Meyerhofs Method
Courses > Foundation Analysis and Design > Bearing Capacity of Shallow Foundations > Meyerhofs MethodIntroduction
In geotechnical engineering, bearing capacity is the capacity of soil to support the loads applied to the ground. The bearing capacity of soil is the maximum average contact pressure between the foundation and the soil which should not produce shear failure in the soil. Ultimate bearing capacity (q_{f}) is the theoretical maximum pressure which can be supported without failure; allowable bearing capacity (q_{a}) is the ultimate bearing capacity divided by a factor of safety. Sometimes, on soft soil sites, large settlements may occur under loaded foundations without actual shear failure occurring; in such cases, the allowable bearing capacity is based on the maximum allowable settlement.
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There are three modes of failure that limit bearing capacity: general shear failure, local shear failure, and punching shear failure.
Concepts and Formulas
Meyerhof's bearing capacity theory:
Meyerhof (1963) proposed a formula for calculation of bearing capacity similar to the one proposed by Terzaghi but introducing further foundation shape coefficients.
He introduced a coefficient s_{q} that multiplies the N_{q } factor, depth factors d_{i} and inclination factors i_{i} depth factors di and inclination factors ii for the cases where the load line is inclined to the vertical.
Meyerhof’s general bearing capacity equations
Meyerhof provided two general equations  one for the case when the resultant load at the bearing level (Qb) is vertical (no horizontal component), and one for the case when Qb is inclined from vertical (can be resolved into vertical and horizontal components) with the horizontal component of load in the direction of the width of the footing.
Vertical load:
Q_{b} = c N_{c} S_{c} d_{c} + q_{0} N_{q} S_{q} d_{q} + 0.5 g B N_{g} S_{g} d_{g}
Inclined load:
Q_{b} = c N_{c} d_{c} i_{c} + q_{0} N_{q} d_{q} i_{q} + 0.5 g B N_{g} d_{g} i_{g}
where q_{0} is the vertical stress at the bearing level outside the footprint of the foundation (called the surcharge pressure). q_{0 }should be the total vertical stress for total stress strength parameters, and the effective vertical stress for effective strength parameters. The value of g to be used in the above equations depends on the depth of the groundwater table relative to the depth of embedment of the foundation. See how ground water level can affect unit weight.
Meyerhof's bearing capacity factors
Nc, Nq, Nr: Meyerhof’s bearing capacity factors depend on soil friction angle, f.
N_{q} = e^{ptanf} tan^{2}(45+f/2)
N_{c }= cot f ( N_{q} – 1)
Ng = (N_{q}1) tan (1.4f)
S_{c}, S_{q}, S_{g}: shape factors
d_{c}, d_{q}, d_{g}: depth factors
i_{c}, i_{q}, i_{g}: incline load factors
Friction angle 
Shape factor 
Depth factor 
Incline load factors 
Any f 
S_{c}=1+0.2Kp(B/L) 
d_{c}=1+0.2ÖK_{p} (D/B) 
i_{c}=i_{q}=(1q/90°)^{2} 
f = 0 
S_{q}=S_{g}=1 
d_{q}=d_{g}=1 
i_{g}=1 
f³10° 
S_{q}=S_{g}=1+0.1Kp(B/L) 
d_{q}=d_{g}=1+0.1ÖKp (D/B) 
i_{g}=(1q/f)^{2} 
C: Cohesion of soil
g: unit weight of soil (See how ground water level can affect it)
D: depth of footing
B, L: width and length of footing
K_{p }= tan^{2}(45+f/2), passive pressure coefficient.
q = tan^{1}(Q_{h}/Q_{v}) = angle of the load in degrees
The table shown below can be used to approximately check your calculation of the main bearing capacity factors (Nc, Nq, and Ng) using Meyerhof’s method. It is, of course, better and more accurate to check it with our free online bearing capacity calculator for shallow foundations.
Table 1: Meyerhof’s bearing capacity factors
f 
Nc 
Nq 
Nr 

0 
5.1 
1 
0 
5 
6.5 
1.6 
0.1 
10 
8.3 
2.5 
0.4 
15 
11 
3.9 
1.2 
20 
14.9 
6.4 
2.9 
25 
20.7 
10.7 
6.8 
30 
30.1 
18.4 
15.1 
35 
46.4 
33.5 
34.4 
40 
75.3 
64.1 
79.4 
Figure 1: Meyerhof’s bearing capacity factors.
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Solved sample problems
Example 1: Strip footing on clayey sand (English units)
Given:

Soil properties:

Soil type: clayey sand.

