Soil Mechanics
Engineering for people who
really "dig" dirt!
Footing Design
Post

Framed Buildings

The first type of footing we will be using is the typical round
concrete footing found in agricultural pole buildings. The Wisconsin Administrative
Building Code requires a minimum footing depth of 48". The maximum footing depth is
generally determined by the local water table and site geology. The
Ultimate Bearing
Capacity (
q
ult
)
of the soil will determine the size and spacing of the required footi
ngs.
Gut Reaction

Soil mechanics is common sense. Big building, big footing. Soft soil,
big footing. All the equation will tell us is "how big?" and "how many?"
Bearing capacity can be calculated with the following equation. (You might want to ta
ke
your glycerin pill before looking at this…)
q
ult
=1.2*cN
c
+
D
N
q
+ 0.6
RN
Don't Get Excited!
We will pick this equation apart one piece at a
time. The first term
q
ult,
is the
ultimate bearing capacity
of the soil.
Essentially, when we find
this quantity we will know how much force
will be required to make the soil
creep
. Creep is a characteristic of
many materials. It is the tendency to deform over time. Creeping
soil beneath the footings of a building will cause the building to
settle.
If the building creeps equally in all locations, there is
generally not a problem. But, when creep is inconsistent, look out!
q
ult
is expressed in kips per square foot. Let's examine that
unit of measure. Kips is short for kilo pounds which means 1,0
00
pounds. Per square foot refers to the area of the bottom of the
footing. What the units are telling us is that it takes a certain
amount of force to make a footing with an area of one square foot
creep. For example if we say that
q
ult
is 5.5 kips/ft
2
we mean that it
would take 5,500 pounds to make a footing with a surface area of 1 square foot creep.
If we think critically about
ultimate bearing capacity (
q
ult
)
,
we realize that this
is the level of force that will cause the soil to creep. Being that
we don't want to take that
chance, we will introduce a
safety factor
. Referencing the
Standard Handbook for Civil
Engineers,
we find that a typical safety factor for footings is in the range of 3 to 4. This
means that the footing will support 3 to 4 tim
es the weight that the structure will place on
it. For example, if we have a structure that places a load of 2 tons on a column that is
supported by a footing, that footing must not begin to creep until we apply 8 tons of
force. This gives us a safety fa
ctor of 4.
The entire goal of this equation is to identify how much weight a footing will
safely support, and then use this information to determine how big each footing must be,
and how many of them we need. The first term on the right side of the equat
ion is a
The leaning tower of Pisa
leans because the soil under
the footings has crept at an
uneven rate.
coefficient that is used to make the formula work for round footings. If we changed the
shape of the footing, we would need to change the coefficient. As long as we keep a
round footing, the first term will always start with 1.2
The next ter
m we need to explore is
cN
c
.
This term is actually made up of two
factors. The first is
c
which refers to the soil's
cohesion
. Cohesion is a property that
measures a soil's tendency to stick together. There are a number of field tests to
determine thi
s, but the recommendation of practicing engineers is to estimate it using the
fact that cohesion is approximately equal to 1/2 of the
unconfined compression strength
of the soil. Unconfined compression strength of soil is commonly measure with a device
ca
lled a penetrometer.
Penetrometers work
by penetrating the
ground with a probe
and measuring the
amount of force
required to maintain a
certain rate of
penetration.
Lets say
that we tested our lot
and the unconfined
compression strength
was 3 kips/ft
2
.
This
tells us that it would take 3000 pounds force to make a footing with a surface area of 1
square foot creep into the soil. Now, remember that cohesion (
c
) is approximately 1/2 of
the unconfined compression strength, or about 1.5 kips. Our equation no
w looks as
follows:
q
ult
= 1.2*1.5N
c
+
D
N
q
+ 0.6
RN
An alternative, often easier, method is to look up the cohesion of the soil in a
chart like the one pictured on the next page. The first thing we need to do is determine
the type of soil on the s
ite. This can be accomplished by referencing an
Agronomic Soil
Map
from the U.S. Department of Agriculture. Soil maps give us a graphic depiction of
the type of soil in an area. In this case, a soil map is not available, but we can use our
knowledge of l
ocal geological history and a sample of the soil to determine its geological
classification. Referencing the chart on the next page, the most logical conclusion we
can draw is that the soil is of the Glacial
–
Till type.
Now that we know the soil type,
we can determine its cohesion by referencing the
soil classification chart. If we look at the soil classification chart, we see that cohesion
ranges from 1,000

