Chapter 16.9 - Analysis and design of Transverse stairs

In the previous section we saw the arrangement of the stair in plan and sectional elevation views. We calculated the loads also. It appears as an ordinary longitudinal stair. In this section we will see how the load distribution is modified when the flight is given an embedment into the side wall. Later in this section, we will begin our discussion about 'transverse stairs'.

From the figs. in the previous section, we can see that the load on the Going is transferred to the ground at the bottom end and to the beams at the top end. But if the waist slab of the Going have some embedment into the side wall of the main building, then some load will be transferred to that wall also. Cl.33.2 of the code gives us the details about the load distribution when such an embedment is provided. For this clause to be applicable, the embedment that is provided, should not be less than 11 cm. Consider the modified plan of the stair given below:

Fig.16.47
Modified plan

The waist slab has an embedment of 12 cm into the wall of the main building. As this is greater than 11cm, we can apply the clause to the stair. The section xx is given in fig.16.48 below. This sectional view helps to calculate the loads based on the clause.

Fig.16.48
Section XX

We know that the load is acting on the whole width 'W' cm of the stair (W =90cm for our stair). But when a minimum embedment of 11cm is provided, the load is assumed to act on a reduced width of (W – 15cm). This is shown in the fig.16.48 above. 

Let us see how we can apply this to our stair: 
We have calculated a load of 20.16 kN/m² on the Going. If there is no embedment, the total load on the whole area of the Going will be equal to 20.16 multiplied by the total area of the Going. This is equal to 20.16 x 0.9 x 4.0 = 72.576 kN. (here 4.0 is the effective span of the Going)

As there is an embedment of 12 cm in our case, the total load can be modified as 20.16 x (0.9 – 0.15) x 4.0 = 60.48 kN.  [Where (0.9 – 0.15) is the reduced width]. So we can discard 72.576 and use 60.48 kN. 

Now, one more modification have to be made:
When a normal longitudinal stair bends, the width that resists the bending is W. But when the embedment mentioned above is available, the effective width of the stair, that resists the bending can be assumed to have an increased value. According to the code, this increased value is (W + 0.075). This is also shown in the fig.16.48 above. 

So the load is acting on an area of [(W +0.075) x effective span ]. Thus, dividing the total load by this new area will give us the new load per 1 x 1 m square area. In our case, this is equal to 60.48 / [(0.9 + 0.075) x 4.0] = 15.51 kN/m². So for the analysis and design, we can use a UDL of 15.51 kN/m instead of 20.16 kN/m. 

The difference is transferred to the side wall. In our case, the wall is resting on the ground. So there is foundation to take up this extra load coming on the wall. But if the wall is resting on a beam, then that beam will have to be designed for an additional UDL per meter length. In our case, this additional load = 20.16 - 15.51 = 4.65 kN/m. [The landing portion of our stair does not have any such embedment. So no modification needs to be applied to the load on the landing].

It should also be noted that, this method of giving an embedment in the side wall for a longitudinal stair causes construction difficulties, and so, it is not generally used in practice.

Transverse Stairs

We have already seen a schematic diagram of a transverse stair here. Now we will discuss the detailed analysis and design procedure. The fig.16.49 below shows the view of an office building. It has a Basement floor, Ground floor and a First floor. The floor level of the Ground floor is at some height above the surrounding Ground surface. So a stair is provided for the access to the ground floor. Also there is open space between the basement of the building and the surrounding earth. So the stair acts as a 'bridge' also.

Fig.16.49
Example of a Transverse stair

The next fig.16.50 below shows the details of the stair. A portion of the earth filling below the Ground level is removed to get a clear picture. A portion of the stair is also removed to see the waist slab.

Fig.16.50
Detailed view of Transverse stair

We can see that the brick work of the steps is resting on the waist slab. The waist slab is resting on two beams on the sides of the stair. These beams are called stringer beams. Each of the stringer beams are resting on columns: 
• Column in the retaining wall support the lower end of the stringer beam
• Column in the main building support the upper end of the stringer beam
• The waist slab spans between the two stringer beams. 
So the waist slab spans in a direction, which is perpendicular to the direction of travel of the pedestrians. In other words, the span is in the direction of the transverse axis of the stair. So this is a transverse stair. 

