Chapter 2 - Analysis of Sections by Working Stress Method

Analysis of SectionsHere we will analyze sections of structural members which are subjected to bending (flexure). Structural members like beams and slabs are subjected to bending. Some other members like columns, walls etc., are also some times subjected to bending due to eccentric loading, lateral pressure etc.,

A structural designer have to deal with the following two types of problems:

(1) Analysis problems
When dealing with an analysis problem, the following details will be given:
● Complete cross sectional dimensions of the member
● Details of reinforcing steel
● Properties of the materials

Based on those given details, the following should be determined:
● Stresses in the materials
● Deflections, crack widths etc.,
● The ultimate bending moment or load that the member can resist

(2) Design problems
For this, the following details will be given:
● The ultimate bending moment or load that the member should resist
● Allowable stresses in the materials

Based on those given details, the following should be determined:
● Complete cross sectional dimensions of the member
● Details of reinforcing steel and required properties of the materials.

Analysis of Beams:
Beam is a commonly used structural element. It may exist independently in a structure as in a simple brick masonry car porch where it is provided above the entrance in the front. It may also exist as the part of a structural frame work with lots of beams, columns and slabs as in the case of multi storeyed buildings.

Usually a beam has to resist bending moments and shear forces. In some cases a twisting moment may exist. There may also be an axial force acting on the beam. This axial force may be tensile or compressive. But often, in the case of beams, this force will not give rise to high stresses and so it is neglected.

Theory of bending:
Reinforced concrete is not a homogeneous material. It is a composite material consisting of concrete and steel. We will first discuss the theory of bending for a homogeneous material.

The most important assumption in the theory of bending is that the plane cross sections taken normal to the longitudinal axis of the beam remain plane even after bending. (This assumption is not valid for Deep beams). Another point that we must consider here is that, when a beam bends due to a sagging moment, the fibres below the neutral axis are in tension and the fibres above are in compression. ( The reverse happens in hogging moments as in cantilevers). With this knowledge and the above assumption about the plane sections, the following point can be deducted:

● The strain in any fibre is proportional to the distance of the fibre from the neutral axis. As the distance increases, the strain also increases. In fig 2.1, the green coloured graph is obtained by plotting the strains along the X axis. The values of strains have opposite signs in the regions above and below the neutral axis. This is because above the neutral axis, the strains are compressive and below the neutral axis, they are tensile. We can see that among all the fibres in the compression zone, the top most fibre is situated at the maximum distance y_c,max    from the neutral axis, and here, the compressive stress has a maximum value of f_c,max. The compressive strain is also maximum here. Similarly,among all the fibres in the tension zone, the bottom most fibre is situated at the maximum distance y_t,max from the neutral axis, and here, the tensile stress has a maximum value of f_t,max  and the tensile strain is also maximum. When the distance from the neutral axis decreases, the stress and strain also decreases linearly, as shown by the blue and green graphs.

Fig.2.1
Stress and strain distribution in a homogeneous section

We can derive some useful equations from this diagram. For that, we first find the areas of the two triangles above and below the neutral axis in the stress diagram.

The basic bending equation is:

Where
M = Bending moment at the section
I = Moment of inertia of the section
E = Modulus of Elasticity of the material
R = Radius of curvature of the neutral layer of the beam
f = stress in the fibre
y = distance of the fibre from the neutral axis of the beam

From the above we get
 

and
 

Thus we obtained the base of each of the triangles in the stress diagram. Now, the altitudes are already known:
y_t,max for the lower triangle and  y_c,max for the upper triangle.

So We get the area of the upper triangle as (1/2) x Base x Altitude =
 

Similarly, the area of the lower triangle is

We have used 2-dimensional figs. for the calculations. But the stress diagrams are in fact, stress blocks, and they have volume. This is shown in the 3-dimensional view given below:

Fig.2.2
Three dimensional view of stress distribution

We know the area of the triangles. If we multiply this area by the width of the beam we will get the volume of the stress block. Volume of the compressive stress block is the total compressive force, and the volume of the tensile stress block is the total tensile force.

So total compressive force C at the section is given by:

Eq.2.1
  

and total tensile force T at the section is given by:

Eq.2.2

The total compression C acts at the centroid of the 'area above neutral axis'. The total tension T acts at the centroid of the 'area below neutral axis'. For the section to be in equilibrium, the condition C=T should be satisfied.

The Resisting moment developed at the section is given by:

Eq.2.3
M = Cz  = TzSince C = TWhere z is the 'lever arm' or the distance between the two centroids.
This M is equal to the moment applied by the external load at this section.

