In
the previous section we completed the analysis and design of singly reinforced beams for flexure. Now we will see the analysis of 'One way slabs'. The bending (flexure) of a one way slab is similar to the bending of a beam.
The following presentation shows some basic details about the behaviour of simply supported slabs :
From the above presentation, we get a clear idea of how to distinguish between a one way slab and a two way slab:
We take the following ratio:
5.1: ly/lx. If this ratio is greater than 2, then the slab is a one way slab. If it is less than or equal to 2, then it is a two way slab.
This method is to be used only on those slabs which are resting on walls on all the four sides. If there are walls only on the opposite sides, the load transfer will occur in that direction only, and it will be a one way slab.
Now we will see how a one way slab can be compared to a beam: We divide the slab into a number of strips as shown in the fig.5.10 below:
Fig.5.10
One way slab divided into strips
The strips extend from one support to the other. Each strip has a width of 1m (shown in the fig. as 100cm). Each of these strips act as individual beams. So each of these beams will have a width of 1m and a total depth D, where D is the total depth of the slab.
Assuming that only 'uniformly distributed load' is acting on the slab, and that the slab is of uniform thickness, all these strips are identical. We can take any one of them for analysis and design. Let us take one strip as shown in the fig.5.11 below:
Fig.5.11
Loads on an individual strip
For designing this strip, we want to know the maximum bending moment acting on it. We can calculate it by using the same method that we would use if it was an individual beam. That is., we can use the formula:
Mmax = wl2/8.
where w is the load per unit length of the strip, and l is the effective span of the strip.
So we want w. For this, in the above fig.5.11, a length of 1m has been marked off on the strip, in a direction parallel to the length of the strip. When we mark off 1m on the strip parallel to it's length, there will be a dimension perpendicular to the length. As we are considering strips of 1m width, this perpendicular dimension will also be 1m. So we are having an area of 1x1m2 on the strip. The load acting on this 'area' is the load acting per meter 'length' of the strip. So we can summarize the above discussion as follows:
• To design the one way slab, we take any one strip of 1m width.
• We want the load per unit 'length' of this strip
• This load load per unit 'length' that we require, is same as the load per unit 'area' of the slab.
So we will now see the details of the load per unit area:
Self weight of the slab in the unit area: - - - - (1)
Volume of the concrete block in the unit area = 1m x1m x'D'm = 'D'm3. ('D' should be in meters.)
So weight of the concrete block in unit area = D x25kN/m3 = 25DkN/m2
Self wt. of finishes: - - - - (2)
This can be obtained from data books or relevant codes. Usually it varies from 0.5 to 1.0 kN/m2.
Live loads: - - - - (3)
This can be obtained from IS code 875. This load depends upon the nature of use of the building. (Residential, office, storage purpose etc.,). Usually it is given in kN/m2
All the relevant clauses of the code must be considered in arriving at the appropriate value of the loads to be used in the design.
Sum of the above three items will give the characteristic load per meter square of the slab. This is equal to the characteristic load per unit length of the strip. We must multiply it by the load factor to obtain the factored load.
When we are given a slab for analysis, we will be given it's steel in terms of 'spacing of bars of a particular diameter'. For example: '#8 @ 180 c/c'. This means that 8 mm bars are provided at a spacing of 180mm c/c. But to analyse the slab as a beam, we want the area Ast within a width of the beam (1000mm). Let us see how we can obtain this from the 'spacing': The fig.5.12 below shows a part sectional elevation of a slab. The spacing of bars is denoted as 's' mm.
Fig.15.12
Area of steel in a slab
From the fig. we can see that the number of bars 'n' in a width of 1m will be equal to 1000/s. If the diameter of bar is denoted as Φ, then the area of one bar = πΦ2 /4. So the total area of bars in the strip = nπΦ2 /4 = (1000/s)( πΦ2 /4) Thus we get
Eq.5.2:
(Where Φ and s are in mm)
When we are required to analyse a given one-way slab section, Eq.5.2 can be directly used to calculate Ast in a 1000 mm wide strip. Then the strip can be treated as a beam of width 1000 mm and can be analysed to find MuR.
