Chapter 16.12 - Design example of a Transverse stair

In the previous section we completed the discussion on 'load calculation on a transverse stair'. Link to a solved example which illustrates the analysis and design is given below:

Solved example 16.3

The reinforcement details according to the above solved example is shown in the figs.16.64 and 16.65 given below:

Fig.16.64
Sectional elevation of transverse stair

Fig.16.65
Section XX

We will now discuss the various features of the above two figs:
The main bars are given as the bottom most layer. The distributor bars are given as the second layer from the bottom. This arrangement will give maximum possible effective depth ‘d’ for the section. From the section XX, we can see that the main bars are given two 90^o bends at both the ends. This will give a 'hook' like arrangement at both ends. The top portion of these 'hooks' will act as the steel required for resisting any possible hogging moment at the supports. The length required for this top steel is 0.15l, where l is the effective span of the waist slab. But the area required for this top steel is only half of that at the mid span. So, as shown in the inset, only alternate bars are given these hooks.
The distributor bars are of mild steel. So the symbol Φ is given to denote them.
In the above example, only a part plan (shown in the fig. below) of the stair was given along with the problem data.

Problem data of solved example 16.3:

This is because, this small portion is sufficient for defining the problem of this type of a transverse stair. But some times, this stair may have a landing also as shown below.

Fig.16.66
Landing in a transverse stair

In this case also the sloping portion can be analysed and designed by the same procedure, and the landing can be analysed and designed as a simply supported one-way slab.

---

Stairs cantilevering from the side of a beam

The following fig.16.67 shows the part view of a stair.

Fig.16.67
View of a cantilever stair

It is projecting from the side of a wall. The wall is shown in a finished state. So we cannot see more details. The fig. below shows the view before the plastering is applied to the wall.

Fig.16.68
View of a stair cantilevering from a stringer beam

We can see that the stair is projecting from the side of a beam, which is concealed inside the wall. In actual construction, the beam (which frames into columns) and the waist slab will be casted first, and after curing and removal of form works, the masonry wall will be constructed above and below the beam. The elevation and section are shown in the figs. below:

Fig.16.69
Part elevation view of cantilever stairs

Fig.16.70
Section XX

The waist slab bends in a direction perpendicular to the direction of travel of the pedestrians, and so it is a transverse stair. The load calculation procedure is the same as that for the transverse stair supported on two stringer beams. So we can use the same Eq.16.25 for the loads. The line diagram for the analysis will be as shown below:

Fig.16.80
Line diagram for cantilever stairs

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/10 . However, this should be finalized only after the complete design and doing the various checks.

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

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Chapter 16.11 - Perpendicular and Tangential loads on Transverse stairs

In the previous section we saw the self weight of the waist slab and steps. In this section we will see the other loads.

3) Wt. of finishes: As in the case of longitudinal stairs, we obtain this from codes or data books. Let us denote this load as w_f,data. We know that this is specified on 1m² horizontal area. But we are having an inclined area. If we take the exact value from the code or data book, we will be using an excess quantity of load as shown below:

Fig.16.57
Horizontal 1m exceeds inclined 1m

[We are using the symbol ‘w_f,data’ for denoting the wt. of finishes that we obtain from codes or data books. In the case of longitudinal stairs, we did not use such a symbol for the wt. of finishes. The reason is that, for longitudinal stairs, the 'values from codes or data books' is applied directly in the calculations. But here in the case of transverse stairs, as we will soon see, a modification have to be applied to it. So it is convenient to denote it using a symbol.]

From the fig.16.57 we can see that the length ac which is equal to 1m is greater than the horizontal projection of the green coloured area. So we need to take a lesser length on the horizontal plane. This lesser length can be determined using the following fig.16.58

Fig.16.58
Lesser length on the horizontal plane, which corresponds to the 1m on the inclined plane

From the two red dotted lines we can see that ab is our required 'reduced length'. This can be calculated from triangle abd, [Note that ad = 1m. This is because ad, the two red dotted lines, and the green strip, are the sides of a Parallelogram] We get ab = 1m x cosθ. Here θ is the angle at a. So our next step is to calculate θ.

By using the 'theorem of alternate interior angles', θ is also the angle that the waist slab makes with the horizontal. By using the same theorem, it is also the angle at the upper corner of a step, as shown in the inset of the above fig.16.58. So we will get the value of θ from the relation: cosθ = T/B, where ‘B’ is the hypotenuse of a step, given by B = √(R² + T²) .

