Chapter 10 (cont..6) - Loads coming on the T-beam and the effective flange width

In the previous section, we calculated the effective spans. Now we will calculate the loads coming on the beam:

We have seen the difference between one-way slabs and two-way slabs. It was discussed in the presentation given at the beginning of chapter 5. Based on that we calculate ly/lx for each of the slab panels of our present case. It can be seen that all the four panels have the same ly and lx. = 4078 / 3586 =1.137

This is less than 2, and so each panel is a two-way slab. Thus some loads are transferred to the supporting beams, and the rest are transferred to the walls.

The fig. 10.17 given below helps us to visualize the calculation of loads coming on to the middle beam CF.

Fig.10.17
Transfer of loads

The slab panels are separated into triangular areas and trapezoidal areas by drawing lines at 45^o from the corners.

From the fig. we can see that the load in the green colored triangular portions will be transferred to the walls and the load from the yellow colored trapezoidal portions will be transferred to the beams. The beam CF will carry two trapezoidal areas, one on each side.  One of them is shown in the fig. 10.18 below:

Fig.10.18
Load on beam CF from one side

This trapezoidal portion is divided into two triangles and a rectangle. Base of the triangles in the above fig. is same as half of the base of the green triangle in the previous fig.10.17. And this will also be equal to the height of the triangle (∵ lines are drawn at 45^o from the corners of slab panels).  Thus we get (for the triangles in fig.10.18), height = base = 3586/2 =1793 mm. From the above discussion, we get the following results:

• Area of one triangle = 1/2 x1793 x1793 = 1607424.5 mm²
• Horizontal distance of the c.g of this triangle from the center of the beam = 492/2 + (1/3) x1793 = 843.67 mm
• Total area of the trapezoid = [(492 +4078)/2] x1793 = 4097005 mm²

So reactions at the supports are:
R_C =R_F = 4097005 / 2 = 2048502.5 

Taking moments about the center of the beam:
[2048502.5 x (4078 / 2)] -[1607424.5 x 843.67] -[1793 x(492/2) x(492/4)] = 4176896597.5 -1356130469.83 -54252594 =2766513533.67

Equating this to the central moment of a simply supported beam, acted upon by a udl, we get:

wl²/8 = 2766513533.67
From this we get w = 1330.84

This is the udl that produces the same effect as produced by the above trapezoidal loading on the beam. In other words, we can remove the trapezoid, and place a rectangle with ht. 1330.84 mm (1.33 m) on the beam. The following fig.10.19 shows this derived udl.

Fig.10.19
UDL on beam CF

So now we know the load coming on the beam from the slab: Each one meter length of the beam CF will carry the load coming from [(1.33 + 1.33) x 1] m² area of the slab. This is shown in fig.10.20 below.

Fig.10.20
Load on 1m length of the beam

Based on this, we can now calculate all the loads coming on one meter length of the beam CF.
• Self wt. of slab coming on the beam =1 x2.66 x0.12 x25 =7.98 kN/m
• Wt. of finishes of the slab = 1 x2.66 x1.3 =3.458 kN/m
• LL from the slab = 1 x2.66 x4.0 =10.64 kN/m
• Self wt. of beam = 1 x 0.25 x 0.23 x 25 =1.44 kN/m

So total load per meter length = 7.98 +3.458 +10.64 +1.44 =23.518 kN/m
Applying the load factor of 1.5, we get the factored load =w_u = 1.5 x 24.05 =35.277 kN/m. So M_u = w_ul²/8 = 35.277 x 4.23²/8 =78.9 kNm

Now we must calculate the 'most probable' d:

Eq.10.5

z is the lever arm which is taken as the larger of the following:

10.6
  ♦ 0.9d
  ♦ d - Df/2

For an initial value of d, we can deduct 50 mm from D. So we get d = 320 mm. So z is the larger of:
  ♦ 0.9d = 0.9 x 320 = 288 mm
  ♦ d - Df/2 = 320 - 120/2 =260 mm

Thus z = 288 mm. Substituting the values in 10.5 we get A_st,reqd = 758.78 mm². Assuming 3 - 18#, we get an area of 763.41 mm². These bars can be arranged in a single layer as shown in the fig.10.21 below:

Fig.10.21
Bars in single layer

The exposure condition is given as 'moderate'. So we can give a concrete cover of C_c = 30 mm. Thus d = 370 -30 -8 -9 =323 mm. This is the 'most probable' d.

Earlier, we assumed d = 370 -50 =320. Even though the d has changed, the effective span of the beam will not change because the lesser value is still the c/c distance between supports.

Check for minimum horizontal distance between bars (details here):
The the total width of the beam at the bottom most layer of bars can be written as:
230 = 2 x C_c + 2 x Φ_l + 3 x Φ + 2 x S_h. 
⇒ 230 = 2 x 30 + 2 x 8 + 3 x 18 + 2 x S_h. 
From this we get S_h = 50 mm
• Bar diameter = 18 mm
• Max. size of aggregate + 5 = 20 + 5 = 25 mm
Larger of the above = 25 mm < 50 mm. Hence OK

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Next we have to calculate the effective flange width b_f. 

Eq.9.5

l₀ =4230 mm; b_w =230 mm; D_f =120 mm; s₁ = s₂ =3500 +230 =3730 mm. Substituting these values we get: (a) 1655 mm & (b) 3730 mm. b_f = smaller of the two = 1655 mm.

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Now we have all the required details to begin the actual design process. We will see this in the next section.

