SOLVED PROBLEMS IN FOUNDATION ENGINEERING

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Aldous Sanders
6 years ago
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1 Probl # (6) rctagular footig is actd by a vrtical load of 8060 kn ad two horizotal forcs of 500 kn i both th log ad th short dirctios with a distac of 0.6 fro th groud surfac. Dtri:. Th bas prssur at th four corrs if th footig was o soil.. Pil loads at th four corrs if th footig was o pils whos pattr as show. Solutio: soil 0 kn 8060 KN coc 4 kn Hp = 500 KN * pdstal Hp = 500 KN 5 D C Hs = 500 KN Pdstal.5 4 = 44 kn Footig = 76 kn Soil (5 7 ).0 0 = 60 kn Colu load 8060 kn Total load = = 0,000 kn. 68

2 . as Prssur at th Four Corrs: q ax i , q ax i 0, ax i q = = 55 kn/ q = = 9 kn/ q C = = 4 kn/ q D = = 0 kn/. Pil oads at th Four Corrs. y y x x x y KN = ( + ) + = 76 pil. = 4 + = 8 pil. 0,000 8 ax i = = 97 KN/pil = = 65 KN/pil C = = 497 KN/pil D = = 4 KN/pil 69

3 Probl # (7) For th group of frictio pils show; dtri th sttlt of th clay layrs. T o ta l = K N 7 S O F T C Y 5.0 / D f = / D f = 8.0 c o r s s ib l c o r s s ib l 0 F r ic tio p il s 0.4 v = K N = /(4 *.8 )=.4 K N / = 8.5 K N / = K N / =.9 K N / = 9.0 K N / c o p r s s ib l p il s g r o u p d d d v r tic a l s tr s s w h ic h c u a s s ttl t v r y d s g r a v l Solutio: S P H v = Prssur ara S is th sttlt of th clay layr = 0.5 ( ).5 = 45 kn/. = 0.5 ( ).5 = 6 kn/. = 0.5 ( ).0 = 66 kn/. 4 = 0.5 (.9 + 9).0 = 79 kn/. S = = S = = 0.06 S = = S 4 = = Total Sttlt = = 75. v Not that alost th half of th sttlt of th pil group is du to coprssio of th uppr.50 thick layr. 6:

4 Probl # (8) Dtri th sttlt of th group of poit barig pils show. T o ta l = K N E d a r i g P il s D f = 8.0 S O F T C Y 0.4 v = K N = k N /.0 p il s g r o u p C Y 6.0 =.9 k N / = 9.0 K N /.0.0 v r y d s g r a v l Solutio: S P H v = Prssur ara S is th sttlt of th clay layr v Not; is usd istad of P i th followig quatios S S 77.0 kn / ( ta 0 ) (.8 ta 0 ).9 kn / ( ta 0 ) (.8 6 ta 0 ) 9 kn / ( ta 0 ) (.8 ( ) ( ) ta 0 ) Total Sttlt S = = = 4.5 6;

5 Probl # (9) riforcd cocrt circular pil is driv to th soil as show blow. Calculat th gativ ski frictio forc actig o th pil. N w s a d fill ( ) 7 KN 5 o C o p r s s ib l c la y la y r ( ) q u 8 KN 0 KN C o p r s s ib l c la y la y r ( ) 0.5 R o c k Solutio: 0.5 R o c k K o si 0.4 C u ta q u 5 KN R s u lta t o f la t r a l a r th p r s s u r a t r s t Po KN KN F P o D F KN F Dh Cu F KN Total gativ ski frictio, F = F + F = 7. kn 75

6 Probl # (0) pil supportd foudatio is giv blow. vrtical forc of 00 kn (icludig all th loads o foudatio) is actig dowward at poit K. Dtri th loads carrid by pils, ad C. x is o f S y try Ө K c.g.4 C Ө P il C a p Solutio: ta.464 o 60 ta ta 0 0 x x 0.8 Ecctrici ty : ( ) 0.5 ot : KN. ot of irtia of pils: x x Pil. Th; th forc o pils,, ad C will b; C kn kn / pil / pil 75

7 Probl # () O of th xtrior colus of a stl fra structur is actd by a vrtical forc of 0,000 kn, a horizotal forc of 000 kn ad a ot of 5000 kn.. Du to a poor foudatio soil; it was dcidd to support that colu by a pil foudatio cosistig of a twlv pil group. Pils ar cocrt filld stl pip pils. llowabl axial coprssiv load o ach pil is 000 kn. t is also prdictd that ach pil ca also rsist a horizotal forc of 60 kn. Passiv rsistac of soil will ot b cosidrd. s th slctd pil group satisfactory? K N K N K N.5 *.5 pdstal 5.5 * 7.5 P il c a p a tt r p il a tt r p il V r tic a l P il s Solutio: Colu load = 0,000 kn Wight of pdstal = = 08 kn Wight of pil cap = =,980 kn Wight of soil = ( ) 0 =,560 kn Total vrtical loads = 0, ,980 +,560 =,648 kn ot = = 000 kn. Horizotal loads = 000 kn 75

8 ( ) 60 pil C D Th; = 7-7 = 486 kn/pil = 7-7 = 90 kn/pil C = 7 +7 = 54 kn/pil D = 7 +7 = 788 kn/pil Total load = 4548 = kN Ubalacd horizotal forc = KN Now; kn / pil 60 O. K. axiu axial load: R D D cos ta cos 0.95 U b a la c h r iz o ta l fo r c D 564 KN RD kn 000 kn O. K. C KN

4.2 Design of Sections for Flexure

4.2 Design of Sections for Flexure 4. Dsign of Sctions for Flxur This sction covrs th following topics Prliminary Dsign Final Dsign for Typ 1 Mmbrs Spcial Cas Calculation of Momnt Dmand For simply supportd prstrssd bams, th maximum momnt