Civil Engineering is vast field from Culverts to Dam , Buildings To Steel Bridges, so as so the Civil Engineering Materials, It includes Steel, Concrete, Timber, Aluminium, Glass etc. Steel Design is much easier in most of the cases than RCC Design. Civil Engineering Broadly has three main phase namely Planning – consisting of Drawing, Estimating etc. Structural Design consisting of Analysis of Structures and Design Of Structures and last of all Execution which means erecting the actual work according to the Planning and Designing. Structural Engineering has always been a field of respect amongst all the Civil Engineers. Today I will discuss about a part of Structural Engineering which will be dedicated toward the Design of Steel Beam. After reading this article you will be able to design a steel beam by your own without any problem. To make the explanation easy for understanding I will elaborate a simple yet considerably good problem which will cover all your needs as far as the Civil Engineering and Structural Design is concerned. I know now a days all the Structural Engineering Analysis and Design is being done by Computer Software like STAAD Pro, Ansys and all that, but without having the knowledge of manual Design of civil and structural engineering all those are useless because you will be not able to understand the Design Data and the outcome from the software.

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. 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 :-

M = Bending
Moment Acting on the Beam due to loads

I = Moment
of Inertia of the Beam section

f = Bending
Stress in the Steel Beam

y = Distance
between the any fibre of section and the Neutral Axis [For getting maximum Bending Stress the value of y must be maximum as the Bending Stress is directly proportional to the distance between fibre at a point and the Neutral Axis, hence we need to consider extreme fibre which gives maximum value of 'y', this equals to h/2 where h is the depth of section, as the Neutral Axis of symmetrical sections like I-Section will fall at its C.G.]

E = Modulus
of Elasticity of the Beam section

R = Radius
of Curvature of the Shape of Bending of the Beam

But we won’t
be needing all these, we just need the (M/I) =(f/y) for the design of steel
beam.

Now we can
rewrite the equation (M/I) =(f/y) as
(M/f) = (i/y) can we? Yes we surely can by cross multiplication. Here the term
obtained (i/y) is known as Section Modulus and this is a very important factor
because upon it the strength of any section depends. Actually When the Maximum distance between Neutral Axis and the fibre is considered, that is the distance between the Neutral Axis and the extreme fibre, meaning the maximum value of 'y', This term (i/ymax) or Section
Modulus is denoted with ‘Z’ , therefore Z=(i/ymax) . I hope up to this level you
have understood, and these things you have already studied in structural
analysis isn’t it? Yes you have for sure. So basically in a Design of Steel
Beam we will design the beam for flexur, that is for Maximum Bending Moment and
then chose a suitable Beam section from the list of available sections which
varies from country to country and this list can obtained from bureau of
standards of your country. And after choosing that section then we will check
for other factors like Shear Stress, Deflection and look if that section can
stand safely. That’s all isn’t it simple? Okay now lets get started, I will
discuss How to Design a Steel Beam in Step By Step that You cannot escape
without Understanding.

Let us consider a problem for explanation, suppose there is a hall room measuring 15m X 6m inside and the walls are 250mm thick. And it is given or you have thought to provide beams at a centre to centre distance of 3m apart. The beams are supporting a R.C.C. roof slab of 150mm thick with finishing on it, the flanges being restrained on slab. The Hall is a commercial type building. So let us design these beams.

###
__Step 1 – Preparing a Neat Sketch from
the Problem :-__

__Step 1 – Preparing a Neat Sketch from the Problem :-__

This is
important in all types of Civil Engineering Structural Design Problem. As You
know Drawing is the Language of Engineers and Line is the Language of Drawing,
so prepare a neat drawing after reading the problem and give all the dimensions
possible in the drawing. A Drawing is a must and it must be 100% correct as all
the designing will be dependent upon this drawing. So prepare the drawing with
caution. Here I’ve prepared one, now study it thoroughly, What Data you are
getting? Yes the inside dimensions of the room that is 15m X 6m, This thickness
of the wall which will act as the bearing of the beam are 250mm thick, and the
Centre to Centre distance between the beam that is 3m.

