The structural calculations are performed by adding several factors, partial safety factors, on the load depending on which type of load that is applied (self weight, cars, wind, snow etc). If there are more than one load, the loads must be combined in different load combinations. There are also additional safety factors included when doing the evaluation of the structure. See one example below.
Example - beam located in a building
One beam of 10m is loaded with three loads: self weight, snow load and wind load. The beam dimension is RHS 80x80x5 with material S355. The beam is simply supported and wind and snow is uniformly distributed over the beam and assumed to be applied in the same direction as self weight (for simplicity). Evaluate the bedning capacity of the beam.
Self weight: 110 N/m
Solution
The section has cross-section class 1, and hence the plastic modulus, Wpl, is used to determine the bending capacity (EN 1993-1-1 sec 6.2.5). Wpl = 4.11*10^4 mm^3.
The beam is located in a building so load combinations are taken from EN 1990, table A1.2(B). Equation 6.10 is used to get 2 different load combinations, one with snow load as leading variable action and one with wind load as leading variable action. The partial factor for self weight is 1,35, factor for leading variable load is 1,5 and factor for the accompanying variable load is 1,5 times a reduction factor taken from table A1.1. The load combination will be as below:
Combination 1 = 1,35*self weight + 1,5*Snow load + 1,5*0,6*wind load
Combination 2 = 1,35*self weight + 1,5*wind load + 1,5*0,7*snow load
gives:
Combination 1 = 1,35*110N/m + 1,5*50N/m + 1,5*0,6*10N/m = 148,5+75+9 = 232,5N/m
Combination 2 = 1,35*110N/m + 1,5*10N/m + 1,5*0,7*50N/m = 148,5+15+52,5 = 216N/m
This tells us that combination 1 gives the highest load and hence the can be used to evaluate the beam bending capacity. The beam is evaluated below:
Note: The partial safety factors and reduction factors may be specified in the national annex.
