Chapter 3. Shallow Foundation Design
1. Bearing capacity problem. Figure 3.26 shows that a square shallow spread footing is
designed to support a column. The load on the column, including the column weight,
is 650 kN. The foundation rests in homogeneous sandy clay. Subsurface exploration
and laboratory testing found the soil’s effective cohesion is 55 kN/m2 and the effective
friction angle is 15. The groundwater table is 1.2 m below the ground surface. The
bulk unit weight above the groundwater table is 18 kN/m3; the saturated unit weight
below the groundwater table is 18.8 kN/m3. Use Terzaghi’s bearing capacity theory
and assume general shear failure.
(1) Determine the foundation embedment and design the dimensions of the spread
footing. The factor of safety for bearing capacity is 3.0.
(2) Choose the foundation dimensions that are acceptable in the construction, then
calculate the allowable bearing capacity based on these dimensions.
Q
Ground surface
Df
Dw
γ
B
Groundwater
table
γsat
Figure 3.26 Shallow foundation design sample problem
Solution:
A spread column footing is to be designed. A square footing is chosen.
Step 1: The loads from the superstructure: Q = 650 kN.
The subsoil condition: c = 55 kN/m2,  = 15,  = 18 kN/m3, sat = 18.8 kN/m3.
Step 2: Based on Table 3.4, the minimum Df = 0.5 m, choose Df = 0.6 m (< Dw = 1.2 m).
Step 3: Groundwater table (GWT) elevation: Dw = 1.2 m. It is assumed the GWT affects
the ultimate bearing capacity.
Step 4: The minimum factor of safety is 3.0.
Step 5: The Terzaghi’s ultimate bearing capacity equation for square footing is:
qult = 1.3c¢N c + qN q + 0.4g BN r
Using  = 15, find the bearing capacity factors for general failure in Table 3.1:
N c = 12.86, N q = 4.45, and Ng = 1.52.
The effective stress at the bottom of the foundation is: q = 180.6 = 10.8 kN/m2.
Dw - D f
g - g ¢ to substitute  in the bearing capacity equation,
Use g = g ¢ +
B
(
)
where,  =  sat   w = 18.89.81= 8.99 kN/m3 and
g = 8.99 +
1.2 - 0.6
4.91
(18 - 9.81) = 8.99 +
B
B
So:
qult = 1.3c¢N c + qN q + 0.4g BN g
æ
4.91 ö
= 1.3´ 55´12.86 +10.8 ´ 4.45+ 0.4 ´ ç8.99 +
÷ B ´1.52
B ø
è
= 964.66 + 4.46B
Step 6: The foundation slab thickness should be determined by structural analysis.
Assume the thickness is 0.3 m. Then the foundation slab’s weight is:
W f = 23.56 ´ 0.3B2 = 7.07B2 (kN) (unit weight of concrete is 23.56 kN/m3)
The stress at the bottom of the foundation is:
Q + W f 650 + 7.07B2 650
p=
=
= 2 + 7.07
A
B2
B
Step 7: The allowable bearing capacity should be at least p.
q
964.66 + 4.46B 650
qall = ult =
= 2 + 7.07
FS
3
B
Solve for B and find B = 1.43 m. Use B = 1.5 m.
The stress at the bottom of the foundation is:
Q + Wf
650
p=
=
+ 7.07 = 296.0 kN/m2
A
1.5 ´1.5
The ultimate bearing capacity is:
qult = 971.35 kN/m2.
The allowable bearing capacity is:
q
971.35
=323.8 kN/m2 > 296.0 kN/m2
qall = ult =
FS
3
2. Bearing capacity problem. The problem statement is the same as in Problem 1 and
shown in Figure 3.26. Use the general bearing capacity theory.
(1) Determine the foundation embedment and design the dimension of the spread
footing. The factor of safety for bearing capacity is 3.0.
(2) Choose the foundation dimensions that are acceptable in the construction, then
calculate the allowable bearing capacity based on these dimensions.
Solution:
Step 1: The loads from the superstructure: Q = 650 kN.
The subsoil condition: c = 55 kN/m2,  = 15,  = 18 kN/m3, sat = 18.8 kN/m3.
Step 2: Based on Table 3.4, the minimum Df = 0.5 m, choose Df = 0.6 m (< Dw = 1.2 m).
Step 3: Groundwater table (GWT) elevation: Dw = 1.2 m. It is assumed the GWT affects
the ultimate bearing capacity.
Step 4: The minimum factor of safety is 3.0.
Step 5: The general ultimate bearing capacity equation for square footings is:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i
2
Using Table 3.3 and given  = 15, the bearing capacity factors are:
N c =10.98, N q = 3.94, and Ng = 2.65.
The foundation is square, i.e., B = L. The shape factors are:
æ Bö æ N ö
æ 3.94 ö
Fcs = 1+ ç ÷ ç q ÷ = 1+1´ ç
= 1.359
è L ø è Nc ø
è 10.98 ÷ø
æ Bö
Fqs = 1+ ç ÷ tanf ¢ = 1+1´ tan15 = 1.268
è Lø
æ Bö
Fg s = 1- 0.4 ç ÷ = 0.6
è Lø
Assuming
Df
B
£ 1 , and since  = 15 > 0, the depth factors are:
æD ö
Fcd = 1+ 0.4 ç f ÷
è Bø
æ 0.6 ö
0.24
= 1+ 0.4 ç
÷ = 1+
è B ø
B
Fg d =1
Since the load is vertical, the load inclination factors Fci, Fqi, Fi are all 1.0.
The effective stress at the bottom of the foundation is: q = 180.6 = 10.8 kN/m2.
Dw - D f
g - g ¢ to substitute  in the bearing capacity equation,
Use g = g ¢ +
B
(
)
where,  = sat  w = 18.89.81= 8.99 kN/m3 and
g = 8.99 +
1.2 - 0.6
4.91
(18 - 9.81) = 8.99 +
B
B
So:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i
2
æ 0.24 ö
æ 0.176 ö
= 55 ´10.98 ´1.395´ ç1+
÷ ´1+18.8 ´ 3.94 ´1.268 ´ ç1+
÷ ´1
è
è
B ø
B ø
æ
4.91 ö
+0.5´ ç8.99 +
÷ B ´ 2.65´ 0.6 ´1´1
è
B ø
Step 6: The foundation slab thickness should be determined by structural analysis.
Assume the thickness is 0.3 m. Then the foundation slab’s weight is:
W f = 23.56 ´ 0.3B2 = 7.07B2 (kN) (unit weight of concrete is 23.56 kN/m3)
The stress at the bottom of the foundation is:
Q +W f 650 + 7.07B 2 650
p=
=
= 2 + 7.07
A
B2
B
Step 7: The allowable bearing capacity should be at least p.
qall =
qult 150
=
+ 7.07
FS B 2
That is:
284.47 + 2.33B +
72.90 150
= 2 + 7.07
B
B
Solve for B and find B = 1.37 m. Use B = 1.5 m.
The stress at the bottom of the foundation is:
Q +W f
650
+ 7.07 = 296.0 kN/m2
A
1.5´1.5
The ultimate bearing capacity is:
p=
=
qult = 961.2 kN/m2.
The allowable bearing capacity is:
q
961.2
= 320.4 kN/m2 > 296.0 kN/m2
qall = ult =
FS
3
3. Bearing capacity problem. A circular foundation is designed to support a water tank.
The diameter of the tank is 8 meter. The total load from the water tank is
approximately 2500 kN. The foundation embedment is 1.0 m. The subsoil is silty sand
with c = 0 and  = 35. The groundwater table is far blow the foundation. The
foundation soil is compacted to 100% of the maximum dry unit weight of 18.8 kN/m3.
Determine the factor of safety for the bearing capacity. Any method can be used.
Solution:
Solution using allowable stress design and total factors of safety:
Use Terzaghi’s bearing capacity theory.
Given: diameter D = 8 m, Q = 2500 kN, Df = 1.0 m,
c = 0,  = 35,  = 18.8 kN/m3, no groundwater effect
Assume general failure, using Table 3.1 and given  = 35, the bearing capacity
N c =57.754, N q = 41.440, and Ng = 45.410.
factors are:
Effective stress: q = 18.8  1.0 = 18.8 kN/m2
For circular footing:
qult =1.3c¢N c + qN q + 0.3g BNg
= 0 +18.8 ´ 41.44 + 0.3´18.8´ 8´ 45.41
= 2828 kN/m 2
Pressure due to super-structure:
Q
2500
p= =
= 49.76 kN/m 2
A 1 ´ 3.14 ´ 82
4
qult 2828
=
= 56
p
49.8
The bearing capacity has a very large factor safety.
