Steel Buildings in Europe

Part 7: Fire Engineering 7 - 64 heat extraction surface of the peripheral construction elements. In this example c is given, for simplicity, the worst value. t ä = q R  c  w = 126  0,25  1,70 = 54,0 min Through interpolation in Table A.2, in the safety category K2 for an equivalent fire duration of 54 minutes, a maximum surface area of 4800 m² can be defined. At this point, some additional work by the designer could be useful in reviewing the input data. Is the fire load case too high? What will happen when the opening surfaces are modified and the ground floor is also modified at the same time? Alternatively, what about the surfaces? Can the surface be reduced by 200m²? The onus is on the designer to present and explain the different opportunities to the client and to list the comparison costs. The second possibility using the full verification method is more precise. The maximum floor surface is calculated using the basic value for the surface of 3000 m² times factors F1 to F5. The factor values are taken from tables of DIN 18230-1 and do not need to be determined. According to table 3 of DIN 18230-1 the factor F1 is: 1,9 According to table 5 of DIN 18230-1 the factor F2 is: 1,5 According to table 6 of DIN 18230-1 the factor F3 is: 1,0 According to table 7 of DIN 18230-1 the factor F4 is: 1,0 According to table 7 of DIN 18230-1the factor F5 is: 0,7. Inserted into the formula: A = 3000  F1  F2  F3  F4  F5 = 3000  1,9  1.5  1,0  1,0  0,7 A = 5989 m². In this method, the fire resistance classification of the structural components has to be calculated with the following equation: Required fire resistance duration t f = t ä     L The design of the fire resistance duration includes the following factors:  the equivalent fire duration of 54 minutes  the safety factor  of 0,6 according to Table 2 of DIN 18230-1, and  the factor alpha L takes into account the fire related infrastructure of 0,9 according to Table 4 of DIN. Hence: t f = 54  0,6  0,9 = 29,16 min => R30