Design of Circular Base Plates
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Dajin Liu1
Abstract: Circular base plates are commonly used for pipe columns, such as pylons in cable-stayed bridges, lighting poles, and electric
power line posts. Although the explicit solutions for rectangular base plates can be found in many textbooks and AISC design procedures
for base plates, a complete design procedure for circular base plates under different loading conditions has not been documented except
for a large eccentric load. In this paper, a complete allowable stress design procedure is presented for circular base plates with axial loads,
small eccentric loads, and moderate eccentric loads. Several design examples are shown for different loading conditions.
DOI: 10.1061/共ASCE兲1084-0680共2006兲11:1共13兲
CE Database subject headings: Plates; Columns; Axial loads; Eccentric loads; Design.
Introduction
Base plates are usually used to distribute column loads to a
supporting concrete foundation. Depending on the column cross
sections, base plates can be rectangular or circular in shape. Rectangular base plates are obviously suitable for steel columns with
I or W sections. Circular base plates are commonly used for pipe
columns, such as pylons in cable-stayed bridges, lighting poles,
and electric power line posts.
Several different loading conditions, such as axial loads and
eccentric loads, are considered for the design of base plates. The
elastic behavior is usually assumed for the bearing stress distribution. For a base plate under axial loads, the design bearing
pressure is uniformly distributed, as shown in Fig. 1. The base
plate size can be easily determined by its design bearing stress
and the allowable bearing stress.
When the axial load is combined with moment, a base plate
could experience small, moderate, and large eccentricities which
equal the moment divided by the axial force. For the case of small
eccentricity, the compressive bearing area is on the full base plate
area and the bearing stress is linearly distributed in a rectangular
shape, as shown in Fig. 2. The maximum bearing stress should
not exceed the allowable bearing stress.
For moderate eccentricity, the compressive bearing area is
greater than one-half of the total base plate area and the bearing
stress is linearly distributed in a triangle shape, as shown in Fig.
3. The maximum bearing stress should not exceed the allowable
bearing stress.
For large eccentricity, the compressive bearing area is less than
one-half of the total base plate area and the compressive bearing
stress is linearly distributed in a triangle shape, as shown in Fig.
4. It is necessary to provide enough anchor bolts to resist the
1
Senior Structural Engineer, Parsons, 10 South Riverside Plaza, Suite
400, Chicago, IL 60606.
Note. Discussion open until July 1, 2006. Separate discussions must
be submitted for individual papers. To extend the closing date by one
month, a written request must be filed with the ASCE Managing Editor.
The manuscript for this paper was submitted for review and possible
publication on January 24, 2005; approved on April 24, 2005. This paper
is part of the Practice Periodical on Structural Design and Construction, Vol. 11, No. 1, February 1, 2006. ©ASCE, ISSN 1084-0680/2006/
1-13–19/$25.00.
tensile component resulting from the moment. The maximum
bearing stress is assumed equal to the allowable bearing stress
and the maximum anchor bolt force should not exceed the allowable anchor bolt load.
The design procedure for rectangular base under different
loading conditions can be easily found in many textbooks and
AISC 共1989, 1990兲 steel design guide series. However, a complete design procedure for circular base plates under different
loading conditions has not been documented except for large eccentric loading 共Liu 2004兲. In this paper, a step by step allowable
stress design 共ASD兲 procedure is developed for circular base
plates with axial loads, small eccentric loads, and moderate eccentric loads. A design example is presented for each loading
condition.
General Assumptions
The following assumptions are used to analyze circular base
plates with different loading conditions:
1. Elastic behavior;
2. The resultant compressive bearing stress is located at the
共center of gravity兲 共c.g.兲 of the compressive bearing area 共Liu
2004兲; and
3. The critical section used to determine the base plate thickness should be based on 0.80 times the outside dimension of
round columns 共AISC 1990兲.
Circular Base Plates with Axial Loads
Under axial load, the bearing pressure is uniformly distributed
between the base plate and the supporting concrete. The base
plate size and thickness can be easily determined based on the
allowable concrete bearing capacity and design bearing stress. A
minimum number of anchor bolts should be provided.
