The process of calculation to determine the amount of constituents during a concrete combine or mortar with outlined proportion may be a a part of the quality text in concrete technology. you will refer any customary textbook to be told and apply identical. One Christian Bible on this course is of AN Indian author Dr M L GAMBHIR that i’d counsel. Anyway, the solution to your question is: Since you’ve got not mentioned whether or not the 1:3 cement sand mortar is as per weight batching or volume batching, here i am going to describe the method for each the categories. Weight batching: Suppose one metric weight unit of mortar is to be ready. Cement:sand magnitude relation is 1:3 and suppose w/c magnitude relation is zero.4 This means of one metric weight unit mortar, 0.4 half is water, one half is cement and three components ar sand. 1 half = one metric weight unit /( zero.4 + 1 + 3) This gives, 1 part = 0.227 Quantities of various constituents ar, water = 0.4 * one half =0.4 * 0.227 kg = 0.09 kg Cement = one half = zero.227 kg Sand = 3* one half = zero.683 kg If the sand used for this combine contained adsorbate water adequate five-hitter of weight of sand, the calculations would modification as follows: Suppose one metric weight unit of mortar is to be ready. Cement:sand magnitude relation is 1:3 and suppose w/c magnitude relation is zero.4 This means of one metric weight unit mortar, 0.4 half is water, one half is cement and three components ar sand. The dry sand weighs solely 1/1.05 times ie: .952 times of dampish sand. Hence, to take care of three components dry sand, dampish sand can got to be added in 3*1.05 half = three.15 part. Of the whole zero.4 half water needed within the mortar, five-hitter * three half = .15 half are contributed from adsorbate water in sand and solely zero.25 half water are needed to be added outwardly. 1 half = one metric weight unit /( zero.25 + one + three + .15) This gives, 1 part = 0.227 Quantities of various constituents ar, water needed = zero.25 * one half =0.25 * 0.227 kg = 0.056 kg Cement = one half = zero.227 kg Sand = 3.15* one half = zero.717 kg Volume batching: Suppose one metric weight unit of mortar is to be ready. Cement:sand magnitude relation is 1:3 by volume and suppose w/c magnitude relation is zero.4 Assume bulk densities of water as one g/c, cement as one.4 g/cc and fine mixture as one.6 g/cc. So, mass magnitude relation of cement: sand is one half * one.4 : three half * one.6 . This means of one metric weight unit mortar, 0.4 half is water, one half is cement and three.43 components ar sand. 1 half = one metric weight unit /( zero.4 + 1 + 3.43) This gives, 1 part = 0.207 Quantities of various constituents ar, water = 0.4 * one half =0.4 * 0.207 kg = 0.083 kg = 0.83 L of water. Cement = one half = zero.207 kg = 0.207/ 1.4 = 0.1478 L of cement Sand = 3.43* one half = zero.71 kg = 0.71/1.6 = 0.444 L of dry sand. If the sand used for this combine contained adsorbate water adequate five-hitter of weight of sand, that conjointly caused bulking of sand volume by 2 hundredth the calculations would modification as follows Suppose one metric weight unit of mortar is to be ready. Cement:sand magnitude relation is 1:3 by volume and suppose w/c magnitude relation is zero.4 Assume bulk densities of water as one g/c, cement as one.4 g/cc and fine mixture as one.6 g/cc. So, mass magnitude relation of cement: sand is one half * one.4 : three half * one.6 . This means of one metric weight unit mortar, 0.4 half is water, one half is cement and three.43 components ar sand. Of the whole zero.4 half water needed within the mortar, 5% * 3.43 part = .172 half are contributed from adsorbate water in sand and solely zero.228 half water are needed to be added outwardly. 1 half = one metric weight unit /( zero.4 + 1 + 3.43) This gives, 1 part = 0.207 Quantities of various constituents ar, water = 0.4 * one half =0.4 * 0.207 kg = 0.083 kg = 0.83 L of water. Cement = one half = zero.207 kg = 0.207/ 1.4 = 0.1478 L of cement Changed bulk density of sand is one.6 * 1.05/1.2 = 1.4 ( as a result of each adsorbate Sand = 3.43* one half = zero.71 kg = 0.71/1.4 = 0.50 L of dampish sand.