Chapter 10 (cont..5) - Second solved example on design of T-beams

In the previous section, we completed the design of a T-beam section. Now we will see another design problem:

Solved example 10.2
A room has internal dimensions 4000 x 14690 mm. (Figs.10.15 & 10.16) The walls of the room are of brick masonry 230 mm thick. The slab of the room has a thickness of 120 mm, and it rests on these outer walls and three equally spaced beams along the 4000 mm direction. The width of the web of these beams is 230 mm. The slab carries a Live load of 4.0 kN/m². Assume wt of finishes as 1.3 kN/m² in addition to the self weights. Assuming Fe 415 steel, design an interior beam of the room. Assume moderate exposure conditions.


Fig.10.15
3D view (The slab is not shown for clarity)

Fig.10.16
Plan view

Solution:
The slab of this room is a continuous one, with four spans. We will find the effective span of the slab for each of these spans. The calculations are given below:
Calculation of effective spans:

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For this problem, it is convenient to mention before hand that, each of the spans have their support widths less than it's ln/12 (∵ 3500/12 = 291.67). So while using the cl.22.2(b) of IS 456, we will not have to look to the portion below the magenta colored dashed line of the chart. (Fig.7a.4)

Total depth of the slab is 120mm. Assume dia. of main bars of slab = 8 mm and Cc = 30 mm. So effective depth d = 120 -30 -4 = 86 mm

First we will consider the end span. The calculations based on IS 456 is given below:
Clear span l_n =3500 mm.
l_n/12 = 3500/12 =291.67. So t₁ < l_n/12 & t₂ < l_n/12

As mentioned above, we only need the portion above the magenta colored dashed line for all spans of the beam. This is shown in the fig. below. This fig. is applicable to all the spans.

Fig.7a.4
Application of chart to end span

Now we calculate the following:
• c/c distance between the supports = 3500 +115 +115 =3730
• clear span + effective depth = 3500 +86 =3586
Effective span = l_eff = Lesser of the above = 3586 mm
Thus we calculated l_eff of the end span.

Now we will consider the second span. The calculations based on IS 456 is given below:
Clear span l_n =3500 mm.
l_n/12 = 3500/12 =291.67. So t₁ < l_n/12 & t₂ < l_n/12

Fig.7a.4 is applicable here also. Now we calculate the following:
• c/c distance between the supports = 3500 +115 +115 =3730
• clear span + effective depth = 3500 +86 =3586
Effective span = l_eff = Lesser of the above = 3586 mm

Thus we calculated l_eff of the second span. From the plan view, we can see that the continuous slab is symmetric. Thus all the spans have l_eff = 3586 mm.

The above calculated value is the effective spans in the long direction. We must calculate l_eff in the 4000 mm direction also:

The effective depth of the slab in this direction = 120 -30 -8 -4 =78 mm. An additional 8 mm is coming here because the layer of bars which take up the loads in the 4000 mm direction will be above the layer of bars which take up the load in the 3586 mm direction. So the effective depth will be less by an amount equal to one bar diameter. We will learn more details about this when we discuss Two-way slabs.

As the slab is simply supported in this direction, the effective span is the lesser of the following:

(a) clear span + effective depth = 4000 +78 =4078 mm
(b) c/c distance between the supports = 4000 +230 =4230 mm

So the effective span in this direction = 4078 mm

This completes all the calculations regarding 'effective spans' of the slab in our problem.

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Next we have to calculate the effective span of the beam:
For this calculation, we will require the effective depth d. But in this problem, the depth of the beam is not given. So we will have to assume some preliminary dimensions for our beam. 

Let us adopt l/D ratio of 13. Where l is the clear span. So D= l/13 =4000/13 =307.7 mm. So the depth of the beam below the slab = 307.7 -120 =187.7 mm.  This can be rounded off to 200 mm. So D = 320 mm. Also we have b_w =230 mm.

Now we have to do the checks that have to be done immediately after fixing the preliminary dimensions. These are given in 4.2, 4.3 and 4.4   

4.2: D/b ratio = 320/230 = 1.39. The recommended range is 1.5 to 2. If it is to be 1.5, D should be 1.5 x 230 =345 mm. So depth below slab =345 -120 =225. For ease of putting up the form work, we will adopt 250 mm. Thus we get D = 250 +120 =370 mm. Now D/b =370/230 =1.61. Hence OK.

