Cost is a major controlling factor in civil engineering construction projects. Different types of floor systems are adopted in reinforced concrete slab designs such as solid slabs, waffle slabs, ribbed slabs, flat slabs, etc. Each floor system has its advantages, applications, and cost implications in construction. This article aims to evaluate the cost comparison of solid and ribbed slabs.

The idea behind the adoption of the ribbed slab system is the need to reduce the volume of concrete in the tension zone of a concrete slab. Theoretically, the tensile strength of concrete is assumed to be zero during the structural design of flexural structural elements such as beams and slabs. By implication, all the tensile stresses from bending are assumed to be resisted by the reinforcements (see the stress block of flexural concrete sections in Figure 1).

If concrete is assumed to do no work in the tension zone, it then makes economical sense to reduce the volume of concrete in that zone (bottom of the slab). To achieve this, beams of relatively shallow depths (ribs) are spaced at intervals to resist the flexural stresses due to the bending of the floor, with a thin topping (of about 50 mm). When this is done, a thick volume of concrete is no longer uniformly provided in the tensile zone of the concrete.

As the span of a floor increases, the thickness of the concrete and the quantity of reinforcements required to satisfy ultimate and serviceability limit state requirements also increase. However, by introducing ribbed slabs, longer spans can be economically spanned.

To reduce the cost and labour of constructing ribbed slabs, clay hollow pots, sandcrete blocks, or polystyrene are usually provided as infills between the ribs. While clay hollow pots and sandcrete blocks will improve the stiffness of the floor, the same, however, cannot be confidently said about polystyrenes.

There are significant cost implications of adopting either solid or ribbed slabs. In a study by Ajema and Abeyo (2018), they observed that frames with solid slabs are more economical than frames with a ribbed slab when subjected to seismic action. In another study by Nassar and Al-Qasem (2020) on the cost of different slab systems, they observed that the flat slab system reduces the total cost of construction by 7% compared to the solid slab system, 4 % compared to the one-way ribbed slab system, and 3.33% compared to the two-way ribbed slab system.

Mashri et al (2020) compared the cost of constructing solid slabs and hollow block ribbed slabs and concluded that ribbed slabs are cheaper than solid slabs. In a study on the assessment of the cost difference between solid and hollow floors, Dosumu and Adenuga (2013) observed that the cost of in-situ solid slabs are higher than that of hollow slab provided the hollow slab is a one-way hollow floor and not a waffle floor.

However, it is important to note that the scenario on the issue of cost can vary depending on the size and geometry of the slab. For short one-way slabs, solid slabs may be cheaper than ribbed slabs, while in large-span two-way slabs, ribbed slabs may be cheaper than solid slabs.

With the aid of design examples and quantity estimation, let us compare the cost of constructing a two-way slab of dimensions 5.225m x 7.426m that is discontinuous at all edges using ribbed slab and solid slab. The slab is to support a live load of 2.5 kN/m^{2}.

**Design of Ribbed Slab**

**Reinforcement details**

Characteristic yield strength of reinforcement; f_{yk} = 410 N/mm^{2}

Partial factor for reinforcing steel – Table 2.1N; γ_{S} = 1.15

Design yield strength of reinforcement; f_{yd} = f_{yk}/γ_{S} = 357 N/mm^{2}

**Concrete details **

Concrete strength class; C25/30

Aggregate type; Quartzite

Aggregate adjustment factor – cl.3.1.3(2); AAF = 1.0

Characteristic compressive cylinder strength; f_{ck} = 25 N/mm^{2}

Mean value of compressive cylinder strength; f_{cm} = f_{ck} + 8 N/mm^{2} = 33 N/mm^{2}

Mean value of axial tensile strength; f_{ctm} = 0.3 N/mm^{2} × (f_{ck})^{2/3} = 2.6 N/mm^{2}

Secant modulus of elasticity of concrete; E_{cm} = 22 kN/mm^{2} × (f_{cm}/10)^{0.3} × AAF = 31476 N/mm^{2}

**Design Information**

Spacing of ribs = 450 mm

Topping = 50 mm

Rib width = 150 mm

Span = 5.225 m (simply supported)

