Technical University of Mombasa
Department of Building and Civil Engineering
ECV 4424
Foundation Engineering II
Lecture notes
By John Chege
2017
Page 1 of 70
ECV 4424 Foundation Engineering II
Contact Hours: 45 hours
Pre-requisites: Foundation Engineering, I
Purpose: This course is intended to equip a student with knowledge and skills for attaining
competence in design of foundations and piles.
Expected Learning Outcomes
By the end of this course the student should be able to;
i.
Describe the different types of foundations
ii.
Describe methods of site investigations and exploration
iii.
Describe different on site methods of sampling and tests
Content
Foundation footings: strip, rafts, piles, piers and caissons.
Foundation types: selection of suitable types of foundations for given sittings such as footings,
mats, strip, rafts, piles, piers.
Site investigations and exploration: planning, geological information, sub-surface exploration.
Retaining walls: design and failure modes, abutments, sheet piling and cofferdams.
Site investigations: boring and sampling, penetration tests, remote sensing, geophysical
exploration.
Laboratory: Penetration tests, Geophysical exploration
Mode of delivery:
Lectures and tutorials, Talks by selected professionals, Assigned reading of relevant materials,
Audio-visual presentation, Group/class discussions, visits to selected projects and firms
Instruction materials and equipment:
White boards, Handouts and LCD/Overhead projectors, E-materials
Course assessment:
Continuous Assessment Tests
30%
Semester Examination
70%
Total
100%
References:
Bowles JE, 1982, Foundation Engineering, McGraw-Hill international book
company, Tokyo.
Tomlinson MJ and Boorman R (1986), Foundation and construction, Longman scientific and
technical, England
Franklin JA and Dussealt MB (1989) Rock Engineering, McGraw-Hill international editions,
London
Chen FH (1975) Foundations on expansive soils, Elsevier scientific Publishing Company
Page 3 of 70
Retaining Walls
Introduction
Retaining walls are used to retain soils between two different elevations in areas of terrain
possessing undesirable slopes or in areas where the landscape needs to be shaped severely and
engineered for more specific purposes like hillside farming or roadway overpasses. The most
important consideration in proper design and installation of retaining walls is to recognize the
tendency of the retained material to move. This creates lateral earth pressure behind the wall
which depends on the angle of internal friction (φ) and the cohesive strength (c) of the retained
material, as well as the direction and magnitude of movement the retaining structure undergoes.
Earth pressures will push the wall forward or overturn it if not properly taken into account. Any
groundwater behind the wall that is not dissipated by a drainage system causes hydrostatic
pressure on the wall. If the wall is not designed to retain water, a proper drainage system behind
the wall in order to limit the pressure to the wall's design value is needed. Drainage materials
will reduce or eliminate the hydrostatic pressure and improve the stability of the material
behind the wall.
Types of retaining walls
1
Gravity walls
Gravity walls as shown in Fig. 1.1, depend on their mass (stone, concrete or other heavy
material) to resist pressure from behind and may have a 'batter' setback to improve stability by
leaning back towards the retained soil. For short landscaping walls, they are often made from
mortarless stone or segmental concrete units (masonry units). Dry-stacked gravity walls are
somewhat flexible and do not require a rigid footing in frost areas. Tall gravity retaining walls
are increasingly built as composite walls such as reinforced earth with precast facing; gabions;
crib walls; or soil-nailed walls
Page 4 of 70
Road way
Mass Stone Wall
Road
way
Reinforced earth wall
Crib Wall
Gabion Mattress Wall
Figure 1.1: Types of gravity retaining Walls
2
Cantilevered and counterfort retaining walls
Cantilevered retaining walls are made from an internal stem of steel-reinforced, cast-in-place
concrete or mortared masonry (often in the shape of an inverted T). These walls consist of a
cantilever stem, cantilever heel and toe. For high walls in excess of eight meters designing
counterfort on the back of the wall, or buttress in the front, improves their strength in resisting high
loads. This type of wall uses less material than a traditional high cantilever walls when designed
carefully. The horizontal load is taken by spanning horizontally.
Bridge Abutment walls are in the category of cantilever or counterfort walls. These are walls for
support to the bridge and the backfill of the end span. They are designed to take traffic loading
and breaking forces from the traffic. The usual practice is to introduce drainage holes at the low
water level and therefore we do not design for the pore water. They may be connected to the deck
by bridge bearings and therefore do not get any moments from the deck or may monolithically be
connected to the deck in which case they take both force reactions and moment.
Basement Walls are also in this category. They retain earth below the ground level. They should
typically, be designed for the at rest case because they are restrained from movement by the
building framework. They are supported by the floor beams and columns. They are to be water
proofed or designed as water retaining structures to protect the basement from dampness and
ingress of water. It is usual to allow water to collect in a sump for pumping to a suitable outlet.
These walls are buttressed by the main structure.
Cantilevered and counterfort retaining walls
Page 6 of 70
Ground
Ground
Floor
Original
ground
Floor
Upper
Upper
Basement
basement
Lower Basement
Basement
wall
Lower
Basement
Bridge Abutment walls
Basement Wall
3
Sheet pile wall
Sheet pile retaining walls are usually used in soft soils. Sheet pile walls are made out of steel, vinyl
or wood planks which are driven into the ground. They are usually driven 1/3 height above ground,
2/3 below ground. This however may be altered depending on the environment. Taller sheet pile
walls will need a tie-back anchor, placed in the soil a distance behind the face of the wall that is
tied to the wall, usually by a cable or a rod. Anchors are then placed behind the potential failure
plane in the soil.
4
Anchored retaining wall
An anchored retaining wall can be constructed in any of the aforementioned alternatives but also
includes additional strength using cables or other stays anchored in the rock or soil behind it. The
anchors are driven into the material with boring; anchors are then expanded at the end of the cable,
either by mechanical means or often by injecting pressurized concrete, which expands to form a
bulb in the soil. This method is very useful where high loads are expected or where the wall itself
has to be slender and would be too weak to retain the soil
Design of retaining walls
The Design of any Retaining Wall is concerned with its stability. The stability of the retaining
wall is due to its self-weight and the dead weight on top of the heel. The wall is designed to obtain
an acceptable factor of safety with respect to:
Page 7 of 70
a.
b.
c.
d.
Overall slope stability failure of the soil around the wall,
Overturning,
Sliding,
Ensuring that allowable soil bearing pressures is not exceeded at the base of the wall.
This is critical at the toe of the wall.
These design stability failure modes are shown in Figure 2.2
a)
Overall slope stability failure
b)
Overturning
d) Bearing capacity failure
c)
Sliding
Figure 2.2: Retaining wall failure modes
The design steps of a retaining wall can be summarized as follows:
i.
Start with an assumed geometry of the wall. For a cantilever wall the following
dimensions are generally good:
a. The footing width, to be about 0.4 to 0.7 of the height of the wall
b. The toe projection is 1/3 to 1/4 of the width of the base
c. The footing thickness and the stem width at the footing is1/10 to 1/14H
of the
height of the wall.
ii.
Compute overturning moments, calculated about the front toe (bottom edge of the
footing).
iii.
Compute resisting moments based upon the assumed footing width, calculated about the
front edge of the footing.
iv.
An overturning factor of safety (resisting moments/ overturning moments) of at least 1.5
is considered safe.
v.
Check sliding. A factor of safety with respect to sliding of 1.5 is considered safe.
vi.
In the absence of more accurate data, the coefficient of friction may be taken as:
Page 8 of 70
Soil
Coefficient of friction
Where the supporting base soil has a strength parameter 𝑡𝑎𝑛 𝛿
2
𝜑 and 𝑐 = 0
𝑤ℎ𝑒𝑟𝑒 𝛿 = ( ) 𝜑
3
For coarse grained soil without silt
0.55
For coarse grained soil with silt
0.45
For silt
0.35
vii.
viii.
Calculate the eccentricity of the total vertical load. Is it within or outside the middle-third
of the footing width?
Calculate the soil pressure at the toe and heel. If the eccentricity, e > B/6 (B = width of
footing), it will be outside the middle third of the footing width (not recommended!), and
because there cannot be tension between the footing and soil, a triangular pressure
distribution will be the result. If this condition cannot be avoided, then adjust the wall
dimensions.
𝑃
6𝑒
𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑞 =
(1 ±
)
𝐵𝐿
𝐿
𝐹𝑜𝑟 𝑡ℎ𝑒 𝑟𝑒𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑤𝑎𝑙𝑙 𝐵 = 1
∴𝑞=
ix.
x.
𝑃
6𝑒
(1 ±
)
𝐿
𝐿
Design the stem. Start at the bottom of the stem where moments and shears are highest.
Then, for economy, checkup the stem to determine if the bar size can be reduced or
alternate bars dropped. The thickness of the stem may vary, top to bottom. The minimum
top thickness for reinforced concrete walls is usually 150 mm.
Design footing for moments and shears.
Example 1
A section of a cantilever retaining wall is shown in Fig. 2.3. The backfill has the following
properties:
𝐶𝑜ℎ𝑒𝑠𝑖𝑜𝑛 𝑐 = 0 𝑘𝑁/𝑚2 , 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∅ = 41𝑜 , 𝛾 = 16 𝑘𝑁/𝑚3
2
𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝛿 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑎𝑙𝑙 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 = ( ) ∅
3
Unit weight of concrete = 24 kN/m3
𝑆𝑎𝑓𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 = 200 𝑘𝑁/𝑚2
Check the stability of the retaining wall.
