Fire Resistance Design

Cover over steel reinforcement is meant to protect the steel from rust and other environmental impacts. It is determined by the level of exposure of the reinforced concrete element, the type of the element and the fire resistance period required for the design. The levels of exposure include low, mild, high and severe exposures to dangerous chemically harmful conditions, such as saline water for severe which requires 40mm minimum cover for beams, and fresh water for mild conditions which requires 25mm minimum cover for similar beams (Arya, 2009). Other conditions such as exposure to corrosive elements for example, in a chemical factory, also determine the amount of cover for steel in an element exposed to that condition. In case of fire, the duration that the concrete element is expected to remain intact is determined by the cover protecting the steel from the fire heat (Housing, 2004). The duration gives room for fire emergency response. The duration ranges from one hour for 25mm cover to three hours for 40mm cover. The class of concrete mix also determines the minimum cover that should be used for the element. High class mixes require more cover than low class mixes. Concrete mix of cube strength below 30N/mm2 requires a minimum of 20mm cover, while that above 40N/mm2 requires a minimum of 30mm cover (ASI, 2001).

Briefly discuss how fire resistance is designed for in reinforced concrete elements. You may use the Standard AS3600 as well as other references. (200 words – 2 marks)

Fire resistance is the amount of time a structure or an element would take before collapse when exposed to continuous fire. To determine the duration, three methods can be used, which are: use of tabulated data, which contains information for general use from research institutions; conduction of fire test, which involves a direct application of a fire resistance test on an element of structure; or using fire engineering calculations, which is a basis for calculating the fire resistance of a structural element (ASI, 2001). To alter the effect of fire on structures, the size and shape of elements; properties of reinforcement; concrete cover provided to reinforcement and the load supported are some of the factors that could be manipulated. Concentration on the proneness of the elements to fire is given to the exposed surfaces of the elements, such as one side of the wall, the soffit of the floor, the sides and soffit of the beams and all sides or one or more sides of the columns, depending on the type of the protection offered by adjacent walls (BSI, 2001). The protection concrete cover given to the reinforcement, and the density of concrete used is often given greater priority in the design, and most designers, especially of small structures, only and sufficiently apply these two factors to design for the element’s fire resistance.

Crack Control in Concrete Slabs

Concrete slabs are designed for crack control on the sides which are under tension, known as the flexural cracking control. Cracks are controlled by minimizing the distance between the reinforcement, without reducing beyond the allowable distance for workability of concrete say by vibration (Housing, 2004). Clearly, the maximum distance between bars in tension to control cracking should be more or equal to four times the diameter of the bars in tension. Where bars of mixed diameters are used, the bar with 0.45 times the diameter of the largest bar used should be ignored, except when considering those in the side faces of beams (BSI, 2001). Tabulated values can be used where the conditions apply. However, the service stress of the reinforcement can be used to determine clear spacing, which according to standards, must be less than 300mm. In other conditions, the distance between the bars to control cracking cannot be more than 750mm. in slabs, specifically, the grade of steel and the reinforcement percentage in concrete are the factors considered (ASI, 2001). For instance, in no case should the clear spacing between bars exceed the lesser of three times the effective depth, or 750mm. for steel grade 460, slab depth of less than 200mm is assumed sufficient; for 250 grade steel and slab depth less than 250mm is also sufficient; and where the reinforcement percentage is less than 0.3% (Housing, 2004).

Discuss the significance of a steel beam being ‘restrained’ or ‘laterally braced’; what is the difference (in structural terms) if a beam is ‘built-in’ to a slab, or if a beam, carrying the same load, is ‘unrestrained’? (150 words – 1.5 marks)

Lateral bracing, or otherwise, restraining of structural elements is employed in a situation where there is a need to provide protection from lateral, or horizontal or torsional forces or loads (BSI, 2001). These may occur when the loading on the beam is unsymmetrical in the cross-section, or where there are other loads such as wind or earthquake loads. Provision and design for lateral restraints is taken as a creation of a rigid supports. The stiffness of the joint is assumed to be infinite (Arya, 2009). The resistance to loading in this case is thus considered to be distributed to the whole structure, rather than the particular beam element. Unrestrained beams carry loads and do not redistribute them to the slabs connected to them. Where a beam is built in to a slab, a percentage of the loading is retrained by the slab, and the beam is designed as a wide flanged beam, rather than a beam section alone. The second moment of inertia is thus increased, and hence the consequent reduced stress in the beam.

