Chapter 3 (cont..10) Limiting Moment of Resistance

In the previous section we saw the details about the 'Ultimate Moment of Resistance'. In this section, we will discuss about another property of beam sections, called the 'Limiting Moment of Resistance'. This is denoted as M_u,lim.  We will discuss this based on an example. The fig.3.34 below shows the comparison between two beam sections, Beam A and Beam B.

Fig.3.34
Comparison between two beam sections

Both sections have the same cross sectional dimensions. The difference is only in the area of steel. The results of analysis are also shown in the fig. Beam A is capable of giving a resisting moment of 64.37 kNm at the ultimate state. Beam B is capable of giving a resisting moment of 88.24 kNm at the ultimate state. Both have the same area of concrete and all other properties are the same. We know that the 'difference in area of steel' is causing this 'difference in M_uR'. (We have seen earlier that when a lower area of steel is provided, the section will reach the ultimate state at lower loads).

Beam A has the 'potential' to give more. But the area of steel provided in it does not allow it to offer a higher resistance. So we can think of increasing the area of steel so that the full potential can be utilized. But we cannot just give any higher area. Beam B shows the ill effect of giving such a higher area of steel: It has become an over reinforced section. So we want to know the limit upto which the steel area can be increased, with out making it an Over reinforced section.

We have seen earlier that x_u,bal (Eq.3.13) is the bench mark for deciding whether a beam section is under reinforced or over reinforced. If the depth of NA is less than x_u,bal, then the section is under reinforced. If it is greater than x_u,bal, then the section is over reinforced. So we can increase the area of steel upto such a level that the depth of NA becomes equal to x_u,bal. We will now see how this increase can be achieved:

We know the equation to calculate the x_u,bal of any given section.

Eq.3.13


Where

When a beam section is having the depth of NA equal to x_u,bal at the ultimate state, the compressive force in concrete (calculated using Eq.3.7) will be equal to 

C_u  = 0.362 f_ck b x_u,BAL

The tensile force in steel will be equal to 0.87f_y A_st. The stress is equal to 0.87f_y because, the section is not an over reinforced one, and so the steel would have yielded at the ultimate state.

The area of steel A_st that we are using here is the 'limiting steel'. Because, if we use an area greater than this, x_u will exceed x_u,bal . So we will denote it as A_st,lim. Thus we get

T_u = 0.87f_y A_st,lim 

For equilibrium,  C_u = 0.362 f_ck b x_u,BAL = T_u = 0.87f_y A_st,lim  

From this we get

Eq.3.21

So we have the area of steel required to obtain the 'full potential' from the beam section. We can now calculate the magnitude of this 'full potential'. We will set up the section (by giving an area of steel equal to A_st,lim) with the depth of NA equal to x_u,BAL.

So the Compressive force = C_u = 0.362 f_ck b x_u,BAL.
The lever arm = z = d - 0.416x_u,BAL . 
Thus the moment =  0.362 f_ck b x_u,BAL( d - 0.416 x_u,BAL ) 

This is the full potential from the beam section. That means, we cannot expect more from the section. In other words, it is the 'Limiting Moment of Resistance' of the given section. So we can write:

Eq.3.22
M_u,lim = 0.362 f_ck b x_u,BAL( d - 0.416 x_u,BAL ) 

Using the above Eq.3.22, we can obtain the M_u,lim of any given beam section. Now we will see another important derivation:

In the course of derivation for M_u,lim, we saw that the calculation of x_u,BAL is an important step. It is the bench mark. If the depth of NA is greater than x_u,BAL, then the beam is an over reinforced one. So we can say that the maximum value that x_u can take is x_u,BAL. In the code, this maximum value that x_u can take, is denoted as x_u,max.

We have seen Eq.3.13 above, which gave x_u,BAL. Now this can be re-written in terms of x_u,max  as:

Eq.3.23

Where


So x_u,max, just like x_u,BAL, depends only on d and f_y. So if we are given a beam section, just by using it's d, and the f_y value of it's steel, we can write the x_u,max of that section. We can bring d to the left side so that f_y is the only variable present on the right side:

Eq.3.24

On the right side, we can give three different values for f_y. They are 250, 415 and 500 for the three grades of steel. If we can make a table with these values, lengthy calculations on the right side can be avoided. The table 3.4 below is prepared in this way:

Table 3.4

These values are given below the Fig.23 of the code.

Sample calculation:

Let us take Fe500 steel. Є^*_st = [500/(1.15 x 2 x 105)] + 0.002 = 0.0041739

Substituting this in Eq.3.24 we get x_u,max /d= 0.0035/(0.0035+0.0041739) = 0.4560

So, if we are given a beam section with steel of Fe500 grade, we can calculate x_u,max of the section directly by multiplying it's effective depth d by 0.4560.

Now let us consider Eq.3.22 above which gives the M_u,lim of any section. We will re-write it in terms of x_u,max:

Eq.3.25
M_u,lim = 0.362 f_ck b x_u,max( d - 0.416 x_u,max ) 

In this case also, if we can make a table, lengthy calculations can be avoided. Let us divide both sides of Eq.3.25 by  bd². We will get:

Eq.3.26

This is the same eq. given in cl.G-1.1(c) of the code.

Let us put the quantity on the right side equal to 'K'. That is:

Eq.3.27

The right side has two variables, which are x_u,max /d  and f_ck . We can put the three values for the three grades of steel (from table 3.4 above) for x_u,max /d. We can also put the commonly used values of f_ck. Thus the table 3.5(b) shown below can be prepared:

Sample calculation:

Let us take Fe415 steel and M25 concrete. For Fe415 steel, x_u,max /d = 0.4791. For M25 concrete, f_ck = 25. So putting these values in Eq.3.27, we get K = 3.472. This is the same value in the table 3.5(b). So if we are given a beam section with f_ck = 25 and f_y = 415, we can directly calculate it's M_u,lim by multiplying 3.472 by it's bd² .

