soil mechanics cheatsheet, Cheat Sheet of Soil Mechanics and Foundations

Download soil mechanics cheatsheet and more Cheat Sheet Soil Mechanics and Foundations in PDF only on Docsity! For a soil, given: D10 = 0.08 mm D30 = 0.22 mm D60 = 0.41 mm Calculate the uniformity coefficient and the coefficient of gradation of the soil. Ex 1) A density test was performed on a soil sample obtained from the field. The sojl sample was obtained from a cylindrical sampler similar to those shown in Figures 3.19 and 3.20. The sample was contained in 4 brass rings, each of which had an inside diameter of 2.43 in. and was a fh ol net weight of the soil in the four rings was 1.35 Ib, Solution: The total volume of the soil V is equal to the inside volume of four brass rings: 1 tie = 1 i243 in)(4 ind) X Vv grl24 in, )*(4 in.) (ik Therefore, the unit weight of the soil is given by: y= fR-— 222, = pomee 0.0107 fe ) = 0.0107 ft? Ex 2) A 27.50 lb soil sample has a volume of 0.220 ft’, a moisture content of 10.2%, and a specific gravity of solids of 2.65. Compute the unit weight, dry unit weight, degree of saturation, void ratio, and porosity. Solution using fundamental and derived equations: a y -4 27.501 _ 125 vee! ~ 0.220 88 125.0 lb/ft? 3 Ya = ie Te pion 7 3th = par ae X 100% = 59% 62.4 Ib/ft 1 113.4 Ibift? 2.65 _ (2.65)(62:4 Ibi?) 113.4 Ib/tt} 0.4582 x 100% = ——— Yo = 31.4% 1 + 0.4582 x 100 31.4% -—1= 0,458 Ex 4) The moist weight of 0.40 cubic ft of soil is 45.6 Ibs. If the soil dry weight (after oven drying) is 37.8 Ibs and the specific gravity of soil solids is 2.65, determine the following: a) Water content b) Moist unit weight c) Void ratio d) Porosity e) Degree of saturation Volume V,= 0.04 ft? v= 0.17 2 . Vy = 0.13 f° V,=0.23 ft Solid w=3780 Water Ww, 45.6lb — 37.8Ib -_ —% 9 ee 100% 1: ea Oe = aa = 20.6% Unit weight W 45.6lb 2, of moist = = 8 = 144. 01/43 soil (y) V0.40rt? — Wy _ 45.6lb - 37.8lb _ ; 3. w= goat = Ot8tt W, 37.81b ~ Gov (2,65)(62.41b/f5) V,= Vv Vi, — Ve = 0.40 ft? — 0.13 #9 — 0.23 #3 = 0,04 #8 V,=V-—V, = 0.40 ft? — 0.23 ft? = 0.17 #3 or Y= Vq + Vy = 0.04 fi? + 0.13 ft? = 0.17 ft? 0.17 ft Void = = = 0,74 ‘oid ratio (e) 0.238 V, ; 4. Porosity (n) = a x 100% = oe x 100% = 42.5% Degree of _ Ww 0.143 ft 5. xX 100% = = X 100% = 76.59 saturation(S) VY, 0.47t8 % = 76.5% Ex 5) The degree of Satiiration OF a Soilisi000% and the moist unitweightsse:20slbsicubiewt. The water content is 36% Determine: a) Specific gravity b) Void ratio = Wy + us 120|b (A) Ww, 7088 wo iwhac z (B) Can tent From Equation (B), W,, = 0.36W,; substitute into Equation (A). 0.36W, + W, = 120 Ib W, = BI2I5) W, = 0.36W, = (0.36)(88.2'b) S118 15) Vy = Vy = 0.51 £7 | Water | Wy=318 1b v, =0.49 £0 Solid Wi, = 88.2 Ib vette GC, > 4 & i1.< Ce < 3. Well graded Gravel G, > 6 & 1<€,< 3 Well graded Sand Results from liquid, plastic and shrinkage limit tests conducted on a soil are given below: Liquid limit test (N, w): (14, 38.4%), (16, 36.5%), (20, 33.1%) and (28, 27.0%) Plastic limit tests: PL=17.2 and 17.8%: use average value Shrinkage limit test: V; = 20.6 cm3, Vp = 13.8 em?, M, = 47.5 gand Mz = 34.6 g (dry soil) a. draw the flow curve and obtain the liquid limit b. what is the plasticity index of the soil c. what is the soil shrinkage limit F a. Refer to the plot of w versus N. ® LL =29.0 _AT24I7.8 | b. Plasg = 173 é PI=LL PL, = 29.0-17.5 =11.5 M,-M, = Vy Kp si (4 000 lam (2, (100) 47,5-34.6 20.6-13.8 = : 100) —| — ——— (19100 [ 34.6 k ) ( 34.6 \ (100) = 17.63 A soil has a liquid limit of 61 and a plastic limit of 30, A moisture content test performed on an undisturbed sample of this soil yielded the following results: Mass of soil + can before placing in oven 96,20 g Mass ofsoil + can after removal from oven 71.90 g Mass of can 20,80 g Compute the following: (a) The plasticity index. (b) The moisture content, (c) The liquidity index. »« Answers? Mass bo) Wb.2-26.8 = 75.4 HA -268 = Ey] i = ee Ls Table 5.2 Unified Soil Classification System (Based on Material Passing 76.2 mm Sieve) Criteria for assigning group symbols Group symbol Gravels Clean Grave's 0,2 4end1 $c, $3 cw More than 50% of coarse Less than 5% fines* Cy <4 end/or Cp <1 or G.> 3 GP Coarse-grained soils factonre:aneconNo.4 — Grave's with Fines PI < 4 or piots below “4’ line (Figure 5.2) GM More than 50% sieve Wore than 12% fings®? ——-P/ > 7 and plots on or above “A” Ine (Figure 5.2) cc retainad on No. 200 Sands Clean Sands Cy2Gand1$0,< 3° sw sleve 80% or more of coarse —_Less then 5% fines® 0, <6 endlor C, <1 or O,> sP fraction passes No. 4 Sands with Fines PI< 4 or pots belo line (Figure 5.2) SM siave More than 12% finas®’__P/> 7 and plots cn or above "A" line (Figure 5.2) sc Inorganic PI> T and plots on or above "A’ line (Figure §.2)° cL Silts and clays PI <4 or pits below “A’ line Figure 5.2)° ML Liquid limit less than 50 i Fine-grained soils eS ae Organic Liquic limitoven died «9 75; see Figure 5.2: OL zone OL 50% or more passes Liguid imit—net dried No. 200 sieve Inorganic PI plots on or above *A’ line (Figure 5.2) cH Silts and clays PI plots below “A” line (Figure 5.2) MH Einuid Ire e0 or more ‘Organic iguid limit oven dried - 9 75: see Figure §.2; OH zone OH Tiquid limit—rot dried Highly organic soils Primarily organic matter, dark in color, and organic oder Pt *Gravels with 5 t0 12% fine require dual symbols OW-CM, GW-GO, OF-GM, GP-SC. Deo : (Diy)? 'Sands with 5 to 12% fines require dual symbols: SW-SW’, SW-SC, SP-SM, SI C.==8; C. = Far Do 0 Dry if 4 $ PI <7 ard plots in the hatched area in Figure 6.2, use dual symbol GC-GM or 3C-SMI ¢if 4 © PI<7 ard plots in the hatched area in Figure 6.2, use dual symbol SL-ML.