Do calculations to establish what is the permanent action $($dead load$),$ G$,$ you should use for design of the floor joists. Floor joists are spaced at $400$ mm centres with $300mm50mm$ sawn timber $($ NZ pine$)$ sections$=4\mathrm{.}6$ KN$/$m$^3.$ The mezzanine floor area has $18$ mm pine plywood screwed on top of the floor joists that can be assumed to weigh $0\mathrm{.}1$ kPa $.$ There is a $10$ mm plasterboard $($fibrous plaster$)$ ceiling screwed to the underside of the floor joists. Insulation and services within the floor structure can be assumed to add $0\mathrm{.}1$ kPa to the floor dead load. Combine all dead load contributions and compute a uniformly distributed line load, $w_G($in $k\frac{N}{m}),$ acting on the joists.Figure $5.$ Main member layout proposed for mezzanine floor structure.

Broken up to answer this question:

$1$a$)$ What is the dead load due to the self$-$weight of the floor joist? $[$kN$/$m$](3$dp$)$

$1$b$)$ What is the dead load due to other superimposed dead loads, such as flooring, ceiling, and service? $[$kN$/$m$](3$dp$)$

$1$c$)$ Combine all dead load contributions and compute a uniformly distributed line load, wG $($in kN$/$m$),$ acting on the joists. $[$kN$/$m$](3$dp$)$