geo tech enginerring part 2 foundations

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worked examples

syllabus: phase relationships(geo technical) (plastic index) Pi=Ll-Pl. ll=liquid limit. Pl=plastic limit. coeffiecient of uniformity Cu= d60/d10.

coeffiecient of concavaity(for curvature) cc=d30^2/(d10 x d60). hydraulic conductivity(also known as permeability). from constant head test: k=q/(iate). i=dh/dl, q=total quantity of water. from falling head test: k=2.303(al/ate) log10(h1/h2). A=cross sectional area of test specimen perpendicular to flow. a=cross sectional area of reservoir tube. te=elapsed time. h1=head at time t=0, h2=head at time. t=te. L=length of soil column. discharge velocity, v=ki.

q7)continuous footing founded 3 feet below ground surface of clay for which y-125pcf(per cubic foot), c=1200psf, (per square foot). and angle of torsion sign/internal friction=0 degrees. if factor of safety is the to be at least 3.0, what is most nearly recommended alowable bearing capacity? for purposes of this problem, take Nc= 5.14, Nq=1,and ny=0. *shallow footing, y unit weight, (c=cohesion), torsion sign= angle of internal friction.

a) syllabus: ultimate bearing capacity. q ult=cNc+y'dfnq+1/2y'bny. *use factor of safety

Nc=bearing capacity factor for cohesion, Nq=bearing capacity factor for depth. ny=bearing capacity for unit weight. b=width of strip footing. Df= depth of footing below ground surface.

9u=cNc + yDfNq+ 1/2y Bny(0, as ny=0)= (1200psf)(5.14)+(125 pcf )(3ft)(1.0)=6543psf.

9u(allowing load)=9u(ultimate load)/fs(factor of safety)=6543psf/3.0=2181psf. (answer)

q8)what is most nearly the overturning moment per foot of lineal wall due to the applied earth pressure? y=120pcf, torsion sign(angle of internal friction)= 32 degrees. earth pushing this way. (wall problem). gravity wall retains lean dry sand.

a)cohesion value for sand=0. 2 modes of failure:earth pushing hence wall is going to slide and overturning pressure increases as a function of depth(fluid mechanics). stress profile pushing up against the wall. passive pressure resulting against active pressure pushing.

ka (yxh) (to determine pressure at bottom of wall. 2 pressures of wall)-> active earth pressure and passive earth pressure. syllabus: active forces on retaining wall per unit wall length(as shown).

ka= rankine active earth pressure coefficient(smooth wall,c=0, level backfill)= tan^2(45^2-internal friction/2)

syllabus: passive forces on retaining wall per unit wall length(similar to the active forces shown). kp=rankine passive earth pressure coefficient(smooth wall,c=0, level backfill) =tan^2(45degrees =internal friction angle/2)

gamma h->affected by pressure of active pressure coefficient.

ka(active pressure coefficient) ka=tan^2(45degrees-internal friction angle/2)=tan^2(45 degrees-32/2degrees)=tan^2(29degrees)=0.3073.

pa(pressure at bottom of wall)=ka(gamma h)=(0.3073)(120pcf)(8ft) = 294.97psf.

pa=1/2pa h=1/2 (294.97psf)(8ft)=1180.0 lb/ft of wall(assumption) pa->magnitude. mo=pa(h/3)=(1180.0 lb/ft)(1/3)(8ft)=3146.7 ft-lb/ft of wall.(answer) (pounds per foot of that consumption.

mo->determine moment of that wall(overturning moment)

*wall is preventing earth pressure from tipping it over. must dethrone 2 factors of safety and summation of forces. if summing moments, the earth pressure is wanting to overturn one way; but wall is acting ib opposite direction.

syllabus: horizontal stress profiles and forces effective vertical stress. pore water pressure. effective horizontal force. pa1=1/2 sigma1'kAH1. sigma'=y1 x h1. pw=1/2yw h2^2. effective horizontal force. pa1=1/2sigma1' kah1. pa2=sigma1'kah2 +1/2sigma2'kah2. sigma2'=(y1 x h1) + (y2x h2) - (yw x h2). y2=saturated unit weight.

*standard case for retaining law is to use sand (granular soil) as it allows water to seep through; as opposed to clay where water is stuck.