Total depth of slab = 250 mm

Design live load q_{k} = 2.5 kN/m^{2}Weight of finishes = 1.2 kN/m^{2}

Partition allowance = 1.5 kN/m^{2}

**Load AnalysisDead Load**Self-weight of topping = 24 × 0.05 × 0.45 = 0.54 kN/m

Self-weight of ribs = 24 × 0.15 × 0.2 = 0.72 kN/m

Weight of finishes = 1.2 × 0.45 = 0.54 kN/m

Partition allowance = 1.5 × 0.45 = 0.675 kN/m

Self-weight of heavy duty EPS = (0.156 × 0.2 × 0.45) = 0.014 kN/m

Total dead load per rib g

_{k}

**= 2.489 kN/m**

**Live Load**Characteristic live load = 2.5 kN/m

^{2}

Total live load per rib = 2.5 × 0.45 = 1.125 kN/m

**At ultimate limit state;**

P_{Ed} = 1.35g_{k} + 1.5q_{k} = 1.35(2.489) + 1.5(1.125) = 5.047 kN/m

**Flexural Design**

Design bending moment; M_{Ed} = 17.2 kNm

Effective flange width; b_{eff} = 2 × b_{eff,1} + b = 450 mm

Effective depth of tension reinforcement; d = 209 mm

K = M / (b_{eff} × d^{2} × f_{ck}) = 0.035

K’ = 0.207

Lever arm; z = 199 mm

Depth of neutral axis; x = 2 × (d – z) / l = 26 mm

lx <= h_{f} – Compression block wholly within the depth of flangeK’ > K – No compression reinforcement is required

Area of tension reinforcement required; A_{s,req} = M / (f_{yd} × z) = 243 mm^{2}

Tension reinforcement provided; 2Y16 (A_{s,prov} = 402 mm^{2})

Minimum area of reinforcement – exp.9.1N; A_{s,min} = max(0.26 × f_{ctm} / f_{yk}, 0.0013) × b × d = 51 mm^{2}

Maximum area of reinforcement – cl.9.2.1.1(3); A_{s,max} = 0.04 × b × h = 1500 mm^{2}

PASS – Area of reinforcement provided is greater than area of reinforcement required

**Deflection control**

Reference reinforcement ratio; ρ_{m0} = (f_{ck} )^{0.5} / 1000 = 0.00500

Required tension reinforcement ratio; ρ_{m} = A_{s,req} / (b_{eff} × d) = 0.00259

Required compression reinforcement ratio; ρ’_{m} = A_{s2,req} / (b_{eff} × d) = 0.00000

Structural system factor – Table 7.4N; K_{b} = 1.0

Basic allowable span to depth ratio ; span_to_depth_{basic} = K_{b} × [11 + 1.5 × (f_{ck})^{0.5} × ρ_{m0} / ρ_{m} + 3.2 × (f_{ck})^{0.5} × (ρ_{m0} / ρ_{m} – 1)^{1.5}] = 39.900

Reinforcement factor – exp.7.17; K_{s} = min(A_{s,prov} / A_{s,req} × 500 N/mm^{2} / f_{yk}, 1.5) = 1.500

Flange width factor; F1 = if(b_{eff} / b > 3, 0.8, 1) = 1.000

Long span supporting brittle partition factor; F2 = 1.000

Allowable span to depth ratio; span_to_depth_{allow} = min(span_to_depth_{basic} × K_{s} × F1 × F2, 40 × K_{b}) = 40.000

Actual span to depth ratio; span_to_depth_{actual} = L_{m1_s1} / d = 25.000*PASS* – Actual span to depth ratio is within the allowable limit

**Shear Design**

Angle of comp. shear strut for maximum shear; θ_{max} = 45 deg

Strength reduction factor – cl.6.2.3(3); v_{1} = 0.6 × (1 – f_{ck} / 250 N/mm^{2}) = 0.540

Compression chord coefficient – cl.6.2.3(3); a_{cw} = 1.00

Minimum area of shear reinforcement – exp.9.5N; A_{sv,min} = 0.08 N/mm^{2} × b × (f_{ck} )^{0.5} / f_{yk} = 146 mm^{2}/m

Design shear force at support ; V_{Ed,max} = 13 kN

Min lever arm in shear zone; z = 199 mm

Maximum design shear resistance – exp.6.9; V_{Rd,max} = a_{cw} × b × z × v_{1} × f_{cwd} / (cot(θ_{max}) + tan(θ_{max})) = 134 kN

PASS – Design shear force at support is less than maximum design shear resistance

Design shear force at 209 mm from support; V_{Ed} = 12 kN

Design shear stress; v_{Ed} = V_{Ed} / (b × z) = 0.407 N/mm^{2}

Area of shear reinforcement required – exp.6.8; A_{sv,des} = v_{Ed} × b / (f_{yd} × cot(θ)) = 69 mm^{2}/m

Area of shear reinforcement required; A_{sv,req} = max(A_{sv,min}, A_{sv,des}) = 146 mm^{2}/m

