For the structural design of any structural member, the structural engineer should be familiar with one or more structural design software such as SAP2000, ETABS, STAAD PRO, SAFE or any other design software. Since these design softwares work on the basis of hit and trials, a structural engineer should guess initiation value. For this he/she should go through the preliminary design as for beam – preliminary deign of beam, then only will reach the target value in first or second trial in most cases.

Before detailed design of beam, it will be the better practice to know about the preliminary design. Primary beam means that beam which is supported by two column and secondary beam means that beam which is supported on beam on one or both end.*Note: All the design procedure are based on Indian Standard codes but if you use other codes than this, that your country accept, then also it would not be wrong for the preliminary case.*

There is no any standard rule for secondary beam. The preliminary sizing of secondary beam is done same as normal (primary) beam but the size should be reduced by 1.5 times of the size obtained.

Table of Contents

**Preliminary design of beam:**

The preliminary design is based on serviceability condition and check for strength in detailed design.

For deflection control, as per IS 456:2000, Clause 23.2.1,

L/d ≤ αβγδλ

Where,

L = length of beam

d = Effective depth of the beam

α = basic value

= 20, for simply supported

= 26, for continuous

β = modification factor for length more than 10m

= 10/span

γ = modification factor for tensile reinforcement

δ = modification factor for compression reinforcement

λ = modification faction for web-flanged ratio

Assumptions:

Percentage of steel = 1.2% so modification factor β = 0.9 (for fe415 )

Modification factor δ and λ = 1

For span less than 10 m;

For simple supported case, αβγδλ = 20**0.9 =18 *

* For continuous case*, *αβγδλ = 26**0.9=23.4

The **width of beam** is normally taken as 1/2 to 2/3 of the depth of the beam but it should not be less than the thickness of wall supported on it. The beam with least width by depth ratio gives more economical design but the ratio should not be less than 0.3. [ **From ductile detailing code**]

*In residential building, If we go through this value, it satisfies the serviceability criteria but this cause some difficulties in strength (flexural and shear) criteria. So, designer mostly prefer the ratio of span to depth in the range from 12 to 16. But it would not be wrong if we go through serviceability criteria and enlarge depth by certain value.*

Let’s take an example of beam with one side discontinuous having span of 5m.

Let’s take an example of beam with one side discontinuous having span of 5m.

From deflection criteria, as per IS456:2000, Clause 23.2.1,

L/d ≤ αβγδλ

Assumptions:

Percentage of steel = 1.2% so modification factor β = 0.9 (for fe415 )

Modification factor δ and λ = 1

For span less than 10 m;

For simple supported case

αβγδλ = 20**0.9 =18 *

*For continuous case αβγδλ = 26**0.9=23.4

Since, one side continuous and one sides discontinuous

l/d =20.7 (average of simple supported and continuous)

d=span/20.7

d= 5000/20.7

d=241mm

Adopt d= 300mm

Diameter of tensile steel bar = 16mm

Nominal clear cover = 25mm

So, overall-depth= effective depth + clear cover + half diameter of bar

= 300 +25+16/2

= 333 mm; adopt 335mm

Width of beam (b) = ½ to 2/3 of overall depth, but not less than thickness of wall

= 335*1/2 to 335*2/3

= 167.5 to 223.3 mm

Since the thickness of brick-wall is 230 mm so we cannot take less than that.

Hence provide width of 230 mm, equal to thickness of wall

Dimension of beam obtained = 230 mm * 335 mm

Let us take another example of continuous beam of 12m span.

Let us take another example of continuous beam of 12m span.

Solution,

From deflection criteria, as per IS456:2000, Clause 23.2.1,

L/d ≤ αβγδλ

Assumptions:

Percentage of steel = 1.2% so modification factor β = 0.9 (for fe415 )

Modification factor δ and λ = 1

For span more than 10 m;

Modification factor γ = 10 / span = 10/12 = 0.833

For continuous case,

αβγδλ = 26**0.9**0.833 = 19.5

since, continuous , l/d =19.5

d= span/19.5

d= 12000/19.5

d=615

Adopt d= 700mm

Diameter of tensile steel bar = 25mm

Nominal clear cover = 25mm

So, overall-depth= effective depth + clear cover + half diameter of bar

= 700 +25+25/2

= 737.5 mm; Take 740mm

Width of beam (b) = ½ to 2/3 to of overall depth, but not less than thickness of wall

= 740/2 to 740*2/3

= 370mm to 493mm

So provide width of beam 425mm

Dimension of beam obtained = 425mm * 740mm

After preliminary design of beam, detailed designed can be done as explained under the topic Design of beam.

**Point should be known before design of beam:**

For beam resisting factored axial compression stress not more than 0.08 fck

**a. Dimension**

i. Beam shall preferably have width-to-depth ratio of more than 0.3.

ii. Beam shall not have not width less than 200 mm.

iii. Beam shall not have depth D more than 1/4th of clear span. This may not apply to the floor beam of frame staging of elevated RC water tanks.

**b. Longitudinal reinforcement**

i. Minimum longitudinal steel ratio ρmin provided on any section is:

ρmin = 0.24 √fck / fy

ii. Maximum longitudinal steel ratio ρmax provided on any face at any section is 0.025 (ie. 2.5%)

iii. The minimum area of tension reinforcement shall not be less than that of As/bd = 0.85/fy

iv. The maximum area of tension or compression reinforcement shall not exceed 0.04bD.

**c. Diameter**

i. The beam shall have at least 12mm diameter bar each at top and bottom faces

ii. The diameter of stirrup (traverse reinforcement) should not be less than 8mm

**d. Spacing of reinforcement**

i. The maximum spacing of traverse reinforcement shall not exceed 0.75 d for vertical stirrups and in no cases spacing shall exceed 300 mm.

ii. The maximum spacing of longitudinal bar near the tension face of a beam depends upon the moment distribution carried out in analysis and characteristic strength of reinforcement. For no moment distribution and for fy = 415 N/mm^{2}, it should not exceed 180 mm and fy = 500 N/mm^{2}, it should not exceed 150 mm.

e. **Clear cover**

The clear cover of main reinforcement for beam should not be less than 25mm and should be increased depending on exposure condition.

f. **Reinforcement arrangement**

i. longitudinal steel on bottom face of the beam framing into a column at the face of column shall be at least half the steel on its top face at the same section.

ii. Longitudinal steel in beam at any section on top or bottom face shall be at least 1/4th of longitudinal steel provided at the top face of the beam at the face of the column.

iii. *Not more than 50 % of reinforcement should be lapped at a place.iv. The spacing of traverse reinforcement at lapping zone shall not exceed 150 mm*.

**All these points are based on Indian codes mostly from IS 13920:2016**

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