4. Shallow Foundations

4.BEARING CAPACITY OF
SHALLOW FOUNDATIONS
Dr MWAJUMA I. LINGWANDA
The most carefully designed structures can fail if they
are not supported by suitable foundations

Definition of Shallow foundations
▪ Shallow foundations are also called spread foundations
▪ But how shallow is shallow?
▪ The common definition of shallow foundations refers to the case where the
founding depth is less than the breadth
▪ i.e. when 𝑫 < 𝑩 the foundation is shallow
▪ What about wide raft foundations?
▪ The above definition is not acceptable for wide rafts
▪ Therefore, a foundation is shallow if the depth is less than 3 m or less than the
breadth of the foundation
CEB 3319 FOUNDATION ENGINERING 2

Types of shallow foundations
• Pad foundations
These are usually provided to support structural columns. They may be circular,
square or rectangular slab section
• Strip foundations
A strip foundation has a length >> than breadth. They are normally provided for
load bearing walls or closely spaced columns where it is not economical to deal
with individual pads CEB 3319 FOUNDATION ENGINERING 3

Types of shallow foundations cont..
• Raft/Mat foundations
Raft foundations are required on soils
of low bearing capacity or where
structural columns are closely spaced
in both directions such that individual
pads would nearly touch each other.

Foundation design criteria
Three main design criteria must be considered for a shallow foundation
1. Adequate depth
✓ The founding depth i.e. the depth from the surface to its underside should
be sufficient to prevent effects due to changes of ground surface
✓ Sufficient depth to prevent overturning moments when the structure is
subjected to horizontal forces
2. Limiting settlement
✓ The amount of tolerable total and differential settlements depends on
functional and user requirements and also economic factors
✓ Most settlement damage may be classified as architectural damage as they
normally affect finishes.

Foundation design criteria cont..
3. Factor of safety against shear failure
✓ In foundations, shear failure occurs when soil divides into separate zones which
move partially or fully along slip surfaces
✓ At this point, a plastic yielding condition has developed whereby the shear stress
along the slip surface has reached a limiting value(ultimate limit state)
✓ The principal criteria for design will be satisfying a given factor of safety against
shear failure CEB 3319 FOUNDATION ENGINERING 6

Three principal modes of shear failure
1. General shear failure
➢ This mode of failure occurs in dense or overconsolidated soils of low compressibility
➢ A clearly defined plastic yield slip forms under the footing and develop outwards to the
surface
➢ Heaving of the ground occurs on both sides of the footing
➢ Failure is sudden and often accompanied by severe tilting on one side

Three principal modes of shear failure cont..
2. Local shear failure
➢ Occurs in compressible soils
➢ A significant vertical movement may take place before any noticeable heaving
➢ Shear planes develop but fail to extend to the ground surface
➢ Very little tilting may take place
➢ The settlement which occurs usually is the principal design criteria
CEB 3319 FOUNDATION ENGINERING
8

Three principal modes of shear failure cont..
3. Punching shear failure
➢ Occurs in weak compressible soils
➢ Considerable vertical movement take place
➢ Slip surfaces restricted to vertical planes adjacent to the sides of the footing
➢ Heaving at the surface is usually absent and may be replaced by a drag down
➢ Normally there will be no tilting
➢ This kind of failure may be observed in highly compressible clays and in loose sands

Important Definitions
• Total foundation pressure (𝒒) or gross loading intensity
Is the intensity of total pressure on the ground beneath the foundation after the
structure has been erected and fully loaded
• The net bearing pressure (𝒒 𝒏) or the net foundation pressure
Is the net change in total stress experienced by the soil at the base of the foundation
Let 𝝈 𝟎 = the stress removed due to excavation (𝝈 𝟎 = 𝜸𝑫)
𝒒 = Total foundation pressure/ contact pressure at the base of the foundation
Then, 𝒒 𝒏 = 𝒒 − 𝝈 𝟎