While designing a beam, we can use the converse of the above: As the result of the design, we will get Ast that has to be provided in one strip of 1000mm width. But we want to express it as 'spacing of bars of a particular diameter'. If Φ is the diameter, area of a single bar = πΦ2 /4 . So the number of bars required to make up Ast = n = Ast /( πΦ2 /4) =4Ast / ( πΦ2) Thus we get
5.3:
(Where Φ is in mm and Ast is in mm2)
Fig.5.13
Lateral deformation of beam section
Now let us consider the strip of slab. In this case, the expansion above the NA and the contraction below the NA is prevented by the strips on the two sides of the design strip. So the portion above the NA will experience a lateral compressive reaction from the adjacent strips, so that the lateral expansion is prevented. The portion below the NA will experience a lateral tensile reaction from the adjacent strips, so that the lateral contraction is prevented. These lateral forces will give rise to secondary moments in the transverse direction as shown in the fig. 5.14 below:
Fig.5.14
Secondary moments in slab
In the above fig., the restraining forces are indicated by the magenta colored arrows.
• The top arrows shows compressive reaction from the adjacent strips. This compressive reaction prevents the expansion (above the neutral axis) of the design strip.
• The bottom arrows shows tensile reaction from the adjacent strips. This tensile reaction prevents the expansion (below the neutral axis) of the design strip .
These two reactions together form a couple. It will try to bend the slab in the transverse direction. Thus we see that the one way slab require reinforcements in the transverse direction to resist the secondary moments. These reinforcements in the transverse direction are called secondary reinforcements.
The secondary reinforcements serve some other purposes also:
• Effects due to concentrated loads:
When a concentrated load is applied on the slab, bending moments in the transverse direction are induced in the slab. So we want secondary reinforcements in the transverse direction to resist these moments.
• Shrinkage and temperature effects:
Freshly placed concrete will shrink when it dries. The extent of this shrinkage can be minimized by using appropriate water cement ratio, and by proper moist curing. But even after taking all necessary measures, some shrinkage will always occur. If a slab of usual dimensions rests freely on it's supports, it can freely shrink. But usually the slab is kept in position by beams, walls above supporting walls etc., So it cannot shrink freely. So when restrained slabs shrink, tensile stresses will develop in it. This will give rise to cracks.
A similar effect is produced due to temperature variations also. There will be thermal expansion and contraction of the slab due to temperature variation. A restrained slab can not expand or contract freely. So this will give rise to cracks.
These cracks can be minimized if we provide steel in a direction perpendicular to the cracks. But shrinkage and temperature effects occur in all direction, and so the cracks can occur in any direction. If bars are provided in two sets, each set perpendicular to the other (in the form of a grid), crack formation in any direction can be resisted. We already have one set which is provided to resist the bending moment. We also learned about the secondary reinforcements which are provided in the transverse direction. So these two sets will resist the formation of cracks due to shrinkage and temperature effects.
It may be noted that during shrinkage, the slab is trying to pull inwards. The bars which try to resist this inward pull will be experiencing compressive stresses.
Thus we can see that the secondary reinforcements are essential not only for resisting the secondary moments, but also to reduce the formation of cracks due to shrinkage and temperature effects. We will discuss about the 'quantity' of secondary reinforcements to be provided as per the code, when we take up the design of one-way slabs.
The following solved examples demonstrates the process of analysis of a singly reinforced one-way slab:
Solved example 5.1
We will now see the solved example that we did earlier based on Fig.3.32. There we analysed the beam and calculated the safe load that the beam can carry. Now let us analyse the slab and find the safe load that it can carry. The fig.3.32 is shown below again. But this time the section of the slab is shown:
Analysis steps are shown here. We get MuR = 14.4kNm
Now we calculate Mu:
Mu = wul2/8. where, l= effective span; wu = load per meter square area of slab.
