Unit 2 Soil Moisture PDF

Unit 2 Soil Moisture Soil found in the earth will typically have moisture or water present. In the lab, the amount of Soil water present in soil is Void space filled with air...

Unit 2 Soil Moisture Soil found in the earth will typically have moisture or water present. In the lab, the amount of Soil water present in soil is Void space filled with air Void space filled with water determined by mass. Determining soil moisture is important because it has a direct impact on a soil's behavior. Water that is present in a highly permeable soil has the potential to transport contaminants.  𝑴𝒂𝒔𝒔 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓 𝑴𝒘𝒂𝒕𝒆𝒓 𝐌𝐂% = 𝟏𝟎𝟎 OR 𝐌𝐂% = 𝟏𝟎𝟎 𝑴𝒂𝒔𝒔 𝒐𝒇 𝒅𝒓𝒚 𝒔𝒐𝒊𝒍 𝑴𝒅𝒓𝒚 Moisture Content: 𝑊𝑒𝑡 𝑠𝑜𝑖𝑙 & 𝑡𝑎𝑟𝑒 − 𝐷𝑟𝑦 𝑠𝑜𝑖𝑙 & 𝑡𝑎𝑟𝑒 MC% = 100 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑠𝑜𝑖𝑙 ** ADD FORMULAS TO CHEAT CHEET** 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑤𝑒𝑡 𝑠𝑜𝑖𝑙 − 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑠𝑜𝑖𝑙 𝑀𝑤𝑒𝑡 − 𝑀𝑑𝑟𝑦 MC% = 100 OR MC% = 100 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑠𝑜𝑖𝑙 𝑀𝑑𝑟𝑦 Mass of Water: Mass of water = (𝑊𝑒𝑡 𝑠𝑜𝑖𝑙 & 𝑡𝑎𝑟𝑒) − (𝐷𝑟𝑦 𝑠𝑜𝑖𝑙 & 𝑡𝑎𝑟𝑒) Mass of water = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑤𝑒𝑡 𝑠𝑜𝑖𝑙 − 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑠𝑜𝑖𝑙 OR Mwater = 𝑀𝑤𝑒𝑡 − 𝑀𝑑𝑟𝑦 Mass of Soil: Mass of dry soil = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑠𝑜𝑖𝑙 & 𝑡𝑎𝑟𝑒 − 𝑇𝑎𝑟𝑒 OR Mdry = 𝑀𝑑𝑟𝑦 & 𝑡𝑎𝑟𝑒 − 𝑇𝑎𝑟𝑒 Mass of wet soil = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑤𝑒𝑡 𝑠𝑜𝑖𝑙 & 𝑡𝑎𝑟𝑒 − 𝑇𝑎𝑟𝑒 OR Mwet = Mwet & tare − Mtare  𝑴𝒂𝒔𝒔 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓 𝐌𝐂% = 𝟏𝟎𝟎 𝑴𝒂𝒔𝒔 𝒐𝒇 𝒅𝒓𝒚 𝒔𝒐𝒊𝒍 Note: the mass of water is always divided by the mass of dry soil, which 𝑴𝒘𝒂𝒕𝒆𝒓 is a constant value. 𝐌𝐂% = 𝟏𝟎𝟎 𝑴𝒅𝒓𝒚 Example 1.1 A soil sample was brought to a lab in order to determine the moisture content. Given the following information calculate the moisture content (MC %): the wet mass of the sample = 1.15 kg and the dry mass of the sample = 1.00 kg. 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑤𝑒𝑡 𝑠𝑜𝑖𝑙 − 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑠𝑜𝑖𝑙 MC% = 100 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑠𝑜𝑖𝑙 1.15kg − 1.00kg Dry Mass! MC% = 100 1.00kg 0.15kg MC% = 100 1.00kg 𝐌𝐂% = 𝟏𝟓. 𝟎% Specimen Containers Also referred to as the moisture tare, pans, tares or moisture cans. Used with an oven or stove to dry back soil samples and remove moisture. Typically made of steel or aluminum. Each container should have a unique identification number. When testing soil in the lab, we need to determine the mass of the clean, empty, try container prior to weighing our soil sample. Example 1.2 Example 1.3 Determine the water content (w %) of an aggregate sample with the A water content determination is required for a silty clay sample. The mass following results. The mass of the container used to hold the sample = 250 of the wet soil plus the container was 17.53 g, and the mass of the dry soil g, the mass of the wet sample and the container = 1550 g, and the mass of plus the container was 14.84 g. The tare of the empty container was 7.84 the dry sample and the same container = 1275 g. g. Calculate the moisture content of the soil sample. 𝑊𝑒𝑡 𝑠𝑜𝑖𝑙 & 𝑡𝑎𝑟𝑒 − 𝐷𝑟𝑦 𝑠𝑜𝑖𝑙 & 𝑡𝑎𝑟𝑒 MC% = 100 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑠𝑜𝑖𝑙 MC%= (17.53-14.84/14.84-7.84)x100 MC%= (1550-1275/ 1275-250) x100 =38.4% =26.83% Example 1.4 Example 1.5 A sample of fully saturated clay has a mass of 1345 g in its natural state and The moisture content equation is used to determine the value of the 985 g after drying. What is the water content of the soil? plastic limit test. The following results were obtained for a plastic limit test. Given the following data, determine the amount of water in the soil. Mass of wet soil + tare = 22.12 g Mass of dry soil + tare = 20.42 g Mass of the container (tare) = 3.50g 36.5 If the moisture content % and wet mass are known, the mass of the dry soil can be determined. Solve for MDRY 𝑀𝑤𝑒𝑡 − 𝑀𝑑𝑟𝑦 MC% = 100 𝑀𝑑𝑟𝑦 Example 1.6 Example 1.7 The mass of a wet soil sample = 1 200 g, the Ten thousand (10 000) kg of C-base aggregate was stockpiled. A sample was taken and the following moisture moisture content was determined to be 10%. Calculate the dry mass of the soil sample. content test results were obtained: the mass of the wet sample = 2.500 kg and the mass of the dried sample = 2.450 kg. The construction specification requires that the compaction moisture content be 10%. Does the 𝑀𝑤𝑒𝑡 current moisture in the stockpile meet the specification? If no, how much additional water would need to be 𝑀𝑑𝑟𝑦 = 𝑀𝐶% added to the stockpile in order to meet the requirement of 10%? 1+ 100 MC%= (2.5-2.45/2.450)x100 =2.04% Mdry= 1090.9 g CURRENT MOISTURE DOES NOT MEET SPECIFICATIONS Mdry = 10 000/1.0204 = 9 800kg 0.1= Mwater/9800 Mwater= 980 kg Mwater=0.0204x9800 =200 Example 1.8 A stockpile of sub grade soil has a wet mass of 400 000 kg. A sample of the wet soil is obtained from the stockpile. The following moisture content measurements were determined: mass of the tare = 0.550 kg; the mass of the tare and the wet sample = 2.550 kg and the mass of the tare and the dry sample = 2.375 kg. Calculate the moisture content of the sample. Determine the dry mass of the sub-grade soil. Would the moisture content be sufficient to meet the construction specifications requirement of 20 % moisture content? If no, determine the amount of water (In liters) the contractor would need to add to meet the requirement.