Chapter 10 (cont..2) - Design of flanged sections

In the previous section, we said that it is important to get a fairly good approximation for the preliminary value of d. Now we will see how this is achieved:

We know the basic equation:
Bending moment = Force x Lever arm  
⇒ Bending moment = Area of steel x stress in steel x Lever arm.
⇒ Area of steel = Bending moment / (stress in steel x Lever arm)

This can be written as:
Eq.10.5

• M_u is the factored bending moment which has to be resisted,
• 0.87f_y is the stress in steel. (Stress is equal to 0.87f_y because, the quantity of steel that we provide must be such that the beam will be under reinforced)
• z is the lever arm

For using in the above Eq.10.5, z can be taken as the larger of the following:
10.6
  ♦ 0.9d
  ♦ d - Df/2

So here another problem arises: 
• We are using Eq.10.5 above to calculate the approximate area of steel that will be required. 
• This is to determine whether the steel will have to be provided in more than one layer. 
• From the number of layers, and the dia. of bars, we will get the effective depth d. 
• Our ultimate aim here is to determine d. 
• But from 10.6 above, we see that d should be known prior to using Eq.10.5.

This forms a loop. The solution to this problem is to use an initial d by deducting 50 mm from the total depth D. This d is to be used in 10.6 to obtain z. Thus we will be able to use Eq.10.5. 

From the Eq.10.5, we will get the 'most probable' area of steel that will be required. This area must be converted into 'number and diameters' of bars. (details here.) Then we will know whether the steel will be in single layer or multiple layers. Based on the number of layers and the diameter of bars, we can calculate d. This will be the 'most probable' d. 

With the calculation of d, we complete all the calculations on 'preliminary dimensions'. That is., we now have a complete beam section to begin our work. 

It may be noted that b_f also needs to be calculated at this stage. But it does not come under the category of 'preliminary dimensions'. This can be explained as follows: We have the required equations for calculating b_f of any type of T-beams or L-beams. Eqs. 9.5, 9.6 etc., are some of those. We can see that in those equations, the only variable is b_w. All others like l₀, D_f etc., are constants. Now, this b_w is not much of a 'variable' because it is a parameter that receives an early 'fix' depending on the width of supports. So b_f itself is not a variable.

There are some checks that we have to perform immediately after fixing up the preliminary dimensions. These are same as those for the rectangular sections. We saw them in 4.2, 4.3 and 4.4. For our present flanged beam sections, b should be changed to b_w.

Let us now see the different steps involved in the design process.
Step A:
We begin the work by analysing the section that we have. We analyse it first, to find the Limiting moment of resistance, M_ulim. If M_ulim is less than M_u, we know that the beam section will not serve it's purpose even if we give high quantities of steel. In that case, the section will have to be improved by increasing the depth or making it 'doubly reinforced'. 

Step B:
If M_ulim is greater than M_u, then we can proceed to the next step. This step also involves an analysis of the section. This time to find x_u. From the value of x_u, we will know whether 
• the NA falls in the flange (case 1) or 
• in the web (case 2). If in the web, we will know (from the value of 0.43x_u) whether the section belongs to 
    ♦ Case 2(1) or 
    ♦ Case 2(2).    

So we get the position of NA. That is., we get the case into which our beam section falls. This will be the 'most probable' case because, we are using the 'most probable' area of steel obtained from Eq.10.5 above.

If it is Case 1, we can design it in just the way we would design a rectangular beam of width b_f. We have learned the design procedure here. But the steps 1 & 2 in that procedure can be avoided. This can be explained as follows: In the case of rectangular sections, step 1 involves the calculation of required d. Step 2 involves finalizing D. These two steps are done to ensure that the final section will be capable to resist the load. But here we already do this in step A. In step A, we check whether M_u < M_ulim. If M_u is indeed less than M_ulim, then what ever be the position of NA, the section will be able to resist M_u if we give the appropriate amount of steel. So Steps 1 and 2 for rectangular sections is equivalent to step A for flanged sections. The remaining steps 3 & 4 are same for both the types. The only change to be made is to change b to b_f.

If it is case 2(1) or case 2(2), much more calculations are involved. Here also, the 'calculation of d required' and 'finalizing D' can be avoided because we do this in step A. So we will proceed to the next step.