Important Definitions cont..
• The net effective bearing pressure (𝒒 𝒏
´ ) or the net effective foundation pressure
The net foundation pressure expressed in terms of effective stress. When the
water table is above the base of the foundation the net effective bearing
pressure is given by;
𝒒 𝒏
´ = 𝒒 − 𝝈 𝟎
´
Where 𝝈 𝟎
´
= 𝜸𝑫 − 𝜸 𝒘 𝒉 𝒘

Important definitions cont..
• The ultimate bearing capacity (𝒒 𝒇)
Is the value of the loading intensity at which
the supporting ground fails in shear
• The net ultimate bearing capacity (𝒒 𝒏𝒇)
• Is the value of net loading intensity at which
the supporting ground fails in shear
• LOAD – SETTLEMENT CURVE
Settlement
Load
0

• The factor of safety against shear failure is defined as the ratio between the net
ultimate bearing capacity and the net bearing pressure
𝑭 =
𝒒 𝒏𝒇
´
𝒒 𝒏
´
=
𝒒 𝒇 − 𝝈 𝒐
´
𝒒 − 𝝈 𝒐
´
For building foundations, the recommended factor of safety is between 3 and 2.5
• Presumed bearing pressure
Is the net loading intensity considered appropriate to the particular type of
ground for preliminary design purposes. The value is based on local experience or
by calculation from strength tests using a factor of safety against shear failure

• The allowable bearing pressure (𝒒 𝒂)
Is the maximum allowable net loading intensity of a ground in a given case,
taking into consideration the bearing capacity and the estimated amount
and rate of settlement that will occur.
Considering shear failure, allowable design bearing capacity is given by
𝒒 𝒂 =
´
𝑭
+ 𝝈 𝒐
´

Simple bearing capacity formula
• Assumptions
i. The footing experiences a bearing
capacity failure
ii. Failure occurs along a circular surface
iii. Undrained conditions prevail, 𝒄 𝒖 > 𝟎, 𝝓 𝒖 = 𝟎
iv. The shear strength of the soil between the ground
and depth D is neglected
v. The soil between the ground and D is considered
as a surcharge and produces a vertical stress = 𝜸𝑫
• Taking moments about point O,
• 𝒒 𝒇 𝑩.
𝑩
𝟐
= 𝒄 𝒖 𝝅𝑩. 𝑩 + 𝜸𝑫𝑩.
𝑩
𝟐
• 𝒒 𝒇 = 𝟐𝝅𝒄 𝒖 + 𝜸𝑫
• 𝒒 𝒇 ≈ 𝟔. 𝟐𝟖𝒄 𝒖 + 𝜸𝑫
Consider a long footing with cross section shown below

Class exercise
With similar assumptions regarding ground conditions as the previous
foundation, derive a simple bearing capacity formula for the following cases
I. The foundation rests on the ground level
II. The foundation rests at a depth D and a long and wide surcharge load due
to soil compacted on each side at the ground level to a height H. The unit
weight of the compacted soil being 𝜸 𝟐
(i) 𝒒 𝒇 = 𝟐𝝅𝒄 𝒖 (ii) 𝒒 𝒇 = 𝟐𝝅𝒄 𝒖 + 𝜸𝑫 + 𝜸 𝟐 𝑯

Terzaghi´s bearing capacity formula
• The formula was developed for long foundations (strip foundations) then
extended to square and circular foundations
• A general shear failure is considered

Terzaghi´s bearing capacity formula cont..
• Assumptions
✓ The depth of the foundation is less than or equal to its width (𝑫 ≤ 𝑩)
✓ No sliding occurs between the foundation and the soil
✓ The soil is homogeneous semi infinite mass
✓ The shear strength of the soil is presented by both c and ϕ
✓ Failure is by general shear mode
✓ The soil between the ground surface and a depth D has no shear strength
i.e. it serves only as a surcharge load
18