The slab is simply supported on the two beams, and the span can be taken as the center to center distance between the two beams. [Though it is assumed to be simply supported, some top steel should be provided at supports to resist any hogging moments that may develop]. We have already discussed the design of beams earlier. So, in our present discussion, we are concerned with the analysis and design of the waist slab only.

It may be noted that, in the above views, finishing works are not yet applied on the stairs, while the main building is shown in a finished condition. So the brick works for the steps have to be done in such a way that, when the tile is laid on the top most step, it's top surface will be in the same level as the top surface of the tile of the floor of the main building. In the next section, we will see more details about this stair.

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Chapter 16.8 - Stairs built into side walls

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.

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Chapter 16.7 - Load transfer from Flights to Landings

In the previous section we saw the loads on the flight of a stair which is supported on landings. In this section we will see the loads on the landings.

Part 2: Loads on the Landings:
In part 1, we have calculated w₂. We have seen that it is calculated by the same method that we would use for calculating the w₂ of an ordinary longitudinal stair. (See Eq.16.10). This is the original load on the landing. It is obtained on a 1 x 1 m square area. 

There is more. w₂ calculated above is the original load on the landing. But the landing is supporting the flight also. So the load from the 'Going' will also come on to the landing. Let us see how this load from the Going is distributed on the landing.

We have the load w₁ on the Going. It is the load per 1 x 1 m sq. area (horizontal projection). So, once we calculate w₁, we will be able to calculate the total load on the going. From the fig.16.41 below, this will be equal to w₁ x F_w x l₁. 

Fig.16.41
Transfer of loads from the Going to the Landing

Where F_w is the width of the flight and l₁ is the clear length of the Going. This is the total load acting in the total length of the Going. Half of this will be transferred to one landing and the other half to the other landing. So in the fig.16.41, the load from pqrs will be transferred to the landing 1. Similarly the load from qruy will be transferred to the landing 2.

So now we know the magnitude of the load which is transferred from the Going to the landing. Next we must know how this much load is distributed on the landing. This is also shown in the fig 16.41. The load from pqrs is assumed to be distributed uniformly on the corresponding area pmns of the landing. As the load is uniformly distributed, we can divide the load by the area to get the load per 1 x 1 m sq.

• Load on pqrs = 0.5w₁F_w l₁ 
• Area of  mnsp = F_w L_w .(Where Lw is the width of the landing). 
So load per unit area = 

This is the load on one half (mnsp) of the landing. The other half will also be subjected to the same load because it is supporting the other flight. Thus the whole landing is carrying the load 0.5w₁l₁ / L_w. 

So we have obtained the two loads: 
• The original load on the landing = [w2] per sq.mt
• The load coming from the Going = [0.5w₁l₁ / L_w] per sq.mt 

The sum of the two will be the total load acting on the landing per sq.mt. Thus:
Eq.16.19: The total load acting on the landing per sq.mt =

What role does this 'total load' play in the analysis and design of the landing slab? Let us find out:
The landing slab is designed as a One-way slab. It spans between the hatched portions shown in fig.16.41. We have already seen the design of one-way slabs here. We have seen that the load on a 1 x 1m square area of the slab is the same UDL per meter length, that is required for the design of a 1 m wide strip. So we conclude that, the above 'total load' is the UDL per meter length that we require to design a 1m wide strip of our landing slab.

So we have the methods to obtain all the loads on the stair. With these loads, we can do the analysis to find the Bending moments and shear forces. Once they are calculated, we can do the design.
  
We will now do the  analysis and design of our stair. It is given as:

Solved example 16.2

The arrangement of bars according to the design in the above solved example is shown in the figs.16.42 given below:

Fig.16.42
Reinforcement details of Going and Landing

Now we will discuss some details about the above fig.
As in previous examples, bar ‘a’ becomes top bar in the top landing, and to compensate for this, bar ‘b’ is provided at the same diameter and spacing as ‘a’. Also, ‘a’ and ‘b’ are given enough embedment as explained earlier on the basis of fig.16.25. At the bottom intermediate landing, ‘c’ and ‘d’ are provided to resist any possible hogging moment. When there is hogging moment, these bars will be in tension, and so in order to prevent them from straightening up, they are given enough embedment in accordance with fig.16.25.