Page 9 of 70
(Rao: pg. 355)
Fig. 2.3: Cantilevered retaining wall
Solution
1 − sin 41𝑜
𝐾𝑎 =
= 0.2077
1 + sin 41𝑜
Horizontal lateral active earth pressure:
Top of the wall:
𝑝𝑎 = (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒)𝐾𝑎 − 2𝑐√𝐾𝑎
𝑝𝑎 = (35)0.2077 − 2(0)√0.2077
𝑝𝑎 = 7.2695 − 0 = 𝟕. 𝟐𝟕 𝒌𝑵/𝒎𝟐
Bottom of the wall:
𝑝𝑎 = (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒)𝐾𝑎 − 2𝑐√𝐾𝑎
𝑝𝑎 = (𝛾𝑧 + 𝑞)𝐾𝑎 − 2𝑐√𝐾𝑎
𝑝𝑎 = [(16)6 + 35]0.2077 − 2(0)√0.2077
𝑝𝑎 = [96 + 35]0.2077 − 0 = 𝟐𝟕. 𝟐𝟏 𝒌𝑵/𝒎𝟐
Page 10 of 70
7.3
6m
𝟏
2
27.2
Fig. 2.3-1: Pressure distribution on the wall back
Force (kN)
𝑃1 = (7.3)6 = 43.8
1
𝑃2 = ( ) (27.2 − 7.3)6 = 59.7
2
Moments about the heel H
Arm of moment (m) Moment (kNm)
3
131.4
2
119.4
𝐇𝐨𝐫𝐢𝐳𝐨𝐧𝐭𝐚𝐥 𝐜𝐨𝐦𝐩𝐨𝐧𝐞𝐧𝐭 𝐑 𝐇 = 𝟒𝟑. 𝟖 + 𝟓𝟗. 𝟕 = 𝟏𝟎𝟑. 𝟓
Weight of the stem = (5.5)(0.4)24 = 52.8
Weight of the base = (3.5)(0.5)24 = 42
Weight of the backfill = (5.5)(2)16 = 176
Surcharge = (35)(2) = 70
2.2
1.75
1.0
1.0
𝐓𝐨𝐭𝐚𝐥 𝐦𝐨𝐦𝐞𝐧𝐭
𝐕𝐞𝐫𝐭𝐢𝐜𝐚𝐥 𝐜𝐨𝐦𝐩𝐨𝐧𝐞𝐧𝐭 𝑹𝒗 = 𝟓𝟐. 𝟖 + 𝟒𝟐 + 𝟏𝟕𝟔 + 𝟕𝟎 = 𝟑𝟒𝟎. 𝟖 𝐤𝐍
Let R v act at x̅ from the heel:
𝑇ℎ𝑒𝑛 (𝑅𝑣 ) x̅ = 686.46
∴ (340.8) x̅ = 686.46
∴ x̅ = 2.01 m
∴ 𝐸𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑒 = x̅ −
= 2.01 −
Direct stress on the base =
𝐵
2
3.5
= 0.26 𝑚 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒
2
Rv
Rv
340.8
=
=
= 97.37 kN/m2
Base area (B)(1) (3.5)( 1)
Page 11 of 70
116.16
73.5
176
70
686.46
Moment on the base = (R v )(e) = 340.8(0.26) = 88.61kNm
𝑀𝑜𝑚𝑒𝑛𝑡
𝑀
𝑀
88.61
Bending stress in the base 𝑓𝑏 =
= =
=
= ±43.4 𝑘𝑁/𝑚2
(1)(3.5)2
𝑏𝑑 2
𝑆𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑍
6
6
Centre line
3.5 m
Base
H
T
e = 0.26 m
R v = 340.8 kN
Direct stress
43.4 𝑘𝑁/𝑚2
97.37 kN/𝑚2
+
+
-
43.4 𝑘𝑁/𝑚2
Bending stresses
𝑞𝑚𝑎𝑥 = 140.77 𝑘𝑁/𝑚2
+
𝑞𝑚𝑖𝑛 = 53.97 𝑘𝑁/𝑚2
Combined stresses
Fig. 2.3-2: Pressure distribution in the base
Stresses in the base:
𝑞𝑚𝑎𝑥 = 97.37 + 43.4 = 140.77 𝑘𝑁/𝑚2
Page 12 of 70
𝑞𝑚𝑖𝑛 = 97.37 − 43.4 = 53.97 𝑘𝑁/𝑚2
𝑆𝑎𝑓𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑞𝑠𝑎𝑓𝑒 = 200 𝑘𝑁/𝑚2
∴ 𝑆𝑜𝑖𝑙 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑂𝐾
Sliding
2
𝑅𝑣 𝑡𝑎𝑛 𝛿 340.8𝑡𝑎𝑛 [(3) 41]
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 𝐹 =
=
= 1.7 > 1.5
𝑅𝐻
103.29
Sliding OK
Stability against overturning about the toe T
Moments about the toe T
Force (kN)
Arm of moment (m) Moment (kNm)
3
130.86
(7.27)(6)
𝑃1 =
= 43.62
1
2
119.64
𝑃2 = ( ) (27.21 − 7.27)(6) = 59.82
2
68.64
Weight of the stem = (5.5)(0.4)24 = 52.8 3.5 − 2.2 =1.3
73.5
Weight of the base = (3.5)(0.5)24 = 42
3.5 − 1.75 = 1.75
440.00
Weight of the backfill = (5.5)(2)16 = 176 3.5 − 1 = 2.5
175
Surcharge = (35)(2) = 70
3.5 − 1 = 2.5
𝑇𝑜𝑡𝑎𝑙 𝑜𝑣𝑒𝑟𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑡𝑜𝑒 = 130.86 + 119.64 = 250.5 𝑘𝑁𝑚
𝑇𝑜𝑡𝑎𝑙 𝑠𝑡𝑎𝑏𝑖𝑙𝑖𝑠𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑡𝑜𝑒 = 68.64 + 73.5 + 440 + 175 = 757.14 𝑘𝑁𝑚
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑜𝑣𝑒𝑟𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑡𝑜𝑒 𝑇
𝑆𝑡𝑎𝑏𝑖𝑙𝑖𝑠𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡
757.14
=
=
= 3.0 > 1.5
𝐷𝑖𝑠𝑡𝑎𝑏𝑖𝑙𝑖𝑠𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡
250.5
∴ 𝑇ℎ𝑒 𝑤𝑎𝑙𝑙 𝑖𝑠 𝑠𝑡𝑎𝑏𝑙𝑒
Reference: Rao: pg. 355
Page 13 of 70
Example 2
400
200
Bridge
reaction =100 900
kN/m
Bridge
breaking force
= 15kN/m
6000
Water side
600
600
1200
2000
Fig. 2.3-3: Bridge abutment
A bridge abutment wall
A typical section of a bridge abutment is shown in Ошибка! Источник ссылки не найден.-3.
1
2
3
4
5
6
7
Design data:
Coefficient of earth pressure, assuming a drained granular backfill
(∅ = 30o ), 𝐾𝑎
Presumed allowable bearing capacity of the soil (weathered rock)
Traffic surcharge
[(8)(loaded length + 250)]
1
Breaking force =
(
)
2
loaded length
(Loaded length = 11.5 m)
Reaction from the deck
Unit weight of the compacted soil
Unit weight of concrete
Check the stability of the abutment
Model solution
Page 14 of 70
1
3
500 𝑘𝑁/𝑚2
10 kN/m2
15 kN/m
100 kN/𝑚
18 kN/m3
24 kN/m3
Top of the wall:
𝐴𝑐𝑡𝑖𝑣𝑒 𝑡ℎ𝑟𝑢𝑠𝑡 𝑏𝑒ℎ𝑖𝑛𝑑 𝑡ℎ𝑒 𝑤𝑎𝑙𝑙 𝑝𝑎 = (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒)𝐾𝑎 − 2𝑐√𝐾𝑎
1
= (10) − 2(0)√𝐾𝑎 ( 𝑔𝑟𝑎𝑛𝑢𝑙𝑎𝑟 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙, 𝑐𝑜ℎ𝑒𝑠𝑖𝑜𝑛 𝑐 = 0𝑜 )
3
∴ 𝑝𝑎 = 𝟑. 𝟑𝟑𝟑 𝒌𝑵/𝒎𝟐
Bottom of the wall:
𝑧 = 0.9 + 6.0 + 0.6 = 7.5 𝑚
𝐴𝑐𝑡𝑖𝑣𝑒 𝑡ℎ𝑟𝑢𝑠𝑡 𝑏𝑒ℎ𝑖𝑛𝑑 𝑡ℎ𝑒 𝑤𝑎𝑙𝑙 𝑝𝑎 = (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒)𝐾𝑎 − 2𝑐√𝐾𝑎
= (10 + 𝛾𝑧)𝐾𝑎 − 2𝑐√𝐾𝑎
1
= [10 + (18)7.5] − 2(0)√𝐾𝑎
3
1
∴ 𝑝𝑎 = [10 + (18)7.5] = 𝟒𝟖. 𝟑𝟑𝟑 𝒌𝑵/𝒎𝟐
3
3.333
7.5 m
𝟏
2
48.333
Fig. 2.3-1: Pressure distribution on the back of the abutment
Moments about the heel
Force (kN)
𝑃1 = (3.333)7.5 = 25
Arm of moment (m) Moment (kNm)
7.5
93.75
= 3.75
2
7.5
(48.333 − 3.333)
421.875
= 2.5
(7.5) = 168.75
𝑃2 =
3
2
Breaking force = 15
6.6
Totals
𝐇𝐨𝐫𝐢𝐳𝐨𝐧𝐭𝐚𝐥 𝐜𝐨𝐦𝐩𝐨𝐧𝐞𝐧𝐭 𝐑 𝐇 = 𝟐𝟓 + 𝟏𝟔𝟖. 𝟕𝟓 + 𝟏𝟓 = 𝟐𝟎𝟖. 𝟕𝟓 kN
Page 15 of 70
99
614.625
Moments about the heel
Force (kN)
Arm of moment (m) Moment (kNm)
Deck reaction = 100
2.4
240
𝐒𝐭𝐞𝐦:
(𝑖): (0.9)(0.2)24 = 4.32
2.1
9.072
(ii): (6)(0.6)24 = 86.4
2.3
198.72
Weight of the base = (3.8)(0.6)24 = 54.72
1.9
103.968
Weight of the backfill = (6.9)(2)18 = 248.4
1
248.4
Surcharge = (10)(2) = 20
1
20
𝑻𝒐𝒕𝒂𝒍 𝒎𝒐𝒎𝒆𝒏𝒕𝒔
𝟖𝟐𝟎. 𝟏𝟔
𝑹𝒗 = 𝟏𝟎𝟎 + 𝟒. 𝟑𝟐 + 𝟖𝟔. 𝟒 + 𝟓𝟒. 𝟕𝟐 + 𝟐𝟒𝟖. 𝟒 + 𝟐𝟎 = 𝟓𝟏𝟑. 𝟖𝟒 𝒌𝑵
Let R v act at x̅ from the heel:
𝑇ℎ𝑒𝑛 (𝑅𝑣 ) x̅ = 820.16 + 614.625 = 1434.785
∴ (𝟓𝟏𝟑. 𝟖𝟒 ) x̅ = 1434.785
∴ x̅ = 𝟐. 𝟕𝟗𝟐𝟑 𝒎
𝐵
∴ 𝐸𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑒 = x̅ −
2
3.8
∴ 𝑒 = 𝟐. 𝟕𝟗𝟐𝟑 −
= 0.8923 𝑚 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒
2
Direct stress on the base =
Rv
Rv
513.84
=
=
= 135.221 kN/m2
Base area (B)(1) (3.8)( 1)
Moment on the base = (R v )(e) = (513.84) (0.8923 ) = 458.5 kNm
Bending stress in the base 𝑓𝑏 =
𝑀𝑜𝑚𝑒𝑛𝑡
𝑀
𝑀
458.5
= =
=
= ±190.513 𝑘𝑁/𝑚2
2
(1)(3.8)2
𝑏𝑑
𝑆𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑍
6
6
Page 16 of 70
Centre line
3.8 m
Base
H
T
e = 0.8923 m
R v = 513.84 kN
Direct stress
190.513 𝑘𝑁/𝑚2
135.221 kN/𝑚2
+
+
-
190.513 𝑘𝑁/𝑚2
Bending stresses
𝑞𝑚𝑎𝑥 = 325.734 𝑘𝑁/𝑚2
𝑞𝑚𝑖𝑛 = 0 𝑘𝑁/𝑚2
+
-
𝑞𝑚𝑖𝑛 = −55.292 𝑘𝑁/𝑚2
Combined stresses
Fig. 2.3-4: Pressure distribution in the base of the abutment
Stresses in the base:
𝑞𝑚𝑎𝑥 = 135.221 + 190.513 = 325.734 𝑘𝑁/𝑚2
𝑞𝑚𝑖𝑛 = 135.221 − 190.513 = −55.292 𝑘𝑁/𝑚2
Soil cannot take tension
Page 17 of 70
∴ 𝑞𝑚𝑖𝑛 = 0
𝑆𝑎𝑓𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑞𝑠𝑎𝑓𝑒 = 500 𝑘𝑁/𝑚2
∴ 𝑆𝑜𝑖𝑙 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑖𝑠 𝑛𝑜𝑡 𝑒𝑥𝑐𝑒𝑒𝑑𝑒𝑑
Sliding
2
𝑅𝑣 𝑡𝑎𝑛 𝛿 513.84 𝑡𝑎𝑛 [(3) 30]
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 𝐹 =
=
= 0.9 < 1.5
𝑅𝐻
208.7𝟓
Sliding not OK: (abutment restrained by the bridge deck against sliding)
Example 3
C
𝟐𝟓𝒐
D
4
𝜸 = 𝟏𝟕. 𝟔 𝒌𝑵/𝒎𝟑
6.1 m
1
2
3
A
B
1.52 m
0.92 m
0.61 m
Fig. 2.3-5: Gravity wall
Fig. 2.3-5 shows a gravity retaining wall:
𝑀𝑎𝑠𝑜𝑛𝑟𝑦 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 = 24 𝑘𝑁/𝑚3
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∅ = 33𝑜
𝐶𝑜ℎ𝑒𝑠𝑖𝑜𝑛 𝑐 = 0 𝑘𝑁/𝑚2
𝑆𝑎𝑓𝑒 𝑠𝑜𝑖𝑙 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑞𝑠𝑎𝑓𝑒 = 215 𝑘𝑁/𝑚2
Page 18 of 70
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑏𝑎𝑠𝑒 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝜇 = 0.65 𝑘𝑁/𝑚2
Investigate the stability of the wall.
