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 I Session AU3

2. Fire in a multi-storey building Fire safety requirements include “fire resistance” or ensuring stability of the structure for a specified time (or specified fire), therefore columns, beams and floor and ceiling of compartment must remain stable For structural stability in a single floor fire (covered by regulation) - Determine response of floor system and whether protection is required - Columns nearly always need protection (from analysis or prescriptive codes) For structural stability in a multiple floor fire (not covered by regulation) -Floor and column interaction must be determined to ensure stability Simple analytical calculation methods are developed to analyse these cases

3. General model of a composite beam in fire Temperature, T(z) Distance, z Ambient Temperature, T o CONCRETE STEEL Structure subjected to the illustrated temperature distribution A A Section AA k r1 k r2 ktkt

4. Permutations of the general case k r 1 = k r 2 = k t = 0 k r 1 = k r 2 = 0 k t = 1 k r 1 = k r 2 = k t = 1 k r 1 6 = 0 k r 2 6 = 0 k t 6 = 0 Simple calculation methods will be presented to estimate responses for the first three cases The calculations can be used to make quick estimates in real structures, or as benchmark solutions to test FE software

5. Unrestrained uniform thermal expansion lTlT l

6. Restrained thermal expansion - yielding

7. Restrained thermal expansion - buckling Buckling strain Post-buckling thermal expansion strain ² b = ¼ 2 ¸ 2 this would lead to deflections  ² p = ® ¢T ¡ ¼ 2 ¸ 2

8. Deflections Assuming the displaced shape to be equivalent to an analytical curve 1- Sin curve (easy) 2- Circular arc (accurate) neglect in calculation ± s i n ( ² p ) = 2 l ¼ r ² p + ² 2 p 2 ± c i rc ( ² p ) = l 2 Ã s 1 6 ² p ¡ s 1 6 ² p ¡ 1 !

9. Thermal expansion against finite restraints L, E, A, I TT ktkt PP ¾ = E ® ¢T ³ 1 + EA k t L ´ ¢T cr = ¼ 2 ® ¸ 2 µ 1 + EA k t L ¶

10. Buckling under different restraint stiffnesses

11. Unrestrained uniform thermal gradient

12. Restrained thermal bowing Assuming no mean temperature increase  T = 0

13. Thermal expansion and bowing combined P P MM L, E, A, I  T, T,z P = EA ( ² T ¡ ² Á ) M = EI Á L, E, A, I  T, T,z z x ± z ± x ± x = ( ² T ¡ ² Á ) l P P ± z L, E, A, I  T, T,z P = EA ( ² T ¡ ² Á ) Deflections assuming ² T > ² Á ± z = ± c i rc ( ² Á ) ¼ ± s i n ( ² Á ) Post-buckling deflections where, ² p = ² T ¡ ² Á ¡ ¼ 2 ¸ 2 a dd t o ± z a b ovea ± s i n ( ² p ) ca l cu l a t i on Pre-buckling deflections where, C = 0 : 3183 l 2 A ¼ I ( ² T ¡ ² Á ) ± z = ± c i rc ( ² Á ) + C ± c i rc ( ² Á ) 1 ¡ C from thermal bowing from P-  moments

14. Buckling temperatures increase with gradient ¢T cr = 1 ® µ ¼ 2 ¸ 2 ¡ ² Á ¶

15. Typical concrete deck slab cross section

16. Typical temperature distributions in slab

17. Divide cross-section into slices to simplify

18. Equivalent mean temperature and gradient

19. Application to a composite section  T=150 o C 70 60 10 284 Dimensions In mm E=200GPa  =12x10 -6 E=12.5GPa  =9x10 -6

20. Midspan deflection (mm) Roller-end disp / rotation Restraining force (kN) Restraining moment I 8.1+51.6 = 59.74.7 / 0.02600 II 0.6+0 = 0.60 / 01540.837+295 III 8.1+51.6+16.7 = 76.40 / 0.0321540.80 IV 8.1+51.6+12.6 = 72.31.2 / 0.0301158.50 Calculations for composite beam of 9m span Equivalent mean temperature  T = 60.6 o C Equivalent thermal gradient T,z = 0.51 o C/mm Equivalent curvature  = 5.1x10 -6 = 0.000607 = 0.000088 ) ² Á Equivalent area of the composite = 14847.8 mm 2 Equivalent second moment of area of the composite = 289210176.1 mm 4 Equivalent elastic modulus of the composite = 200 GPa Translational spring restraint stiffness (for case IV) = 1000kN/mm This represents all the data required to obtain all the items of response for the following four cases (also including a udl of 5.48 kN/m) ) ² T

