By :Dr. Aeid A. Abdulrazeg Structural Design I Course Code: CIVL312 Design of solid slabs By :Dr. Aeid A. Abdulrazeg Dr. Aeid A. Abdulrazeg

Design of solid slabs

Types of Slab Slabs are plate elements forming floors and roofs in buildings which normally carry uniformly distributed loads. Slabs maybe simply supported or continuous over one or more supports and are classified according to the method of support as follows: Spanning one way between beams or walls Spanning two ways between the support beams or walls Flat slabs carried on columns and edge beams or walls with no interior beams Slabs may be solid of uniform thickness or ribbed with ribs running in one or two directions. Slabs with varying depth are generally not used.

Solid Slab

Flat slab without drop panel

Flat slab with drop panel

Waffle slab

Pre-Cast slab

Analysis of Slab Slabs may be analysed using the following methods. Elastic analysis covers three techniques: (a)Idealization into strips or beams spanning one way or a grid with the strips spanning two ways (b)Elastic plate analysis (c)Finite element analysis—the best method for irregularly shaped slabs or slabs with non-uniform loads •For the method of design coefficients use is made of the moment and shear coefficients given in the code, which have been obtained from yield line analysis. •The yield line and Hillerborg strip methods are limit design or collapse loads methods

Design Procedure RC slab behave primarily as flexural members with the design similar to that for beams. The breadth of the slab is already fixed and a unit breadth of 1 m is used in the calculations. The compression reinforcement are often not required The shear stress are usually low except when there are heavy concentrated load.

A design procedure for carrying out detail design of slab may be list out as follows Step Task Standard 1 Select size of thickness of the slab. Clause 3.3.6 2 Calculate min cover for durability fire and bond requirements. Clause 3.3.7 3 Analysis structure to obtain critical moment and shear forces Clause 3.5 4 Design flexural reinforcement. Clause 3.4.4 5 Check Shear Clause 3.5.5 6 Check deflection Clause 3.4.6 7 Detailing

Simplified Analysis BS8110 permits the use of a simplified load arrangement for all slabs of maximum ultimate design load throughout all spans or panels provided that the following conditions are met: in a one-way spanning slab, the area of each bay exceeds 30 m2, In this context, a bay means a strip across the full width of a structure bounded on the other two sides by lines of support as shown in Figure. the ratio of the characteristic imposed load to the characteristic dead load does not exceed 1.25, the characteristic imposed load does not exceed 5 kN/m2 excluding partitions

Fig. Slab definitions

Shear in Slab Experimental work has indicated that, compared with beams, shallow slabs fail at slightly higher stresses and this incorporated into the values of design ultimate shear stress vc given in table (BS8110). The shear stress at a section in a solid is given by : V is the shear force due to the ultimate load, d effective depth of the slab and b is the width of the section considered. The code requires that for a solid slab: v can not be greater than the lesser of v can not be greater than vc for slab thickness less than 200 mm. If v > vc , shear reinforcement is required

Punching Shear in Slab Punching shear is a type of failure of reinforced concrete slabs subjected to high localized force. A concentrated load (N) on a slab causes shearing stress around the load, this effect is referred to punching shear and it is given by:

Punching Shear failure in Slab

The initial critical section for shear is shown in the figure below:

Example 1 A slab 175 mm thick, d = 145 mm, is constructed with grade 30 concrete and is reinforced with 12 mm bars at 150 mm centres one way and 10 mm bars at 200 mm centres in other direction. Determine the maximum load that can be carried on an area, 300 × 400 mm, without exceeding the ultimate shear stress. For 12 mm bars at 150 mm centres Average For 10 mm bars at 200 mm centres From table 3.8(BS8110) vc = 0.62 N/mm2

Punching shear perimeter = (2a+2b+ 12d) = 600 + 800 +12 × 145= 3140 mm 1.5 d b 1.5 d Maximum load = vc × perimeter × d = 0.62 × 3140 × 145 = 282 × 103 N At the face of the loaded area, the shear stress Which less than

Punching Shear- Reinforcement Design If the reinforcement is required for the initial critical section, this steel should be located within the failure zone lying between the face of the loaded area and the perimeter checked. The amount of reinforcement required is given by: If In either case

Provision of shear reinforcement

Example 2 A 260 mm thick slab of grade 30 concrete is reinforced by 12 mm high- yield bars at 200 mm centres in each directions. The slab is subjected to a mild environment and must be able to support a localised concentrated load of 650 kN over a square of 300 mm side. Determine the shear reinforcement required for fyv = 250 N/mm2 For mild exposure, nominal cover required by grade 30 concrete is 25 mm, thus average effective depth allowing for 8 mm links is 260 – (25+8+12) = 215 mm Check shear stress at face of loaded area Perimeter u = 4 × 300 = 1200 mm Which less than

