left(d-\frac{a}{2}\right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +A'_s(f_s\;or\;f_y)(d-d')

$V_u=\phi(V_c+V_s)

$\downdownarrows w/o\;axial\downdownarrows$
$V_c=0.17\lambda\sqrt{{f'}_c}b_wd$
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$\downdownarrows w/\;axial \downdownarrows$
$V_c=0.17\lambda\sqrt{{f'}_c}b_wd\left(1+\frac{N_u}{14A_g}\right)$

CHOOSE SMALLEST VALUE
$V_c=\left(\dfrac{1}{6}\lambda\sqrt{{f^\prime}_c}+17\rho_w\frac{V_ud}{M_u}\right)b_wd$
$V_c=\left(\dfrac{1}{6}\lambda\sqrt{{f^\prime}_c}+17\rho_w\right)b_wd$
$V_c=0.29\lambda\sqrt{{f'}_{c\ }}b_wd

CHOOSE SMALLEST VALUE
$V_c=\left(\dfrac{1}{6}\lambda\sqrt{{f^\prime}_c}+17\rho_w\frac{V_ud}{M_m}\right)b_wd$
$V_c=0.29\lambda\sqrt{{f'}_c}b_wd\sqrt{1+\frac{0.29N_u}{A_g}}

$V_c=\ 0.17\ \lambda\ \sqrt{{f'}_c}{\ b}_w\ d\ \left(1+\frac{N_u}{3.5\ A_g}\right)$

$\sigma=\frac{F}{A}$
$f_{yt}=\frac{V_s}{A_v(\frac{d}{S})}$
$S=\frac{A_vf_{yt}d}{V_s}

$V_s\le0.33\sqrt{{f\prime}_c}\,\,b_w\,\,d$
$\ \ \ \ \ \ \ then\rightarrow\ \ \ \ \ \ \ {\ S}_{max}=\frac{d}{2}\ or\ 600$
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$V_s\ >0.33\sqrt{{f\prime}_c}\,\,b_w\,\,d$
$\ \ \ \ \ \ \ then\rightarrow\ \ \ \ \ \ \ {\ S}_{max}=\frac{d}{4}\ or\ 300$
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$S_{max}=\frac{A_vf_{yt}}{0.062\; \sqrt{{f'c}}\; b_w}$
$S_{max}=\frac{A_vf_{yt}}{0.35b_w}

$A_{vmin}=0.062\sqrt{f'c}\frac{b_w}{f_{yt}}S$
$A_{vmin}=0.35\frac{b_w}{f_{yt}}S

$\dfrac{1}{2}\phi\ V_c\le\,V_u\ <\phi\,V_c$
$therefore $
$MINIMUM\ SHEAR$$REINFORCEMENT$$AREA$

$\phi\,V_c< V_u $
therefore
$V_u=\phi\left(V_c+V_s\right)

$I_e=\left(\frac{M_{cr}}{M_{a}}\right)^3I_g+\left[1-\left(\frac{M_{cr}}{M_{a}}\right)^3\right]I_{cr}\le\ I_g$

$\dfrac{1}{2}\phi\ V_c>\,V_u\ $
$\therefore$
$S_{max}

$T_{Uth}=0.75 \left(\frac{1}{12}\right) \lambda\ \sqrt{{f'}_c}\ \left(\frac{{A_{cp}}^2}{P_{cp}}\right)$

$T_{Uth}=0.75 \left(\frac{1}{12}\right)\ \lambda\ \sqrt{{f'}_c}\ \left(\frac{{A_{cp}}^2}{P_{cp}}\right)\ \sqrt{1+\frac{3f_{pc}}{ \lambda\ \sqrt{{f'}_c}\ }}$

$T_{Uth}=0.75 \left(\frac{1}{12}\right)\ \lambda\ \sqrt{{f'}_c}\ \left(\frac{{A_{cp}}^2}{P_{cp}}\right)\ \sqrt{1+\frac{3N_u}{\lambda\ A_g\ \sqrt{{f'}_c}\ }}$