ASD Procedure
1. Determine the maximum allowable stress 共AISC 1990兲
F p = 0.35f ⬘c
冑
A2
艋 0.7f ⬘c
A1
共1兲
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Fig. 1. Circular base plates with axial loads
2.
The required base plate area Acir is
Acir =
␲N2 P
=
4
Fp
共2兲
where P = axial load.
The diameter of the circular base plate, N, is determined
by
N=
3.
冑
4P
␲F p
共3兲
Determine the base plate thickness, based on the elastic bearing stress distribution
tp =
冑
6M pl
Fb
共4兲
where M pl = moment for a 1-in. 共or 1-mm兲 wide strip at the
critical section, i.e., the total moment at the critical section
Fig. 2. Circular base plates with small eccentric loads
divided by the chord at the critical section. Fb = allowable
bending stress, equal to 0.75Fy.
Example 1
Design a circular base plate for an axial load of 200 kips
共889.6 kN兲, as shown in Fig. 5. The outside diameter of the steel
column pipe is 10 in. 共254 mm兲. The ratio of concrete to plate
areas A2 / A1 = 1.0. The allowable stresses for the base plate is
50 ksi 共345 MPa兲. The concrete strength, f ⬘c ⫽3 ksi 共20.7 MPa兲.
The ASD is used in this design example.
1. The maximum design bearing stress is
F p = 0.35共3兲冑1.0 = 1.05 ksi 共7.24 MPa兲
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Fig. 3. Circular base plates with moderate eccentric loads
2.
The diameter of the circular base plate is
N=
冑
4共200兲
= 15.57 in. 共396 mm兲
␲共1.05兲
use 16 in. 共406 mm兲
The actual bearing stress, f 1 is
f1 =
3.
Fig. 4. Circular base plates with large eccentric loads
200
= 0.99 ksi 共6.86 MPa兲
␲共16兲2/4
The critical section is located 4 in. from the centerline of the
base plate. Consider the circular segment at the critical section and calculate the section properties of the segment as
follows:
M pl =
f 1AsegC
B
where d = diameter of the column
␣ = 共␲/180兲共arc cos共4兲/8兲 = 1.047 rads = 60°
M pl = 共0.99兲共39.31兲共1.64兲/13.86
= 4.60 in. kips 共523 mm kN兲
The chord of the segment is
B = 2共8兲sin共60兲 = 13.86 in. 共352 mm兲
and the plate thickness is determined by the following:
and the area of the segment is
Aseg = 0.0087266共16/2兲 共2兲共60兲 − 13.86共4兲/2
2
tp =
冑
6共4.60兲
0.75共50兲
= 0.86 in. 共21.86 mm兲
= 39.31 in.2 共25,360 mm2兲
use 7/8 in. 共22 mm兲
Since ␣ = 1.047艌 ␲ / 4 = 0.785, the c.g. of the segment is
C=
冋
2 sin3 60
16
− cos 60
2 3共1.047 − sin 60 cos 60兲
= 1.64 in. 共42 mm兲
The moment at the critical section is
册
Circular Base Plates with Small Eccentric Loads
For small eccentric loads, the bearing stress occurs on the full
base plate. A linear bearing stress distribution is usually assumed,
as shown in Fig. 2. The maximum bearing stress must not exceed
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Fig. 5. Design Example 1—Circular base plates with axial loads
共1 in.= 25.4 mm, 1 ksi= 6.895 MPa兲
the allowable bearing stress. The resultant for the bearing stress
must equal the axial load. A minimum number of anchor bolts
should be provided.
The bearing stresses are calculated as if the represents the
cross section of a beam. At the edge of the base plate, they are
f 1,2 =
P
Mc
±
Acir
I
Fig. 6. Design Example 2—Circular base plates with small eccentric
loads 共1 ft kip= 1.356 m kN, 1 in.= 25.4 mm, 1 ksi= 6.895 MPa兲
4.
Determine the base plate thickness, based on the elastic bearing stress distribution:
tp =
冑
6M pl
Fb
共8兲
where M pl = moment for a 1-in. 共or 1-mm兲 wide strip at the
critical section, i.e., the total moment at the critical section
divided by the chord at the critical section. Fb = allowable
bending stress, equal to 0.75Fy.
