Chapter 10 - Integral action between slab and beam for flanged beams

In the previous section we completed the analysis of flanged sections. In this section we will consider the design. At the beginning of chapter 9 we saw that two conditions should be satisfied to ensure the integral action between the slab and the beam. There we saw the first condition: The slab and the beam should be cast together. In most cases, the slab and beam are indeed cast together. The form works consists of horizontal board for the bottom side (soffit) of the beam, vertical boards for the sides of the beam, and horizontal boards for the under side of the slab. After tying the reinforcement bars and keeping them in position, fresh concrete is poured from the bottom of the beam. Concrete is poured continuously in such a way that, first the beam is filled up, and then the slab. Necessary compaction is ensured by using a needle vibrator or other such devices.

It may be noted that the stirrups of the beam also play a major role in ensuring integral action between the slab and the beam. These stirrups confine the main bars of the beam. They extend from the beam into the slab, upto the top surface of the slab. They are able to extend upto the top surface because the 'stirrup suspender bars' of the beam are near the top surface of the slab. So sufficient stirrups should be provided for the beam to ensure the integral action. We will learn about the design of stirrups when we discuss 'shear design'.  

We will now see the second condition. Consider a slab and beam system used as the roof of a room, as shown in the plan view below:

Fig.10.1
Main bars of a slab given parallel to beam does not contribute towards flange action

We can see that the main bars extend from wall AB to wall CD. This is because, in the design of this particular slab, the slab loads are assumed to be transferred to the walls AB and CD only. No loads are transferred to the other two walls BC and AD. So no loads are transferred to the beam either. Thus all the main bars of the slab are parallel to the beam. In such a case, even if the slab and beam are cast together, the slab will not make a 'contribution' as the flange of the beam.

For the slab to make the 'contribution', a specific quantity of main bars of the slab should be available in a direction perpendicular to the beam. So in the above case, we must provide some 'special bars' perpendicular to the beam. The details of this are given in the cl.23.1.1.b of the code. The requirements of this clause can be illustrated if we draw the section XX shown in the plan view above. This is shown in the fig.10.2 below:

Fig.10.2
Section XX

We can see that the length of each of the 'special bars' on either side of the beam, should not be less than l/4, where l is the span of the beam. The area of these bars should not be less than 60% of the area provided at the midspan of the slab. Also, these bars should be provided near the top surface of the slab. By providing the specified quantity and length of these 'special bars', at the correct position, we can ensure that the slab will act as the flange.

In the fig.10.1 above, distributor bars will be present in a direction perpendicular to the beam. But they are not available near the top surface of the slab through out the length of the beam. Also their quantity may not satisfy the 60% requirement. So we must provide the 'special bars' even though the distributor bars are present in the perpendicular direction to the beam.

Now consider the slab shown in the fig.10.3 below:

Fig.10.3
Main bars of a slab given perpendicular to beam will contribute towards flange action

In this case, the main bars of the slab are provided in a direction perpendicular to the beam. It is a continuous slab. We have already discussed the analysis and design of such slabs. We know that the slab will experience a hogging moment at the beam. So there will be top bars to resist the hogging moment. And these bars will be perpendicular to the slab. So we do not need any 'special bars'. The top bars that we designed at an intermediate support for a slab can be seen here. It may be noted that even if we provide bent-up bars instead of straight bars, the required effect will be obtained because, the bent-up bars will become top bars at supports. In any case, the required length of l/4, and the quantity of 60% mentioned above should be available.

So now we know how to ensure integral action between slab and beam. We must extend this discussion to isolated T-beams and L-beams also. In those cases, the stirrups of the beam should extend to opposite sides of the beam, and travel upto the edges of the slab. This is shown in the fig.10.4 below.

Fig.10.4
Ensuring integral action between slab and beam for isolated flanged beams 

A 3D view can be seen below:
Fig.10.5
3D view of bars in isolated T-beams

Fig.10.6
3D view of bars in isolated L-beams

From the above figs., we can see that the number of stirrups will be equal to the number of slab bars. The number of stirrups is calculated using shear force considerations. But the number of slab bars (obtained as spacing in 1m) is calculated from bending moment considerations. We must correlate both these and the larger number should be provided. The diameter of these bars should also be checked.

The presentation given below shows some examples of flanged beams, and the method of their performances.





From the above presentation, it is clear that each structure should be examined carefully to determine the locations where we can design a beam as a 'flanged beam'. 

We can now take up the discussion on design procedure. In the next section, we will see the basic requirements like concrete cover, minimum steel etc., 

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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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Chapter 16.13 - Stairs cantilevering on the sides of Stringer beam

In the previous section we saw the details of transverse stairs which cantilever from the side of a stringer beam.  The following solved example illustrates the design of  such a stair.

Solved example 16.4

The reinforcement details according to the above solved example is shown in the figs.16.81, 82 and 83 given below. The cantilever waist slab has to be properly anchored into the beam. For this, the stirrups of the beam are extended into the slab as shown below in fig 16.81:

Fig.16.81
Extending the stirrups into the slab

When such an arrangement is given, we must take care of the diameter of bars and the spacing requirements. The spacing of the main bars of the slab will be the same as the spacing of the stirrups. So the stirrups of the beam should be having a uniform spacing throughout it's length. And for our case, this spacing should not be greater than 250mm c/c, and the diameter should not be less than 10mm.

If the spacing of 10mm dia. stirrups required for the beam is less than 250mm c/c, then the spacing of bars in the slab will also become less than 250mm c/c. In that case check should be done to ensure that p_t is less than p_t,lim. Other wise, the slab will become over reinforced.

The elevation and sectional views are given below:

Fig.16.82
Sectional elevation

Fig.16.83
Section XX

Now we will discuss some more features of the above figs: In cantilever members, the main tensile steel is given at the top. So in our case, the main bars are given as the top most layer. The distributor bars are given as the second layer from the top. This arrangement will give maximum possible effective depth ‘d’ for the section. So while bending the bars for the stirrups of the beam at the site, care should be taken to see that the extension given to the stirrup will be at the exact required level in the slab. A bar bending schedule should be prepared to give the exact measurements.

As the slab has a thickness of 110mm, a bottom layer consisting of #8 bars at 250mm c/c should be given. This is indicated by the green bars in section XX. In this layer, alternate bars should be given two 90 degree bends at the end. This is for tying them with the top layer. This bottom layer is not shown in the sectional elevation in fig.16.82.

Transverse stairs cantilevering from both sides of a stringer beam:

The following fig. 16.84 shows the part view of a stair

Fig.16.84
Part view of a stair
 

The waist slab is cantilevering on both sides of the stringer beam. The load calculation procedure is same as that of the other types of transverse stairs that we saw before. So we can use the same Eq.16.25 for the calculation of the loads. From this eq., we will get w₁ , the load per 1m length, on a 1m wide strip. If the cantilevering length on both the sides are equal, we need to do the design of only one side as shown in the line diagram below:

Fig.16.85
Line diagram for stairs cantilevering on both sides of a stringer beam

The effective span l of the cantilever (cl.22.2.c) is the length of the cantilever up to the face of the support plus half the effective depth. For initial proportioning, we can assume the thickness of waist slab to be clear length of cantilever/5 . However, this should be finalized only after the complete design and various checks.

In the next section, we will see a solved example of the above type of stairs.

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