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 90o 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.
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 wf,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 ‘wf,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, wf,data is the load in 1m2. So the load in the new reduced area = wf,data x cosθ. So we can write:
Eq.16.22: The load from finishes on 1m2 area on the inclined surface = wf,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 wLL,data , and write:
Eq.16.23: The load from LL on 1m2 area on the inclined surface = wLL,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.
In the previous section we saw an example which gives the basic details about transverse stairs. In this section we will discuss more details. The fig.16.51 below shows the sectional elevation of the stair in the example that we saw.
Fig.16.51
Sectional elevation of stair
In the above figs., the waist slab is resting on the top of the stringer beams. Another arrangement possible is that in which the beams are projecting above the stair. The view and sectional elevation of this arrangement are shown in the figs.16.52 and 16.53, below:
Fig.16.52
View of stairs with projecting stringer beams.
Fig.16.53
Sectional elevation of stairs with projecting stringer beams
In the above fig. we cannot expect flange action because the flange will be in tension. So we must design it as rectangular beams.
Yet another arrangement that is possible, is to use masonry walls instead of beams. This can be used if adequate foundation can be provided for the walls, and also if the height of walls can be kept within permissible limits.
In all the above methods, the waist slab will be spanning in a direction, perpendicular to the direction of travel, and so it is a transverse stair.
Loads on a Transverse Stair
Let us now analyse the loads acting on this transverse stair.
In the case of a longitudinal stair, we determined the load on a horizontal projection having an area of 1 m2, as shown earlier in fig.16.16. But for a transverse stair, we use the loads on a 1 m2 area, on the inclined slab itself. This is shown in the fig.16.54 below:
Fig.16.54
1 m2 area on inclined surface
The method of using such a load is used for the 'analysis and design of the stair' will be explained later. At present let us determine the magnitude of this load. It has four items as given below:
1) Self wt. of waist slab: This can be calculated based on the sectional view in fig.16.55 given below:
Fig.16.55
Self wt. of waist slab
• The area of the portion coloured in green in the above fig. is 1 x t m2. (‘t’ is in m). Note that the '1m' here is the same 1m marked perpendicular to the red strip in the previous fig.16.54
• The length of the block (perpendicular to the plane of the paper) is 1m. This '1m' is the same 1m marked parallel to the red strip in the fig.16.54
So volume = 1 x t x 1 = t m3.
Thus the wt. of this block having an area of 1m2 = t x 25 = 25t kN. where 25 is the unit wt of concrete in kN/m3.
Thus we can write:
Eq.16.20: Self wt. of waist slab = 25t kN/m2
2) Self wt. of steps:
• Area of cross section of 1 step = [0.5 x R(Rise) x T(Tread)] (∵area of the triangle = 0.5 x Base x Altitude)
• Length of the triangular prism perpendicular to the plane of the paper = 1m. This '1m' is the same 1m marked parallel to the red strip in the fig.16.54
• So volume = 0.5 x R x T x 1 =0.5RT
• Thus wt. of one step with a length of 1m perpendicular to the plane of the paper = 0.5RTγs Where γs is the unit wt. of the material of the step. (For brick masonry, γs = 20 kN/m3)
Now we have to calculate the no. of such steps with in the length of 1m shown in fig.16.55. For that, we will use the fig.16.56 given below:
Fig.16.56
No. of steps in 1m along the slope
From the fig., it can be easily seen that the no. of steps within the two red dotted lines will be equal to 1m divided by 'B' where B = √(R2 + T2) (∵ B is the hypotenuse of the triangle with base T and altitude R)
So we can write
Eq.16.21
The total wt. of all the steps within the 1m2 area =
It may be noted that 1/B will be a 'whole number + a fraction' in most cases. The fractional part indicates that 'parts of steps' will also be coming in the 1m2 area.
The next item is the ‘weight of finishes’. We will see the details about it in the next section.
