HOMEWORK SOLUTION

HW #3

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 3.2, 3.3b, 3.4, & 3.5 ..Braja M. Das textbook)

3.2 Gross allowable load = 1805 kN, FS = 3, Df = 1.5 m, g = 15.9 kN/m3, f = 34o, c = 0. Width of square foundation (B)?

Terzaghi general shear failure equation:

qu = 1.3 cNc + qNq + 0.4 gBNg, from table 3.1 Nq = 36.5, Ng = 38.04

Allowable bearing capacity (qall) = 1805/B2 = qu/FS

qu = 0 + 15.9*1.5*36.5+ 0.4*15.9*B*38.04

qall = qu/3 = 290.175 + 80.6448*B

1805/B2 = 290.175 + 80.6448*B

1805 = 290.175*B + 80.6448*B2

B = 2 m

3.3b B = 1.5 m, Df = 1.2 m, f = 35o, c = 0, g = 17.8 kN/m3, FS = 4, continuous foundation

qu = c Nc Fcs Fcd Fci + q Nq Fqs Fqd Fqi + ½ g B Ng Fgs Fgd Fgi

From table 3.4: Nq = 33.3, Ng = 48.03, tan f = 0.7 and from table 3.5:

Fqs = 1 + B/L tan f ; B/L » 0

Fqs = 1 + 0 = 1,

We have Df/B = 0.8 < 1

Fqd = 1 + 2 tan f (1 – sin f)2 Df/B = 1 + 2*0.7*(1 – 0.5736)2*0.8 = 1.2037

Fgd = 1

Fqi = Fgi = 1  no inclination

Allowable bearing capacity = qu/FS

= (17.8*1.2*33.3*1*1.2037*1+ ½ +17.8*1.5*48.03*1*1*1)/4 = 374.34kN/m2

3.4 Shallow square foundation, angle of load inclination b = 15o

B = 5.5 ft, Df = 4 ft, g = 107 lb/ft3, f = 25o, c = 350 lb/ft2, FS = 4

By using eq. 3.25:

qu = c Nc Fcs Fcd Fci + q Nq Fqs Fqd Fqi + ½ g B Ng Fgs Fgd Fgi

B/L = 1.0 and Df/B = 0.727

From table 3.4 for f = 25o

Nc = 20.72, Nq = 10.66, Ng = 10.88, Nq/Nc = 0.51, tan f = 0.47

From table 3.5:

Fcs = 1 + 0.51 = 1.51

Fqs = 1 + 0.47 = 1.47

Fgs = 1 – 0.4 = 0.6

Fcd = 1 + 0.4*0.727 = 1.2908

Fqd = 1 + 2*tan f (1 – sin f)2*Df/B = 1.228

Fgd = 1

Fci = Fqi = (1 – bo/90o) = 0.694

Fgi = (1 – b/f)2 = 0.16

Qu = 350*20.72*1.51*1.2908*0.694+107*4*10.66*1.47*1.228*0.694

+0.5*107*5.5*10.88*0.6*1*0.16= 980964+5715.79+307.34 = 15832.77 lb/ft2

Allowable bearing capacity = qu/FS = 3958.1925 lb/ft2

Allowable load = qall*B2 = 119,735.32 lb

3.5 B = 2 m, L = 3 m, Df = 1.5 m, f = 25o, c = 50 kN/m2, FS = 4

qall = (qu – q)/FS

From eq. 3.25:

qu = c Nc Fcs Fcd Fci + q Nq Fqs Fqd Fqi + ½ g B Ng Fgs Fgd Fgi

from table 3.4:

Nc = 20.72, Nq=10.66, Ng = 10.88, Nq/Nc = 0.5, tan f = 0.47

From table 3.5:       Df/B = 0.75

Fcs = 1 + B/L * Nq/Nc = 1 + 2/3 * 0.5 = 1.33

Fqs = 1 + B/L * tan f = 1 + 2/3 * 0.47 = 1.313

Fgs = 1 – 0.4 * B/L = 1 – 0.4 * 2/3 = 0.733

Fcd = 1 + 0.4 * Df/B = 1 + 0.4 * 0.75 = 1.3

Fqd = 1 + 2 * tan f (1 – sin f)2 * Df/B = 1.235

Fgd = 1

Fci = Fqi = Fgi =1

We have q = D1*g + D2 * (gsatgw)

= 16.8 + (19.4 – 9.81) = 26.39 kN/m2

qu = 50*20.72*1.33*1.3*1+26.39*10.66*1.313*1.235+0.5(19.4–9.81)

*2*10.88*0.733

= 1791.244+456.172+76.48=2323.896 kN/m2

qall(net) = (2323.896-26.39) = 574.3765 kN/m2

Qall(net) = qall(net) * B * L = 3446.3 kN