Model solution
𝐾𝑎 = 𝑐𝑜𝑠𝛽
𝑐𝑜𝑠𝛽 − √𝑐𝑜𝑠 2 𝛽 − 𝑐𝑜𝑠 2 ∅
𝑐𝑜𝑠𝛽 + √𝑐𝑜𝑠 2 𝛽 − 𝑐𝑜𝑠 2 ∅
∴ 𝐾𝑎 = 𝑐𝑜𝑠25𝑜
∴ 𝐾𝑎 = 0.91
𝑐𝑜𝑠25𝑜 − √𝑐𝑜𝑠 2 25𝑜 − 𝑐𝑜𝑠 2 33𝑜
𝑐𝑜𝑠25𝑜 + √𝑐𝑜𝑠 2 25𝑜 − 𝑐𝑜𝑠 2 33𝑜
0.91 − √0.82 − 0.703
0.91 + √0.82 − 0.703
0.91 − √0.82 − 0.703
∴ 𝐾𝑎 = 0.91 (
)
0.91 + √0.82 − 0.703
0.57
∴ 𝐾𝑎 = 0.91 (
) = 0.41
1.252
𝑡𝑎𝑛 25𝑜 =
𝐶𝐷
0.61
∴ 𝐶𝐷 = 0.61𝑡𝑎𝑛 25𝑜 = 0.285 𝑚
𝐵𝐶 = 𝐵𝐷 + 𝐶𝐷 = 6.1 + 0.285 = 6.39 𝑚
1
𝑇𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑛 𝐵𝐶 𝑃𝑎 = ( ) [(𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒)𝐾𝑎 − 2𝑐√𝐾𝑎 ][𝐴𝑟𝑒𝑎]
2
1
∴ 𝑃𝑎 = ( ) [(𝛾𝑧)𝐾𝑎 − 2𝑐√𝐾𝑎 ][𝐴𝑟𝑒𝑎]
2
1
∴ 𝑃𝑎 = ( ) [(17.6)(6.39)0.41 − 2(0)√𝐾𝑎 ][6.39]
2
∴ 𝑃𝑎 = 147.32 𝑘𝑁/𝑚 𝑟𝑢𝑛
The thrust acts parallel to the top surface of the backfill: Therefore:
1)𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑃𝑎ℎ = 𝑃𝑎 𝑐𝑜𝑠25𝑜 = 147.32 𝑐𝑜𝑠25𝑜 = 133.52 𝑘𝑁
Page 19 of 70
2) 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑃𝑎𝑣 = 𝑃𝑎 𝑠𝑖𝑛25𝑜 = 147.32 𝑠𝑖𝑛25𝑜 = 62.26 𝑘𝑁
Moments about B
Part
Vertical force
(kN)
1
111.264
2
134.688
3
44.652
4
34.302
𝑃𝑎𝑣
62.26
𝑃𝑎ℎ
Totals
Horizontal force
(kN)
133.52
Lever arm (m)
2.037
1.07
0.407
0.203
0
1
( ) 6.39 = 2.13
3
133.52
387.166
Moment about B
(kNm)
226.65
144.12
18.173
6.963
0
284.4
680.306
Let vertical resultant R v act at x̅ from the heel B:
𝑇ℎ𝑒𝑛 (387.166) x̅ = 680.306
∴ x̅ = 𝟏. 𝟕𝟓𝟕 𝒎
∴ 𝐸𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑒 = x̅ −
∴ 𝑒 = 𝟏. 𝟕𝟓𝟕 −
𝐵
2
3.05
= 0.232 𝑚 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒
2
Direct stress on the base =
Rv
Rv
387.166
=
=
= 126.94 kN/m2
(B)(1)
(3.05)(
Base area
1)
Moment on the base = (R v )(e) = 387.166(0.232 ) = 89.823 kNm
Bending stress in the base 𝑓𝑏 =
𝑀𝑜𝑚𝑒𝑛𝑡
𝑀
𝑀
89.823
= =
=
= ±57.935 𝑘𝑁/𝑚2
2
(1)(3.05)2
𝑏𝑑
𝑆𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑍
6
6
Page 20 of 70
Centre line
3.05 m
Base
H
T
e = 0.232 m
R v = 387.166 kN
Direct stress
57.935 𝑘𝑁/𝑚2
126.94 kN/𝑚2
+
+
-
57.935 𝑘𝑁/𝑚2
Bending stresses
𝑞𝑚𝑎𝑥 = 184.875 𝑘𝑁/𝑚2
+
𝑞𝑚𝑖𝑛 = 69.01 𝑘𝑁/𝑚2
Combined stresses
Fig. 2.3-2: Pressure distribution in the base
Stresses in the base:
𝑞𝑚𝑎𝑥 = 126.94 + 57.935 = 184.875 𝑘𝑁/𝑚2
Page 21 of 70
𝑞𝑚𝑖𝑛 = 126.94 − 57.935 = 69.01𝑘𝑁/𝑚2
𝑆𝑎𝑓𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑞𝑠𝑎𝑓𝑒 = 215 𝑘𝑁/𝑚2
∴ 𝑆𝑜𝑖𝑙 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑂𝐾
Sliding
𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑠𝑙𝑖𝑑𝑖𝑛𝑔 𝐹 =
𝑅𝑣 𝑡𝑎𝑛 𝛿 (387.166)0.65
=
= 1.9 > 1.5
𝑅𝐻
133.52
Sliding OK
Reference: Smith (pg. 249)
CANTILEVER SHEET PILES
When considering the stability of a cantilever sheet pile wall driven into soil, it is usual to
assume that, at the moment of failure, the pile rotates about a point O, some distance above the
base: Illustration Fig. 2.3-3:
O
𝑃𝑝1
𝑷𝒂
𝑃𝑝2
Fig. 2.3-3: Theoretical pressure distribution
𝑃𝑎 = 𝐴𝑐𝑡𝑖𝑣𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑡ℎ𝑟𝑢𝑠𝑡 𝑐𝑎𝑢𝑠𝑖𝑛𝑔 𝑜𝑣𝑒𝑟𝑡𝑢𝑟𝑛𝑖𝑛𝑔
𝑃𝑝1 = 𝑃𝑎𝑠𝑠𝑖𝑣𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑖𝑛𝑓𝑟𝑜𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒 𝑎𝑏𝑜𝑣𝑒 𝑂
𝑃𝑝2 = 𝑃𝑎𝑠𝑠𝑖𝑣𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑏𝑒𝑙𝑜𝑤 𝑂
To simplify calculations:
a) point O is assumed to be at the foot of the pile,
b) 𝑃𝑝2 is assumed to act as a point load at O: Illustration Fig. 2.3-4
Page 22 of 70
H
D
𝑷𝒑𝟏
𝑃𝑎
𝑷𝒑𝟐
Fig. 2.3-4: Pressure distribution assumed for design
Moments about base:
𝐷
𝐻+𝐷
(𝑃𝑝1 ) ( ) = 𝑃𝑎 (
) + (𝑃𝑝2 )(0)
3
3
The depth D obtained by this method is increased by 20% to provide a factor of safety.
Example
An excavation 5.5 m deep in cohesionless soil is supported by a vertical cantilever sheet pile wall.
The pile extends to a depth of 3.6 m below the bottom of the excavation.
Soil properties:
𝑈𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝛾 = 19.2 𝑘𝑁/𝑚3
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∅ = 33𝑜
Investigate the stability of the wall.
Model solution
𝐾𝑎 =
1 − 𝑠𝑖𝑛∅ 1 − 𝑠𝑖𝑛33𝑜 0.455
=
=
= 0.295
1 + 𝑠𝑖𝑛∅ 1 + 𝑠𝑖𝑛33𝑜 1.545
𝐾𝑝 =
1
1.545
=
= 3.4
𝐾𝑎 0.455
𝑯 = 𝟓. 𝟓 𝒎, 𝑫 = 𝟑. 𝟔 𝒎
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑎𝑐𝑡𝑖𝑣𝑒 𝑡ℎ𝑟𝑢𝑠𝑡 𝑝𝑎 = (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒)𝐾𝑎 − 2𝑐√𝐾𝑎
∴ 𝑝𝑎 = (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒)𝐾𝑎 − 2(0)√𝐾𝑎
∴ 𝑝𝑎 = (𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒)𝐾𝑎
Page 23 of 70
1
∴ 𝑇𝑜𝑡𝑎𝑙 𝑎𝑐𝑡𝑖𝑣𝑒 𝑡ℎ𝑟𝑢𝑠𝑡 𝑃𝑎 = [( ) (𝛾)(𝐻 + 𝐷)(𝐾𝑎 )] (𝐻 + 𝐷)
2
1
= [( ) (19.2)(5.5 + 3.6)(0.295)] (5.5 + 3.6)
2
∴ 𝑃𝑎 = 234.52 𝑘𝑁/𝑚
Moments about the toe
3.6
9.1
) = 234.52 ( ) + (𝑃𝑝2 )(0)
3
3
9.1
3
∴ 𝑇ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑃𝑝1 = 234.52 ( ) ( ) = 592.81 𝑘𝑁
3
3.6
1
𝑇ℎ𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑝𝑎𝑠𝑠𝑖𝑣𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑃𝑝 = [( ) 𝛾𝐷𝐾𝑝 ] 𝐷
2
1
∴ 𝑃𝑝 = [( ) (19.2)(3.6)3.4] 3.6 = 423 𝑘𝑁/𝑚 < 592.81 𝑘𝑁
2
(𝑃𝑝1 ) (
The wall is unstable and the depth should be increased
Reference: Sutton (pg. 148)
ANCHORED SHEET PILES
(i)Anchor blocks
Anchoring is done to reduce the penetration depth required
Tie rod
Anchor block
Sheet pile
𝜃=
∅
− 45𝑜
2
Possible failure planes
𝜃=
∅
+ 45𝑜
2
Fig. 2.3-5: Anchored sheet pile
Page 24 of 70
Tie rods are anchored in beams; plates or concrete blocks some distance behind the wall:
Illustration in Fig.2.3-5. The anchor block should be outside the possible failure plane. The tie rod
force is resisted by the passive resistance mobilized by the anchor.
(ii) Anchor piles
(a)Single anchor pile
When space is limited, an anchor pile is used
Sheet Piling
Tie rod
Anchorage Pile
𝜃=
∅
+ 45𝑜
2
Fig. 2.3-6: Anchored sheet pile: additional pile when space is limited
(b)Raking piles
To avoid bending in the anchor pile, a pair of raking piles is used
Sheet Piling
Tie rod
Raking Piles
𝜃=
∅
+ 45𝑜
2
Fig. 2.3-7: Anchored sheet pile: Raking piles installed
Page 25 of 70
PENETRATION OF PILING FOR ANCHORED SHEET PILES
Free earth support method (depth d is just sufficient to balance lateral forces)
A
h
T
D
d
𝑃𝑎
𝑷𝒑𝟏
𝑑/3
ℎ+𝑑
3
B
Fig. 2.3-8: Pressure distribution assumed for design
1) The wall is considered free to rotate about the base B,
2) Moments about the tie give an expression for (d),
3) Actual penetration depth = (1.2) d
4) Horizontal resolution of forces gives tension T
Example
A
1m
T
D
4m
C
𝛾 = 21 𝑘𝑁/𝑚3
𝑐 = 0,
∅ = 30𝑜
d
B
Fig. 2.3-9
Determine the minimum penetration depth d of the pile to achieve free earth support conditions.
