Two way slabs are supported on unyielding supports on all four sides and the ratio of longer span to shorter span is less than 2. All the design procedure are explained under the topic design of two-way slab. Here we discuss about the example of two-way slab design.
Example of two-way slab:
Let’s take an example of two-way slab design;
Q.Design of reinforced slab for a room of clear dimension 4m*6m. The slab is supported on monolithically casted beam all-around of width 230mm. Assume live load and floor finish. Also select suitable grade of steel and concrete.
Assume, the building is institutional so, the live load should be equal to 3 kN/mm2 for normal room. [From IS 875 part 2]
Floor finish with partition load = 1.5 kN/mm2
For the structural member, the code stated that we can’t use grade of concrete less than 20MPa and from ductility criteria for seismic resistance, the grade of normal HYSD steel should not be greater than 415 N/mm2.
So, assume grade of concrete = 25 N/mm2
And grade of steel = 415 N/mm2
Note: Initially while designing, we don’t know the width of beam. For this case, we should calculate width of beam from preliminary design of beam.
Let’s explain the example of two-way slab design in step wise:
Step 1. Calculate aspect ratio and determine either it is one-way or two-way slab.
Longer span/ shorter span = 6/4 = 1.5 < 2
Since, the ratio of longer span to shorter span is less than 2,it is two-way slab.
Step 2. Determine the minimum thickness of slab from serviceability criteria of deflection control.
[As it is already discussed in preliminary design of slab, you can go there just by a single click for further details.]
Assume, shorter span/ depth = 30
Effective depth = 4000/30 = 133.33mm
Assume, diameter of bar = 10mm and clear cover = 15mm
So, overall depth of slab = effective depth + clear cover + half diameter of bar
= 133 +15 + 5
For construction easiness take overall depth 150mm and check.
So, effective depth = 150 – 15 – 5 = 130 mm
Step 3. Calculate the effective span of slab from IS 456 clause 22.2 which is least of following values
- Center to center of supports or clear span + width of support
- Clear span + effective depth of slab
Since width of support (ie. 230mm) is greater than effective depth of slab ( ie. 130mm) so, the point (b) gives least value.
Effective span of slab along shorter side = Clear span + effective depth of slab
= 4000+ 130 = 4130mm
Effective span of slab along longer side = Clear span + effective depth of slab
= 6000 + 120 = 6120 mm
[Reinforcement in longer span lies above main reinforcement in shorter span so effective depth is less by a diameter of bar]
Step 4. Determine the total factored load due to live and dead load considering 1m wide strip.
Self-weight of slab = depth of slab * unit width * unit weight of RCC
= 0.15 * 1 * 25 = 3.75 KN/m
Dead load due to floor finish and partition = 1.5 * 1 = 1.5 KN/m
Live load on slab = 3 * 1 = 3 KN/m
Total load = 3.75 + 1.5 + 3 = 8.25 KN/m
Factored load (w) = 1.5 * total load = 1.5 * 8.25 = 12.375 KN/m
Step 5. Calculate the factored design moment Mu and factored design shear force Vu.
If you are designing your building, you simply know about the boundary condition ie. types of panel.
Here, take two adjacent edges discontinuous.
Aspect ratio = longer span/shorter span = 6120/4130 = 1.48
Take 1.5 for simplicity since, bending moment coefficient increases with increase in aspect ratio, it will not cause any problems.
The factored bending moment can be calculated by using the coefficient of bending moment from table 26 of IS 456:2000.
|Short span coeff. αx||Long span coeff. αy|
The maximum bending moment per unit width is given by,
Mx = αx w lx2
My = αy w lx2
|Mx (kN-m)||My (kN-m)|
The factored shear force can be calculated by using the given formula:
Vu = w (r⁴ /(1+r⁴) ) * lx /2
= 12.375 * (1.48⁴/(1+1.48⁴)) * 4.13/2
= 21.15 KN
Step 6. Check the thickness of slab.
