strength of materials used in this article
weight of concrete = 24kN/m³
Characteristic strength of concrete fcu = 25N/mm²
Characteristic strength of steel ?y = 460N/mm²
General procedure for design of solid slabs to BS Code.
1.Determine a suitable slab depth; you can easily estimate this by using the formula stated below;
Effective depth = span/( basic ratio × modification factor)
For a simply supported design, the basic ratio is = 20 ( table 3.9, page 35 of BS 8110 part 1 : 1997). Use an initial modification factor (m.f) of 1.4.
Also the architect/ engineer may specify the slab thickness for you.
However, your slab design must satisfy deflection requirements otherwise you will have to redesign.
Ways to make your design slab satisfy deflection requirements if failed initially is to either increase slab thickness or amount of reinforcement or both.
2. Calculate the main and secondary reinforcement areas; you can do this by using the formulas stated in the BS code. ( Details later)
3. Check for excessive deflection ; as stated earlier in number 1, your design must satisfy deflection requirements.
4. Detailing requirements; the economical arrangement of steel reinforcement. You can refer to clause 126.96.36.199, BS 8110. An explanatory diagram is shown below for a simply supported case.
It means that 40% of the reinforcement placed around the center of the slab ( which is critical) only, should extend to the edges of the slab ( which is less critical).
Explanation of other important points in solid slab design to BS 8110
Effective span of slab:
Referring to the diagram above, the effective span of slab refers to ‘A’ the distance between the centers of bearings, or the clear distance between supports ‘D’, plus effective depth of slab ‘d’.
Calculating steel areas.
In calculating the slab self weight, the overall depth of the slab referred to as ‘h’, should be used. h is the effective depth of slab plus allowance for cover to reinforcement plus half the assumed main steel diameter bar.
The self weight of the slab together with the dead and live load is used to determine the design moment ‘M’.
The value of M must be checked against the value of Mú which is the ultimate moment of resistance
Mú = 0.156fcubd²
Where b=1000mm and fcu = strength of concrete.
If Mú ≥ M then the slab doesn’t need compression reinforcement which is usually the case.
Main reinforcement steel areas
The area of reinforcement Aₛ can be determined using Aₛ= M÷(0.87?yz) where,
M = design moment= WL²/8 (for simply supported slab)
W = design load of slab in kN/m²
L = span of slab in meters.
?y= strength of steel
z = d[0.5 + √ (0.25 – K/0.9)] and
K= M / fcubd²
Secondary reinforcement steel areas
Secondary reinforcement also refers to distribution steel. As per BS 8110, it is calculated as follows;
Aₛ min = 0.24% Ac when ?y = 250N/mm²
Aₛ min = 0.13% Ac when ?y = 460N/mm²
Check that your slab design meets deflection requirements
Design service stress, fs = (2 ?y As req.) ÷ ( (3 As prov), (table 3.10 page 36 of BS 8110 part 1 1997.)
Where As req & As prov is the area of steel calculated and the area of steel provided, respectively.
m.f = 0.55 + (477 – fs) ÷[120 ( 0.9 + M/bd²)] ≤ 2
Now this m.f is your actual m.f since it is a function of the area of steel calculated and area of steel provided.
Use this new m.f to replace the m.f you initially assumed to be 1.4 in the equation;
Effective depth; d = span/( basic ratio × modification factor).
you can also determine m.f from the table below
If the assumed value of d is greater than actual d, then deflection requirements are satisfied.
The BS code specifies that crack width should not exceed 0.3mm. except by calculation, the following rules should ensure crack width requirements are satisfied;
fy = 250N/mm² and depth of slab ≤ 250mm
fy = 460N/mm² and depth of slab ≤ 200mm
the percentage of reinforcement ( 100As/bd) ≤ 0.3%
Maximum bar spacing of reinforcement
BS code specifies that the clear distance between tension bars should be less than 3d or 750mm whichever is the lesser.
Example design of a simply supported solid slab
From the diagram of part of a floor plan shown above, I will be designing the panel labeled A.
The first thing to check is if the slab is a one or two way slab. By observing the dimensions relating to the slab panel, the longer side is
3401 + 587 + 1225 = 5213mm
The shorter side of the panel is
1150 + 1225 = 2375mm
Therefore longer side ÷ shorter side =
5213÷2375= 2.19 ≥ 2;
Panel A is a one way slab.