Page 26 of 70
Model solution
A
1m
T
D
4m
mh
d
C
𝑷𝒑
B
7d
35 𝑘𝑁/𝑚2
Fig. 2.3-10
1
𝐴𝑐𝑡𝑖𝑣𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝐶 = ( ) (21)5 = 35 𝑘𝑁/𝑚2
3
𝐴𝑡 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒:
(𝑖) 𝑃𝑝 = 3 × 21 × 𝑑 = 63𝑑
1
(𝑖𝑖) 𝑃𝑎 = ( ) × 21 × 𝑑 + 35 = 7𝑑 + 35
3
Moments about D:
1
2
𝑑
35 × 5 2
(63 − 7)𝑑 ( ) 𝑑 (4 + 𝑑) = 35𝑑 (4 + ) +
( × 5 − 1)
2
3
2
2
3
∴ 𝑑 3 + 5.06𝑑 2 − 7.51𝑑 − 10.92 = 0
Points of inflection:
𝐿𝑒𝑡 𝑦 = 𝑑3 + 5.06𝑑 2 − 7.51𝑑 − 10.92
∴
𝑑𝑦
= 3𝑑2 + (2)(5.06)𝑑 − 7.51
𝑑𝑑
𝐴𝑡 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛:
3𝑑2 + (2)(5.06)𝑑 − 7.51 = 0
Page 27 of 70
∴ 𝑑 2 + 3.373𝑑 − 2.5 = 0
(𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑑)
∴𝑑=
−𝑏 ± √(𝑏 2 − 4𝑎𝑐)
2𝑎
∴𝑑=
−3.373 ± √3.3732 − 4(1)(−2.5)
2(1)
∴ 𝑑 = −1.6865 ± 2.312
∴ 𝑑 = −4,
+0.63
Minima and maxima:
𝐿𝑒𝑡 𝑦 = 𝑑 2 + 3.373𝑑 − 2.5
𝒑𝒐𝒊𝒏𝒕 𝒅 = −𝟒:
𝑊ℎ𝑒𝑛 𝑑 = −4.1,
𝑦 = (−4.1)2 + 3.373(−4.1) − 2.5 = +0.4807
𝑊ℎ𝑒𝑛 𝑑 = −3.9,
𝑦 = (−3.9)2 + 3.373(−3.9) − 2.5 = −0.4447
-4
gradient +ve
-ve
∴ 𝐴𝑡 − 4 , 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑦 = 𝑑3 + 5.06𝑑 2 − 7.51𝑑 − 10.92 ℎ𝑎𝑠 𝑎 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
𝑷𝒐𝒊𝒏𝒕 𝒅 = +𝟎. 𝟔𝟑:
𝑊ℎ𝑒𝑛 𝑑 = +0.62,
𝑦 = (0.62)2 + 3.373(0.62) − 2.5 = −0.02434
𝑊ℎ𝑒𝑛 𝑑 = +0.64,
𝑦 = (0.64)2 + 3.373(0.64) − 2.5 = +0.06832
gradient -ve
+ve
+0.63
∴ 𝐴𝑡 + 0.63 , 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑦 = 𝑑 3 + 5.06𝑑 2 − 7.51𝑑 − 10.92 ℎ𝑎𝑠 𝑎 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
Page 28 of 70
Table 1
d
-6.5
-6.1
-6
-5
-4
-3
-2
-1
0
1
2
3
y
-22.99
-3.8484
0.26
28.1
36.06
30.14
16.34
0.66
-10.9
-12..34
2.34
39.14
50
40
Function of d
30
20
10
0
-8
-6
-4
-2
0
-10
-20
Depth d (m)
-30
Fig. 2.3-11: Plot of (𝑦 = 𝑑 3 + 5.06𝑑2 − 7.51𝑑 − 10.92) 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑑𝑒𝑝𝑡ℎ 𝑑
From the chart Fig.2.3-11
𝑫𝒆𝒑𝒕𝒉 𝒅 = 𝟏. 𝟗𝟎 𝒎
Force in the tie:
1
𝑃𝑎 = ( ) × 21 × 6.9 = 48.3 𝑘𝑁/𝑚2
3
𝑃𝑎𝑠𝑠𝑖𝑣𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝐵,
𝑃𝑝 = 3 × 21 × 1.9 = 119.7 𝑘𝑁/𝑚2
1
1
∴ 𝐹𝑜𝑟𝑐𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑖𝑒 𝑇 = ( ) (48.3) × 6.9 − ( ) (119.7 ) × 1.9
2
2
𝐴𝑐𝑡𝑖𝑣𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝐵,
= 166.635 − 113.715 = 52.92 𝑘𝑁/𝑚 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑤𝑎𝑙𝑙
FIXED EARTH SUPPORT METHOD
Page 29 of 70
2
4
This method gives full fixity at the base of the pile with an accompanying increase in penetration
depth and reduction in the bending moments on the pile.
(b) Upper part
A
T
D
T
h
𝒙
d
𝑃𝑎
𝑃𝑝
R
O
𝐾𝑝 𝛾𝑥
𝐾𝑎 𝛾(ℎ + 𝑥)
B
(a) The problem
O
R
𝑑−𝑥
𝑃𝑝1
(𝑑 − 𝑥)/3
𝑃𝑝2
𝐾𝑝 𝛾(𝑑 − 𝑥)
B
𝐾𝑎 𝛾(𝑑 − 𝑥)
(c) Lower part
Fig. 2.3-12: The pressure distribution assumed for design
The wall is regarded as two walls, AO and OB, entirely separate from each other (Equivalent
beam method of analysis)
Upper part
Taking moments about D gives R
Lower part
𝑇𝑎𝑘𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡𝑠 𝑎𝑏𝑜𝑢𝑡 𝐵 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑒𝑠 𝑃𝑝2 𝑎𝑛𝑑 𝑔𝑖𝑣𝑒𝑠 𝑑
Position of point O
∅𝒐 20
𝒙
25
30
35
𝟎. 𝟐𝟓𝒉 𝟎. 𝟏𝟓𝒉 𝟎. 𝟎𝟖𝒉 𝟎. 𝟎𝟑𝟓𝒉
Page 30 of 70
For most backfills, the average value of ∅ is 30o and if 𝑥 is assumed as (0.1h), little error will be
involved.
Example
A
1m
T
D
4m
𝛾 = 21 𝑘𝑁/𝑚3
C
∅ = 30𝑜
𝑐 = 0,
d
B
Fig. 2.3-13
Determine the minimum penetration depth d of the pile to achieve fixed earth support conditions
Model solution
A
T
D
T
h
𝟎. 𝟓
d
𝑃𝑎
𝑃𝑝
R
O
31.5
(a)The problem
38.5
(b) Upper part
B
O
𝑅 = 55.4
𝑘
𝑃𝑝1
𝑘/3
𝑃𝑝2
63𝑘
Page 31 of 70
B
7𝑘
(c) Lower part
Fig. 2.3-14: The pressure distribution assumed for design
Upper part
𝐴𝑠𝑠𝑢𝑚𝑒 𝑥 = 0.1ℎ = 0.5 𝑚
1 − 𝑠𝑖𝑛∅ 1 − 𝑠𝑖𝑛30𝑜 0.5 1
=
=
=
1 + 𝑠𝑖𝑛∅ 1 + 𝑠𝑖𝑛30𝑜 1.5 3
1
𝐴𝑐𝑡𝑖𝑣𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝑂 = ( ) 21 × 5.5 = 38.5 𝑘𝑁/𝑚2
3
𝐾𝑎 =
𝑃𝑎𝑠𝑠𝑖𝑣𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 𝑂 = (3)21 × 0.5 = 31.5 𝑘𝑁/𝑚2
Moments about D:
38.5
2
31.5
0.5
× 5.5 ( × 5.5 − 1) =
× 0.5 (4.5 −
) + 4.5𝑅
2
3
2
3
282.33 = 34.125 + 4.5𝑅
∴ 𝑅 = 55.4 𝑘𝑁
Force in the tie T
38.5
𝑇𝑜𝑡𝑎𝑙 𝑎𝑐𝑡𝑖𝑣𝑒 𝑡ℎ𝑟𝑢𝑠𝑡 = (
) 5.5 = 105.875 𝑘𝑁
2
31.5
𝑇𝑜𝑡𝑎𝑙 𝑝𝑎𝑠𝑠𝑖𝑣𝑒 𝑡ℎ𝑟𝑢𝑠𝑡 = (
) 0.5 = 7.875 𝑘𝑁
2
∴ 𝑇 = 105.875 − 7.875 − 55.4 = 42.6 𝑘𝑁
Lower part
𝐿𝑒𝑡 (𝑑 − 0.5) = 𝑘
∴ 𝐾𝑝 𝛾(𝑑 − 𝑥) = 3 × 21 × 𝑘 = 63𝑘
1
∴ 𝐾𝑎 𝛾(𝑑 − 𝑥) = ( ) × 21 × 𝑘 = 7𝑘
3
Moments about B:
𝑘
63𝑘
1
7𝑘
1
∴ 55.4𝑘 + (38.5 − 31.5)𝑘 ( ) − (
)𝑘( )𝑘 +( )𝑘( )𝑘 = 0
2
2
3
2
3
𝑘2
𝑘2 𝑘
∴ 55.4𝑘 + (38.5 − 31.5) ( ) − (63 − 7)
=0
2
2 3
7
𝑘3
∴ 55.4𝑘 + 𝑘 2 − 56
=0
2
6
∴ 𝑘 3 − 0.375𝑘 2 − 5.94𝑘 = 0
Page 32 of 70
Factorizing:
∴ 𝑘(𝑘 2 − 0.375𝑘 − 5.94) = 0
∴ 𝐸𝑖𝑡ℎ𝑒𝑟 𝑘 = 0, 𝑜𝑟 𝑘 2 − 0.375𝑘 − 5.94 = 0
Considering the function,
𝑘 2 − 0.375𝑘 − 5.94 = 0
(𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑘)
−𝑏 ± √(𝑏 2 − 4𝑎𝑐)
∴𝑘=
2𝑎
∴𝑘=
−(−0.375) ± √0.3752 − 4(1)(−5.94)
2(1)
∴𝑘=
0.375 ± 4.889
2
∴ 𝑘 = 2.632, −2.257
∴ 𝑘 = 0, 2.632, −2.257
(i) 𝑊ℎ𝑒𝑛 𝑘 = 0:
𝑑 − 0.5 = 0
𝑑 = 0.5 𝑚
(ii) 𝑊ℎ𝑒𝑛 𝑘 = 2.632
𝑑 − 0.5 = 2.632
𝑑 = 3.132 𝑚
(iii) 𝑊ℎ𝑒𝑛 𝑘 = −2.257
𝑑 − 0.5 = −2.257
𝑑 = −1.757 𝑚
∴ 𝑑𝑒𝑝𝑡ℎ 𝑑 = 3.132 𝑚 = 3.2 𝑚, 𝑠𝑎𝑦
STRUTTED EXCAVATIONS
When excavating a deep trench, the insertion of shuttering to hold up the sides becomes
necessary. The excavation is carried down first to some point, X and rigidly strutted timbering is
inserted between the levels D and X: Illustration Fig. 1(a).
As further excavation is carried out, timbering and strutting are inserted in stages. The resulting
pressure on the back of the wall is roughly parabolic and is illustrated in Fig. 1(b). For design
purposes, pressure distribution diagrams shown in Fig.2 are used.
Page 33 of 70
Pressure distribution for sands
This is illustrated in Fig. 2 (a)
Pressure distribution for clays
The pressure distribution depends on the stability number (N).
A
D
X
H
C
B
(b) Pressure distribution
(a) Excavation
Fig.1: Strutted excavation
(b)
𝝈 = 𝟎. 𝟔𝟓𝑲𝒂 𝜸𝑯
𝜎
(c)
𝜎
0.25H
0.25H
H
0.5H
0.75H
(a)
0.25H
Fig.2: Strutted excavation: Lateral earth pressure distribution diagrams
Page 34 of 70
Pressure distribution for soft to medium clay is illustrated in Fig. 2 (b).
