BEARING CAPACITY EQUATION
TERZAGHI’S EQUATION
PROBLEM No. 1
A clayey soil has an unconfined compressive strength of 48kN/m 2 and an angle of internal friction of
25o.
a) Compute the Terzaghi’s bearing capacity factor Nq.
b) Compute the Terzaghi’s bearing capacity factor Nc.
c) Compute the Terzaghi’s bearing capacity factor Ny.
Solution
a) Terzaghi’s bearing capacity factor Nq
∅
𝑁𝑞 = 𝑡𝑎𝑛2 (45 + ) 𝑒 𝜋𝑡𝑎𝑛∅
2
𝑁𝑞 = 𝑡𝑎𝑛 (45 +
2
25
2
) 𝑒 𝜋𝑡𝑎𝑛25
𝑁𝑞 = ____________
b) Terzaghi’s bearing capacity factor Nc
𝑁𝑐 = (𝑁𝑞 − 1)𝑐𝑜𝑡∅
𝑁𝑐 = (𝑁𝑞 − 1)cot (25)
𝑁𝑐 = ____________
c) Terzaghi’s bearing capacity factor Ny
𝑁𝑦 = 2(𝑁𝑞 + 1)𝑡𝑎𝑛∅
𝑁𝑦 = 2(𝑁𝑞 + 1)𝑡𝑎𝑛25
𝑁𝑦 = ____________
PROBLEM No. 2
A 1.2m x 1.2m square footing has its bottom 2.5m below the ground surface. The soil properties are:
angle of friction of soil is 28 o , unit weight of 16kN/m3 and cohesion of 22 kPa. Terzaghi’s bearing
capacity factors are Nq = 14.72, Nc = 25.80 and Ny = 11.19
a) Compute the ultimate bearing capacity considering general sh ear failure.
b) Compute the ultimate bearing capacity considering local shear failure.
c) Compute the ultimate bearing capacity considering general shear failure if c=0.
Solution
a) Square footing bearing capacity considering general shear failure
𝑞𝑢 = 1.3𝑐𝑁𝑐 + 𝛾𝐷𝑓 𝑁𝑞 + 0.4𝛾𝐵𝑁𝛾
𝑞𝑢 = 1.3(22)(25.80) + (16)(2.5)(14.72) + 0.4(16)(1.2)(11.19)
𝑞𝑢 = _______________𝑘𝑃𝑎
b) Square footing bearing capacity considering local shear failure
𝑞𝑢 = 0.867𝑐𝑁𝑐 + 𝛾𝐷𝑓𝑁𝑞 + 0.4𝛾𝐵𝑁𝛾
𝑞𝑢 = 0.867(22)(25.80) + (16)(2.5)(14.72) + 0.4(16)(1.2)(11.19)
𝑞𝑢 = _______________𝑘𝑃𝑎
c) Square footing bearing capacity considering general shear failure if c = 0
𝑞𝑢 = 𝛾𝐷𝑓 𝑁𝑞 + 0.4𝛾𝐵𝑁𝛾
𝑞𝑢 = (16)(2.5)(14.72) + 0.4(16)(1.2)(11.19)
𝑞𝑢 = _______________𝑘𝑃𝑎
PROBLEM No. 3
A square foundation is 2m x 2m in plan. The soil supporting the foundation has a friction angle of 25 o
and cohesion of 20kN/m2. The unit weight of soil is 16.5kN/m3. Determine the allowable gross load on
the foundation with a factor of safety of 3. Assume that the depth of the foundation is 1.5m and that
general shear failure occurs in the soil. Use Nq = 12.72, Nc = 25.13 and Ny = 8.34
Solution
a. Solve for the ultimate bearing capacity
𝑞𝑢 = 1.3𝑐𝑁𝑐 + 𝛾𝐷𝑓 𝑁𝑞 + 0.4𝛾𝐵𝑁𝛾
𝑞𝑢 = 1.3(20)(25.13) + (16.5)(1.5)(12.72) + 0.4(16.5)(2)(8.34)
𝑞𝑢 = 1078.29𝑘𝑃𝑎
b. Solve for the allowable load per unit area of the foundation
𝑞
1078.29
𝑞𝑎𝑙𝑙 = 𝐹𝑆𝑢 =
= 359.5 𝑘𝑁/𝑚2
3
c. Solve for the total allowable gross load
𝑄𝑎𝑙𝑙 .𝑔𝑟𝑜𝑠𝑠(𝑙𝑜𝑎𝑑 ) = 𝑞𝑎𝑙𝑙 (𝐴𝑟𝑒𝑎 𝑜𝑓 𝑓𝑜𝑜𝑡𝑖𝑛𝑔)
𝑄𝑎𝑙𝑙 .𝑔𝑟𝑜𝑠𝑠(𝑙𝑜𝑎𝑑 ) = 3.59.5(2𝑥2)
𝑄𝑎𝑙𝑙 .𝑔𝑟𝑜𝑠𝑠(𝑙𝑜𝑎𝑑 ) = 1438 𝑘𝑁
PROBLEM No. 4
Refer to Problem No.3. Assume that the shear-strength parameters of the soil are the same. A square
foundation measuring BxB will be subjected to an allowable gross load of 1000kN with FS of 3 and
depth of foundation of 1m. Determine the size of the foundation.
