area of plaster = 3.05 x 3.05 = 9.3 m2
(convert thickness of plaster into m divide by 1000)
volume of plaster = 9.3 x (10/1000) = 0.09 m3
quantity of cement = 0.09/1+5 =0.09/6 = 0.02 m3
as cement comes in bags of 50 kg therefore volume of one
bag cement = o.0347 m3 (exact)
(or it may be taken 0.034 or 0.035 m3)

no. of cement bags = 0.02/0.0347 = 0.58 bags (almost 1 bag
of cement required for plaster )

quantity of sand = 0.02 x 5 = 0.10 m3

this is exact calculation
you can add 30% extra for wastage etc.

Area to be plastered = 10'*10' = 100'
= 100/10.76 = 9.29sqm
(ie; 1 sqm = 10.76 sqft)
Thickness of mortar layer = 10mm = 0.010m
Volume of mortar required = 9.29*0.010 = 0.09cum
Add 15% for allallowance for calculation of surface and
wastages = (15/100)*.09 = 0.013 cum
Therefore, total volume of mortar required = .09+0.013 =
0.103 cum, say 0.10cum
Sand required for 0.10cum of CM 1:5 = 0.10 cum
Cement required for 0.10 cum of CM 1:5 = (1440/5)*0.10
= 288*0.10 = 28.8 kg
(ie; 1 cum of cement weighs 1440 kg)

10' x 10' x 0.0333'
3.33 cft
Dry metarial in one cft = 1.27 cft
3.33 x 1.27
4.229 cft
sum of ratio = 1+5 = 6
quantity of cement = 1*4.229/6
= 0.7048 cft
volume of one bag = 1.25 cft
=0.7048/1.25
=0.56 cement bag
quantity of sand = 5*4.229/6
= 3.524 cft

TOTAL QUANTITY OF PLASTERING=3.05*3.05*.01=.093 cu.m
by thumb rule
.75 cu.m of plastering requires =350kg of cement
1750 kg of sand for
the ratio of 1:5 (the weights are calculated with specific
gravities) from this

add 30% extra for wastage and uneveness=.121 cu.m
so for .121 cu.m of plaster requires = 56.5 kg of cement
282.5 kg of sand

Dear I would like discuss in deep beams problem in design
and serviceability. please contribute to prepare a
comprehensive discussion. first deep beam definition:

1cracks found in brick wall in service building, how to
rectify it,with low cost ?
how to find propagation of cracks (in depthwise) and arrest
on that stage?
Is it require to design for Expansion joint in Brick work?
if so what is the required distaces?