𝛾𝐻
𝑇ℎ𝑒 𝑠𝑡𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑛𝑢𝑚𝑏𝑒𝑟 𝑁 =
>4
𝑐
The pressure is the larger of:
(𝑖) 𝜎 = 𝛾𝐻 [1 − (
4𝑐
)]
𝛾𝐻
(𝑖𝑖) 𝜎 = 0.3𝛾𝐻
𝑊ℎ𝑒𝑟𝑒 𝑐 = 𝑑𝑟𝑎𝑖𝑛𝑒𝑑 𝑐𝑜ℎ𝑒𝑠𝑖𝑜𝑛, 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∅ = 0𝑜 , 𝛾 = 𝑈𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑙𝑎𝑦
Pressure distribution for stiff clay is illustrated in Fig. 2(c)
𝜎 = 0.2𝐻 𝑡𝑜 0.4𝐻 𝑤𝑖𝑡ℎ 𝑎𝑛 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑓 𝜎 = 0.3𝛾𝐻
It is applicable to clays with
𝑆𝑡𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑁𝑢𝑚𝑏𝑒𝑟 𝑁 =
𝛾𝐻
≤4
𝑐
DESIGN OF VARIOUS COMPONENTS OF A BRACED CUT
Struts
In construction work, struts should have a minimum vertical spacing of about 3.0 m or more. For
braced cuts in clayey soils, the depth of the first strut below the ground surface should be less than
2𝑐
the depth of the tension crack. 𝐹𝑜𝑟 ∅ = 0𝑜 , 𝑑𝑒𝑝𝑡ℎ 𝑡𝑜 𝑡ℎ𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑐𝑟𝑎𝑐𝑘 𝑧0 = 𝛾
A simplified conservative procedure may be used to determine the strut loads:
Step 1
Draw the pressure envelope for the braced cut. Also show the proposed strut levels. Fig. 2 shows
pressure envelope of a sandy soil. The strut levels are marked A, B, C and D. The sheet piles (or
soldier Beams) may be assumed to be hinged at all strut levels except for the top and bottom ones.
Step 2
Determine the reactions for the two simple cantilever beams (top and bottom) and all the simple
beams in between.
Step 3
Calculate the strut loads as follows:
𝑃𝐴 = (𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐴)(𝑠𝑝𝑎𝑐𝑖𝑛𝑔 𝑠)
𝑃𝐵 = (𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐵1 + 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐵2 )(𝑠𝑝𝑎𝑐𝑖𝑛𝑔 𝑠)
𝑃𝐶 = (𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐶1 + 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐶2 )(𝑠𝑝𝑎𝑐𝑖𝑛𝑔 𝑠)
𝑃𝐷 = (𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐷)(𝑠𝑝𝑎𝑐𝑖𝑛𝑔 𝑠)
Page 35 of 70
Step 4
Knowing the strut loads at each level, selection of the suitable sections from the steel tables is
done.
(b) Method for determining the strut loads
Fig. 3: Conservative method of design
Page 36 of 70
SHEET PILES
Step 1
For each of the sections shown in Fig.(b), determine the maximum bending moment.
Step 2
Determine the maximum value of the maximum bending moments obtained in step 1
Step 3
Obtain the required section modulus of the sheet piles:
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑀 = (𝑠𝑡𝑟𝑒𝑠𝑠 𝜎)(𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑑𝑢𝑙𝑠 𝑧)
∴𝑧=
𝑀𝑚𝑎𝑥
𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
𝑊ℎ𝑒𝑟𝑒 𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 = 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑓𝑙𝑒𝑥𝑢𝑟𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑒𝑒𝑡 𝑝𝑖𝑙𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
Step 4
Choose a sheet pile that has a section modulus greater than or equal to the required section
modulus from the steel tables
WALES
They may also be treated as though they are pinned at the struts.
(𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐴)(𝑆 2 )
∴ 𝐴𝑡 𝑙𝑒𝑣𝑒𝑙 𝐴, 𝑀𝑚𝑎𝑥 =
8
𝑊ℎ𝑒𝑟𝑒 𝑠 = 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑢𝑡𝑠
𝐴𝑡 𝐵, 𝑀𝑚𝑎𝑥 =
(𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐵1 + 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐵2 )(𝑆 2 )
8
𝐴𝑡 𝐵, 𝑀𝑚𝑎𝑥 =
(𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐶1 + 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐶2 )(𝑆 2 )
8
𝐴𝑡 𝐷, 𝑀𝑚𝑎𝑥 =
(𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐷)(𝑆 2 )
8
Section modulus of the wales
∴𝑧=
𝑀𝑚𝑎𝑥
𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
Page 37 of 70
Example
Fig. 4: Braced cut
Fig.4 shows a braced cut. The struts are located at 4.0 m center- to - center on plan:
(i)Draw the earth pressure diagram and determine the strut loads at levels A, B and C.
(ii)Determine the sheet pile section modulus, 𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 = 170 × 103 𝑘𝑁/𝑚2
(iii)Determine the required section modulus of the wales at level A,
(𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 = 173 × 103 𝑘𝑁/𝑚2 )
Model solution
𝐾𝑎 = (
1 − 𝑠𝑖𝑛∅
1 − 𝑠𝑖𝑛32𝑜
0.47
)=(
)=
= 0.307
𝑜
1 + 𝑠𝑖𝑛∅
1 + 𝑠𝑖𝑛32
1.5299
(i) 𝐸𝑎𝑟𝑡ℎ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚:
𝜎 = 0.65𝛾𝐻𝐾𝑎 = 0.65 × 17.6 × 0.307 × 9 = 31.6 𝑘𝑁/𝑚2
Page 38 of 70
Fig. 5: Earth pressure distribution
Referring to Fig.(b):
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑀𝐵1 = 0
5
∴ 3 × 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐴 = 31.6 × 5 ×
2
5
31.6 × 5 × 2
∴ 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐴 =
= 131.67 𝑘𝑁/𝑚
3
Page 39 of 70
∴ 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐵1 = (31.6)(5) − 131.67 = 26.33 𝑘𝑁/𝑚
Referring to Fig.(c):
𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑀𝐵2 = 0
∴ 3 × 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐶 = 31.6 × 4 ×
∴ 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐶 =
4
31.6 × 4 × 2
3
4
2
= 84.27 𝑘𝑁/𝑚
∴ 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝐵2 = (31.6)(4) − 84.27 = 42.13 𝑘𝑁/𝑚
𝑆𝑡𝑟𝑢𝑡 𝑙𝑜𝑎𝑑 𝑎𝑡 𝐴 = (131.67)(𝑠𝑝𝑎𝑐𝑖𝑛𝑔) = (131.67)(4) = 526.68 𝑘𝑁
𝑆𝑡𝑟𝑢𝑡 𝑙𝑜𝑎𝑑 𝑎𝑡 𝐵 = (𝐵1 + 𝐵2 )(𝑠𝑝𝑎𝑐𝑖𝑛𝑔) = (26.33 + 42.13)(4) = 273.84 𝑘𝑁
𝑆𝑡𝑟𝑢𝑡 𝑙𝑜𝑎𝑑 𝑎𝑡 𝐶 = (84.27)(𝑠𝑝𝑎𝑐𝑖𝑛𝑔) = (84.27)(4) = 337.08 𝑘𝑁
(ii)Sheet pile section modulus
Referring to Figures (b) and (c):
Based on the load diagrams, the shear force diagrams are given in Fig.6:
Fig. 6: Shear force diagram
68.47
𝑥1 =
= 2.17 𝑚
31.6
Page 40 of 70
𝑥2 =
52.67
= 1.67 𝑚
31.6
𝑀𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝐴 = (𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑚𝑒𝑛𝑡) − (𝑙𝑜𝑎𝑑 𝑚𝑜𝑚𝑒𝑛𝑡)
2
= 0 − (31.6)(2) ( ) = −63.2 𝑘𝑁𝑚
2
𝑀𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝐶 = (𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑚𝑒𝑛𝑡) − (𝑙𝑜𝑎𝑑 𝑚𝑜𝑚𝑒𝑛𝑡)
1
= 0 − (31.6)(1) ( ) = −15.8 𝑘𝑁𝑚
2
𝑀𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝐵 ′ = (𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝐴) − (𝑙𝑜𝑎𝑑 𝑚𝑜𝑚𝑒𝑛𝑡)
2 + 2.17
= (131.67)(2.17) − (31.6)(2 + 2.17) (
) 𝑘𝑁/𝑚
2
= 285.7239 − 274.74462
= 10.98 𝑘𝑁𝑚
𝑀𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝐵 ′′ = (𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝐶) − (𝑙𝑜𝑎𝑑 𝑚𝑜𝑚𝑒𝑛𝑡)
(1 + 1.67)
= (84.27)(1.67) − (31.6)(1 + 1.67)
2
= 140.7309 − 112.63662
= 28.1 𝑘𝑁𝑚
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝑀𝐴 = 63.2 𝑘𝑁𝑚
∴ 𝑆ℎ𝑒𝑒𝑡 𝑝𝑖𝑙𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑍 =
=
𝑀𝑚𝑎𝑥
𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
63.2
= 37.2 × 10−5 𝑚3
170 × 103
Wales at level A
(𝑅𝑒𝑎𝑐𝑡𝑖𝑛 𝑎𝑡 𝐴)(𝑠𝑝𝑎𝑐𝑖𝑛𝑔)2 (131.67)(4)2
𝑀𝑚𝑎𝑥 =
=
= 263.34 𝑘𝑁𝑚
8
8
∴ 𝑊𝑎𝑙𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑍 =
=
𝑀𝑚𝑎𝑥
𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
263.34
= 1.522 × 10−3 𝑚3
173 × 103
FOUNDATIONS
Page 41 of 70
Shallow foundations
Types of shallow foundations
Foundations that are encountered in practice may be classified into two broad categories namely
shallow and deep foundations. Under shallow foundations the following categories are usually
encountered: a)
b)
c)
Strip foundations for wall and closely spaced columns
Spread or isolated footings for individual columns. In this category it is usual to
consider combined foundations for two or three closely spaced columns as spread
or isolated footings
Raft foundations covering large sections of the foundation area
Under deep foundations the following two types of foundations are encountered: a)
Piles
b)
Caissons
In the selection of the foundations to adopt for a structure it is usually necessary to consider the
function of the structure, its loads, the subsurface conditions and the cost of the foundation being
adopted in comparison to other possible types of foundations.
Foundations for common buildings
These are single and double storied buildings with structural walls as the main form of support.
The buildings are generally on good bearing soils. The bearing soils include red coffee soils,
gravelly soils and firm sandy, gravelly clays. The footing for these common buildings is shown
on Figure 1.1. The 600 mm width is a practical width, which allows masons to maneuver in the
trench.
200-150
mm
200-150
mm
masonry
masonry wall
wall
thick
thick
100mm
100mm slab
slab with
with BRC
BRC:no
A 65
65 at
at the
the topface
top face
200-150 DPC
DPC
200-150
Damp proof
membrane
Damp
proof
membrane
quary dust blinding
150
mm
minimum
150
mm
minimum
dropdropa
dropdropasountonsd
100-200
mmthick
thick
hardcore
100-200 mm
hardcore
(treated against termites)
A
A minimum
minimum of
of 1000
1000 mm
mm
depth
depth of
of foundations
foundations
200
600mm
wide xx 200mm
200mm thick
deep
600mm wide
mass
mass concrete
concrete foundation
foundation
600 mm
Figure 1.1: Typical strip footing for an ordinary building
Page 42 of 70
The following are the general considerations in the usage of the standard footing.
a)
No reinforcement is needed for strips where the load can be distributed through 45o.
b)
The foundations should be excavated and the last 150 mm excavation be finalized when
the concreting can be done without further delay. This minimizes the softening of the
foundation
c)
The mass concrete is usually by volume batching to achieve grade 15 concrete. A ratio of
1:3:6; cement: sand: coarse aggregates (ballast), respectively, is generally sufficient.
d)
Reinforced concrete foundations are done for areas with concentrated loads. These are
usually column supports. Grade 25 concrete is the lowest class of concrete allowed in the
new BS 8110, but grade 20 of concrete can be considered.
Isolated column footing
Example 1: (Axial load and moment)
Fig. 1.2 shows a column carrying 900 kN dead load , 450 kN live load and major axis moment
𝑀𝑥𝑥 = 150 𝑘𝑁𝑚. The safe bearing capacity of the soil 𝑞𝑠𝑎𝑓𝑒 = 300 𝑘𝑁/𝑚2 . Investigate the
base against bearing failure.