One more type of bar, named as ‘e’ is provided at the top landing. The reason for providing these bars can be explained as follows: The stair is supported on the landings. The support for the flight comes from the interior portion of the Landing. But the inner edge of the Landing (the line of intersection between horizontal landing portion and inclined flight portion) is also capable of providing some support. So the edge will act as an intermediate support of a continuous system. We know that hogging moments will develop at intermediate supports, and so top steel will have to be provided. Thus ‘e’ is provided to resist this hogging moment.

Another point that we have to note is the position of the main bars of the Landings. We can see that it is provided as the bottom most layer. This will give maximum effective depth, and so the ‘Moment of resistance’ MuR of the section of the Landing will be more. But in any case , these bars should indeed be the bottom most layer because, they have to provide support to the stair by carrying the bars(‘a’, ‘b’, ‘c’, ‘d’ and ‘e’) of the stairs. And in order to carry them, the bars of the Landing should be the bottom most layer.

Distributor bars for the Landing need not be designed because the bars of the stairs are coming in a transverse direction to the main bars of the Landing.

A small amount of steel (#8 @ 200 c/c) is also provided as top bars for the Landings. These bars will resist any hogging moment developed at the supports (the hatched portions in fig.16.41 above) of the Landings, and will also assist in tying the top bars of stairs, and thus to keep them in position.
It may also be noted that in the above fig., bars of different types are shown separately only for clarity. This was explained earlier based on the fig.16.25(b).

In the next section, we will discuss the load distribution in a particular type of longitudinal stairs.

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Chapter 16.6 - Stairs supported on Landings

Towards the end of the previous section we saw that the supports provided to the flights of the stair is unclear from the plan and sectional views. To get a clear picture of how the flights are supported, we will now see some 3D views of the building. The direction of views can be understood from the 'North direction' which is indicated in the key plan given in the previous section.

Fig.16.36
View from South East

Fig.16.37
View from North East

Fig.16.38
View from North West

We can see that, two brick walls are built, one on Bm1 and the other on Bm5. The brick wall over Bm5 can be seen in the sectional view that we saw earlier in fig.16.35. These two walls are parallel. The intermediate landing is supported by these two walls. That is., the intermediate landing is a one-way slab, spanning between these two walls. So at B, the flight AB can rest on this landing. In the same way, at C, the flight CD can also rest on this landing. Thus there is no need to take the flights up to Bm 3 at the rear of the building.

What about the supports at A and D? We can see that at A, there is a landing (at Floor1 level) in between Bm1 and Bm5. This landing is designed as a one-way slab spanning between these two beams. Then, the flight AB is supported on this landing at A. 
There is an exact duplicate of this landing, directly above it (at Floor2 level). It spans between Bm1 and Bm5. Flight CD is supported at D on this landing.  Thus there is no need to take the flights up to Bm 4 in the front of the building.

So we can summarize the support details as follows:
Flight AB is supported between two landings. One landing at A and another landing at B
The landing at A is at level 1. It is a one way slab spanning between Bm1 and Bm5
The landing at B is the intermediate landing of the stair. It is a one way slab spanning between walls built over Bm1 and Bm5
Flight CD is supported between two landings. One landing at C and another landing at D
The landing at C is the same intermediate landing on which end B of the first flight AB is supported
The landing at D is at level 2 It is a one way slab spanning between Bm1 and Bm5. This landing is an exact duplicate of the landing at A of the first flight.

---

It must be noted that in this type of stairs, special care should be taken while fixing up the arrangement and position of bars of both the sloping portions and the landings. We will see the details about it when we design the reinforcements of this type of stairs.