This means that the main reinforcement bars will span along the y- axis ( shorter side) and the distribution bar will span along the x – axis ( longer side).
Note that in a one way slab, the span of the slab is the shorter side. The span of the slab panel A being designed is 2375mm.
Further observation of the slab panel will show that its thickness (overall depth of slab) has been given as 150mm;
h = 150mm
Your duty as the designer is to calculate the required steel reinforcement and check if the design satisfies deflection requirements. If the slab thickness wasn’t given then you are free to assume a thickness and then check that it works.
Calculating the Design load.
The formula stated by BS code is
1.4Gk + 1.6Qk
where Gk = dead load
Qk = live load
dead load is the self weight of the slab plus finishes
Live load refers to variable or movable loads such as people, furniture etc, the slab carries. Live loads for different categories of buildings are stated in BS 6399 part 1: 1997.
Gk = weight of concrete x slab thickness
= 24kN/m³ x 0.15m ( slab thickness of 150mm)
= 3.6kN/m², plus finishes of say 1.2 kN/m² which gives a total gk of
Gk = 4.8kN/m²
For private dwellings, Qk = 1.5kN/m² (see table 1 of BS 6399 part 1: 1997)
Design load = 1.4gk + 1.6qk
=( 1.4x 4.8 + 1.6 x 1.5) kN/m²
= 9.12kN/m² per m width ( slabs are designed per m width.)
Design moment M = WL²/8
where W = design load and L = span of slab
W=[ 9.12kN/m² x ( 2.375m)² ] / 8
ultimate moment of resistance
Now d = effective depth of slab which can be estimated as overall depth of slab minus concrete cover to reinforcement minus half of main steel diameter
I.e. d = 150mm – 25mm – 6mm = 119mm.
( 25mm is concrete cover to reinforcement and 6mm is half the diameter of 12mm steel rod)
b = 1000mm ( slabs are designed per m width)
fcu = 25N/mm² ( strength of concrete)
Mu = 0.156 x 25N/mm² x 1000mm x (119mm)² = 55.2279 x 10⁶ Nmm
Since M < Mu no compression reinforcement is required.
K= M / fcubd²
= 6.43 x 10⁶/ ( 25 x 1000 x 119²)
z = d[0.5 + √ (0.25 – K/0.9)]
= 119[ 0.5 + ✓ ( 0.25 – 0.0182/0.9)]
= 116.54mm ≤ 0.95d
Limiting to 0.95d
0.95d = 113.05; use z= 113.05
= 6.43 x 10⁶ ÷ ( 0.87 x 460 x 113.05)
Aₛ required. = 142.12 mm²/m width of slab
From fig. 1 shown above, use Y10 bars at 200mm spacings as main steel reinforcement
Aₛ provided = 393 mm²/m
Minimum Steel reinforcement specified by code = 0.13%bh
= 0.13% x 1000 x 150
Aₛ minimum required.= 195 mm²/m
With reference to fig 1, use Y10 – 250 bars. Aₛ = 314mm²/m
Note that the calculated required reinforcement is less than the minimum reinforcement specified by code, in this case the minimum reinforcement specified by code supersedes and should be used to select provided steel. Also using Y10- 200 as main and distribution steel is OK. It’s your design, you are in charge, just make sure you follow the code requirements and design laying emphasis on economy.
Design service stress,
fs = 2 fy As req./ (3 As prov)
fs = 110.9N/mm²
m.f = 0.55 + (477 – fs) ÷[120 ( 0.9 + M/bd²)] ≤ 2
= 0.55 + (477- 110.9)÷[120(0.9 + (6.43 x 10⁶/100 x119²)]
=2.80 ≤ 2
Since m.f is limited to 2,
Hence, m.f = 2
dmin = span / basic ratio x m.f
= 2375mm / (20 x 2)
59.375mm ≤ 119mm hence deflection is satisfied.
Crack width check
Since the slab is less than or equal to 200mm thick, crack width check is satisfied
Maximum spacing check
The reinforcement spacing of 200mm for main steel and 250 mm for distribution steel is less than 3d ( 3 x 119 = 357mm) hence maximum spacing check is satisfied.