HOMEWORK SOLUTION

 

HW #9

 

CE 455

FOUNDATION AND EARTH STRUCTURE

 

(Question no. 8.3, 8.8 ..Braja M. Das textbook)

 

 


Sand

g=17.3 kN/m3

f=30o

 

L1 = 3 m

 

W.T

 
8.3

 

 

 

 

 

 

 


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 = P1L1+P1L2+ (P2-P1)L2+ P2L3

= *17.3*3+17.3*6+ (36.48-17.3)*6+ *36.48*1.42 =213.19 kN/m

- Calculate

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

+ (P2-P1)L2*(L3+1/3 L2)+ 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+gL2)Kp+gL3(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)/g2(Kp-Kp)2 = 1134.8

From equation 8.20 A4 = P(6P5+4P)/g2(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

From equation 8.22

Mmax = P(+z)-( gz2(Kp-Ka))(1/3)z = 1478.93 kN.m

 

 

 

8.8

Sand

g=103 lb/ft3

f=36o

 

L1 = 5 ft

 

W.T

 

 

 

 

 

 

 

 


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 = P1L1+P1L2+ (P2-P1)L2 =2535 lb/ft

- Calculate

= =( P1L1*( L2+ 1/3 L1)+ P1L2+1/6 (P2-P1)L2)/P

= 5.4 ft

- Calculate D:

From equation 8.48

D2(4c-(gL1+gL2))-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)-P6z2/2= 15274.113 lb.ft