Load P
𝑴𝒙𝒙 = 𝟏𝟓𝟎 𝒌𝑵𝒎
1000 mm
3.5 m
Fig. 1.2: Column pad foundation: load and uniaxial moment
Model solution
𝑇𝑜𝑡𝑎𝑙 𝑎𝑥𝑖𝑎𝑙 𝑙𝑜𝑎𝑑 𝑃 = 900 + 450 = 1350 𝑘𝑁
Page 43 of 70
𝐴𝑟𝑒𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 =
𝑃
𝑞𝑠𝑎𝑓𝑒
=
1350
= 4.5 𝑚2
300
𝑊𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑢𝑛𝑑𝑎𝑡𝑖𝑜𝑛 =
4.5
= 1.3 𝑚
3.5
∴ 𝑇𝑟𝑦: 3.5 𝑚 × 1.75 𝑚
𝐴𝑟𝑒𝑎 𝐴 = 3.5 × 1.75 = 6.125 𝑚2
Stresses in the base:
𝑚𝑎𝑥
𝑞𝑚𝑖𝑛
=
𝑃 𝑀
±
𝐴 𝑧
∴ 𝑞𝑚𝑎𝑥 =
𝑃 𝑀 𝑃
𝑀
+ = + 2
𝐴 𝑧
𝐴 𝑏𝑑
6
∴ 𝑞𝑚𝑎𝑥 =
1350
150
+
= 220.41 + 42 = 262.41 𝑘𝑁/𝑚2 < 300𝑘𝑁/𝑚2
6.125 (1.75)3.52
6
∴ 𝑺𝒂𝒇𝒆 𝒔𝒐𝒊𝒍 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒏𝒐𝒕 𝒆𝒙𝒄𝒆𝒆𝒅𝒆𝒅
Combined footings
When a column is near a property boundary, a square or a rectangular footing axially loaded under
the column could extend to the adjoining property. In such a case there are three alternatives which
are illustrated in Fig. 2.1.
1 Cantilever footing
A cantilever footing or strap footing is normally comprised of two footings connected by a beam
called a strap.
2 Combined footing
A combined footing is a long footing supporting two or more columns in one row.
3 Mat or raft foundation
A mat or raft foundation is a large footing, usually supporting several columns in two or more
rows.
Page 44 of 70
Property boundary
Fig. 2.1: Schematic plan showing mat, strap and combined footings
Example 2: Strap beam
Fig.2.2 (a) gives a foundation beam with vertical loads and moment acting thereon. The beam
measures 700 mm width x 500 mm depth. Length of the beam is 8.0 m. A uniform load of 16
kN/m (including self-weight) is imposed on the beam. Draw:
(i)The base pressure distribution,
Page 45 of 70
(ii)The shear force diagram,
(iii)The bending moment diagram.
320 kN
400 kN
160 kNm
(a)
𝑹𝟏
1m
6m
V
𝑹𝟐
1m
𝑥 = 3.53 𝑚
143.52 kN
134.13 kN
(b) 68.47 kN
77.85 kN
Base pressure distribution
277.14
𝑥 = 2.95 𝑚
57.16
S.F.
(c)
-122.86
-262.84
62.2
27.8
(d)
BMD
438.3
Fig. 2.2
Page 46 of 70
Model solution:
Resultant vertical force v = 320 + 400 + (16)(8) = 848 kN
Location of the resultant vertical force:
Moments about the right hand edge of the beam:
8
𝑉𝑥 = (320)(7) + (400)(1) + (16)(8) ( ) − 160
2
8
∴ 848𝑥 = (320)(7) + (400)(1) + (16)(8) ( ) − 160
2
∴ 𝑥 = 3.5283 𝑚
∴ 𝑒𝑐𝑐𝑒𝑛𝑡𝑟𝑖𝑐𝑖𝑡𝑦 𝑒 = 4 − 3.5283 = 0.472 𝑚 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑒
𝑃
6𝑒
∴ 𝑞𝑚𝑎𝑥 = ( ) (1 + )
𝐴
𝐿
848
6 × 0.472
∴ 𝑞𝑚𝑎𝑥 = (
) (1 +
) = 205.03 𝑘𝑁/𝑚2
0.7 × 8
8
848
6 × 0.472
∴ 𝑞𝑚𝑖𝑛 = (
) (1 −
) = 97.823 𝑘𝑁/𝑚2
0.7 × 8
8
∴ 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑙𝑜𝑎𝑑 = (0.7)(205.03) = 143.52 𝑘𝑁
∴ 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑙𝑜𝑎𝑑 = (0.7)(97.823) = 68.47 𝑘𝑁
Shear force diagram
68.47 + 77.85
− (16)(1) = 57.16 𝑘𝑁
2
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝑅1 (𝑤𝑖𝑡ℎ 320 𝑘𝑁):
𝑆ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝑅1 =
𝑉 = −320 + 57.2 = −262.84 𝑘𝑁
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑜𝑓 𝑅2 :
77.85 + 134.13
𝑉 = −262.8 + (
) (6) − (16)(6) = 277.14 𝑘𝑁
2
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝑅2 (𝑤𝑖𝑡ℎ 400𝑘𝑁)
𝑉 = 277.32 − 400 = −122.86 𝑘𝑁
Page 47 of 70
Bending moment diagram
Determination 𝑥: point of zero shear between 𝑅1 and 𝑅2 (graphically determined or calculated)
Graphical method:
400
300
Moments (kNm)
200
100
0
0
1
2
3
4
5
6
7
8
9
-100
-200
-300
Span (m)
Fig. 2.3: Shear diagram (determination of x: point of zero shear to the right of 𝑅1 )
From Fig. 2.3: x = 2.95 m to the right of R1
𝑦1
𝑦2
𝑦4
𝑦3
143.52
68.48
𝑥
1 m 𝑅1
𝑅2
Fig. 2.4
𝑦4 = 143.52 − 68.48 = 75.04 𝑘𝑁
𝑦3 75.04
=
7
8
∴ 𝑦3 = 65.66 𝑘𝑁
Page 48 of 70
1m
𝑦2
75.04
=
(1 + 2.95)
8
∴ 𝑦2 = 37.051 𝑘𝑁
𝑦1
75.04
=
(1)
8
∴ 𝑦1 = 9.38 𝑘𝑁
1
9.38
1
1
𝑀𝑅1 = (68.47)(1) ( ) + (
) (1) ( ) − (16)(1) ( ) = 34.235 + 1.563 − 8 = 27.8 𝑘𝑁𝑚
2
2
3
2
𝐹𝑟𝑜𝑚 𝑙𝑒𝑓𝑡 𝑒𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚 𝑡𝑜 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑧𝑒𝑟𝑜 𝑠ℎ𝑒𝑎𝑟 = 1 + 𝑥 = 3.95 𝑚
3.95
37.051
3.95
3.95
𝑀3.95 = (68.47)(3.95) (
)+(
) (3.95) (
) − (320)(2.95) − (16)(3.95) (
)
2
2
3
2
= 534.152 + 96.348 − 944 − 124.82 = −438.32 𝑘𝑁𝑚
7
1
7
7
𝑀𝑅2 = (68.47)(7) ( ) + (65.66) ( ) (7) ( ) − (320)(6) − (16)(7) ( ) + 160
2
2
3
2
= 1677.515 + 536.223 − 1920 − 392 + 160 = 61.738 𝑘𝑁𝑚
Alternatively, considering the beam from the RHS end:
68.47 + 𝑦3 = 68.47 + 65.66 = 134.13 𝑘𝑁𝑚
1
143.52 − 134.13
2
1
𝑀𝑅2 = (134.13 )(1) ( ) + (
) (1) ( ) − (16)(1) ( )
2
2
3
2
= 67.065 + 3.13 − 8 = 62.195 𝑘𝑁𝑚
Reference: Murthy II, pg. 248.
Example 3: Trapezoidal footing
Fig. 2.5 shows an end column along a property line connected to an interior column by a
trapezoidal footing. The following data is provided:
Column loads:
𝑄1 = 2016 𝑘𝑁, 𝑄2 = 1560 𝑘𝑁
𝑆𝑖𝑧𝑒 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛𝑠: 0.460 𝑚 × 0.460 𝑚
𝐿𝐶 = 5.48 𝑚
Page 49 of 70
𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑞𝑎 = 190 𝑘𝑁/𝑚2
Determine the dimensions “a” and “b” of the trapezoidal footing.
Fig. 2.5: Combined footing (Trapezoidal footing)
Model solution
Moments about centerline of column 1:
(2016 + 1560)𝑥2 = (1560)(5.48)
∴ 𝑥2 = 2.391 𝑚
0.46
𝑥1 = 2.391 + (
) = 2.621 𝑚
2
2016 + 1560 2016 + 1560
𝐴𝑟𝑒𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝐴 =
=
= 18.8211 𝑚2
𝑞𝑎
190
Page 50 of 70
0.46
𝐿 = 𝐿𝑐 + (
) 2 = 5.48 + 0.46 = 5.94 𝑚
2
𝑎+𝑏
𝑎+𝑏
𝐴𝑟𝑒𝑎 𝐴 = (
)𝐿 = (
) 5.94 = 18.8211 𝑚2
2
2
∴ (𝑎 + 𝑏) = 6.3371 𝑚
𝐴𝑟𝑒𝑎 𝐴 = (
(𝑖)
𝑎+𝑏
)𝐿
2
𝑎+𝑏
) 𝐿𝑥1
2
𝐿
𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 = (𝑏𝑙) ( )
2
1
𝐿
𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = (𝑎 − 𝑏) ( ) 𝐿 ( )
2
3
𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑒𝑑𝑔𝑒 = (
𝑎+𝑏
𝐿
1 𝐿2
∴(
) 𝐿𝑥1 = (𝑏𝐿) ( ) + (𝑎 − 𝑏) ( ) ( )
2
2
2 3
6.3371
5.94
1 5.942
(𝑏)(5.94)
(𝑎
∴(
) 5.94𝑥1 =
(
) + − 𝑏) ( ) (
)
2
2
2
3
18.8212𝑥1 = 17.6418𝑏 + 5.8806(𝑎 − 𝑏)
∴ 𝑥1 = 0.9373𝑏 + 0.3125(6.3371 − 𝑏 − 𝑏)
∴ 2.621 = 0.9373𝑏 + 0.3125(6.3371 − 2𝑏)
∴ 2.621 = 0.9373𝑏 + 1.98034 − 0.625𝑏
∴ 2.621 = 0.3123𝑏 + 1.98034
∴ 𝑏 = 2.0514 𝑚
∴ 𝑎 + 2.0514 = 6.3371 𝑚
∴ 𝑎 = 4.2857 𝑚
Reference: Murthy II, pg. 249.
Rectangular combined footing
Page 51 of 70
(𝑖𝑖)
(𝑖𝑖𝑖)
(𝑖𝑣)
Fig. 2.6: Rectangular combined footing
(a) Determine the area of the foundation:
Q1 + 𝑄2
𝑞𝑎𝑙𝑙
Where
Q1 , Q 2 = column loads
𝐴=
𝑞𝑎𝑙𝑙 = 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑠𝑜𝑖𝑙 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
(b) Determine the location of the resultant of the column loads:
From fig. 2.6,
𝑋=
𝑄2 𝐿3
Q1 + 𝑄2
(c) For a uniform distribution of soil pressure under the foundation, the resultant of the column
Page 52 of 70
loads should pass through the centroid of the foundation. Thus,
𝐿 = 2(𝐿2 + 𝑋)
Where L = length of the foundation.
(d) Once the length L is determined, the value of 𝐿1 can be obtained:
𝐿1 = 𝐿 − 𝐿2 − 𝐿3
Note that the magnitude of 𝐿2 will be known and depends on the location the property line.
(e) The width of the foundation is then
𝐵=
𝐴
𝐿
Example 4: Rectangular combined footing
Refer to Fig. 2.6:
Given:
Q1 = 400 𝑘𝑁
Q 2 = 500 𝑘𝑁
qall = 140 𝑘𝑁/𝑚2
L3 = 3.5 𝑚
Based on the location of the property line, it is required that 𝐿2 be 1.5 m. Determine
the sizes of 𝐵 and 𝐿 of the rectangular combined footing.