Solution
Ultimate Gross load
𝑄𝑢𝑙𝑡 .𝑔𝑟𝑜𝑠𝑠 (𝑙𝑜𝑎𝑑) = 𝑄𝑎𝑙𝑙 .𝑔𝑟𝑜𝑠𝑠𝑙𝑜𝑎𝑑 (𝐹𝑆)
𝑄𝑢𝑙𝑡 .𝑔𝑟𝑜𝑠𝑠 (𝑙𝑜𝑎𝑑) = 1000(3)
𝑄𝑢𝑙𝑡 .𝑔𝑟𝑜𝑠𝑠 (𝑙𝑜𝑎𝑑) = 3000𝑘𝑁
Ultimate bearing capacity
𝑞𝑢 = 1.3𝑐𝑁𝑐 + 𝛾𝐷𝑓 𝑁𝑞 + 0.4𝛾𝐵𝑁𝛾
𝑞𝑢 = 1.3(20)(25.13) + (16.5)(1.0)(12.72) + 0.4(16.5)(𝐵 )(8.34)
𝑞𝑢 = 863.26 + 55.04𝐵
Size of footing
𝑄𝑢𝑙𝑡 .𝑔𝑟𝑜𝑠𝑠(𝑙𝑜𝑎𝑑 )
𝑞𝑢 =
𝐵2
= 863.26 + 55.04𝐵
𝐵 = 1.77𝑚 ≈ 1.8𝑚
PROBLEM No. 5
A square footing carries a gross mass of 30,000 kg. Using a factor of safety of 3, determine the size of
the footing—that is, the size of B if it has soil density of 1850kg/m3, internal friction angle of 35 o and
cohesion of 0. Depth of footing is 1.0m.
Solution
Unit weight of soil
𝛾=
1850 𝑥 9.81
1000
= 18.15 𝑘𝑁/𝑚3
Allowable gross load
𝑄𝑎𝑙𝑙 =
30,000 𝑥 9.81
1000
= 294.3 𝑘𝑁
Solving for B
𝑞𝑎𝑙𝑙 =
𝑄𝑎𝑙𝑙
294.3
1
𝐵2
294.3
𝐵2
𝐵2
=
𝑞𝑢
𝐹𝑆
= 3 (1.3𝑐𝑁𝑐 + 𝑞𝑁𝑞 + 0.4𝛾𝐵𝑁𝑦 )
1
= [1.3(0)(57.75) + (18.15)(1)(41.44) + 0.4(18.15)𝐵(45.41)]
3
𝐵 = 0.95𝑚
BEARING CAPACITY EQUATION
MEYERHOF’S EQUATION
Problem No.1
A square foundation is 2m x 2m in plan. The soil supporting the foundation has a friction angle of ϕ =
25o and c’ = 20 kN/m2. The unit weight of soil, y, is 16.5 kN/m3. Determine the allowable bearing
capacity on the foundation with a factor of safety (FS) of 3. Assume that the depth of the foundation,
Df, is 1.5 m and N q = 10.66, N c = 20.72 and N y = 10.88
Solution
a) Inclination Factor: Since the load is vertical then β o = 0o
Solving for Fci and Fqi
Solving for Fyi
𝛽𝑜
2
0
2
𝐹𝑐𝑖 = 𝐹𝑞𝑖 = (1 − 90 )
𝐹𝑐𝑖 = 𝐹𝑞𝑖 = (1 − 90 )
𝐹𝑐𝑖 = 𝐹𝑞𝑖 = 1
𝐵
𝑁𝑞
2
10.66
)
∅
0
𝐹𝑦𝑖 = (1 − 25 )
𝐹𝑦𝑖 = 1
b) Shape Factor
