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Workshop at Indian Institute of Science 9-13 August, 2010 Bangalore India Fire Safety Engineering & Structures in Fire Organisers:CS Manohar and Ananth.

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Presentation on theme: "Workshop at Indian Institute of Science 9-13 August, 2010 Bangalore India Fire Safety Engineering & Structures in Fire Organisers:CS Manohar and Ananth."— Presentation transcript:

1 Workshop at Indian Institute of Science 9-13 August, 2010 Bangalore India Fire Safety Engineering & Structures in Fire Organisers:CS Manohar and Ananth Ramaswamy Indian Institute of Science Speakers:Jose Torero, Asif Usmani and Martin Gillie The University of Edinburgh Funding and Sponsorship: Structural behaviour – simple calculations II Session AU4

2 Point of highest tensile stress (reinforcement rupture) Short span Long span Tensile membrane action in slabs

3 Deflected shape of BS Corner test

4 Analysis of floor slabs ♦Main steps of analysis –Determine design fire and slab/unprotected beam temperature distributions Chose fire definition (ISO834, BS476, Eurocode parametric, etc.) 1D heat transfer over slab depth Determine equivalent thermal actions (  T and T,Z ) –Determine structural response to thermal actions for the assumed restraint conditions Deflections Mechanical stress and strain in reinforcement –Determine the deflection that will cause mechanical strain at any point in the reinforcement to equal rupture strain –Determine ultimate load by equating internal and external work all load is carried by membrane action all membrane resistance is provided by reinforcement strain hardening of the reinforcement is ignored –Add the catenary contribution of any unprotected steel beams

5 Plates/slabs subjected to thermal effects Laterally restrained to end translations (end rotations free) Temperature distribution T(z) represented by  T and T,z

6 Plate equilibrium with membrane forces

7 Stress-strain relationship

8 Membrane and bending strains in a plate Under large displacements

9 Stress resultants: bending D = E t 3 12 ( 1 ¡ º 2 ) D is the flexural rigidity of a plate

10 Stress resultants: membrane Using Airy stress function F to describe membrane stress ¾ xx 2 x 2 ¾ yy 2 y 2 ¾ xy 2 y Therefore, N x 2 x 2 h N y 2 y 2 h N xy 2 y h

11 Equilibrium equation: membrane Differentiating and adding all membrane strains, Writing strains in terms of membrane forces and then the Airy stress function Substituting in the compatibility equation above

12 Equilibrium equation: bending From equilibrium of moments in X and Y directions and forces in Z direction Eliminating Q x and Q y from the last equation using the previous two Using the stress resultants derived earlier for bending and membrane forces,

13 Stress-strain relationship (inc. thermal effects)

14 Membrane/bending strains (inc. thermal effects) Under large displacements

15 Equation of equilibrium for membrane forces Equation of equilibrium for bending moments w is the deflection M T is the thermal moment N T is the thermal force F is an Airy stress function describing the membrane stress E is the Young’s modulus of the plate material h is the depth of the plate Final governing equations

16 Two approximations were made in finding a solution for the deflection profile and stress-strain distribution in the slab due to a thermal field (varying in z only)  xx can vary in y but not x and  yy can vary in x but not y the analysis is geometrically nonlinear but material behaviour is linear In obtaining a solution to the governing differential equations the deflection of the slab w and the thermal moment M T are represented as double sine series (using a single Fourier series term) Solution of the governing equations

17 ♦The coefficients M T mn are calculated by performing the integration: ♦Thisallows the thermal moment to be written as: Solution

18 ♦The solution of the Airy stress function F consists of a homogeneous solution F H and a particular solution F P ♦The particular solution is obtained by solving the equation for equilibrium of membrane forces using only the first Fourier term: ♦The general solution of the Airy stress function F can therefore be described as: Solution

19 ♦As the plate is assumed to be laterally restrained along all of its edges the values of P and Q can be derived to be: ♦The final solution for the Airy stress function can therefore be written: Solution

