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.
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.
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.
No comments:
Post a Comment