Solving for Fcs
2
𝛽𝑜
𝐹𝑦𝑖 = (1 −
2
Solving for Fqs
𝐹𝑐𝑠 = 1 + (𝐿 ) (𝑁 )
Solving for Fys
𝐵
𝐹𝑦𝑠 = 1 − 0.4 (𝐿 )
𝐵
2
𝐹𝑦𝑠 = 1 − 0.4 (2 )
𝐹𝑦𝑠 = 0.6
𝐹𝑞𝑠 = 1 + (𝐿 ) 𝑡𝑎𝑛∅
𝑐
𝐹𝑐𝑠 = 1 + (2 )(20.72 )
𝐹𝑐𝑠 = 1.514
2
𝐹𝑞𝑠 = 1 + (2 )𝑡𝑎𝑛25
𝐹𝑞𝑠 = 1.466
c) Depth Factor:
𝐷
Condition 𝐵𝑓 ≤ 1 ; 0.75 ≤ 1
For ∅ > 0.0 ; 25 > 0.0
Solving for Fqd
)2
𝐹𝑞𝑑 = 1 + 2𝑡𝑎𝑛∅(1 − 𝑠𝑖𝑛∅
Solving for Fcd
𝐷𝑓
𝐹𝑐𝑑 = 𝐹𝑞𝑑 −
( )
𝐵
𝐹𝑞𝑑 = 1 + 2𝑡𝑎𝑛25(1 − 𝑠𝑖𝑛25)2 (
1.5
2
)
𝐹𝑞𝑑 = 1.233
Solving for Fyd
1−𝐹𝑞𝑑
𝑁𝑐 𝑡𝑎𝑛∅
1−1.233
𝐹𝑐𝑑 = 1.233 −
𝐹𝑐𝑑 = 1.257
20.72𝑡𝑎𝑛25
d) Meyerhof Ultimate Bearing Capacity:
𝑞𝑢 = 𝑐𝑁𝑐 𝐹𝑐𝑠 𝐹𝑐𝑑 𝐹𝑐𝑖 + 𝑞𝑁𝑞 𝐹𝑞𝑠 𝐹𝑞𝑑 𝐹𝑞𝑖 + 0.5𝐵𝛾𝐹𝑦𝑠 𝐹𝑦𝑑 𝐹𝑦𝑖
𝑞𝑢 = (20)(20.72)(1.514)(1.257)(1) + (16.5 𝑥 1.5)(10.66)(1.466)(1.233)(1) +
(0.5)(2)(16.5)(10.88)(0.6)(1)(1)
𝑞𝑢 = 1373.2 𝑘𝑁/𝑚2
e) allowable bearing capacity
𝑞
𝑞𝑎𝑙𝑙 = 𝐹𝑆𝑢 =
1373 .2
3
= 457.71373.2 𝑘𝑁/𝑚2
𝐹𝑦𝑑 = 1
Problem No.2
A square foundation (B x B) has to be constructed as shown below. Assume that У = 105 lb/ft3, Уsat =
118 lb/ft3, 𝜙 = 34o , D f = 4ft, and D 1 = 2ft. The allowable bearing capacity, Qall , with FS = 3 is 150,000
lb. Determine the size of the foundation. Nq = 29.44 and N y = 41.06
Solution
a) Allowable Bearing Capacity
𝑞𝑎𝑙𝑙 =
𝑄𝑎𝑙𝑙
𝑞𝑎𝑙𝑙 =
𝑞𝑢
𝐵2
𝐹𝑆
150000
=
𝐵2
𝑙𝑏/𝑓𝑡 2
(eq’n. 1)
𝑞𝑁𝑞 𝐹𝑞𝑠 𝐹𝑞𝑑 + 0.5𝛾 ′ 𝐵𝑁𝛾 𝐹𝑦𝑠 𝐹𝑦𝑑
=
3
b) Shape Factor
Solving for Fqs
𝐵
𝐹𝑞𝑠 = 1 + ( ) 𝑡𝑎𝑛∅
𝐿
𝐵
(eq’n.2)
Solving for Fys
𝐵
𝐹𝑦𝑠 = 1 − 0.4 ( )
𝐿
𝐵
𝐹𝑞𝑠 = 1 + (𝐵 ) 𝑡𝑎𝑛34
𝐹𝑦𝑠 = 1 − 0.4 (𝐵 )
𝐹𝑞𝑠 = 1.67
𝐹𝑦𝑠 = 0.6
c) Depth Factor:
𝐷
Condition 𝑓 ≤ 1 ;
𝐵
For ∅ > 0.0 ; 34 > 0.0
Solving for Fqd
𝐷
𝐹𝑞𝑑 = 1 + 2𝑡𝑎𝑛∅(1 − 𝑠𝑖𝑛∅)2 ( 𝑓 )
𝐹𝑞𝑑 = 1 +
𝐹𝑞𝑑 = 1 +
𝐵
4
2
2𝑡𝑎𝑛34(1 − 𝑠𝑖𝑛34) (𝐵 )
