The data available for load calculation are:
R = 150mm, T = 300mm, t= 190mm, wf,Data = 0.8 kN/m2, w LL,Data = 5 kN/m2,
Weight density of the brick masonry for steps = γs = 20 kN/m3
Weight density of the brick masonry for steps = γs = 20 kN/m3
Loads on sloping portions:
(1) Self wt. of waist slab =
= 5.31 kN/m2
(2) Self wt. of steps =
Eq.16.5
0.5Rγs = 1.5 kN/m2
Eq.16.5
0.5Rγs = 1.5 kN/m2
(3) Finishes = 0.8 kN/m2
(4) LL = 5 kN/m2
Total of the four items = 12.61 kN/m2Factored load (Load factor = 1.5) = 18.92 kN/m2
Loads on Landings:
(1) Self wt. of slab =
25t = 4.75 kN/m2
25t = 4.75 kN/m2
(2) Finishes = 0.8 kN/m2
(3) LL = 5 kN/m2
Total of the three items = 10.55 kN/m2Factored load (Load factor = 1.5) = 15.83 kN/m2
We have discussed the case of right angled stairs on the basis of fig.16.30 There we saw that only half of the load should be taken on the landing of such stairs. Here we have a similar situation. The landing at B is common to flights AB and BC. The load on this landing will contribute to the BM in both the flights. And also, both these flights are independent. That is., they have their own supports. None of them are supported on any landing.
The same situation is present at the landing at C. So we can see that at both the landings, only half the load (15.83 x 0.5 = 7.92 kN/m2) should be considered. Thus we can add loads to the line diagram in the previous fig.105, as shown below:
Fig.16.106
Line diagram showing the loads
Line diagram showing the loads
We can see that CD is the exact mirror image of AB. So we need to do the analysis and design of AB and BC only. The design of AB can be applied to CD also.
First we will analyse AB:
Eq.16.12:
The following data is available to us for using in the above equations:
w1 = 18.92 kN/m, w2 = 7.92 kN/m, l1 = 2.81 m, l2 = 1.36 m
So we get RA = 37.01 kN and RB = 26.93 kN
The Bending moment at any point at a distance x from the 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 A:
Eq.16.14
The maximum bending moment occurs at a point where the shear force is equal to zero. So we equate the above equation 16.14 to zero and solve for x
Thus we get x = 1.96m from support A
Substituting this value in Eq.16.13, we get Maximum bending moment = 36.2 kNm
Thus we determined the reactions and bending moment in flight AB. Now we will analyse BC:
Reactions at supports B and C is given by
The following data is available to us for using in the above equations:
w1 = 18.92 kN/m, w2 = 7.92 kN/m, l1 = 1.2 m, l2 = 1.36m
So we get RB = RC = 22.12 kN
The maximum bending moment occurs at midspan and is given by
Eq.16.16
= 26.17 kNm.
Thus we determined the reactions and bending moment in flight BC also. So now we can design the steel and do the various checks. The design part is given as two separate pdf files the links of which are given below:
The reinforcement details according to the above designs are shown in figs.16.107,108 and 109 below:
Fig.16.109
Reinforcement details of flight CD
Reinforcement details of flight CD
The bar diameters and their spacing in the last flight CD will be same as that of the first flight AB. But the bending of the bars at the supports is different for the two flights. So they are shown separately.
The second flight BC does not require distributor bars in the landings B and C. This is because the main bars of flights AB and CD will act as distributors. So we need to provide distributor bars for BC in it’s sloping portion only.
It should be noted that the main bars of the two flights AB and CD pass above the main bars of BC. Flight BC is not required to carry the loads from the other flights. But even then it’s bars are given beneath the others. This is because it is always better to give the bars of the shorter span as the bottom most layer.
This completes the analysis and design of the open well staircase. In the next chapter we will discuss about the design of Two way slabs.
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