4000. It might seem like 1.5 is a totally unreasonable number, but
remember, we are working
in kips and the chart is in pounds, so we will need to divide
the numbers on the chart by 1,000. This gives us a cohesion range of 1

4, in which 1.5
fits nicely.
A Pocket Penetrometer can be used to
test the unconfined
compression strength of soil at the site of excavation.
Slowly, but surely, we are moving forward. Now to define
N
c
(pronounced N su
b c).
N
c
is a soil

bearing capacity factor, whose values relate to the
angle of internal friction,
(the Greek letter psi

pronounced si)
.
These values can be calculated when
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ranges from 35
–
45 degrees. We will
guess in the middle at 40 degrees. Now that we know the friction ang
le, we can reference
the Soil Bearing Capacity chart on the next page to find the value of
N
c
Now our equation looks as follows:
q
ult
= 1.2*1.5*75.3 +
D
N
q
+ 0.6
RN
Next, we can work on the second term. The first part of this
term is
(the Gre
ek letter gamma), which denotes the
weight of the soil per cubic foot. A typical value is 130
pounds per cubic foot. Because we are expressing forces in
kips, we need to be consistent and express 130 pounds as
.130 kips. Our equation now looks like this
:
q
ult
= 1.2*1.5*75.3 + .130 D
N
q
+ 0.6
RN
The following term,
D
(pronounced D sub f) represents the
depth of the footing in feet. As you will recall, we will be
using 4 foot deep footings. Our equation now looks as
follows.
q
ult
=
1.2*1.5*75.3
+ .130*4 N
q
+ 0.6
RN
To finish the second term, we need to define
N
q
(pronounced
N sub q)
.
N
q
is a bearing factor of the soil. Once again we
reference the chart at the left. Assuming a friction angle of
40 degrees, we find that
N
q
has a value of
64.2. Now our
equation appears as:
q
ult
= 1.2*1.5*75.3 + .130*4*64.2+ 0.6
RN
The third term starts out with a constant of
0.6
, which we will leave alone. The next term
as you will recall represents the weight of the soil per cubic foot. We will s
ubstitute
the number .130 kips per square foot for this number. Our equation now appears:
q
ult
= 1.2*1.5*75.3 + .130*4*64.2+ 0.6*.130 RN
There are now only two quantities left to define. R denotes the radius of the
footing we are using. For our purpo
ses, we need to be able to substitute a variety of
values for
R,
so we will not fill this in right now.
The only other quantity is the bearing
factor
N
(N sub gamma)
.
Referencing our friction angle chart, we find the value of
N
to
be 109.4 Our equat
ion appears now as:
q
ult
= 1.2*1.5*75.3 + .130*4*64.2+ 0.6*.130*R*109.4
Multiplying within terms, we find the following:
q
ult
= 135.54 + 33.38+ 8.53R
Adding all of the terms together gives us the ultimate bearing capacity of the footing in
kips.
q
ul
t
= 168.92 + 8.53R
This is an interesting problem, because it creates what is called a circular reference.
Basically, to know the radius of the footing, we have to know
q
ult
but to find
q
ult
we have
to know the radius of the footing. To end this chicken
and egg scenario, we will pick
values for
R
and calculate the corresponding
q
ult
.
This process is known to
mathematicians as
iteration
.
Directions
: Place your calculated values for
q
ult
in the table below.
Diameter of
Footing in
Inches
Convert to
rad
ius (
㈩
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(
ㄲ1
Ca灡c楴渠
䭩灳⁰K爠獱畡re
景潴
晲潭o
景f浵ma)
a晥t Fac瑯t
(
㐩
㐢
㈢
⸱㘷.
ㄷ〮㌴楰猯晴
2
42.5 kips/ft
2
6"
8"
10"
12"
14"
16"
18"
20"
22"
24"
As you may have noti
ced, the values do not change all that greatly even as the footings
become very large. This is because as you will recall,
q
ult
is measured in kips per square
foot. To get the actual bearing capacity of the footing, we need to multiply our value by
the a
rea of the footing. Fill in the table below to calculate the
actual bearing capacity
of
the footings.
Diameter of
footing in
inches
Area of footing
in square inches
(diameter/2)
2
*
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楮煵a牥 e琠
⡩(
2
ㄴ㐩
q
ult
in kips per
square foot
(from chart
above)
Actual bearing
capacity (Area in
ft
2
*
q
ult
in kips/ft
2
)
4"
12.57 in
2
.087
42.5 kips/ft
2
3.71 kips
6"
8"
10"
12"
14"
16"
18"
20"
22"
24"
Looking at the table above, you can now select
the appropriately sized footing for your
building. The first thing you will need is the
Load on Each Footing
value from your
Estimating Foundation Loads
worksheet. Remember that this load is in pounds and we
are working in kips, so you will need to divid
e by 1,000 before we use that value here.
Using the chart above, you can determine the correct footing. The
actual bearing
capacity
should be the smallest value that exceeds your
load on each footing
value.
This saves materials. We know that we can
get away with using small footings because
we have carefully calculated the size needed. Also, our safety factor of 4:1 gives us some
cushion just in case our measurements are not spot