---

From the above discussion, it is clear how flights are supported on landings. The whole width of the landing slabs (1.2m in our case) is taking part in supporting the flights. But when we calculate the effective span of the flight, we can consider only a certain portion of the width.  To determine this width, we will use the diagram shown below:

Fig.16.39
Portions of landings that can be considered for calculating Effective span

The width of the first landing is 2X and that of the second landing is 2Y. (Here X and Y are in m). Such a notation using 'twice X and Y' is only for the ease of presenting the formula. The widths need not be exact multiples of 2. It only implies that, whatever be the widths of the landings, half of those widths should be used for the calculations. (The hatched portions shows the support available for the landing slabs. These portions are not of any significance in our present discussion about effective span). The steps involved in the calculations are as shown below:
Step 1:
Determine X, half the width of the first landing.
Then, a certain length (denoted as 'Quantity 1') is determined as follows:
IF X is less than 1 m, THEN, Quantity 1 = X 
IF X is greater than 1 m, THEN, Quantity 1 = 1 m
This simply means that Quantity 1 = the lesser of {X ; 1 m}. It can be presented in a graphical form as shown below:

Step 2:
Determine G, the Going.
Let Quantity 2 = G
Step 3:
Determine Y, half the width of the second landing.
Then, a certain length (denoted as 'Quantity 3') is determined as follows:
IF Y is less than 1 m, THEN, Quantity 3 = Y 
IF Y is greater than 1 m, THEN, Quantity 3 = 1 m
This simply means that Quantity 3 = the lesser of {Y ; 1m} 
Step 4:
This is the final step in which we just add the above three lengths. The sum thus obtained is the effective span. Thus:
Eq.16.17
Effective span = Quantity 1 + Quantity 2 + Quantity 3
From the method of calculation of Quantity 1 and Quantity 3, we can say that, 
only half the width of a landing can become part of the effective span. 
Also, if this half width is greater than 1m, then, only 1m can become part of the effective span.

---

If we analyse the above calculations, we can obtain one more inference:
Eq.16.18
If width of each of the two landings is less than 2m, then, we can calculate the effective span directly as: 
Effective span = c/c distance between the landings.

---

So we can add these details to the above fig.16.39, and modify it as shown below:Fig.16.40
Effective span of flights supported on Landing slabs

Load calculations
The design of this type of stair has two parts. 
In the first part, we design the flight which has an effective span obtained using 16.17. 
But this flight has to be provided with adequate support on the landing slabs. So the design of landings is done as the second part. 

So we will do the calculations of the loads also in two parts. In part 1, we will calculate the loads on the flight, and in part 2, we will calculate the loads on the landing.
Part 1: Loads on the flight.
From the fig.16.40 above, we can see that when we determine the effective span of the flight, we are getting a line diagram with two landings of lengths l2 and l3 , and a Going with length l1 . This is same as the line diagram (fig.16.21) of an ordinary longitudinal stair supported on walls or beams, that we saw at the beginning of the chapter. 

So the methods for load calculations are also the same: 
• The sloping portion has 4 load components which make up w₁ (Eq.16.9), 
• and the landing portions have 3 load components which make up w₂ (Eq.16.10). 
Thus we can easily compute w₁ on the sloping portion and w₂ on the landing portions. [It may be noted that even if the flight is not symmetrical (the widths of landings not being equal), w₂ will be the same on both the landings, as w₂ is the load per unit area. But for this, both the landings should have the same thickness.]

---

After computing w₁ and w₂, a modification has to be applied to w₂. Recall that a modification to w₂ was made in the case of 'stairs at right angles' which we saw in the previous section. Here also the modification is required because, w₂ is resisted by both landing and the flight. But there is an important difference. It can be explained on the basis of the following fig:

Fig(a) shows the landing of a right angled stair and Fig(b) shows the flights supported on landings. Consider the upper flight in (b). 
• The load w₂ on the landing will contribute towards the bending of this flight. 
• The load w₂ will cause the bending of the landing also.
• So some portion of w₂ is resisted by the flight and the remaining portion is resisted by the landing. 
• But the portion resisted by the flight will 'come back on the landing' because, the landing itself is a support for the flight. 
• This is different from the situation in fig(a), in which the flight is not supported on the landing, but on a separate wall or beam.