STEEL BEAM PLAN |

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

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

The
Influence Area of a Beam means the area of which the loads are acting that
beam, or simply you can say that the beams have to be designed for the loads on
that area that is the influence are. Here in this particular problem we see
that all the beams are in a similar situation, that means they are within a
same room and same direction, they are supporting the same roof with same
finishing, and these beams are spaced at a same distance, that means the
condition of all these beams are identical to each other, so we will do
structural design for any one of the beams and that design will be fit for
other beams. If the conditions were not identical then we have to design each
of them individually for economy or we have to design the beam which is having
the greatest load on it. Let us consider the Beam B for Design

Here all
beams are 3m apart C/C distance from each other, hence a single beam B is
having Beam A on the left at 3m distance apart and Beam C on the right side is
also 3m apart, so the Beam B is supporting half of the load of the area between
Beam B and Beam A on the left side and on the right side the Beam B is
supporting half of the load of the area between Beam B and Beam C. Hence it
means on the left side supporting a strip of (3/2) = 1.5m width and on the
right side again supporting a strip of (3/2) = 1.5m width. So the influence
area becomes a strip of width 1.5m + 1.5 m = 3m and its length being from
centre to centre of the bearing at the each end of the Beam that is the
Effective Length of the Beam.

STEEL BEAM I - SECTION |

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

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

Now we have
to calculate the loads acting on that influence area as calculated above as
Civil Engineering Structural Design will be based on these Loads. These loads
can be broadly classified as Dead Loads and Live Loads. Dad loads means the
loads coming from all unmovable Objects like Slab, Beam itself, Flooring etc.
and Live Loads Means the load coming due to the movable objects such as we
humans, furniture and other movable loads. Generally Dead Loads are calculated
and Live Loads are specified according to the Type of the structure, varying in
intensities with different types. Like for residential building generally it is
2 KN/ m

^{2}and for Commercial Building it is 4 KN/m^{2}. The Dead Loads which are to be calculated are as follows:-
Dead Loads
– 1) Self weight of beam(Assumed 1KN per
m)

2) Load of
slab supported @25 KN /m

^{3}for R.C.C.
3) Load of
floor finishing (generally 0.5KN/m

^{2})
4) Load of
Brickwork if any @19.2 KN/m

^{3}
All the
loads are to be calculated on per m run basis so that it becomes a U.D.L. The
total load acting on the beam will be
the sum of the Dead Loads and Live Loads

Therefore,
Total Load per metre run, w = Dead Loads + Live Loads

Here in this
case the load calculation will be as follows :

__A) Dead Loads –__
I.
Due
to the self weight of the beam – 1 KN/m

II.
Due
to the 150mm thick slab = (1 x 3 x 0.15) x 25 = 11.25 KN/m that
is [Length x Breadth x Thickness] x Density

III.
Due
to Floor Finishing @ 0.5 KN/m

^{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 IntensityTherefore, Total Load per metre run, w = (1 + 11.25 + 1.5) +12 = 13.75 + 12 = 25.75 KN/m

###
__Step 4 – Calculation of Effective
Length :-__

__Step 4 – Calculation of Effective Length :-__

It is taken
as the length between the centre of bearings of the beam at each end.

Therefore in
our case having a clear span of 6m and 250mm support at each end by means of
wall we get,

Effective length, l = 6 + (0.25/2) + (0.25/2)
= 6+0.125+0.125 = 6.25m

###
__Step 5 – Calculation of Maximum
Bending Moment :-__

__Step 5 – Calculation of Maximum Bending Moment :-__

Here we will
introduce the Steel Beam Design Formulae for the first time. Now calculate the
maximum bending moment acting on the beam by adopting suitable formula which
are as follows :-

i)
For
point load at the mid span of the beam: –
M = (w.l/4)

ii)
For
U.D.L. Throughout the span of the beam :- M= (w.l

^{2}/8)
iii)
For
Point load at any point of beam :- M =
(w.a.b/l)

For any
unusual loading you have to calculate the maximum bending moment and shear
force by shear force bending moment diagram drawing procedure.

Here in this
case as we are having a U.D.L. of 25.75 KN/m throughout the span hence we will
use the second formula that is M = (w.l

^{2}/8)
Therefore,
in our problem we get,

Maximum
Bending Moment, M = (w.l

^{2}/8) = ((25.75 x 6.25^{2})/8) = 125.73 KN-m
= 125.73
x1000 x 1000 N-mm = 125730000 N-mm

###
__Step 6 – Calculation of Section
Modulus Required :- __

__Step 6 – Calculation of Section Modulus Required :-__

Here we will
use another Civil and Structural Engineering Design Formula
for steel beam design. Here we will determine the Section Modulus
required in order to resist the Maximum Bending Moment acting on the beam. At
the star of this article we had found that (M/I) = (f/y) or (M/f) = (i/y) again
Z=(i/ymax) therefore we can rewrite it as Z=(M/f). The value of ‘f’ depends upon
the grade of steel and factor of safety. Considering Fe250 Grade of steel, and
according to code of practice of the different country the factor safety will
vary for obtaining the permissible stress (f) from the Yield Strength of Steel.
Here I will follow the IS 800 Code and according to it Maximum Permissible Stress =
0.66 X Yield Strength. But in case of I beam and channel with equal flanges the permissible bending compressive stress shall be calculated from the table given in the code by knowing the value of ((D/T)/(l/ryy)) where,