First we calculate effective span:
• c/c distance between supports = 2950 +250 =3200mm
• clear span + effective depth = 2950 +126 = 3076mm
l = lesser of the above = 3076mm
wu is the unknown. Equating Mu and MuR we get:
wul2/8 = 14.4 ⇒ wu x 3.0762 x (1/8) = 14.4 ⇒ wu =12.18 kN/m
But wu is the load per m2 on the slab. This load consists of the following three components:
(1) Self wt = 25D = 25 x .16 = 4kN/m2
(2) Wt. of finishes = 1 kN/m2 (assumed)
(3) Wt. of partitions = 1.25 kN/m2 (assumed)
So total DL = 6.25 kN/m2. But there may be uncertainties in the DL. so we must multiply it by 1.5 Thus we get 6.25 x1.5 = 9.375kN/m.
So we can write: (9.375 + 1.5 x LL) = 12.18 ⇒LL = 1.87 kN/m2
Thus we can specify that the maximum characteristic LL that can be applied on the slab is 1.87 kN/m2
The following presentation shows some basic details about the behaviour of simply supported slabs :
From the above presentation, we get a clear idea of how to distinguish between a one way slab and a two way slab:
We take the following ratio:
5.1: ly/lx. If this ratio is greater than 2, then the slab is a one way slab. If it is less than or equal to 2, then it is a two way slab.
This method is to be used only on those slabs which are resting on walls on all the four sides. If there are walls only on the opposite sides, the load transfer will occur in that direction only, and it will be a one way slab.
Now we will see how a one way slab can be compared to a beam: We divide the slab into a number of strips as shown in the fig.5.10 below:
Fig.5.10
One way slab divided into strips
The strips extend from one support to the other. Each strip has a width of 1m (shown in the fig. as 100cm). Each of these strips act as individual beams. So each of these beams will have a width of 1m and a total depth D, where D is the total depth of the slab.
Assuming that only 'uniformly distributed load' is acting on the slab, and that the slab is of uniform thickness, all these strips are identical. We can take any one of them for analysis and design. Let us take one strip as shown in the fig.5.11 below:
Fig.5.11
Loads on an individual strip
For designing this strip, we want to know the maximum bending moment acting on it. We can calculate it by using the same method that we would use if it was an individual beam. That is., we can use the formula:
Mmax = wl2/8.
where w is the load per unit length of the strip, and l is the effective span of the strip.
So we want w. For this, in the above fig.5.11, a length of 1m has been marked off on the strip, in a direction parallel to the length of the strip. When we mark off 1m on the strip parallel to it's length, there will be a dimension perpendicular to the length. As we are considering strips of 1m width, this perpendicular dimension will also be 1m. So we are having an area of 1x1m2 on the strip. The load acting on this 'area' is the load acting per meter 'length' of the strip. So we can summarize the above discussion as follows:
• To design the one way slab, we take any one strip of 1m width.
• We want the load per unit 'length' of this strip
• This load load per unit 'length' that we require, is same as the load per unit 'area' of the slab.
So we will now see the details of the load per unit area:
Self weight of the slab in the unit area: - - - - (1)
Volume of the concrete block in the unit area = 1m x1m x'D'm = 'D'm3. ('D' should be in meters.)
So weight of the concrete block in unit area = D x25kN/m3 = 25DkN/m2
Self wt. of finishes: - - - - (2)
This can be obtained from data books or relevant codes. Usually it varies from 0.5 to 1.0 kN/m2.
Live loads: - - - - (3)
This can be obtained from IS code 875. This load depends upon the nature of use of the building. (Residential, office, storage purpose etc.,). Usually it is given in kN/m2
All the relevant clauses of the code must be considered in arriving at the appropriate value of the loads to be used in the design.
