Dr Badorul Hisham Abu Bakar cebad@eng.usm.my SOLID SLAB DESIGN Dr Badorul Hisham Abu Bakar cebad@eng.usm.my

SOLID SLAB DESIGN Topics are: Classification of solid slab Analysis and design of one way and two way slabs Torsion reinforcement in slab Check for serviceability limit state Detailing in slab Problems and solutions to slab design

Classification of Solid Slabs The design of solid slab is similar to beam design with the following EXCEPTION: Compression reinforcement is normally not required. Shear reinforcement is usually not applied. A unit breath of 1m is applied in calculation Concrete solid slabs can be classified into two: One way slab Two way slab

One way slab If one of the requirements is complied: Supported at two edges Ratio of longer side to shorter side (Ly/Lx) more than 2 So, consider as ONE WAY SLAB The reinforcement in one way slab is designed in one direction only.

Ly Lx Ly/Lx > 2.0 – one way slab Ly/Lx ≤ 2.0 – two way slab

Figure 1: Slab supported at two edges only

Analysis of One Way Slab The analysis is similar to a simply supported beam with maximum load case of 1.4GK + 1.6QK. For CONTINUOUS ONE WAY SLAB, simplified analysis is recommended in Table 3.12, BS 8110: Part 1: 1997. Requirements also outlined in Clause 3.5.2.3, BS 8110: Part 1: 1997 as follows: Area of each bay less than 30m2 The ratio of the characteristic imposed load to the characteristic dead load does not exceed 1.25 3. Characteristic imposed load, QK not more than 5 kN/m2 excluding partitions

Fig.2 Effective width of solid slab carrying a concentrated load near an supported edge

Method of analysis 3 cases of loading The 3 loads can be analysed similar to continuous beam The bending moment and shear force diagrams for the 3 load cases are then super-imposed together to determine the maximum design bending moments and shear forces at mid-span supports.

Figure 3: Bay and Panel in Slab

Figure 4: 3 Load Cases Arrangement

Two way slab Ratio of longer side to shorter side (Ly /Lx) is less than or equal to 2.0. Refer to Clause 3.5.3, BS 8110: Part 1: 1985. Figure 5 illustrates the 9 types of panel as mentioned in Table 3.15, BS 8110: Part 1: 1985. So, considered as TWO WAY SLAB.

Figure 5: 9 Types of Panels

Analysis of two way slab SIMPLY SUPPORTED UNRESTRAINED SLAB is defined as slabs which do not have sufficient provision to resist torsion at corners and to prevent the corner from lifting. can be refer to Figure 3.6 provides an illustration of above mentioned condition. Required equations and coefficients are given in equations 10 to 13 and Table 3.14, BS 8110, Part 1: 1985.

Figure 6: Two Way Slab Ly /Lx

Figure 7: Illustration of a Simply Supported Unrestrained Slab

SIMPLY SUPPORTED RESTRAINED SLAB where the corners are prevented from lifting and provision for torsion (Figure 8) Required equations and coefficients are found in equations 14 to 15 and Table 3.15: BS 8110, Part 1: 1985.

Figure 8 : Moments and their corresponding directions

Simply Supported Slabs lx nyly nxl x Consider a central strip of unit width of slab in each direction. Let nx and ny be the loads per unit length carried by the strips of length lx and ly, respectively, where lx is the shorter span and the uniformly distributed load on the slab is n = nx + ny. The central deflection of the two strip must be the same. Hence, neglecting the restraint imposed by adjacent part of the slab:

BS 8110 Pt 1 1997

Shear Check Required checking are described in Clause 3.5.5, BS 8110, Part 1: 1997. Design shear stress, v, is less than the design concrete shear, vc, resulting the following expression: Maximum shear can be derived using equations 19 and 20 of BS 8110: Part 1: 1985.Maximum shear in panel is given in Figure 3.8. Related tables for checking are Table 3.9 and 3.17 of BS 8110: Part 1: 1985.

Figure 9 : Maximum Shear Arrangements

Torsion Reinforcement Specified in Clause 3.5.3.5(5), (6) and (7) of BS 8110: Part 1: 1997 Torsion reinforcement provisions are to be as follows (Figure 10): Provide both top and bottom layers. Each of the layers with reinforcements on both span directions. For discontinuous slab in both directions, area of torsion reinforcement in each of the layers should be ¾ of the reinforcement area required at mid span. For discontinuous in either in one direction, the area of torsion reinforcement in each of the layer should be ½ of the reinforcement area required at mid span. Provide torsion reinforcement at the distance of at least Lx/5 from the boundary.

Figure 10: Torsion Reinforcement Provision

BS 8110 Pt 1 1997

Loads on supporting beams Applying two methods Based on empirical formula. Based o equation 19 and 20, and Figure 3.10 of BS 8110: Part 1: 1997.

A B C D 0.75 lx lx ly Vs U.D.L = n A2 A1 Area, A1 = ½ *lx* lx/2 = lx2/4 Load on beam AB = nlx/4 Moment about AB = 1/3* lx/2 *nlx2/4

For the same moment AB, the value of an equivalent distributed load, Vs over a length equal to 0.75 Lx is given by ½*Vs*n*Vs*0.75lx= 1/3* lx/2*nlx2/4 Vs= lx/3 Therefore, for calculation of the load on the supporting beam the loading from the slab may be considered as approximately equivalent to a uniformly distributed load of nlx/3 per unit length on shorter span = βvx nlx… Similarly, nlx/6[3-(lx/ly)2] per unit length on the longer span = βvx nlx For a square slab when lx = ly, the values are given as Βvx = 1/3 = 0.33 Βvy = 1/3 = 0.33

Serviceability limit state checking Checking for deflection is referred to Table 3.10, BS 8110: Part 1:1985 and corresponding modification factor for tension reinforcement are applied. Only the reinforcement at the centre of the span in the width need to be considered. Checking for cracking is to adhere to Clause 3.12.11.2.7. Clear spacing between bars should be less than 3d or 750 mm which ever lesser. If slab thickness > 200mm so, further checking is need according to Clause.

Detailing Should comply to Figure 3.25 of BS 8110: Part 1: 1997.

Summary to solid slab design

Example 1. Design the floor slab for the floor as shown below. All columns have cross-sectional areas of 300 x 300 mm, the main beams and respectively. The weight of the finishes and partitions is 1.5 kN/m2 and the imposed load is 3 kN/m2 of floor area. fcu=30 MPa and fy=410 N/mm2, fire resistance for two hours. All columns 300 x 300 , Main beams 500 x 300, Edge beams 350 x 300 5 @ 5 m 14 m

Example 2. Compute the thickness of the floor slab and reinforcement required for the one-way slab shown below. Check for deflection and cracking at service loading and detail the reinforcement arrangement. Weight for finish and partition = 1.5 kN/m2 of floor area Imposed loading = 3.0 kN/m2 fcu = 25 N/mm2 fy = 250 N/mm2 12 m 6 @ 5 m

Example 3. Design the two way slab with the following data. Characteristic D.L (Finishes & partitions) = 1.5 kN/m2 Characteristic I.L = 4.5 kN/m2 fcu = 25 MPa, fy =250 N/mm2 Assume minimum cover to reinforcement = 20 mm Allow for 12 mm bars and try an overall depth of 150 mm. The reinforcement in the shorter span will be placed in the outer layer and d will be different for the tywo spans. 6 m 5 m