■ As the 'portion of w₂ resisted by the flight', comes back to the landing, no reduction should be applied to w₂ while analysing the landing.
■ A reduction of 0.5 can be applied to w₂ while analysing the flight because, the flight will not be resisting a full w₂. 

---

With this modification applied, part 1, the load calculation on flight is complete. When those loads are calculated, we can find the moments and reactions. If the flight is symmetrical, we can use the same equations Eq.16.15 and 16.16 (which we derived for simple longitudinal stairs) to find the reactions and the moments. In the next section, we will see part 2, the load calculation on landing.

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Chapter 16.5 - Right angled stairs

In the previous section we discussed the special precautions to be taken in the arrangement of reinforcements at Landings. In this section we continue the discussion on stairs in general. We will now discuss about a different type of stair.

Right angled stairs
The stair shown in fig.16.30 below has two flights. The second flight is deviating from the first flight at an angle of 90^o.

Fig.16.30
Stair taking a right angled turn

The first flight AB is supported on the ground at A, and on the wall-2 at B. The second flight CD is supported on wall-1 at C and on a beam at D. [At the time of construction, the walls 1 and 2 are stopped at the level of the bottom surface of the intermediate landing, so that the landing can be extended into the walls. After the casting and curing of landing and flight AB, the wall construction is resumed. The bars for CD are extended from the landing because CD can be casted only after completing wall-2, as it rests on the beam at D. The safety and stability of the beam at D should be carefully checked.] The bearing of the landing on the two walls can be seen in the plan view given below:

Fig.16.31
Plan view of right angled stair

For analysis, we have to draw the line diagram of each of the flights separately. The line diagrams are drawn from support to support. That is., from A to B, and from C to D. This is shown in figs.16.32 and 16.33 below.

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We have to make a modification to the loads on the intermediate landing. It can be explained as follows: • We are designing the flights separately. 
• When we assign loads to each flights separately, the loads coming on the intermediate landing will be assigned to both the flights.
• If we assign the full landing load w₂ to one flight, it will mean that, the full load w₂ acting on the landing is resisted by that flight only. It will also mean that the other flight does not have to resist w₂
• But we know that both flights will take part in resisting w₂. In other words, w₂ contributes towards the bending of both the flights.
• So if we assign the full w₂ to any one stair, there will be under estimation of the load on the other stair
• On the other hand, if we assign full w₂ to both the flights, there will be over estimation. This is because, neither of the flights have to resist the full w₂.
• So we give a part of w₂ for one flight, and the remaining part for the other flight. In usual design practice, this 'part' is exact 50%. That is., 0.5w₂ is given to one flight and 0.5w₂ is given to the other flight.
• Thus, we first calculate w₂ on the landing as usual, and then assign 0.5w₂ for the landing common to each flight. This is shown in the line diagrams below.

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Fig.16.32Line diagram for flight AB

Fig.16.33
Line diagram for flight CD

It may be noted that the modification factor of 0.5 is not applied to w2 on the topmost landing. This is because, it is not a common landing, and so, duplication does not occur. Also note that, the last step of flight AB should be of reinforced concrete as shown in fig.16.32. This is to ensure the smooth continuation of the main bars of flight CD into wall-1.
Once the line diagrams are prepared, we can do the analysis to determine the maximum bending moments, and then do the design as usual.

Stairs supported on landings
In the cases that we saw so far, the flights were supported on walls or beams. Now we will consider a type of longitudinal stair in which the the flights are supported by 'landing slabs'. To fully understand such a support, we have to get more details of the building. So let us start from the 'Key plan'. Fig.16.34 below shows a part of the key plan of a two storey building. The plan of the stairs in Floor 2 is shown within the key plan.

Fig.16.34
Portion of the key plan showing the Staircase portion

The sectional view is shown in the fig.16.35 below:

Fig.16.35
Sectional elevation

The beams and columns in the above figs. are named according to the specifications given in SP34. For example, COL 2 Q1 indicates the column Q1 (the column at the North East corner of the building) in Floor 2.