4.3: l/D = 4000/370 = 10.81. This is greater than 2. So it is not a Deep beam

4.4: • 60b = 60 x230 =13800. 
• 250b²/d = (250 x230²/320) =41328.125 (Assuming d = 370 -50 =320 mm)
Lesser of the above = 13800. The clear distance of 4000 mm available to the beam is less than 13800. Hence OK.

So we have done the three checks, and the preliminary D can be taken as 370 mm.   d can be taken as 370 -50 =320 mm. The beams are simply supported. So the effective span is the lesser of the following:
(a) clear span + effective depth =4000 +320 =4320 mm
(b) c/c distance between supports = 4000 +230 =4230 mm

So the effective span of the beam = 4230 mm. Thus we have completed all the effective length calculations. It may be noted that the above results will change if we change any of the dimensions of the beam later in this design process.

In the next section, we will calculate the loads coming on the beam.

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Chapter 10 (cont..1) - Deflection control and preliminary dimensions of flanged beams

In the previous section, we have seen the conditions for ensuring integral action between the slab and the beam. We also saw different types of flanged beams and their performances. Now we will see the basic requirements that a newly designed 'flanged beam' should satisfy.

The basic concepts of the process of design of singly reinforced rectangular sections were discussed earlier. There, in the introduction part, we discussed about:
1.concrete cover 
2.Minimum distance to be provided between the bars of beams
3.Maximum spacing allowable between bars of beams
4.Beams with overall depth greater than 750 mm
5.Minimum area of flexural reinforcements in beams
6.Maximum allowable area of flexural reinforcements in beams
7.Deflection control of singly reinforced beams and
8.Guide lines for fixing up the dimensions of beams


Links to each of the above items can be seen here.

All the above eight topics are applicable to Flanged beams also. However, some minor modifications have to be made to some of them. We will now look at the required modifications:

No.5: Minimum area of flexural reinforcements in flanged beams

We have seen that the minimum area of flexural reinforcement required in the case of a rectangular section is given by Eq.4.1

In the case of flanged sections, the width b should be replaced by b_w, the width of the web. So we get

Eq.10.1

No.6: Maximum allowable area of flexural reinforcements in flanged beams

Similar to the above,  the width b should be replaced by b_w, the width of the web. So we get

10.2
A_st ≯ 0.04b_wD 

No.7: Deflection control of singly reinforced flanged beams

We have seen the modification factors that have to be applied to the basic l/d ratio. There we mentioned that the reduction factor k_f will be discussed when we take up the design of flanged sections. In fact, we have to do two things: 
• we have to see some modifications to the factor k_t and 
• we have to learn about the new factor k_f

So we now look at cl 23.2.1(e) of the code. When we do the deflection calculations for a singly reinforced 'flanged beam', we have to determine the values of three quantities: Modification factor α, Modification factor k_t, and the reduction factor  k_f. The final equation can be written as:

(l/d)_actual  ≤  [(l/d)_basic] α k_t k_f 

Modification factor α can be determined by the same procedure as for a rectangular section. It may be noted that we do not need to calculate α if the span is less than 10 m, because then it's value is equal to 1.

Let us first see k_t: To calculate k_t, we have to first calculate the percentage of tension reinforcement by using the equation:

Where A_st,p is the actual area of steel provided

But in our present case of flanged beams , the effective width of flange, b_f has to be used instead of b . So the equation becomes:

The rest of the procedure for calculating k_t  is same as that for a rectangular beam.

Now we come to the factor  k_f.  For it's calculation, first we calculate the ratio of 'web width' to 'flange width'. That is, the ratio: b_w / b_f  . Then we use this ratio to obtain k_f from the fig 6 of the code.

Thus we can write the deflection control equations in the final form:

10.3
For singly reinforced flanged beams with span less than 10m,

(l/d)_actual  ≤  [(l/d)_basic] k_t k_f

10.4
For singly reinforced flanged beams with span greater than 10m,

(l/d)_actual  ≤  [(l/d)_basic] α k_t k_f

If the beam is a cantilever with span greater than 10 m, actual deflection calculations should be made.

So we have Seen the code procedure for deflection control of a flanged beam. But this method has been found to give results that are not normally expected. So it is recommended that for calculations regarding deflection control of a flanged beam, the over hanging portions should be ignored, and it should be considered as a rectangular section of width b_w and effective depth d . That is., when we have to do the deflection control calculations of a singly reinforced flanged section, we must consider it as a rectangular section of width b_w, and effective depth d, and then do the calculations.

