HOMEWORK SOLUTION

HW #10

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 8.12, 8.13,8.17,8.18,8.20,8.22 ..Braja M. Das textbook)

 l2

 l1= 4 ft

 Sand g=120 lb/ft3  f=40o

 L1 = 10 ft

 W.T

8.12

a) The theoretical embedment D:

Ka = tan2(45-f/2) = 0.217

Kp = tan2(45+f/2) =  4.6

Kp-Ka = 4.383

g = gsat-gw = 129.4-62.4 = 67 lb/ft3

P1 = gL1Ka = 120*1-*0.217=260.4 lb/ft2

P2 = (gl1+gL2) Ka = (120*10+67*250*0.217=623.875 lb/ft2

L3 = P2/(g(Kp-Ka)) = 623.875/(67*4.383)=2.12 ft

P = ½ P1L1+P1L2+ ½ (P2-P1)L2+ ½ P2L3=13016.745 lb/ft

= =( ½ P1L1*(L3+L2+ 1/3 L1)+P1L2(L3+ ½ L2)

+ ½ (P2-P1)L2*(L3+1/3 L2)+ ½ P2L3- 2/3 L3)/P

= 14.08 ft

L43+1.5L42(l2+L2+L3)-3P((L1+L2+L3)-( +l1))/g(Kp-Ka) = 0

L43 + 49.68 L42  2531.89 = 0

L4 = 6.7 ft

Dtheory = L3 + L4 = 2.12 + 6.7 = 8.82 ft

From equation 8.65

 27 ft

 25 ft

 10 ft

 2.12 ft

 623.875 lb/ft2

 260.4 lb/ft2

 1967.53 lb/ft2

Pd= g (Kp-Ka) L4 = 1967.53 lb/ft2

c) From equation (8.66) :

F = P  ½ (g (Kp-Ka))L42

= 6425.52 lb/ft  (Anchor force)

8.13

We have           D theory = 8.82 ft

D actual = 1.3 * 8.82 = 11.466 ft

Also, we have,

E = 29*106 lb/in2, sall = 25 kip/in2

H = L1 + L2 + Dactual = 46.466 ft

To obtain Mmax:

From equation 8.72

Mmax = (GM)(CML1)ga(L1+L2)3

From fig. 8.20 for l1/(L1+L2) = 0.11 and f = 40o

GM = 0.016

From fig. 8.23 for l1/(L1+L2) = 0.11 and L1/(L1+L2) = 0.286

CML1 = 1.022

From equation 8.73

ga = (gl12+(gsat-gw)L22+2gL1L2)/(L1+L2)2 = 93.98 lb/ft3

Mmax = 65888.63 lb.ft

By taking section Pz-27:

s=30.2 in3/ft, I = 184.2 in4/ft

Md = sall.S = (2500*30.2)/12 = 62916.67 lb.ft/ft

From equation 8.75

r = H4/EI = 4.8*10-4

Log r = -3.06

Md/Mmax = 0.95

From fig 8.24 The section is safe.

The pile section is Pz  27.

 l2

 l1= 4 ft

 Sand g=115 lb/ft3  f=40o

 L1 = 8 ft

 W.T

8.17

Free earth support method

a)  Ka = tan2(45-f/2) = 0.217

g = gsat-gw = 65.6 lb/ft3

P1 = gL1Ka = 199.64 lb/ft2

P2 =(gl1+gL2) Ka = 484.344 lb/ft2

P1 = ½ P1L1+P1L2+ ½ (P2-P1)L2 =7638.4 lb/ft

= =( ½ P1L1*(L2+ 1/3 L1)+ ½ P1L22 +1/6 (P2-P1)L22)/P1

= 10.8 ft

P6 = 4c-(gL1+gL2) = 3768 lb/ft2

To determine D:

From equation 8.77

P6D2+2P6D(L1+L2-l1)-2P1 (L1+L2-l1-1)  = 0

3768D2+180864D+201653.76=0

D2+48D+53.52=0

D = 1.15 ft

b) from equation 8.76

F = P1  P6D = 3305.2 lb/ft

 l2

 l1= 5 ft

 Sand g=108.5 lb/ft3  f=35o

 L1 = 9 ft

 W.T

8.18

Fixed earth support method:

a) Maximum moment:

Ka = tan2(45-f/2) = 0.27

Kp = tan2(45+f/2) =  3.69

g = gsat-gw = 66.1 lb/ft3

Kp-Ka = 3.69-0.27 = 3.42

For f = 35o    L5/(L1+L2) =0.03

L5 = 0.03*35 = 1.05 ft

L = l2+L2+L5 = 4 + 26 + 1.05 = 31.05 ft

Intensity of load at anchor level:

Pa=gl1Ka = 108.5*5*0.27=146.475 lb/ft2

Intensity of load at depth = L5 below dredge line:

Pd = g(Kp-Ka)L5 = 237.365 lb/ft2

P1 = gL1Ka = 263.655 lb/ft2

P2 = (gL1+gL2) Ka = 727.677 lb/ft2

W = ½ (P1-Pa)l2+ ½ (P2-P1)L2+ ½ (P2-Pd)L5=6524.06 lb/ft

Mmax = WL/8 = 25321.5 lb.ft/ft

b) Depth of Embedment D:

P = SMo/L = ( ½ P1L1(2/3L1-L1)+P1L2(L2/2+L1-l1)+ ½(P2-P1) L2(2/3L2+l2)+PdL5(L5/2+L2+l2)+ ½ PdL5(L5/3L2+l2)

= 8302.76 lb

From equation 8.87

c) Anchor Force F:

F = SMI/L = ( ½P1L1(1/3L1+L2+L5)+PdL52/2+ ½ (P2-Pd)L5*2/3*L5)/L

= 6147.85 lb/ft

8.20

From equation 8.96

a)      B = 0.3 m

Pu = 22.015 kN/m2

b)      B = 0.6 m

Pu = 36.26 kN/m2

c)      B = 0.9 m

Pu =48.56 kN/m2

8.20

From equation 8.101

Pu = 9*h2*c

= 9*32*700

= 56700 lb