Chapter 3 (cont..7) Details of a Balanced section

In the previous section we said that in a balanced section, during the loading, the attainment of ultimate state in the topmost compression fibre and the yielding of steel happens simultaneously. Before analyzing a balance section, let us show the features of all the three types in a single fig., so that they can be compared easily. Fig.3.25 below shows such a comparison between the three sections at the ultimate state. For convenience, we will assume that all the sections have the same grade of steel, denoted as f_y. So the combinations will be altered only by the differences in area of steel.

Fig.3.25
Comparison between Balanced, Under reinforced and over reinforced sections

Let us see the various features of the above fig.:

• The strain in concrete at the topmost compression fibre is 0.0035 in all the three cases.
• So the stress in concrete at the topmost compression fibre is 0.447f_ck in all the three cases.
• The strain in steel ε_st,U for an under reinforced section is greater than ε*_st .(ε_st,U falls to the right of the yellow dashed line which passes through ε*_st)
• The strain in steel ε_st,O for an over reinforced section is less than ε*_st.(ε_st,O falls to the left of the yellow dashed line)
• The stress in steel f_st,U for an under reinforced section is equal to 0.87f_y.(f_st,U falls on the magenta dashed line which passes through 0.87f_y)
• The stress in steel f_st,O is less than 0.87f_y in the case of Over reinforced section. (f_st,O falls to the left of the magenta dashed line)

We have already seen how the above features will be obtained while conducting experiments on Under reinforced and Over reinforced beams.

For doing the analysis, we will now isolate the Balanced section in the middle row of the above fig.3.25. This is shown in the fig.3.26 below:

Fig.3.26
Stresses and strains in a balanced section at ultimate state 

We are going to use the principle of similar triangles. In order to clarify the shapes, an imaginary line ab is drawn. So we have two similar triangles: abc and ab'c'.

• altitude of abc = ab = d

• altitude of ab'c' = b'c' = x_u,BAL

• base of abc = bc = 0.0035 + ε*_st

• base of ab'c' = ab' = 0.0035

So we get

Eq.3.13

From the above equation, x_u,BAL can be easily calculated because d is a given section property, and ε*_st is a constant (Eq.3.12) given by:

In the above two equations, the only variables are d and f_y. So we can make an important conclusion: All beams having a particular value of d and a particular grade of steel f_y will be having a particular value of x_u,BAL . In other words, if we are given any beam section, just by using it's d and f_y, we will be able to say what it's depth of NA would be, if it were a balanced section.

Similarly, let us try to calculate x_u,U the depth of NA of an Under reinforced section. For this, we will isolate the first row of fig.3.25 above:

Fig.3.27
Stresses and strains in an under reinforced section at ultimate state

Here also, using similar triangles, we will get:

Eq.3.14

But here we do not know the value of ε_st,U . We know that ε_st,U will be greater than ε*_st . But the exact value is not known. 

This same problem is present in the case of over reinforced beams also: Using similar triangles we will get

Eq.3.15

We know that ε_st,O will be less than ε*_st . But the exact quantity is not known. So the above method of using the similar triangles is applicable to Balanced sections only. To find the depth of NA of Under reinforced and Over reinforced sections, we have to look for alternate methods. 

But before we see those alternate methods, we will make an important inference. We have seen earlier (based on fig.3.22) that, when an Under reinforced beam is loaded to failure, the NA moves upwards. Similarly, when an Over reinforced beam is loaded to failure, the NA moves downwards. Also, we have discussed the reasons for these upward or downwards movements. We will get a mathematical presentation of this, if we arrange the strain diagrams of the three beams in a single row. So we will take the three strain diagrams of fig.3.25 and arrange them as shown in fig.3.28 below:

Fig.3.28
Comparison of strains

The strain at the top most compression fibre is same in all the three cases. In the case of balanced section, the strain graph which is the line pq cuts the vertical axis at 'o'. In the case of under reinforced section, pq will cut the vertical axis at a point above 'o' because ε_st,U > ε*_st . In the case of Over reinforced sections, pq will cut the vertical axis at a point below 'o' because ε_st,O < ε*_st .  

We know that the point where the strain graph cuts the vertical axis is the position of the NA because there, the strain is equal to zero. So the above fig.3.28 is a mathematical proof that:
• Depth of NA of an Under reinforced section will be less than that of a Balanced section and 
• Depth of NA of an Over reinforced section will be greater than that of a Balanced section. 

So some steps that can be done at the beginning of an analysis procedure of any given beam section can be out lined now:

When a beam section is given to us, we will be given it's 
• width b, 
• total depth D, 
• effective depth d, 
• Area of steel A_st, 
• grade of steel (characteristic yield strength) f_y, and 
• grade of concrete f_ck. 

The first step is to write down what it's depth of NA would be, if it were a balanced section. This can be done using eq.3.13 above. The only quantities we need are d and f_y. We will denote this calculated value as x_u,BAL. 

Next step is to calculate it's actual depth of NA. If this actual depth is less than x_u,BAL, then it is an Under reinforced section. If it is greater than x_u,BAL , then it is an Over reinforced section.

But we have seen just now that the principle of similar triangles cannot be used for calculating the depth of NA of Under reinforced sections and Over reinforced sections. So in the next section we will see the alternate methods.

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