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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