Shear reinforcement provided; 2Y 8 legs @ 150 c/c

Area of shear reinforcement provided; A_{sv,prov} = 670 mm^{2}/m

PASS – Area of shear reinforcement provided exceeds minimum required

Maximum longitudinal spacing – exp.9.6N; s_{vl,max} = 0.75 × d = 157 mm*PASS* – Longitudinal spacing of shear reinforcement provided is less than maximum

However, calculations have shown that shear reinforcements are not required since V_{Ed} (13 kN) is less than V_{Rdc} (25.2 kN). According to clause 6.2.1(4) of EN 1992-1-1:2004, when, on the basis of the design shear calculation, no shear reinforcement is required, minimum shear reinforcement should nevertheless be provided according to clause 9.2.2. The minimum shear reinforcement may be omitted in members such as slabs (solid, ribbed or hollow core slabs) where transverse redistribution of loads is possible.

Therefore to save cost, let us provide a triangular pattern link spaced at 300mm c/c, and 1Y12 (A_{sprov} = 113 mm^{2}) at the top of the rib.

**Design of Solid Slab**

RC slab design

In accordance with EN1992-1-1:2004 incorporating corrigendum January 2008 and the UK national annex

**Slab definition **

Type of slab; Two way spanning with restrained edges

Overall slab depth; h = 200 mm

Shorter effective span of panel; l_{x} = 5225 mm

Longer effective span of panel; l_{y} = 7426 mm

Support conditions; Four edges discontinuous

**Loading**

Characteristic permanent action; G_{k} = 6.5 kN/m^{2}

Characteristic variable action; Q_{k} = 2.5 kN/m^{2}

Partial factor for permanent action; γ_{G} = 1.35

Partial factor for variable action; γ_{Q} = 1.50

Quasi-permanent value of variable action; ψ_{2} = 0.30

Design ultimate load; q = γ_{G} × G_{k} + γ_{Q} × Q_{k} = 12.5 kN/m^{2}

Quasi-permanent load; q_{SLS} = 1.0 × G_{k} + ψ_{2} × Q_{k} = 7.2 kN/m^{2}

**Concrete properties**

Concrete strength class; C25/30

Characteristic cylinder strength; f_{ck} = 25 N/mm^{2}

Partial factor (Table 2.1N); γ_{C} = 1.50

Compressive strength factor (cl. 3.1.6); a_{cc} = 0.85

Design compressive strength (cl. 3.1.6); f_{cd} = 14.2 N/mm^{2}

Mean axial tensile strength (Table 3.1); f_{ctm} = 0.30 N/mm^{2} × (f_{ck} )^{2/3} = 2.6 N/mm^{2}

Maximum aggregate size; d_{g} = 20 mm

**Reinforcement properties**

Characteristic yield strength; f_{yk} = 410 N/mm^{2}

Partial factor (Table 2.1N); γ_{S} = 1.15

Design yield strength (fig. 3.8);f_{yd} = f_{yk} / γ_{S} = 356.5 N/mm^{2}

**Concrete cover to reinforcement**

Nominal cover to outer bottom reinforcement; c_{nom_b} = 25 mm

Fire resistance period to bottom of slab; R_{btm} = 60 min

Axial distance to bottom reinft (Table 5.8); a_{fi_b} = 10 mm

Min. btm cover requirement with regard to bond; c_{min,b_b} = 12 mm

Reinforcement fabrication; Not subject to QA system

Cover allowance for deviation; Dc_{dev} = 10 mm

Min. required nominal cover to bottom reinft; c_{nom_b_min} = 22.0 mm*PASS* – There is sufficient cover to the bottom reinforcement

**Reinforcement design at midspan in short span direction (cl.6.1)**

Bending moment coefficient; b_{sx_p} = 0.0881

Design bending moment; M_{x_p} = b_{sx_p} × q × l_{x}^{2} = 29.9 kNm/m

Reinforcement provided; 12 mm dia. bars at 200 mm centres

Area provided; A_{sx_p} = 565 mm^{2}/m

Effective depth to tension reinforcement; d_{x_p} = h – c_{nom_b} – f_{x_p} / 2 = 169.0 mm

K factor; K = M_{x_p} / (b × d_{x_p}^{2} × f_{ck}) = 0.042

Redistribution ratio; δ = 1.0

K’ factor; K’ = 0.598 × d – 0.18 × d^{2} – 0.21 = 0.208

K < K’ – Compression reinforcement is not required

Lever arm; z = min(0.95 × d_{x_p}, d_{x_p}/2 × (1 + √(1 – 3.53 × K))) = 160.5 mm

Area of reinforcement required for bending; A_{sx_p_m} = M_{x_p} / (f_{yd} × z) = 523 mm^{2}/m