(b) Check first critical shear perimeter at 1.5d from the load face. Perimeter u = 4 × 300 + 12 × 215 = 3780 mm From table 3.8(BS8110) vc = 0.5 N/mm2 Shear reinforcement is required is les than the minimum 0.4, thus we take : Total number of 8 mm links required =

The links must be distributed evenly between two perimeters within the failure zone. Position the links on two perimeters (215 × 0.5 = 110 mm & 215 × 1.25 = 270 mm) from the face of the load. The lengths of these perimeters are Number of links on perimeter, Number of links on perimeter,

(c) Check second critical shear perimeter at (1. 5 +0 (c) Check second critical shear perimeter at (1.5 +0.75)d from the load face. Perimeter u = 4 × 300 + 8 × 2.25 × 215 = 5072 mm Shear reinforcement is required Total number of 8 mm links required = In part b, on the perimeter at270 mm from the load face 8 links are already provided, thus at least 11 links are required.

(d) Check third critical shear perimeter at (1. 5 +1 (d) Check third critical shear perimeter at (1.5 +1.5)d from the load face. Perimeter u = 4 × 1590= 6360 mm no further reinforcement is required

Span/effective depth ratio Excessive deflection of slab will cause damage to ceiling, floor finishes and other architectural details.

Consider a simply-supported beam of rectangular cross-section supporting a distributed load as shown in Figure : The deflection can be expressed as a fraction of the span by dividing both sides by L.

The limitations on deflection are governed by satisfying the basic (span/effective depth) ratio from Table 3.9 (BS8110), modified accordingly for tension and compression steel using Table 3.10 (BS8110) and Table 3.11 (BS8110). The ratios are given for both rectangular and flanged sections and are based on limiting the total deflection to ≤ (span/250). This should ensure that any deflection occurring after construction of finishes and partitions ≤ (span/500) and ≤ (20 mm).

Example 2 A rectangular concrete beam 250 mm wide × 475 mm overall depth is simply supported over a 6.0 m span. Using the data given, check the suitability of the beam with respect to deflection. Data: Characteristic strength of concrete (fcu) 40 N/mm2 Characteristic strength of main steel (fy) 460 N/mm2 Area of reinforcement steel required 897 mm2 Assume the distance to the centre of the main steel from the tension face is 50 mm Design ultimate moment at mid- span 150 kN. m

291.7

Detailing of Sections The success of any reinforced concrete element is dependent on efficient and practical techniques being adopted during casting of the concrete when detailing the type of steel, diameter of bar, shape of reinforcement and its location within the formwork. Minimum percentage of reinforcement Minimum area

(b) Maximum spacing of the bars In the case of slabs, rules are given in Clause 3.12.11.2.7 which should ensure adequate control of cracking; these are summarized in Figure:

(b) Minimum Spacing of Bars (Clause 3.12.11.1) Guidance is given in the code for minimum bar spacing to ensure that members can be constructed achieving adequate penetration and compaction of the concrete to enable the reinforcement to perform as designed. (i) to develop sufficient bond between the concrete and the bars such that the required forces are transferred between the steel and the concrete, and (ii) to provide protection to the steel against corrosion, fire, etc.

Example 1 (Design of slab) A simply-supported reinforced concrete beam supports two discontinuous slabs as shown in Figure. Using the design data given determine: (i)the required slab reinforcement (ii) sketch typical reinforcing arrangements indicating the curtailment and anchorage of the steel at the support. Effective depth of slab d = 174 mm

Example 2 (Design of slab) A floor system consisting of a solid in-situ reinforced concrete slab cast integrally with the support beams is supported over four spans of 6.0 m as shown in Figure. given determine: (a) design suitable slab reinforcement, (b) check the suitability of the slab with respect to shear and deflection, (c) prepare a sketch indicating all reinforcement; use the simplified rules indicated in Clause 3.12.10 for curtailment.

Effective depth of slab d = 250 mm

Design of Solid Slab(Two way- Span) When simply-supported slabs do not have adequate provision to resist torsion at the corners, and to prevent the corners from lifting, the maximum moments per unit width are given by the following equations: n: total design ultimate load per unit area : length of shorter side. :length of longer side.

Design of Solid Slab(Two way- Span)

Design of Solid Slab(RESTRAINED SOLID SLABS) The design method for restrained slabs is given in BS8110: Part 1, clause 3.5.3.5. In these slabs the corners are prevented from lifting and provision is made for torsion. The maximum moments at mid-span on strips of unit width for spans lx and ly are given by