No.8: Guide lines for fixing up the dimensions of beams

Fig.10.11 below shows the view of a part of a structure. It shows a slab supported over some beams and masonry walls. If the slab is cast integrally with the beams, and the condition of the arrangement of slab bars mentioned in the previous section is satisfied, they can be considered as T-beams. As these T-beams are supported over a number of masonry walls, they are 'continuous T-beams.'

Fig.10.7
3D View of Continuous T-beams

Let us consider any one of these continuous T-beams. We can analyse it and draw a bending moment diagram. Fig 10.8 below shows a portion of a typical bending moment diagram of a continuous system. It is shown here to get an understanding about the 'nature of bending moments' in continuous systems. And also to see a comparison between the 'values' of the bending moments. A video about continuous systems can be seen here.  

Fig.10.8
Portion of a Bending moment diagram

We can see that at the continuous supports, the Bending moment has values of 19.3 and 18. But at mid span, it has a lower value of 12.9. In this way, the bending moment at continuous supports will normally be higher than the bending moments at mid spans. So the higher bending moments at these supports should be used to fix up the dimensions of the beams.

However, each structure should be examined and analysed carefully to see if this general rule is true for that structure.

It should also be noted that the bending moment at supports are hogging moments. So the beam will experience tension at the fibres above the neutral axis. So the flange, which is at the top portion of the beam will be under tension, and the concrete in the flange will have cracked. So the flange cannot be considered in the design. Thus, at the supports, the beam section should be designed as a rectangular section. Towards the midspan, the moment becomes sagging type. Here the beam will act as a proper flanged beam. But the width of the web b_w and effective depth d are already fixed by considering the moments at the supports. So the only calculation to be done here is that of the tension steel. But the effective width b_f of the flange has to be calculated prior to the calculation of steel. So in the case of continuous T-beams, the preliminary dimensions are fixed by the methods that we discussed for 'rectangular sections'. And these methods should be applied at the supports. We can say that there are no new methods to be learnt for fixing up the preliminary dimensions of a 'continuous T-beam'.

But this is not the case of 'simply supported T-beams'. Here we do have to learn some new methods. Especially for fixing up the 'preliminary depth'. Fig.10.13 below shows the view of some simply supported T-beams. (The slab above the beams is not shown in the fig. for clarity)

Fig.10.13
3D view of Simply supported T-beams

In this case, the bending moment at supports is equal to zero. Towards the mid span, there will be sagging moment in the beam, and the flange will take up compressive force. So it will act as a proper T-beam. The traditional method for fixing up the preliminary width is same as that for a rectangular section.  We saw those details here. Based on that discussion, b_w is selected from general values such as  200, 230, 250, 300 etc.,

Now we come to the depth. We have earlier seen (in the discussion on rectangular beams) that, the strength (to resist bending) of a beam can be increased by increasing  either it's depth or width. So a beam having a large width will have considerable strength. A T-beam or L-beam will indeed be having a larger width, on account of their flange. This will mean that these beams can withstand the bending moments even if they have smaller depths. But in reality, we must not provide smaller depths under any circumstances. This is because, the increased strength obtained from the increased width may be sufficient to resist bending. But the increased strength obtained in this way will not help the beam to resist 'Shear'. Also it will not help the beam to 'control deflection'. That is., even if smaller depth is sufficient to resist 'bending', we do have to provide sufficient depth to resist other effects like 'shear', 'torsion' etc., Also sufficient depth is required to control 'deflection'. 

So the method that we will use to fix up the preliminary depth should take into account the above points. The range of 13 to 16 for l/D ratio, gives satisfactory results for flanged sections. From the value of D, we usually deduct 50 mm to obtain d. We know that the quantity of '50 mm' is obtained by simple calculations on some assumed values: [Concrete cover 30 mm] + [dia. of links 10 mm] + [half the dia. of main bars = half of 20 = 10 mm] = 50 mm. Here the main bars of 20 mm dia. are assumed to be placed in a single layer. But when heavy loads are acting, a large quantity of bars will have to be provided. Such large quantities can be accomodated only if we arrange the bars in 'layers'.

When the bars are arranged in layers, the above quantity of '50 mm' will change significantly. Consider the scenario:
• We start off by simply deducting 50 mm
• The beam actually have a much different 'd' from that calculated using 50 mm
Then the subsequent calculations will not give correct results. This is particularly important for flanged beams because, as we will soon see, there are a lot of calculations involving 'd'. So it is important to get a fairly good approximation of 'd' in the initial stage itself. In the next section we will see how this can be done.


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