21. Limitations of the method ♦Not suitable for very high temperatures (much over 550 o C) where the geometric and material nonlinearities become much stronger and catenary resistance becomes more important than bending resistance ♦For very high thermal gradient problems where thermal bowing dominates the response as this creates tensile membrane strains in the beam and cannot be dealt with using this method (for realistic structures in fire situations this is hardly ever the case)

22. Mean temperature Thermal gradient C C T C T y T C Gravity load EI   T M load local buckling beam bottom flange?  Beam bottom flange “buckling” phenomenon

23. 127 Mpa160 Mpa286 Mpa573 Mpa >> 318 Mpa w = 16.5 kN/m  80 o C T,z = 0.5 o C/mm 524 Mpa  Plastic yielding! Bottom flange stress (when steel is at 150 o C)

24. Restrained beam test “buckling” and cracking

25. Local buckling (BS Corner Test)

26. Local buckling (BS Demonstr. Test)

27. x ; u z ; w u ( x ) = z d w d x ) ² x = d u d x = z d 2 w d x 2 w ( x ) d w d x New method: Analysis of beams (in bending)

28. x ; u z ; w M F F M y ; v w ( x ) d w d x ² x = ² xm ¡ z d 2 w d x 2 ) ¾ x = E ² xm ¡ E z d 2 w d x 2 F x = R h 2 ¡ h 2 ¾ x d z = EA ² xm M x = R h 2 ¡ h 2 ¾ x z d z = ¡ EI d 2 w d x 2 Analysis of beams (add membrane strain)

29. Analysis of beams (equilibrium)

30. ) EI d 4 w d x 4 = p + F x d 2 w d x 2 Equilibrium of Moments Equilibrium of forces in the z direction F xz d x ¡ ( M x + d M x ) + M x = 0 ) F xz = d M x d x d F xz + p d x + F x d 2 w d x 2 d x = 0 ) ¡ d F xz d x = ¡ d 2 M x d x 2 = p + F x d 2 w d x 2 Analysis of beams (Equilibrium)

31. ¾ x = E ² xm ¡ E z d 2 w d x 2 ¡ E ® ¢T ( z ) ² x = ² xm ¡ z d 2 w d x 2 ¡ ® ¢T ( z ) ² T = ® ¢T ( z ) F x = EA ² xm ¡ N T M x = ¡ EI d 2 w d x 2 ¡ M T N T = E ® R h = 2 ¡ h = 2 ¢T d z = EA ® ¢T f orun i f orm ¢T M T = E ® R h = 2 ¡ h = 2 ¢T z d z = EI ® T ; z f orun i f ormgra d i en t Analysis of beams (Thermal Loading)

32. EI d 4 w d x 4 = p + d 2 M T d x 2 + EA ² xm d 2 w d x 2 ¡ N T d 2 w d x 2 Analysis of beams (complete equation)

33. Application to laterally restrained beam w = u = M x = 0 a t x = 0 an d x = L M T ( x ) = M T s i n ¼x L w ( x ) = w T s i n ¼x L Represent w and M as, ² xm = ¼ 2 w 2 T 4 L 2 Also use the sin formula for deflection to obtain

34. Solution Midspan deflection can be calculated by solving the following cubic equation w 3 T + µ 4 I A ¡ 4 N T L 2 ¼ 2 EA ¶ w T + 16 M T L 2 ¼ 3 EA = 0 Axial force in the beam is then given by F x = EA ¼ 2 w 2 T 4 L 2 ¡ N T One solution of a cubic equation (in standard form of Vieta) defining, then the roots are, ; ;;;