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But before going to start You need to understand few things which I should tell you first or at the later stage you may find some problem.

__What is a Structural Engineer and Job Profile :- __

As a Civil Construction material steel was vastly used in early days for building works and in cases where the loads are heavy but now a days most of the buildings are made of RCC however Steel is used in the places like Steel Bridge Design, Railways, Docks , Over bridges and etc. places where the loads are very heavy because the Bearing and Shearing capabilities of Steel is more than Concrete. And many a buildings in places like United Kingdom, United States, Australia and many other modern countries Slab Beam and Column Design are still being done with Steel as a Civil Construction material. Steel which are used in structural design as a construction material are used in the form of Rolled Steel Sections of different shapes like I Sections, Channels, Tubes, and also in built up sections. For Design of Steel beam mainly I sections are used. Now remember that great equation of the Theory of Bending which is (M/I) = (f/y) = (E/R) , I know you can steel recall this don’t you?. Okay let us just recap it in an easy way, in this equation the denotations are as follows :-

__Step 1 – Preparing a Neat Sketch from the Problem For Steel Beam Design :-__

__Step 1 – Preparing a Neat Sketch from the Problem For Steel Beam Design :-__

STEEL BEAM PLAN |

__Step 2 – Calculation of the Influence Area of the Structural Steel Beam :-__

__Step 2 – Calculation of the Influence Area of the Structural Steel Beam :-__

STEEL BEAM I – SECTION |

__Step 3 – Calculation Of Loads acting on the Steel Beam :-__

__Step 3 – Calculation Of Loads acting on the Steel Beam :-__

^{2}and for Commercial Building it is 4 KN/m

^{2}. The Dead Loads which are to be calculated are as follows:-

^{3}for R.C.C.

^{2})

^{3}

__A) Dead Loads –__

^{2}= (1 x 3) x 0.5 = 1.5 KN/m that is [Length x Breadth] x Load Intensity

__B)Live Load –__@ 4 KN/m

^{2}= (1 x 3) x 4 = 12 KN/m [Length x Breadth] x Load Intensity

__Step 4 – Calculation of Effective Length Steel Beam :-__

__Step 4 – Calculation of Effective Length Steel Beam :-__

__Step 5 – Calculation of Maximum Bending Moment On The Steel Beam :-__

__Step 5 – Calculation of Maximum Bending Moment On The Steel Beam :-__

^{2}/8)

^{2}/8)

^{2}/8) = ((25.75 x 6.25

^{2})/8) = 125.73 KN-m

__Step 6 – Calculation of Section Modulus Required For Steel I Beam :- __

__Step 6 – Calculation of Section Modulus Required For Steel I Beam :-__

D = Overall depth of the beam

T = Mean thickness of gthe compression flange, which equals to the area of horizontal portion of flange divided by width

l = Effective length of compression flange

ryy = Radius of gyration of section about its axis of minimum strength (y-y axis)

However the value of permissible compressive stress shall never exceed 0.66fy, where fy = Yield Strength of Steel. Here for easy understanding considering that the permissible compressive stress has got the same value that of 0.66fy.

in case of Fe250 the Yield Strength is 250 N/mm

^{2}, Hence Maximum Permissible Stress, f = 0.66 x 250 = 165 N/mm

^{2}.

_{required}= (M/f) = (125730000/165) = 762000 mm

^{3}[Unit Derivation, Z = (i/ymax) = (mm

^{4}/mm) = mm

^{3}]

^{3}, Hence 762000 mm3 = (762000/(10x10x10)) = 762 cm

^{3}

__Step 7 – Steel I Beam Selection of Suitable Section from Steel Table :-__

__Step 7 – Steel I Beam Selection of Suitable Section from Steel Table :-__

_{required}), and also have to note all necessary properties of that Steel Section. In our case I will use SP-6 Steel Table, and I have found the following section to be suitable in case of our Civil Engineering Design Problem:-

_{w}= 8.1mm

_{xx }= 13630.3cm

^{4}

_{xx}= 778.9 cm

^{3}

__Step 8 – Check For Shear Of Steel Beam :-__

__Step 8 – Check For Shear Of Steel Beam :-__

_{va}= (V/h.t

_{w}), After getting this Average Shear Stress we have to calculate the permissible Shear Stress which depends upon the grade of steel and also upon the factor of safety which is varying according to Code of Practice of different country, according to IS 800 Permissible Shear Stress = 0.4 x Yield Strength of Steel, Hence for Fe250 Grade Steel we get, Permissible Shear Stress, T

_{vm}= 0.4 x 250 = 100 N/mm

^{2}. Now if the permissible stress is greater than or equal to that of the Average Shear Stress on Beam then the Section is Safe In Shear, or else it is unsafe therefore, we have to select the next higher section in terms of Section Modulus and area of Web (h.t

_{w}) and give trial for shear check, until it becomes safe.

_{va}= (V/h.t

_{w}) = (80450/(350 x 8.1)) = 28.38 N/mm

^{2}< 100 N/mm

^{2}

__Step 9 – Check for Deflection Of Steel Beam :-__

__Step 9 – Check for Deflection Of Steel Beam :-__

^{3}/E.I)

^{5}N/cm

^{2}

^{4}[Notation = Capital Eye]

^{7}N/cm

^{2}

^{4}

^{3}/E.I) = (5/384) X ((160940 X 625

^{3})/((2×10

^{7}) X 13630.3)) = 1.877 cm

Apart from these checks there are also other checks like, Check for Vertical Buckling [Important for Point Loads], Check for Direct Compression in Web, Check for Diagonal Buckling.This check procedure have not been included in this article now, but will be Updated Soon.

For making this Article Universal I’ve used only Common Terms and Denotations which are well known in all of the Countries Like United Kingdom, United States, Australia, India and other places, as the Denotations may vary Code to Code of Different Countries.

If you like this Article then I am sure you will like the following well researched articles too, come on check them out also

Great work Mr.Ranadip

I am doing a research on the difference between codes for designing simple beam to see how it will be the differences and i wonder if you could help me for providing the codes like Indian code, uk us and if you know any other resources

If you could provide the table for sections for the other countries

Again i'd love to thank you very much for that example, have a nice day.

Mr. Ranadip, the code used for the type of beam design is perhaps IS:800-2000.Pl.explain the method using IS:800-2007

Mr.Ranadip,

You have made beam design calculations lucidly simple. Many thanks and compliments. Had I come across such clear & step by step explanation of beam design 35 years ago I would have certainly opted to be a civil engineer. Incidentally I have a steel/Pre-stressed RCC I beam design problem for which I need advise. Is it possible to connect with you by e-mail. Please inform. Best regards. Uday Karandikar