FS =
Solution using limit state design and partial factors of safety:
It is assumed that the water level in the tank is a variable and unfavorable action hence
the design effect of the actions, Ed, can be found as:
𝐸𝑑 = 𝑉𝑄,𝑘 × 𝛾𝑄 = 2500 × 1.5 = 3750 kN
Using Terzaghi´s equation for circular footings, and material partial factors of safety
equal to 1.0:
𝑉𝑑 =
𝑉𝑑 =
𝑉𝑑 =
𝐴×(1.3𝑐´𝑁𝑐 +𝑞𝑁𝑞 +0.3𝛾𝐵𝑁_𝛾 )
𝛾𝑅𝑣
50.26×(18.8×1.0×41.44+0.3×18.8×8×45.41 )
1.4
50.26×(18.8×1.0×41.44+0.3×18.8×8×45.41 )
1.4
=101535 kN
Since 𝐸𝑑 ≤ 𝑉𝑑 the limit state is satisfied and the foundation is safe.
4. Bearing capacity problem. A continuous footing is designed to support a bearing
wall with a load of 300 kN/m. The groundwater table is 1.0 m below the ground
surface. The subsoil is silty sand with effective cohesion of 15 kN/m2 and effective
internal friction angle of 35. The bulk unit weight above the groundwater table is 17.5
kN/m3; the saturated unit weight below the groundwater table is 18.5 kN/m3. Use
Terzaghi’s bearing capacity theory and assume local shear failure.
(1) Determine the embedment depth and the dimensions of the wall footing. The factor
of safety for bearing capacity is 3.0.
(2) Choose the foundation dimensions that are acceptable in the construction, then
calculate the allowable bearing capacity based on these dimensions.
Solution:
A continuous footing is designed.
Step 1: The loads from the superstructure: Q = 300 kN/m.
The subsoil condition: c = 15 kN/m2,  = 35,  = 17.5 kN/m3, sat = 18.5 kN/m3.
Step 2: Based on Table 3.4, the minimum Df = 0.6 m, choose Df = 1.0 m (=Dw ).
Step 3: Groundwater table (GWT) elevation: Dw = 1.0 m. The GWT affects the ultimate
bearing capacity.
Step 4: The minimum factor of safety is 3.0.
Step 5: The Terzaghi’s ultimate bearing capacity equation for continuous footing is:
qult = c¢Nc + qNq + 0.5g BNg
Assume local failure,
2
2
c¢ = c¢ = ´15 = 10 kN/m 2
3
3
Using Table 3.2 and  = 25, the bearing capacity factors are:
N c =14.809, N q = 5.60, and Ng = 1.911.
The effective stress at the bottom of the foundation is: q = 17.51.0 = 17.5 kN/m2.
Use   to substitute  in the bearing capacity equation,
where,  =  sat   w = 18.59.81= 8.69 kN/m3
So:
qult = c¢N c + qN q + 0.5g ¢ BN g
= 10 ´14.809 +17.5 ´ 5.6 + 0.5 ´ 8.69B ´ 1.911
= 246.1+ 8.6B
Step 6: The foundation slab thickness should be determined by structural analysis.
Assume the thickness is 0.3 m. Then the foundation slab’s weight is:
W f = 23.56 ´ 0.3B = 7.07B (kN/m) (unit weight of concrete is 23.56 kN/m3)
The stress at the bottom of the foundation is:
Q + W f 300 + 7.07B 300
p=
=
=
+ 7.07
A
B ´1
B
Step 7: The allowable bearing capacity should be at least p.
q
246.1 + 8.6B 300
qall = ult =
=
+ 7.07
FS
3
B
Solve for B and find B = 3.52 m. Use B = 3.6 m.
(It is noted the wall footing width is very large, it is caused by the high load.)
The stress at the bottom of the foundation is:
Q + W f 300
p=
=
+ 7.07 = 90.4 kN/m2
A
3.6
The ultimate bearing capacity is:
qult = 277.1 kN/m2.
The allowable bearing capacity is:
q
276.2
=92.3 kN/m2 > 90.4 kN/m2
qall = ult =
FS
3
5. Bearing capacity problem. The problem statement is the same as in Problem 4. Use
the general bearing capacity theory. Assume general failure mode.
(1) Determine the embedment depth and the dimension of the wall footing. The factor
of safety for bearing capacity is 3.0.
(2) Choose the foundation dimension that is acceptable in the construction, then
calculate the allowable bearing capacity based on the actual dimension of the
foundation.
Solution:
Step 1: The loads from the superstructure: Q = 300 kN/m.
The subsoil condition: c = 15 kN/m2,  = 35,  = 17.5 kN/m3, sat = 18.5 kN/m3.
Step 2: Based on Table 3.4, the minimum Df = 0.6 m, choose Df = 1.0 m (=Dw ).
Step 3: Groundwater table (GWT) elevation: Dw = 1.0 m. The GWT affects the ultimate
bearing capacity.
Step 4: The minimum factor of safety is 3.0.
Step 5: The general ultimate bearing capacity equation for square footings is:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i
2
It is noted that the general bearing capacity equation is based on general failure.
So, general failure is assumed, and the soil’s strength is not reduced.
Using Table 3.3 and based on  = 35, the bearing capacity factors are:
N c =46.124, N q = 33.296, and Ng = 48.029.
The foundation is continuous footing, i.e., B << L. The shape factors are:
æ Bö æ N ö
æ 3.94 ö
Fcs = 1+ ç ÷ ç q ÷ = 1+ 0 ´ ç
= 1.0
è L ø è Nc ø
è 10.98 ÷ø
æ Bö
Fg s = 1- 0.4 ç ÷ = 1.0
è Lø
Assuming
Df
B
£ 1 , and since  = 35 > 0, the depth factors are:
æD ö
Fcd = 1+ 0.4 ç f ÷
è Bø
0.4
æ 1ö
= 1+ 0.4 ç ÷ = 1+
è Bø
B
Fg d =1
Since the load is vertical, the load inclination factors Fci, Fqi, Fi are all 1.0.
The effective stress at the bottom of the foundation is: q = 17.51.0 = 17.5 kN/m2.
Use   to substitute  in the bearing capacity equation,
where,  = sat  w = 18.59.81= 8.69 kN/m3 and
So:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g ¢ BNg Fg s Fg d Fg i
2
æ 0.4 ö
æ 0.255 ö
= 15 ´ 46.124 ´ 1´ ç 1+
´ 1+ 17.5 ´ 33.296 ´ 1´ ç 1+
÷
÷ ´1
è
è
B ø
B ø
+0.5 ´ 8.69B ´ 48.029 ´ 1´ 1´ 1
425.3
= 1044.5 +
+ 208.7B
B
Step 6: The foundation slab thickness should be determined by structural analysis.
Assume the thickness is 0.3 m. Then the foundation slab’s weight is:
W f = 23.56 ´ 0.3B = 7.07B (kN/m) (unit weight of concrete is 23.56 kN/m3)
The stress at the bottom of the foundation is:
p=
Q + Wf
A
=
300 + 7.07B 300
=
+ 7.07
B ´1
B
Step 7: The allowable bearing capacity should be at least p.
qall =
qult
=p
FS
1274.5+
That is:
425.3
+ 208.7B 300
B
=
+ 7.07
3
B
Solve for B and find B = 0.36 m. Use B = 0.5 m.
The stress at the bottom of the foundation is:
Q + Wf
300
+ 7.07 = 607.1 kN/m2
A
0.5
The ultimate bearing capacity is:
p=
=
qult = 2229.4 kN/m2.
The allowable bearing capacity is:
q
2229.4
= 743.1 kN/m2 > 607.1 kN/m2
qall = ult =
FS
3
(Note, the solution of Problem 5 should not be compared with that of Problem 4, because
Problem 5 assumes general failure and the soil’s strength is not reduced. This causes the
foundation width much smaller than that in Problem 4).
6. Bearing capacity problem. A rectangular footing is used to support a column. The
dimensions of the foundation are 3 m by 4m, and the foundation embedment depth is
1.5 m. The soil’s strength parameters are c = 38 kN/m2,  = 25,  = 18 kN/m3. The
groundwater table is 25 m below the ground surface. Using FS = 3.0, determine the
allowable bearing capacity.
Solution:
Solution using allowable stress design and total factors of safety:
Given: The subsoil condition: c = 38 kN/m2,  = 25,  = 18 kN/m3. The groundwater
table (GWT) depth is Dw = 25 m > (B + Df = 3+1.5 = 4.5 m). The GWT has no effect on
bearing capacity.
Since it is rectangular footing, the general ultimate bearing capacity equation is used:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i
2
Using Table 3.3 and based on  = 25, the bearing capacity factors are:
N c =20.721, N q = 10.662, and Ng = 10.876.