We can think about the supports of the stair in this way:
Flight AB can be supported at A on Bm 4 (of Floor1). At B, it can be supported on a brick wall built above Bm 3 (parallel to the Risers). Flight CD can be supported at C on this brick wall. At D, it can be supported on Bm 4 (of Floor 2) So the flights will be spanning between these beams and the brick wall on Bm 3. That is.,
• Flight AB spans between Bm4 and wall above Bm3 [Both Bm4 and Bm3 are at level 1]
• Flight CD spans between wall above Bm3 and Bm4 [Bm3 is at level 1 and Bm4 is at level 2]
But such a brick wall above Bm3 is not shown in the sectional view. Also, at A and D, the the flights are not shown to have any bearing into the beams Bm4 on either levels. So how are the flights being supported? For the answer, we will look at the views given in the next section.

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Chapter 16.4 - Bars for landings in stairs

In the previous section we designed a stair and saw it's reinforcement details. In this section we will discuss more details about the arrangement of bars.

In the fig.16.22 which shows flight AB, the bottom layer bar (bar type 'a') in the sloping portion becomes the top layer in the landing portion. For the bottom layer of landing, extra bars are given. Similar arrangement can be seen in flight CD (fig.16.23) also. We will now discuss the reason for giving such an arrangement. Consider fig. 16.24 below:

Fig.16.24
Stair bars without embedment

In the fig.16.24, the bottom bar from the sloping portion continues into the landing in such a way that it is the bottom bar in the landing also. When the loads are applied on the slab, the bar will be in tension, and it will try to straighten up. Only the concrete cover is present there to resist this tendency of the bar to straighten up. This concrete cover does not have enough thickness to adequately resist this tendency, and cracks may develop. So we must extend this bar to embed it into a 'mass of concrete'. To achieve this embedment, the bar is taken up to near the top surface of the landing, and then a bend is given to make it horizontal. The measurements required for this embedment is shown in the fig.16.25 below:

Fig.16.25
Length of embedment required by the bars

We can see that, when the type 'a' bars become the top bars of the landing, the bottom portion of the landing is left with out any bars. So some extra bars (denoted as bar type 'b') are given at the bottom layer of the landing. These bars should have the same diameter and spacing as 'a' type bars. The 'b' bars also should have sufficient embedment. So they are taken upto near the top surface of the sloping slab, and then given a bend, to make them parallel to the slope. 

The point of intersection of 'a' and 'b' is taken as the 'critical' point. The specified embedment should be measured from this point. In the fig., the length required is specified as 'Ld(min)'. Why is it specially mentioned as 'min'? The explanation is as follows: The bars we are considering are 'top bars'. Their main purpose is to resist the hogging moment (that can possibly arise even at a simple support due to partial fixity). So they must have the specified length which is more than the length over which the hogging moment can possibly act. Thus we have two lengths to consider:
• The length required for resisting the hogging moment. Which is taken as 0.25l for general cases
• The length required for the embedment to prevent straightening up. Which is Ld
The largest of the above two lengths should be used. This will satisfy both the requirements. So the mention of 'min' tells us to take both criteria into consideration. It may be noted that in Limit state design, Ld is the unique value that we saw in a previous chapter. Also, Ld  should be provided on both sides of the critical point.

Now we consider the landing portion at ‘C’ (the intermediate landing) for the flight CD in fig.16.23. Here the bar type ‘a’ will not try to straighten up. So it does not require any extra embedment. So these bars continue as bottom bars into the landing. But at the support at this intermediate landing, hogging moments can occur if a wall is constructed above the landing (causing partial fixity), as shown in the fig. below:

Fig.16.25 (a)
Hogging moment at support

We can give top bars in the landing which will continue as top bar in the sloping portion also. But when the hogging moment occurs, these bars will be in tension, and will try to straighten up. So we must give two sets of bars (types ‘c’ and ‘d’) as shown in fig.16.25(a) above.
It must be noted that in fig.16.23, the different sets of bars are shown in separate layers only for clarity. In the actual structure, they will be in same layer as shown in the fig. below:

Fig.16.25(b)
Types of bars in same layers

---

Stairs with overhanging Landings
Now, we can discuss the arrangement of bars in another type of longitudinal stairs. In this type, the supports are at the ends of the sloping slab as shown in the fig.16.26 below:

Fig.16.26
Supports at the ends of sloping slab

From the fig., we can see that the supports are at the ends of the sloping slabs. The intermediate landing and the top landing are overhanging beyond the supports. In other words, the landings are cantilevers. This type of stairs are more economical. Let us see how this economy is achieved: As shown earlier, the thickness of the waist slab is taken as 1/20 of the effective span. In the fig. above, the effective span is reduced because, the supports are now nearer to each other. So the waist slab thickness can be reduced. The bending moment will also be reduced because of the reduced effective span.

A cantilever structure will produce more bending moment than a simply supported structure. So at a glance, we may feel that more steel will be required for resisting the bending moment from the cantilevers. But here, the cantilevering span is small when compared to the simply supported span between the supports. Also the load on the cantilever landings is less than the load on the sloping portion.
However, it is important to note that hogging moments will be produced at the supports because of the cantilever action. This is shown in the figs. below:

Fig.16.27
Bending moment diagram for flight AB

Fig.16.27
Bending moment diagram for flight CD

From the above figs., it is clear that hogging moments will be present at the supports in this type of stairs. The values of maximum hogging moments and sagging moments can be easily calculated from basic principles. Then we can determine the steel required to resist these moments. The steel required to resist the hogging moments should be given as top steel at the supports. The arrangement of bars for this type of stairs is shown in the figs. below:

Fig.16.28
Arrangement of bars for flight AB

Fig.16.29
Arrangement of bars for flight CD

In the above figs., a new type of bar denoted as 'd' is given as top bars at all supports where overhang is present. These are the bars which resist the hogging moment. They must have sufficient embedment as indicated by 'y' in the figs. Also note that these bars are taken to the farther face of the sloping slab, and then bent to make them parallel to the slope. This is to give maximum embedment inside concrete.

In the next section, we will discuss about another type of longitudinal stairs in which, the second flight takes a right angled turn from the first flight.

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Chapter 16.3 - Reinforcement details of staircase

In the previous section we saw the self wt. of stair slab and the steps. Now we will see the other loads coming on the stairs.

Self weight of finishes (16.6)
The self weight of finishes applied over the steps should be considered in the design. It can be obtained from data books or relevant codes. Usually it varies from 0.5 to 1.0 kN/m². This load obtained from data books or codes, is assumed to act vertically on a horizontal plane. So there is no need to make any modifications, and we can apply it directly.

Live Loads (16.7)
The Live loads acting on the stairs can also be obtained from the data books or relevant codes. IS 875: 1987(part II) recommends a uniformly distributed load of 5 kN/m². This load is to be applied on both the sloping portion and the horizontal landing. In buildings such as residences, where the specified Live loads on the floors do not exceed 2 kN/m², and the stairs are not liable to be overcrowded, the Live load can be taken as 3kN/m². As in the case of self wt. of finishes, the LL obtained from data books or codes, is assumed to act vertically on a horizontal plane, and so there is no need make any modifications, and we can apply it directly.

Loads on Landing
The above items 16.3, 16.5, 16.6 and 16.7 are the four loads that come on the sloping slab (Going) of a flight. Let us now see the loads that come on the landing.

The landing is horizontal, and so there is no need to make any modifications for slope. The volume of a 1 x1m square block of landing slab is 1 x 1 x t = t m³. Multiplying this by the weight density of reinforced concrete will give the weight. So we get 25t kN/m². (16.8)
• The loads coming from the finishes can be taken as the same value (16.6) that we saw for the sloping portion.
• The LL can also be taken as the same load (16.7) for sloping portion.

