## Learn :

Design of singly reinforced rectangular section for flexure, factored moment, ultimate moment of resistance, limiting moment of resistance factor, fixing dimension of the section, Area of tension steel.

## Design of singly reinforced rectangular section for flexure

The design problem is generally of determining dimensions (cross-sectional) of a beam (b X D) and the area of steel for a known moment or load. The basic requirement for safety at the limit of collapse (flexure) is that the factored moment Msbecause of loads should not exceed the ultimate moment of resistance Mulim of the section and the failure should be ductile.

therefore $M_{u}\leq M_{u lim}$
Taking equality $M_{u}= M_{u lim}$
$=0.36f_{ck}.\frac {x_{umax}}{d}\left ( 1-\frac {0.42 x_{u max}}{d} \right )bd^{2}$
$M_{u}=R_{u}bd^{2}$
For the given material i.e., grade of concrete and type of steel, Ru is constant and is called as limiting moment of resistance factor.
$R_{u}=0.36f_{ck}\frac {x_{u max}}{d}\left ( 1-\frac {0.42x_{umax}}{d} \right )$

Table gives value of Ru for various grades of concrete and different types of steel.

### TABLE Values of Ru for Balanced Design.

 Concrete Grade Type of Steel Reinforcement Fe 250 $(R_{u} = 0.149 f_{ck})$ Fe 415 $(R_{u} = 0.138 f_{ck})$ Fe 500 $(R_{u} = 0.133 f_{ck})$ M15 2.23 2.07 1.99 M20 2.98 2.76 2.66 M25 3.72 3.45 3.32 M30 4.47 4.14 3.99 M35 5.21 4.83 4.65 M40 5.96 5.52 5.32

### Fixing Dimensions of the Section

The value of b may be suitably fixed as $\frac {d}{2} to \frac {d}{3}$ or 200 mm, 250 or 300 mm etc. The effective depth of the section is obtained as :
$d_{reqd}=\sqrt{\frac {M_{u}}{R_{u}.b}}$
This depth corresponds to the minimum depth required for a balanced section. It is desirable to adopt a value of d which is larger than dreqd, in order to obtain an under-reinforced section.
Overall depth of the beam should be expressed in rounded figures.
Having fixed the rounded off value of D, correct value of effective depth d can be obtained as follows :
$d = D – Clear cover -\phi’-\frac {\phi}{2}$
where \phi’ is the diameter of shear stirrups and \phi is the diameter of main tensile bars.

### Area of Tension Steel

Determining area of tension steel : Now, the dimensions of the beam are fixed and the required area of tension steel (Ast) can be calculated such that the ultimate moment of resistance of section is equal to the factored moment Mu
therefore $M_{u}=M_{ulim}$
$=0.87f_{y}.A_{st}(d-0.42 x_{u max})$
$A_{st}=\frac {M_{u}}{0.87f_{y}(d-0.42x_{u max})}$
The above formula is applicable for balanced section.
However, if under reinforced section are designed then Ast is calculated from the following eqn.
$M_{u}=0.87f_{y}A_{st}.d\left ( 1-\frac {f_{y}.A_{st}} {bd f_{ck}} \right )$ [Cl. G.l. (b) of IS 456]
This will give a quadratic equation in terms of Ast which can be solved easily and the smaller value of root is applicable as solution.