Sum of the above three items will give the characteristic load per meter square of the slab. This is equal to the characteristic load per unit length of the strip. We must multiply it by the load factor to obtain the factored load.
Calculation of Area of steel and spacing of bars
Now that we know how to obtain the bending moment acting on a strip of a one-way slab, we can analyze/design it as a beam, whose width is equal to 1m (or 1000mm), and whose effective depth is equal to the effective depth of the slab.When we are given a slab for analysis, we will be given it's steel in terms of 'spacing of bars of a particular diameter'. For example: '#8 @ 180 c/c'. This means that 8 mm bars are provided at a spacing of 180mm c/c. But to analyse the slab as a beam, we want the area Ast within a width of the beam (1000mm). Let us see how we can obtain this from the 'spacing': The fig.5.12 below shows a part sectional elevation of a slab. The spacing of bars is denoted as 's' mm.
Fig.15.12
Area of steel in a slab
Eq.5.2:
(Where Φ and s are in mm)
When we are required to analyse a given one-way slab section, Eq.5.2 can be directly used to calculate Ast in a 1000 mm wide strip. Then the strip can be treated as a beam of width 1000 mm and can be analysed to find MuR.
While designing a beam, we can use the converse of the above: As the result of the design, we will get Ast that has to be provided in one strip of 1000mm width. But we want to express it as 'spacing of bars of a particular diameter'. If Φ is the diameter, area of a single bar = πΦ2 /4 . So the number of bars required to make up Ast = n = Ast /( πΦ2 /4) =4Ast / ( πΦ2) Thus we get
5.3:
Transverse moments in one-way slabs
In the above discussions, we considered a 1m wide strip, and it was assumed to act as an independent beam. But there is a difference between the bending of a beam and the bending of a strip of slab. First let us consider the bending of a beam. When the beam is subjected to a sagging moment, the portion above the NA is under compression. Due to this compression, there will be a lateral expansion for the portion above the NA. This is due to the poisson effect. In the same manner, the portion below the NA is under tension, and hence there will be a lateral contraction. So after bending, the cross section of the beam will have a nearly trapezoidal shape as shown in fig. 5.13 below:Fig.5.13
Lateral deformation of beam section
Now let us consider the strip of slab. In this case, the expansion above the NA and the contraction below the NA is prevented by the strips on the two sides of the design strip. So the portion above the NA will experience a lateral compressive reaction from the adjacent strips, so that the lateral expansion is prevented. The portion below the NA will experience a lateral tensile reaction from the adjacent strips, so that the lateral contraction is prevented. These lateral forces will give rise to secondary moments in the transverse direction as shown in the fig. 5.14 below:
Fig.5.14
Secondary moments in slab
In the above fig., the restraining forces are indicated by the magenta colored arrows.
• The top arrows shows compressive reaction from the adjacent strips. This compressive reaction prevents the expansion (above the neutral axis) of the design strip.
• The bottom arrows shows tensile reaction from the adjacent strips. This tensile reaction prevents the expansion (below the neutral axis) of the design strip .
These two reactions together form a couple. It will try to bend the slab in the transverse direction. Thus we see that the one way slab require reinforcements in the transverse direction to resist the secondary moments. These reinforcements in the transverse direction are called secondary reinforcements.
The secondary reinforcements serve some other purposes also:
• Effects due to concentrated loads:
When a concentrated load is applied on the slab, bending moments in the transverse direction are induced in the slab. So we want secondary reinforcements in the transverse direction to resist these moments.
• Shrinkage and temperature effects:
Freshly placed concrete will shrink when it dries. The extent of this shrinkage can be minimized by using appropriate water cement ratio, and by proper moist curing. But even after taking all necessary measures, some shrinkage will always occur. If a slab of usual dimensions rests freely on it's supports, it can freely shrink. But usually the slab is kept in position by beams, walls above supporting walls etc., So it cannot shrink freely. So when restrained slabs shrink, tensile stresses will develop in it. This will give rise to cracks.