共5兲
where c = N / 2 and I = ␲N4 / 64. Let e = N / 8, Eq. 共5兲 yields f 2 = 0.
Therefore, e = N / 8 is the limit for small eccentricity.
Example 2
ASD Procedure
1. Determine the maximum allowable stress 共AISC 1990兲:
F p = 0.35f ⬘c
2.
3.
冑
A2
艋 0.7f ⬘c
A1
共6兲
Assume a trial base plate size, N;
Calculate the maximum bearing stress at the edge
f1 =
If f 1 ⬎ F p, go to Step 2.
P
Mc
+
Acir
I
Design a circular base plate for an axial load of 200 kips
共889.6 kN兲 and a moment of 500 in. kips 共56.5 m kN兲 as shown
in Fig. 6. The outside diameter of the steel column pipe is 16 in.
共406 mm兲. The ratio of concrete to plate areas A2 / A1 = 1.0. The
allowable stresses for the base plate is 50 ksi 共345 MPa兲. The
concrete strength, f ⬘c is 3 ksi 共20.7 MPa兲. The ASD is used in this
design example.
1. The maximum design bearing stress is
共7兲
F p = 0.35共3兲冑1.0 = 1.05 ksi 共7.24 MPa兲
2.
Assume N = 22 in. 共559 mm兲
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e=
One equilibrium equation is used to determine the unknowns.
The sum of the forces yields:
M 500
N 22
=
= 2.5 in. 共64 mm兲 ⬍ =
8
P 200
8
= 2.75 in. 共70 mm兲
3.
This is the case for small eccentricity.
Calculate the maximum and minimum bearing stresses at the
edges
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500共22/2兲
200
= 1.004 ksi 共6.93 MPa兲
f1 =
2 +
␲共22兲
␲共22兲4
4
64
⬍ 1.05 ksi 共7.24 MPa兲
f2 =
4.
where Rc = resultant for the compressive bearing stress which can
be broken into
500共22/2兲
200
= 0.048
2 +
␲共22兲
␲共22兲4
4
64
R1, the resultant for the compressive bearing stress at the half of
the circle, can be expressed as
R1 = f 1
ksi 共0.33 MPa兲
␣ = 共␲/180兲共arc cos共6.4兲/11兲 = 0.95 rad = 54.42°
冋
冋
M pl = f 2 +
N/2 + 0.4d + C
共f 1 − f 2兲
N
where d = diameter of the column
冋
M pl = 0.048 +
冉
冊
C2 =
册冉 冊
AsegC
B
=
C2 =
⫻共1.88兲/17.89 = 5.37 in. kips 共606 mm kN兲
tp =
冑
= 0.93 in. 共23.53 mm兲
For moderate eccentricities, the bearing stress occurs on the partial base plate. A linear triangle bearing stress distribution is assumed, as shown in Fig. 3. The maximum bearing stress should
not exceed the allowable bearing stress. The resultant for the
bearing stress must be equal to the axial load. A minimum number
of anchor bolts should be provided.