It may be noted that the neutral axis of a section passes through the centre of gravity or centroid of the section. It does not depend on the external load applied on the beam. If the external load increases or decreases, the change that is happening in the beam is that the force experienced by the fibres increases or decreases but the position of the neutral axis remains the same. For a beam of homogeneous material, the position of the neutral axis can be changed only by changing the dimensions of the cross section. Thus for a homogeneous beam of rectangular cross section, the position of neutral axis can be changed only by changing the width or height of the cross section. In other words, a homogeneous beam of particular cross section have a definite position of neutral axis.

Analysis of  composite Sections

The analysis described above is not suitable for non homogeneous or composite materials. Composite materials are made up of two or more materials. eg: Reinforced concrete which is made up of two materials, concrete and steel. In such cases the theory has to be modified. Fig 2.3 below shows the representation of a composite material

Fig.2.3
Representation of a Composite material

The two materials, Material 1 and Material 2  should be bonded together and should act as an integral unit without any slip at their interface. If this condition is satisfied, the strain in material 1 situated at a distance y from the neutral axis will be same as the strain in the surrounding material 2, at the same distance y from the neutral axis. So the graph showing the strains will be same for both the materials as shown in fig 2.4. (But the stresses will be different because, stress depends on the property of the material)

Fig.2.4
Stress and Strain distribution for a composite Material

In the above fig.2.4, the stresses are plotted along the X direction to a suitable scale.

The stresses can be calculated using the hookes law, which states that

Let
f_1y = Stress in a layer in Material 1, situated at a distance y from the Neutral axis
ε_y = Strain at this layer
E₁ = Young's modulus of Material 1
f_2y = Stress in a layer in Material 2, situated at a distance y from the Neutral axis
ε_y = Strain at this layer (Note that strain at this distance y  is same for both the materials, as there is no slip between the materials)
E₂ = Young's modulus of Material 2
From Eq 2.4 above, we get

and

 

From the above, it follows that

Which is same as

 

The ratio  E₂ / E₁ is denoted by m, and is called modular ratio

So we get
Eq.2.4m = E₂ / E₁

Eq.2.5
f_2y = m f_1y

We can remove Material 2 from this composite material, and replace it with Material 1. Then the composite material will once again become a homogeneous material. We must replace the Material 2 in such a way that the strength of the section of the resulting homogeneous material is same as that of the composite material that we were having. We must find the quantity of Material 1 required to achieve this.

Transformed Section:The fig 2.5 given below represents a section which is completely made up of Material 1.

Fig.2.5Representation of a Transformed section

We can see that Material 2 which was colored yellow in the previous fig 2.3, has been replaced by Material 1. But the width has been changed to b₃ and the depth has been changed to d₃, giving an area A_m3 = b₃d₃

So we started with a composite beam. We have replaced Material 2 in that composite beam with an area A_m3 of  Material 1. When this change is made, the resultant force in the beam should remain unaltered. In other words, when loaded, the new beam should give the same effect as the old composite beam.

● First let us consider fig 2.4. An infinitesimal area in Material 2 is marked off. The distance of this small region from the neutral axis is y● The area of this small region da₂  is given by da₂ = 2(b₂ / 2)d_y =  b₂d_y ● The tensile stress in this small region is f_2y 

The resultant force df₂ in this small region can be obtained by the equation:  Force = Stress x Area.

So we get
Eq.2.6.
df₂ = f_2y x da₂ 

Now we again consider the basic bending equation:

From this it follows that:

Eq.2.7

Substituting 2.7 in 2.6, we get

Total force F₂ in the Material 2 is given by summing the forces in all the infinitesimal regions. Thus:

But

Where y₂ = Distance of the center of gravity of Material 2 from the NA, and
A_m2 = Total area of Material 2

Thus we get
Eq.2.8

Similarly from fig 2.5, we can show that
Eq.2.9

Where F₃ = Total force in the newly placed Material 1
y₃ = Distance of the centroid of the newly placed Material 1 from the NA
A_m3 = b₃d₃ = Total area of the newly placed Material 1

R remains the same in both the cases because of the following reason: When we make a new beam with Material 1 only, we want it to give the same effect as the composite beam. So, when loaded, the radius R should be the same.

So we want F₂ = F₃ .
So equating Eqs.2.8 and 2.9 we get

But from 2.4, we have m = E₂ / E₁   So we get
Eq.2.10

We can simplify the above Eq.2.10 further. For this, we have to arrange the area A_m3 in the new beam, in a particular way. Fig 2.6 below shows both the sections.

