**The Dimension Of This Slab**

Length – 6m or 20 feet

Breadth – 6m or 20 feet

Depth – 0.150m or 0.5 feet

Grade of concrete – M20

**M20 Mix Ratio = 1 : 1.5 : 3**

**The Mix Ratio Denotes The Following**

**1**

**– Cement**

**1.5**– Sand (Fine aggregate – 1.5 Times of Cement Quantity)

**3**– Blue metal (Coarse aggregate – 3 Times of Cement Quantity)

Volume of concrete we need = length x breadth x depth =

**L x b x h = 6 x 6 x 0.15 = 5.4 m**

^{3 }

^{ }**O**

**K**

**First, we can find the volume of concrete for**

**1m3**

**so we can use on different volume**

**The volume of 1m**

^{3}concrete

We find out the volume of concrete in wet condition 1m3 (After the consolidation of Cement + sand + water). So in dry conditions, we have to add 30 to 35% sand bulkage and add 20% for wastage.

Accounting bulkage of sand - 34%

Accounting wastage of material -20%

**Hence 1 cum becomes 1.54 (1+0.34+0.2) cum**

**Total Part of the Concrete**= 1+1.5+3 = 5.5 Parts

**Calculation of Volume of Cement in 1m3 of Concrete**

**Cement Quantity**= (Cement Part / Concrete Parts) x Concrete Volume

Cement part = 1 (1:1.5:3)

Concrete parts = 1+1.5+3 = 5.5 Parts

Density of cement - 1440 kg/cum

Cement = (1/5.5) x 1.54 = 0.2798

Density of Cement = 1440 kg/ m3 = 0.2798 X 1440 = 403.19 kg

**1 Cement Bag Weight = 50 Kg**

Hence 403.29/50 =

**8.06 bags**

**LCET**

**ED**

**TRIVIA**

To be simple

Volume of,

One cement bag = 0.0347 cum

One cement bag = 1.225 cft

Hence Required cement = ((1/5.5) x (1.54))/(0.0347) = 8.06 bags

**Calculation of The volume of sand in 1m3 of Concrete**

**Sand Quantity**= (sand Part / Concrete Parts) x Concrete Volume

Sand part = 1.5 (1:1.5:3)

Concrete parts = 1+1.5+3 = 5.5 Parts

Density of sand = 1500–1550 kg/cum

Sand = (1.5/5.5) x 1.54 = 0.42

Density of Cement = 1500 kg/ m3 = 0.42 X 1500 = 630 kg

But we all are buying sand in cft

**Therefore,**

**1 Cum = 35.31 Cft**

Sand Quantity = 0.42

= 0.42 x 35.31=

**14.83 Cft**

100 cft = 1 Unit

2.83 cum = 1 unit

1 load - 4 unit

Generally on lifting for sand, 1 unit sand equals 35 cement bags of sand

1 cft box = 1 x 1 x 1.25 ft

**Calculation of Volume of aggregate in 1m3 of Concrete**

**Aggregate Quantity**= (sand Part / Concrete Parts) x Concrete Volume

Aggregate part = 3 (1:1.5:3)

Concrete parts = 1+1.5+3 = 5.5 Parts

Density of aggregate = 1520 - 1680 kg/cum

Aggregate = (3/5.5) x 1.54 = 0.84

But we all are buying aggregate in cft

Therefore,

**1 Cum = 35.31 Cft**

Aggregate Quantity = 0.82

= 0.84 x 35.31 = 29.66 Cft

**Total Materials Required for 1m**

^{3}

^{ }
Cement - 8 bags

Sand - 14.83 Cft

Aggregate - 29.66 Cft

**Therefore,**

We need the volume of concrete for 5.4m

^{3}^{ }

Cement = 5.4 x 8 ( 8 bag of cement for 1m

^{3}) = 43.2 – 43 bags
Sand = 5.4 x 14.83 (14.83 cft of sand for 1m

^{3}) = 80.08 cft
Aggregate = 5.4 x 29.66 (29.66 cft of aggregate for 1m

^{3}) = 160.164 cft**Total Materials Required for 5.4m**

^{3}

^{ }**Cement**- 43 bags

**Sand**– 80.08 Cft

**Aggregate**– 160.164 Cft

Smaller size aggregate's price little high than the larger size

Generally, we use 40 mm aggregate instead of 20mm in PCC for cost-effective

**Calculation of Volume of water in 1m3 of Concrete**

Water cement ratio 0.5 for a bag = 0.5 x 50 ( weight of a bag) = 25 litre

**Therefore,**

We have 43 bags so 25 litre x 43 = 1075 litre

1 drum = 200 (208) litre

**So,**

For 43 bags = 1075/208 = 5.16 drums

**As LCETED civil engineer, you should know how to use the Excel sheet to find this quantity. so use this sheet for your uses**

**You Can Change This Value As Per Your Specification**

**Download it by clicking the link**

**LCET**

**ED**

**TRIVIA**

.

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