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Figure 1.1: Shear stresses based on Terzaghi’s soil bearing capacity theory
Based on Terzaghi’s bearing capacity theory, column load P is resisted by shear stresses at edges of three zones under the footing and the overburden pressure, q (=gD) above the footing. The first term in the equation is related to cohesion of the soil. The second term is related to the depth of the footing and overburden pressure. The third term is related to the width of the footing and the length of shear stress area. The bearing capacity factors, Nc, Nq, Ng, are function of internal friction angle, f.
Terzaghi's Bearing capacity equations:
Strip footings:
Qu = c Nc + g D Nq + 0.5 g B Ng [1.1]
Square footings:
Qu = 1.3 c Nc + g D Nq + 0.4 g B Ng [1.2]
Circular footings:
Qu = 1.3 c Nc + g D Nq + 0.3 g B Ng [1.3]
Where:
C: Cohesion of soil, g : unit weight of soil, D: depth of footing, B: width of footing
Nc, Nq, Nr: Terzaghi’s bearing capacity factors depend on soil friction angle, f.
Nc=cotf(Nq –1) [1.4]
Nq=e^{2}(3p/4-f/2)tanf / [2 cos2(45+f/2)] [1.5]
Ng=(1/2) tanf( Kpr /cos^{2} f -1) [1.6]
K_{pr}=passive pressure coefficient.
(Note: from Boweles, Foundation analysis and design, "Terzaghi never explained..how he obtained Kpr used to compute Ng")
Table 1: Terzaghi’s Bearing Capacity Factors
f |
Nc |
Nq |
Nr |
0 |
5.7 |
1 |
0 |
5 |
7.3 |
1.6 |
0.5 |
10 |
9.6 |
2.7 |
1.2 |
15 |
12.9 |
4.4 |
2.5 |
20 |
17.7 |
7.4 |
5 |
25 |
25.1 |
12.7 |
9.7 |
30 |
37.2 |
22.5 |
19.7 |
35 |
57.8 |
41.4 |
42.4 |
40 |
95.7 |
81.3 |
100.4 |
Figure 2 Terzaghi’s bearing capacity factors
Given:
Soil properties:
Soil type: cohesionless soil.
Cohesion: 0 (neglectable)
Friction Angle: 30 degree
Unit weight of soil: 100 lbs/ft^{3}
Expected footing dimensions:
3 ft wide strip footing, bottom of footing at 2 ft below ground level
Factor of safety: 3
Requirement: Determine allowable soil bearing capacity using Terzaghi’s equation.
Solution:
From Table 1 or Figure 1, Nc = 37.2, Nq = 22.5, Nr = 19.7 for f = 30 degree
Determine ultimate soil bearing capacity using Terzaghi’s bearing
capacity equation for strip footing
Qu = c Nc + g D Nq + 0.5 g B Ng
= 0 +100x2x22.5+0.5x100x6x19.7
= 10410 lbs/ft^{2}
Allowable soil bearing capacity,
Qa = Qu / F.S. = 10410 / 3 = 3470 lbs/ft^{2} @ 3500
lbs/ft^{2}
Given:
Soil type: Clay
Soil properties:
Cohesion:2000 lbs/ft^{2}
Friction Angle: 0 (neglectable)
Unit weight of soil: 120 lbs/ft^{3}
Expected footing dimensions:
6 ft by 6 ft square footing, bottom of footing at 2 ft below
ground level
Factor of safety: 3
Requirement: Determine allowable soil bearing capacity using Terzaghi’s equation.
Solution:
From Table 1 or Figure 1, Nc = 5.7, Nq = 1.0, Nr = 0 for f = 0 degree
Determine ultimate soil bearing capacity using Terzaghi’s bearing capacity equation for square footing
Qu = 1.3 c Nc + g D Nq + 0.4 g B Ng
= 1.3x1000x5.7 +120x2*1+ 0
= 7650 lbs/ft^{2}
Allowable soil bearing capacity,
Qa = Qu / F.S. = 7650 / 3 = 2550 lbs/ft^{2} @ 2500
lbs/ft^{2}
Given:
Soil properties:
Soil type: sandy clay
Cohesion: 500 lbs/ft^{2}
Friction Angle: 25 degree
Unit weight of soil: 100 lbs/ft^{3}
Expected footing dimensions:
10 ft diameter circular footing for a circular tank, bottom of
footing at 2 ft below ground level
Factor of safety: 3
Requirement:
Determine allowable soil bearing capacity using Terzaghi’s equation.
Solution:
From Table 1 or Figure 1, Nc = 17.7, Nq = 7.4, Nr = 5.0 for f = 20 degree
Determine ultimate soil bearing capacity using Terzaghi’s bearing capacity equation for circular footing
Qu = 1.3 c Nc + g D Nq + 0.3 g B Ng
= 1.3x500x17.7 +100x2x7.4+0.3x100x10x5.0
= 17985 lbs/ft^{2}
Allowable soil bearing capacity,
Qa = Qu / F.S. = 17985/ 3 = 5995 lbs/ft^{2} @ 6000 lbs/ft^{2}