HOMEWORK SOLUTION
HW #9
CE 455
FOUNDATION AND EARTH STRUCTURE
(Question no. 8.3, 8.8 ..Braja M. Das textbook)
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Sand g=17.3 kN/m3
f=30o L1 = 3 m W.T
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8.3
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a)
- Calculation of Ka and Kp:
Ka = tan2(45-f/2) = tan2(45-15)=0.33
Kp = tan2(45+f/2) = tan2(45+15)=3.0
- Calculate P1 and P2,
from equation 8.1
P1=gL1Ka = 17.3*3*0.33=17.3kN/m2
from equation 8.2
P1=(gL1+gL2)Ka = (17.3*3+(19.4-9.81)*6)*0.33=36.48kN/m2
- Calculate L3:
From equation 8.6
L3 = P2/(g’(Kp-Ka)) =36.48/((19.4-9.81)*(3-0.33))=1.42m
- Calculate P:
P = ½ P1L1+P1L2+ ½ (P2-P1)L2+ ½ P2L3
= ½ *17.3*3+17.3*6+ ½ (36.48-17.3)*6+ ½ *36.48*1.42 =213.19 kN/m
- Calculate _files/image007.gif){kind=small}
= _files/image010.gif){kind=small}
=( 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
= (0.5*17.3*3*(1.42+6+1)+17.*6*(1.42+3)+0.5*(36.48+17.3)*6(1.42+2)
+0.5*36.48*2/3*1.422)/213.19
= 4.215 m
- Calculate P5:
From equation 8.11
P5=(gL1+g’L2)Kp+g’L3(Kp-Ka)=(17.3*3+9.59*6)*3+9.59+1.42(3-0.33)
= 364.63 kN/m2
- Calculate A1, A2, A3, A4:
From equation 8.17 A1 = P5/g’(Kp-Ka) = 364.63/9.59(3-0.33)=14.26
From equation 8.18 A2 = 8P/g’(Kp-Ka) =66.7
From equation 8.19 A3 = 6P(2g’(Kp-Ka)+P5)/g’2(Kp-Kp)2 = 1134.8
From equation 8.20 A4 = P(6P5+4P)/g’2(Kp-Kp)2
= 3284
- Find Lu:
From equation 8.16 L44+A1L43-A2L42-A3L4-A4=0
L44+14.26L43-66.7L42-1134.8L4-3284=0
L = 9.45 (trial and eror)
Theoritical depth D = L3+L4 =1.42+9.45=10.87 m
b) Total Length of Sheet Pile L1+L2+1.3(L3+L4)=3+6*1.3*10.87=23.131 m
c) From equation
8.21 _files/image012.gif){kind=small}
From equation 8.22
Mmax = P(+z’)-( ½ g’z’2(Kp-Ka))(1/3)z’
= 1478.93 kN.m
8.8
Sand g=103 lb/ft3
f=36o L1 = 5 ft W.T
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a)
- Calculation of Ka :
Ka = tan2(45-f/2) =0.26
- Calculate P1 and P2,
from equation 8.1
P1=gL1Ka = 108*5*0.26=140.4lb/ft2
from equation 8.2
P1=(gL1+gL2)Ka =296.4 lb/ft2
- Calculate P1:
P1 = ½ P1L1+P1L2+ ½ (P2-P1)L2 =2535 lb/ft
- Calculate
= =( ½ P1L1*( L2+
1/3 L1)+ ½ P1L2+1/6 (P2-P1)L2)/P
= 5.4 ft
- Calculate D:
From equation 8.48
D2(4c-(gL1+g’L2))-2DP1-(P1(P1+D
)/(gL1+gL2)+2c)=0
2060D2-5070D-50395.6=0
Theoritical Depth (D) = 6.02 ft.
b) Actual Length of Sheet Pile L1+L2+1.4D=23.43 ft
c) Maximum moment:
From equation 8.42
P6=4c-(gL1+gL2)=2060
From equation 8.49
=P1/Pt=1.23 ft
From equation 8.50
Mmax = P1(+z’)-P6z’2/2= 15274.113 lb.ft