1 M. Sc.
Steel Structures
Lecture # 6
By
Dr. Qasim Shaukat Khan
2
COLUMN BASES
• The function of a column base is to safely
transfer column forces to the reinforced
column footing underneath.
• The same concept may also be used to design
bearing plates for the beams resting on
reinforced concrete or masonry.
• The column base may be subjected to only
axial load, axial load plus moment or axial
load plus shear.
3
a) Axial Load b) Axial Load Plus Moment c) Axial Load Plus Shear
Basic Types of Column Bases.
• 4 In the first case, the load is applied through
the centroid of the column at the centroid of
the base plate.
• The column end in this case is considered as a
hinge for the analysis.
• In the second case, either the load acts at some
eccentricity from the column centroid or
moment is also transferred to the foundation
making the column end either a fixed or a
partially fixed end.
• Anchor bolts are needed to resist the
5
developed tension in case of heavy moments.
• Although some shear force is also present in
this second type, a separate design for this
shear force is not generally required.
• The third type is usually more important when
the bracing is connected at the base.
• The shear is resisted through the friction
between the column and the base plate due to
heavy axial loads or bearing in the horizontal
direction by the provision of bolts or shear
lugs.
6
In case of bearing plates with moments, three
different combined bearing stress distributions
are possible depending upon the magnitude of
the eccentricity.
Case I: e N/6
• In Figure, this case represents the application of load in the
kern of the section (middle one - third dimension) and the
bearing stresses will be compressive throughout with no
anchor bolts required for the moment.
• However, minimum anchor bolts may be used for shear and
for extra safety in the horizontal direction.
• The kern or core of the contact area is defined as the area
around the centroid of the footing inside which if load is
applied no tension is developed anywhere in the area.
e
7 Pu
f2
f1
N
N/6
Figure 10.14.Base Plates with Low Eccentricity of Load.
8
A Common Hinged Column Base
Anchor bolts should be provided very close to the centroid and
should be minimum in number. The support at the base plate’s end
is assumed to be hinged. However, providing anchor bolts at the
end will change the support condition from hinge to fix.
9
Front Side View
A True Hinged Column Base.
10Case II: N/6 < e N/2 Without Anchor
Bolts
• For this moderate eccentricity without
anchor bolts ,bearing occurs over a portion
of the plate denoted by
A = 3 (N/2 – e), for equilibrium of applied
and resistive forces and moments.
Greater is the value of e, smaller is A and
the bearing pressure increases quickly.
e
11 Pu
f1
N/6 A
Figure. Base Plates Having Moderate Eccentricity Without Anchor Bolts (Case
2)
12 Case III: Large Eccentricity With Anchor
Bolts
• Figure shows the case of base plate with
large eccentricity, in which anchor bolts are
to be used providing the developed tensile
resistive forces.
• Linear distribution of pressure is still
assumed even at the ultimate stages as an
approximation.
e
Pu
13
A
Tu f1
A
N
Figure: Base Plates Having Large Eccentricity With Anchor Bolts (Case
3)
14
(a) Base With Moderate Moment
(a) Base With High Moment
Figure 10.18.Column Bases With Moments.
15
Axially Loaded Base Plates
• W-section columns are considered here that
are centered over the base plate and the
reinforced concrete footing.
• The following nomenclature will be used
for the design:
tp16 = base plate thickness,
fp = bearing stresses under the base plate,
B = base plate size parallel to flange of the W
section,
N = base plate size parallel to web of the W
section,
0.80 bf = supported base plate dimension parallel to
the flange,
0.95 d = supported base plate dimension parallel to
the depth of the section,
n = overhang of base plate parallel to the flange,
= (B 0.80 bf) / 2,
m = overhang of base plate parallel to the
17
column depth,
= (N 0.95 d) / 2,
fc’ = concrete compressive strength, MPa,
A1 = area of the base plate, B N,
A2 = area of the supporting concrete
foundation that is geometrically
similar to the base plate
c = resistance factor for bearing on
concrete,
= 0.60,
P = ultimate capacity of the concrete in
18p bearing,
A2
= 0.85 f c A1 1.7 f c A 1
A1
when the concrete area is greater than the plate
area, bearing stress is increased due to
confinement provided by the extra concrete,
and Pu = factored axial load from the column.
19
• The base plate is assumed to bend about the critical
sections (perimeter of central 0.85 bf 0.95 d
portion of the base plate) as a cantilever beam
subjected to uniformly distributed bearing stress.
• The critical cantilever span is the greater of m and n.
