The commonest type of slab used in building construction is the solid slab. It is also the one that easy step in structural design. On special occasions, there are other types of slab used in buildings for one or more of the following reasons:
1. Circulation
2. Size of live load
3. Control of deflection
4. Aesthetics
5. Function
6. Client’s appeal, etc
Among these slabs, we have:
FLAT SLABS
The new European unified code for the design of reinforced concrete element, Eurocode 2 (EC 2) defines flat slab as a slab supported by columns, that is beamless reinforced concrete slab supported directly by columns. They may be of uniform thickness throughout or have deeper thickness (called drops) around the vicinity of the supporting columns (see Figure 1). They are used in floors to buildings and in flat roofs to buildings and reservoirs. The stability of the flat slabs depends on the monolithic interaction between the slab and the supporting column. The column drops are effective in reducing shearing stresses especially where large live loads are involved and because of its enhanced thickness, provide higher moment of resistance for the negative moment occurring at the column areas. Since flat slabs are beamless, they allow both light and air circulation within the premises and offer reduced storey height.
Description and sizing of column head and column drop
Column head
EC2 recommends that the effective dimension of a column head is the lesser of the actual column head or the column dimension plus 2 (dh – 40), where dh is the depth of the head. For flared head, the actual depth is measured 40 mm below the soffit of the slab. The effective diameter of a column or column head is the diameter of a circle whose area equals the cross-sectional area of the column or if column heads are used, the area of the column head based on the effective dimensions.
Column drops
These are effective by the standard if the smaller dimension is at least 1/3 of the panel smaller dimension. Smaller drops are, however, useful when considering shear.
Analysis of flat slabs
The analysis of flat slabs is similar to solid slabs and according to the conditions set out in section 3.5.2.3 of BS 8110. Mathematical analysis of flat slabs is very complex because of the behaviour of such slabs. Empirical methods are usually recommended for design. One such empirical methods is the Hardy Cross or other suitable elastic methods. In the case, the slab may be divided longitudinally and transversely into frames consisting of column and middle strips. The width of the slab should be the full width of the panel (for stiffness calculation) when considering vertical loads and one-half for the horizontal loading. The stiffness should be based on the concrete section alone. Flat slabs panels should be assumed to be divided into column strips and middle strips. for slabs without drop, the column strip is 1/4 of the shorter span on both sides of the column centre-line and the remainder forms the middle strip while with slabs with drops, the full width of the drop forms the column strips provided its dimension is not less than 1/3 of the panel smaller dimensions. If it does, it should be treated as slab without drop.
The analysis of a flat slab structure may be carried out by dividing the structure into a series of equivalent frames. The moments in these frames may be determined by:
- A method of frame analysis using the moment distribution or the stiffness method on a computer (Try Tekla Tedds software).
- A simplified method using the moment and shear coefficients of Table 1 below subject to the following conditions listed below
i. The lateral stability is not dependent on the slab-column connection.
ii. The conditions of Table 1 are satisfied.
iii. There are at least 3 rows of panels of approximately equal span in the direction being considered
iv. The size exceeds 30 m2
Table 1; Distribution of Design Moments in Panels of First Slabs
Unless there are perimeters beams, which are adequately designed for torsion, moments transferred to the edge or corner columns should be limited to the moment of resistance of a rectangular section equal to 0.17 bed2fck (See figure 9.9 for EC 2) for the deflection of be. The positive moment in the end span should be adjusted accordingly.
Design approach of flat slabs
EC 2 does not have any particular section that deals specifically with the analysis and design of flat slabs. But section 9.4 of the code gives general guidance on the reinforcement as well as the definition of effective width of flat slab and both relate to slabs at internal and edge column (Clause 9.4.1 and 9.4.2). Clause 9.4.3 discuss the details of punching shear requirements. In addition, Annex 1 of EC 2 treats in good details the distribution of the moments on both the column and middle strips.
