Chapter 3 (cont..9) Ultimate Moment of resistance of Rectangular beam sections

In the previous section we saw two solved examples. The first solved example is that of an under reinforced section, and the second one is that of an over reinforced section. After solving the two problems, we can say that we can find the depth of NA x_u, the stress, and the strain in the steel (all at the ultimate state), for any given beam section. The stress and strain in concrete at the ultimate state for both under reinforced and over reinforced beam sections will be 0.447f_ck N/mm² and 0.0035 respectively. Also the stress in steel at ultimate state for an under reinforced section will always be 0.87f_y. These do not need any calculation.

But the ability to find the stress and strain in steel and x_u for any given beam section is not sufficient. We must be able to 'specify the load' that can be safely applied on the beam.

For this purpose, we must calculate a quantity called 'Ultimate Moment of Resistance'. This is denoted as M_uR. We must calculate this quantity for every beam section that is given to us for analysis. So let us see the details about it:

'Ultimate Moment of Resistance', as the name indicates, is the internal Moment of resistance, that the section is able to offer at the ultimate state. So this is the 'maximum' resistance that we can expect from the section because, after the ultimate state, the beam itself will not serve any purpose. We have already seen the various quantities that are required to calculate this internal resistance:

From the fig.3.21 which we have seen earlier, it can be seen that the internal resistance is given by M_uR = T_u . z = C_u . z

We have seen that z = d - 0.416x_u. (Eq.3.8)

After finding the stress in steel, we get T_u = [Stress in steel x Area of steel]

After finding the value of x_u, we get C_u =  0.362 f_ck b x_u  (Eq.3.7)

We can use either C_u or T_u for calculating M_uR. Both will give the same result because, for equilibrium, C_u will be equal to T_u.

Thus we can say that we are able to calculate M_uR for any beam section. If we apply an external load such that a moment of M_uR is produced at the beam section, an equal internal moment of M_uR will develop at that section, and so it will reach the ultimate state.

For every beam that is given to us for analysis, we must calculate M_uR. Let us calculate M_uR for the beams in each of the solved examples 3.1 and 3.2 that we saw in the previous section:

First solved example:

• The beam is Under reinforced. So the stress in steel = 0.87 x 415 = 361.05 N/mm²

• Force T_u = stress x area = 361.05 x 603.19 = 217781.7495 N = 217.78 kN

• The depth of NA = x_u,U = 120.32 mm. So z = d - 0.416x_u = 400 – (0.416 x 120.32) = 349.95 mm.

• Thus M_uR = T_u . z = 217.78 x 10³ x 349.95 = 76212111 Nmm = 76.2 kNm

Second solved example:

• The beam is over reinforced. The force in steel T_u, and the value of x_u,O can be taken from the final (fifth) cycle. So T_u = 387.04 kN.

• x_u,O = 232.28 mm. So z = d - 0.416x_u,O = 400 – 0.416 x 232.28 = 303.37 mm.

• Thus M_uR = T_u . z = 387.04 x 10³ x 303.37 = 117416324.8 Nmm = 117.42 kNm

The maximum resisting moment that the first beam can offer is 76.2 kNm. If a moment greater than this is applied, the section will fail. But as the steel has already yielded, there will be large deflections, wider cracks etc., at the time of failure. So we will get enough warning.

The maximum resisting moment that the second beam can offer is 117.42 kNm. If a moment greater than this is applied, the section will fail. Here the steel has not yielded at ultimate state. So there will not be any deflections, wider cracks etc., at the time of failure. Thus the failure will be sudden with out any warning.

Now let us see how M_uR can be used for specifying safe loads:

We will discuss this with the help of an example. The fig. below shows the plan and part section of a slab and beam.

Fig.3.32 

Plan and part section of slab and beam

The slab has a total width of 250 + 2950 + 250 = 3450mm. It rests on two of it's opposite edges. On the left edge, the support is a masonry wall, and on the right edge, it is a beam having an effective span of 4000mm. The beam is simply supported on two columns. We are asked to analyze this beam. The cross sectional dimensions are given in the section XX. The reinforcement details are: 3-#16 at an effective depth of 410mm. f_ck = 20 N/mm² and f_y = 415 N/mm²

We can analyze it as a rectangular section with b= 250, D = 460 and d = 410mm. The results of the analysis are as follows: (calculation steps are not shown here)

It is an under reinforced section with M_uR = 78.39kNm

Now we will discuss about how this result can be used:

If the factored load acting per meter length of the beam is denoted as w_u, then we can write: The factored moment M_u acting on the beam at mid section is given by:

So M_u= w_u x 4² x (1/8) = 2w_u - - - (1)

Now we will split w_u into it's two components:

w_u = w_u,DL + w_u,LL - - - (2)

Where w_u,DL is the factored DL and w_u,LL is the factored LL

Characteristic load w_DL:

Self wt. of beam per meter length = .25 x .3 x 25 = 1.875 kN/m - - - (3)

Self wt. of slab: for this we will use the fig.3.33 below:

Fig.3.33 

Loads from slab

The tributary area has a width of 1725mm. So a 1m wide strip will have an area of 1.725 x 1 = 1.725 m².

So self wt. of slab = 1.725 x .16 x 25 = 6.9 kN/m - - - (4)

Self wt. of finishes over slab assuming @ 1kN/m² = 1.725 x 1 = 1.725 kN/m - - - (5)

DL of partitions assuming @ 1.25 kN/m² = 1.725 x 1.25 = 2.16 kN/m - - - (6)

So w_DL = sum of (3) to (6) = 12.66 kN/m - - - (7)

This is the characteristic DL. There can be uncertainties in the magnitude of possible DL. So we use the load factor 1.5. Thus w_u,DL = 12.66 x 1.5 = 18.99 ≈ 19.0 kN/m - - - (8)

Substituting (8) in (2), we get

w_u = 19.0 + w_u,LL - - - (9)

Substituting (9) in (1) we get :

Mu = 2 (19.0 + w_u,LL ) = 38 + 2 w_u,LL - - - (10)

But the maximum value that M_u can take = M_uR = 78.39kNm because, this is the maximum resistance that the beam can offer.

So we can write M_u = M_uR = 78.39 = 38 + 2 w_u,LL - - - (11)

From this, we get w_u,LL = 20.195 kN/m

This is the LL from 1.725 m² area. So the LL from 1 m² area = 20.195/1.725 = 11.7 kN/m².

But there may be uncertainties in the LL that can possibly occur during the life time of the structure. So we cannot specify that this much load LL can be applied on the slab. We must apply the load factor of 1.5. So we get 11.7/1.5 = 7.8 kN/m²

Thus we can specify that characteristic LL of 7.8 kN/m² can be safely applied on the slab, as far as the safety of the beam is concerned. But this load may not be safe for the slab. M_uR of the slab should be determined, and similar calculations should be done separately.

So by calculating the M_uR of a given beam section, we are able to specify the characteristic loads that can be applied on the beam.

It may be noted that the above example is only for a demonstration of the calculation steps. In an actual case, we have to do further checks like the safety of the slab section, Safety of the masonry wall and columns, check for deflection, check for development length, check for shear, check for torsion, other possible loads like wind loads and seismic loads, stability of the structure etc., In short, it should be checked that all the elements of the structure satisfy the clauses of all relevant codes.

In the next section we will see some more properties of beam sections.

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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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