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+\chapter{Rectangular Beams}
+\section{Introduction}Ductile reinforced concrete beams under ultimate loads fail by tension failure modes with
+concrete fully or partially crushed. These modes of failure fall in the regions 3, 4 and 4A
+(\fig 1.6 and 1.7 of Chapter 1). When moment at a section is small, only tension reinforcement
+will suffice. But when compressive force in concrete compression zone is more than the concrete
+capacity, compression steel will be required at the compression face of section. Thus reinforced
+concrete beams may have two types of sections—singly or doubly reinforced sections.
+
+\section{Singly Reinforced Rectangular Section}
+When the position of neutral axis is such that xu, min S xu S. xu, max \fig \ref{Singly reinforced section 1}, the concrete
+compression zone gets subjected to the full concrete stress diagram as given by \fig 21 of the
+Code. The equilibrium of the section \fig \ref{Singly reinforced section}. gives,
+\begin{equation}
+T=Cw
+\label{Singly Reinforced}
+\end{equation}
+\begin{equation}
+M_u=C_w.(d-0.42x_u)
+\label{Reinforced Section}
+\end{equation}
+with
+\hspace{2cm} $$T=A_st \times 0.87f_y$$
+$$C_w=0.36f_ck\hspace{0.1cm}.\hspace{0.1cm} b\hspace{0.1cm} .\hspace{0.1cm} x_u$$
+and introducing two non-dimensional parameters,
+$$\mu=\frac {A_st}{bd} \times 0.87\frac {f_y}{f_ck}$$
+$$k=\frac{M_u}{f_ckbd^2}$$
+\begin{figure}
+\centering
+\includegraphics[width=0.8\textwidth]{images/ch2-1.png}
+\caption{Singly reinforced rectangular section with concrete fully crushed at collapse.}
+\label{Singly reinforced section 1}
+\end{figure}
+%---------------------------------------------------------------------------------------
+\newpage
+Equations \eqn \ref{Singly Reinforced}  and \eqn \ref{Reinforced Section}  give, on simplification,
+\begin{equation}
+\mu = 0.36n
+\label{Concrete}
+\end{equation}
+\begin{equation}
+k=\mu(1-0.42n)
+\label{Crushed at collapse}
+\end{equation}
+These equations are the same as given by the Code in its Appendix G-1.1. The Code has,
+however, not made clear that these equations do not apply for low values of n, When the tension
+steel strain exceeds the maximum tension steel strain, for the simple reason that the Code
+has made no stipulation in this regard. The Code has instead given the provision of the
+minimum area of tension reinforcement,
+$$\frac{A_s(min)}{bd} = \frac{0.85}{f_y}$$
+which will greatly limit the tension steel strain and thereby restrict cracking and spalling of
+concrete of tension zone.\\
+When the position of neutral axis is such that O S xu S xu, min, the concrete compression
+zone is subjected to only a part of the standard concrete stress diagram given by the Code and
+the above expressions are not valid for this range. The conditions of equilibrium of the section
+\fig \ref{Singly reinforced section}give on simplification,
+\begin{figure}
+\centering
+\includegraphics[width=0.8\textwidth]{images/ch2-2.png}
+\caption{Singly reinforced rectangular section with concrete not fully crushed at collapse.}
+\label{Singly reinforced section}
+\end{figure}
+\begin{figure}
+\centering
+\includegraphics[width=0.8\textwidth]{images/ch2-3.png}
+\caption{Values for non-dimensional parameters ${\alpha}$,${\beta}$ and ${\gamma}$ \fig \ref{Singly reinforced section}.}
+\label{Values for parameters}
+\end{figure}
+%--------------------------------------------------------------------------------------------
+\newpage
+\begin{equation}
+\mu=\beta.n
+\label{non-dimensional}
+\end{equation}
+\begin{equation}
+k=\mu(1-\gamma.n)
+\label{parameters}
+\end{equation}
+where
+\begin{equation}
+\hspace{2cm} \beta = \frac{C_w}{f_ck\hspace{0.1cm}.\hspace{0.1cm}b\hspace{0.1cm}.\hspace{0.1cm}x_u}
+\label{Beta value}
+\end{equation}
+\begin{equation}
+\gamma = \frac{\bar{x}}{x_u}
+\label{stress}
+\end{equation}
+The concrete stress at the extreme fibre of section \fig \ref{Singly reinforced section} is given by
+\begin{equation}
+f_c = \alpha.f_ck
+\label{extreme fibre section}
+\end{equation}
+The non-dimensional parameters on $\alpha$, $\beta$, and $\gamma$ can be read from \fig \ref{Values for parameters}, which gives curves for
+these parameters with respect to the concrete compression strain at the extreme fibre of section
+(81) derived simply by rules of Geometry.