Generally, analysis and design criteria will be based on BS 8110. The code gives recommendation on the dimension of the column head to be used, the need to divide the panel to column and middle strips and general recommendation for shear which is slightly altered by the provisions of EC 2. BS 8110 recommends a minimum thickness of 125 mm for flat slabs.
Example
Design a flat slab with internal column 250 mm in diameter and column heads are flared in cross-section. The external columns are 400 x 400 mm with drops. Design the slab using grade 25 – 500 concrete assuming the column head has a diameter of 1.5 m diameter and bay of 7 m x 7 m.
Solution
Loading: given 200 mm slab thickness and characteristic strength of concrete, γc = 25 kN/m3
Concrete own weight = 0.2 x 25 = 5 kN/m2
Finishes = 1.2 kN/m2
Partition allowance = 0.5 kN/m2
Total permanent action, Gk = 5 + 1.2 + 0.5 = 6.7 kN/m2
Total Gk (kN) = 6.7 x area of bay = 6.7 x 7 x 7 = 328.3 kN
Assuming depth of drop = 100 mm (0.1 m), weight of drop = 0.1 x 25 = 2.5 kN/m2
Assuming dimension of drop = 2.5 x 2.5 m, weight of drop (kN) = 2.5 x 2.5 x 2.5 = 15.625 kN
Total permanent action = 328.3 + 15.625 = 343.925 kN
Assuming variable action, Qk of 2.5 kN/m2, total Qk = 2.5 x 7 x 7 = 122.5 kN
Design action, n = 1.35 Gk + 1.5 Qk = 1.35 x 343.925 + 1.5 x 122.5 = 464.3 + 183.75 = 648.05 kN
Equivalent distributed load = Design action/ area of bay = 648.05/ (7 x 7) = 648.05/49 = 13.23 kN/m2
Table 2; Minimum dimensions and axis distance for flat slabs for fire resistance
The effective span, L = clear span between column heads + ((slab thickness/2) at either ends) = (7 – 1.5) + ((0.2+0.1)/2) x 2) = 5.5 + (0.15 x 2) = 5.5 + 0.3 = 5.8 m
Assuming a 25 mm cover and allowing for two equal 16 mm bars orthogonal to each other, the effective depth is taken as the mean that is,
Support, deff = 300 – 25 – 16 = 259 mm
Span, deff = 200 – 25 – 16 = 159 mm
The drop width is 2.5m and one-third span (7/3) = 2.333m, which is less than the drop size. Therefore, the drop panel width is taken as the column strip width of the drop panel which is 2.5 m.
Table 3; The ultimate bending moment and shear force coefficients in one-way spanning slabs
F = total ultimate action = 1.35 Gk + 1.5 Qk; l = effective span
Area of bay ≥ 30 m2 for the table above to apply
Bending reinforcement
Since the variable action is less than the permanent action and bay size, 7 x 7 = 49 m2 ≥ 30 m2.
From the table 3 above,
Centre of Interior Span
Positive moment = 0.063 Fl = 0.063 x 648.05 x 5.8 = 236.8 kNm
The width of the middle strip = 7 – 2.5 = 4.5 m which is greater than half the panel dimension (3.5 m). Therefore, the properties of this moment taken by the middle strip can be taken as 0.45 from Table 1 above.
The value is adjusted as follows, 0.45 x (4.5/ (7/2)) = 0.45 x (4.5/3.5) = 0.579
Thus, the middle strip positive moment = 0.579 x 236.8 = 137.1072 kNm
The column strip positive moment = (1-0.579) x 236.8 = 99.693 kNm
For the middle strip, k = M/fckbd2 = 137.1072 x 106/ (25 x 4500 x 1592) = 0.0482
Lever arm, la = 0.5 + √0.25 – (k/1.134) = 0.5 + √0.25 – (0.0482/1.134) = 0.956, use 0.95
Z = lad = 0.95 x 159 = 151.05
Asreq = M/0.87fykZ = 137.1072 x 106/ (0.87 x 500 x 151.05) = 2086.6532 mm2
Provide 20 H12 bars (Asprov = 2260 mm2) each way in the span distributed evenly across the 4.5 m width of the middles strip (spacing = 225 mm = maximum allowable for a slab in an area of maximum moment).