冉冊
2
共2␣2兲 − B2共A − N/2兲/2
冕
冕 冑冉 冊
冏 冉 冊冋冉 冊 册 冏
1
Aseg
册
b
xBxdx
0
b
0
N
2
2
2x
2
1
−
3
Aseg
N
2
2
1
Aseg
− x2dx
3/2 x=b
共14兲
− x2
x=0
冉 冊再冉 冊 冋冉 冊 冉 冊 册
1 2
Aseg 3
N
2
3
−
N
2
2
− A−
N
2
2 3/2
冎
共15兲
ASD Procedure
1. Determine the maximum allowable stress 共AISC 1990兲:
F p = 0.35f ⬘c
use 7/8 in. 共24 mm兲
Circular Base Plates with Moderate Eccentric Loads
共12兲
Let b = A − N / 2, Eq. 共14兲 yields
11 + 6.4 + 1.88
共1.004 − 0.048兲 共57.67兲
22
6共5.37兲
0.75共50兲
冊
C2 = distance from the c.g. of the remaining portion of the bearing
area to the centerline of the base plate. C2 is determined by the
following integration 共Swokowski 1983兲:
=
and the plate thickness is determined by the following:
共11兲
共13兲
册
册
␲N2
8
A − N/2 − C2
Aseg
A
冉冊 冋
= 1.88 in. 共48 mm兲
The moment at the critical section is
冉
N
␲N2 1
− 0.0087266
4 2
2
and the area of the segment is
2 sin3 54.42
22
− cos 54.42
2 3共0.95 − sin 54.42 cos 54.42兲
冊冉 冊
where Aseg = area of the remaining portion of the bearing area
which is determined by
Aseg =
Since ␣ = 0.95艌 ␲ / 4 = 0.785, the c.g. of the segment is
A − N/2 + C1
A
R2 = f 1
B = 2共11兲sin共54.42兲 = 17.89 in. 共455 mm兲
= 57.67 in.2 共37208 mm2兲
冉
where C1 = distance from the c.g. of one-half of the circle to the
centerline of the base plate; and C1 = 2N / 3␲ 共AISC 1989兲 R2, the
resultant for the compressive bearing stress at the remaining portion of the bearing area, can be expressed as
The chord of the segment is
Aseg = 0.0087266共22/2兲2共2兲共54.42兲 − 17.89共6.4兲/2
共10兲
Rc = R1 + R2
O.K.
The critical section is located at 6.4 in. 共163 mm兲 from the
centerline of the base plate. Consider the circular segment at
the critical section and calculate the section properties of the
segment as follows:
C=
共9兲
P = Rc
2.
3.
4.
冑
A2
艋 0.7f ⬘c
A1
共16兲
Assume a trial base plate size, N;
Pick a trial bearing length A; and
Calculate the resultant for the compressive bearing stress at
one-half of the circle:
R1 = f 1
冉
A − N/2 + C1
A
冊冉 冊
␲N2
8
共17兲
Calculate the resultant for the compressive bearing stress at
the remaining portion of the bearing area:
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Pract. Period. Struct. Des. Constr. 2006.11:13-19.
e=
M 600
N 22
=
= 3 in. 共76 mm兲 ⬍ =
= 11 in. 共279 mm兲
2
P 200
2
and
e = 3 in. 共76 mm兲 ⬎
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3.
N 22
=
= 2.75 in. 共70 mm兲
8
8
This is the case for moderate eccentricity.
Assume A = 14.3 in. 共363 mm兲 Determine the section properties of Area 2, as shown in Fig. 7:
␣2 = 共␲/180兲关arc cos共14.3 − 11兲/11兴 = 1.266 rad = 72.54°
B2 = 2共22/2兲sin 72.54 = 20.99 in. 共533 mm兲
and the area of Area 2 is
Aseg = ␲共22兲2/8 − 关0.0087266共22/2兲2共2兲共72.54兲
− 20.99共14.3 − 22/2兲/2兴
= 71.50 in.2 共46,126 mm2兲
The distance from the c.g. of Area 2 to the centerline of the
plate is
冉冊
2
1
兵共11兲3 − 关共11兲2 − 共14.3 − 11兲2兴3/2其
71.50 3
C2 =
Fig. 7. Design Example 3—Circular base plates with moderate
eccentric
loads
共1 ft kip= 1.356 m kN,
1 in.= 25.4 mm,
1 ksi= 6.895 MPa兲
R2 = f 1
5.
冉
= 1.64 in. 共42 mm兲
Determine the distance from the c.g. of one-half of the circle
共Area 1兲 to the centerline of the plate as follows:
C1 =
冊
A − N/2 − C2
Aseg
A
共18兲
4.
If P ⫽ R1 + R2, go to Steps 2 or 3.
