practically for slabs steel is caluclated in this manner:
for a given room of size 6.5x5.2,first we provide main reinforcement in shorter span and distribution steel in longer span.so,here we provide main reinforcement in breadth side(i.e... 5.2m side).and distribution in other side.so,we preapre a tabular form for a given spacing and dia of steel.
then we caluclate no of bars.therefore,no.of bars=length/spacing.next bar length is caluclated according to sections given.i mean there will be given different sections for rooms and given a brief diagram for steel placing.so both top mat and bottom mat we caluclate the bar length.for both main and distribution steel.to find qty=bar lengthxno.of barsx((dxd)/162)xunit weight of steel.here note that length and bar lengths are different.length means room size and bar length means length of bar with projections inside the room.
l spacing no dia barlength qty
main steel:
top mat 6.5 0.15 44 10 this is acc to
section
bottom mat 6.5 0.15 44 10 5.2+0.23+0.2
distribution steel:
top mat 5.2 0.2 26 8 6.5+0.23+0.2
bottom mat = 0.2 8 6.5+0.23+0.2
=same as main steel top mat
here 0.23 is wall thickness.so both sides 0.23 means we take one 0.23 and add bending value as 0.2 in most cases

it will be depend on the area. just calculate the area then
spacing.,.,after than total length/ spacing = no of bar. use
the formula d*d/162 = total steel waight per meter.,.,.
multiply total to total length= total steel in the rcc slab

The above are to design steel requirments in a slab.
However to calculate amount of steel in a slab u simply need
drawings. A simple assumtion of 10d (i.e:- 10 * diameter )
is ok for bent up calculations.

a)tell density of steel , concrete, bricks , sand and
aggregates.? b) what is knife test for wooden doors ? c)
what is the meaning of term abrasion? what is tested in it?
d) why pile foundation is provided in BCS?e) By total
station what activities can be performed and on what
principle it works? f) what is creep?