The foundation is rectangular footing, B = 3 m, L = 4 m. The shape factors are:
3 æ 10.662 ö
æ Bö æ N ö
Fcs = 1+ ç ÷ ç q ÷ = 1+ ´ ç
÷ = 1.386
è L ø è Nc ø
4 è 20.721 ø
æ Bö
Fg s = 1- 0.4 ç ÷ = 0.7
è Lø
Assuming
Df
B
£ 1 , and since  = 25 > 0, the depth factors are:
æD ö
Fcd = 1+ 0.4 ç f ÷
è Bø
æ 1.5 ö
= 1+ 0.4 ç ÷ = 1.2
è 3ø
Fg d =1
Assume the load is vertical, the load inclination factors Fci, Fqi, Fi are all 1.0.
The effective stress at the bottom of the foundation is: q = 181.5 = 27 kN/m2.
So:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i
2
= 38 ´ 20.721 ´ 1.386 ´ 1.2 ´ 1+ 27 ´ 10.662 ´ 1.35 ´ 1.165 ´ 1
+0.5 ´ 18 ´ 3 ´ 10.876 ´ 0.7 ´ 1´ 1
= 1968.0 kN/m2
The allowable bearing capacity is
qall =
qult 1968.0
=
= 656.0 kN/m 2
FS
3
Solution using limit state design and partial factors of safety:
The bearing capacity of the footing can be found using the general theory
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i
2
In other problems involving limit design material partial factors of safety have been used
with a value of 1.0. This is the case for Design Approach 2 of EN 1997-1:2004. In that
case the bearing capacity would be the same as in the allowable stress approach but then
divided by a partial factor of safety on the bearing resistance of 1.4 (as in Problem 3). To
illustrate other set of partial factors of safety, Design Approach 3 of EN 1997-1:2004 is
used here. Hence,
Design values of geotechnical parameters:
𝑐´
38
𝑐´𝑑 = 𝛾 𝑘 = 1.25 = 30.4 kPa
𝑐´
𝜑´𝑑 = 𝑎𝑡𝑎𝑛
𝛾𝑑 =
𝛾𝑘
𝛾𝛾
=
(𝑡𝑎𝑛𝜑´𝑘 )
17.5
1.0
𝛾𝜑´
= 𝑎𝑡𝑎𝑛
𝑡𝑎𝑛25
1.25
= 20.5
= 17.5 kN/m3 (Assuming bulk unit weight in problem 4)
Hence,
𝑁𝑐 = 15.325
𝑁𝑞 = 6.735
𝑁𝛾 = 5.791
𝐵
3
𝐹𝛾𝑠 = 1 − 0.3 ( 𝐿 ) = 1 − 0.3 (4) = 0.78
𝐵
3
𝐹𝑞𝑠 = 1 + 0.3 ( 𝐿 ) 𝑠𝑖𝑛𝜙 = 1 + 0.3 (4) 𝑠𝑖𝑛20.5 = 1.08
𝐹𝑐𝑠 =
(𝐹𝑞𝑠 𝑁𝑞 −1)
(𝑁𝑞 −1)
=
1.08×6.735−1
6.735−1
= 1.09
And then, the design resistance can be calculated as (ignoring depth factors):
𝑉𝑑 =
𝑉𝑑 =
𝐴×(𝑐´𝑁𝑐 𝐹𝑐𝑠 +𝑞𝑁𝑞 𝐹𝑞𝑠 +0.5𝛾𝐵𝑁𝛾 𝐹𝛾𝑠 )
𝛾𝑅𝑣
12×(30.4×15.325×1.09+17.5×1.5×6.735×1.08+0.5×17.5×3×5.791×0.78)
1.0
𝑉𝑑 = 9807.8 kN
7. Bearing capacity problem with inclined load. A continuous footing is designed to
support a bearing wall with a load of 300 kN/m. The load on the wall footing is
inclined with respect to the vertical direction at 15. The groundwater table is 10 m
below the ground surface. The subsoil is silty sand with effective cohesion of 15
kN/m2 and effective internal friction angle of 35. The bulk unit weight above the
groundwater table is 17.5 kN/m3; the saturated unit weight below the groundwater
table is 18.5 kN/m3. Use FS = 3.0. Determine the foundation’s embedment depth,
dimensions, and use the actual dimensions that are generally accepted in field
construction to determine the allowable bearing capacity.
Solution:
Step 1: The loads from the superstructure: Q = 300 kN/m.
The subsoil condition: c = 15 kN/m2,  = 35,  = 17.5 kN/m3, sat = 18.5 kN/m3.
Step 2: Based on Table 3.4, the minimum Df = 0.6 m, choose Df = 1.0 m (=Dw ).
Step 3: Groundwater table (GWT) elevation: Dw = 10 m. It is assumed the GWT does not
affect bearing capacity.
Step 4: FS= 3.0.
Step 5: The general ultimate bearing capacity equation for continuous footings is:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i
2
It is noted that the general bearing capacity equation is based on general failure.
So, general failure is assumed, and the soil’s strength is not reduced.
Using Table 3.3 and based on  = 35, the bearing capacity factors are:
N c =46.124, N q = 33.296, and Ng = 48.029.
The foundation is continuous footing, i.e., B << L. The shape factors are:
æ Bö æ N ö
æ 3.94 ö
Fcs = 1+ ç ÷ ç q ÷ = 1+ 0 ´ ç
= 1.0
è L ø è Nc ø
è 10.98 ÷ø
æ Bö
Fg s = 1- 0.4 ç ÷ = 1.0
è Lø
Assuming
Df
B
£ 1 , and since  = 35 > 0, the depth factors are:
æD ö
Fcd = 1+ 0.4 ç f ÷
è Bø
0.4
æ 1ö
= 1+ 0.4 ç ÷ = 1+
è Bø
B
Fg d =1
The inclination of the vertical load respect to the vertical direction is  = 15.
The effective stress at the bottom of the foundation is: q = 17.51.0 = 17.5 kN/m2.
Use   to substitute  in the bearing capacity equation,
where,  = sat  w = 18.59.81= 8.69 kN/m3 and
So:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i
2
æ 0.4 ö
æ 0.255 ö
= 15 ´ 46.124 ´ 1´ ç 1+
´ 0.694 + 17.5 ´ 33.296 ´ 1´ ç 1+
÷
÷ ´ 0.694
è
è
B ø
B ø
+0.5 ´ 17.5B ´ 48.029 ´ 1´ 1´ 0.326
295
= 884 +
+ 137B
B
Step 6: The foundation slab thickness should be determined by structural analysis.
Assume the thickness is 0.3 m. Then the foundation slab’s weight is:
W f = 23.56 ´ 0.3B = 7.07B (kN/m) (unit weight of concrete is 23.56 kN/m3)
The stress at the bottom of the foundation is:
p=
Q + Wf
A
=
300 + 7.07B 300
=
+ 7.07
B ´1
B
Step 7: The allowable bearing capacity should be at least p.
qall =
qult
=p
FS
884 +
That is:
295
+137B 300
B
=
+ 7.07
3
B
Solve for B and find B = 0.64 m. Use B = 0.8 m.
The stress at the bottom of the foundation is:
Q + Wf
300
+ 7.07 = 382.1 kN/m2
A
0.8
The ultimate bearing capacity is:
p=
=
qult = 1362.3 kN/m2.
The allowable bearing capacity is:
q
1262.3
= 454.1 kN/m2 > 382.1 kN/m2
qall = ult =
FS
3
8. Bearing capacity problem with one-way eccentricity. A rectangular footing is to
support a column. The load on the column, including the weight of the foundation, is
Q = 1200 kN. The foundation is also subjected to an overturning moment of 150
kNm. The foundation rests in homogeneous silty sand with c = 0 and  = 35. The
foundation embedment depth is 1 m, and the dimensions are LB=2 m  1.5 m. The
one-way eccentricity is in the L (longer dimension) direction. The groundwater table is
at the same level of the bottom of the foundation. The bulk unit weight above the
groundwater table is 17 kN/m3; the saturated unit weight below the groundwater table
is 18 kN/m3. Determine the factor of safety for the bearing capacity.
Solution:
Solution using allowable stress design and total factors of safety:
Use the Meyerhof’s effective area method.
M 150
B 1.5
=
= 0.125m < =
= 0.25m . OK. No tension is
Q 1200
6
6
developed at the bottom of the foundation.
Step 2: Determine the effective dimensions. The one-way eccentricity is in the L (longer
dimension) direction.
So,
effective width: B = B = 1.5 m;
effective length: L = L  2e = 2.0  0.25 = 1.75 m.
Step 3: Use the general bearing capacity equation to determine the ultimate bearing
capacity.
Step 1: Eccentricity: e =
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i
2
Using Table 3.3 and given  = 35, the bearing capacity factors are:
N c =46.124, N q = 33.296, and Ng = 48.029.