• Terzaghi proposed a three-term expression for ultimate bearing capacity of a
long strip footing of breadth B as
𝒒 𝒇 = 𝒄𝑵 𝒄 + 𝝈 𝒐 𝑵 𝒒 +
𝟏
𝟐
𝑩𝜸𝑵 𝜸
• 𝒒 𝒇 = ultimate bearing capacity
• 𝒄 = cohesion of soil beneath the foundation
• 𝝈 𝟎 = overburden pressure at the foundation level
• 𝑩 = breadth of the foundation
• 𝜸 = unit weight of the soil beneath the foundation
• 𝑵 𝒄, 𝑵 𝒒 and 𝑵 𝜸 areTerzaghi´s bearing capacity factors which are function of ϕ only
19

• 𝑵 𝒒 = 𝒆 𝝅𝒕𝒂𝒏𝝓
. 𝒕𝒂𝒏 𝟐
𝟒𝟓 +
𝝓
𝟐
• 𝑵 𝒄 =
𝑵 𝒒−𝟏
𝒕𝒂𝒏𝝓
for ϕ > 0
• 𝑵 𝒄 = 𝟓. 𝟕 for ϕ =0
• 𝑵 𝜸 ≈ 𝟏. 𝟖 𝑵 𝒒 − 𝟏 𝒕𝒂𝒏𝝓
• Under undrained conditions, ϕ =0 and c = cu
• 𝑵 𝒒 = 𝟏
• 𝑵 𝒄 = 𝟓. 𝟕
• 𝑵 𝜸 = 𝟎
• 𝒒 𝒇 = 𝟓. 𝟕𝒄 𝒖 + 𝝈 𝟎
Note that
Because of the shape of the failure
surface, the values of 𝑐 and 𝜙 only
need to represent the soil between the
bottom of the footing and a depth B
below the bottom. The soils between
the ground surface and a depth D are
treated simply as overburden.

Bearing capacity factors in chart
21

Example 1
A strip footing is 2.8 m wide and is founded at a depth of 2.4 m in a soil having
the following properties: 𝒄´
= 𝟏𝟐𝒌𝑵/𝒎 𝟐
𝝓´
= 𝟐𝟎 𝒐
𝜸 = 𝟏𝟗𝒌𝑵/𝒎 𝟑
. Determine
the net ultimate bearing capacity.

Solution to example 1
Sketch the problem and indicate the known parameters (soil properties and
dimensions)
𝒒´ 𝒏 = 𝒒´ 𝒇 − 𝝈´ 𝒐
𝒒´ 𝒏 = 𝒄´𝑵 𝒄 + 𝜸´𝑫𝑵 𝒒 +
𝟏
𝟐
𝑩𝜸´𝑵 𝜸 − 𝜸´𝑫
determine the bearing capacity coefficients
𝑵 𝒒 = 𝒆 𝝅𝒕𝒂𝒏𝝓
. 𝒕𝒂𝒏 𝟐
𝟒𝟓 +
𝝓
𝟐
= 𝒆 𝝅𝒕𝒂𝒏𝟐𝟎
× 𝒕𝒂𝒏 𝟐
𝟒𝟓 +
𝟐𝟎
𝟐
= 𝟔. 𝟑𝟗𝟗

Solution to example 1 cont..
𝑵 𝒄 =
𝑵 𝒒 − 𝟏
𝒕𝒂𝒏𝝓
=
𝟔. 𝟑𝟗𝟗 − 𝟏
𝒕𝒂𝒏𝟐𝟎
= 𝟏𝟒. 𝟖𝟑𝟓
𝑵 𝜸 = 𝟏. 𝟖 𝑵 𝒒 − 𝟏 𝒕𝒂𝒏𝝓 = 𝟏. 𝟖 𝟔. 𝟑𝟗𝟗 − 𝟏 𝒕𝒂𝒏𝟐𝟎 = 𝟑. 𝟓𝟑𝟕
Substituting to the net bearing capacity equation
𝒒´ 𝒏 = (𝟏𝟐 × 𝟏𝟒. 𝟖𝟑𝟓) + (𝟏𝟗 × 𝟐. 𝟒 × 𝟔. 𝟑𝟗𝟗) + (
𝟏
𝟐
× 𝟐. 𝟖 × 𝟏𝟗 × 𝟑. 𝟓𝟑𝟕) − 𝟏𝟗 × 𝟐. 𝟒
𝒒´ 𝒏 = 𝟓𝟗𝟎. 𝟖𝟓 − 𝟒𝟓. 𝟔
𝒒´ 𝒏 = 𝟓𝟒𝟓. 𝟐𝟓 ൗ𝒌𝑵
𝒎 𝟐