A similar effect is produced due to temperature variations also. There will be thermal expansion and contraction of the slab due to temperature variation. A restrained slab can not expand or contract freely. So this will give rise to cracks.
These cracks can be minimized if we provide steel in a direction perpendicular to the cracks. But shrinkage and temperature effects occur in all direction, and so the cracks can occur in any direction. If bars are provided in two sets, each set perpendicular to the other (in the form of a grid), crack formation in any direction can be resisted. We already have one set which is provided to resist the bending moment. We also learned about the secondary reinforcements which are provided in the transverse direction. So these two sets will resist the formation of cracks due to shrinkage and temperature effects.
It may be noted that during shrinkage, the slab is trying to pull inwards. The bars which try to resist this inward pull will be experiencing compressive stresses.
Thus we can see that the secondary reinforcements are essential not only for resisting the secondary moments, but also to reduce the formation of cracks due to shrinkage and temperature effects. We will discuss about the 'quantity' of secondary reinforcements to be provided as per the code, when we take up the design of one-way slabs.
The following solved examples demonstrates the process of analysis of a singly reinforced one-way slab:
Solved example 5.1
We will now see the solved example that we did earlier based on Fig.3.32. There we analysed the beam and calculated the safe load that the beam can carry. Now let us analyse the slab and find the safe load that it can carry. The fig.3.32 is shown below again. But this time the section of the slab is shown:
Analysis steps are shown here. We get MuR = 14.4kNm
Now we calculate Mu:
Mu = wul2/8. where, l= effective span; wu = load per meter square area of slab.
First we calculate effective span:
• c/c distance between supports = 2950 +250 =3200mm
• clear span + effective depth = 2950 +126 = 3076mm
l = lesser of the above = 3076mm
wu is the unknown. Equating Mu and MuR we get:
wul2/8 = 14.4 ⇒ wu x 3.0762 x (1/8) = 14.4 ⇒ wu =12.18 kN/m
But wu is the load per m2 on the slab. This load consists of the following three components:
(1) Self wt = 25D = 25 x .16 = 4kN/m2
(2) Wt. of finishes = 1 kN/m2 (assumed)
(3) Wt. of partitions = 1.25 kN/m2 (assumed)
So total DL = 6.25 kN/m2. But there may be uncertainties in the DL. so we must multiply it by 1.5 Thus we get 6.25 x1.5 = 9.375kN/m.
So we can write: (9.375 + 1.5 x LL) = 12.18 ⇒LL = 1.87 kN/m2
Thus we can specify that the maximum characteristic LL that can be applied on the slab is 1.87 kN/m2
When we analysed the beam based on fig.3.32, we found that a characteristic LL of 7.8 kN/m2 can be applied on the slab, as far as the safety of the beam is concerned. But now we find that such a load can never be applied on the slab. As far as the safety of the slab is concerned, we can specify that a characteristic LL of only 1.87 kN/m2 can be applied on the slab.
This completes the details about the procedure of analysis of a one way slab. In the next chapter we will see the design part.
This completes the details about the procedure of analysis of a one way slab. In the next chapter we will see the design part.
Hello, shouldn't the number of bars, written below figure 5.12 be equal to ((1000/s) + 1), if you divide it just by 1000/s, would it not give the number of SPACINGS instead of number of REBARS. It would be great if you could provide me a reply.
ReplyDeleteThis is related to the analysis of an existing slab. We will be taking a 1000 mm strip randomly. In such a strip we do not have to add extra one bar. If we take the strip starting exactly from a bar, then extra one will have to be added. Please discuss this concept with your professor and post a reply.
DeleteWhen we do the analysis, we are interested in finding the least possible moment of resistance. So the number of bars should be the least possible number. Further more, the bar at the edge of the strip cannot be taken into account. Because, it will not have the required concrete cover. It will not contribute towards the moment of resistance. Please post your thoughts about this aspect.
Delete@Irusha Costa - You are right!
ReplyDelete