Step C:
If it is case 2(1):
The BM offered by the section is given by Eq.9.11 

Eq.9.11

This should be equal to or greater than M_u. For obtaining a solution, We will first equate it to M_u. 
For solving, we must simplify it. Note that the last term is constant. We will put it equal to K. Thus we get:

Let us simplify it further:
Eq.10.7

There is a term in (xu/d)² and another term in (xu/d). It is a quadratic equation. We can write it as:
A X² + B X + C = 0

Where A = 0.1506 b_w d²f_ck ; B=-0.362b_wd²f_ck  ; C = M_u -K and X = x_u/d

A, B and C contain only the known values b, d, f_ck , M_u and K. So if we are using a spread sheet program, it will calculate A, B and C just when the input data are given. The program will then directly give the two solutions of the quadratic equation. That is., we will get two values for x_u/d. Out of these, one will be greater than 1, and the other will be less than 1. We must choose the lesser value because x_u is always less than d. If we choose the value which is greater than 1, when we multiply it with d, we will get a value for x_u which is greater than d.

This step comes to an end with the calculation of x_u . So step C can be summarized as:
• Calculate the values of A, B and C and then solve x_u
The next step is the calculation of A_st.

Step D: With the calculation of x_u, we can split the section into an upper part and a lower part. The compressive force in the upper part is given by Eq.9.12

Eq.9.12:
C_uw + C_uf = C_u = 0.362f_ck b_w x_u + 0.447f_ck D_f (b_f -b_w)

The tensile force in the steel in the lower part is given by :

T_u = 0.87 f_y A_st

Where A_st is the area of steel. The stress in steel is equal to 0.87 f_y because, it is an under reinforced section, and so the steel would have yielded at the ultimate state.

For equilibrium, C_u = T_u. So, equating the two, we get
Eq.10.8

This completes the design procedure if the beam section belongs to case 2(1). In the next section, we will now discuss the case 2(2). 


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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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Chapter 3 (cont..5) Tension failure in Limit state method

In the previous section we have seen that f_st can take different values. To explain this behaviour, we will discuss about the stages that lead to the failure of a beam section.

Consider a beam being loaded. When the load increases, the stresses in concrete and steel also increase. The internal moment which resist the applied external bending moment is given by Force x Lever arm. This is shown in the fig. below:

Fig.3.21
Internal moment at the beam section

The force that we use for computation of the internal moment can be the compressive force C in concrete or the tensile force T in steel. Both will give the same result. So we can write:

Internal moment developed = T x z = C x z

z is the lever arm, which is the distance between the point of application of C and T. C acts at the centroid of the concrete stress block and T acts at the centroid of steel.

Let us now see the different stages that the beam section will pass through, while it is being loaded to the ultimate state:

Stage 1: C and T increases normally to resist the external load. The stresses and strains are in the linear elastic state. This is shown in the fig.3.22 below. The strain in steel at this stage is denoted as ε_st,1. This is less than ε*_st .

Here ε*_st is the 'quantity of strain' which when obtained, confirmation can be made that the steel has yielded. This is the last point of the curved portion, which is obtained by drawing the dashed parallel line through 0.002 strain. (The last row in the table for Fe415 and Fe500) So we can write:


Eq.3.12

   

In the stage 1, ε_st,1 < ε*_st , because the steel has not yielded. It is in the initial linear elastic state. Also ε_c,1 the strain in concrete is far less than 0.0035, the ultimate strain. The stress f_st,1 in steel is less than 0.87f_y . We know that 0.87f_y is the design yield stress (f_y/1.15 = 0.87f_y). That is., the stress corresponding to the strain  ε*_st. 

The stress f_c,1 in the top most compression fibre is far less than 0.447f_ck , the ultimate stress in concrete.  

Fig.3.22
Stages before failure of beam section

Stage 2: The steel enters the non-linear stage and has begun to yield. But once the steel has yielded, it will continue to elongate at constant stress. That is., the strain will increase at constant stress. (The final horizontal line of the stress-strain curve).  As the stress in steel is constant, the force T in it is also constant. But as the external load is being increased, the applied bending- moment is also increasing.