+
+
+Equations \eqn \ref{non-dimensional}and \eqn \ref{parameters} apply when n $<$ $(n_{min})$ and Eqs. \eqn \ref{Concrete} and \eqn \ref{Crushed at collapse} apply when $(n_{min})$ ${\leq}$ n ${\leq}$ $(n_{max})$.
+When n $>$ $(n_{max})$, the Code prescribes that n = $(n_{max})$ must be assumed for
+ensuring ductility. With n =$(n_{max})$, equation \eqn \ref{Section4} gives the limiting moment of resistance
+(M u, am) of singly reinforced rectangular section as,
+\begin{equation}
+M_{u,lim} = 0.36 n_{max}(1-0.42n_{max})\hspace{0.1cm}.\hspace{0.1cm}f_ck\hspace{0.1cm}.\hspace{0.1cm}bd^2
+\label{Singly R Section}
+\end{equation}
+\section{Doubly Reinforced Rectangular Section}
+When the applied moment $M_u$ is greater than the limiting moment of resistance $(M_{u,lim})$
+of a given singly reinforced rectangular section, compression reinforcement is required to be
+provided giving a doubly reinforced section. The equilibrium of the section \fig \ref{reinforced section} gives,
+\begin{equation}
+T = C_w + C_s
+\label{Equillibrium of section}
+\end{equation}
+\begin{equation}
+M_u = C_w(d-0.42x_{u,max}) + C_s(d-d ')
+\label{EOSII}
+\end{equation}
+
+\begin{figure}
+\centering
+\includegraphics[width=0.8\textwidth]{images/ch2-4.png}
+\caption{Doubly reinforced rectangular section.}
+\label{reinforced section}
+\end{figure}
+Introducing an additional non-dimensional parameter,
+$$\mu' = \frac{A_{sc}}{bd} \times \frac{f_{sc}}{f_{ck}}$$
+These equations give, on simplification,
+\begin{equation}
+\mu=0.36n_{max} + \mu'
+\label{Additional parameter}
+\end{equation}
+%-----------------------------------------------------------------------------------------
+\newpage
+\begin{equation}
+k=0.36n_{max}(1-0.42n_{max}) + \mu'(1-\frac{d'}{d})
+\label{APII}
+\end{equation}
+Value of $(f_{sc})$ is to be read from \fig 23 of the Code corresponding to compression steel strain
+($\varepsilon_{sc}$), which is given by \fig \ref{Doubly reinforced rectangular section.},
+
+\begin{equation}
+\varepsilon_{sc} = 0.0035(1-\frac{1}{n_{max}}.\frac{d'}{d})
+\label{Compression steel strain}
+\end{equation}
+\section{Design Charts}
+Equations \eqn \ref{Concrete} to \eqn \ref{parameters} are used to develop charts for singly reinforced rectangular
+sections. For an assumed value of n, relevant equations from those given above are used. to
+give values for k and $\mu$, which are plotted to give \chartm 2.1 and 2.2, one each for steel types
+Fe 250 and Fe 415. For doubly reinforced sections, equations \eqn \ref{Additional parameter} and \eqn \ref{APII} are used to get
+values for $\mu$ and k for an assumed value of $\mu’$ and d'/d. These give a series of straight lines for
+k, which depend on d'/d, while the straight line for$\mu’$is independent of d'/d . \chartm 2.1 and 2.2
+give all these curves with Y-Y axis representing n or $\mu’$, where n applies for singly reinforced
+sections and $\mu’$ for doubly reinforced sections. Values for$(f_{sc})$ are given against values of d'/d in
+each chart for ready use. There is an advantage in plotting curves for singly and doubly
+reinforced sections on the same co-ordinate axes, in that, the (capacity of a singly reinforced
+section need not be specially checked when the applied moment is greater than it. Use of design
+charts is illustrated by the following numerical examples.
+\section{Numerical Examples}
+\begin{example}
+ Singly Reinforced Rectangular Section (Design)
+
+\given 
+b = 30 cm, $(f_{ck})$ = 1.5 $\knpcs$
+
+ D = 60 cm, $f_y$ = 41.5 $\knpcs$
+
+d = 56.25 cm, $M_u$  = 170 $\knm$
+
+This example is taken from Design Aids$^{8}$ (Page 11).
+
+\required$(A_{st})$
+
+\solution
+$$k= \frac{170 \times 100}{1.5 \times 30 \times (56.25)^2} =0.119$$
+Chart 2.2 gives,
+$$n=0.40$$, 
+$$\mu=0.145$$
+$$A_{st}=\frac{0.145 \times 30 \times 56.25 \times 1.50}{0.87 \times 41.5} = 10.17 cm^2$$
+\end{example}
+
+\begin{example} Singly Reinforced Rectangular Section (Investigation).
