CE 333
Geotechnical Engineering II
Sultan Mohammad Farooq
Sheikh Sharif Ahmed
Department of Civil Engineering
Chittagong University of Engineering & Technology
Bearing Capacity of Shallow Foundation
EXAMPLE 01
A rectangular footing is shown in
figure.
Determinea) The ultimate bearing capacity
of the soil,
b) The net bearing capacity, and
c) The net allowable bearing
pressure.
Assume a factor of safety, FS = 3.
Bearing Capacity of Shallow Foundation
EXAMPLE 01 : SOLUTION
Using Terzaghi’s Equation
For Rectangular Footing,
𝑞𝑢 = 𝑐𝑁𝑐
𝑓𝑜𝑟 ∅ = 28°
𝑁𝑐 = 31.61
𝑁𝑞 = 17.81
𝑁𝛾 = 15.03
𝐵
1
𝐵
1 + 0.3
+ 𝑞𝑁𝑞 + 𝛾𝐵𝑁𝛾 1 − 0.2
𝐿
2
𝐿
Bearing Capacity of Shallow Foundation
∅
Nc
Nq
Ny
∅
Nc
Nq
Ny
∅
Nc
Nq
Ny
0°
5.70
1.00
0.00
17°
14.56
5.45
3.63
34°
52.64
36.50
35.23
1°
6.00
1.10
0.08
18°
15.52
6.04
4.13
35°
57.75
41.44
41.08
2°
6.30
1.22
0.18
19°
16.56
6.70
4.70
36°
63.53
47.16
48.11
3°
6.62
1.35
0.28
20°
17.69
7.44
5.34
37°
70.07
53.80
56.62
4°
6.97
1.49
0.39
21°
18.92
8.26
6.07
38°
77.50
61.55
67.00
5°
7.34
1.64
0.51
22°
20.27
9.19
6.89
39°
85.97
70.61
79.77
6°
7.73
1.81
0.65
23°
21.75
10.23
7.83
40°
95.66
81.27
95.61
7°
8.15
2.00
0.80
24°
23.36
11.40
8.90
41°
106.81
93.85
115.47
8°
8.60
2.21
0.96
25°
25.13
12.72
10.12
42°
119.67
108.75
140.65
9°
9.09
2.44
1.15
26°
27.09
14.21
11.53
43°
134.58
126.50
172.99
10°
9.60
2.69
1.35
27°
29.24
15.90
13.15
44°
151.95
147.74
215.16
11°
10.16
2.98
1.58
28°
31.61
17.81
15.03
45°
172.29
173.29
271.07
12°
10.76
3.29
1.84
29°
34.24
19.98
17.21
46°
196.22
204.19
346.67
13°
11.41
3.63
2.12
30°
37.16
22.46
19.75
47°
224.55
241.80
451.29
14°
12.11
4.02
2.43
31°
40.41
25.28
22.72
48°
258.29
287.86
600.15
15°
12.86
4.45
2.79
32°
44.04
28.52
26.21
49°
298.72
344.64
819.32
16°
13.68
4.92
3.19
33°
48.09
32.23
30.33
50°
347.51