$T_{cr}=0.33\ \lambda\ \sqrt{{f'}_c}\ \left(\frac{{A_{cp}}^2}{P_{cp}}\right)$

$T_{cr}=0.33\ \lambda\ \sqrt{{f'}_c}\ \left(\frac{{A_{cp}}^2}{P_{cp}}\right)\ \sqrt{1+\frac{3f_{pc}}{ \lambda\ \sqrt{{f'}_c}\ }}$

$T_{cr}=0.33\ \lambda\ \sqrt{{f'}_c}\ \left(\frac{{A_{cp}}^2}{P_{cp}}\right)\ \sqrt{1+\frac{3N_u}{ \lambda\ A_g\ \sqrt{{f'}_c}\ }}$

Lesser of the Two
$T_n=\ \frac{2\ A_o\ A_t\ f_{yt}}{S}\ cot\theta$
$T_n=\ \frac{2\ A_o\ A_l\ f_{yt}}{P_h}\ cot\theta$

$A_l\ =\ P_h\frac{A_t\ }{S}\frac{f_{yt}\ }{f_y}\ {cot}^2\ \theta$

CHOOSE SMALLEST

300 or $\frac{P_h}{8}$

CHOOSE LARGER
At = area of one shear reinforcement
(one leg only).

CONSIDERED AS SHORT COLUMN

$f_r=\ 0.62\lambda\sqrt{{f\prime}_c}\ =\ \frac{M_{cr}\ y_{tension}}{I_g}$
$f_r=\ 7.5\lambda\sqrt{{f\prime}_{c(psi)}}\ =\ \frac{M_{cr}\ y_{tension}}{I_g}$

$trian\ to\ distd\ load=\ \frac{qs}{3}$
$trape\ to\ distd\ load=\ \frac{qs}{3}\ \left(\frac{3-\ m^2}{2}\right)$

$E_c=\ 0.043\ {w_c}^{1.5}\sqrt{{f'}_c}$

$E_c=\ 4700\ \sqrt{{f'}_c}$

LOADS used on FINDING the
SUSTAINED DEFLECTION
are SERVICE LOADS (NOT FACTORED)\*

$\lambda_\Delta=\frac{\xi}{1+\ 50\ \rho'}$

$I_e^{conti}=\frac{I_e^{+critical}+I_e^{-critical}}{2}$

$b_f=\left[\begin{matrix}\frac{L}{4}\\b_w\ +\ 16\ t_f\\b_w\ +\ \ \frac{S_1}{2}+\frac{S_2}{2}\\\end{matrix}\right]$

$b_f=\left[\begin{matrix}b_w\ +\frac{L}{12}\\b_w\ +\ 6\ t_f\\b_w\ +\ \ \frac{S_1}{2}\\\end{matrix}\right]$

$b_f=\left[\begin{matrix}b_w\ +\frac{l_n}{12}\\b_w\ +\ 6\ t_f\\b_w\ +\ \ \frac{S_1}{2}\\\end{matrix}\right]$

$b_f=\left[\begin{matrix}\frac{l_n}{4}\\b_w\ +\ 16\ t_f\\b_w\ +\ \ \frac{S_1}{2}+\frac{S_2}{2}\\\end{matrix}\right]$

$P_u=\ 0.75\ \left(0.85\right)\left(0.85\ {f'}_c\left(A_g-A_{st}\right)+A_{st}f_y\right)$

$P_u=\ 0.65\ \left(0.80\right)\left(0.85\ {f'}_c\left(A_g-A_{st}\right)+A_{st}f_y\right)$

choose smallest
$\frac{4}{3}A_g\ ,\ 16\ d_B\ ,\ 48\ d_{st}$
$and\ $
$Least\ Column\ Dimension$

$V_c=\ \frac{1}{6}\lambda\ \sqrt{{f\prime}_c}\left(0.8D^2\right)$

choose largest
$\rho_s=0.45\left(\frac{D^2}{{D_c}^2}-1\right)\frac{{f'}_c}{f_{yh}}$
$\rho_s=\dfrac{{0.12\ f'}_c}{f_{yh}}\ $

$S\ =\ \frac{\pi\left(D_c-d_{st}\right)}{\rho_s}\left(\frac{d_{st}}{D_c}\right)^2$