+
+\given Same as Ex.2.1 with $(A_{st})$= 6.40 cm$^{2}$
+
+\required $(M_{u})$
+
+\solution
+$$\mu=\frac{6.40 \times 0.87 \times 41.5}{30 \times 56.25 \times 1.5} =0.09$$
+Chart 2.2 gives,
+$$n=0.245, \mu=0.0775$$
+$$M_{u}=0.0775 \times 1.50 \times 30 \times (56.25^2)/100= 110\knm.$$
+\end{example}
+%------------------------------------------------------------------------------
+\newpage
+\begin{figure}
+\centering
+\includegraphics[width=0.95\textwidth]{images/ch2-5.png}
+\caption{}
+\label{fig:Values for parameters}
+\end{figure}
+%-------------------------------------------------------------------------------
+\newpage
+\begin{figure}
+\centering
+\includegraphics[width=0.95\textwidth]{images/ch2-6.png}
+\caption{}
+\label{Values for parameters}
+\end{figure}
+%------------------------------------------------------------------------------------
+\newpage
+\begin{example} Doubly Reinforced Rectangular Section (Design).
+
+\given Same as in Ex. 2.1 but with $(M_{u})$= 320 kNm and d’ = 3.75 cm. This example is taken
+from Design Aids$^{8}$(Page 13).
+
+\required $(A_{st})$, $(A_{sc})$
+\solution
+$$k=\frac{320 \times 100}{1.50 \times 30 \times (56.25)^2}=0.225$$
+Chart 2.2 gives for
+$$\frac{d'}{d}=\frac{3.75}{56.25}=0.067$$ 
+$$\mu'=0.0925.\mu=0.265$$
+$$A_{st}=\frac{0.265 \times 30\times 56.25 \times 1.50}{0.87 \times 41.5}=18.58cm^2$$
+For
+$$\frac{d'}{d}=0.067$$
+Chart 2.2 gives by interpolation
+$$f_{sc}=35.48\knpcs$$
+$$A_{sc}=\frac{0.0925 \times 30 \times 56.25 \times 1.50}{35.48}=6.60cm^2$$
+\end{example}
+
+
+\begin{example} Doubly Reinforced Rectangular Section (Implementation).
+
+\given  Same as in Ex. 2.1 but with $(A_{sc})$=4.0cm$^{2}$
+$$A_{st}=18.0cm^2 \hspace{0.1cm}and\hspace{0.1cm} d'=3.75$$
+\required $M_u$
+
+\solution For
+$$\frac{d'}{d}=0.067$$ \chartm 2.2 gives,
+$$f_{sc}=35.48\knpcs$$
+$$\mu'=\frac{4.0 \times 35.48}{30 \times 56.25 \times 1.50}=0.056$$
+Chart 2.2 gives for $\frac{d'}{d}$=0.067
+$$k=0.19$$
+$$M_u= 0.19 \times 1.50 \times 30 \times (56.25)^2/100=270\knm.$$ 
+A check on the adequacy of tension steel area is further required. For $\mu'$ = 0.056, \chartm 2.2
+gives
+$$\mu=0.23$$
+$$A_{st}(required)=\frac{0.23 \times 30 \times 56.25 \times 1.50}{0.87 \times 41.5}=16.12cm^2$$
+which is less than the provided $(A_{sc})$= 18.0 cm$^{2}$
+
+The capacity of the given section is, therefore, correctly computed to be 270 $\knm$.
+\end{example}
+%-----------------------------------------------------------------------------------------
+\newpage
+\section{Conclusion}
+Design charts have been furnished for singly and doubly reinforced concrete rectangular
+beams which are independent of concrete quality and are based on non-dimensional
+parameters. These charts are quite concise and incorporate the modifications necessary for
+low values of n. Further, these charts can be used with equal ease for design as well as
+investigation of a given singly or doubly reinforced rectangular section.
+
+
+
+
+
diff --git a/main.tex b/main.tex
index 64774ca..a3854c9 100755
--- a/main.tex
+++ b/main.tex
@@ -13,5 +13,8 @@
 
       	%inputing chapter1 file%
         \input{input/LimitStateDesign.tex}
+	
+	%inputing chapter2 file%
+	\input{input/RectangularBeams.tex}
         \printbibliography
 \end{document}
diff --git a/usepackage.tex b/usepackage.tex
index bae2c58..ff634cf 100755
--- a/usepackage.tex
+++ b/usepackage.tex
@@ -25,6 +25,7 @@
 \newcommand{\fetwofivezero}{$Fe$ 250}
 \newcommand{\kn}{kN} 
 \newcommand{\knpms}{kN/m^2}
+\newcommand{\knpcs}{kN/cm^2}
 \newcommand{\knm}{kNm}
 \newcommand{\mm}{mm}
 \newcommand{\npmms}{N/mm^2}