The thickness of slab is calculated for given value of factored moment then check with initial value. The thickness calculated from factored design moment should be less than initial value if not, revise with greater depth.
Mu = Mu,lim = 0.36 fck b xu [d – 0.42 xu]
For Fe 415, xu/d = 0.48 [from IS 456:2000 clause 38.1]
Therefore, Mu = Mu,lim = 0.138 fck b d²
[Since, width of slab b is taken as 1m ]
15.83 * 106 = 0.138 * 25 * 1000 * d²
d = 67.74mm < 130mm, OK
Step 7. Determine the area and spacing of main reinforcement in both direction.
The area can be calculated from given formula;
Mu = 0.87 fy As [d – fy As/ fck b]
Mu = 0.87 * 415 * As [130 – 415* As/ 25*1000]
|Asx (mm2)||Asy (mm2)|
And the minimum area of main steel reinforcement is equal to 0.12% of bD.
As,min = 0.12 * 1000 * 150/100 = 180 mm2
The area of reinforcement at mid-span of longer reinforcement is less than min. so increase it to 180 mm2.
Since diameter of bar = 10 mm
Area of bar = 3.14 * 10 * 10 / 4 = 78.54 mm2
Spacing of bar = area of a bar * width of strip / area of reinforcement required
|Along short span mm||Along long span mm|
The spacing of main reinforcement bar should not be more than 3 times the effective depth of solid slabs or 300 mm whichever is smaller.
If you want to use straight curtailed bar, then you can keep different spacing for all. But if you want bend-up bar, then you shall keep same spacing only along one side.
For construction easiness, you also go forward by using min spacing for all.
Let’s choose bent-up case,
Provide spacing along shorter span = 200 mm
Provide spacing along longer span = 300 mm
So area provided is given by,
Along shorter span = 78.54 * 1000 / 200 = 392.7 mm2
Along longer span = 78.54 * 1000 / 300 = 261.8 mm2
Step 7. Check the adequacy of depth from deflection control criteria.
For deflection control criteria, using IS 456:2000 clause 23.2.1
For modification factor for tension reinforcement,
fs = 0.58 fy * area required / area provided
= 0.58 * 415 * 353.2/392.7
Percentage of steel = 392.7 *100 / 1000*150 = 0.26 %
From fig.4, modification factor for tension reinforcement = approx. 1.8
Allowable l/d ratio = basic value * modification value
= 26 * 1.8 = 46.8 > Assumed (30)
Note: for calculating fs, area required and area provided along shorter span should be used such that this gives greater value. Here, reinforcement at support is used. But you can also check for mid-span
Step 8 : Check for shear
It is not necessary to check shear in two-way shear if you design through serviceability criteria.
But if you have doubt then you can check by using code IS456:2000 clause 40.2 and table 19 and 20.
Step 9. Calculate the torsional reinforcement.
At the discontinuous corner, torsional reinforcement should be provided in two layer. Each layer should have area per unit width equal to 75% of mid-span tension reinforcement for adjacent edge of a slab is discontinuous. But if only one edge is discontinuous then 37.5% will be okay.
Length of torsional reinforcement = shorter span / 5 = 4130/5 = 826 mm
- At corner of adjacent edge discontinuous,
Spacing of bar = 4/3 * spacing provided in mid-span along shorter direction
= 4/3 * 200 = 266mm
So provide 5 bar of 10 mm diameter and 1000 mm long @ 250 mm c/c
2. At other corners of one edge discontinuous
Spacing of bar = 8/3 * spacing provided in mid-span along shorter direction
= 8/3 * 200
= 533 mm
So provide 3 bar of 10mm diameter and 900mm long @ 450 mm c/c
Note: Here we have calculated spacing directly. But you can calculate area then spacing also if you feel uneasy or have doubt on it.
For this example of two-way slab, the arrangement of reinforcement is not shown but can be done as mentioned on SP 34 Section 9.