If the slab is more than 200 mm thick, the maximum spacing can be taken from the Table 4 below and steel stress calculated from fs = (fyk/1.15) x [(Gk + 0.3Qk)/ (1.35Gk + 1.5Qk)] x (1/δ) x (Asreq./Asprov.)
Where δ = ratio of distributed moment to the undistributed moment. E.g for 20% redistribution, δ = 0.80, else, δ = 1.0 (for no redistribution)
Table 4: Maximum bar spacing (mm) for high bond bars in tension caused predominantly by loading (slab thickness ≥ 200 mm)
For column strip moment, M = 99.693 kNm
K = M/fckbd2 = 99.693 x 106/ (25 x 2500 x 2592) = 0.0238
Lever arm, la = 0.5 + √0.25 – (k/1.134) = 0.5 + √0.25 – (0.0238/1.134) = 0.979, use 0.95
Z = lad = 0.95 x 259 = 246.05
Asreq = M/0.87fykZ = 99.693 x 106/ (0.87 x 500 x 246.05) = 931.4339 mm2
Provide 10 H12 bars (Asprov = 1130 mm2) in the span distributed across the 2.5 m width of the column strip (spacing = 250 mm).
Interior support
Negative moment = -0.063 Fl = -0.063 x 648.05 x 5.8 = 236.8 kN and this can also be divided into
Middle strip = 0.25 x (4.5)/ (7/2) x 236.8 = 0.25 x (4.5/3.5) x 236.8 = 0.3214 x 236.8 = 76.114 kNm
And column strip = (1 – 0.3214) 236.8 = 160.69 kNm
For the middle strip,
K = M/fckbd2 = 76.114 x 106/ (25 x 4500 x 1592) = 0.027
La = la = 0.5 + √0.25 – (k/1.134) = 0.5 + √0.25 – (0.027/1.134) = 0.976, use 0.95
Z = lad = 0.95 x 159 = 151.05
Asreq = M/0.87fykZ = 76.114 x 106/ 0.87 x 500 x 151.05 = 1158.79 mm2
Provide evenly spaced 12 H12 bars as top steel (Asprov = 1356 mm2) to 375 mm maximum spacing
For the column strip
K = M/fckbd2 = 160.69 x 106/ 25 x 2500 x 2592 = 0.0383
la = 0.5 + √0.25 – (k/1.134) = 0.5 + √0.25 – (0.0383/1.134) = 0.965, use 0.95
z = lad = 0.95 x 259 = 246.05
Asreq = M/0.87 fykZ = 160.69 x 106/ 0.87 x 500 x 246.05 = 1501.33 mm2
Provide 15 H12 bars as top steel at 180 mm spacing over the full 2.5m width of the column strip (Asprov = 1695 mm2).
Checks
Punching shear
At the column head,
Perimeter, Uo = π x diameter of column head = π x 1500 = 4712.39 mm
Shear force, VEd = F – (π/4) lh2 n where lh = length of the column head and n = equivalent load = total load divided by the bay area.
Thus, VEd = 648.05 – (π/4) 1.52 x 13.23 = 624.494 kN
To allow for the effects of moment transfer, V is increased by 15% for an internal column, thus
VEd,eff = 1.15 x 624.494 = 718.1681 kN
Maximum permissible shear force, VRd,max = 0.5 Uod [0.6 (1 – (fck/250))] fck/1.5 = 0.5 x 4712.39 x 259 x [0.6 (1 – (25/250) ]25/1.5 x 10-3 = 610254.505 x 0.54 x 37.5 x 10-3 = 12357.654 kN
Thus, VEd,eff is far less than VRd,max which is ok
Note: when the converse is the case, the depth at the column head should be increased by increasing the slab thickness or column head capital or both.
