1. Find: Bearing Capacity, qult [lb/ft2]
infinite strip
footing
3,200
4,800
C) 7,600
D) 8,200
4 [ft]
8 [ft]
Clay
Find the ultimate bearing capacity, q ‘ultimate’, in pounds per square feet. [pause] In this problem ---
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

2. Find: Bearing Capacity, qult [lb/ft2]
infinite strip
footing
3,200
4,800
C) 7,600
D) 8,200
4 [ft]
8 [ft]
Clay
an infinite strip footing 8 feet wide, is buried to a depth of 4 feet beneath the soil surface.
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

3. Find: Bearing Capacity, qult [lb/ft2]
infinite strip
footing
3,200
4,800
C) 7,600
D) 8,200
4 [ft]
8 [ft]
Clay
The soil is a stiff clay with the given soil properties. [pause] Since the groundwater table is not mentioned, it is assumed to be very deep.
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

4. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
4 [ft]
8 [ft]
Clay
We begin by writing out our equation for the ultimate bearing capacity of an infinite strip footing.
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

5. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
cohesion
term
4 [ft]
8 [ft]
Clay
This equation sums the resisting forces from cohesion, ---
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

6. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
cohesion
term
wedge
4 [ft]
weight
term
8 [ft]
Clay
the wedge weight, ---
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

7. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
cohesion
term
wedge
overburden
4 [ft]
weight
term
term
8 [ft]
Clay
and the overburden, of the soil. In this problem we are neglecting the footing thickness, assuming it to be zero.
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

8. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft]
8 [ft]
Clay
We begin by determining which variables we need to calculate.
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

9. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft] Nc
8 [ft]
Clay
Let’s begin by solving for the bearing capacity factors, for the cohesion, N sub c, ---
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

10. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft] Nc
8 [ft] Nγ
Clay
for the wedge weight, N sub gamma, ---
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

11. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft] Nc
8 [ft] Nγ
Clay
and for the overburden weight, N sub q. In a drained condition, --- Nq
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

12. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft]
Nc = (Nq-1) * cot φ
8 [ft]
Nγ = 2 * (Nq+1) * tan φ
Clay
these bearing capacity factors are all a function of the friction angle, phi. Here, I’m using the wedge weight multiplier, N sub gamma, developed by Vesic.
Nq = tan2(45+φ/2) * eπ * tanφ
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

13. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft]
Nc = (Nq-1) * cot φ
8 [ft]
Nγ = 2 * (Nq+1) * tan φ
Clay
We first solve for N sub q, the overburden multiplier, ---
Nq = tan2(45+φ/2) * eπ * tanφ
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

14. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft]
Nc = (Nq-1) * cot φ
8 [ft]
Nγ = 2 * (Nq+1) * tan φ
Clay
by plugging in our phi value, of 16 degrees, ---
Nq = tan2(45+φ/2) * eπ * tanφ
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

15. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft]
Nc = (Nq-1) * cot φ
8 [ft]
Nγ = 2 * (Nq+1) * tan φ
Clay
N sub q equals The other two bearing capacity factors are solved ---
Nq = tan2(45+φ/2) * eπ * tanφ
γT=120 [lb/ft3]
= 4.34
OCR>1
c=400[lb/ft2]
φ=16˚

16. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft]
Nc = (Nq-1) * cot φ
8 [ft]
Nγ = 2 * (Nq+1) * tan φ
Clay
by plugging in N sub q, and phi. [pause] N sub c equals ---
Nq = tan2(45+φ/2) * eπ * tanφ
γT=120 [lb/ft3]
= 4.34
OCR>1
c=400[lb/ft2]
φ=16˚

17. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft]
Nc = (Nq-1) * cot φ = 11.63
8 [ft]
Nγ = 2 * (Nq+1) * tan φ
Clay
11.63, and N sub gamma equals, ---
Nq = tan2(45+φ/2) * eπ * tanφ
γT=120 [lb/ft3]
= 4.34
OCR>1
c=400[lb/ft2]
φ=16˚

18. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ? ? ? ?
4 [ft]
Nc = (Nq-1) * cot φ = 11.63
8 [ft]
Nγ = 2 * (Nq+1) * tan φ= 3.06
Clay
3.06.
Nq = tan2(45+φ/2) * eπ * tanφ
γT=120 [lb/ft3]
= 4.34
OCR>1
c=400[lb/ft2]
φ=16˚

19. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ?
4 [ft]
Nc = 11.63
8 [ft]
Nγ = 3.06
Clay
We have now solved our bearing capacity factors.
Nq = 4.34
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

20. Find: Bearing Capacity, qult [lb/ft2]
infinite Nq Nc Nγ φ 4 8 12 16 20 24 28 32
1.00
1.43
2.06
2.97
4.34
6.40
9.60
14.72
23.18
5.14
6.19
7.53
9.28
11.63
14.83
19.32
25.80
35.49
0.00
0.34
0.86
1.69
3.06
5.39
9.44
16.72
30.21
strip
footing
4 [ft]
8 [ft]
Clay
As a time saving strategy, it may be helpful to print out a table of these bearing capacity factors at various values of phi.
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

21. Find: Bearing Capacity, qult [lb/ft2]
infinite Nq Nc Nγ φ 4 8 12 16 20 24 28 32
1.00
1.43
2.06
2.97
4.34
6.40
9.60
14.72
23.18
5.14
6.19
7.53
9.28
11.63
14.83
19.32
25.80
35.49
0.00
0.34
0.86
1.69
3.06
5.39
9.44
16.72
30.21
strip
footing
4 [ft]
8 [ft]
Clay
Here are the factors we just computed, for the friction angle of 16 degrees.
γT=120 [lb/ft3]
OCR>1
c=400[lb/ft2]
φ=16˚

22. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ?
4 [ft]
8 [ft]
Clay
We next determine the effective unit weight, and the overburden stress, of the soil. Since the groundwater table is not mentioned, ---
γT=120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
c=400[lb/ft2]
Nq = 4.34
φ=16˚

23. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ?
4 [ft]
γ’= γT
8 [ft]
Clay
we’ll set the the effective unit weight equal to the total unit weight,which equals ---
γT=120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
c=400[lb/ft2]
Nq = 4.34
φ=16˚

24. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ?
4 [ft]
γ’= γT
γ’= 120 [lb/ft3]
8 [ft]
Clay
120 pounds per cubic feet. [pause] Next, q, the overburden stress, is calculated. In the context of bearing capacity problems, ---
γT=120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
c=400[lb/ft2]
Nq = 4.34
φ=16˚

25. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ?
4 [ft]
γ’= γT
γ’= 120 [lb/ft3]
8 [ft]
Clay
q is the same as the vertical effective stress calculated in geostatic stress problems, ---
γT=120 [lb/ft3]
q= σ’v
Nc = 11.63
Nγ = 3.06
OCR>1
c=400[lb/ft2]
Nq = 4.34
φ=16˚

26. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ?
4 [ft]
γ’= γT
γ’= 120 [lb/ft3]
8 [ft]
Clay
which is equal to the effective unit weight, times the thickness of the soil layer, for all soil layers above the footing base.
γT=120 [lb/ft3]
q= σ’v
Nc = 11.63
q= γ’*h
Nγ = 3.06
OCR>1
c=400[lb/ft2]
Nq = 4.34
φ=16˚

27. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ?
4 [ft]
γ’= γT
γ’= 120 [lb/ft3]
8 [ft]
Clay
This unit weight value corresponds to the 4 feet of soil above the footing base, which is 120 pounds per cubic foot, ---
γT=120 [lb/ft3]
q= σ’v
Nc = 11.63
q= γ’*h
Nγ = 3.06
OCR>1
c=400[lb/ft2]
Nq = 4.34
φ=16˚

28. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing ? ?
4 [ft]
γ’= γT
γ’= 120 [lb/ft3]
8 [ft]
Clay
and the height of this soil layer is 4 feet. The overburden stress equals ---
γT=120 [lb/ft3]
γT=120 [lb/ft3]
q= σ’v
Nc = 11.63
q= γ’*h
Nγ = 3.06
OCR>1
c=400[lb/ft2]
Nq = 4.34
φ=16˚

29. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
4 [ft]
γ’= γT
γ’= 120 [lb/ft3]
8 [ft]
Clay
480 pounds per square feet.
γT=120 [lb/ft3]
q= σ’v
Nc = 11.63
q= γ’*h
Nγ = 3.06
OCR>1
q= 480 [lb/ft2]
c=400[lb/ft2]
Nq = 4.34
φ=16˚

30. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
4 [ft]
γ’= γT
γ’= 120 [lb/ft3]
8 [ft]
Clay
Having solved for the effective unit weight and overburden stress, ---
γT=120 [lb/ft3]
q= σ’v
Nc = 11.63
q= γ’*h
Nγ = 3.06
OCR>1
q= 480 [lb/ft2]
c=400[lb/ft2]
Nq = 4.34
φ=16˚

31. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
4 [ft]
8 [ft]
Clay
the ultimate bearing capacity due to general shear failure can be computed. The values are plugged in for the ---
γT=120 [lb/ft3]
γ’= 120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
q = 480 [lb/ft2]
c=400[lb/ft2]
Nq = 4.34
φ=16˚

32. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
4 [ft]
8 [ft]
Clay
cohesion term, ---
γT=120 [lb/ft3]
γ’= 120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
q = 480 [lb/ft2]
c=400[lb/ft2]
Nq = 4.34
φ=16˚

33. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
4 [ft]
8 [ft]
Clay
wedge weight term, ---
γT=120 [lb/ft3]
γ’= 120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
q = 480 [lb/ft2]
c=400[lb/ft2]
Nq = 4.34
φ=16˚

34. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
4 [ft]
8 [ft]
Clay
and overburden term.
γT=120 [lb/ft3]
γ’= 120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
q = 480 [lb/ft2]
c=400[lb/ft2]
Nq = 4.34
φ=16˚

35. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
ft2 lb
ft2 lb
qult=4,652
+1,469
4 [ft]
ft2 lb
+2,083
8 [ft]
Clay
Finally the, bearing capacity computes to, ---
γT=120 [lb/ft3]
γ’= 120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
q = 480 [lb/ft2]
c=400[lb/ft2]
Nq = 4.34
φ=16˚

36. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
strip
footing
ft2 lb
ft2 lb
qult=4,652
+1,469
4 [ft]
ft2 lb
+2,083
8 [ft]
ft2 lb
qult=8,204
Clay
8,204 pounds per square foot.
γT=120 [lb/ft3]
γ’= 120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
q = 480 [lb/ft2]
c=400[lb/ft2]
Nq = 4.34
φ=16˚

37. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
3,200
4,800
C) 7,600
D) 8,200
strip
footing
ft2 lb
ft2 lb
qult=4,652
+1,469
4 [ft]
ft2 lb
+2,083
8 [ft]
ft2 lb
qult=8,204
Clay
Looking at the possible solutions, ---
γT=120 [lb/ft3]
γ’= 120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
q = 480 [lb/ft2]
c=400[lb/ft2]
Nq = 4.34
φ=16˚

38. Find: Bearing Capacity, qult [lb/ft2]
infinite
qult=c*Nc + 0.5*γ’*B*Nγ+q*Nq
3,200
4,800
C) 7,600
D) 8,200
strip
footing
ft2 lb
ft2 lb
qult=4,652
+1,469
4 [ft]
ft2 lb
+2,083
8 [ft]
ft2 lb
qult=8,204
AnswerD
Clay
the answer is D.
γT=120 [lb/ft3]
γ’= 120 [lb/ft3]
Nc = 11.63
Nγ = 3.06
OCR>1
q = 480 [lb/ft2]
c=400[lb/ft2]
Nq = 4.34
φ=16˚

39. ( ) ? γclay=53.1[lb/ft3] Index σ’v = Σ γ d H*C σfinal ρcn= log
Find: σ’v ρc d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4
( )
H*C σfinal
ρcn=
1+e σinitial
log ‘