geotechnical engineering-civil engineering

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geo tech engineering part 1 (soils)

q1) wet weight =220 lb. dry weight 205 lb. total volume = 1.73 ft^3. specific gravity =2.65(gs). what is void ratio?

a) Ww= Wair+Wwater. Wair=0. Ws=solids. Ww+Ws=W. Va+Vw=Vv. Va=Vair. Vw=Vwater. Vs=Vsolids. Vv+Vs= V. void ratio=Vv/Vs. wa=0. W=220 lb. wair=0. Ws=25 lb. Ww=15 lb.

specific gravity= r/s between material's unit weight and the unit weight of water.

fundamentals properties: volume of voids->vv=va+vw. total unit weight y=w/v. saturated unit ysat=(gs+e)yw/(1+e). Yw=62.4 lb/ft^3 or 9.81 kn/m3

effective(submerged) unit weight=y=ysat-yw. Ww=220-205=15 lb. Vw=Ww/&w=15 lb/62.4^15/ft^3= 0.2403 ft. dry unit weight->yd=Ws/v. unit weight of solids->ys=Ws/Vs

Vs=Ws/ys=Ws/gsyw=205/(2.65)(62.4^16/ft^3)=1.240 ft^3. V=1.73ft^3. vw=0.240ft3. Va=V-(Vs+Vw)=1.73-(0.240ft^3+1.24ft^3)=0.25ft^3. Vv=Va+Vw+0.490ft^3. e=Vv/Vs=0.4903/1.240ft^3=0.40(answer)

q2)d60=0.49mm, d10=0.19mm,coeffiecient of curvature(cc) ? d10 X d60=denominator.

a)go past number 4 sip->sand/silk. anything retained on 4 sip->gravel. difference between sand and silk->what goes through 200 sip. cu=d60/d10

d30^2=numerator. cu=d60/d10 ; Cc=d30^2/d10 x d60 = 0.4^2(from graph)/0.19x 0.49 =1.71.(answer)

q3)percentage of fines=69%(fine grained soil). liquid limit=72. plastic limit=48. (water contents) unified soil classification system. what is soils's classification?

a)water content%->W=Ww/Ws X 100. specific gravity of soil solids->ws/vs / yw. void ratio->e=Vv/Vs.

PI=LL-PL=72-48=24(range of water behaviour where we can observe plastic behaviour of soils.

plastic index->difference of liquidity limit. liquid limit>plastic limit>shrinkage limit.

liquid limit->water content where water behaves in certain way as plastic; between liquid and plastic(like transition).

plastic limit->water content which define difference behaving like a plastic and like semi solid.

*std practice fr classification of soils for engineering purposes(unified soil classification system) plastic index->difference of liquidity limit.

coarse grained soils->gravel,sands,. fine grained soil->silt,clay. void ratio e=Vv/Vs. pi plots below a line(elastic salt).(answer)

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q4)for impermeable dam, what is rly the total flow of water under the dam per foot of dam width?

a)ground->o elevation. dam is 20ft from ground. soil under dam is permeable; water can flow under dam. k had->coefficient of permeability; how much water the soil is allowing to pass through over a given period of time. flow net->way engineers can map the flow of water under a dam.

fig flow net; equal potential generating drop lines->hit right angles at surface. *4 flow paths; each potential line >representing total amount of pressure drop. 10 drops. k hyd(coefficient of hydration)=0.01 ft/min.

Nd=20.;Nf=4. ->Nd=10;Nf=2

discharge velocity v=ki. h=20ft. k=0.01. flow net: kh(Nf/Nd), where q=flow per unit time. Nf=number of flow channels, Nd=number of equipotential drops. h=total hydraulic head differential.

q=k(change/delta h)(Nf/Nd)=0.01ft/min0(20')(4/10)(1 ft)=0.08 ft^3/mm(answer)

q5) for soil profile shown, what is most nearly the effective stress at point c?

a)dry sand y dry=16.5/m^3

saturated sand y sat=19.25 kN/m^3. *deeper u go , higher the pressure. saturated unit weight can be computed, if have dry unit weight/void unit weight.

sigma c=gamma dry(6m) + gamma salt(13m) -gamma h20(13m) = 16.5kn/m^3)(6m) + (19.25k/m3)(13m) - 9.81k/m3(unit weight of water)(13m) =221.72kpa (answer)*compute effective stress->add stresses going down.

anything under ground water table, deal with saturated unit weights. subtract out water pressure however far point is away from ground water table.

q6)direct shear test performed on dry silty sand. normal stress=150kpa, diameter=71mm,height=25mm. shear force at failure=276n. what is most nearly the angle of internal friction?

a)soil mechanics->is the soil going to behave approximately or is it going to fail if applied with a given amount of stress.

different failure crterion. eg,steel design->von mesis/j2 criterion for ductile materials like metals->mohr coulomb favoured criterion more appropriate for brittle materials like soil,concrete.

mohr coulomb failure: towel=sigma (torsion sign) = c. (sheer stress equals cohesion adds normal stress tangent of angle of internal friction.)

difference between granular soils like silt,clay is idea of cohesion. *dry silt sand->sand is defined as cohesionless soils.

dry silt sand->c=0. (towel-0 + tan torsion sign. sigma=150kps(normal stress) direct shear stress->make cylinder of soil and just shearing it(perform on cylinder).

v=276n=0.276kn. A=pi/4 d^2(= pi/4( 0.071m)^2=0.00396m^2.

towel=V/A=0.276kn/0.00396m^2=69.7kpa. 69.7=0+ 150tan torsion sign.

tan torsion sign=69.7/150. torsion sign=tan^-1(69.7/150)= 24.9^degrees. (answer)