Model solution
𝐴𝑟𝑒𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝐴 =
Q1 + 𝑄2 400 + 500
=
= 6.43 𝑚2
𝑞𝑎𝑙𝑙
140
𝐿𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑋 =
(500)(3.5)
𝑄2 𝐿3
=
= 1.95 𝑚
Q1 + 𝑄2 400 + 500
For a uniform distribution of soil pressure under the foundation, the resultant of the
column loads should pass through the centroid of the foundation. Thus,
𝐿 = 2(𝐿2 + 𝑋) = 2(1.5 + 1.95) = 6.9 𝑚
Page 53 of 70
𝐿1 = 𝐿 − 𝐿2 − 𝐿3 = 6.9 − 1.5 − 3.5 = 1.9 𝑚
Thus
𝐵=
𝐴 6.43
=
= 0.93 𝑚
𝐿
6.9
MAT FOUNDATION
Mat (raft) foundations are used in the following cases:
1) If the area of isolated and combined footing > 50% of the structure area, because this
means the loads are very large and the bearing capacity of the soil is relatively low,
2) If the bearing capacity of the soil is low,
3) If the soil supporting the structure is:
(a) Expansive soil: Expansive soils are characterized by clayey materials that shrinks and
swells as it dries or becomes wet respectively. It is recognized from high values of Plasticity
Index, Plastic limit and Shrinkage limit.
(b) Compressible soil: It contains a high content of organic material and not exposed to great
pressure during its geological history, so it will undergo significant settlement. So mat
foundation is used to avoid differential settlement,
(c) Collapsible soil: Collapsible soils are those that appear to be strong and stable in their
natural (dry) state, but they rapidly consolidate under wetting, generating large and often
unexpected settlements. This can yield disastrous consequences for structures built on such
deposits.
Page 54 of 70
Section
Plan
Fig. 2.6: Mat Foundation: Flat plate of uniform thickness
Design of mat foundations
Fig. 2.7: Mat Foundation
Page 55 of 70
Step 1: Fig. 2.7 shows a mat foundation of dimensions 𝐿 × 𝐵 and column loads of
𝑄1 , 𝑄2 , 𝑄3 , … … … ..
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑:
𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑 𝑄 = 𝑄1 + 𝑄2 + 𝑄3 + ⋯
𝑒𝑞𝑛. 1
Step 2: Determine the pressure on the soil, q, below the mat points, A, B, C, D, ….by using the
equation:
𝑞=
𝑄
𝐴
±
𝑀𝑦 𝑥
𝐼𝑦
±
𝑀𝑥 𝑦
𝑒𝑞𝑛. 2
𝐼𝑥
Where:
𝐴 = 𝐵𝐿
𝐵𝐿3
𝐼𝑥 =
𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑎𝑏𝑜𝑢𝑡 𝑥 − 𝑎𝑥𝑖𝑠
12
𝐿𝐵 3
𝐼𝑦 =
𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑎𝑏𝑜𝑢𝑡 𝑦 − 𝑎𝑥𝑖𝑠
12
𝑀𝑥 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠 = 𝑄𝑒𝑦
𝑀𝑦 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑦 − 𝑎𝑥𝑖𝑠 = 𝑄𝑒𝑥
The load eccentricities, 𝑒𝑥 𝑎𝑛𝑑 𝑒𝑦 , in the 𝑥 and 𝑦 directions can be determined by using (𝑥 ′ , 𝑦 ′ )
coordinates:
𝑄1 𝑥1′ + 𝑄2 𝑥1′ + 𝑄3 𝑥3′ +. .
𝑄
𝐵
𝑒𝑥 = 𝑥 ′ −
2
𝑥′ =
𝑒𝑞𝑛. 3
𝑒𝑞𝑛. 4
Similarly,
𝑄1 𝑦1′ + 𝑄2 𝑦1′ + 𝑄3 𝑦3′ +. .
𝑦 =
𝑄
𝐿
𝑒𝑦 = 𝑦 ′ −
2
′
𝑒𝑞𝑛. 5
𝑒𝑞𝑛. 6
Page 56 of 70
Example 5: Mat foundation
13.8695 m
13.75 m
9.685 m
10.25 m
Fig.2.8: Mat Foundation
The plan of a mat foundation is shown in fig. 2.8. Calculate the soil pressure at points A, B, C,
D, E, and F. (Note: All column sections are 0.5 𝑚 × 0.5 𝑚). All loads shown are factored loads.
Model solution
𝑞=
𝑄 𝑀𝑦 𝑥 𝑀𝑥 𝑦
±
±
𝐴
𝐼𝑦
𝐼𝑥
𝐴 = 𝐵𝐿 = (20.5)(27.5) = 563.75 𝑚2
9.685 m
Page 57 of 70
𝐵𝐿3 (20.5)(27.53 )
𝐼𝑥 =
=
= 35528 𝑚4
12
12
𝐿𝐵 3 (27.5)(20.53 )
=
= 19743 𝑚4
12
12
𝑄 = 470 + (2)(55) + 600 + (2)(660) + (2) + (2)(1600) + (4)(2000) = 14690 𝑘𝑁
𝐼𝑦 =
𝑀𝑦 = 𝑄𝑒𝑥 ; 𝑒𝑥 = 𝑥 ′ −
𝐵
2
Moments about the left edge:
Take left bottom corner as the origin of a coordinated field:
Let Q locate at the point (𝑥 ′ , 𝑦′ ):
𝑄1 𝑥1′ + 𝑄2 𝑥1′ + 𝑄3 𝑥3′ +. . . . +𝑄𝑛 𝑥𝑛′
∴𝑥 =
𝑄
′
∴ 𝑥′ =
(20.25)(470 + 1600 + 1600 + 600)
86467.5
1
1
[+(10.25)(660 + 2000 + 2000 + 660)] =
[+54530] = 9.685 𝑚
14690
14690
+1275
+(0.25)(550 + 2000 + 2000 + 550)
𝐵
20.5
= 9.685 −
= −0.565 ≈ −0.57 𝑚 (0.57 𝑚 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑜𝑓 𝑐𝑒𝑛𝑡𝑟𝑒)
2
2
Hence, the resultant line of action is located to the left of the centre of the mat.
𝑒𝑥 = 𝑥 ′ −
∴ 𝑀𝑦 = 𝑄𝑒𝑥 = (14690)(0.57) = 8373.3 𝑘𝑁𝑚
Moments about the bottom edge:
𝑦′ =
𝑄1 𝑦1′ + 𝑄2 𝑦1′ + 𝑄3 𝑦3′ +. . . . +𝑄𝑛 𝑦𝑛′
𝑄
(0.25)(470 + 660 + 550)
420
1
1
+(9.25)(1600
+
2000
+
2000)
+51800
∴ 𝑦′ =
[
]=
[
] = 13.8695 𝑚
14690 +(18.25)(1600 + 2000 + 2000)
14690 +102200
+49322.5
+(27.25)(600 + 660 + 550)
𝑒𝑦 = 𝑦 ′ −
𝐿
27.5
= 13.8695 −
= 0.1195 𝑚
2
2
∴ 𝑀𝑥 = 𝑄𝑒𝑦 = (14690)(0.1195) = 1755.455 𝑘𝑁𝑚
∴ the resultant locates at the point (9.685, 13.8695) while the centre is the point (10.25, 13.75)
Page 58 of 70
𝑄 𝑀𝑦 𝑥 𝑀𝑥 𝑦
±
±
𝐴
𝐼𝑦
𝐼𝑥
14690 8373.3𝑥 1755.455𝑦
∴𝑞=
±
±
563.75
19743
35528
𝑞=
∴ 𝑞 = 26.06 ± 0.42𝑥 ± 0.05𝑦
∴ 𝒒 = 𝟐𝟔. 𝟎𝟔 ± 𝟎. 𝟒𝟐𝒙 ± 𝟎. 𝟎𝟓𝒚
Therefore,
𝐴𝑡 𝐴: 𝑞 = 26.06 + (0.42)(10.25) + (0.05)(13.75) = 26.06 + 0.305 + 0.6875 = 31.05 𝑘𝑁/𝑚2
𝐴𝑡 𝐵: 𝑞 = 26.06 ± (0.42)(0) + (0.05)(13.75) = 26.06 ± 0 + 0.6875 = 26.75 𝑘𝑁/𝑚2
𝐴𝑡 𝐶: 𝑞 = 26.06 − (0.42)(10.25) + (0.05)(13.75) = 26.06 − 4.305 + 0.6875 = 22.44 𝑘𝑁/𝑚2
𝐴𝑡 𝐷: 𝑞 = 26.06 − (0.42)(10.25) − (0.05)(13.75) = 26.06 − 4.305 − 0.6875 = 21.07 𝑘𝑁/𝑚2
𝐴𝑡 𝐸: 𝑞 = 26.06 ± (0.42)(0) − (0.05)(13.75) = 26.06 ± 0 − 0.6875 = 25.37 𝑘𝑁/𝑚2
𝐴𝑡 𝐹: 𝑞 = 26.06 + (0.42)(10.25) − (0.05)(13.75) = 26.06 + 4.305 − 0.6875 = 29.68 𝑘𝑁/𝑚2
Plate 1: Mat foundation under construction
Page 59 of 70
Example 5.1: Mat foundation
The mat in example 5 is to be constructed in a soil layer having the following properties:
∅ = 15𝑜 , 𝑐 = 72 𝑘𝑁/𝑚2 , 𝛾 = 20 𝑘𝑁/𝑚3
A footing of dimensions 20.5 𝑚 × 27.5 𝑚, is to be founded at a depth of 2.5 m into this layer.
Assuming a factor of safety of 3, determine the adequacy of the mat against bearing.
Model solution
For φ = 15o ,
N𝑐 = 12.9,
N𝑞 = 4.4,
N𝑦 = 2.5
𝐅𝐨𝐫 𝐚 𝐫𝐞𝐜𝐭𝐚𝐧𝐠𝐮𝐥𝐚𝐫 𝐟𝐨𝐨𝐭𝐢𝐧𝐠:
Ultimate bearing capacity:
B
B
q ult = cNc (1 + 0.3 ) + γzNq + 0.5γBNγ (1 − 0.2 )
L
L
Page 60 of 70
∴ q ult = (72)(12.9) (1 + 0.3
20.5
20.5
) + (20)(2.5)(4.4) + (0.5)(20)(20.5)(2.5) (1 − 0.2
)
27.5
27.5
∴ q ult = 1136.51 + 220 + 436.1 = 1792.61 kN/m2
𝑂𝑣𝑒𝑟𝑏𝑢𝑟𝑑𝑒𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = γz = (20)(2.5) = 50 𝑘𝑁/𝑚2
q safe =
q ult − Overburden pressure
+ Overburden pressure
Factor of safety
∴ q safe =
1792.61 − 50
+ 50 = 630.87 = 630 kN/m2 > 31.05 𝑘𝑁/𝑚2
3
∴ Soil bearing OK
Deep Foundations
Deep foundation can be categorized into three major types. These are:
a) Pile foundations
b) Drilled piers
c) Caisson foundations.
The ground and structural conditions which require the use of the three types are discussed under
each of the sections dealing with the three types of the foundations.
Pile foundations
5.1.1 Introduction
Pile foundations are structural members used to transmit surface loads to lower levels in the soil
mass. They are used when soil beneath the level at an appropriate raft or conventional footing is
too weak or too compressible to provide adequate support to the structure load. The piles have
small cross-sectional area compared to their lengths. The pile materials generally include timber,
steel or concrete. The transfer is by vertical distribution of load along the pile surface and at the
pile end point.