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q9)consolidation problem. a normally consolidated clay has a liquid limit of 45 and an initial effective stress(at the midpoint of the layer) of 2000psf. placement of a sand fill over the layer causes a vertical stress increase of 800 psf. most nearly, how much will the void ratio at the midpoint of the layer increase?

a). clay layer>retains water(like a sponge).

if clay layer in a soil profile->water will escape. consolidation is a primary setlement of soils.

pre consolidation->pre loads sides by putting stores on it to get the settlement out of the way before construction.

syllabus :Normally consolidated->in virgin compression range : Cc=delta e/delta log p.

compression index->in virgin compression range:Cc= delta/delta log p. by correlation to liquid limit: Cc=0.009(LL-10).

recompression index->in recompression range : Cr=delta e/delta log p. ->by correlation to compression index Cc:Cr=Cc/b. p2->final pressure. p1->initial pressure.

void ratio(e)=Vv/Vs. Cc=0.009(LL-10)=0.009(45-10)=0.315.

compression index(cc)=delta e(void ratio)/delta log p(pressure) =Cc(log10p2-log10p1)=Cc log10(p2/p1) =0.315 log10(2800psf/2000psf) = 0.046(answer)

q10)initial void ratio=0.900. compression index=0.35. 5 ft which consolidated clay layer located between 2 sand layers. surface load increases existing average stress in the layer from 1000psf to 1500psf. most nearly, what is the settlement of the clay layer?

syllabus: delta p =Iqs. I=stress influence value at center of consolidating stratum. qs=applied surface stress causing consolidation.

if (po and delta +po)< pc then delta h=( ho/(1+po) ) (Cr log(po+deltap) /po )

if (po and (po+delta p)>pc, then delta h=( ho/(1+po) ) (Cc log (po+delta p) /po).

if po

where: delta h=change in thickness of soil layer. *soil layer has stress and internal water. apply some load. its going to settle. apply virgin compression range(cc) (soil as it is is & can press it). compression index: Cc=delta e/ delta log p. by correlation to liquid limit : Cc=0.009 (LL-10)->apply this concept to qn.

compression index(press it and then press again). delta p->change in stress. po->initial stress.

(po+delta p)-> final stress. *1feet=12 inches

a)delta h (change in thickness of soil layer =( h(height)/(1+e^o)(void ratio) )( cc(virgin compression range)log10 ( (po+delta p)/po ) = ( h0/(1+e^0) ) ( cclog10(p2/p1) ) =( 5ft/(1+0.900) ) (0.35) ( log10(1500psf/1000psf) ) = 0.162ft1.946" (inches) (answer)

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q11) 10 ft thick clay layer drained by sand layers at top and the bottom. it is predicted to settle a total of 2.0 inches under the stress increase due to a foundation load. the coefficient of consolidation is 0.03 ft^2/day. what is most nearly the settlement at the end of one year?

a)*2 primary sources of settlement in civil engineering: 1)primary settlement of consolidation ; put load. immediately get settlement right there & then. 2)secondary settlement consolidation ; over time. put load on foundation. let it sit there for 5 years etc. going to settle over time.

syllabus : (relate to parameters given in problem). Time factor with degree of consolidation-> related to percent of settlement that u achieve.

time factor tv= Cvt (coefficient of consolidation multiply time) / hdr2(drainage, height square). double drainage.

Cv=0.03 ft^2/day (given). t=365 days (1 year). Hdr is farthest distance water has to travel to get out of sand layer. r=5ft. tv=( (0.03ft^2/day)(365 day) )/(5ft)^2=0.438.

*assumption: water could escape from top or bottom of sand layers. in this scenario, we take drainage height to be halved of 10ft = 5ft.

syllabus. different types of drainage with h0 constraint. 3 types. 2 way drainage(1x). 1 way drainage (2x).

*variation of time factor with degree of consolidation. based on formula sheet, u%=72->0.431

u%=73->0.446. ans is 0.438 so u% closer to 72%. cv=0.03ft^2/day. t=365 day. Hdr=5ft. u =72.5%.

therefore, s=(72.5%)(2"(inches)) = 1.45 inches(answer).

over more time, will get more settlement. 1 other way of looking at it is looking on the flipside; how much time would it then take to generate settlement.