Question

IUF QUESTION 1 2 a) Based on BS5930: 2015: Code of Practice how the particle size distribution curve is det sieving exercise. Elaborate the main concept in of Practice for Ground Investigation, the code explains urve is determined from conducting the wet and de ain concept in conducting these tests. (C01-P01)(C2) (3 marks) b) A partially saturated soil was taken from Cameron Highlanus properties below: taken from Cameron Highlands remediation site having the Initial weight of sample Final weight of sample after oven dried Volume of sample extracted (i.e. overall volume) Specific gravity of soil tested = 2.75 kg = 2.32 kg = 1280 cm = 2.55 Calculate the moisture content, bulk and dry unit weights, void ratio, porosity and degree of saturation for the given soil. (CO2-PO2)(C4) 14

Answer 1

Tally concept o conducting wet ay dry Seving to parbide size distribut on wive: Particle size distribution analysis determine the Seve analysis sedimentation analyse wed for coorse grained sols used for fine grained souls size > 754 - size <i5H The grain size analysis is very much usque in the classificaton a soils. It helps us in determining the swtability o the soil for various civil Engineering purposes such as in the Construction a reade, earthen dams, airfields &c. It also helps us in predicting the soil- water movement on the first hand. the parti de siae distribution of soil may be determined by either q the following methods or a Combination of both. Sieve Analysis I dry sieve analysis a wet sieve analyes Wed Size <2mm → on conducting wet cy dry sieve analysis, we can find % finer results from various calculations a plot a graph between finer ay si ze in mm (particle size distribution curve). then we can estimate whether the soil is well graded on tunipormly graded cor) Gap graded

gop grodd % finer ting ouve well grades unipmly proded Size (mm) - Particle siãe distribution is drawn 1 on lo scale The curve also known as grading curve with the help of this cure, we will be able to grade the soil. A soil may be well graded poolly graded G arde ? => poolly grader ® Given data: Jnital weight g Sample: W - 2.75 kg - 2450 gm Final wt g sample covendred: Wd 2.32 kg - 2320 gm volume e Sample Total V = 1980 cm3 specific grauity & soil G=2.55 To find w,d, dd, e, n, s = ? Bulk oven- drying method to and Moisture content: W- W - Wd non (2.75 -2.327. 2 x 100 = 18.53 %. 2.32 0.1853 Bulk unit weight of soil: 8= W = 2.75x1000 - 2.148 gmic.ca 1280 → Dry unit weight of soil Id = 2 = 2.148 = 1.812 gmice, unit weight of water w = 1gmlcc → Po wa → e- VG-1(.55) - 1 = 0.407 - vod rato d 1.812 0.907 a povesty na fe - 0.400 0.8893 - 28.93 → Degree of Saturadon es - wG - SE WG - 0.1853 (P:55) 0.407 S = 1.561 I+W + 0.1853