Since c = 0, no need to calculate Nc, Fcs, Fcd, Fci
Use effective dimensions to calculate the shape factors:
æ B¢ ö
æ 1.5 ö
Fg s = 1- 0.4 ç ÷ = 1- 0.4 ´ ç
= 0.657
è L¢ ø
è 1.75 ÷ø
Use the original dimensions to calculate the depth factors:
Df
Since
B
=
1
< 1 and  >0
1.5
Fg d =1
Since the load is vertical, the load inclination factors Fci, Fqi, Fi are all 1.0
The effective stress at the bottom of the foundation is: q = 1.017 = 17 kN/m2.
The groundwater table is at the bottom of the foundation,
use   = sat - w = 18 – 9.81 = 8.19 kN/m3
So:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g ¢ B¢Ng Fg s Fg d Fg i
2
= 0 + 17 ´ 33.296 ´ 1.600 ´ 1.170 ´ 1+ 0.5 ´ 8.19 ´ 1.5 ´ 48.029 ´ 0.657 ´ 1´ 1
= 1059.6 +193.8 = 1253.4 kN/m 2
Step 4: Calculate the ultimate load that the foundation can sustain:
Qult = qult ( L ¢B¢ ) = 1253.4 ´ (1.75 ´1.5 ) = 3290.2 kN
Step 5: The factor of safety for the entire foundation:
Qult 3290.2
=
= 2.74
Q
1200
Step 6: Calculate the factor of safety for local failure.
FS=
qmax =
FS =
Q 6M
1200
6 ´150
+ 2 =
+ 2
= 600.0 kN/m 2
BL B L 1.5 ´ 2.0 1.5 ´ 2.0
qult 1253.4
= 2.09
=
qmax 600.0
The smaller FS is the FS of the design. So: FS = 2.09 < 3.0. The dimension of the
foundation should be increased to satisfy FS  3.0.
Solution using limit state design and partial factors of safety:
Assuming that both the vertical load and the moment are unfavorable but permanent
loads. Hence,
FG,d = 1200(1.35) = 1620 kN
MQ,d = 150(1.35) = 202.5 kNm
Design effect of the actions
Vd = FG,d = 1620 kN
Design eccentricity
ed = Md/Vd = 202.5/1620 = 0.125 m< 1.2/1.5 = 0.8 m
Design values of geometrical data
B’ = 1.5 m
L’ = L – 2ed = 2.0 – 2(0.125) = 1.75 m
Design values of geotechnical parameters
cd = c’k/c’ = 0/1.00 = 0 kN/m2
d = atan(tan’k/’) = atan(tan35/1.00) = 35o


d = k/ = 17/1.00 = 17 kN/m3 (above ground water level)


d = k/ = 18/1.00 = 18 kN/m3 (below ground water level)
The general ultimate bearing capacity equation for square footings is:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g B¢N r Fg s Fg d Fg i
2
Using Table 3.3 and given d = 35, the bearing capacity factors are: N c =46.124,
N q = 33.296, and Ng = 48.029.
Use effective dimensions to calculate the shape factors:
𝐵′
1.5
𝐹𝛾𝑠 = 1 − 0.3 ( 𝐿′ ) = 1 − 0.3 (1.75) = 0.7
𝐵′
1.5
𝐹𝑞𝑠 = 1 + ( 𝐿′ ) sin  = 1 + (1.75)sin 35 = 1.49
𝐹𝑐𝑠 =
(𝐹𝑞𝑠 𝑁𝑞 −1)
(𝑁𝑞 −1)
=
1.49×33.296−1
33.296−1
= 1.51
Use the original dimensions to calculate the depth factors:
𝐷
1.0
Since 𝐵𝑓 = 1.5 = 0.67 < 1:
𝐹𝑐𝑑 = 1.26
𝐹𝑞𝑑 = 1.15
Fd = 1.0
Since the load is vertical, the load inclination factors Fci, Fqi, Fi are all 1.0.
The effective stress at the bottom of the foundation is: q = 117 = 17.0 kN/m2.
Use g = g ¢ +
Dw - D f
B
(g - g ¢ ) to substitute  in the bearing capacity equation,
where,𝛾´ = 𝛾𝑠𝑎𝑡 − 𝛾𝑤 = 18 − 9.8 = 8.2 kN/m3 and
So:
Rk = (cNcFcsFcdFci+qNqFqsFqdFqi+0.5BNFsFdFi)A
= (1733.2961.491.151
+0.58.21.048.0290.711)(1.51.75)
= 2907.8 kN
Design resistance
Rd = Rk/R = 2907.8/1.40 = 2077 kN
Verification of ultimate limit state
Vd = 1620 kN ≤ Rd = 2077 kN
Design is satisfactory
9. Bearing capacity problem with one-way eccentricity. A continuous footing is
designed to support a bearing wall with a vertical load of 340 kN/m. In addition, the
foundation is subjected to an overturning moment of 170 kNm per unit length (meter)
of the wall. The groundwater table is 15 m below the ground surface. The bulk unit
weight above the groundwater table is 17.5 kN/m3; the saturated unit weight below the
groundwater table is 18.5 kN/m3. The subsoil is silty sand with effective cohesion of
25 kN/m2 and effective internal friction angle of 35. Use FS of at least 3.0. Determine
the foundation’s embedment depth, dimensions, and use these dimensions to
determine the allowable bearing capacity. The trial-and-error method may be used to
obtain the dimensions.
Solution:
Use the Meyerhof’s effective area method, and trial-and-error method.
Assume the groundwater table does not affect the bearing capacity.
Based on Table 3.4, load P = 340 kN/m, minimum Df = 60 cm, use Df = 1.0 m.
Step 1:
Eccentricity: e = M/Q = (170 kNm/m) / (340 kN/m)= 0.5 m
In order to avoid tension at the bottom of the foundation, B6e = 3.0 m
Use trial-and-error, and use a preliminary B = 3.0 m
Step 2:
Determine the effective dimensions. Effective width: B = B  2e = 3 – 1.0 = 2.0 m;
effective length: L = L =  (for continuous footing).
Step 3:
Use the general bearing capacity equation to determine the ultimate bearing capacity.
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i
2
Using Table 3.3 and given  = 35, the bearing capacity factors are:
N c =46.12, N q = 33.30, and Ng = 48.03
Use effective dimensions to calculate the shape factors:
æ B¢ ö æ N ö
2.0 23.17
Fcs = 1+ ç ÷ ç q ÷ = 1+
´
=1
¥ 35.49
è L ø è Nc ø
æ B¢ ö
2.0
Fqs = 1+ ç ÷ tanf ¢ = 1+
´ tan32 = 1
¥
è Lø
æ B¢ ö
2.0
Fg s = 1- 0.4 ç ÷ = 1- 0.4 ´
=1
¥
è Lø
Use the original dimensions to calculate the depth factors:
Since Df/B = 1/3 < 1, and  = 35 > 0, the depth factors are:
æD ö
Fcd = 1+ 0.4 ç f ÷
è Bø
æ 1ö
= 1+ 0.4 ç ÷ = 1.133
è 3ø
Fg d =1
Since the load is vertical, the load inclination factors Fci, Fqi, Fi are all 1.0.
The effective stress at the bottom of the foundation is: q = 1.017.5 = 17.5 kN/m2.
So:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g B¢N r Fg s Fg d Fg i
2
= 25 ´ 46.12 ´ 1´ 1.133 ´ 1+ 17.5 ´ 33.30 ´ 1´ 1.085 ´ 1
1
+ ´ 17.5 ´ 2.0 ´ 48.03 ´ 1´ 1´ 1
2
= 2779 kN/m 2
Step 4: Calculate the ultimate load that the foundation can sustain:
Qult = qult ´ ( B¢ ) = 2779 ´ 2.0 = 5558 kN per linear meter of the footing
Step 5: The factor of safety for the entire foundation:
Qult 5558 kN/m
=
= 16.3 > 3.0
Q
340 kN/m
Step 6: Calculate the factor of safety for local failure.
FS =
qmax =
Q 6 M 340 kN/m 6 ´170 kN im/m
+ 2 =
+
= 425 kN/m 2
2
2
B¢ B¢
2.0 m
2.0 (m )
qult 2779
=
= 6.54 >3.0
qmax 425
The smaller FS is the FS of the design. So: FS = 6.53> 3.0.
Conclusion: the width of the continuous foundation is B = 3.0 m, the embedment is Df =
1.0 m.
FS =
10. Concepts of mat foundations. Select the correct answer(s) in the following multiplechoice questions.
Answer: the correction answers are marked by 
(1) A mat foundation is:
 (A) A type of shallow foundation.
(B) A type of deep foundation.
(C) Neither a shallow nor a deep foundation.
 (D) A footing supporting an array of columns in several rows in each direction.
 (E) Also known as a raft foundation.
 (F) Required to cover at least 75% of the total footprint of the structure.
(2) The following conditions may warrant the consideration of a mat foundation:
 (A) The superstructure load is high or the subsoil is erratic or weak.