Reference	Calculation	Output
AS 1170.1 Clause 2.3.1.4 AS 4100 Cl 5.2.5.1 AS 4100 Cl 5.2 AS 4100 Cl 6.3.2 AS4100 Cl 6.1.3 AS4100 Cl 6.5.1	Details and assumptions · Beam type: Simply supported, internal UB · Use: holds gymnasium concrete floor · Span of beam: 6m · Span of floor: 5m · Thickness of concrete floor: 200mm · Restrained · Concrete density: 24kN/m3 · Design code: AS 4100, AS 1170.1 · Assume steel strength = 275N/mm2 Loading Calculation Dead load due to the concrete floor DL = 24×0.2×5 = 24kN/m Live load (gymnasium use), use 5kN/m2, hence = 5×5 = 25kN/m Factored loading = 1.4×24 + 1.6×25 = 73.6kN/m Moments and Reaction Reactions at the ends; Due to symmetricity, RA=RB = 73.6×6/2 = 220.8Kn Moments M at the center for simply supported beam: = wl2/8 = 73.6x6x6/8 = 331.2kNm Selection of UB Taking py = 275 Plastic modulus Sx > M/Py = 331.2×106/275 = 1204cm3 Sections available: 1. 356 x 171 x 67 UB; Sx = 1210cm3 2. 406 x 178 x 60 UB; Sx = 1190cm3 3. 457 x 152 x 60 UB: Sx = 1280cm3 Using the d/t criteria, option (2) has d/t = 46.2 <70?, thus is Plastic Hence, use UB 406 x 178 x 60 Shear capacity check Shear capacity Pv = 0.6pvAv = 0.6 pytD  = 0.6x275x7.8×406.4 = 523Kn Fv = 220.8kN should be less or equal to 0.6Pv = 313.8Kn Hence, OK in shear Bending moment check Moment Capacity Mc = PySx = 275 x 1190 x 103 = 327kNm 1.2PyZ = 1.2x175x1060x1000 = 349.8kNm, hence Ok Moment due to loading + self-weight of beam 307.5+1.4(60×9.81/1000)x6x6/8 = 311.21kNm < 327kNm Hence, OK For moments Deflection check Deflection at center = 5wl4/384EI E=205kN/mm2, I = 21500×10-8 , w = 49Kn, L=6m Hence, deflection = 18.78mm Maximum deflection = span/200 = 6000/200 = 30mm Hence, OK

 

Significance of Restraint and Lateral Bracing

Reference	Calculation	Output
AS 3600 Cl 5.2.5.1 AS 1170.1 Clause 2.3.1.4 AS 3600 Clause 6.1.2.4 AS 3600 AS 3600 Cl. 9.9.1.2 Cl. 9.9.1.2 AS 3600 Cl. 7.3.4.2 Cl. 5.2.1.2	Design details and assumptions · Beam type: Simply supported, internal rectangular beam · Use: holds gymnasium concrete floor · Span of beam: 6m · Span of floor: 5m · Thickness of concrete floor: 200mm · Restrained · Concrete density: 24kN/m3 · Concrete strength: 30MPa · Design code: AS 3600, AS 1170.1 · Assume steel strength = 500N/mm2 Loading Load due to the supported slab:  = 0.2x24x5 = 24kN/m Load due to the beam: = 0.6x24x0.35 = 5.04kN/m Total dead load = 24+5.04 = 29.04kN/m Live load, = 5×5 = 25kN/m Factored loading = 25×1.6+29.04×1.4 = 80.656kN/m Design check for bending As provided = 2463mm2 by the 4N28 d=560mm As required: = M/(0.87xfyxZ) M = WL2/8 = 80.656x6x6/8 = 362.952kNm As = 362.952×106 / (0.87x500x532) = 1568.37mm2 hence ok Design check for shear v=V/bd V=80.656×6/2 = 241.968Kn Thus, v = 241.968×1000/(350×560) = 1.245N/mm2 vc=100As/bd = 100×2463/(325×560) = 1.35 > 0.67, the nominal vc hence, provide nominal links, as below: Asv/Sv = 0.4×350(0.87×500) = 0.322 Use N8 links @ 250mm spacing. Deflection check For rectangular beams, span/d ratio < 20 for adequacy. Hence, 6000/560 = 10.7 <20 Ok Thus, the beam is sufficient in its current design

