Chapter 9 (cont..8)

In the previous section we completed the analysis of an L-beam section. In all the solved examples that we saw so far in this chapter, the effective width b_f of the sections were already given. We will now see an example in which we have to calculate b_f ourselves. The following presentation shows the basic details of a structure. Using those details, we can calculate the b_f of the T beams in that structure.

Solved example 9.4:



Assume f_ck = 20  N/mm² and Fe 415 steel.

Solution:
We will first write the following data:
D_f =110 mm, D =510 mm, b_w =250 mm, d =464 mm, A_st =603.19 mm² (3-16Ф).

Clear span of beam = cs =3500 mm
Widths of supports = ws =230 mm
c/c spacing of beam s₁,s₂ =4250 mm

Now we can start the analysis procedure:
Our first task is to calculate the effective flange width b_f. For this we can use Eq.9.5 

b_f is the lesser of (a) and (b). But we have to calculate l₀ first. For simply supported beams, l₀ is the effective span.

Effective span is the lesser of the following:
(1) Clear span + effective depth = 3500 + 464 = 3964 mm
(2) c/c distance between supports = 3500 + 230/2 + 230/2 =3730 mm

So l₀ = effective span =3730 mm. Substituting all the values in Eq.9.5 we get: 

(a) =1531.67 ; (b) =4250 mm. b_f = Lesser of the two = 1531.67 ≈ 1532 mm
So our aim is to analyse the beam section shown in the fig. below:

Fig.9.31
Section of T-beam

  
For Fe 415 steel, x_u,max / d = 0.4791 (Table 3.4)
So x_u,max for our beam section = 0.4791 x 464 =222.30 mm - - - - (1) 

We know that the given beam section may fall into any of the three cases shown in fig.9.17. To find x_u and M_uR, we first need to know the particular case into which our beam will fall. For this, we have to do some tests. 

The first test will be to check whether the beam falls into the category of 'Case 1: D_f ≥ x_u'. For doing this test, we have to make the following assumption:

• x_u has the maximum value possible within case 1. 

This maximum possible value is equal to D_f. So we put x_u = D_f =110 mm

We need to get a good understanding about this assumption. We can see a '≥' symbol in the name of this case 1. This means that x_u can have a value less than or equal to D_f. The maximum value possible is D_f. In the above assumption, we are giving this maximum possible value to x_u. In this situation, the stress distribution will be as shown in the fig.9.32 below:

Fig.9.32
Stress distribution when x_u = D_f

In the above fig., x_u = 110 mm. So the NA coincides with the bottom edge of the flange.

• Also, when we assume the above value for x_u, it automatically implies that: 'we are assuming the section to be under reinforced'. This is because, the value of 110 mm for x_u is less than x_u,max which is obtained in (1)

If the above two assumptions that we make are true, at ultimate state, the compressive force in concrete will be equal to

C_u = 0.362 f_ck b_f x_u = 1220084.8 N

Note that the above equation is obtained by changing b to b_f in Eq.3.7  

And the tensile force in steel (steel would have yielded at ultimate state if it is an under reinforced section) will be equal to

T_u = 0.87f_y A_st = 217781.75 N 

From the above results, we find that C_u is larger than T_u. This is not allowable. So we want a decreased value for C_u. Let us assess this situation:

• If we want a decreased value of C_u, then concrete area should be decreased.
• This will imply that the position of NA should be raised, which is same as saying: value of x_u should decrease.
• We have given the maximum possible value of x_u in Case 1.
• If we now decrease x_u, then, Point 'A' in the above fig.9.32 will remain inside the flange.
• This will imply that our beam belongs to case 1.

• Also, the value of 110 mm for x_u is already lesser than x_u,max. We are now going to decrease it further. So the beam section will indeed be under reinforced.

Thus the 'tests' are complete. We can proceed to do the analysis. We know that 'Case 1' is same as a beam of rectangular cross section. We have earlier seen a solved example demonstrating the calculation of xu of an under reinforced rectangular section here. Let us now follow the same procedure and calculate the x_u of our present beam:  

x_u is given by 

= 19.635 mm < x_u,max. Hence OK

Note that the above equation is obtained by replacing b by bf in Eq.3.17

Now we have to calculate M_uR. This is given by:
M_uR = T_u x z. 
Where T_u = 0.87f_y x A_st = 217781.75 mm
z = d - 0.416x_u = 464 - 0.416 x 19.64 = 455.83 mm

Thus we get M_uR = 99271454.87 Nmm = 99.27 kNm

This completes the analysis of the given T-beam. In the next section, we will see the limiting moment of resistance of T-beams and L-beams.

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