﻿ Meyerhof CE-REF.COM

Custom Search

## Meyerhof's general bearing capacity equations

### Meyerhof’s general bearing capacity equations

Qu = c Nc Sc Dc + g D Nq Sq Dq + 0.5 g B Ng Sg Dg                         [1.7]

Qu = c Nc Sc Dc Ic + g D Nq Sq Dq Iq + 0.5 g B Ng Sg Dg Ig          [1.8]

Where:

Nc, Nq, Nr:  Meyerhof’s bearing capacity factors depend on soil friction angle, f.

Nc = cot f ( Nq – 1)                                                                       [1.9]

Nq = eptanf tan2(45+f/2)]                                                            [1.10]

Ng = (Nq-1) tan (1.4f)                                                                   [1.11]

Sc, Sq, Sg: shape factors

Dc, Dq, Dg: depth factors

Ic, Iq, Ig: incline load factors
 Friction angle Shape factor Depth factor Incline load factors Any f Sc=1+0.2Kp(B/L) Dc=1+0.2ÖKp (B/L) Ic=Iq=(1-q/90°)2 f = 0 Sq=Sg=1 Dq=Dg=1 Ig=1 f³10° Sq=Sg=1+0.1Kp(B/L) Dq=Dr=1+0.1ÖKp (D/B) Ig=(1-q/f)2

C: Cohesion of soil

g : unit weight of soil

D: depth of footing

B, L: width and length of footing

Kpr = tan2(45+f/2), passive pressure coefficient.

q = angle of axial load to vertical axis

Table 2: 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 2: Meyerhof’s bearing capacity factors #### Example 4: Strip footing on clayey sand

Given:

• Soil properties:

• Soil type: clayey sand.

• Cohesion: 500 lbs/ft2

• Cohesion: 25 degree

• Friction Angle: 30 degree

• Unit weight of soil: 100 lbs/ft3

• 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 = tan2(45+f/2) = tan2(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 2 or Figure 2, 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/ft2

Allowable soil bearing capacity,

Qa = Qu / F.S. = 15257 / 3 = 5085 lbs/ft2 @ 5000 lbs/ft2

#### Example 5: Rectangular footing on sandy clay

Given:

• Soil properties:

• Soil type: sandy clay

• Cohesion: 500 lbs/ft2

• Friction Angle: 20 degree

• Unit weight of soil: 100 lbs/ft3

• 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 = tan2(45+f/2) = tan2(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 2 or Figure 2, 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/ft2

Allowable soil bearing capacity,

Qa = Qu / F.S. = 13217 / 3 = 4406 lbs/ft2 @ 4400 lbs/ft2

#### Example 6: Square footing with incline loads

Given:

• Soil properties:

• Soil type: sandy clay

• Cohesion: 1000 lbs/ft2

• Friction Angle: 15 degree

• Unit weight of soil: 100 lbs/ft3

• 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 = tan2(45+f/2) = tan2(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

q = tan-1 (20/100) = 11.3°

 Ic=Iq=(1-q/90°)2=(1-11.3/90)2= 0.76 Ig=(1-q/f)2=(1-11.3/15)2=0.06

From Table 2 or Figure 2, 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/ft2

Allowable soil bearing capacity,

Qa = Qu / F.S. = 8179 / 3 = 2726 lbs/ft2 @ 2700 lbs/ft2