So we can easily calculate cosθ. Now, 1 x Cosθ = cosθ = the required reduced length. The length perpendicular to the plane of the paper is 1m. So the reduced area = cosθ x 1m = cosθ m²

So we determined the 'reduced area' that is to be used. Now, w_f,data  is the load in 1m². So the load in the new reduced area = w_f,data x cosθ. So we can write:
Eq.16.22: The load from finishes on 1m² area on the inclined surface = w_f,data x cosθ

4) Live load: This is also specified in the codes and data books as load per 1m²  on horizontal area. So following the above discussion, we will denote it as w_LL,data , and write:
Eq.16.23: The load from LL on 1m² area on the inclined surface = w_LL,data x cosθ.

Thus we get all the four items that constitute the load on the waist slab. We add these four, and denote the sum as W. Thus:
Eq.16.24: W= Sum of the four quantities calculated using Eq.16.20, 21, 22 and 23

Now we need to understand the effects of this load on the stair. Let us analyse:
The W acts vertically as shown below:

Fig.16.59
Total load W acting vertically downwards

This W can be resolved into two components as shown below:
Fig.16.60
Components of the vertical force W

One component acts perpendicular to the slab and the other acts parallel to the slab. From the triangle shown in the inset, we can see that the angle between W and the component perpendicular to the waist slab is equal to θ. So the magnitude of this perpendicular component will be equal to W cosθ. And the magnitude of the component which is tangential to the sloping surface of the waist slab will be equal to W sinθ.

Now we will see the effect of these two components. We will see the effects separately. The effect of the perpendicular component is shown in the view given below:

Fig.16.61
Effect of perpendicular component

From the above view, we can see that the perpendicular component will bend the plane surface of the waist slab into a three dimensional surface. That is., the plane surface bends to become the surface of a cylinder. The effect of the tangential component is shown in the view below:

Fig.16.62
Effect of Tangential component

We can see that tangential component does not bend the slab into any three dimensional surface. Instead, it causes the slab to bend in it's own plane. Here we must understand the difference between these two types of bending. 
• In the first case, when the slab is bent into a cylindrical shape, only the thickness 't' of the slab is available to resist the bending. 
• But in the second case, when the slab bends in it's own plane, a very large depth is available to resist the bending. For example, if we are considering a 1m wide strip, the whole 1m will be available to resist this bending. 

So the tangential component is ignored in the design. We provide steel to resist the perpendicular component only.

We have discussed earlier about the design of simple horizontal slabs. 
• In simple horizontal slabs, the slab member is horizontal, and the loads acting on them are vertical. So the loads are acting perpendicular to the slab surface.
• In the transverse stairs, when the tangential component is ignored, they become an inclined slab acted upon by loads perpendicular to it's surface. 

Both are same: A slab surface acted upon by forces perpendicular to it. So we can design transverse stairs in the same way as a simple horizontal slab.

Thus we reach a conclusion: We calculate W and multiply it with cosθ. The product 'Wcosθ' is the load which will act perpendicular to the slab surface. We provide reinforcements to resist this Wcosθ.

This Wcosθ is the load acting on 1 m² area on the inclined surface. We are designing the slab as a 1m wide strip. So this is the same load on 1m length of the strip. (See the explanation based on fig.5.11.) So we can write:
Eq.16.25
The UDL per meter length on 1m wide strip of a transverse stair = w₁ = Wcosθ.
• where W= Sum of the four quantities calculated using Eq.16.20,21,22 and 23
• and cosθ = T/B
When the above calculations are done, we can draw the line diagram of the 1m wide strip as shown below:
Fig.16.63
Line diagram for a 1m wide strip of a Transverse stair

In the above line diagram, the slab is shown to be simply supported. So the maximum bending moment at midspan can be obtained as w₁l²/8. If the ends are fixed, or if the slab is continuous over a number of supports, detailed analysis should be done.

To begin the analysis and design, we must know the self weight of the waist slab. For calculating the self wt., we must know the cross sectional dimensions. The width is fixed at 1000 mm, as we are considering a 1m wide strip. We want the depth. 

For a preliminary design, we can use a thump rule: Upto a clear span of 2m, a value of 100mm can be assumed for ‘t’ of a transverse stair supported on two stringer beams or walls. However, this value should be finalised only after doing the various checks like check for deflection, shear check etc.,

In the next section we will see a solved example based on the above discussion.

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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 16.13 - 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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