structural design…designing slabs?
hey im just trying to work out the design load for a slab
Its to be for a proposed library which is to accommodate reading rooms with book storage
Imposed loadings extract from BS 6399
Reading rooms with book storage = 4.0 KN/m^2
so to work out the self weight of the slab
its the
density x span x effective depth
but to get the design load do i need to multiply it by 1.4 for the dead load or by 4.0 for the library imposed load?
Not familiar with that particular code, but, you have a bit of an error elsewhere. The loading given for the library is probably a live load, to be multiplied appropriately and added into the slab self weight. How you add them depends on the load combination you are using, which might be where you put that 1.4 to use. The 4.0 is not a factor, it is another load.
Not familiar with that particular code, but, you have a bit of an error elsewhere. The loading given for the library is probably a live load, to be multiplied appropriately and added into the slab self weight. How you add them depends on the load combination you are using, which might be where you put that 1.4 to use. The 4.0 is not a factor, it is another load.
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