 

The design process is an iterative sequence of steps required to achieve the most feasible design for the particular problem (Seyyed, 2005). A change in circumstances prompts a change in solving the problem in that particular setting, and hence a new design is to be realised. To this end, the process has five key steps upon which decisions hang and are applied. These steps are general and can be applied in any design activity in the order given.

1. Identification of the problem.

This is also called the brainstorming stage. It involves clear identification of the need to be solved through the engineering design to be developed. It should be precise in its expression and articulation, especially considering the client’s expected outcome (IoE, 2012). The problem statement is thus developed, from a vague state in the mind of the designer to a concrete statement of what is required to be dealt with. Any special requests from the client are also collected at this stage, to have a comprehensively defined problem. The information from the client stands as the main criteria for the design, such as need for low cost design, simple and manual, or automated design (Seyyed, 2005).

1. Gathering of the pertinent information

Before design, the relevant information concerning the problem stated is reviewed. This may include the review of the clarity of the problem definition; the understanding of the need of the client; the identification of the factories or materials that may be involved in the process, whether the design needs originality of the items, or use of the existing materials and inventions (IoE, 2012). Traditional and current resources of information such as articles and journals are reviewed at this stage to identify the possible history and solutions about the problem before. It may involve generation of possible solutions from historical archives and scholarly articles.

At this point, the designer comes up with several possible ways that can be viable solutions to the identified problems, with regard to the client’s design criteria. The designer begins with the available and commonly applied solutions to the problem as found in the literature review in stage two, then proceeds to the generation of alternative solutions that he/she may be having in mind for the problem (Seyyed, 2005). The solutions are weighed in accordance to the criteria, and then are arranged in the order of suitability or possible feasibility in solving the existing problem identified.

Feasibility studies and analysis of the possible solutions are conducted to be compared. The solutions must be financially or economically, practically, technologically, environmentally and administratively feasible. The designer outlines the requirements for each solution to be undertaken, the advantages of the solution and the challenges or the disadvantages that the solution may have in solving the problem. The best of the solution with regard to the problem may not be essentially the cheaper, acceptable or aesthetically considered solution (Seyyed, 2005). The need for complete feasibility study is to weigh the outcomes in terms of the advantages and disadvantages, with regard to the maintenance and operation of the design.

The designer is then required to develop a working plan/prototype for the design idea selected. A prototype is a working model of the real design, developed to imitate, simulate or predict its functioning and behaviour in the natural circumstances as much as possible. It may be in form of a computer model, or a tangible creation that can be tested under simulated ‘natural’ conditions. A pass in the test results takes the designer tto the implementation stage, while a fail in the test results returns the designer to the alternatives, or any other stage as appropriate (Seyyed, 2005).

The stages are somehow inseparable, with continuous search in literature, brainstorming and identification of alternatives going on even in later stages. The design process is cyclic, with adjustments being done at every required moment or stage (Seyyed, 2005).

References

Arya, C. (2009). Design of Structural Elements. London: Spon Press.

ASI. (2001). Australian Standard. 

BSI. (2001). Structural use of Concrete. BSI.

IoE. (2012). Engineering Design Process. Retrieved from TeachEngineering: https://www.teachengineering.org/k12engineering/designprocess

Seyyed, K. (2005). Engineering Design Process.