We have to learn one more property of the beam section: The 'percentage of steel' provided in it. This is denoted by the symbol p_t. As the name indicates, it is the area of steel in a section, expressed as a percentage of the area of cross section. It is important to note that this area of cross section is equal to bd and NOT bD. That is., the effective depth d is used in calculating the area of cross section. Not the Total depth D. So we get 

Eq.3.28

Let us now consider the limiting value of steel given by Eq.3.21. We will re-write it in terms of x_u,max. 

Eq.3.29

Let us multiply both sides by 100, and at the same time divide both sides by bd. Then we will get:

Eq.3.30

100A_st,lim / bd is the percentage of limiting steel, and is denoted as p_t,lim. The right side has two variables: f_ck and f_y. (x_u,max /d depends on f_y). So we can give the commonly used values for f_ck, and f_y make a table. This is shown in table 3.5(a) below.

Sample calculation:

Let us take M25 concrete and Fe500 steel. Then from Eq.3.30,  p_t,lim = 41.61 x (25/500) x .4560 = 0.949. This is the same value given in Table 3.5.

So, if we are given a beam section having f_ck = 25 and f_y = 500, then we can say that it's p_t,lim is equal to 0.949. That means, the maximum area of steel that the section can have, with out making it Over reinforced is 0.949 x(bd/100)

Table 3.5

Let us now calculate the values of M_u,lim and A_st,lim of the beam section in fig.3.34 above. We have the following data: b =230mm, d =350mm, f_ck = 20 N/mm², f_y = 415 N/mm²

Calculation of M_u,lim:
• From Eqs.3.26 and 3.27, M_u,lim / bd² = K 
• From Table 3.5(b), corresponding to  f_ck = 20N/mm² and f_y = 415 N/mm², K = 2.777
• So we get M_u,lim = 2.777 x 230 x 350² = 78241975 Nmm = 78.24 kNm

Calculation of A_st,lim:
• From Eq.3.30, 100A_st,lim / bd = p_t,lim
• From Table 3.5(a), corresponding to f_ck = 20N/mm² and f_y = 415 N/mm², p_t,lim = 0.961
• So we get A_st,lim = (0.961 x 230 x 350)/100 = 773.605 mm²

Now we will add this section to the fig.3.34 and obtain the fig.3.35 below:

Fig.3.35
Comparison between three sections 

In the fig., A_st,lim is given as such. It is not converted into bars. The reason is that, we have only a certain number of bar diameters available in the market. Also, we can only provide whole number of bars. Fractions of bars cannot be provided. So it is difficult to provide the exact area. In this case, however, we will not be needing to give the exact A_st,lim in real beams. A_st,lim is used only as a bench mark.

Based on the above fig.3.35, we can now try to learn an important aspect about M_ulim. We see that the beam in the middle has it's 'capacity' denoted as M_ulim. The beams on either sides of it have their 'capacities' denoted as M_uR. From the discussions that we had so far, we can say that, 'M_ulim' is a particular value of 'M_uR' of a beam section. That is: Keeping all the properties same, if we change A_st only, then for each value of A_st, there will be a corresponding value for M_uR. But there will be only one value for M_ulim., and it corresponds to A_st,lim. 

When we are given a beam section, we know what all data (like cross sectional dimensions, grade of concrete etc.,) we will need to calculate it's M_uR. All those data will be needed to calculate it's M_ulim also, except the area of steel A_st. The reason for this is: M_ulim does not depend on A_st.

• We calculate M_ulim using all other details
• And then specify the A_st (using Eq.3.29 given above) that the beam will require to develop this M_ulim.
• This 'required A_st' is denoted as A_st,lim.

Starting from a low value of A_st, if we keep on increasing it's value, the ultimate moment of resistance M_uR of the beam section will also keep on increasing. At a particular value of A_st,lim, It's M_uR will become equal to M_ulim. If we still increase A_st, the section will become over reinforced. This can be presented in a graphical form as shown below:

Fig.3.36
Variation of M_uR and x_u

The above fig. shows four conditions of a beam section. In all the four conditions, the section have the same cross sectional dimensions and effective depth d. The grade of concrete and steel are also the same. But A_st, is different. 
• (1) is such that the A_st provided in it is far less than A_st,lim.  So the resisting moment M_uR that the beam section can offer at the ultimate state will be far less than it's M_ulim. We can say that the beam section in (1) has a low 'capacity'.  
• In (2), A_st provided is greater than in (1) but still less than A_st,lim. So it will have a greater capacity than in (1). 
• In (3), A_st = A_st,lim. So it will have the 'full possible capacity'. 
• In (4), the section will have a higher capacity than in (3). But it is over reinforced. So the failure will not be ductile, and it is undesirable.

The above graph shows the variation of x_u also.
• x_u will also keep on increasing as we increase A_st. 
• When the particular quantity of exact A_st,lim is given, x_u will be equal to x_u,max. 
• When A_st is still increased, x_u will become greater than x_u,max, indicating that the section has become over reinforced.

From the above discussions, it is clear that M_ulim is a particular value of M_uR. In fact, it is clear in the equation for M_ulim (Eq.3.25 given above) itself. We can see that in that equation, we are multiplying the compressive force by the lever arm. Just as we obtain M_uR by multiplying compressive force by the lever arm. The only difference is that, in Eq.3.25, the compressive force is that which the beam section will be experiencing when x_u = x_u,max. And also the lever arm is that when x_u = x_u,max. So we can write:

3.31

We will learn it's application when we discuss the design of rectangular beams in the next section.

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