Fig.2.6
Composite section and Transformed section

In the fig, the height d₃ of the Material 1 on the sides of the transformed section is kept same as that of Material 2 in the composite section. The area is adjusted by changing only the width. When only the width is changed, the vertical distance y₃  of the centroid from the NA remains unchanged at y₂ . Thus y₂ = y₃ . Substituting this in 2.10, we get Eq.2.11
A_m3 = mA_m2  

So from the initial composite material, we can remove all the area A_m2 of Material 2 and replace it with an area mA_m2 .of Material 1 in such a way that , and centroidal distance of both A_m2 and mA_m2 from the NA are the same.

Now we can apply the above principles to Reinforced concrete which is a composite material made of Concrete and Steel.

Transformed section of a Reinforced concrete Beam:
In the working stress method (WSM) of analysis of reinforced beams, we transform the given section into an equivalent concrete section by removing the area of steel A_st and replacing it by an area of concrete equal to m A_st  We know that m = E₂ / E₁ . The subscript ‘2’ denotes that material of the composite section, which is replaced by the original material. For reinforced concrete beam, steel is the material which is being removed, and concrete is the orginal material. So m for reinforced concrete beam is given by: E_s / E_cWhere E_s = Elastic modulus of Steel and,
E_c = Elastic modulus of Concrete
The elastic modulus of steel is a constant. It can be taken as E_s = 200 kN/mm²

But that of concrete depends upon long term effects like creep. So the code gives us a formula for determining the modular ratio in the case of reinforced concrete.

Eq.2.12

(Clause B 1.3(d) of Annex B of the code)
where σ_cbc  is the permissible compressive strength of concrete in bending. It's value can be obtained from Table 21 of the code. Note that these are the permissible values upto which compression can be applied to concrete, when the concrete is subjected to bending. As in the case of a horizontal beam in which the concrete above the NA will be subjected to compression, due to the bending of the beam. It is different from the condition where the concrete is subjected to direct compression, as in the case of a column.

The following Table 2-1 gives the values of σ_cbc and m for the different grades of concrete:

We can do a sample calculation with a random value of concrete grade. Let us take M35. From table 21, the value of σ_cbc for this concrete grade is 11.5 N/mm² . So m = 280 / (3 x 11.5) = 8.11

We have seen earlier that to convert a reinforced concrete section into a transformed section, we remove the entire steel A_st  from the section and add extra concrete, and the area of this extra concrete is m A_st . The following fig. 2.7 explains the process in detail:

Fig.2.7
Reinforced concrete section and it’s Transformed section

In the above fig.2.7, fig (a) shows the beam section. In fig(b), all the steel has been removed and extra concrete of area m A_st  has been added. In this fig., hollow regions are present where the steel was previously placed. Also note that the new concrete is placed at the same level of steel. This is to ensure that the vertical distance of the centroids of steel and the newly placed concrete are the same. In fig (c), the hollow regions has been filled with concrete. The hollow regions have an area of A_st . So that much can be reduced from the extra concrete on the sides. So the area of the extra concrete on the sides becomes m A_st  - A_st = (m-1)A_{st .} Fig (d) shows the stress distribution in the transformed section. The stress in concrete at the level of steel (denoted as f_cs ) can be easily obtained by using the 'principles of similar triangles'.  Knowing f_cs  , we can calculate the stress in steel f_st by using Eq 2.5 , by which, the stress in Material 2 (steel) is m times the stress in Material 1 (concrete). So we get:

Eq.2.13
f_st  = mf_cs

In the above fig.2.7, two vertical dimensions are marked: D and d. D is the total depth of the beam section. d is the distance of the centroid of the steel reinforcing bars from the top most compression fibre of the beam section. It is known as the ‘effective depth’ of the beam.

Now we will try to understand the behaviour of a newly constructed reinforced concrete beam. If the load is applied for the first time on the newly constructed beam, and this load is increased gradually from zero, the bottom most concrete fibres, which are in tension, will crack for the first time, when the tensile stress in them reaches a particular value. This value of the tensile stress is called the modulus of rupture and is denoted by f_cr . The code (cl 6.2.2) gives the following empirical formula for calculating f_cr .

Eq.2.14
  

We know that the basic bending equation is

From this equation we can get the moment at first crack or the Cracking moment

Eq.2.15

Where M_cr  is the cracking moment,
I_T is the moment of inertia of the transformed section and
y_t is the distance of the extreme tension fibre from the NA.
f_cr  is calculated from Eq.2.14

Following is the link to a solved example which shows the application of the topics that we discussed above :
Solved example 2-1

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