20 ♦The final solution is obtained by substituting the equations for the deflection function, thermal moment function and Airy stress function into the equation for flexural equilibrium and applying a Galerkin procedure ♦The Galerkin procedure integrates the equation of flexural equilibrium over the plate such that: Solution

21 ♦In finding a solution for the deflection and stress and strain distribution in a plate under thermal loading only the first term in the series will be considered so that, if the deflection due to thermal loading is defined as w T, then the solution is obtained by solving: Solution: deflection

22 ♦If the deflection of the slab w T is known then the membrane stress distribution along the boundary of the slab can be calculated using the Airy stress function F calculated previously: Solution: stresses

23 ♦Rearranging the stress functions mechanical strains can be obtained. Solution: strains

24 Typical boundary membrane stress distribution

25  A 5m x 5m plate was analysed using the method presented and also a finite element analysis  The details of the plate were:  slab depth h = 100mm  Young ’ s modulus E = 40,000N/mm 2  Poisson ’ s ratio  = 0.3  thermal expansion coefficient  = 8x10 -6  thermal gradient T,z = 5 o C/mm  thermal expansion  T = 200 o C  The deflections obtained were:  Theoretical=136mm  Finite element=148mm Example

26 Comparison with numerical solution

27 Ultimate load calculation  The ultimate load capacity of the slab will be determined using an energy method  The following assumptions are made:  all of the load is carried through membrane action  any tensile strength in the concrete is ignored (all load carried by reinforcing bars)  strain hardening of the reinforcement is ignored

28 Assumptions the slab is assumed to be rectangular boundaries are assumed to be vertically and laterally restrained in translation perimeter beams are assumed to displace much less than the centre of the slab anchorage to tensile membrane forces is available at the boundaries (this is fulfilled if anchorage to reinforcement is sufficient to enable tensile yielding) tensile membrane resistance is provided only by the reinforcement the temperature distribution varies only through the depth of the slab during the heating phase failure occurs in a ductile manner with no localisation of strain in the reinforcement the pre-failure deflected shape is governed entirely by the temperature distribution in the slab (consistent with the assumed restraint conditions) the temperature of the reinforcement is identical to the surrounding concrete and so is its thermal expansion coefficient i.e. no slippage occurs

29 Failure criterion Mechanical strain at any point in the slab reinforcement reaches its rupture strain. In practice this can be taken as EC2 specified ductility limit for the type of steel used In rectangular slabs this will usually occur in the middle of the longest side (given the reinforcement density is the uniform)

30 Deflections ♦The total deflection of the slab will be assumed to consist of two components: w T : the deflection due to the thermal load w q : the deflection due to the ultimate load ♦The total deflection is defined:

31 Procedure  Failure is assumed to occur when a limiting value for the mechanical strain in the reinforcement is reached  Based on this value the total deflection w t, at which point the slab is said to have reached failure, can be calculated  w q can therefore be calculated  The ultimate load q ult will be calculated by equating the internal and external work done as the slab moves through the deflection w q

32  If V is the volume of a reinforcing bar then the total internal work done is:  Using the assumption that the deflected shape forms a double sine surface, the external work done for an incremental load  q ult moving through an increment of deflection  w is: Internal and external work

33 Ultimate load  Comparing the internal and external work done, the load at increment n (q n ) is:

34 Including unprotected beams  Unprotected secondary beams can be included by considering them in the same manner as the reinforcing bars  For each beam calculate the stress and strain due to the thermal load and also at the ultimate load  The equation describing the internal work done is then modified to:

35 British Steel Cardington Tests

36 Corner Test failure envelope

37 Corner Test - deflections

38 Corner Test - mechanical strains

39 BRE Cardington Fire Tests

40 BRE Large Compartment - failure envelope

41 BRE Large Compartment - deflections

42 BRE Large Compartment - mech. strains

43 BRE Large Comp. - including beams

44 Conclusions  Method seems to confirm that the Tests were far away from failure as inferred from the computational modelling work  Method provides a very details solution and picks up all the key features of the floor system  Analyses can be carried out at a fraction of the cost of an FE model  Highly feasible for use in performance based design


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