on.
Load on Each Footing ______ pounds
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Dinner Table Conversation Words
Actual Bearing Capacity

The
actual amount of weight that a footing can be expected to support safely.
Calculated by dividing the
ultimate bearing capacity
by the
safety factor
.
Agronomic Soil Map

A map published by the U.S. Department of Agriculture that graphically depicts
the
characteristics of soil in a given area.
Cohesion

The intermolecular attraction by which the elements of a body are held together. (how "sticky"
a soil is)
Creep

The tendency of an amorphous material to experience permanent deformation over long per
iods of
time.
D
(pronounced D sub f)

The depth of the footing in feet. Wisconsin building code limits this value to
not less than 4 feet.
Iteration

A computational procedure in which the desired result is approached through repeated cycles of
suc
cessively better approximation. This is a very technical way of saying "guess and check".
Kips

Kilopounds, or 1,000 pounds.
Load on Each Footing

A value that predicts the amount of weight that the building will place on each
footing. This is calcul
ated on the
Estimating Foundation Loads
worksheet.
N
c
(pronounced N sub c)

soil

bearing capacity factor, dimensionless term, whose values relate to the
angle of internal friction,
.
N
q
(pronounced N sub q)

soil

bearing capacity factor, dimensionless term, whose values relate to the
angle of internal friction,
.
N
(N sub gamma)

soil

bearing capacity factor, dimensionless term, whose values relate to the angle of
internal fr
iction,
.
R

Denotes the radius of the footing in feet.
Rated Bearing Capacity

The actual amount of weight a footing can be expected to safely support. It is
calculated by dividing the ultimate bearing capacity by the safety factor.
Safety Factor

A mathematical algorithm used to compensate for uncertainties in the engineering process.
Ultimate Bearing Capacity

The load per unit area that a soil will support without creeping. Usually
expressed in kips per square foot.
Unconfined Compression S
trength

The force needed to cause an unconstrained column of soil to
crumble, or shrink more than 15% in length, whichever comes first.
(the Greek letter gamma)

The unit weight of the soil expressed in pounds or kips
per cubic foot.
Foundation
Engineering Worksheet
NAME_______________________
Directions:
Using the procedures explained in the Soil Mechanics packet, determine the
ultimate bearing capacities of the following peer and footing type foundations. The
equation is given below.
q
ult
=
1.2cN
c
+
D
N
q
+ 0.6
RN
where:
q
ult
= ultimate bearing capacity,
c
= cohesion of soil (measured with a shearvane

as a rule of thumb, the
unconfined compressive strength is about twice the cohesion of the soil),
= effective unit weight of soil,
D
f
= depth of footing, or distance from ground surface to base of footing,
R
= radius of circular footing,
N
c
,
N
,
N
q
= soil

bearing capacity factors, dimensionless terms, whose values
relate to the angle of internal fri
ction,
.
1.
The illustrated foundation is placed in a sandy, silty soil that was deposited by a slow
moving stream. The unconfined compressive strength is unknown, but you may refer
to the chart in the packet to find the average cohesion of this soil. Th
e density of this
soil has been measured at 120 pounds per cubic foot. What is its ultimate bearing
capacity? Assuming a safety factor of 4, how much weight could we safely place on
this footing?
2.
The illustrated foundation i
s placed in a sandy, silty soil at the bottom of a drained
lake. The unconfined compressive strength is found to be .5 kips. The density of this
soil has been measured at 140 pounds per cubic foot. What is its ultimate bearing
capacity
? Assuming a safety factor of 3, how much weight could we safely place on
this footing?
1'
4'
2'
8"
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