# Design of Singly reinforced Beam with Numerical Examples

We know concrete is strong in compression and weak in tension. When the load is applied to the beam, it gets compressed on the upper part and tensioned at the bottom part. Hence to overcome this tension force, we need a steel rod at the bottom of the beam.

So, when we provide steel rods in the bottom of the beam then, this type of beam is known as a singly reinforced beam but when we provide steel rods on the upper and bottom part of the beam then it will be called a doubly reinforced beam.

Here, we will discuss singly reinforced beams, Please read the article below for steps of designing singly reinforced beams.

Singly reinforced beam design- Step by Step Numerical

#### Singly reinforced Numerical Example -1

1. A rectangular reinforced concrete beam is to be simply supported on two walls of 125 mm width with a clear span of 6.0 m. The characteristics live load is 12 kN/m and the grade of concrete is M20. Also, use Fe 415 steel. What is the effective span of the Beam? Design a suitable section for bending and determine the necessary tensile steel.

Solution:

From given,

• Beam is simply supported with width of support = 125 mm
• Clear span = 6.0 m
•  Live load = 12 kN/m
•  fck = 20 N/mm²
•  fy = 415 N/mm²
• Then, effective span (leff) = ?

Design suitable section, also check in deflection:

Assume depth of beam (d) =

d = 500 mm

• Calculation of limiting value of bending moment (Muilim)

From ANNEX G of code IS 456:2000;

Muilim  =

Or,    [Muilim = 0.36 fck bxuimax  X (d-0.42 xuimax)] -96-P

also, from clause 38.1 of code IS 456:2000;

For Fe 415 steel; —–> (P-70)

Xuimax = 0.48d

= 0.48 X 500

= 240 mm

Or, Muilim   = 0.36 X 20 X 250 X 240 X (500-0.42 X 240)

• Muilim = 172.454 kN-m
• Comparison of Mu with Muilim

Since, Mu < Muilim

Hence, the section can be designed as singly reinforced section.

Calculation of area of reinforcement

From, Annex ‘G’ of Code IS 456:2000;

or, 601.5233 = Ast-1.66 × 10¹ (Ast)²

By solving: Ast=677.782 mm

Provide, diameter of bar = 20 mm

Then, number of bars required =

= 2.1585

= 3

Provide 3-20 mm ∅ bars with

Ast, provided       =

= 942 mm² > Ast, calculated ok

Check:

Minimum reinforcement required is

Ast,min

.’. Ast,min = 256.024 mm2 < 942 mm2

O.K.

Maximum reinforcement

(Ast) max = 4% of bD

= 4/100 x 250 x 550

= 5500mm2  > 942mm2

O.K.

Design summary:

b = 250mm

d = 500mm

D = 550mm

3-20 mm ∅ bars

#### Singly reinforced Numerical Example -2

2. A rectangular Beam is to be simply supported on supports of 230 mm width. The clear span of the beam is 6m. The Beam is to have a width of 300 mm. The characteristics superimposed load is 12 kN/m. Using M20 concrete and Fe 415 steel, design the beam.

Solution:

From given:

• Beam is simply supported on supports of width = 230 mm
• clear span of beam   = 6 m
• width of beam (b)     = 300 mm
• fck                                = 20 N/mm²
• fy                 = 415 N/mm²

Assumption of depth of beam (d)

Assume, depth of beam (d) =  1/12th  to  1/15th of span

= 1/12 X 6000 to 1/15 X 6000

= 500 to 400

Take, d = 400

Provide effective cover of the beam   = 500

Then, overall depth of beam       = d + eff. Cover

= 400+50

.’. D = 450 mm

Effective span calculation:

From clause 22.2 (a) of code IS 456:2000

For supply support Beam:

The effective span of the beam should be taken as smaller of following two:

1. Clear span + effective depth of beam

= 6000+400

or,

b) c/c of support

= 6000+ 230/2 + 230/2

= 6230mm

= 6.23m

.’. Effective span  (𝜗

eff.)  = 6.23m

• Calculation of total loas, total factored load and factored bending moment

From given,

Self wt. Of beam = 𝛾

RCC bD x 1

= 20×0.3×0.45×1

= 3.375

= 15.375 kN/m

.’. factored total load = 1.5 x 15.375

.’. Wu    =   23.06 kN/m

Now ,

Factored bending moment (mu) is given by

Mu    = {Wu (𝜗

eff.)}/8

= {23.06 x (6.23)2}/8

.’. Mu = 111.878 kM/m

Also, the maximum shear force occurs at the support and is given by

Vu =    (wu 𝜗

eff. )/2    =   (23.06 x 6.23)/2

.’. V= 71.832kN

• Calculation of limiting value of Bending moment (muilim)

From ANNEX “G” of Code IS 456:2000;

Also, from clause 38.1 of code 456:2000;

For, Fe 415 steel

Xui max  = 0.48 x d

= 0.48 x 400

Xui max  = 192mm

.’. Mui lim = 0.36 x 20 x 300 x 192 x (400 – 0.42 x 192)

.’. Mui lim = 132.445kN/m

• Comparison of Mu with  Mui lim

Since, Mu (111.878) < Mui lim (132.445)

Hence, the section can be designed as singly reinforced section

• Calculation of area of Reinforcement:

From ANNEX “G” of Code IS 456:2000;

Or, 774.671 = Ast – 1.729 x 10-4 (Ast)2

By solving, we get     Ast = 921.487mm2

Provider diameter of bars = 20mm

Then number of bars required is

= 2.9346

= 3

.’. Provide 3 bars of 20 mm diameter with Ast,

Provided = 3 x (π/4) x (20)2

= 942mm2 > 921.487mm2

• Design of  shear reinforcement

From clause 40.1 of code IS 456:2000;

Nominal shear stress in beam is

Where, b = total width

d = effective depth

= 0.598 N/mm

Also, percentage of reinforcement in beam is

Pt = 0.785

Then, from table 19 of code IS 456:2000;

Design shear strength of concrete

For M20 concrete and Pt = 0.785

By using linear interpolation

.’. = 0.5684 N/mm2

Again, from table 20 of code IS 456:2000; maximum shear stress              (max) for M20 concrete is

= 2.8 N/mm2

Thus,

< <max

Hence, shear force resisted by concrete

= x bd

= 0.5684 x 300 x 400

= 68208 N

.’. Shear force to be resisted by reinforcement is

Vus      = Vux bd

= 71832 – 68208

Vus = 3624 N

Now, provide 2 legged 6mm dia vertical stirrups and Fe 250 steel

Then, from clause 40.4 of code 456:2000,

For vertical stirrups

Where, fy = 250 N/mm2

Vus    = 3624 N

Asv    = 2 x π/4 x (6)2

= 56.52 mm2

d       = 40 mm (depth of beam)

sv      = spacing of stirrups to be calculated then

3624 = (0.87 x 250 x 56.25 x 400)/2

.’. sv  = 1356.854 mm

But, maximum spacing is permitted at a inimum of two:

1. 0.75 d = 0.75 x 400 = 300 mm
2. 300 mm

.’. provide Sv = 300 mm

Hence, provide 2 legged 6mm vertical stirrups @300 mm c/c spacing

• Check for deflection:

From clause 23.2.1 of code IS 456:2000;

For simply supported beam

Basic value of span of effective depth ratio = 20

Also, factors for no compressive steel (k1) = 1

for not flanged section (k2) = 1

And for modification factor (k3) for tensil steel

From figure 4 of same clause,

fs   = 265.4585 N/mm2

And also , Pt = 0.785

Then, from figure 4 of same clause,

For fs      =      235.4585 and

Pt     =      0.785

Modification factor (k3)  = 1.18

Then, maximum permitted span to effective depth ratio I.e;

= 1 x 1 x 1.18 x 20

= 23.6

Also,

=  15.575

Thus

Hence, deflection control is satisfactory.

• Check for minimum reinforcement

The minimum reinforcement is given by

Astmin = (0.85 bd)/fy

= (0.85 x 400 x 300) / 415

= 245.783 mm < 942 mm2

• Check for maximum reinforcement

The maximum reinforcement is given by

Astmax  = 4% of bD

= 4/100 x 300 x 450

= 5400mm2 > 942mm2

• Details of reinforcement are given below: