In the previous section we completed the analysis and design of a stair in which flights are supported on the landings. In this section we will discuss about a different kind of 'load distribution', that occurs in special type of longitudinal stair. Fig.16.43 below shows the external view of a building.
Fig.16.43
External stair of a building
A room is constructed as an 'extension' to the Main building. The stair climbs upto the roof of this extension room. This roof is at a lower level than the roof of the main building. From the top of this lower roof, another smaller stair is provided, which can be used to climb up to the roof of the main building. This smaller stair is built at the rear of the building, separately from the main stair, and is not seen in the view. We are concerned about the main stair in our present discussion. The section and plan of the stair are shown below:
Fig.16.44
Section of stair
Fig.16.45
Plan of Stair
We can see that the stair is supported on the ground at the bottom end, and on a beam at the top end. This is a longitudinal stair. The lower roof acts as the landing. The loads on the Going and the Landing are calculated below:
Four items constitute the load on the Going, and is given by Eq.16.9.These four items are:
1) Self wt. of waist slab on a horizontal 1m2 area (in kN/m2) =
= 5.991 kN/m2 .
2) Self wt of steps per 1m2 of horizontal area (in kN/m2) = (R γs) ⁄2 = 0.5Rγs
The value of γs for brick masonry is 20kN/m3
= 1.65 kN/m2
3) Self wt. of finishes per 1m2 of horizontal area (in kN/m2)
= 0.8 kN/m2 (obtained from code)
4) Live Load per 1m2 of horizontal area (in kN/m2)
= 5 kN/m2 (obtained from loading code)
Sum of above four = 13.441 kN/m2
So factored load = w1 = 13.441 x 1.5 = 20.16 kN/m2
Similarly, Three items constitute the load on the Landing, and is given by Eq.16.10. These three items are:
1) Self wt of steps per 1m2 of horizontal area (in kN/m2) = 25t (where ‘t’ is the thickness of the landing)
= 25 x 0.12 = 3.0 kN/m2
2) Self wt. of finishes per 1m2 of horizontal area (in kN/m2)
= 0.8 kN/m2 (same as for Going)
3) Live Load per 1m2 of horizontal area (in kN/m2)
= 5 kN/m2 (same as for Going)
Sum of above three = 8.8 kN/m2
So factored load = w2 = 8.8 x 1.5 = 13.2 kN/m2
Half of this load (0.5 x 13.2 = 6.6 kN/m2) can be assigned in the design of stairs because the full load will be used for the design of the landing slab. The line diagram of the stair is shown in the fig.16.46 below. Note that it is a continuous structure. This because of the intermediate support at B.
Fig.16.46
Line diagram for the stair
It can be seen that the above line diagram is incomplete. The length l2 is missing. We can easily obtain l2 from the architectural drawings. Once l2 is known, we can analyse the stair as a strip of a continuous slab, and the design can be done. But as mentioned earlier, the aim of our present discussion is to learn the details about a special kind of load distribution in longitudinal stairs. We are not doing the analysis and design of this stair. We are not concerned about the portion BC in our present discussion. In fact, we are going to focus our attention only on the portion from A to B, the ‘Going’ portion. We will discuss about it in the next section.
In
the previous chapters we saw the analysis and design of simply supported beams and slabs. In this chapter we will discuss about the analysis of continuous one-way slabs. When we complete this discussion about continuous one way slabs, we will be able to do the analysis of continuous beams also.
A video of a small scale experiment, showing the behaviour of continuous beams can be seen here
A 3D view of a continuous beam is given in fig.7.1 below. A view of a continuous slab can be seen in fig.7.2
Fig.7.1
View of a continuous beam
Before we proceed with the discussion on continuous members, we must know how to calculate the 'Effective span' of the various spans. A detailed discussion is given here.
We have already discussed about the effective span of simply supported members here.
To design a member like a slab or beam, we have to first find the bending moments (BM) and shear forces (SF) at the various points (supports, midpoints of spans etc.,) of the member. In the case of simply supported members, these can be found out easily using the three equations of equilibrium. But in the case of continuous members, we have to use one of the methods like Moment distribution method, Slope deflection method or kani's method.
Following solved example demonstrates the method of obtaining the BM and SF at various points of a continuous slab.
Solved example 7.1:
Consider a continuous one-way slab ABCD shown in the fig.7.2 below. It has a thickness of 200mm. It carries a Dead load of 1.5 kN/m2 in addition to it's self weight. Also there is a LL of 4 kN/m2. The effective spans of AB, BC and CD are 4.5, 4.0 and 4.2 m respectively. Our aim is to find the bending moments and shear forces at different points of the slab.
Fig.7.2
View of continuous slab ABCD
Solution:
Just as we saw in the case of a simply supported one-way slab, in the case of a continuous one-way slab also, we consider a strip of 1m width. This strip extends from one end support 'A' to the other end support 'D', as shown in the fig.7.3 below:
Fig.7.3
Strip of 1m width in a continuous slab
Now it has become a continuous beam of width 1m, and total depth D which is equal to the thickness of the slab. We know how to calculate the load per meter length of this beam: It is the load per meter area of the slab. (details here) So the next step is to calculate this load per unit area.
DL per unit area:
• Self weight = 25D = 25 x0.2 = 5.0 kN/m2
• Additional DL (given) = 1.5 kN/m2
• Total DL, wDL = 6.5 kN/m2
Thus wu,DL = 1.5 x6.5 = 9.75 kN/m2
• LL per unit area (given) wLL = 4.0 kN/m2
Thus wu,LL = 1.5 x4.0 = 6.0 kN/m2
We have computed DL and LL separately. The reason for this will be explained soon.
These loads per square area on the slab is same as the load per meter length of the 1m wide beam. So we can write:
• wu,DL per meter length of the beam = 9.75 kN/m
• wu,LL per meter length of the beam = 6.0 kN/m
With the above information, we can draw the line diagram for analysis. This is shown in the fig.7.4 below:
Fig.7.4
Line diagram for analysis
Based on the above fig.7.4, we can do the structural analysis of the beam ABCD. We can use methods like slope deflection method, moment distribution method, or kani's method. (The steps of the structural analysis are not shown here). Based on the results of the analysis, we can draw the BM and SF diagrams. These are shown in the figs.7.5 and 7.6 below.
Fig.7.5
Bending moment diagram
Fig.7.6
Shear force diagram
We can tabulate the values at different points as shown in the fig.7.7 below:
Fig.7.7
Tabulated results
So from the above table, we will get the required values of BM and SF at the various points. It may seem that we can now start the design process. But we are not there yet. The values in the above table, are not the 'maximum values'. Even though the whole beam ABCD has been loaded with all the maximum LL and DL, there are some 'possibilities' by which, the BM and SF at the various points along the beam will exceed the values shown in the table. Let us now examine such possibilities:
Consider a situation in which the first span AB is loaded with maximum possible DL and LL. But it's adjacent span BC and the span CD after that does not carry any LL. The line diagram of this situation is shown in the fig.7.8 below:
Fig.7.8
LL on span AB only
Note that in the above fig., the DL on spans BC and CD, are the 'unfactored' DL. This means they are the characteristic DL. The reason for this can be explained as follows: We want to know the result when 'full load' is applied on AB and 'reduced load' is applied on BC and CD. So we make a reduction on the loads on BC and CD. While making such a reduction, we must make a 'maximum possible reduction'. To achieve this, we must remove not only the LL, but also the 'Extra DL that can possibly occur'. So we avoid the application of the 'Load factor' to the DL, thus using only it's characteristic value. In short, we are using the 'lowest possible loads' on BC and CD.
Based on the above line diagram, we can do a structural analysis. The results are shown in the table in fig.7.9 below:
Fig.7.9
Tabulated results
We can see that the sagging moment at the midspan in AB is now 27.13. But in the fig.7.7, it was 24.95. So the sagging moment has increased due to the reduction in loads on other spans. In this way, the moments at other points also may increase depending on the pattern of loading. We must consider all possibilities. That is., we must consider all possible load combinations. We must do the structural analysis for each of these combinations, and then note down the maximum value at various points. All possible combinations are analysed, and the results are shown in tabular form in the fig.7.10 below:
Fig.7.10
Possible combinations and results
Now from the above fig., we have to take out the maximum values at each point along the length of the slab. This is shown in the fig.7.11 below:
Fig.7.11
Extreme values
In the above figs., '1' denotes the maximum BM at supports; '2' denotes the maximum BM at midspans, and '3' denotes the maximum SF at supports.
*2.a denotes the minimum BM at midspans. We have to take special care about minimum BM in BC because, for one particular combination (case 6), this mid span is bending upwards. This usually happens when a lightly loaded short span comes in between two heavily loaded long spans.
We can examine the various extreme values in the above fig. and make the following conclusion:
(a) Extreme values of the negative (hogging) BM at supports:
• -32.45 kNm for the first interior support 'B' is obtained from case 4 where LL is applied on the spans on either side of that support.
• -29.08 kNm for the second interior support 'C' is obtained from case 5 where LL is applied on the spans on either side of that support.
In general, the extreme values of the negative BM at a support occurs when the LL is applied on the spans on either side of that support and also on every alternate span thereafter.
'and also on every alternate span thereafter' in the above statement cannot be demonstrated in our slab because it has only three spans. On a beam or slab which has more spans, it can be shown as in fig.7.12 below:
Fig.7.12
Loading arrangement for maximum hogging moment at support B
(b) Extreme values of positive BM at center of spans:
• 28.44 kNm at the center of span AB is obtained from case 6 where the LL is applied on span AB itself and also on span CD which is an alternate span.
• 12.9 kNm at the center of span BC is obtained from case 2 where LL is applied on span BC only.
• 25.45 kNm at the center of span CD is obtained from case 6 where the LL is applied on span CD itself and also on span AB which is an alternate span.
In general, The maximum positive (sagging) moment in a span occurs when LL is applied on that span and every other alternate span. On a beam or slab which has more spans, it can be shown as in fig.7.13 below:
Fig.7.13
Loading arrangement for maximum sagging moment at midspan BC
(c)Minimum value of positive BM at center of spans:
• 6.8 kNm at the center of span AB is obtained from case2 where
1. LL is applied on the adjacent span which is BC and
2. No LL is applied on AB & CD.
• -7.7 kNm at the center of span BC is obtained from case6 where
1. LL is applied on the adjacent spans which are AB and CD
2. No LL is applied on BC.
• 5.33 kNm at the center of span CD is obtained from case2 where
1. LL is applied on the adjacent span which is BC and
2. No LL is applied on AB & CD.
In general, The minimum positive moment in a span occurs when
1. No LL is applied on that span but
2. LL is applied on the adjoining spans and every alternate span thereafter.
On a beam or slab which has more spans, it can be shown as in fig.7.13 below:
Fig.7.13
Loading arrangement for minimum sagging moment at midspan BC
It should be noted that this minimum positive moment may turn out to be the maximum negative moment at the midspan. That is, the beam will be bending upwards. This has occurred in case 6 above where a lightly loaded short span is placed in between two heavily loaded long spans
(d) Extreme values of total shear force at supports:
• 30.36 kN for the first end support 'A' is obtained from case6 where:
1. LL is applied on the span AB of which this support is a part and
2. LL is applied on the alternate span CD
• 78.56 kN (42.65 from the outer side plus 35.91 from the inner side) for the first interior support 'B' is obtained from case4 where LL is applied on the spans on either side of that support.
• 74.6 kN (40.0 from the outer side plus 34.6 from the inner side) for the second interior support 'C' is obtained from case5 where LL is applied on the spans on either side of that support.
• 28.66 kN for the last end support 'D' is obtained from case6 where
1. LL is applied on the span CD of which this support is a part and
2. LL is applied on the alternate span.
In general, the extreme values of the shear force at a support occurs when the LL is applied on the spans on either side of that support and also on every alternate span thereafter. This is the same condition for obtaining the extreme values of negative BM at supports (fig.7.12).
So we obtained some general conditions for the extreme values of BM and SF. We did this using an example. But these general conditions can be obtained by using the principles of 'Influence lines' also.
In the next section we will discuss a simplified procedure given by the code.
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In
the previous section we completed the analysis of one way slabs. Now we will see the design process. As in the case of beams, here also, we will first see the procedure for fixing up the 'preliminary dimensions'. We have seen that for analysis and design, the slabs are considered as strips, and we can take any one strip. As we are dealing with 1m wide strips, the width of the beams that we have to design is fixed at 1000mm. We have to fix up only the preliminary depth. Let us see how this is done:
We know that the depth provided to a beam must be sufficient for it to resist the bending moment acting on it. In addition to this, the depth must be sufficient to control the deflection also. So the depth depends on two criteria:
• To resist the bending moment, and
• To control deflection.