 (B) Excessive differential settlement is likely to occur.
 (C) The foundation elevation is at the groundwater fluctuation zone.
 (D) The structure is subjected to non-uniform lateral loads.
(E) A shallow firm stratum is present in the subsoil.
11. Vertical stress increase problem. As shown in Figure 3.27, a continuous footing is
designed to support a bearing wall with load of 300 kN/m. The groundwater table is
far below the ground surface. The subsoil is homogeneous silty sand with bulk unit
weight of 17.5 kN/m3. The width of the wall foundation is B = 1 m, the foundation
embedment depth is Df = 0.5 m. Use the following methods to determine the vertical
stress increases at z = 0, z = 2.5B, and z = 5B, and calculate the average vertical stress
increase in the soil layer below the center of the footing from z = 0 to 5B. Note, z starts
from the bottom of the foundation.
(1) Boussinesq method.
(2) Poulos and Davis method.
(3) 2:1 method.
Q
Ground surface
Df
B
z
H = 5B
Figure 3.27 Illustration of shallow foundation with subsoil stratum
Solution:
Given parameters:
Continuous footing.
The loads from the superstructure: Q = 300 kN/m.
 = 17.5 kN/m3, no groundwater present.
Df = 0.5 m, B = 1.0 m
(1) Based on Table 3.6, the Boussinesq equation for vertical stress increase below a
uniform strip load is:
Ds z =
(
)
pé
a + sina ×cos a + 2g ùû
pë
Q 300
=
= 300 kN/m2
B
1
For vertical stress increase below the centerline of a strip loading:
æ 0.5B ö
and  = 2
g = tan -1 ç
è z ÷ø
Where: p =
(2) Use the simplified equation (3.51) given by Poulos and Davis (1974). For continuous
foundations with length B, the vertical stress increase at depth z below the center of the
foundation is:
2.60
é æ
ö ù
ê ç
÷ ú
1
ê ç
÷ ú
Ds z = ( q - s 0¢ ) ê12
ç æ ö ÷ ú
ê ç 1+ ç B ÷ ÷ ú
êë è è 2z ø ø úû
where:
q = uniform stress at the bottom of the foundation.
Q 300
q= =
= 300 kN/m2
B
1
0 = effective stress due to self-weight of soil at the bottom of the foundation.
0 = 0.517.5 = 8.75 kN/m2.
(3) The 2:1 method for vertical stress increase below a continuous foundation is:
Ds z =
p0 ´ B
( B + z)
p0 =
Q 300
=
= 300 kN/m2
B
1
The average vertical increase uses equation (3.55):
Ds z(av) =
(
1
Ds z(top) + 4Ds z(mid ) + Ds z(bot )
6
)
The vertical stress increases using the three methods are calculated at the top, middle, and
bottom of the soil layer and listed in the following table.
z (m)
z (Boussinesq)
z (Poulos and
2:1 method
(kN/m2)
0
2.5 (2.5B)
5 (5B)
Average vertical
stress increase
(kN/m2)
(kN/m2)
300.0 (=2=180)
63.5 (=2=22.6)
36.4 (=2=11.4)
Davis, 1974)
(kN/m2)
291.2
28.2
7.4
98.4
68.5
115.5
300
85.7
50
12. Vertical stress increase problem. As shown in Figure 3.27, a square spread footing
is designed to support a column. The load on the column, including the foundation
weight, is 600 kN. The dimensions of the foundation are BB=1.2 m  1.2 m, the
foundation embedment is Df = 0.6 m. The subsoil is homogeneous silty sand with bulk
unit weight of 18 kN/m3. Use the following methods to determine the vertical stress
increases at z = 0, z = 2.5B, and z = 5B, and calculate the average vertical stress
increase below the center of the footing in the soil layer from z = 0 to 5B. Note, z starts
from the bottom of the foundation.
(1) Boussinesq method.
(2) Poulos and Davis method.
(3) 2:1 method.
Solution:
Given parameters:
Square footing.
The loads from the superstructure: Q = 600 kN.
 = 18 kN/m3, no groundwater present.
Df = 0.6 m, BB=1.2 m  1.2 m
(1) The Boussinesq equation for vertical stress increase below the corner of a rectangular
foundation is:
é
2
2
æ 2 2
öù
ê 2mn m + n +1 ç m + n + 2 ÷ú
ç 2 2 ÷
2
2
2 2
p êê m + n + m n +1 è m + n +1 øúú
Ds z =
æ
ö
2
2
4p ê
ú
-1 2mn m + n +1 ÷
ê+tan ç 2 2
ú
ç m + n - m 2 n 2 +1 ÷
êë
úû
è
ø
Divide the square footing into four equal sections, can calculate the vertical stress
increase under each quadruple; the result should be multiplied by 4 to obtain the stress
increase beneath the center of the foundation. In the above equation, m = B/z, and n =
L/z, where B = L = 1.2/2 = 0.6 m (divide the foundation into four sections).
Q
600
p=
=
= 416.7 kN/m2
B ´ B 1.2´1.2
(2) Use the simplified equation (3.50) given by Poulos and Davis (1974). For square
foundations with length B, the vertical stress increase at depth z below the center of the
foundation is:
1.76
é æ
ö ù
ê ç
÷ úú
ê ç
1
÷ ú
Ds z = q - s 0¢ ê12
ç
æ Bö ÷ ú
ê ç
1+
çè 2z ÷ø ÷ ú
ê è
ø
ë
û
(
)
where:
q = uniform stress at the bottom of the foundation.
Q
600
q=
=
= 416.7 kN/m2
B ´ B 1.2´1.2
0 = effective stress due to self-weight of soil at the bottom of the foundation.
0 = 0.618 = 10.8 kN/m2.
(3) The 2:1 method for vertical stress increase below a rectangular foundation is:
Ds z =
p0 ´ B ´ L
( B + z) ´ ( L + z)
where B = L = 1.2 m.
The average vertical increase uses equation (3.55):
Ds z(av) =
(
1
Ds z(top) + 4Ds z(mid ) + Ds z(bot )
6
)
The vertical stress increases using the three methods are calculated at the top, middle, and
bottom of the soil layer and listed in the following table.
z (m)
0
3 (2.5B)
6 (5B)
Average vertical
stress increase
(kN/m2)
z (Boussinesq)
z (Poulos and
416.7
36.6
9.7
Davis, 1974)
(kN/m2)
406.2
27.1
7.1
95.5
87.0
2
(kN/m )
2:1 method
(kN/m2)
416.7
34.0
11.6
94.1
13. Vertical stress increase problem. The problem statement is the same as in Problem
12, except that the foundation is rectangular with LB=1.5 m  1.2 m. Use the
following methods to determine the vertical stress increases at z = 0, z = 2.5B, and z =
5B, and calculate the average vertical stress increase in the soil layer from z = 0 to 5B.
Note, z starts from the bottom of the foundation (Figure 3.27).
(1) Boussinesq method.
(2) Poulos and Davis method.
(3) 2:1 method.
Solution:
Given parameters:
Square footing.
The loads from the superstructure: Q = 600 kN.
 = 18 kN/m3, no groundwater present.
Df = 0.6 m, BL=1.2 m  1.5 m
(1) The Boussinesq equation for vertical stress increase below the corner of a rectangular
foundation is:
é
2
2
æ 2 2
öù
ê 2mn m + n +1 ç m + n + 2 ÷ú
ç 2 2 ÷
2
2
2 2
p êê m + n + m n +1 è m + n +1 øúú
Ds z =
æ
ö
2
2
4p ê
ú
-1 ç 2mn m + n +1 ÷
ê+tan
ú
ç m 2 + n 2 - m 2 n 2 +1 ÷
êë
úû
è
ø
Divide the square footing into four equal sections, can calculate the vertical stress
increase under each quadruple; the result should be multiplied by 4 to obtain the stress
increase beneath the center of the foundation. In the above equation, m = B/z, and n =
L/z, where B = 1.2/2 = 0.6 m, L = 1.5/2 = 0.75 m (divide the foundation into four
sections).
Q
600
p=
=
= 333.3 kN/m2
B ´ L 1.2´1.5
(2) Use the simplified equation (3.50) given by Poulos and Davis (1974). For square
foundations with length B, the vertical stress increase at depth z below the center of the
foundation is:
1.76
é æ
ö ù
ê ç
÷ úú
ê ç
1
÷ ú
Ds z = q - s 0¢ ê12
ç
æ Bö ÷ ú
ê ç
1+
çè 2z ÷ø ÷ ú
ê è
ø
ë
û
(
where:
)
q = uniform stress at the bottom of the foundation.
Q
600
q=
=
= 333.3 kN/m2
B ´ L 1.2´1.5
0 = effective stress due to self-weight of soil at the bottom of the foundation.