Example 2
Determine the net ultimate bearing capacity of a strip footing of breadth
2.5 m which is to be founded 1.25 m below the surface of a thick layer of
clay having 𝒄 𝒖 = 𝟖𝟖𝒌𝑵/𝒎 𝟐

Solution to example 𝒒 𝒏 = 𝒒 𝒇 − 𝝈 𝒐
𝒒 𝒏 = 𝒄 𝒖 𝑵 𝒄 + 𝜸𝑫𝑵 𝒒 +
𝟏
𝟐
𝑩𝜸𝑵 𝜸 − 𝜸𝑫
Bearing capacity coefficients for 𝜑 = 0;
𝑵 𝒒 = 𝟏 𝑵 𝒄 = 𝟓. 𝟕 𝑵 𝜸 = 𝟎
Substituting the coefficients we have
𝒒 𝒏 = 𝟓. 𝟕𝒄 + 𝜸𝑫 − 𝜸𝑫
𝒒 𝒏 = 𝟓. 𝟕𝒄
𝒒 𝒏 = 𝟓. 𝟕 × 𝟖𝟖
𝒒 𝒏 = 𝟓𝟎𝟏. 𝟔 ൗ𝒌𝑵
𝒎 𝟐

Example 3
A strip foundation presented in the figure below is required to transmit an
inclusive load of 650kN/m to the supporting soil. Adopting a factor of safety
of 3, determine the breadth of the footing required.

The supported inclusive load is an allowable load with a factor of safety F = 3
𝒒 𝒂 =
´
𝑭
+ 𝝈 𝒐
´
𝒒 𝒇 = 𝒄𝑵 𝒄 + 𝜸𝑫𝑵 𝒒 +
𝟏
𝟐
𝑩𝜸𝑵 𝜸
Since the soil has no cohesion, the ultimate bearing capacity equation reduces to
𝒒 𝒇 = 𝜸𝑫𝑵 𝒒 +
𝟏
𝟐
𝑩𝜸𝑵 𝜸
Bearing capacity coefficients
𝑵 𝒒 = 𝒆 𝝅𝒕𝒂𝒏𝝓. 𝒕𝒂𝒏 𝟐 𝟒𝟓 +
𝝓
𝟐
= 𝒆 𝝅𝒕𝒂𝒏𝟑𝟎. 𝒕𝒂𝒏 𝟐 𝟒𝟓 +
𝟑𝟎
𝟐
= 𝟏𝟖. 𝟒𝟎𝟐
𝑵 𝜸 = 𝟏. 𝟖 𝑵 𝒒 − 𝟏 𝒕𝒂𝒏𝝓 = 𝟏. 𝟖 𝟏𝟖. 𝟒𝟎𝟐 − 𝟏 𝒕𝒂𝒏𝟑𝟎 = 𝟏𝟖. 𝟎𝟖𝟓
28

