## CALCULATION OF QUANTITY OF MORTAR AND QUANTITY OF MATERIAL REQUIRED FOR THE PREPAIRATION OF MORTAR

The
quantity of mortar for a uniform surface shall be obtained by multiplying the
thick of the plaster and the total area to be plastered (Area × Thickness). But
for filling up the joints and to make up the un-uniformity of the surface the
quantity may be increased by 30% of the quantity of mortar required for a
uniform surface.

To
obtain the total dry volume of constituent materials or mortar the wet volume
may further increased 25%.

**FOR THE CALCULATION WE CONSIDERING 12MM THICK PLASTER FOR 100 SQM. WALL AREA**

__QUANTITY OF MORTAR__**:**

**To obtain –**

Quantity
of Mortar for wet mix of uniform layer

**= (12/1000) × 100 = 1.2 Cum.**
Adding
30% to fill up joints and uneven surface etc., the Quantity of Mortar for wet
mix

**= [1.2 + (1.2×30%)] = 1.56 Cum.**
Again
adding 25% for the total dry volume

**= [1.56 + (1.56 × 25%)] = 1.95 Cum.; Says 2.0 cum**

__QUANTITY OF MATERIAL__**:**

**To obtain, considering 1:6 cement sand mortar.**

Dry
volume of mortar = 2.0 cum

Cement: Sand = 1:6

Cement + Sand = 1 + 6 = 7

**For Cement**–

**For Sand –**

**SIMILARLY, MATERILA FOR 20MM THICK PLATERING OF 100SQM WALL AREA**

__QUANTITY OF MORTAR__**:**

**To obtain –**

Quantity
of Mortar for wet mix of uniform layer

**= (20/1000) × 100 = 2.0 Cum.**
As
the thickness of the plaster is more, so in case of 20mm thick plastering adding
20% to fill up joints and uneven surface etc., the Quantity of Mortar for wet
mix

**= [2.0 + (2.0×20%)] = 2.40 Cum.**
Again
adding 25% for the total dry volume

**= [2.40 + (2.40 × 25%)] = 3.0 Cum.**

__QUANTITY OF MATERIAL__**:**

**To obtain, considering 1:6 cement sand mortar.**

Dry
volume of mortar = 3.0 cum

Cement: Sand = 1:6

Cement + Sand = 1 + 6 = 7

**For Cement**–

**For Sand -**

__CALCULATIONS FOR RICH MORTAR PLASTERING__**:**

Rich
Mortar Plastering generally used for R.C.C Roof ceiling.

The
quantities of material will be less for rich mortar plastering because the use
of cement quantity will be more then the quantities of other material, so the
voids in sand and the reduction in volume of dry mortar will be less.

**FOR THE CALCULATION WE CONSIDERING 12MM THICK PLASTERING OF 100 SQM. WALL AREA FOR R.C.C CEILING:**

__QUANTITY OF MORTAR__**:**

**To obtain –**

Quantity
of Mortar for wet mix of uniform layer

**= (12/1000) × 100 = 1.2 Cum.**
In
case of R.C.C ceiling plastering the unevenness of surface will be less, so
considering 20% for the un-uniform surface,
the Quantity of Mortar for wet mix

**= [1.2 + (1.2×20%)] = 1.44 Cum.**
Again
adding 25% for the total dry volume

**= [1.44 + (1.44 × 25%)] = 1.80 Cum.**

__QUANTITY OF MATERIAL__**:**

**To obtain, considering 1:4 cement sand mortar (generally 1:3 to 1:4 considered for R.C.C ceiling plastering.**

Dry
volume of mortar = 1.80 cum

Cement: Sand = 1:4

Cement + Sand = 1 + 4 = 5

**For Cement**–

**For Sand –**

**SIMILARLY, MATERILA FOR 6MM THICK PLATERING OF 100 SQM WALL AREA FOR R.C.C CEILING:**

__QUANTITY OF MORTAR__**:**

**To obtain –**

Quantity
of Mortar for wet mix of uniform layer

**= (6/1000) × 100 = 0.6 Cum.**
In
case of R.C.C ceiling plastering the unevenness of surface will be less, so
considering 20% for the un-uniform
surface, the Quantity of Mortar for wet mix

**= [0.6 + (0.6×20%)] = 0.72 Cum.**
Again
adding 25% for the total dry volume

**= [0.72 + (0.72 × 25%)] = 0.9 Cum. Says 1.0 cum**
Generally
in practice for 6mm thick plastering of R.C.C ceiling the quantity of dry
volume of mortar may be considered as 1.0 cum.

__QUANTITY OF MATERIAL__**:**

**To obtain, considering 1:4 cement sand mortar.**

Dry
volume of mortar = 1.0 cum

Cement: Sand = 1:4

Cement + Sand = 1 + 4 = 5

**For Cement**–

**For Sand –**

__FOR NEAT CEMENT FLOORING__**:**

The
neat cement finishing, in floor, dado, or skirting, the thickness of the cement
layer over the finished uniform surface may be taken as 1.5mm (1/6”) thick.

**To calculate the quantity of cement considering a uniform floor area of 100 Sqm with 1.5mm thick cement finishing –**

Quantity
of cement paste = (1.5/1000) × 100 = 0.15 Cum.

Adding
25% for the total dry volume = [0.15 + (0.15 × 25%)] = 0.187 Cum. Says 0.19 Cum.

Considering
1 cum of cement = 1600 kg, for 0.19 cum = 0.19 × 1600 = 304 kg = 6.08 Bags Says
6 bags.

For
neat cement finishing of 100 Sqm. floor areas with thickness of 1.5mm 6 bags of
cement required.

## No comments:

## Post a Comment