1.05
Solving for Fyd
𝐹𝑦𝑑 = 1
𝐵
d) For the second term “q”
𝑞 = 𝛾𝐷1 + (𝛾𝑠𝑎𝑡 − 𝛾𝑤 )(𝐷𝑓 − 𝐷1 )
𝑞 = (2)(105) + (118 − 62.4)(4 − 2)
𝑞 = 321.2 𝑙𝑏/𝑓𝑡 2
e) For Equation 2
𝑞𝑎𝑙𝑙 =
𝑞𝑎𝑙𝑙
𝑞𝑁𝑞 𝐹𝑞𝑠 𝐹𝑞𝑑 + 0.5𝛾 ′ 𝐵𝑁𝛾 𝐹𝑦𝑠 𝐹𝑦𝑑
3
1.05
(321.1)(29.44)(1.67)(1 + 𝐵 ) + 0.5(118 − 62.4)𝐵(41.06)(0.6)(1)
=
3
𝑞𝑎𝑙𝑙 = 5263.9 +
5527.1
𝐵
+ 228.3𝐵
(eq’n 3)
f) Equate Eq’n 1 and 3
150000
5527.1
= 5263.9 +
+ 228.3𝐵
2
𝐵
𝐵
𝐵 = 4.5𝑓𝑡
Problem No.3
A square column foundation is to be constructed on a sand deposit. The allowable load Q will be
inclined at an angle β = 20° with the vertical. Determine the allowable load. Nq = 18.4 and N y = 22.4
Solution
a) Solving for the second term “q” and third term “y”
𝑞 = 𝛾𝐷𝑓 = 18 (0.7) = 12.6 𝑘𝑁/𝑚2
𝛾 = 18 𝑘𝑁/𝑚3
) Shape Factor
Solving for Fqs
𝐵
𝐹𝑞𝑠 = 1 + (𝐿 ) 𝑡𝑎𝑛∅
𝐵
Solving for Fys
𝐵
𝐹𝑦𝑠 = 1 − 0.4 (𝐿 )
𝐵
𝐹𝑞𝑠 = 1 + ( ) 𝑡𝑎𝑛30
𝐹𝑦𝑠 = 1 − 0.4 ( )
𝐹𝑞𝑠 = 1.577
𝐹𝑦𝑠 = 0.6
𝐵
𝐵
c) Depth Factor:
𝐷
Condition 𝐵𝑓 ≤ 1 ;
For ∅ > 0.0 ; 30 > 0.0
Solving for Fqd
𝐷
𝐹𝑞𝑑 = 1 + 2𝑡𝑎𝑛∅(1 − 𝑠𝑖𝑛∅)2 ( 𝑓 )
𝐹𝑞𝑑 = 1 +
Solving for Fyd
𝐹𝑦𝑑 = 1
𝐵
0.7
2
2𝑡𝑎𝑛30(1 − 𝑠𝑖𝑛30) (1.25 )
𝐹𝑞𝑑 = 1.162
d) Inclination Factor: Since the load is vertical then β o = 20o
Solving for Fci and Fqi
Solving for Fyi
𝛽𝑜
2
𝛽𝑜
𝐹𝑐𝑖 = 𝐹𝑞𝑖 = (1 − 90 )
𝐹𝑦𝑖 = (1 −
𝐹𝑐𝑖 = 𝐹𝑞𝑖 = (1 − 90 )
𝐹𝑐𝑖 = 𝐹𝑞𝑖 = 0.605
𝐹𝑦𝑖 = (1 − 30 )
𝐹𝑦𝑖 = 0.11
20 2
2
)
∅
20 2
e) Ultimate Bearing Capacity
𝑞𝑢 = (12.6)(18.4)(1.577)(1.162)(0.605) + 0.5(18)(1.25)(22.4)(0.6)(1)(0.11)
𝑞𝑢 = 273.66 𝑘𝑁/𝑚2
f) Allowable Bearing Capacity
𝑞𝑎𝑙𝑙 =
𝑞𝑢
𝐹𝑆
=
273 .66
3
= 91.22 𝑘𝑁/𝑚2
g) Allowable Load
𝑄𝑐𝑜𝑠20 = 𝑞𝑎𝑙𝑙 𝐵 2
𝑄𝑐𝑜𝑠20 = (91.22)(1.25)2
𝑄 = 151.7 𝑘𝑁