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)
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l2 l1= 4 ft Sand g=120 lb/ft3
f=40o L1 = 10 ft W.T
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8.12
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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+gL2) 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
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=( 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
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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 = H4/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
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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+gL2) Ka = 484.344 lb/ft2
P1 = ½ P1L1+P1L2+ ½ (P2-P1)L2 =7638.4 lb/ft
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=( ½ P1L1*(L2+
1/3 L1)+ ½ P1L22 +1/6 (P2-P1)L22)/P1
= 10.8 ft
P6 = 4c-(gL1+gL2)
= 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
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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
Total load of span:
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+gL2) 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
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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
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From equation 8.96
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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
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From equation 8.101
Pu = 9*h2*c
= 9*32*700
= 56700 lb