𝒒 𝒇 = (𝟏𝟗 × 𝟏. 𝟓 × 𝟏𝟖. 𝟒𝟎𝟐) + (
𝟏
𝟐
× 𝑩 × 𝟐𝟎 × 𝟏𝟖. 𝟎𝟖𝟓)
𝒒 𝒇 = 𝟓𝟐𝟒. 𝟒𝟔 + 𝟏𝟖𝟎. 𝟖𝟓𝑩
Substituting 𝑞 𝑓 into 𝑞 𝑎 equation..The value of 𝑞 𝑎 given is converted to a stress in ൗ𝑘𝑁
𝑚2
𝟔𝟓𝟎
𝑩
=
𝟓𝟐𝟒. 𝟒𝟔 + 𝟏𝟖𝟎. 𝟖𝟓𝑩 − 𝟏𝟗 × 𝟏. 𝟓
𝟑
+ 𝟏𝟗 × 𝟏. 𝟓
1950 = 180.85𝐵2 + 495.69𝐵 + 85.5𝐵
𝟏𝟖𝟎. 𝟖𝟓𝑩 𝟐
+ 𝟓𝟖𝟏. 𝟏𝟗𝑩 − 𝟏𝟗𝟓𝟎 = 𝟎
𝑩 = 𝟐. 𝟎𝟓 𝒎

Effect of water table
• POSITION I:
𝒒 𝒇 = 𝒄𝑵 𝒄 + 𝜸𝑫 𝟏 + 𝜸´𝑫 𝟐 𝑵 𝒒 +
𝟏
𝟐
𝑩𝜸´𝑵 𝜸
• POSITION II: (water table near the base of the footing)
𝟏
𝟐
𝑩𝜸´𝑵 𝜸
• POSITION III: (water table far from the base of the footing)
𝟏
𝟐
𝑩𝜸𝑵 𝜸
30

Example 4
A strip footing of breadth 1.75 m is required to transmit a net load of 650kN/m to a
compacted sand of the following properties: : 𝒄´ = 𝟎 𝝓´ = 𝟑𝟖 𝒐 𝜸 𝒔𝒂𝒕 = 𝟐𝟎. 𝟓𝒌𝑵/𝒎 𝟑.
Assume that the water table may rise to the surface and adopting a factor of safety
of 3, determine the required depth of the footing.

From the definition of factor of safety
𝑭 =
𝒒 𝒏𝒇
´
𝒒 𝒏
´
𝒒´ 𝒏𝒇 = 𝒄´𝑵 𝒄 + 𝜸´𝑫𝑵 𝒒 +
𝟏
𝟐
Bearing capacity coefficients
𝑵 𝒒 = 𝒆 𝝅𝒕𝒂𝒏𝟑𝟖
. 𝒕𝒂𝒏 𝟐
𝟒𝟓 +
𝟑𝟖
𝟐
= 𝟒𝟖. 𝟗
𝑵 𝜸 = 𝟏. 𝟖 𝟒𝟖. 𝟗 − 𝟏 𝒕𝒂𝒏𝟑𝟖 = 𝟔𝟕. 𝟒
Substituting the known variables and observing the effect
of water table

𝒒´ 𝒏𝒇 = 𝟐𝟎. 𝟓 − 𝟗. 𝟖𝟏 × 𝑫 × 𝟒𝟖. 𝟗 +
𝟏
𝟐
× 𝟏. 𝟕𝟓 × 𝟐𝟎. 𝟓 − 𝟗. 𝟖𝟏 × 𝟔𝟕. 𝟒 − (𝟐𝟎. 𝟓 − 𝟗. 𝟖𝟏)𝑫
𝒒´ 𝒏𝒇 = 𝟓𝟏𝟐. 𝟎𝟏𝟓𝑫 + 𝟔𝟑𝟎. 𝟒𝟒
Recall that
𝑭 =
𝒒 𝒏𝒇
´
𝒒 𝒏
´
Substituting we have
𝟑 =
𝟓𝟏𝟐. 𝟎𝟏𝟓𝑫 + 𝟔𝟑𝟎. 𝟒𝟒
ൗ𝟔𝟓𝟎
𝟏. 𝟕𝟓
𝑫 = 𝟎. 𝟗𝟒 𝒎 ~𝟏 𝒎 33