• The most economical base plate may be designed when
the cantilever lengths in the two directions are equal
and the concrete area is equal to or greater than four
times the plate area.+
•
20If m or n is lesser than either bf /2 or d/2, the
area between the column flanges is to be
checked for bending.
• The Murray-Stockwell method assumes a
maximum permitted bearing pressure over
an H-shaped contact area under the
column cross-section between the plate
and the concrete.
• The second method assumes a uniform
bearing stress distribution and results in a
conservative but easy design.
21
The procedure of second method involves
calculation of equivalent cantilever length,
l, for bending to occur within the flanges as
follows:
1
n = dbf
4
+The critical section used to calculate bending
moment is at 0.95 times the outside column
dimension for rectangular tubes and at 0.80 times
the outside dimension for round pipes.
Procedure For Design of Axially
22
Loaded Base Plate
• The procedure is started with the known
values of the concrete pad size A2, factored
column load Pu, depth of the section d and
the flange width of the section bf.
• If the concrete area is not known in the
start, it is reasonably assumed and then is
revised later if required.
• The required area of the plate is calculated
23 as follows:
A1 = larger of
Pu
2
1 Pu 1
1. or
A2 0.60 0.85 fc A2 / A1 0.60 0.85 fc
Pu
2.
0.60 1.7fc
3. Minimum size, which may be (d + 18) x (bf + 18) mm.
• One way to find the minimum size of the base
24
plate, after calculating A1 is to find x as under and
then B and N as follows:
x = 0.5x(0.95d – 0.80bf)
N = SQRT(A1) + x >= d + 18
B = A1 / N >= bf + 18
• The plate size B N is determined such that m and
n are approximately equal. The sizes are rounded to
the nearest 10 mm multiples.
• The area of the base plate A1 is then evaluated.
• The exact values of the cantilever lengths in
25 the two directions and the equivalent
cantilever length for bending within the
flanges are determined.
• The greatest value out of the three is chosen
as the design cantilever length.
n = (B 0.80 bf) / 2
m = (N 0.95 d) / 2
n = 1
dbf
4
l =max (m, n, n )
26
• The required plate thickness is calculated and is
rounded to upper whole number millimeters up to
10 mm and then in the sequence 12, 15, 18, 20,
etc.
• Required plate thickness is then evaluated as
follows:
2 Pu
tp l
0.90 Fy B N
27
Example 10.3
• Design a hinged base plate for a W360314
column having a total factored load of 5000
kN. Use A36 steel and concrete of 17 MPa
compressive strength.
Solution:
28
Pu = 5000 kN
d = 399 mm
bf = 401 mm
= 17 MPa
Using A2 2.0 A1
A
29 1
= larger of
1 5,000,000
407,790 mm 2
2 0.60 0.85 17
5,000,000
288,351mm 2
0.60 1.7 17
(d + 18) x (bf + 18) = 174,723 mm2
= 407,790 mm2
(say 640 640mm = 409,600 mm2)
A
30 2
= 2 407,790 = 815,580 mm2
(say 910 910 mm = 828,100 mm2)
n = (640 0.80 × 401) / 2 = 160 mm
1
n = 399 401 = 100 mm
4
m = (640 0.95 × 399) / 2 = 131 mm
l = max (m, n, n ) = 160 mm
31
2 5,000,000
tp = 160
0.90 250 (640 x 640)
= 52.7 mm
(say 55 mm thick plate of size 640 640 mm)
32 Base plates with moments
The bending on the base plate with axial load Pu and
Moment Mu may be changed into an equivalent load
Pu acting at an eccentricity.
e = Mu / P u
Three different combined bearing stress distributions
are possible depending upon the magnitude of the
eccentricity.
Base
33
Plates With Moments
Case I: e N/6
e
Pu
f2
f1
N
N/6
Figure 10.14. Base Plates with Low Eccentricity of Load.
34
• This case, as in Figure 10.14, represents the
application of load in the kern of the section
(middle one-third dimension) and the bearing
stresses will be compressive throughout with no
anchor bolts required for the moment.
• However, minimum anchor bolts may be used
for shear and for extra safety in the horizontal
direction.
• The bearing stresses may be calculated by
considering the plate area as cross-section of a
beam subjected to eccentric loads.
Pu M u N /2
35f1 and f2 = BN
B N 3 /12
Pu 6M u
=
BN B N2
Case II: N/6 < e N/2 Without Anchor Bolts
For this moderate eccentricity without anchor bolts
(Figure 10.15), bearing occurs over a portion of the
plate denoted by A = 3 (N/2 – e), for equilibrium of
applied and resistive forces and moments.
