# Beam Design: Reinforced Concrete Beam, Beam Steel Design

By BYJU'S Exam Prep

Updated on: September 25th, 2023

Design of Beams: Beams are flexural members, i.e., the beam takes to load by bending, and flexural members carry bending moment and shear force. Therefore, the design of beams is for bending moment, shear force, and torsion. The design of beams can be done by various methods. The most prominent ones are Working Stress Method and Limit State Method.

The design of beams is done taking into consideration many factors. We will see all the major factors and the principles used in the design of beams. We will also learn about the moment of resistance, singly reinforced and doubly reinforced beams.

Table of content

## How to Design a Beam?

When external loads are applied, moments are generated, and the maximum moment developed varies in each beam depending on the support conditions and loading combinations. Let M be the maximum moment, also termed as unfactored moment/service moment/working moment.

Factored External Moment = 1.5 x Working Moment

Mu = 1.5M

Design of Beams is done for the factored external moment (Mu)

Maximum internal resistance without failure or ultimate moment of resistance is called Moment of Resistance (MOR). The externally applied moment should be always less than or equal to the internal moment of resistance during the design of beams.

Mu≤MOR

Where,

• MOR = CZ or TZ
• C- Compressive Force
• Z- Lever Arm
• T-Tensile Force

## Singly Reinforced Beam

If Mu > MORlim, the beam design principle will not be satisfied. Therefore, for the design of beams as singly reinforced beams of under reinforced type, Mu < MORlim. Limiting the moment of resistance can be increased by increasing fck or by increasing the depth, but both will increase the cost of construction. Therefore, in such cases, beams are designed as doubly reinforced beams.

### Area of Steel

Mu = MOR

Mu = TZ

Mu = 0.87 fyAst(d-0.42xu)

Ast = 0.5 × fck/fy [1- √(1- 4.6Mu/bd2fck)]bd

Where,

• fck – Characteristic Compressive Strength of Concrete
• fy – Yield Strength of Steel

The area of steel provided should be greater than the area of steel required.

Astmin < Astmin < Astmax

Maximum area of tension steel = Astmax = 0.04 x bD

The area of tension steel provided should be less than the maximum area of tension to avoid congestion during concreting.

The minimum area of tension steel = Ptmin % = Percentage minimum area of tensile steel.

• Ptmin = Ast min/bd ×100
• Ptmin = 85%/fy
• Ast min = 0.85bd/fy

The area of tension steel provided should be more than the minimum area of steel to avoid sudden failure.

 fy Pt min 250 0.34 415 0.205 500 0.17

For the design of beams (singly reinforced and under reinforced beam).

Ast < Astlim

Ast < Astmax

### Some special cases

1. When Xu < Xu,lim
It is an under-reinforced section
2. When Xu = Xu,lim
It is a balanced section
3. When Xu > Xu,lim
It is over reinforced section. In this case, keep Xu limited to Xu,I’m, and the moment of resistance of the section shall be limited to limiting moment of resistance (Mu,lim)

## Doubly Reinforced Beam

A beam is made doubly reinforced beam when a singly reinforced beam becomes over the reinforced beam, i.e., MOR > MORlim. An over-reinforced beam is sudden. In such a case, a beam is designed as a doubly reinforced beam. Doubly reinforced beams are beneficial in case of stress reversal. The compression steel helps in bearing additional strain due to creep and shrinkage. Compression steel is found to reduce deflections. They are helpful in case of shock or impact loads. Overall more ductile compared to single reinforced beams.

Total Compressive Force C =C1 + C2

C=0.36fckbxu+(fsc-fcc)Asc

Where

• fsc = Compressive Stress in compression steel
• fcc = Stress in concrete at the level of compression steel
• fcc = 0.45 fck

Total Tensile Force T = 0.87 fy Ast

Doubly reinforced beams are not economical because compression steel is under stress.

## Design of Doubly Reinforced Beam

Doubly reinforced beams are designed for a moment equal to

Mu = MORlim+(Mu– MORlim)

Area of tension steel, Ast = Ast1 + Ast2

Ast2 = Mu-MORlim/0.87 fy (d-d’)

Area of compressive steel, Asc = Mu-MORlim/(fsc-fcc) (d-d’)

### Flanged Beam

In the case of the monolithic casting of beams and slabs, part of the slab behaves as a beam to take the compression; such beams are called flanged beams.

### Monolithically Casted T Beam

Effective flange width of T beam

bf = bw+ [6Df+ Lo/6]

Where,

• bw = Web width
• bf = Effective flange width
• Df = Flange depth/Depth of slab
• Lo = Distance between points of contra flexure

Maximum effective flange width (bfmax) = bw + Lc

bf < bfmax

Where

• Lc = Clear distance between adjacent beams

Monolithically Cast L Beam

bf = bw+ [3Df+ Lo/12]

(bfmax) = bw + Lc

## Design of Beams Sample Question for Gate

Question: Minimum area of tension steel depends on

a.) Grade of concrete b.) Grade of steel c.) Both a and b d.)None of these