Example 5
Determine the design load on a strip footing indicated in the figure below. The soil
characteristic are 𝒄´
= 𝟏𝟎𝒌𝑵/𝒎 𝟐
𝝓´
= 𝟐𝟖 𝒐
𝜸 = 𝟏𝟖𝒌𝑵/𝒎 𝟑
𝜸 𝒔𝒂𝒕 = 𝟏𝟗. 𝟓 ൗ𝒌𝑵
𝒎 𝟑

𝒒´ 𝒏𝒇 = 𝒄´𝑵 𝒄 + 𝜸´𝑫𝑵 𝒒 +
𝟏
𝟐
Bearing capacity factors
𝑵 𝒒 = 𝒆 𝝅𝒕𝒂𝒏𝝓
. 𝒕𝒂𝒏 𝟐
𝟒𝟓 +
𝝓
𝟐
= 𝒆 𝝅𝒕𝒂𝒏𝟐𝟖
× 𝒕𝒂𝒏 𝟐
𝟒𝟓 +
𝟐𝟖
𝟐
= 𝟖. 𝟖𝟒
𝑵 𝒄 =
𝑵 𝒒 − 𝟏
𝒕𝒂𝒏𝝓
=
𝟖. 𝟖𝟒 − 𝟏
𝒕𝒂𝒏𝟐𝟖
= 𝟏𝟒. 𝟕𝟒
𝑵 𝜸 = 𝟏. 𝟖 𝑵 𝒒 − 𝟏 𝒕𝒂𝒏𝝓 = 𝟏. 𝟖 𝟖. 𝟖𝟒 − 𝟏 𝒕𝒂𝒏𝟐𝟖 = 𝟕. 𝟓𝟎
𝒒´ 𝒏𝒇
= 𝟏𝟎 × 𝟏𝟒. 𝟕𝟒 + 𝟏𝟖 × 𝟏. 𝟓 + (𝟏𝟗. 𝟓 − 𝟗. 𝟖𝟏) × 𝟏 × 𝟖. 𝟖𝟒
+ 𝟎. 𝟓 × 𝟑 × 𝟏𝟗. 𝟓 − 𝟗. 𝟖𝟏 × 𝟕. 𝟓𝟎 − 𝟏. 𝟓 × 𝟏𝟖 + (𝟏𝟗. 𝟓 − 𝟗. 𝟖𝟏) × 𝟏
𝒒´ 𝒏𝒇 = 𝟓𝟒𝟒. 𝟎𝟔 ൗ𝒌𝑵
𝒎 𝟐 or 𝒒´ 𝒏𝒇 = 𝟏𝟔𝟑𝟐. 𝟏𝟖 Τ𝒌𝑵
𝒎
35

Generalized formula for bearing capacity
• When the assumptions stated by Terzaghi are not fulfilled, the following
expression should be used for ultimate bearing capacity
𝒒 𝒇 = 𝒄𝑵 𝒄 𝒔 𝒄 𝒅 𝒄 𝒍 𝒄 + 𝝈 𝟎 𝑵 𝒒 𝒔 𝒒 𝒅 𝒒 𝒍 𝒒 +
𝟏
𝟐
𝑩𝜸𝑵 𝜸 𝒔 𝜸 𝒅 𝜸 𝒍 𝜸
𝒔 𝒄, 𝒔 𝒒, 𝒔 𝜸 are shape factors, applied for non continuous footing
𝒅 𝒄, 𝒅 𝒒, 𝒅 𝜸 are depth factors, applied when D > B
𝒍 𝒄, 𝒍 𝒒, 𝒍 𝜸 are load inclination factors, applied when the load is inclined

Generalized formula for bearing capacity cont..
Some proposed factors
Shape factors:
𝒔 𝒄 = 𝒔 𝒒 = 𝟏 + 𝟎. 𝟐
𝑩
𝑳
𝒔 𝜸 = 𝟏 − 𝟎. 𝟒
𝑩
𝑳
for rectangular footings
𝒔 𝒄 = 𝟏. 𝟑 𝒔 𝒒 = 𝟏. 𝟐 𝒔 𝜸 = 0.8 for square footings
Depth factors:
𝒅 𝒄= 𝟏 + 𝟎. 𝟐
𝑫
𝑩
𝒅 𝒒 = 𝟏 + 𝟎. 𝟏
𝑫
𝑩
𝒅 𝜸= 𝟏