In the case of slabs, where the depth is relatively low, the criterion for deflection control becomes more critical. That is., a slab with a certain depth may effectively resist the bending moment acting on it. But this same depth may not be sufficient to control the deflection. In other words, the slab generally requires more depth to control it's deflection than it requires to resist bending.
So to fix up the preliminary depth, we must use the deflection control criterion. Let us take the case of a simply supported slab (whose effective span is less than 10m). We have already discussed the procedure for deflection control here. The effective depth that the slab actually have in the final structure should satisfy the following relation:
(l/d)actual ≤ [(l/d)basic] α kt - - - (1)
• For simply supported spans, (l/d)basic = 20
• As we are considering spans less than 10m, α need not be taken into account.
Thus we get :
(l/d)actual ≤ 20 x kt - - - (2)
From this expression, we want to calculate 'd'. But kt is also an unknown. So we assume a value for kt. The assumed value should be as accurate as possible. For this we look at the numerous design examples of simply supported slabs that are done in the past by different designers. By examining them, we can find that the percentage of reinforcement pt generally falls in the range 0.4–0.5.
Sample calculation:
Let a slab have the following properties:
D =150mm; d =115mm; Ф =10mm; s =150mm
Then from Eq.5.2,
= 490.87mm2. So pt = 100Ast /bd = [490.87 x100]/[1000 x115] = 0.427
Based on this information, we adopt the following procedure:
• Assume that the area of steel provided is equal to the area of steel required,
• Then fst will be equal to 0.58fy.
Take the value of fy that is going to be used. If it is 415, then fst = 240.7
So we have the values required to find kt: They are:
(a) pt in the range 0.4–0.5, and
(b) fst = 240.7
Corresponding to these values, we will get kt = 1.25 from Fig.4 of the code.
So substituting in (2) we get
6.1: (l/d)actual ≤ 20 x 1.25 ⇒ (l/d)actual ≤ 25. So d that we actually provide in the final slab should be greater than or equal to l/25. This is for simply supported slabs.
For continuous slabs, we can write:
6.2: (l/d)actual ≤ 26 x 1.25 ⇒ (l/d)actual ≤ 32.5. So d that we actually provide in the final slab should be greater than or equal to l/32.5.≈ l/32
From the above expressions, what we get is the effective depth d. To this we must add Cc and half of Ф to get the total depth D. Ф can be assumed to be equal to 10mm.
As in the case of beams, we must do all the design checks to ensure that the final section satisfies all requirements. If the section fails to do so, the whole process should be repeated with improved dimensions.
So we have fixed the preliminary dimensions. The next step is to find the effective depth required from bending moment considerations. Once this is calculated, we must compare it with the d that we have from the preliminary depth. If the required value is less, we can proceed to find the steel required to resist this bending moment. These two steps are same as that for a beam.
Now we must know the 'rules' for distributing the calculated steel into the slab.
Concrete cover and grade of concrete:
The details about the concrete cover that has to be provided to the bars of a slab are same as those of a beam. They were discussed in the section about the design of beams, and we saw table 4.1.
So, if we are to design a slab which will be subjected to 'moderate' exposure conditions, from table 4.1,we get the value of Cc to be provided as 30 mm , and the minimum grade of concrete to be used as M25
Minimum spacing to be provided between the bars of a slab:
The clear space provided between parallel reinforcing bars should not be less than the minimum value specified in cl 26.3.2 of the code. We have seen the details of this clause when we discussed about the minimum spacing between bars of beams here. The same is applicable to slabs also. So we can show the application of this clause to slabs as shown in the fig.6.1 below:
Fig.6.1
Minimum spacing between bars of slab
Maximum spacing allowable between bars of a slab:
When we design the slab we will get the area of steel that should be provided. Sometimes, when we convert it into 'spacing of bars of a particular diameter', we may get a large spacing s. This often happens when the bending moment that the slab has to resist, is low. Large values of s may be obtained on other occasions also: When we choose to provide large diameter bars. This can be explained as follows: From 5.3 we know that
In the above equation, for the same value of Ast, we will get a larger s for a larger Ф.
It may seem to be economical to provide bars at large spacing. But bars at large spacing will not be able to control cracks effectively. Also the bond between steel and concrete will become lesser when bars are at a large distance apart.
