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Meyerhof's general bearing capacity equations

Meyerhof’s general bearing capacity equations

Vertical load:    

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

Inclined load:    

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

Incline load factors:

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

 
 
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