Example 6
A rectangular footing of breadth 6 m and length 15 m is founded at a depth
of 4.5 m in a fine soil having the following properties:
𝒄 𝒖 = 𝟒𝟎 𝒌𝑵/𝒎 𝟐 𝜸 = 𝟏𝟖𝒌𝑵/𝒎 𝟑
Determine the ultimate bearing capacity of the footing

• Determine 𝒒 𝒇
𝒒 𝒇 = 𝒄𝑵 𝒄 𝒔 𝒄 + 𝝈 𝟎 𝑵 𝒒 𝒔 𝒒 +
𝟏
𝟐
𝑩𝜸𝑵 𝜸 𝒔 𝜸
Bearing capacity coefficients for 𝜑 = 0;
𝑵 𝒒 = 𝟏 𝑵 𝒄 = 𝟓. 𝟕 𝑵 𝜸 = 𝟎
𝒒 𝒇 = 𝟓. 𝟕𝒄 𝒖 𝒔 𝒄 + 𝜸𝑫𝒔 𝒒
𝒔 𝒄 = 𝒔 𝒒 = 𝟏 + 𝟎. 𝟐
𝑩
𝑳
= 𝟏 + 𝟎. 𝟐
𝟔
𝟏𝟓
= 𝟏. 𝟎𝟖
𝒒 𝒇 = 𝟓. 𝟕 × 𝟒𝟎 × 𝟏. 𝟎𝟖 + 𝟏𝟖 × 𝟒. 𝟓 × 𝟏. 𝟎𝟖
𝒒 𝒇 = 𝟑𝟑𝟑. 𝟕𝟐 ൗ𝒌𝑵
𝒎 𝟐

Example 7
A footing 2.25 by 2.25m is located at a depth of 1.5 m in a sand, the shear
strength parameters to be used in design being 𝒄´ = 𝟎, and 𝝓´= 𝟑𝟖 𝒐.
Determine the ultimate bearing capacity
(a) if the water table is well below foundation level and
(b) if the water table is at the surface. (qf = 1515.04 kN/m2)
The unit weight of the sand above the water table is 18 kN/m3; the saturated
unit weight is 20 kN/m3.

(a) Water table well below the
foundation base
𝒒 𝒇 = 𝝈 𝟎 𝑵 𝒒 𝒔 𝒒 +
𝟏
𝟐
𝑩𝜸𝑵 𝜸 𝒔 𝜸
𝑁𝑞 = 48.9 𝑁𝛾 = 67.4
𝒔 𝒒 = 𝟏. 𝟐 𝒔 𝜸 = 0.8
𝒒 𝒇 = 𝟏𝟖 × 𝟏. 𝟓 × 𝟒𝟖. 𝟗 +
𝟏
𝟐
× 𝟐. 𝟐𝟓 × 𝟏𝟖 × 𝟔𝟕. 𝟒
𝒒 𝒇 = 𝟐𝟔𝟕𝟔. 𝟐𝟒 ൗ𝒌𝑵
𝒎 𝟐

(b)Water table at the surface Bearing capacity as well as shape factors remains the
same as in part (a)
𝒒 𝒇 = 𝜸 𝒔𝒂𝒕 − 𝜸 𝒘 𝑫𝑵 𝒒 𝒔 𝒒 +
𝟏
𝟐
𝑩 𝜸 𝒔𝒂𝒕 − 𝜸 𝒘 𝑵 𝜸 𝒔 𝜸
𝑞 𝑓
= 20 − 9.81 × 1.5 × 48.9 +
1
2
× 2.25
× 20 − 9.81 × 67.4 × 0.8
𝑞 𝑓 = 896.92 + 618.12
𝒒 𝒇 = 𝟏𝟓𝟏𝟓. 𝟎𝟒 ൗ𝒌𝑵
𝒎 𝟐