As T is remaining constant, the resisting internal moment can be increased only by increasing z. 'z' is the distance between the centroid of concrete stress block and the centroid of steel. The centroid of steel cannot change as it is held in position. But centroid of concrete stress block can change. For this, the effective concrete area taking part in the compression should decrease. For this, the NA should move upwards. Then, the total height of the concrete stress block will decrease, and it's centroid will move upwards. When this happens, the lever-arm z will increase.

These details can be seen in the fig.3.22 above. In the stage 2, the strain ε_st,2 in steel has become equal to ε*_st . So the steel has yielded. The NA has shifted upwards. The strain ε_c,2 in the top most compression fibre is still less than 0.0035.

The stress f_st,2 in steel is at the constant value of 0.87f_y. The stress in concrete f_c,2 is less than 0.447f_ck.

So even though the steel has yielded, the section manages to resist the external bending-moment by increasing z. But as part of the experiment, we are increasing the load further. So the beam moves on to the next stage.

Stage 3: The reduced area of concrete has to take more load now. As a result, stress in it begins to increase. The strain also increase. But the concrete cannot take large strains as in the case of steel. 

We have seen the difference in the strain values of concrete and steel (fig.2.15) when we discussed the working stress method. Here also in the limit state method, this difference in strains has much significance. So we will write it down again:

The reinforcing steel have a greater ductility. It can undergo greater strains than concrete. Yielding of steel occurs at higher stresses, and the amount of yield, which is measured as strain can be in the range 0.12 to 0.20, before failure of steel occur. But for concrete, this range is 0.003 to 0.0045. To get a better understanding of these ranges, we can think of two specimens, one of concrete and the other of steel. Let both of them have the same length of say 20 cm, then the maximum increase in length that the steel specimen can have is calculated as follows:

Strain =  Δl / l = 0.12 to 0.20. So Δl = 0.12 x 20 to 0.2 x 20cm = 2.4 to 4.0cm

Similarly, the decrease in length that the concrete specimen can have is calculated as follows:

Strain =  Δl / l = 0.003 to 0.0045. So Δl = 0.003 x 20 to 0.0045 x 20cm = 0.06 to 0.09cm

From the above, we can see that, when concrete and steel are combined to form a composite material, at higher stresses, the concrete will fail before failure of steel occurs. This happens in our beam also: The load is being increased even after the yielding of steel has occured. Concrete cannot take this large strain, and it will soon reach the ultimate strain of 0.0035, and will fail by crushing.

But before this failure of concrete occur, the steel has yielded to a greater extent. This cause the beam to deflect more and the cracks at the bottom side of the beam become wider. Such deflections and wider cracks give sufficient warning about the beam failure that is about to take place.

So the failure of the beam is by the crushing of concrete. But it was initiated by the yielding of steel under tension. This type of failure is called tension failure.

The details of stage 3 are shown in the fig.3.22 above In the stage 3, the strain ε_st,3 in steel has become greater than ε*_st . This is because the steel is elongating at constant stress. The NA has shifted further upwards. The strain ε_c,3 in the top most compression fibre has reached 0.0035, which is the ultimate strain.

The stress f_st,3 in steel is at the constant value of 0.87f_y. The stress in concrete f_c,3 has reached 0.447f_ck.

It may be noted that in the above experiment, we say that the beam has reached the ultimate state (the state of impending failure), when the strain in the top most compression fibre reached 0.0035. The steel began to yield long before this strain was obtained in concrete. But yielding of steel cannot be taken as a criterion for reaching the ultimate state. Because, as stated earlier, steel can continue to perform through larger strains. The concrete cannot go through these larger strains and so will fail before steel. The beam will serve it's purpose only if both steel and concrete are performing satisfactorily at all the sections. But here concrete has reached it's ultimate state. So the beam as a whole has reached it's ultimate state. Thus strain value of 0.0035 in the extreme compression fibre of concrete is taken as the criterion for deciding whether the beam has reached ultimate state or not.      

The beam shown in fig.3.19 of the previous section is an example of this type of beam. In it, the stress and strain in steel at the ultimate state is in the magenta colored straight line region. So the steel has yielded.

Suppose in the experiment, the beam that we used, has a larger area of steel and/or the steel is of high strength grade, then the results would be different. In the next section we will see the details of such a beam.


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