Determine the base plate thickness, based on the elastic bearing stress distribution:
tp =
冑
6M pl
Fb
Let f 1 = F p = 1.75 ksi 共12.07 MPa兲, the resultant for the compressive bearing stress at one-half of the circle is
R1 = 1.75
冉
14.3 − 22/2 + 4.67
14.3
冊冋
␲共22兲2
8
册
= 185.35 kips 共824 kN兲
The resultant for the compressive bearing stress at the remaining portion of the bearing area is
共19兲
where M pl = moment for a 1-in. 共or 1-mm兲 wide strip at the
critical section, i.e., the total moment at the critical section
divided by the chord at the critical section. Fb⫽allowable
bending stress, equal to 0.75Fy.
R2 = 1.75
冉
冊
14.3 − 22/2 − 1.64
共71.50兲 = 14.55 kips 共65 kN兲
14.3
R1 + R2 = 185.35 + 14.55 = 199.90 kips
共889 kN兲 ⬇ P = 200 kips 共890 kN兲
Example 3
Design a circular base plate for an axial load of 200 kips
共889.6 kN兲 and a moment of 600 in. kips 共67.8 m kN兲 as shown
in Fig. 7. The outside diameter of the steel column pipe is 16 in.
共406 mm兲. The ratio of concrete to plate areas A2 / A1 = 1.0. The
allowable stresses for the base plate is 50 ksi 共345 MPa兲. The
concrete strength, f ⬘c ⫽5 ksi 共35 MPa兲. The ASD is used in this
design example.
1. The maximum design bearing stress is
F p = 0.35共5兲冑1.0 = 1.75 ksi 共12.07 MPa兲
2.
2共22兲
= 4.67 in. 共119 mm兲
3␲
Assume N = 22 in. 共559 mm兲
5.
The difference is only 0.05%. Therefore, the vertical loads
are balanced.
The critical section is located at 6.4 in. 共163 mm兲 from the
centerline of the base plate. Consider the circular segment at
the critical section, and calculate the section properties of the
segment as follows:
␣ = 共␲/180兲共arc cos共6 . 4兲/11兲 = 0.95 rad = 54.42°
The chord of the segment is
B = 2共11兲sin共54.42兲 = 17.89 in. 共455 mm兲
and the area of the segment is
18 / PRACTICE PERIODICAL ON STRUCTURAL DESIGN AND CONSTRUCTION © ASCE / FEBRUARY 2006
Pract. Period. Struct. Des. Constr. 2006.11:13-19.
Aseg = 0.0087266共22/2兲2共2兲共54.42兲 − 17.89共6.4兲/2
= 57.67 in. 共37208 mm 兲
2
2
Since ␣ = 0.95艌 ␲ / 4 = 0.785, the c.g. of the segment is
C=
冋
2 sin3 54.42
22
− cos 54.42
2 3共0.95 − sin 54.42 cos 54.42兲
册
= 1.88 in. 共48 mm兲
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The moment at the critical section is
M pl =
冋
A − N/2 + 0.4d + C
A
册冉
F1AsegC
B
冊
where d = diameter of the column.
M pl =
14.3 − 4.6 + 1.88
共1.75兲共57.67兲共1.88兲/17.89
14.3
= 8.57 in. kips 共969 mm kN兲
and the plate thickness is determined by the following:
tp =
冑
6共8.57兲
0.75共50兲
= 1.17 in. 共29.75 mm兲
use 1 1/4 in. 共32 mm兲
Conclusions
A complete ASD procedure is developed for circular base plates
under axial loads, small eccentric loads, and moderate eccentric
loads. Several design examples are given to show the implementation of these design procedures.
References
American Institute of Steel Construction 共AISC兲. 共1989兲. Manual of steel
construction-ASD. 9th Ed., Chicago.
American Institute of Steel Construction 共AISC兲. 共1990兲. “Column base
plates.” Steel design guide series, Chicago.
Liu, D. 共2004兲. “Circular base plates with large eccentric loads.” Pract.
Period. Struct. Des. Constr., 9共3兲, 142–146.
Swokowski, E. W. 共1983兲. Calculus with analytic geometry, alternate Ed.,
Prindle, Weber, and Schmidt, Boston.
PRACTICE PERIODICAL ON STRUCTURAL DESIGN AND CONSTRUCTION © ASCE / FEBRUARY 2006 / 19
Pract. Period. Struct. Des. Constr. 2006.11:13-19.