This expression is derived based on the fact that A
must be equal to N when e = N / 6 and A must be zero
when e = N / 2.
Greater is the value of e, smaller is A and
36 the bearing pressure increases quickly.
e
Pu
f1
A
N/6
Figure 10.15. Base Plates Having Moderate Eccentricity Without Anchor Bolts.
37 For equilibrium,
1 2Pu
f 1 A B = Pu f1 =
2 AB
Case III: Large Eccentricity With Anchor Bolts
Figure 10.16 shows the case of base plate with
large eccentricity, in which anchor bolts are to
be used providing the developed tensile
resistive forces.
Linear distribution of pressure is still assumed
even at the ultimate stages as an approximation.
e
38 Pu
A
Tu f1
A
N
Figure 10.16. Base Plates Having Large Eccentricity With Anchor Bolts.
Tu = tensile force developed in the anchor
39
bolts,
A = distance of anchor bolt from the
column center,
A = dimension of the portion in
compression,
and
N = centroidal distance of the anchor bolt in
tension from the compression face.
For equilibrium of forces,
Tu + Pu = f1 A B / 2
For equilibrium of moments at the location of anchor bolts:
40
AB A
Pu A + Mu = f 1 N Mu = Pu e
2 3
B 2 BN
f1 A f1 A (Pu A M u ) 0
6 2
F F 2 4 1 Pu A M
f B
6
A =
f1 B / 3
where F = f1 (B N ) / 2
41
After knowing the value of A as above, the
equation for equilibrium of forces presented
earlier gives
Tu = f1 A B / 2 – Pu.
For evaluating this equation, the value of f1 may
be taken equal to the maximum allowable
bearing stress.
The step-by-step procedure for the design of
column base plate subjected to moment may be
summarized as follows:
1.Calculate the eccentricity e from the known
values
42 of Pu and Mu.
2.Determine the allowable bearing stress
under the base plate by selecting some
reasonable A2/A1 ratio.
Fp = 0.85c fc A2 / A1 1.7c fc
where A1 = A B and c = 0.60
3. Calculate minimum area of the plate required,
A1, by assuming zero eccentricity as before.(Pg.
464)
4. Assume a plate size, N x B, reasonably bigger than
43
A1 depending upon the amount of eccentricity of the
load.
Usually N is taken closer to two times the
eccentricity. For no bolt case, N must be larger
than two times the eccentricity.
5. If e N / 6, design by steps 6 and 7, otherwise
move to step 8.
6. Calculate f1, and if f1 > Fp, increase the plate
size and revise the calculations.
7. Calculate the cantilever length m and
calculate bending moment acting on 1-mm strip
of the plate (Mplu) due to the bearing pressure.
Bearing stress at the critical section,
44
f1 f 2
m
fc = f1
N
f c m2
f 1 f c m2
Mplu = Nmm / mm
2 3
4M plu
tp =
0.90 Fy
8. Calculate l (equal to lesser of n and n) just
like an axially loaded base plate with pressure
f1 and find tp as under:
2 f1
tp = l
0.90 Fy
45 9. If e > N / 2, anchor bolts are a must.
However, for e > N / 6 and e < N / 2, anchor bolts
may or may not be used.
The distance A is then determined using the
applicable equation, using f1 equal to Fp.
If its value is closer to N', revise plate size such that
the anchor bolt may develop tension.
f1 BN
F =
2
Pu A M u
f1 B
F F 2
4
6
A = f1 B /3
10.
46
Calculate the value of A1 = A x B and
A2 = (A + extra dimension on one side) x width of
pedestal and check for the assumption of A2 / A1 in
step 2.
11.Determine the total force in the anchor bolts. If
the values are unreasonable, revise the plate size.
12. Calculate m, Mplu and tp as follows:
m = (N x 0.95 d) / 2
47
Moment From Plate Bearing Side
Case I: A ≥ m
f1
fc = f1 m
A
f c m2
f1 f c m2
Mplu = Nmm / mm
2 3
Case II: A < m
f1 A A
Mplu = (m ) Nmm / mm
2 3
48 Moment From Anchor Bolt Side
The plate width over which the anchor bolt force is
spread is assumed based on the load spreading out at
45o angle towards the critical plane (Figure 10.17).
This width is equal to the distance from the bolt to the
critical section for each bolt (l 1 ) plus smaller of this
distance and the bolt edge distance (w).
Total effective width for the anchor bolts:
We = l 1 + smaller of l 1 and w
49
l1
l1
w 45o
Figure 10.17. Spreading of Anchor Bolt Force on the Bearing Plate.