The first critical section for shear is 2.0 x effective depth from the face of the column head, that is, a section of diameter 1.5 + (2 x 2 x 0.259) = 2.54 m (this is outside the drop panel).
Thus, perimeter = U1 = πD = π x 2.5 = 7853.98 mm
From the design steel, ρ1 = Asprov/bwd where Asprov = area of steel at interior support of the column strip, bw = width of the drop and d = depth of the drop.
Thus, ρ1 = 1695/ (2500 x 259) = 0.00262
The shear stress, VRd,c can be calculated from; VRd,c = 0.12 k (100 ρ1 x fck)0.3333 where k = 1 + √(200/d) ≤ 2.0
k = 1 + √ (200/259) = 1.878
thus, VRd,c = 0.12 x 1.878 (100 x 0.00262 x 25)0.3333 = 0.422 N/mm2
and VRd,c = 0.422 x 7853.98 x 259 x 10-3 = 858.424 kN
Since the value is greater than VEd,eff of 718.1681 kN, the section is adequate and shear reinforcement is not required.
At the drop panel, the critical section is 2d where d is the effective depth of the flat slab from the edge of the drop panel with a perimeter given by U = 2a + 2b + 2π x 2d; but 2d = 2 x 159 = 318 mm
U = (2 x 2500 + 2 x 2500 + 2 x π x 318) = 10000 + 1998.053 = 11998.053 mm
The area within this perimeter can be calculated from, Actr = (a + 3d)2 – (4 – π) (2d)2 = [(2.5 + 3(0.159)]2 – (4-π) (2 x 0.159)2 = 8.863 – 0.0868 = 8.7762 m2
Ultimate shear force, VEd = 648.05 – (8.772 x 13.23) = 531.062 kN
VEd,eff = 1.15 x 531.063 = 610.723 kN
ρ1 = (1695)/ (2500 x 159) = 0.0043
k = 1 + √ (200/d) = 1 + √ (200/159) = 2.26 ˃ 2.0, use 2.0
Thus, VRd,c = 0.12 x 2 (0.0043 x 100 x 25)0.3333 = 0.53 N/mm2, and
VRd,c = vRd,c x u x d = 0.53 x 11998.053 x 159 x 10-3 = 1011.08 kN ˃ VEd,eff.
Thus, the section is very adequate
RIBBED SLABS
Ribbed and hollow block floor are economical for buildings where there are long spans of 5 m and above and light or moderate variable actions, such as a hospital wards or equivalent building. It is suitable for heavy loading such as warehouse and garages. Ribbed floor slabs (see Figure 2) are usually made solid near supports, under partitions and concentrated loads in order to achieve greater shear strength.
Structural Analysis
Moment: where practicable, the slab may be analysed using the continuous spanning method. However, if it is impracticable to provide sufficient reinforcement to develop the full design support moment, the slabs may be designed as series of simply supported spans. Where the slabs are ribbed in two directions, they may be designed as two-way spanning or as flat slabs, whichever is more appropriate. This is the design method adopted for waffle floor.
Shear: a rib behaves like a beam and hence is designed as such. The shear force is calculated and the shear resistance checked as below:
The calculation of the reinforcement requires treatment of the section as a T-beam and the effective flange width should calculated as it is done for T-beams. But here, the effective flange width is centre to centre of the ribs. The slab should be designed for shear and the width of the rib should be taken as the bw. The span-effective depth ratio will be based on the shorter span with the basic values taken from (Figure 4.23 of Oyenuga) multiplied by 0.8 where the ratio of the flange width to the rib width exceeds 3. No specific guidance on this is given in EC 2. Where ribs are supported by steel beams, top reinforcements in the region of 25% of the span reinforcements should be provided and extend a distance not less than 15% of the clear span into adjoining span. A light reinforcing mesh in the topping flange can give added strength and durability to the slab, particularly if there are concentrated or moving loads or if cracking is envisaged, an area of mesh equivalent to 0.13% of the topping flange would be adequate
No shear reinforcement is required when VRd,c ˃ VEd. The calculation of both VRd,c and VEd are duly explained in the flat slab design.