0 = 0.618 = 10.8 kN/m2.
(3) The 2:1 method for vertical stress increase below a rectangular foundation is:
Ds z =
p0 ´ B ´ L
( B + z) ´ ( L + z)
where B = 1.2 m, L = 1.5 m.
The average vertical increase uses equation (3.55):
Ds z(av) =
(
1
Ds z(top) + 4Ds z(mid ) + Ds z(bot )
6
)
The vertical stress increases using the three methods are calculated at the top, middle, and
bottom of the soil layer and listed in the following table.
z (m)
0
3 (2.5B)
6 (5B)
Average vertical
stress increase
(kN/m2)
z (Boussinesq)
z (Poulos and
333.3
29.3
7.8
Davis, 1974)
(kN/m2)
322.8
23.7
5.6
76.4
70.5
(kN/m2)
2:1 method
(kN/m2)
333.3
33.1
11.7
79.6
14. Vertical stress increase problem. The problem statement is the same as in Problem
12, except that the foundation is circular with diameter B=1.5 m. Use Poulos and
Davis method to determine the depth at which the vertical stress increase is 10% of the
in-situ effective stress.
Solution:
Given parameters:
Square footing.
The loads from the superstructure: Q = 600 kN.
 = 18 kN/m3, no groundwater present.
Df = 0.6 m, circular footing with diameter B = 1.5 m
(2) Use the simplified equation (3.49) given by Poulos and Davis (1974). For circular
foundations with diameter B, the vertical stress increase at depth z below the center of the
foundation is:
1.50
é æ
ö ù
ê ç
÷ ú
1
ê ç
÷ ú
Ds z = ( q - s 0¢ ) ê1ç æ B ö2 ÷ ú
ê ç 1+ ç ÷ ÷ ú
êë è è 2z ø ø úû
where:
q = uniform stress at the bottom of the foundation.
Q
600
q=
=
= 339.7 kN/m2
1 2 0.25´ 3.14 ´1.52
pB
4
0 = effective stress due to self-weight of soil at the bottom of the foundation.
0 = 0.618 = 10.8 kN/m2.
Effective stress beneath the footing at depth z (z starts from the bottom of the
footing) is:
0.1 s ¢ = s 0¢ + g z
Use trial-and-error method to determine the depth (z) at which the vertical stress
increase (z) is (0.1). The process is presented in the table below.
Depth, z
z
s ¢ = s 0¢ + g z
0.1 
(m)
(kN/m2)
(kN/m2)
(kN/m2)
5
9.56
90
9.0
5.2
8.85
93.6
9.36
5.1
9.20
91.8
9.18
So, at z = 5.1 m, vertical stress increase is 10% of the in-situ effective stress.
15. Elastic settlement problem. As shown in Figure 3.27, a square spread footing is
designed to support a column. The load on the column, including the foundation
weight, is 600 kN. The dimensions of the foundation are BB=1.2 m  1.2 m, the
foundation embedment depth is Df = 0.6m. The subsoil is heterogeneous silty sand
with bulk unit weight of 18 kN/m3. The soil’s Poisson’s ratio is 0.5. The soil’s elastic
modulus increases linearly with depth: Es = 10,000 (kN/m2) + 40 (kN/m2/m)  z, and z
starts from the bottom of the foundation. The thickness of the foundation slab is 0.3m,
the elastic modulus of the foundation is 1.2107 kN/m2. Determine the elastic
settlement of the soil layer from z = 0 to 5B, using the Mayne and Poulos method.
Solution: Use the Mayne and Poulos (1999) method.
The elastic settlement is:
(1- m )
2
Se = q0 Be I G I F I E
E0
For a square footing with L  B = 1.2 m  1.2 m: Be =
4B 2
p
= 1.35 m
The following parameters are given:
Foundation slab thickness: t = 0.3 m
Poisson’s ratio:  = 0.5
Soil’s elastic modulus at z=0: E0 = 10,000 kN/m2
Rate of elastic modulus increase with depth: kE = 40 kN/m2/m
Elastic modulus of foundation slab: Ef = 1.2107 kN/m2
Q 600
q0 = 2 = 2 = 416.7 kN/m 2
B 1.2
The displacement influence factor, IG, is determined using Figure 3.20 in the lecture note.
The normalized Gibson modulus is: b = E0 = 10000 = 185
k E Be
40 ´ 1.35
And H = 5 ´ 1.2 = 4.44
Be
1.35
From Figure 3.20, and use extrapolation, approximately find IG  0.92
The rigidity correction factor, IF, is:
IF =
p
4
+
1
æ E öæ 2t ö
4.6 +10 çç f ÷÷ç ÷
è Es(av) øè Be ø
3
Es(av) = E0 +
=
p
4
+
1
4.6 +10K F
H
6
kE = 10000 + ´ 40 = 10120 kN/m 2
2
2
æ E f ö æ 2t ö æ 1.2 ´107 ö æ 2 ´ 0.3 ö 3
KF = ç
÷ç ÷ =ç
÷ = 630
֍
è Es( av ) ø è Be ø è 10120 ø è 1.35 ø
3
IF =
p
4
+
1
= 0.785
4.6 + 10 ´ 630
The settlement correction factor, IE, is:
I E = 1-
1
1
= 1= 0.940
æ
ö
æ
ö
1.35
B
1.22´0.5-0.4
3.5e
´ ç 1.6 +
3.5e1.22 m-0.4 ´ ç 1.6 + e ÷
0.6 ÷ø
è
D
è
f ø
So the elastic settlement is:
(1- m ) = 416.7 ´ 1.35 ´ 0.92 ´ 0.785 ´ 0.940 ´ æ 1- 0.5 ö
2
Se = q0 Be I G I F I E
E0
2
çè 10000 ÷ø
= 0.029 m = 2.9 cm
16. Elastic settlement problem. A circular spread footing is designed to support a water
tank. The load of the tank is 1800 kN. The diameter of the foundation is 6 m, the
foundation embedment depth is Df = 1.5 m. The subsoil is homogeneous silty sand
with bulk unit weight of 18 kN/m3. The soil’s Poisson’s ratio is 0.4. The soil’s elastic
modulus increases linearly with depth: Es = 30,000 (kN/m2) + 300 (kN/m2/m)  z, and
z starts from the bottom of the foundation. The thickness of the foundation slab is 1 m;
the elastic modulus of the foundation is 1.5107 kN/m2. Bedrock is at 10 m below the
ground surface. Determine the elastic settlement of the soil layer beneath the
foundation, using the Mayne and Poulos method.
Solution: Not provided.
17. Consolidation settlement problem. As shown in Figure 3.28, a square shallow
foundation is built in a clayey soil. The saturated clay layer is 4 m thick. The
preconsolidation stress (c) is determined to be 70 kN/m2. Other parameters are given
in the figure. Determine the primary consolidation settlement in the clay layer beneath
the foundation.
Q=900 kN
Ground surface
Df = 1 m
γ = 17 kN/m3
Groundwater table
2m x 2m
z
H= 4m
γsat = 18 kN/m3
c’ = 50 kN/m2
φ = 15°
Es= 28,000 kN/m2
μ = 0.35
cc = 0.3
cs = 0.06
e0 = 0.55
Figure 3.28 Diagram for problem 17.
Solution:
As shown in the figure below, the 4-m saturated clay soil stratum is divided into two
layers to obtain better prediction of the consolidation settlement. It is assumed that the
soil properties in both layers are the same.
The stress at the bottom of the foundation due to external load is:
p0 =
Q 900
= 2 = 225 kN/m 2
2
B
2
To simplify the calculation and to be conservative, the 2:1 method is used to calculate the
vertical stress increases in each layer.
Ds z =
p0 ´ B ´ L
( B + z) ´ ( L + z)
The average vertical stress increase in each layer is determined using:
Ds z(av) =
1
( Ds z(top) + 4Ds z(mid ) + Ds z(bot ) )
6
The average effective stress of each layer is the effective stress at the middle of the soil
layer. The results are summarized in the following tables.
z (m)
0
0.5
1.0
2.5
4.0
Layer
1
2
In layer 1:
Vertical stress
increase
(kN/m2)
225
144
100
44.4
25
Average vertical stress
increase, z(av)
(kN/m2)
150.2
50.5
Average in-situ effective
stress, 0(av)
(kN/m2)
21.1
37.5
c > 0(av), the soil is overconsolidated.
And c < 0(av) + z(av) = 171.3 kN/m2.
Sc =
s ¢ + Ds (z¢ )av
cs H
s¢
cH
log c + c log 0(av)
1+ e0
s 0(av)
1+ e0
s c¢
¢
0.06 ´ 1
70
0.3 ´ 1
21.1 + 150.2
log
+
log
1+ 0.55
21.1 1+ 0.55
70
= 0.020 + 0.075 = 0.095 m
=
In layer 2:
c > 0(av), the soil is overconsolidated.