Minimum area of reinforcement required; A_{sx_p_min} = max(0.26 × (f_{ctm}/f_{yk}) × b × d_{x_p}, 0.0013 × b × d_{x_p}) = 275 mm^{2}/m

Area of reinforcement required; A_{sx_p_req} = max(A_{sx_p_m}, A_{sx_p_min}) = 523 mm^{2}/m

**Check reinforcement spacing**

Reinforcement service stress; s_{sx_p} = (f_{yk} / g_{S}) × min((A_{sx_p_m}/A_{sx_p}), 1.0) × q_{SLS} / q = 190.7 N/mm^{2}

Maximum allowable spacing (Table 7.3N); s_{max_x_p} = 262 mm

Actual bar spacing; s_{x_p} = 200 mm*PASS* – The reinforcement spacing is acceptable

**Reinforcement design at midspan in long span direction (cl.6.1)**

Bending moment coefficient; b_{sy_p} = 0.0560

Design bending moment; M_{y_p} = b_{sy_p} × q × l_{x}^{2} = 19.0 kNm/m

Reinforcement provided; 12 mm dia. bars at 250 mm centres

Area provided; A_{sy_p} = 452 mm^{2}/m

Effective depth to tension reinforcement; d_{y_p} = h – c_{nom_b} – f_{x_p} – f_{y_p} / 2 = 157.0 mm

K factor; K = M_{y_p} / (b × d_{y_p}^{2} × f_{ck}) = 0.031

Redistribution ratio; d = 1.0

K’ factor; K’ = 0.598 × d – 0.18 × d^{2} – 0.21 = 0.208

K < K’ – Compression reinforcement is not required

Lever arm; z = min(0.95 × d_{y_p}, d_{y_p}/2 × (1 + √(1 – 3.53 × K))) = 149.2 mm

Area of reinforcement required for bending; A_{sy_p_m} = M_{y_p} / (f_{yd} × z) = 358 mm^{2}/m

Minimum area of reinforcement required; A_{sy_p_min} = max(0.26 × (f_{ctm}/f_{yk}) × b × d_{y_p}, 0.0013 × b × d_{y_p}) = 255 mm^{2}/m

Area of reinforcement required; A_{sy_p_req} = max(A_{sy_p_m}, A_{sy_p_min}) = 358 mm^{2}/m

PASS – Area of reinforcement provided exceeds area required

**Check reinforcement spacing**

Reinforcement service stress; s_{sy_p} = (f_{yk} / γ_{S}) × min((A_{sy_p_m}/A_{sy_p}), 1.0) × q_{SLS} / q = 163.1 N/mm^{2}

Maximum allowable spacing (Table 7.3N); s_{max_y_p} = 296 mm

Actual bar spacing; s_{y_p} = 250 mm*PASS* – The reinforcement spacing is acceptable

**Basic span-to-depth deflection ratio check (cl. 7.4.2)**

Reference reinforcement ratio; ρ_{0} = (f_{ck})^{0.5} / 1000 = 0.0050

Required tension reinforcement ratio; ρ = max(0.0035, A_{sx_p_req} / (b × d_{x_p})) = 0.0035

Required compression reinforcement ratio; ρ’ = A_{scx_p_req} / (b × d_{x_p}) = 0.0000

Structural system factor (Table 7.4N); K_{d} = 1.0

Basic limit span-to-depth ratio (Exp. 7.16);

ratio_{lim_x_bas} = K_{d} × [11 +1.5 × (f_{ck})^{0.5} × ρ_{0}/ρ + 3.2 × (f_{ck})^{0.5} × (ρ_{0}/ρ -1)^{1.5}] = 26.20

Mod span-to-depth ratio limit;

ratio_{lim_x} = min(40 × K_{d}, min(1.5, (500 N/mm^{2}/ f_{yk}) × (A_{sx_p} / A_{sx_p_m})) × ratio_{lim_x_bas}) = 34.54

Actual span-to-eff. depth ratio; ratio_{act_x} = l_{x} / d_{x_p} = 30.92*PASS* – Actual span-to-effective depth ratio is acceptable

**Reinforcement summary**

Midspan in short span direction; **12 mm dia. bars at 200 mm centres B1**

Midspan in long span direction; **12 mm dia. bars at 250 mm centres B2**

Discontinuous support in short span direction; **12 mm dia. bars at 200 mm centres B1**

Discontinuous support in long span direction; **12 mm dia. bars at 250 mm centres B2**

**Cost Comparison of Solid and Ribbed Slabs**

In this section, we are going to consider the cost of constructing ribbed slab and the cost of constructing solid slabs (considering the cost of materials only). In this case, the cost of labour is assumed to be directly proportional to the quantity of materials.