Example
Design a ribbed floor slab for a room space 6m x 6m. design the slab using an in-fill burnt bricks of 350 x 225 x 200 mm high. Take topping slab as 50 mm. use 25 – 500 concrete.
Solution
The width of the block is 350 mm and assuming a rib width of 150 mm, total centre to centre of the ribs = 500 mm.
Total depth of rib = 200 + 50 mm (topping) = 250 mm.
Loading (per rib of 500 mm c/c)
Topping = 0.05 x 0.5 x 25 = 0.625 kN/m
Ribs = 0.2 x 0.15 x 25 = 0.75 kN/m
Finishes = 1.2 x 0.5 = 0.6 kN/m
Partition allowance = 1 x 0.5 = 0.5 kN/m
Block, say = 1.0 kN/m
Total permanent action, Gk = 3.475 kN/m
Assuming a variable action of 2.5 kN/m2, Gk = 0.5 x 2.5 = 1.25 kN/m
Ultimate design action, n = 1.35 Gk + 1.5 Qk = 1.35 (3.475) + 1.5 (1.25) = 4.692 + 1.875 = 6.567 kN/m
Span = 6 + 0.225 = 6.225 m
Assuming concrete cover of 20 mm, and bar diameter of 10 mm
Effective depth = 250 – 20 – 10 = 220 mm
Breadth of rib, bw = 150 mm
Moment = wl2/8 = (6.567 x 6.2252)/8 = 31.809 kNm
Effective flange width
beff = bw + Ʃ beff,i
Where beff,i = 0.2 bi + 0.1 lo ≤ 0.2 lo; also beff,i ≥ bi
bi = the clear distance between the webs of adjacent beams.
lo = the distance between the points of zero moments along the beam. For continuous beams, lo = 0.85l for the external support and lo = 0.70l for the inner support. For simply supported beam, lo = l and for cantilever, lo = 0.15 l2 + l1, where l2 = the span before the cantilever and l1 = cantilever span.
In this case, bw = 0.15m and b1 = half the clear distance between adjacent ribs; b1 = (500 -150)/2 = 350/2 = 175 mm; lo = 1.0 x 6000 = 6000 mm.
» beff,1 = beff,2 since the ribs are regularly spaced; beff,1 = beff,2 = 0.2 bi + 0.1; lo ≤ 0.2; lo ≤ b1
beff,1 = beff,2 = 0.2 x 175 + 0.1 x 6000 = 35 + 600 = 635
lo = 0.1 x 6000 = 600 mm
b1 = 175 mm
beff = bw + beff,1 + beff,2 = 150 + 2(0.2 x 175 + 0.1 x 6000) = 150 + 2(35 + 600) = 150 + 1270 = 1420 mm or 0.2 x 6000 = 1200 mm which both exceed the rib spacing of 500 mm, which governs k.
K = M/fckbd2 = (31.809 x 106)/ 25 x 500 x 2202 = 0.0526
la = 0.5 + √0.25 – k/1.134 = 0.951, 0.95
Z = lad = 0.95 x 220 = 209 mm
Asreq = M/0.87fyk Z = (31.809 x 106)/ 0.87 x 500 x 209 = 350 mm2
Provide 2 H16 bars (Asprov = 402 mm2) bottom
At the end interior support design as a rectangular section for the solid slab
M = 31.809 kNm, Asreq = 350 mm2
Provide 2 H16 bars in each 0.5m width of slab, Asprov = 402 mm2
Span – effective depth ratio
At the centre of the span, ρ = 100 Asreq/bd = (100 x 350)/ 500 x 220 = 0.3182 %
The limiting span-effective depth ratio is 28 and k-value = 1.0. for a flange width greater than 3 times the web (500 ˃ (3 x 150) or 450). This should be multiplied by 0.8, that is, the limiting span-effective depth ratio is 0.8 x 28 = 22.4 and actual = 6225/220 = 28.29. Deflection is not satisfied in this case. Increasing the top to 75 mm, we have, d = 275 – 20 – 10 = 245 mm.