Piles may be used in the following circumstances
Page 61 of 70
a)
To transfer loads to a suitable bearing layer when weak strata is ignored and the load is
transferred to an underlying strong bedrock or compact layer,
b)
To transfer load through the shaft friction when compact layer is very deep and would be
impractical to reach,
c)
To support structures over water where conventional excavation and construction of the
foundation is not possible or very expensive to achieve,
d)
To reduce settlement and in particular differential settlement,
e)
Based on cost. It might prove economical to drive piles down the strata and then build on
top of the piles instead of having to excavate deep layers and then construct ordinary
foundations
f)
In structures which have considerable uplift, horizontal and/or inclined forces. This is
especially true for marine and harbor works,
g)
To increase the bearing capacity by vibration and compaction of granular layers of soil,
h)
In soils where deep excavations would result in damage of existing buildings,
Piles can be distinguished by the function they are intended to perform or by the material and
construction procedures used in their construction. The various types of piles by function are shown on
Figure 2.1. The main function of the piles is to take the loads by end bearing or by friction or by
combination of the two. Other functions exist and two which can be cited here include tension piles and
fender piles. The tension piles take lateral forces in place of traditional retaining walls while fender piles
also referred to as dolphin piles are marine structures principally for taking horizontal loads from vessels
in the docking areas. Fig. 2.2 is presentation of piles by their material and construction procedures.
Page 62 of 70
Soft soil
Soft soil
Soft soil
Friction
resistance
Firm
strata
Hard
strata
End bearing pile
Friction pile
Combination
Impact from floating
object
Tension resistance
Tension pile
Dolphin or fender pile
Figure 2.1: Types of piles by function
5.1.2 Classification of Piles by materials and construction
Piles are constructed in a variety of materials, construction methods and functions. This makes a simple
classification difficult. Notwithstanding these difficulties they are classified in accordance with the pile
materials and method of construction (Figure 2.2). This classification also identifies the pile materials.
The principal materials are timber, concrete and steel.
Page 63 of 70
Types of piles
Driven piles
Large displacement
Preformed. Solid
or hollow tubes
closed at the end
and left in position
Solid
Pre-cast concrete or
Timber. Formed to
required lengths as
units with mechanical
a)
H and
pipe piles
Cast in place formed
by driving closed
tubular sections
and then filling the
void as the tube is
withdrawn
Bored piles
Small displacement
Replacement
Steel sections
H Piles
Open ended tubes
unless a plug forms
during driving
A void is formed
by excavation.
the void is filled
with concrete
sides may be
Supported or
unsupported
Hollow
Steel or concrete
tubes closed at the
bottom. Filled or
unfilled after driving
b)
RC
Precast pile
The supporting may effected permanently
by casing or
Temporarily by casing or drilling mud
(Betonite) or
By soil on a continuous auger
c)
Shell
Pile
Figure 2.2: Principal Types of piles
Page 64 of 70
d) Cast in-situ
tube withdrawn
e) Bored pile
Factors governing the choice of pile type
Vital factors to be taken into consideration when determining the type of pile:
1
2
3
4
The location and type of the structure,
Ground conditions and the position of ground water table,
Durability:
i Timber piles are subject to decay (above the water table) and attack by marine borers,
ii Concrete is liable to chemical attack (in presence of salts and acids in the ground),
iii Steel piles may suffer from corrosion
Overall cost: consider consultants and contractors with local experience.
Large displacement piles
Driven and cast in place piles
Advantages include:
1
2
3
4
Pile lengths are readily adjustable,
Reinforcement is not determined by handling stresses,
Can be driven with a closed end to exclude effects of ground water,
Noise and vibration can be eliminated (cast in place piles).
Disadvantages include:
1 Heave of neighbouring ground surface, which could affect nearby structures or services,
2 Disturbance of the soil, which could lead to development of negative skin friction,
3 Displacement of nearby retaining walls,
4 Concrete cannot be inspected after completion,
5 Light steel sections can be can be damaged during driving,
6 Cannot be driven where headroom is limited
Pile lengths of up to 24 m and pile loads of about 1500 kN are common.
Timber Piles
Timber piles are light, easy to handle and cheap (in some countries). They can be joined together
and can be fitted with driving shoes. The timber should be preserved to prevent decay. Untreated
timber embedded below the ground water table has a long life. If the timber is exposed to
alternating wetting and drying is subject to decay. Pile lengths of up to 20 m and loads of up to
600 kN are usual.
Small displacement piles
Examples include:
Page 65 of 70
Rolled Steel sections, open ended steel tubes. They should be treated against corrosion
Bored and cast-in-place non-displacement piles
Advantages include:
1
2
3
4
5
6
No risk of ground heave,
Length can be readily varied,
Soil can be inspected and compared with site investigation,
Can be installed in long lengths with large diameters,
Reinforcement is not dependent on handling or driving conditions,
Can be installed without appreciable noise or vibration and under conditions of
limited headroom.
Disadvantages include:
1 Boring methods may loosen soil,
2 Difficulties with concreting under water,
3
An inflow of water may cause damage to the unset concrete
Pile lengths of up to 45 m with loads of up to 10,000 kN are not unusual.
BEARING CAPACITY OF A PILE
A pile is supported in the soil by the resistance of the toe to further penetration plus the frictional
or adhesive forces along its embedded length.
∴ Ultimate bearing capacity = ultimate base resistance + ultimate skin friction
∴ 𝐐𝒖 = 𝐐𝒃 + 𝐐𝒔
𝒆𝒒𝒏. 𝟏
Cohesionless soil
Q𝑏 can be calculated can be calculated from Terzaghi’s equation for a square or circular
foundation:
q 𝑢𝑙𝑡 = cN𝑐 + γzN𝑞 + 0.4γBN𝛾
(square foundation)
𝑒𝑞𝑛. 2.1
q 𝑢𝑙𝑡 = cN𝑐 + γzN𝑞 + 0.3γBN𝛾
(circular foundation)
𝑒𝑞𝑛. 2.2
For a cohesionless soil, 𝑐 = 0, hence
q 𝑢𝑙𝑡 = γzN𝑞 + 0.4γBN𝛾
(square foundation)
𝑒𝑞𝑛. 3.1
q 𝑢𝑙𝑡 = γzN𝑞 + 0.3γBN𝛾
(circular foundation)
𝑒𝑞𝑛. 3.2
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Usually, the pile breadth or diameter is small compared with its length and the last terms may be
ignored.
Hence,
q 𝑢𝑙𝑡 = γzN𝑞
(𝑓𝑜𝑟 𝑏𝑜𝑡ℎ 𝑠𝑞𝑢𝑎𝑟𝑒 𝑎𝑛𝑑 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑢𝑛𝑑𝑎𝑡𝑖𝑜𝑛𝑠)
𝑒𝑞𝑛. 3.3
Therefore,
Q𝑏 = p𝑒 N𝑞 𝐴𝑏
𝑒𝑒𝑞𝑛. 4
Where:
A𝑏 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑝𝑖𝑙𝑒 𝑏𝑎𝑠𝑒,
p𝑒 = 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 𝑎𝑡 𝑡ℎ𝑒 𝑙𝑒𝑣𝑒𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒 𝑏𝑎𝑠𝑒.
Meyerhof (1953) suggested that the ultimate skin friction, fs may be calculated from the
equation:
f𝑠 = 𝐾𝑠 ̅̅̅𝑡𝑎𝑛𝛿
𝑝𝑒
𝑒𝑞𝑛. 5
Where:
𝐾𝑠 = 𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑝𝑎𝑠𝑠𝑖𝑣𝑒 𝑒𝑎𝑟𝑡ℎ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝑝𝑒 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑜𝑣𝑒𝑟𝑏𝑢𝑟𝑑𝑒𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑣𝑒𝑟 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒
̅̅̅
𝛿 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒
Hence,
𝐐𝒔 = 𝒇𝒔 𝑨𝒔
𝒆𝒒𝒏. 𝟔
Where:
As = surface area of embededded pile
Typical design values:
𝐾𝑠 = 1.0 (𝑓𝑜𝑟 𝑙𝑜𝑜𝑠𝑒 𝑠𝑎𝑛𝑑𝑠)
𝐾𝑠 = 2.0 (𝑓𝑜𝑟 𝑑𝑒𝑛𝑠𝑒 𝑠𝑎𝑛𝑑𝑠)
𝛿 = 0.75∅ (𝑓𝑜𝑟 𝑑𝑟𝑖𝑣𝑒𝑛 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑝𝑖𝑙𝑒𝑠),
𝛿 = 0.6∅ (𝑓𝑜𝑟 𝑑𝑟𝑖𝑣𝑒𝑛 𝑠𝑡𝑒𝑒𝑙 𝑝𝑖𝑙𝑒𝑠)
𝛿 = 0.9∅ (𝑓𝑜𝑟 𝑐𝑎𝑠𝑡 − 𝑖𝑛 − 𝑝𝑙𝑎𝑐𝑒 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑝𝑖𝑙𝑒𝑠)
∅ = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛
Determination of ultimate pile loads from in-situ tests
Page 67 of 70
Meyerhof (1976) suggested the following formulae to be used in conjunction with the standard
penetration test:
Driven piles:
Fine and medium sands:
𝑞𝑏 =
40𝑁𝐷
𝐵
≤ 400 𝑁 (𝑘𝑁/𝑚2 )
Coarse sand and gravel:
𝑞𝑏 =
40𝑁𝐷
𝐵
≤ 300 𝑁 (𝑘𝑁/𝑚2 )
Bored piles:
Any type of granular soil:
𝑞𝑏 =
14𝑁𝐷
𝐵
𝑘𝑁/𝑚2
Large diameter driven piles:
̅
f𝑠 = 2𝑁
𝑘𝑁/𝑚2
Average diameter driven piles:
̅
f𝑠 = 𝑁
𝑘𝑁/𝑚2
Bored piles:
̅
f𝑠 = 0.67𝑁
𝑘𝑁/𝑚2
Where:
𝑁 = 𝑢𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 𝑏𝑙𝑜𝑤𝑠 𝑎𝑡 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒 𝑏𝑎𝑠𝑒
̅ = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑢𝑛𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 𝑏𝑙𝑜𝑤𝑠 𝑜𝑣𝑒𝑟 𝑡ℎ𝑒 𝑒𝑚𝑏𝑒𝑑𝑑𝑒𝑑 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒
𝑁
𝐷 = 𝑒𝑚𝑏𝑒𝑑𝑑𝑒𝑑 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒
𝐵 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒
Cohesive soils
Q𝑏 for piles in cohesive soils, is based on Meyerhof’s equation (1951):
Q𝑏 = 𝑁𝑐 × 𝑐𝑏 × 𝐴𝑏
𝑒𝑞𝑛. 7
Where:
𝑁𝑐 = 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 (𝑡𝑎𝑘𝑒𝑛 𝑎𝑠 9.0),
Page 68 of 70
𝑐𝑏 = 𝑢𝑛𝑑𝑖𝑠𝑡𝑢𝑟𝑏𝑒𝑑 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒,
𝐴𝑏 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑙𝑒.
𝐐𝒔 is given by the equation:
Q𝑠 = 𝛼 × 𝑐̅ × 𝐴𝑠
𝑒𝑞𝑛. 8
Where:
𝛼 = 𝑎𝑑ℎ𝑒𝑠𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟,
𝑐̅ = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑢𝑛𝑑𝑖𝑠𝑡𝑢𝑟𝑏𝑒𝑑 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑎𝑑𝑗𝑜𝑖𝑛𝑖𝑛𝑔 𝑝𝑖𝑙𝑒
Hence,
Q𝑢 = 𝑐𝑏 𝑁𝑐 𝐴𝑏 × +𝛼𝑐̅𝐴𝑠
𝑒𝑞𝑛. 8.1
𝑻𝒉𝒆 𝒂𝒅𝒉𝒆𝒔𝒊𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓 𝜶
In soft clays, 𝛼 ≥ 1.0,
𝐼𝑛 𝑜𝑣𝑒𝑟𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑒𝑑 𝑐𝑙𝑎𝑦𝑠 𝛼 = 0.45
Action of pile groups
Fig. 2.3: End bearing piles
Piles are usually driven in groups (Fig. 2.3). In the case of end bearing piles (where the spacing
< 5𝑑), the bearing capacity of the group equal to the summation of the individual strengths of
the piles.
Page 69 of 70
Pile groups in cohesionless soils
Pile driving in sands and gravels compacts the soil between the piles. This compactive effect can
make the bearing capacity of the pile group greater than the sum total of individual piles.
Drilled piers
Caisson foundations.
Page 70 of 70