**Cost analysis of solid slab**

**Concrete**

Volume of concrete required for the slab = 5.45 × 7.65 × 0.2 = 8.3385 m^{3}

Current unit cost of concrete materials = ₦55,000/m^{3}

Cost of concrete materials = 8.3385 × 55000 = ₦ 458,618

**Quantity of steel required**

Bar mark 1 = 37 × 7.645 × 0.888 = 251.184 kg

Bar mark 2 = 21 × 10.485 × 0.888 = 195.524 kg

Bar mark 3 = 12 × 4.225× 0.617 = 31.28 kg

Bar mark 3 = 8 × 6.625 × 0.617 = 32.701kg

Total = 510.689 kg

Unit cost of reinforcement = ₦ 450/kg

Cost of reinforcement = 510.689 × 450 = ₦ 229,810

**Formwork Required **

Soffit of slab = 36 m^{2} (treated as a constant)

Cost of constructing solid slab = ₦ 229,810 + ₦ 458,618 = ₦ 688,428

**Cost Analysis of Ribbed Slab**

**Concrete**

Volume of concrete required for the topping = 5.45 × 7.65 × 0.05 = 2.084 m^{3}

Volume of concrete required for the ribs = 12 × 0.2 × 0.15 × 5 = 1.8 m^{3}

Total volume of concrete = 3.884 m^{3}

Cost of concrete materials = 3.884 × 55,000 = ₦ 213,620

**Quantity of steel required**

Bar mark 1 = 2 × 12 × 5.7 × 1.579 = 216 kg

Bar mark 2 = 1 × 12 × 5.8 × 0.888 = 61.8 kg

Bar mark 3 (triangular links) = 17 × 12 × 0.612 × 0.395 = 49.314 kg

Total = 327.114 kg

Cost of reinforcement = 327.114 × 450 = ₦ 147,205

BRC Mesh for Topping (A142) – 41.7 m^{2}

Unit cost of BRC mesh = ₦ 1580/m^{2}

Cost of BRC mesh = 41.7 × 1580 = ₦ 65,886

Total cost of reinforcement works = 147,205 + 65,886 = ₦ 213,620

**Clay hollow pot/Sandcrete Blocks/Polystyrene**

Number of block units required = 240 units

Unit price of hollow blocks = NGN 400 per unit

Cost of hollow blocks = 240 × 400 = ₦ 96,000

**Formwork Required **

Soffit of slab = 36 m^{2} (treated as a constant)

Cost of constructing ribbed slab = ₦ 213,620 + ₦ 213,091 + ₦ 96,000 = ₦ 522,711

The Table for comparison is shown below;

Material | Solid Slab | Ribbed Slab | Percentage Reduction |

Concrete | ₦ 458,618 | ₦ 213,620 | 53.42% |

Reinforcement | ₦ 229,810 | ₦ 213,620 | 7.04% |

Hollow Pots/Blocks | – | ₦ 96,000 | – |

Total | ₦ 688,428 | ₦ 522,711 | 24.07% |

**Conclusion**

From the analysis of the two-way slab carried out, it can be seen that the volume of concrete required for a ribbed slab is 53.42% less than that required for a solid slab. Furthermore, the reinforcement required in the ribbed slab is 7.04% less than that required for a solid slab. If the same type of formwork is adopted (completely flat soffit supported with props), and if the cost of labour is directly related to the quantity of materials, then the adoption of the ribbed slab is expected to save cost by about 24.07% compared to solid slab.

**References**

[1] Ajema D. and Abeyo A. (2018): Cost Comparison between Frames with Solid Slab and Ribbed Slab using HCB under Seismic Loading. *International Research Journal of Engineering and Technology* 05(01):109-116

[2] Dosumu O. S. and Adenuga O. A.(2013): Assessment of Cost Variation in Solid and Hollow Floor Construction in Lagos State. *Journal of Design and Built Environment* 13(1):1-11

[3] Mashri M., Al-Ghosni K., Abdulrahman A., Ismaeil M., Abdussalam A. and Elbasir O. M. M. (2020):Design and cost comparison of the Solid Slabs and Hollow Block Slabs.* GSJ* 8(1):110-118 www.globalscientificjournal.com

[4] Reema R. Nassar 1, Imad A. Al-Qasem 2 (2020): Comparative Cost Study for A residential Building Using Different Types of Floor System. *International Journal of Engineering Research and Technology* 13(8): 1983-1991

due to the weight reduction, the net saving by using ribbed slab will be more than 24.07% due to the reduction in sizes of column, beam and foundation members.