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

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}.
Thus
returning to our problem we get that,

Z

_{required}= (M/f) = (125730000/165) = 762000 mm^{3}[Unit Derivation, Z = (i/ymax) = (mm^{4}/mm) = mm^{3}]
For selecting a suitable section from the steel
table that is the chart of Rolled Steel Section we have to get the value in
terms of cm

^{3}, Hence 762000 mm3 = (762000/(10x10x10)) = 762 cm^{3}###
__Step 7 – Selection of Suitable
Section from Steel Table :-__

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

Now we have to
use the steel table which has the standard Rolled Steel Section List and their
properties written on. These Tables are country specific and will vary from
United Kingdom to United States to Australia to India. So you need to use your
country specific steel table. We have to choose a section, preferably a I
Section from the Steel Table, which will have a Section Modulus (use section
modulus about –x-x axis) or ‘Z’ equal to
or greater than what is found to be required (Z

_{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:-
Let us Try
ISMB 350 @52.4 kg/m Having the following properties,

Sectional
Area, a =66.71 cm2

Depth of
Section, h = 350mm

Thickness of
Web, t

_{w}= 8.1mm
Moment of
Inertia, I

_{xx }= 13630.3cm^{4}
Section
Modulus, Z

_{xx}= 778.9 cm^{3}###
__Step 8 – Check For Shear :-__

__Step 8 – Check For Shear :-__

As of now we
have made a design based on flexural strength that is based on Maximum Bending
Moment, and selected such a section which will be safe in Bending. Now we have
to check if that section will be safe in shear or not. For this we have to
calculate the Maximum Shear Force acting on the beam, as for U.D.L. this can be
calculated by using the Structural Engineering Analysis Formula V=(w.l/2) ,
where,

V = Maximum Shear Force

w = Load per
metre run

l =
Effective Length of the Beam

Then by this
shear force we have to calculate the average shear stress on the beam due to
the Maximum Shear Force, by using the formula T

_{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.
In Our case
of problem it will be as follows :-

Maximum
Shear Force, V = (w.l/2) = ((25.75 x 6.25)/2) = 80.45 KN = 80450 N

Average
Shear Stress in Beam, T

_{va}= (V/h.t_{w}) = (80450/(350 x 8.1)) = 28.38 N/mm^{2}< 100 N/mm^{2}
Hence the section
is Safe In Shear.

###
__Step 9 – Check for Deflection :-__

__Step 9 – Check for Deflection :-__

We are
almost done in the Design of Steel Beam, this the last check we have to
perform, the rule is same, if the section satisfies the check then it is safe,
or otherwise we have to check with another section having more depth (h). For
this check we have to calculate the Actual Deflection on the beam by using the
Structural Analysis Formula in case of U.D.L. the formula is :-

Del(Symbolic)
= (5/384) x (W.l

^{3}/E.I)
Where,

Del =
Deflection in cm

W = Total
Load = w.l in N

E = Modulus
of Elasticity, For Steel E = 2 x 10

^{5}N/cm^{2}
l =
Effective Length in cm [ Small L]

I = Moment
of Inertia in cm

^{4}[Notation = Capital Eye]
And
Permissible Deflection = l/300

Here, In the
case of our problem the calculations will be as follows :-

W = Total
Load = w.l = 25.75 x 6.25 = 160.94 KN = 160940 N

E = 2 x 10

^{7}N/cm^{2}
l = 6.25 m =
625 cm

I = 13630.3
cm

^{4}
Actual
Deflection, Del = (5/384) x (W.l

^{3}/E.I) = (5/384) X ((160940 X 625^{3})/((2x10^{7}) X 13630.3)) = 1.877 cm
Permissible
Deflection = l/300 = 625/300 = 2.083 cm > Actual Deflection, Hence Safe.

Therefore
The Section selected by us is Safe in All Respects.

Therefore
Let Us Provide ISMB 350 @ 52.4 kg/m

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.

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.

Now the
Structural Design of Steel Beam has successfully completed in all respect. I
hope You understood all along with ease, I’ve made it as much as easy as
possible. Do Share the Link of this Article On Your Network Of People of All
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