allowable bearing capacity
For a given foundation in a given soil, the allowable design value for the applied
bearing pressure must satisfy the two criteria listed below:
I. An ultimate limit state value (shear strength)
Allowable design bearing capacity:
𝒒 𝒂 =
´
𝑭
+ 𝝈 𝒐
´
II. A serviceability limit state (settlement)
Allowable design bearing capacity 𝒒 𝒂 = bearing pressure corresponding to a
specific limit value 𝑺 𝑳 of undrained or drained settlement
43

allowable bearing capacity cont..
(a) Immediate or undrained
settlement
𝒒 𝒂 =
𝑺 𝑳 𝑬 𝒖
𝑩(𝟏 − 𝝂 𝟐)𝑰 𝒑
+ 𝝈 𝒐
(b) Consolidation or drained
settlement
𝒒 𝒂 =
𝑺 𝑳
𝒎 𝒗(𝑰+𝟏)𝑯 𝒐
+ 𝝈´ 𝒐
Where
• 𝑞 𝑎 = allowable design bearing capacity
• 𝑆 𝐿 = a limit value of undrained or drained stability
• 𝐸 𝑢 = undrained elastic modulus of the soil
• 𝐵 = least lateral dimension of the foundation
𝜈 = Poisson's ratio for the soil
• 𝐼 𝑝 = influence factor for vertical displacement
• 𝑚 𝑣 = coefficient of volume compressibility for the soil
• 𝐻 𝑜 = original thickness of the compressed layer
• 𝐼 = stress influence coefficient
44

allowable bearing capacity cont..
Allowable bearing capacity and settlements of sands can alternatively be
determined from results of in-situ tests such as standard penetration test and
cone penetration test

Design process
The following is the usual procedure followed in foundation design
1. Proposing the type, shape and size of foundation using presumed bearing
values. Suggested in BS 8004 for example, for medium dense sand presumed
bearing values are typically 100 – 300 kN/m2 and for stiff clays between 150 –
300 kN/m2
2. Evaluation of the loading conditions and calculation of expected contact
pressure
3. Determination of shear strength and settlement characteristics of the soil
using laboratory or field tests
4. Evaluation of bearing capacity and compare with the expected pressures
5. Adjusting the size and depth of the footing to arrive to a design which
represents both structural and economic efficiency
46

Exercises
1. A strip footing 2.4 m wide is founded at a depth of 2.8 m in a soil having the
following properties: 𝒄´ = 𝟏𝟐 𝐤𝐏𝐚, and 𝝓´= 𝟐𝟎 𝒐, 𝜸 = 𝟏𝟗𝒌𝑵/𝒎 𝟑
Determine (a)The net ultimate bearing capacity
(b) the design bearing capacity adopting a factor of safety of 3
2. A square footing is to be founded to a depth of 2.6 m and will transmit a
uniform load of 2.6 MN. Adopting a factor of safety of 3, determine the required
size of the footing. Soil properties are:
𝒄 𝒖 = 𝟔𝟒 𝐤𝐍/𝒎 𝟐
𝛄 = 𝟏𝟗 𝐤𝐍/𝒎 𝟑

Exercises
3. A rectangular foundation 10 x 5 m is to be designed, using a factor of safety of 3 to
carry a uniform load of 86.6 MN, in fully drained conditions in the following soil:
𝒄´
= 𝟎, and 𝝓´
= 𝟑𝟓 𝒐
, 𝜸 = 𝟐𝟎 𝒌𝑵/𝒎 𝟑
, 𝜸 𝒔𝒂𝒕 = 𝟐𝟐 𝒌𝑵/𝒎 𝟑
a) Determine a suitable founding depth
b) Determine the percentage reduction in load carrying capacity that results from
the ground water level rising
i. To the underside of the footing
ii. To the ground surface
c) Calculate the factor of safety when the 86.6 MN load is applied and the two
conditions in (b) occur

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