So the code specifies some upper limits to the spacing. The c/c spacing provided between parallel reinforcing bars of a slab should not be greater than that specified in cl 26.3.3 (b) of the code. The requirements stated in this clause can be shown as in fig 6.2 given below:
Fig.6.2
Maximum spacing between bars of slab
In the above fig., it is indicated that the diameter of the main bars should not exceed D/8. This is because 'small diameter bars spaced closer together' is effective in reducing cracks and in improving bond than 'large diameter bars spaced at larger distances'. However, bars smaller than 8mm in diameter are generally not used as main bars of slabs.
Minimum Area of reinforcement for slabs:
The area of reinforcement provided in slabs should not be less than that specified in cl 26.5.2 of the code. The requirements stated in this clause can be written as follows:
6.3: For Fe 250 steel:
Ast ≮ 0.0015Ag
6.4: For Fe 415 steel:
Ast ≮ 0.0012Ag
Where Ast is the area of steel provided in the slab, and Ag is the gross area of cross section of the slab. Ag is equal to bD where b is the width of the slab and D is the total depth of the slab. When we consider a strip of slab for the design, b is equal to the width of the strip.
Transverse reinforcement for one way slabs :
We have seen in the previous chapter that transverse reinforcement is required for one-way slabs. These bars are also called distribution bars of a One-way slab. The quantity of this reinforcement that we have to provide, is given by the code, and is same as that given by 6.3 and 6.4 above.
The c/c spacing between the distribution bars should not exceed the smallest of the following (cl 26.3.3.b (2)):
• 5 times the effective depth d of the slab
• 450 mm
This is shown in fig. 6.1 above
Deflection control for one-way slabs
We have seen some details about deflection control checks in one way slabs when we discussed about the 'preliminary depth' at the beginning of this chapter. There we assumed a value for pt. But now we are discussing the actual 'check'. This is the 'last but one check' that we have to perform after the design of a one way slab (The last one being the check for MuR). So here we will be using the actual pt that is provided. The procedure is same as that for a singly reinforced beam. So we can write the same expressions that we wrote for the beam:
4.19
For singly reinforced rectangular slabs with span less than 10m,
(l/d)actual ≤ [(l/d)basic] kt
4.20
For singly reinforced rectangular slabs with span greater than 10m,
(l/d)actual ≤ [(l/d)basic] α kt
Partial fixity at supports:
Another topic that we must consider, is the design at supports. If the slab is simply supported, the bending moment will be zero at the supports. The details to be considered in such a situation is demonstrated in the following presentation:
(The first five slides are same as the ones we saw in an earlier presentation about the behaviour of slabs in chapter 5)
This completes the discussion about the design of a simply supported one way slab. We will now see a solved example which demonstrates the actual process of design.
Solved example 6.1
In the above solved example, we have determined the reinforcement requirements of a simply supported one way slab. Based on that design we must draw a plan and a sectional view, showing the layout of bars. This is given in the fig.6.3 and 6.4 below:
Fig.6.3
Plan view showing reinforcement details of slab
Fig.6.4
Sectional elevation
In the plan view, we can see that alternate bars are bent up at supports. The reason for this can be explained as follows: We have designed the slab as 'simply supported'. So the bending moments at the support will be equal to zero. But if in the future, partial fixity (shown in the presentation above) is introduced at any of the supports, bending moments will develop at that support. These bending moments will be hogging in nature. So tension will develop near the top surface of the slab at such supports. Thus we have to provide top steel. This can be achieved by bending up the bars. The requirements for this top reinforcement is given by cl.D-1.6 of the code. According to this clause, the bars must extend a distance of 0.1l from the support into the slab. Also, these bars should have an area equal to 50% of that provided at midspan. Thus, when we bend up alternate bars, we will get 50% at the top.
Technically, we need to show only two bars of a particular set. But if we show more bars, the symmetry and pattern will become more clear. Such a plan view is given here.
In the sectional view in fig.6.4 above, the two bars are shown separately only for clarity. The two bars are in fact provided in the same layer as indicated by the '0'mm dimension.
This completes the design and detailing of a 'simply supported' one way slab. Next we will see 'continuous' beams and one way slabs. But to analyse continuous beams and slabs, we must know how to calculate the 'effective spans' in continuous members. So in the next chapter we will see the details about this effective span also.
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= 5.991 kN/m2 .