And c < 0(av) + z(av) = 88.0 kN/m2.
Preconsolidation
pressure, c
(kN/m2)
70
70
Sc =
s ¢ + Ds (z¢ )av
cs H
s¢
cH
log c + c log 0(av)
1+ e0
s 0(av)
1+ e0
s c¢
¢
0.06 ´ 3
70
0.3 ´ 3
37.5 + 50.5
log
+
log
1+ 0.55
37.5 1+ 0.55
70
= 0.031 + 0.058 = 0.089 m
=
So, total consolidation settlement of the 4-m clay layer is:
Sc = 0.095+0.089=0.184 m
18. Consolidation settlement problem. A rectangular shallow foundation and the
subsoil condition are shown in Figure 3.29. The foundation rests on a saturated clayey
soil. Above the groundwater table (GWT):  = 18 kN/m3; below the GWT: sat = 19
kN/m3.
(1) Divide the soil layer into sub-layers based on Figure 3.24; then calculate the
average vertical stress increase in each of the sub-layers. Use the 2:1 method.
(2) Determine the primary consolidation settlement in each of the sub-layers and then
the entire saturated clay layer.
Q = 900 kN
Ground surface
Groundwater table
Df = 1m
L x B = 3m x 2m
z
H=9m
e0 =0.5, cc = 0.35, cs = 0.06, σ’c = 75 kN/m2
e0 =0.48, cc = 0.33, cs = 0.05, σ’c = 100 kN/m2
H1 = 1 m
H2 = 2 m
H3 = 6 m
e0 =0.45, cc = 0.30, cs = 0.05, σ’c = 180
kN/m2
Firm layer
Figure 3.29 Diagram for problem 18.
Solution:
(1) The 9-m saturated clay soil stratum is divided into three layers to obtain better
prediction of the consolidation settlement. The three layers with their respective
characteristics are shown in the figure.
The stress at the bottom of the foundation due to external load is:
p0 =
Q
900
=
= 150 kN/m 2
2
B
3´ 2
To simplify the calculation and to be conservative, the 2:1 method is used to calculate the
vertical stress increases in each layer.
Ds z =
p0 ´ B ´ L
( B + z) ´ ( L + z)
The average vertical increase in each layer is determined using the following equation:
Ds z(av) =
1
( Ds z(top) + 4Ds z(mid ) + Ds z(bot ) )
6
The average effective stress of each layer is the effective stress at the middle of the soil
layer. The results are summarized in the following tables.
Note: the following two tables are very important. Calculation errors in the stresses will
affect the final results.
z (m)
0
0.5
1
2
3
6
9
Layer
1
2
3
Vertical stress
increase
(kN/m2)
150
102.9
75
45
30
12.5
6.8
Average vertical stress
increase, z(av)
(kN/m2)
106.1
47.5
14.5
Average in-situ effective
stress, 0(av) (kN/m2)
22.6
36.4
73.1
Preconsolidation
pressure, c
(kN/m2)
75
100
180
In layer 1:
c > 0(av), the soil is overconsolidated.
And c < 0(av) + z(av) = 128.7 kN/m2.
Sc =
s ¢ + Ds (¢z ) av
cs H
s¢
cH
log c + c log 0( av )
1+ e0
s 0(¢ av ) 1+ e0
s c¢
0.06 ´ 1
75
0.35 ´ 1
22.6 + 106.1
log
+
log
1+ 0.5
22.6 1+ 0.5
75
= 0.075 m = 7.5 cm
=
In layer 2:
c > 0(av), the soil is overconsolidated.
And c > 0(av) + z(av) = 83.9 kN/m2.
Sc =
s ¢ + Ds (¢z ) av
cs H
log 0( av )
1+ e0
s 0(¢ av )
0.05 ´ 2
36.4 + 47.5
log
1+ 0.48
36.4
= 0.025 m = 2.5 cm
=
In layer 3:
c > 0(av), the soil is overconsolidated.
And c > 0(av) + z(av) = 87.6 kN/m2.
Sc =
s ¢ + Ds (¢z ) av
cs H
log 0( av )
1+ e0
s 0(¢ av )
0.05 ´ 6
73.1+ 14.5
log
1+ 0.45
73.1
= 0.016 m = 1.6 cm
=
So, total consolidation settlement of the 9-m clay layer is:
Sc = 7.5+2.5+1.6=11.6 cm
19. Comprehensive shallow foundation design. A square spread footing is to be
designed to support a column. The load on the column, including the column weight,
is 800 kN. The foundation rests in homogeneous sandy clay. Geotechnical
investigation found the soil’s strength parameters are c = 38 kN/m2 and  = 25.
The groundwater table is at the ground surface. The saturated unit weight is 19 kN/m3.
The soil’s Poisson’s ratio is 0.5. The soil’s elastic modulus increases linearly with
depth: Es = 4,000 (kN/m2) + 230 (kN/m2/m)  z, and z starts from the bottom of the
foundation. The thickness of the foundation slab is chosen to be 0.5m, the elastic
modulus of the foundation is 1.2107 kN/m2. The initial void ratio, compression index,
swell index, and preconsolidation pressure are assumed to be consistent throughout the
depth. Their values are: e0 =0.45, cc = 0.33, cs = 0.05, c= 100 kN/m2. Geotechnical
investigation also found the homogeneous sandy clay extends to significant depth.
Design the shallow foundation by performing the following tasks:
(1) Determine the foundation embedment depth. Use the general bearing capacity
theory to determine foundation dimensions to satisfy a minimum factor of safety of
3.0.
(2) Calculate the allowable bearing capacity based on the actual dimensions of the
foundation that are generally accepted in field construction.
(3) Determine the depth at which the vertical stress increase is 10% of the in-situ
effective stress. Any method can be used to calculate the vertical stress increase.
(4) Determine the elastic settlement of the soil layer until the depth where vertical
stress increase is 10% of the in-situ effective stress. Use the Mayne and Poulos
method.
(5) Determine primary consolidation settlement of the saturated clay layer until the
depth where vertical stress increase is 10% of the in-situ effective stress. When
calculating the settlement, first divide the soil layer into sub-layers based on Figure
3.24; then calculate the average vertical stress increase in each of the sub-layers;
then primary consolidation settlement of each layer can be calculated.
If any parameter is needed but not provided in the problem statement, make
appropriate assumptions and explicitly state them.
Solution:
(1)
Step 1: The loads from the superstructure: Q = 800 kN.
The subsoil condition: c = 38 kN/m2,  = 25, sat = 19 kN/m3.
Step 2: Based on Table 3.4, the minimum Df = 0.6 m, choose Df = 1.0 m, considering the
foundation slab thickness is 0.5 m.
Step 3: Groundwater table (GWT) is at the ground surface and affects the ultimate
bearing capacity.
Step 4: The minimum factor of safety is 3.0.
Step 5: The general ultimate bearing capacity equation for square footings is:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g ¢ BNg Fg s Fg d Fg i
2
Using Table 3.3 and given  = 25, the bearing capacity factors are:
N c =20.721, N q = 10.662, and Ng = 10.876.
The foundation is square, i.e., B = L. The shape factors are:
æ Bö æ N ö
æ 10.662 ö
Fcs = 1+ ç ÷ ç q ÷ = 1+1´ ç
= 1.98
è L ø è Nc ø
è 20.721 ÷ø
æ Bö
Fqs = 1+ ç ÷ tan f ¢ = 1+ tan 25 = 1.47
è Lø
æ Bö
Fg s = 1- 0.4 ç ÷ = 0.6
è Lø
Assume
Df
B
£ 1 , and since  = 25 > 0, the depth factors are:
æD ö
0.4
Fcd = 1+ 0.4 ç f ÷ = 1+
B
è Bø
2 æ Df ö
2 1
0.31
Fqd = 1+ 2 tan f ¢ (1- sin f ¢ ) ç
= 1+ 2 ´ tan 25(1- sin 25)
= 1+
÷
B
B
è B ø
Fg d =1
Since the load is vertical, the load inclination factors Fci, Fqi, Fi are all 1.0.
The effective stress at the bottom of the foundation is:
q =  Df = (sat  w) Df =(199.81) 1.0 = 9.19 kN/m2.
So:
1
qult = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g ¢ BNg Fg s Fg d Fg i
2
æ 0.4 ö
æ 0.31 ö
= 38 ´ 20.721 ´ 1.98 ´ ç 1+
´ 1+ 9.19 ´ 10.663 ´ 1.47 ´ ç 1+
÷
÷ ´1
è
è
B ø
B ø
+0.5 ´ 9.19 ´ B ´ 10.876 ´ 0.6 ´ 1´ 1
574.7
= 1703 + 30B +
B
Step 6: The foundation slab’s weight is:
W f = 23.5 ´ B2 ´ 0.5 = 11.75B2 (kN) (unit weight of concrete is 23.56 kN/m3)
The stress at the bottom of the foundation is:
Q + W f 800 +11.75B2 800
p=
=
= 2 +11.75
A
B2
B
Step 7: The allowable bearing capacity should be at least p.
qult ö æ
800
æ
ö
çè qall =
÷ø = çè p = 2 +11.75 ÷ø
FS
B
1æ
574.7 ö 800
+11.75
çè 1703 + 30B +
÷=
B ø B2
That is: 3
Solve for B and find B = 1.16 m. Use B = 1.2 m.
