Weight of Sand, Cement and Water for Mortar Mix Ratio 1:3

Home/ Questions/Q 32473
Next
In Process

Do you need to remove the ads? Become VIP Member

Asked: 2019-05-03T17:47:26+05:30 In:
Sanjay
Sanjay

Sanjay

View Profile
User

How to calculate exactly the weight of sand cement and water in mix ratio 1:3?

4 Answers

  1. poojan

    poojan

    View Profile
    Learner

    Quantity is calculated for 1 Cu.m of mortar:

    For 1 Cu.m mortar
    Cement Sand W/C ratio Remarks
    1 3 0.5
    Total 4 =1+3
    For 1 Cu.m
    Considering 10% Wastage
    Total Quantity of wet mortar 1.1 10% of  1 Cu.m
    Shrinkage factor 1.25
    Total Quantity of dry mortar (Wet mortar x  Shrinkage factor) 1.375 Multiplying with wet mortar volume=1.25*1.1
    Cement
    Cement 0.34 =(1/4)(1.375)
    Cement Bag                   9.96 =0.375/0.0345 (0.0345 is volume of 1 bag)
    Considering 2% Wastage                   9.96 =10.87*1.02
    Total Cement Bag                   9.96
    Sand
    Sand 1.03 =(3/4)*1.5
    Considering 10% Wastage 1.24 =1.125*1.10
    Total sand 1.23 Cum
    Water
    W/C ratio 0.5
    Cement 498 9.96*50 (50kg is the weight of one bag of cement
    Water 249 =498*0.5
    Total water 249 ltr
    • 0
  2. fathima

    fathima

    View Profile
    Beginner

    the quantity calculated for 1m3 mortar

    the ratio is 1:3 for cement and sand

    assume shrinkage factor 1.33

    now we can calculate for cement for 1m3 mortar=1*1.33*1/4

    =0.3325m3

    50kg cement bag volume=50/1440=0.0347m3

    so cement weight=0.3325*50kg/0.0347=479.1kg

    add 10% wastage=479.1+47.91=527.01kg

    sand volume=1*1.33*3/4=0.9975cum

    add 10% wastgae=0.9975+0.0997=1.0972cum

    now assume water-cement ratio=0.6

    so water =0.6*527.01=316.2kg=316.2ltr

    • 1
  3. aviratdhodare

    aviratdhodare

    View Profile

    Since you have not mentioned whether the 1:3 cement sand mortar is as per weight batching or volume batching, here I’ll describe the process for both the types.

    Weight batching:

    • Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 and suppose w/c ratio is 0.4
    • This means of 1 kg mortar, 0.4 part is water, 1 part is cement and 3 parts are sand.
    • 1 part = 1 kg /( 0.4 + 1 + 3)
    • This gives, 1 part = 0.227

    Quantities of different constituents are,

    • water = 0.4 * 1 part =0.4 * 0.227 kg = 0.09 kg
    • Cement = 1 part = 0.227 kg
    • Sand = 3* 1 part = 0.683 kg

    If the sand used for this mix contained adsorbed water equal to 5% of weight of sand, the calculations would change as follows:

    • Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 and suppose w/c ratio is 0.4
    • This means of 1 kg mortar, 0.4 part is water, 1 part is cement and 3 parts are sand.
    • The dry sand weighs only 1/1.05 times ie: .952 times of moist sand. Hence, to maintain 3 parts dry sand, moist sand will have to be added in 3*1.05 part = 3.15 part.
    • Of the total 0.4 part water required in the mortar, 5% * 3 part = .15 part will be contributed from adsorbed water in sand and only 0.25 part water will be required to be added externally.
    • 1 part = 1 kg /( 0.25 + 1 + 3 + .15)
    • This gives, 1 part = 0.227

    Quantities of different constituents are,

    • water required = 0.25 * 1 part =0.25 * 0.227 kg = 0.056 kg
    • Cement = 1 part = 0.227 kg
    • Sand = 3.15* 1 part = 0.717 kg

    Volume batching:

    • Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 by volume and suppose w/c ratio is 0.4
    • Assume bulk densities of water as 1 g/c, cement as 1.4 g/cc and fine aggregate as 1.6 g/cc.
    • So, mass ratio of cement: sand is 1 part * 1.4 : 3 part * 1.6 .
    • This means of 1 kg mortar, 0.4 part is water, 1 part is cement and 3.43 parts are sand.
    • 1 part = 1 kg /( 0.4 + 1 + 3.43)
    • This gives, 1 part = 0.207

    Quantities of different constituents are,

    • water = 0.4 * 1 part =0.4 * 0.207 kg = 0.083 kg = 0.83 L of water.
    • Cement = 1 part = 0.207 kg = 0.207/ 1.4 = 0.1478 L of cement
    • Sand = 3.43* 1 part = 0.71 kg = 0.71/1.6 = 0.444 L of dry sand.

    If the sand used for this mix contained adsorbed water equal to 5% of weight of sand, which also caused bulking of sand volume by 20% the calculations would change as follows

    • Suppose 1 kg of mortar is to be prepared. Cement:sand ratio is 1:3 by volume and suppose w/c ratio is 0.4
    • Assume bulk densities of water as 1 g/c, cement as 1.4 g/cc and fine aggregate as 1.6 g/cc.
    • So, mass ratio of cement: sand is 1 part * 1.4 : 3 part * 1.6 .
    • This means of 1 kg mortar, 0.4 part is water, 1 part is cement and 3.43 parts are sand.
    • Of the total 0.4 part water required in the mortar, 5% * 3.43 part = .172 part will be contributed from adsorbed water in sand and only 0.228 part water will be required to be added externally.
    • 1 part = 1 kg /( 0.4 + 1 + 3.43)
    • This gives, 1 part = 0.207

    Quantities of different constituents are,

    • water = 0.4 * 1 part =0.4 * 0.207 kg = 0.083 kg = 0.83 L of water.
    • Cement = 1 part = 0.207 kg = 0.207/ 1.4 = 0.1478 L of cement
    • Changed bulk density of sand is 1.6 * 1.05/1.2 = 1.4 ( because of both adsorbed
    • Sand = 3.43* 1 part = 0.71 kg = 0.71/1.4 = 0.50 L of moist sand.
    • 34
  4. Komal Bhandakkar

    Komal Bhandakkar

    View Profile
    AUTHOR
    This answer was edited.

    Weight of different ingredient of Mortar 1:3 –

    Here, I will give you a simple way to calculate the weight of different ingredients in a mortar.

    • The mortar ratio is 1:3, it simply means that the 1 part of cement is present and 3 part of sand is present in that type of mortar.
    • Consider, w/c ratio = 0.4
    • We are doing all calculation for 1kg of mortar.

    1 part = 1/(0.4+1+3) = 0.227

    Water Required =0.25 x 0.227 = 0.056kg

    Cement= 0.2247 kg

    Sand =3.15 x0.227 = 0.717 kg

    Thank You.  

     

     

     

    • 23

You must login to add an answer.

Join for free or log in to continue reading...

Need An Account,