To buy exactly so much how many it is required – not less and no more – means to show reasonable profitability. Especially, if business concerns such material as cement to which contraindicated long storage because of its hygroscopicity. May be that, having opened bag after a while, you instead of soft fine fraction will find firm stone. How to calculate exact amount of cement which can be required for laying of tie, filling of the base or plaster of walls?
It is required to you
- - calculator;
- - ruler or tape measure.
Instruction
1. Choose grade of solution which is suitable for that type of works which you are going to carry out. Depending on the nature of the performed work, solution (sand-cement, concrete, shlakotsementny and so forth) various durability (brand) can be required. In particular: - apply solution of M200 brands to couplers of floor – M300; - for laying of brick walls or blocks – M50 – M100; - for production of slag stones – M35 – M50; - for production of keramzitny blocks – M50 – M100; - for plastering of walls of the house – M50 – M100; - for the bases – concrete solution M 200-M 300.
2. Define grade of cement which will be required for providing the required solution grade. For sand-cement slurries the grade of cement has to exceed approximately by two-three times design grade of solution. If you need M200 brand solution, you have to buy cement of M400 or M500 brand. For casting of blocks from foam mortar (shlakobeton, keramzitobeton and so forth) the grade of cement has to exceed solution grade by 6-8 times. So for receiving foam mortar of M50 brand you need M300 or M400 cement.
3. Count the necessary volume of solution. For tie of floor, wall plaster, the base and other similar monolithic concrete structures necessary volume is counted by multiplication of the corresponding parameters (thickness, length, width) at each other. The required amount of solution for brickwork is accepted equal 0.2 - 0.25 m³ by 1 m³ of laying, for wall construction from blocks of standard dimension - 0.05 m³ of solution on 1 m³ of laying.
4. Find on the Internet compounding for mixing of mortar of the brand necessary to you with use of certain grade of cement (step 2). M400 cement is the most widely applied. With its use it is possible to prepare multipurpose solutions. In particular, for preparation of 1 m³: - the keramzitobetona of M75 is required to 250 kg of M400 cement; - M50 shlakobetona – 200 kg; - M300 concrete for the base – 380 kg; - sand-cement finishing or masonry mortar of M100 – 340 kg.
5. Proceeding from the required solution volume (step 3) and amounts of the cement necessary for preparation of 1 m³ of solution (step 4), count amount of cement which is required to you. For example, if it is necessary to put brick garage which volume of walls is 10 m³, the necessary amount of cement pays off as follows: 10*0.25*340=850 kg, where: - 10 – brickwork volume in m³;-0.25 - solution consumption on 1 m³ of laying in m³;-340 – the amount of cement demanded on preparation of 1 m³ of solution in kg.