(2)
The ultimate bearing capacity is:
qult = 1703 + 30 ´1.2 +
574.7
= 2218 kN/m2.
1.2
The allowable bearing capacity is:
q
2217
= 739 kN/m2
qall = ult =
FS
3
(3)
Pressure at the bottom of the footing:
Q + W f 800
p=
= 2 +11.75 = 567.3 kN/m2
A
1.5
Use 2:1 method for vertical stress increase ( Ds v¢ ):
pB 2
(z starts from the bottom of the footing)
( B + z )2
Effective stress:
s 0¢ = g ¢ z + D f
Ds v¢ =
(
)
Let : Ds v¢ = 0.1s 0¢
567.3 ´1.2 2
= 0.1´ (19 - 9.81) ´ (z +1.0)
(1.2 + z )
Find z = 8.55 m
That is:
(4)
(1- m )
2
Elastic settlement: Se = q0 Be I G I F I E
q0 = p = 567.3 kN/m
E0
2
For a square footing with B  B = 1.2 m  1.2 m: Be =
4B 2
p
= 1.35 m
The following parameters are given:
t = 0.5 m
 = 0.5
E0 = 4,000 kN/m2
kE = 230 kN/m2/m
Ef = 1.2107 kN/m2
The displacement influence factor, IG, is determined using Figure 3.20.
E0
4,000
=
= 99
k E Be 320 ´1.35
Given: H = 8.55 m
H 8.55
Find:
=
= 6.33
Be 1.35
From Figure 3.20, find IG = 0.93
IG depends on b =
The rigidity correction factor, IF, is:
p
1
IF = +
4 4.6 +10K F
H
8.55
k E = 4,000 +
´ 320 = 4138 kN/m 2
2
2
3
æ E f ö æ 2t ö æ 1.2 ´107 ö æ 2 ´ 0.5 ö 3
KF = ç
÷ç ÷ =ç
÷ = 1179
֍
è Es( av ) ø è Be ø è 4138 ø è 1.35 ø
Es( av ) = E0 +
IF =
p
4
+
1
p
1
= +
= 0.785
4.6 +10K F 4 4.6 +10 ´1179
The settlement correction factor, IE, is:
1
1
= 1= 0.921
æ
æ
1.35 ö
Be ö
1.22´0.4-0.4
1.22 m -0.4
3.5e
´ ç 1.6 +
3.5e
´ ç 1.6 +
1.0 ÷ø
è
D f ÷ø
è
So the elastic settlement is:
I E = 1-
(1- m ) = 567.3 ´1.35 ´ 0.93 ´ 0.785 ´ 0.921´ æ 1- 0.5 ö
2
Se = q0 Be IG I F I E
E0
= 0.10 m
(5) To simplify the calculation, consider the soil layer with H = 9.0 m
The following schematic is drawn:
2
çè 4,000 ÷ø
Q = 800 kN
Ground surface
Groundwater table
Df = 1m
z
H=9m
Layer 1
BxB= 1.2mx1.2m
H1 = B/2=0.6 m
Layer 2
H2 = B = 1.2 m
Layer 3
H3 = 2B=2.4 m
Layer 4
H3 = 4B=4.8 m
pB 2
(z starts from the bottom of the footing), to
( B + z )2
calculate vertical stress increases at the top, middle, and bottom of each layer. The
following table is prepared.
Use 2:1 method: Ds v¢ =
z (m)
0
0.3
0.6
1.2
1.8
3.0
4.2
6.6
9.0
Vertical stress
increase
(kN/m2)
567.30
370.56
257.33
144.75
92.64
47.26
28.59
13.70
8.01
The average effective stress and average vertical stress increase in each layer are
1
calculated as follows. Ds z(av) = ( Ds z(top) + 4Ds z(mid ) + Ds z(bot ) )
6
Layer
1
Average vertical stress
increase, z(av)
(kN/m2)
386.43
Average in-situ effective stress, 0(av)
(kN/m2)
9.19(1.0+0.3)=11.95
2
3
3
154.83
51.71
15.24
 =sat  w =199.81 = 9.19 kN/m3
9.19(1.0+0.6+0.6)=20.22
9.19(1.0+0.6+1.2+1.2)=36.76
9.19(1.0+0.6+1.2+2.4+2.4)=69.84
In all layers: c = 100 kN/m2.
In layer 1:
c > 0(av), the soil is overconsolidated.
And c < 0(av) + z(av) = 11.95+386.43=398.38 kN/m2.
Sc =
s ¢ + Ds (z¢ )av
cs H
s¢
cH
log c + c log 0(av)
1+ e0
s 0(av)
1+ e0
s c¢
¢
0.05 ´ 0.6
100 0.35 ´ 0.6
398.35
log
+
log
1+ 0.45
11.95 1+ 0.45
100
= 0.019 + 0.082 m
=
= 0.101 m
In layer 2:
c > 0(av), the soil is overconsolidated.
And c < 0(av) + z(av) = 20.22+154.83=174.05 kN/m2.
Sc =
s ¢ + Ds (z¢ )av
cs H
s¢
cH
log c + c log 0(av)
1+ e0
s 0(av)
1+ e0
s c¢
¢
0.05 ´ 1.2
100 0.33 ´ 1.2
174.05
log
+
log
1+ 0.45
20.22 1+ 0.45
100
= 0.029 + 0.066 m
=
= 0.095 m
In layer 3:
c > 0(av), the soil is overconsolidated.
And c > 0(av) + z(av) = 36.76+51.71=88.47 kN/m2.
Sc =
s ¢ + Ds (z¢ )av
cs H
log 0(av)
1+ e0
s 0(av)
¢
0.05 ´ 2.4
88.47
log
1+ 0.45
36.76
= 0.031 m
=
In layer 4:
c > 0(av), the soil is overconsolidated.
And c > 0(av) + z(av) = 69.84+15.24=85.08 kN/m2.
Sc =
s ¢ + Ds (z¢ )av
cs H
log 0(av)
1+ e0
s 0(av)
¢
0.05 ´ 4.8
85.08
log
1+ 0.45
69.84
= 0.014 m
=
So, total consolidation settlement of the 9-m clay layer is:
Sc = 0.101+0.095+0.031+0.014=0.24 m
The consolidation settlement is very large, due to a high load. The load should be reduced
or pre-consolidation should be induced in the soil before foundation construction.
20. Comprehensive mat foundation design. A mat foundation is to be designed. The
dimensions of the mat are 30 m by 30 m. The stress caused by the superstructure and
the mat at the bottom of the foundation is 480 kN/m2. The embedment depth of the
mat foundation is determined to be 10 m below the ground surface, and the thickness
of the mat is 3.0 m. The groundwater table (GWT) is at the bottom of the foundation.
The subsoil is homogeneous clayey soil with the following parameters.
Soil strength parameters: c = 75 kN/m2,  = 25.
Above GWT:  = 17.5 kN/m3; below GWT: sat = 18.5 kN/m3.
Poisson’s ratio  = 0.4.
The soil’s elastic modulus increases linearly with depth: Es (kN/m2) = 15,000
(kN/m2) + 750 (kN/m2/m)  z and z starts from the bottom of the foundation.
The elastic modulus of the foundation is Ef = 1.25107 kN/m2.
Compressibility parameters: e0 = 0.55, cc = 0.3, cs = 0.06, c= 100 kN/m2.
Geotechnical investigation also found the homogeneous sandy clay extends to
significant depth. Design the mat foundation by performing the following tasks:
(1) Calculate the factor of safety for bearing capacity.
(2) Determine the depth at which the vertical stress increase is 10% of the in-situ
effective stress. Any method can be used to calculate the vertical stress increase.
(3) Determine the elastic settlement of the soil layer until the depth where the vertical
stress increase is 10% of the in-situ effective stress. Use the Mayne and Poulos
method.
(4) Determine primary consolidation settlement of the saturated clay layer until the
depth where the vertical stress increase is 10% of the in-situ effective stress. To do so,
first divide the soil layer into sub-layers based on Figure 3.24; then calculate the
average vertical stress increase in each of the sub-layers; then primary consolidation
settlement of each layer can then be calculated.
If any parameter is needed but not provided in the problem statement, make appropriate
assumptions and explicitly state them.
Solution: Not provided.