50 Mplu = (Tu l 1 ) / We
The maximum Mplu value out of the above two
values is used to calculate the required plate
thickness.
4M plu
tp =
0.90 F y
13. The plate thickness is taken as the larger
value out of the calculated values.
51
Design Of Hooked Anchor Bolts
For the design of anchor bolts, AISC Specification
refer to the Appendix-D of ACI – 318 (2005) Code.
Anchor bolts are required for all base plates to
prevent accidental column overturning and to
resist moments and uplift.
The tensile load is resisted by bond between
anchor bolts and concrete and the hook at the
bottom.
The lower hook may also consist of a welded nut but
its design procedure will become different.
a)
52
Select the diameter of the anchor bolt
arbitrarily, or as follows ignoring the shear in the
anchor bolts. If shear is also significant, Fu is
reduced accordingly.
Tu
/4 d 2 =
0.75t Fu Nbt
where Nbt = number of bolts on the
tension side
and t = 0.75
Tu
d =
0.44 Fu Nbt
53
b) Determine the bolt tensile capacity
Tbu = 0.75 t Fu Ag where t = 0.75.
c)Check that sufficient tensile capacity is provided
for the moment.
d)Calculate the required hook length according to
the ACI Code provisions.
54
Example 10.4:
A W 200 x 26.6 column is assumed fixed at the bottom
for the analysis and design.
The service axial dead and live loads are 100 and 165
kN, respectively. The bending is about the strong axis
and the service dead and live load moments at the
bottom of the column are 20 and 35 kN-m, respectively.
The ratio of the concrete column to base plate area (A2 /
A1) is to be kept closer to a value of 2.0. Design the
base plate for the column using A36 steel for both the
base plate and the anchor bolts and concrete having fc'
equal to 20 MPa.
Solutio
55
n:u = 1.2 (100) + 1.6 (165)
P = 384 kN
Mu = 1.2 (20) + 1.6 (35) = 80 kN-m
e = Mu / Pu = (80 / 384) 1000 = 208 mm
Fp = 0.85 0.60 20 2
1.7 0.60 20
= 14.42 MPa
A1,min = Pu / Fp = 384 1000 / 14.42
= 26,630 mm2
Assuming a base plate size of 350 350 mm:
N / 6 = 58.3 mm : N / 2 = 175 mm
e = 208 mm
Because e > N / 2, anchor bolts are definitely
56
required. Assume that the anchor bolts are
placed at an edge distance of 40 mm.
N Dist. Of the anchor bolt = 350 – 40 = 310 mm
from the comp. face
A Dist. Of the anchor bolt = 350 /2 – 40= 416 mm
from the col. centre
Let f1 = Fp =
14.42 MPa
F = f1 (B N )/2
= 14.42 350 310 / 2
= 782,285 N
14.42 350
782,285 782,2852 4 384,000 135 80 10
6
57 6
A =
14.42 350 / 3
= 221 mm (sufficiently lesser than N)
Tu = f1 A B / 2 – Pu
Tu= 14.42 221 350 /2 – 384,000
Tu=173,694 N (or 173.7 kN)
reasonable for the bolt sizes available
For the column section W200 x 26.6, d = 207 mm
bf = 133 mm.
Calculate l (equal to lesser of n and n) justlike
58 an axially loaded base plate with pressure f1 and
find tp as under:
2 f1
tp = l
0.90 Fy
n = (350 0.80 × 133) / 2 = 121.8 mm
1
n = 4 207 133 = 41.5 mm
l = max (n, n ) = 121.8 mm
2 14.42
tp = 121.8
0.90 250
= 43.61 mm
59 Moment From Plate Bearing
Side
m = (N – 0.95 d) / 2
= (350 – 0.95 207) /2 = 76.7 mm
A>m
fc = 14.42 – (14.42 / 221) 76.7 = 9.42 MPa
Mplu = 9.42 76.72 / 2 + (14.42 9.42) 76.72 / 3
= 37,513 N-mm/mm
Moment
60 From Anchor Bolts Side
Distance of the anchor bolt from critical section,
l1 = m – 40
= 36.7 mm
Width on which anchor bolt force of one side is
spread = (36.7 + 36.7)
= 73.4 mm
Mplu = total moment / total width, for two
anchor bolts on tension side
= (173,694 36.7) / (2 73.4)
= 43,424 N-mm/mm
61 Mplu = larger of the above two values
= 43,424 N-mm/mm
4M plu
tp =
0.90 F y
4 43,424
= 0.90 250
= 27.8 mm
say 45 mm
(larger of all of above calculated values)
Use a 350 350 45 mm base plate