Cohesion: 500 lbs/ft^{2}

Cohesion: 25 degree

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 Meyerhof’s equation.
Solution:
Determine ultimate soil bearing capacity using Meyerhof’s bearing capacity equation for vertical load.
Passive pressure coefficient
Kpr = tan^{2}(45+f/2) = tan^{2}(45+25/2) = 2.5
Shape factors:
Sc=1+0.2Kp(B/L) = 1+0.2*2.5*(0)=1 
Sq=Sg=1+0.1Kp(B/L) = 1+0.1*2.5*(0) = 1 
Depth factors:
Dc=1+0.2ÖKp (B/L) = 1+0.2*Ö2 (0) = 1 
Dq=Dg=1+0.1ÖKp (D/B) = 1+0.1*Ö2.5 (3/3) = 1.16 
From Table 1 or Figure 1, Nc = 20.7, Nq = 10.7, Nr = 6.8 for f = 25 degree
Qu = c Nc Sc Dc + g D Nq Sq Dq + 0.5 g B Ng Sg Dg
= 500*20.7*1*1 +100*3*10.7*1*1.16+0.5*100*3*6.8*1*1.16
= 15257 lbs/ft^{2}
Allowable soil bearing capacity,
Qa = Qu / F.S. = 15257 / 3 = 5085 lbs/ft^{2} @ 5000 lbs/ft^{2}
Example 2: Rectangular footing on sandy clay (English units)
Given:

Soil properties:

Soil type: sandy clay

Cohesion: 500 lbs/ft^{2}

Friction Angle: 20 degree

Unit weight of soil: 100 lbs/ft^{3}

Expected footing dimensions:
8 ft by 4 ft rectangular footing, bottom of footing at 3 ft below ground level.

Factor of safety: 3
Requirement:
Determine allowable soil bearing capacity using Meyerhof’s equation.
Solution:
Determine ultimate soil bearing capacity using Meyerhof’s bearing capacity equation for vertical load.
Passive pressure coefficient
Kpr = tan^{2}(45+f/2) = tan^{2}(45+20/2) = 2.
Shape factors:
Sc=1+0.2Kp(B/L) = 1+0.2*2*(4/8)=1.2 
Sq=Sg=1+0.1Kp(B/L) = 1+0.1*2*(4/8) = 1.1 
Depth factors:
Dc=1+0.2ÖKp (B/L) = 1+0.2*Ö2 (4/8) = 1.14 
Dq=Dg=1+0.1ÖKp (D/B) = 1+0.1*Ö2 (3/4) = 1.1 
From Table 1 or Figure 1, Nc = 14.9, Nq = 6.4, Nr = 2.9 for f = 20 degree
Qu = c Nc Sc Dc + g D Nq Sq Dq + 0.5 g B Ng Sg Dg
= 500*14.9*1.2*1.14 +100*3*6.4*1.1*1.1+0.5*100*4*2.9*1.1*1.1
= 13217 lbs/ft^{2}
Allowable soil bearing capacity,
Qa = Qu / F.S. = 13217 / 3 = 4406 lbs/ft^{2} @ 4400 lbs/ft^{2}
Example 3: Square footing with incline loads (English units)
Given:

Soil properties:

Soil type: sandy clay

Cohesion: 1000 lbs/ft^{2}

Friction Angle: 15 degree

Unit weight of soil: 100 lbs/ft^{3}

Expected footing dimensions:
8 ft by 8 ft square footing, bottom of footing at 3 ft below ground level.

Expected column vertical load = 100 kips

Expected column horizontal load = 20 kips

Factor of safety: 3
Requirement:
Determine allowable soil bearing capacity using Meyerhof’s equation.
Solution:
Determine ultimate soil bearing capacity using Meyerhof’s bearing capacity equation for vertical load.
Passive pressure coefficient
Kpr = tan^{2}(45+f/2) = tan^{2}(45+15/2) = 1.7
Shape factors:
Sc=1+0.2Kp(B/L) = 1+0.2*1.7*(8/8)=1.34 
Sq=Sg=1+0.1Kp(B/L) = 1+0.1*1.7*(8/8) =1.17 
Depth factors:
Dc=1+0.2ÖKp (B/L) = 1+0.2*Ö1.7 (8/8) = 1.26 
Dq=Dg=1+0.1ÖKp (D/B) = 1+0.1*Ö1.7 (3/8) = 1.05 
Incline load factors:
q = tan^{1} (20/100) = 11.3°
Ic=Iq=(1q/90°)^{2}=(111.3/90)^{2}= 0.76 
Ig=(1q/f)^{2}=(111.3/15)^{2}=0.06 
From Table 1 or Figure 1, Nc = 11, Nq = 3.9, Nr = 1.2 for f = 15 degree
Qu = c Nc Sc Dc Ic + g D Nq Sq Dq Iq + 0.5 g B Ng Sg Dg Ig
= 500*11*1.34*1.26*0.76+100*3*3.9*1.17*1.05*0.76+0.5*100*8*1.17*1.05*0.06
= 8179 lbs/ft^{2}
Allowable soil bearing capacity,
Qa = Qu / F.S. = 8179 / 3 = 2726 lbs/ft^{2} @ 2700 lbs/ft^{2}
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