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

      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+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