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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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Chapter 9 - Effective flange width of T-beams and L-beams

In the previous section we completed the design of continuous slabs and beams. All the beams that we considered until now had a rectangular cross section. In this section, we are discussing about beams with another cross section: The 'flanged beams'.

We know that the slabs rest on walls or beams. Consider the case when a slab rests on a beam. In this case, if two 'conditions' are satisfied, a portion of the slab will act as the 'flange' of the beam. That is., when the beam bends, the slab will also bend in the same direction. This is shown in the fig.9.1 below.

Fig.9.1
Slab acting as flange of a beam 

The beam will become stronger when it becomes a 'flanged beam' because there is greater concrete area in the upper portion to take up the compression. The lateral stability of the beam also increases. But in order to achieve a flanged section, the following two conditions should be satisfied: The first condition is that the slab and beam on which it rests, should be cast together. The second condition is related to the arrangement of the bars of the slab. This is explained in detail at the beginning of chapter 10 - Design of flanged sections.

T-beams and L-beams are the commonly encountered flanged beams. They are mostly seen in framed structures.  Intermediate beams which have slabs on both sides, act as T-beams. The end beams and the beams around staircase or lift openings, which have slabs only on one side, act as L-beams. In both T-beams and L-beams, a portion of the slab acts integrally with the beam. So when the beam bends, this portion of the slab also bends in the same direction as the beam. This portion of the slab is called the Flange and the portion of the beam below the flange is called the Web.


The following presentation gives the basic idea about flanged beams and the effective width of flanges:

   

An actual building under construction is shown below. The T-beams and L-beams can be clearly distinguished

Equivalent flange width

From the above presentation, we have a basic idea about the concept of 'equivalent flange width' b_f . The following properties are associated with b_f:

• It increases when the span of the beam increases

• It increases when the width of the web  increases.

• It increases when the thickness of the flange increases

• It depends on the type of loading (concentrated, distributed etc.,)

• It depends on the type of support (simply supported, continuous etc.,)

Considering the above factors, the cl 23.1.2 of the code gives the following approximate formulae for calculating b_f.

Eq.9.1: for T-beams

Eq.9.2: for L-beams

In the above formulae, we know that b_w and D_f denotes width of the web and depth of the flange respectively. But l₀ is a new term for us. l₀ is defined as 'the distance between points of zero moments in a beam'. When we consider a simply supported T- beam or L-beam, the distance between the points of zero moments in the beam is obviously equal to it's effective span. Because at the supports, the bending moment is zero for a simply supported beam or slab. But this is not the case when we consider a T-beam or L-beam which is continuous over supports, or when it is part of a framed structure. Consider the fig 9.2 below:

Fig.9.2

Portion of the BM diagram of a continuous member

The above fig. shows a portion of the Bending moment diagram drawn for a continuous beam. The moments at supports are of hogging type, taken as positive and so are marked above the X axis. The moments near midspans are of sagging type, taken as negative and so are marked below X axis. Thus there is a transition from positive to negative and then again back to positive when we move along the length of the beam. In the fig, these transitions occur at points P And Q. Each of these points is called a Point of contraflexure, and at these points bending moment is equal to zero. So for the case shown in the fig, l₀ is equal to the distance PQ.

The above method to calculate l₀ is tedious. So the code permits us to use a much simpler procedure to calculate l₀ . This is given as the note below cl. 23.1.2 (c). According to this note, l₀ may be assumed as 0.7 times the effective span in continuous beams and slabs. We have already discussed the methods for calculating the effective spans of continuous members here. 

So now we know how to calculate b_f using eq 9.1 and 9.2. But when we obtain b_f of a beam using the above formulae, there is a possibility that we get a value which is greater than what is actually available for the beam in the structure. Given below is fig 9.3 which shows the top right side portion of the structure that we discussed in the presentation. It shows the details of the T-beam P and the L-beam Q.

Fig.9.3

Maximum possible width of flange

From the fig, we can deduct the following:

Eq.9.3: The maximum share available for the T-beam P (marked with yellow colour) is

Eq.9.4: The maximum share available for the L-beam Q (marked with white colour) is

So the effective flange width calculated using 9.1 should be checked to see that it is not greater than 9.3 and effective flange width calculated using 9.2 should be checked to see that it is not greater than 9.4. 

In other words, the lesser value from the two methods of calculation should be used.

If we denote the center to center (c/c) span on either side of a T-beam as s₁ and s₂ , and the c/c span on the side of a L-beam as s₁ , we can write 9.1 and 9.2 in the general form as:

Eq.9.5: For T-beams, b_f is the lesser of the following two values:

Eq.9.6: For L- beams, b_f is the lesser of the following two values:

Isolated T-beams and L-beams

In some cases, isolated T-beams or L-beams are encountered. Fig 9.4 below shows an example of an isolated T-beam used for a foot bridge. The fig. shows the view from under side of the bridge.

Fig.9.4
Isolated T-beam in a foot bridge

From the fig. we can see that the slab above the beam which acts as the flange, is discontinuous at the sides. The stringer beam of a staircase is another example of an isolated T-beam. A part view of such a staircase is shown in fig.16.84 in the section on transverse stairs. Cl 23.1.2(c) of the code gives the following formulae for calculating for isolated T-beams and L-beams.

Eq.9.7: For isolated T-beams, take lesser of the following two values:

Eq.9.8: For isolated L-beams, take lesser of the following two values:

Where b is the actual width of the flange. Obviously, b_f cannot be greater than b.

So at this stage we are able to calculate the effective width b_f of the flange for different types of T-beams and L-beams. In the next section we will start the discussion on the analysis of these flanged sections.

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