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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Chapter 16 (cont..12)

The following solved example illustrates the design of a transverse stair.

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.16.65 below) of the stair was given along with the problem data.

Fig.16.65
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. 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 cast 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. 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 12 - Analysis of Doubly reinforced beams

In the previous section, we completed the discussion on design of a doubly reinforced beam section. In this section we will discuss the analysis. When we are given a doubly reinforced beam section for analysis, we have to find the depth of the neutral axis x_u , and the ultimate moment of resistance M_uR.

We know that when we analyse the section using the Limit State Method, we are calculating the forces in the section at the point of impending failure. The concrete at the extreme compression fibre will be about to crush. And also, if it is an under reinforced section, the steel in the tension zone would have yielded.

So now we are going to find the forces in a doubly reinforced section, at the above mentioned point of impending failure. We can base our discussions on the fig.12.1 given below:

Fig.12.1
Stresses, strains and forces

The above fig. shows all the stresses and forces at the ultimate state. We will now see the details of each component:

The depth of NA is denoted as x_u. It's value can be less than or greater than x_u,max, depending on the quantities of steel. It is one of our aims to calculate this x_u of the given beam section.

We have derived the expression for ε_sc in the previous chapter. We used the triangles above the NA in the strain diagram. We have the same strain diagram here also. The only difference is that x_u,max is now changed to x_u. So from Eq.11.1, we get:

Eq.12.1:

When we know ε_sc, we can calculate the stress f_sc in A_sc. But here we cannot use the table 11.1 because, the values in it correspond to the condition of x_u = x_u,max. Here we are having x_u, which may be having a value higher or lower than x_u,max. So we have to use the original tables for calculating f_sc.

The net compressive force in steel is obtained as:
C_us = (f_sc - 0.447f_ck) A_sc
The compressive force in concrete C_uc = 0.362f_ckbx_u 

So the total compressive force is given by:
C_u = 0.362f_ckbx_u  + (f_sc - 0.447f_ck) A_sc

The total tensile force is given by T_u = f_st A_st
equating the above two, we get 

0.362f_ckbx_u  + (f_sc - 0.447f_ck) A_sc = f_st A_st

From this we get
Eq.12.2

But from Eq.12.1, we can see that f_sc depends on x_u. So the above Eq.12.2 will not give a closed form solution. We have to use strain compatibility method.

Another point to note is that, f_st also depend on x_u. If the tension steel has not yielded at ultimate state, we cannot take f_st as equal to 0.87f_y. So for such sections also, we will have to use strain compatibility method. We will discuss the details of the method when we do a solved example.

It may be noted that the above Eq.12.2 will give a closed form solution if both the tension steel and compression steel have yielded. Because, then f_sc and f_st will be having a constant value of 0.87f_y 

Once we determine x_u, we can calculate M_uR, the ultimate moment of resistance of the doubly reinforced section. For this we will use the compressive forces. We have two compressive forces:

• C_u = 0.362f_ckbx_u. Lever arm of this force = d - 0.416x_u. So the contribution from this force = 0.362f_ckbx_u (d - 0.416x_u).
• C_us = (f_sc - 0.447f_ck) A_sc.   Lever arm of this force = d-d'. So the contribution from this force = (f_sc - 0.447f_ck) A_sc (d-d') 

Sum of the above contributions will be the M_uR. So we get:
Eq.12.3: M_uR = 0.362f_ckbx_u(d - 0.416x_u) + ( f_sc - 0.447f_ck) A_sc(d-d')

Under reinforced or Over reinforced Doubly reinforced sections

In the case of singly reinforced sections, we have seen the details of 'balanced sections'. We discussed it based on the fig.3.26. This fig. is shown again below:

Fig.3.26
Stresses and strains in singly reinforced balanced sections

From the strain diagram in this fig., we saw that x_u,Bal, (which is same as x_u,max) is the particular depth of NA at which the strain in tension steel is equal to ε*_st, the yield strain. If in a beam section, x_u is less than x_u,max, the strain ε_st will be greater than ε*_st. Then it is an under reinforced section. If x_u is greater than x_u,max, ε_st will be less than ε*_st. That means the steel will not yield at ultimate state, and it is an over reinforced section.

We can extend the above discussion to Doubly reinforced sections also. As the plane sections remain plane after bending (Details here), the strain diagram for a doubly reinforced section will also be a straight line similar the strain diagram in the above fig. Thus the ε_st of 'tension steel' in a doubly reinforced section will be greater than ε*_st if x_u is less than x_u,max. It will be less than ε*_st if x_u is greater than x_u,max. 

So we can say that while designing a doubly reinforced section, just as in the case of a singly reinforced section, we must ensure that x_u is less than x_u,max. So that, if over loading occurs, and the beam fails, the failure will be a tension failure.

In the next three sections we will see analysis examples based on the above discussions. Links to these sections are given below.

• Solved example 12.1  • Solved example 12.2  • Solved example 12.3 (over reinforced)  

After the above solved examples, a comparison between 'Under reinforced and Over reinforced' doubly reinforced beam sections is given here. 


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