BETWEEN
$\frac{4}{3}A_{g\ }\ or\ \left(25mm+d_{st}\right)$
to
$\left( 75mm+d_{st}\right)$

$\rho_{min}\ \ \ to\ \ \ \rho_{max}\ $
$0.01\ -\ \ 0.08$

$\rho_{min}\ \ \ to\ \ \ \rho_{max}\ $
$0.02\ -\ \ 0.08$

$P_{c\ }=\ \frac{\pi^2\ \left(EI\right)_{eff}}{\left(kL_u\right)^2}$

$\left(EI\right)_{eff}=\ \frac{0.40\ E_c\ I_g}{1+\ \beta_{dns}}$
$\left(EI\right)_{eff}=\ \frac{0.20\ E_c\ I_g\ +\ E_s\ I_{se}}{1+\ \beta_{dns}}\ $
$\beta_{dns}\ =\frac{factored\ sustained\ load\ (usually\ DL)\ }{total\ factored\ load}$

$\ G_{Above\ or\ Below}\ =\frac{\sum\left(\frac{EI}{L}\right)_{all\ conn.\ \ columns}}{\sum\left(\frac{EI}{L}\right)_{all\ conn.\ \ beams}}$

$\delta\ =\ \frac{C_m}{1\ -\frac{P_u}{0.75P_c\ }}\ \geq\ 1.00$

Moment of Inertia about N.A

$\frac{f_s}{n}=\ \frac{M\ y_{sb}}{I_T}$

$f_c=\ \frac{M\ y_{c}}{I_T}$

$\frac{f'_s}{2n}=\ \frac{M\ y_{st}}{I_T}$

$\frac{E_s}{E_c}\geq\ 6$

$f_c=\ 0.45{f'}_c$

$for\ G280\ \ steel\ \ \ \ \ \ \ \ \ f_s=\ 140\ MPa $
$for\ G420\ \ steel\ \ \ \ \ \ \ \ \ f_s=\ 170\ MPa$

ALWAYS
$f_s\ = f_y$
FOR
SINGLY-REINFORCED BEAM (SRB)
ONE-WAY SLAB
FOOTINGS

$\rho_{min}\ \ \ to\ \ \ \rho_{max}\ $
$0.01\ -\ \ 0.06$

EQUATE to S, and CHOOSE SMALLEST

$\frac{A_{sh}\ }{S\ \bullet\ b_c}\ =\ 0.3\ \left(\frac{D^2}{{D_c}^2}-1\right)\frac{{f'}_c}{f_{yt}}$
$\frac{A_{sh}\ }{S\ \bullet\ b_c}\ =\ 0.09\left(\frac{{f'}_c}{f_{yt}}\right)$

$\rho_{max}^{TC}=\ \frac{0.85\ {f\prime}_c\ }{f_y}\ \bullet\beta_1\ \left(\frac{3}{8}\right)\ \left(\frac{d_t}{d}\right)$

$\rho_{max}^{TR}=\ \frac{0.85\ {f\prime}_c\ }{f_y}\ \bullet\beta_1\ \left(\frac{3}{7}\right)\ \left(\frac{d_t}{d}\right)$

in DETERMINING the
MAXIMUM NUMBER of BARS
using $\rho_{max}$
ROUND DOWN to NEAREST WHOLE NUMBER of the obtain PIECES

MINIMUM STRAIN ALLOWED by the CODE
$\epsilon = 0.004$
to have
ALLOWABLE MAX REINFORCEMENT

TRANSITION CONTROLLED

IN EULER's CRITICAL AXIAL LOAD
$P_{cr}$
the WEAK MOMENT of INERTIA
with reduction factor
MUST be USED
in REINFORCED CONCRETE

$\left(\dfrac{A_v}{s}\right)^{combined}=\left(\dfrac{A_v}{s}\right)^{shear} + \left(\dfrac{A_t}{s}\right)^{torsion}$
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$\left(\dfrac{n A_b}{s_{req}}\right)=\left(\dfrac{A_v}{s_v}\right) + \left(\dfrac{n A_t}{s_t}\right)$