415.15
1155.97
Bearing Capacity of Shallow Foundation
EXAMPLE 01 : SOLUTION
Using Terzaghi’s Equation
𝑞𝑢
3
1
= 30 × 31.61 1 + 0.3 ×
+ 17 × 2 × 17.81 + × 17 × 3
4
2
3
× 15.03 × 1 − 0.2 ×
4
𝑞𝑢 = 𝟐𝟎𝟗𝟑 𝒌𝑵/𝒎𝟐
𝑞𝑢 2093
𝑞𝑎𝑙𝑙𝑜𝑤 =
=
= 𝟔𝟗𝟖 𝒌𝑵/𝒎𝟐
𝐹𝑆
3
𝑞𝑢 − 𝛾𝐷𝑓 2093 − 17 ∗ 2
𝑞𝑎𝑙𝑙𝑜𝑤(𝑛𝑒𝑡) =
=
= 𝟔𝟖𝟔 𝒌𝑵/𝒎𝟐
𝐹𝑆
3
Bearing Capacity of Shallow Foundation
EXAMPLE 01 : SOLUTION
Using General Bearing Capacity Equation (With Meyerhof’s
Bearing Capacity Factors)
1
𝑞𝑢 = 𝑐𝑁𝑐 𝐹𝑐𝑠 𝐹𝑐𝑑 𝐹𝑐𝑖 + 𝑞𝑁𝑞 𝐹𝑞𝑠 𝐹𝑞𝑑 𝐹𝑞𝑖 + 𝛾𝐵𝑁𝛾 𝐹𝛾𝑠 𝐹𝛾𝑑 𝐹𝛾𝑖
2
𝑓𝑜𝑟 ∅ = 28°,
𝑁𝑐 = 25.80
𝑁𝑞 = 14.72
𝑁𝛾 = 11.19
Bearing Capacity of Shallow Foundation
∅
Nc
Nq
𝑵𝜸 (M)
𝑵𝜸 (V)
𝑵𝜸 (H)
∅
Nc
Nq
𝑵𝜸 (M)
𝑵𝜸 (V)
𝑵𝜸 (H)
0°
1°
2°
3°
4°
5°
6°
7°
8°
9°
10°
11°
12°
13°
14°
15°
16°
17°
18°
19°
20°
21°
22°
23°
24°
5.10
5.38
5.63
5.90
6.19
6.49
6.81
7.16
7.53
7.92
8.34
8.80
9.28
9.81
10.37
10.98
11.63
12.34
13.10
13.93
14.83
15.81
16.88
18.05
19.32
1.00
1.09
1.20
1.31
1.43
1.57
1.72
1.88
2.06
2.25
2.47
2.71
2.97
3.26
3.59
3.94
4.34
4.77
5.26
5.80
6.40
7.07
7.82
8.66
9.60
0.00
0.00
0.01
0.02
0.04
0.07
0.11
0.15
0.21
0.28
0.37
0.47
0.60
0.74
0.92
1.13
1.37
1.66
2.00
2.40
2.87
3.42
4.07
4.82
5.72
0.00
0.07
0.15
0.24
0.34
0.45
0.57
0.71
0.86
1.03
1.22
1.44
1.69
1.97
2.29
2.65
3.06
3.53
4.07
4.68
5.39
6.20
7.13
8.20
9.44
0.00
0.00
0.01
0.02
0.05
0.07
0.11
0.16
0.22
0.30
0.39
0.50
0.63
0.78
0.97
1.18
1.43
1.73
2.08
2.48
2.95
3.50
4.13
4.88
5.75
25°
26°
27°
28°
29°
30°
31°
32°
33°
34°
35°
36°
37°
38°
39°
40°
41°
42°
43°
44°
45°
46°
47°
48°
49°
20.72
22.25
23.94
25.80
27.86
30.14
32.67
35.49
38.64
42.16
46.12
50.59
55.63
61.35
67.87
75.31
83.86
93.71
105.11
118.37
133.87
152.10
173.64
199.26
229.93
10.66
11.85
13.20
14.72
16.44
18.40
20.63
23.18
26.09
29.44
33.30
37.75
42.92
48.93
55.96
64.20
73.90
85.37
99.01
115.31
134.87
158.50