So we have seen all the loads coming on the stair. The loads that we determine using the above methods are characteristic loads. They must be multiplied by the appropriate load factors to obtain the factored loads.
All these loads are calculated for a 1 x1m square area. But we are considering a beam (strip of slab) of width 1m, as shown in fig.16.16. So the above load will also be the load acting on every 1m length of the beam. In other words, the load on a 1 x1 m sq. area is also the UDL acting per meter length of the beam. So we can represent our beam as shown in the fig. below:

Fig.16.20
Loads acting on the stair

In the above fig.,
Eq.16.9: w₁ (Load per unit length on Sloping portion) = Load factor x {sum of values obtained from (16.3, 16.5, 16.6 and 16.7)}
Eq.16.10: w₂ (Load per unit length on Landing)= Load factor x {sum of values obtained from (16.8, 16.6 and 16.7)}
Note that in Eq.16.10,
• The loads from finishes (16.6) and the LL (16.7) remains the same as that of the sloping portion.
• 16.5 is absent because there are no steps in the landing.
• 16.3 (wt. of inclined slab) is replaced by 16.8 (wt. of horizontal slab)
So w₂ will have a lesser value than w₁ because 16.8 will always be lesser than 16.3.

Thus we can calculate the loads and draw the diagram shown in fig.16.20 with all the details. Based on that fig., we can draw the BM and Shear force diagrams.

In fig.16.14 of the previous section, we have considered a strip from support to support. The fig. is shown again for convenience.

Fig.16.14
1 m wide strip

One support is a masonry wall. The other is the end of the slab which is given an increased thickness. It rests on the foundation. The strip bends between these two supports. The risers (not shown in the fig.) of this stair are all 'parallel to the two supports'. Such stairs which bend between 'supports which are parallel to the risers' are called Longitudinal stairs. The other type is the stairs which bend between 'supports which are perpendicular to the risers' are called Transverse stairs. These two types are shown schematically in the figs. below:

Schematic diagram for Longitudinal stairs

Schematic diagram for Transverse stairs

We will discuss about transverse stairs in later sections.

So we have a longitudinal stairs with two flights AB and CD. Flight AB has one Going and one Landing. Many times in practice, we will come across this type with one Going and one landing, which can be represented by the line diagram shown in fig.16.20 above. So it will be convenient if we derive the general equations for BM and shear forces. These equations are given below:
Reaction at support A is given by:

Eq.16.11

Reaction at support B is given by:
Eq.16.12

The bending moment at any point at a distance x from support A is given by:
Eq.16.13

If we differentiate this equation, we will get the equation for the shear force at any point at a distance x from the support. This is given below:

Eq.16.14

Maximum bending moment occurs at a place where the shear force is equal to zero. So we equate the above equation 16.14 to zero and solve for x. Then we put this value of x in 16.13 and find the value of maximum BM. The reinforcement is then designed for this maximum BM.

We have just seen the procedure for the analysis and design of a flight with one Going and one Landing. Now we will see the procedure for a flight with one Going and two landings. The method of calculation of loads are the same. When the loads are calculated, we can draw the line diagram as shown below:

Fig.16.21
Loads acting on a flight with one Going and two landings

The calculations of BM and SF are easy in this case because the beam and the loadings are symmetrical. (Note that, the lengths of both the landings are taken as l₂. If they are not equal, there will not be symmetry). As before, we will write the equations as given below:
Reactions at supports C and D is given by

Eq.16.15

As the beam and the loadings are symmetrical, the maximum BM will be at the mid span. So there is no need to find the point where the SF is equal to zero. The maximum BM is given by:
Eq.16.16

Where l = l₁ + 2l₂

When the maximum bending moment is calculated, we can design the reinforcement required to resist that bending moment. So we will now do the detailed analysis and design of our stair. It is given as:

Solved example 16.1
The arrangement of bars according to the design in the above solved example is shown in the figs.16.22 and 16.23 given below:

Fig.16.22
Reinforcement details of flight AB

Fig.16.23
Reinforcement details of flight CD

For flight AB, the main bars of 12 mm dia. are provided at 250 mm c/c. For the flight CD, the main bars of 12 mm dia. are provided at 150 mm c/c. In the next section, we will discuss some details about the above two figs.

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