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.
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
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.
thankyou fro sharing a good knowledge
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