187.21
222.30
265.50
6.77
8.00
9.46
11.19
13.24
15.67
18.56
22.02
26.17
31.15
37.15
44.43
53.27
64.07
77.33
93.69
113.99
139.32
171.14
211.41
262.74
328.73
414.33
526.46
674.92
10.88
12.54
14.47
16.72
19.34
22.40
25.99
30.21
35.19
41.06
48.03
56.31
66.19
78.02
92.25
109.41
130.21
155.54
186.53
224.64
271.75
330.34
403.66
496.00
613.15
6.76
7.94
9.32
10.94
12.84
15.07
17.69
20.79
24.44
28.77
33.92
40.05
47.38
56.17
66.76
79.54
95.05
113.96
137.10
165.58
200.81
244.65
299.52
368.67
456.41
Bearing Capacity of Shallow Foundation
Author
Factor
Condition
𝑓𝑜𝑟 ∅ = 0°
Relationship
𝐹𝑐𝑠 = 1 + 0.2
𝐹𝑞𝑠 = 𝐹𝛾𝑠 = 1.0
Shape
𝑓𝑜𝑟 ∅ ≥ 10°
𝐹𝑐𝑠 = 1 + 0.2
𝐵
𝐿
Meyerhof
𝐹𝑞𝑠 = 𝐹𝛾𝑠 = 1 + 0.1
𝑓𝑜𝑟 ∅ = 0°
𝑡𝑎𝑛2 45 +
𝐵
𝐿
∅
2
𝑡𝑎𝑛2 45 +
𝐹𝑐𝑑 = 1 + 0.2
∅
2
𝐷𝑓
𝐵
𝐹𝑞𝑑 = 𝐹𝛾𝑑 = 1.0
Depth
𝑓𝑜𝑟 ∅ ≥ 10°
𝐹𝑐𝑑 = 1 + 0.2
𝐷𝑓
𝑓𝑜𝑟 𝑎𝑛𝑦 ∅
𝑡𝑎𝑛 45 +
𝐵
𝐹𝑞𝑑 = 𝐹𝛾𝑑 = 1 + 0.1
Inclination
𝐵
𝐿
𝐷𝑓
𝐵
𝐹𝑐𝑖 = 𝐹𝑞𝑖 = 1 −
𝑡𝑎𝑛 45 +
𝛼° 2
90°
𝛼° 2
∅°
𝑓𝑜𝑟 ∅ > 0°
𝐹𝛾𝑖 = 1 −
𝑓𝑜𝑟 ∅ = 0°
𝐹𝛾𝑖 = 0
∅
2
∅
2
Bearing Capacity of Shallow Foundation
EXAMPLE 01 : SOLUTION
Using General Bearing Capacity Equation (With Meyerhof’s
Bearing Capacity Factors)
𝐵
∅
3
28
2
2
𝐹𝑐𝑠 = 1 + 0.2
𝑡𝑎𝑛 45 +
= 1 + 0.2
𝑡𝑎𝑛 45 +
= 𝟏. 𝟒𝟐
𝐿
2
4
2
𝐵
∅
3
28
2
2
𝐹𝑞𝑠 = 𝐹𝛾𝑠 = 1 + 0.1
𝑡𝑎𝑛 45 +
= 1 + 0.1
𝑡𝑎𝑛 45 +
𝐿
2
4
2
= 𝟏. 𝟐𝟏
𝐷𝑓
∅
2
28
𝐹𝑐𝑑 = 1 + 0.2
𝑡𝑎𝑛 45 +
= 1 + 0.2
𝑡𝑎𝑛 45 +
= 𝟏. 𝟐𝟐
𝐵
2
3
2
𝐷𝑓
∅
2
28
𝐹𝑞𝑑 = 𝐹𝛾𝑑 = 1 + 0.1
𝑡𝑎𝑛 45 +
= 1 + 0.1
𝑡𝑎𝑛 45 +
𝐵
2
3
2
= 𝟏. 𝟏𝟏
𝐹𝑐𝑖 = 𝐹𝑞𝑖 = 𝐹𝛾𝑖 = 𝟏. 𝟎
Bearing Capacity of Shallow Foundation
EXAMPLE 01 : SOLUTION
Using General Bearing Capacity Equation (With Meyerhof’s
Bearing Capacity Factors)
𝑞𝑢
= 30 × 25.80 × 1.42 × 1.22 × 1 + 17 × 2 × 14.72 × 1.21
1
× 1.11 × 1 + × 17 × 3 × 11.19 × 1.21 × 1.11 × 1
2
𝑞𝑢 = 𝟐𝟑𝟗𝟔 𝒌𝑵/𝒎𝟐
𝑞𝑢 2396
𝑞𝑎𝑙𝑙𝑜𝑤 =
=
= 𝟕𝟗𝟗 𝒌𝑵/𝒎𝟐
𝐹𝑆
3
𝑞𝑢 − 𝛾𝐷𝑓 2396 − 17 ∗ 2
𝑞𝑎𝑙𝑙𝑜𝑤(𝑛𝑒𝑡) =
=
= 𝟕𝟖𝟕 𝒌𝑵/𝒎𝟐
𝐹𝑆
3