100 Asreq/bd = 100 (350)/ 500 (245) = 0.2857
The limiting span – effective depth ratio will be in the region of 32. Thus, using the 0.8 factor, we have 0.8 x 32 = 25.6
Actual span – effective depth ratio = 6225/245 = 25.4 mm
Thus, the overall depth of 225 mm is adequate.
Note: Ribbed floor slab can also be designed for continuous slab which is more practical and common and often involves the use of moment coefficients.
Check for shear
Maximum shear in the rib would occur at 0.6 m from the support centerline (end span)
VEd = 0.5 F – 0.6 x 6.567 = 0.5 x 39.402 – 0.6 x 6.567 = 19.701 – 3.9402 = 15.761 kN; ρ1 = Asprov./bd = 402/ (150 x 220) = 0.0122. Thus, vRd,c = 0.522 N/mm2
VRd,c = vRd,c x bd = 0.522 x 150 x 220 = 17.23 kN
Since VRd,c is greater than VEd, then no shear reinforcement is required provided that the bars in the ribs are securely located during the construction.
Construction principles
During construction, the hollow tiles should be well soaked in water prior to placing the concrete, otherwise shrinkage cracking of the top concrete flange is liable to occur.
The thickness of the concrete flange should not be less than;
i. 40 mm or 1/10 of the clear distance between ribs, whichever is the greater for slabs with permanent blocks.
ii. 50 m or 1/10 of the clear distance between ribs, whichever is the greater, for slabs without permanent blocks.
In case these requirements are not met, then a check of longitudinal shear between web and flange should be made to see if additional transverse steel is needed.
The ribs should be spaced no further apart than 1.5 m and their depth below the flange should not be greater than 4 times their width. Transverse ribs should be provided at spacing no greater than ten (10) times the overall slab depth.
WAFFLE SLAB
The waffle slab is an extension of the ribbed slab in which the slab is ribbed in two directions. Hence, an inverted pot-like hollow if formed which serve as the ceiling for the floor below.
Waffle slab (see Figure 3) is all concrete and slab thickness may be up to 500 mm. Hence, they are expensive and can be used with large spans carrying heavy variable action at least 5.0 kN/m2 or more. There are usually two type of waffle slab: slab and beam waffle; mushroom waffle. In the slab and beam waffle, the ribs terminate into an end beam of the same thickness or deeper. In the case of mushroom waffle, the end beams are totally absent. However, the ribs terminate into a solid slab ‘capital’ around the columns.
Design approach for waffle slab
The design of waffle slab is two-way spanning because the spans are usually very large. Prior to the design, it is necessary to obtain the proposed mould properties which include the various sizes, depths and volume of voids per mould. These are usually presented in tabular form by the mould manufacturers and are required for the design. The coefficient of two-way spanning slabs presented in the code can be used for the design. Bottom reinforcement can be placed within the ribs while top reinforcements are either placed within the ribs or spread as in solid slabs. In mushroom waffle slab, the column capital must be checked for punching shear as in flat slabs.
Example
Design a waffle slab with rectangular grid of 6 m x 6 m as shown on your drawing. The live load on the floor is 5 kN/m2 and the mould is 900 mm and 500 mm thick slab. The live load on the floor is 5 kN/m2 and the mould is 900 mm and 500 mm thick. Use grade 25 – 500 concrete and column size of 400 x 400 mm.
The design approach of waffle is not very far removed from the approach used for ribbed slabs. However, you can find an article here on detailed design of waffle slab. You can use the processes provide to design the example given above.
For Further reading, consult
Simplified Reinforced Concrete Design (Consultant’s approach to Eurocode 2), third edition by Engr Dr V.O. Oyenuga.
