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PDHengineer.comCourse C-2023
Beam Bearing Plates and Column Base
Plates
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Beam-bearing Plates and Column Base Plates by
ASD/LRFD Steel Construction Manual
13th Edition
Column Base Plates
When a steel column is supported by a footing, it is necessary for the column load
to be spread over a sufficient area of the footing. We do this by a steel base plate.
The base plate can be welded or by some type of welded or bolted lug angles.
OSHA requires that you use no less than four anchor bolts for each column base
plate. The lengths and widths of column base plates are usually selected in even
inches, like 8 X 10. The thickness is in 1/8 increments up to 1.25 inches and1/4 inch increments thereafter.
The design bearing strength, cPp, and the allowable bearing strength, Pp/c for
column bases and bearing on concrete are found in J8 of the specification.
Note:
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Geometry:
As you can see:
B = 2n + 0.8bfand N = 2m + 0.95d
Remember, B and N are usually in even inches. Also m and n should be about
equal. Base plates should usually be designed with ASTM A36 material. For most
wide-flange columns subject to axial compression only, a 5/16 inch fillet weld on one
side of each flange will provide adequate strength.
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AISC, Design Guide 1, 2nd edition mentions design procedures for three general
cases of base plates subjected to axial compressive loads.
Example 1
Design a base plate for a W12 X 152 column (Fy=50 ksi) that supports a dead load
of 220 kips and a live of 440 kips. Use and A36 plate (Fy=36 ksi) to cover the entire
area of the 3 ksi concrete pedestal.
This would be Case I: A2=A1
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We see that d=13.7 inches and bf=12.5 inches
LRFD
Pu= 1.2D + 1.6L = (1.2 X 220 kips)+(1.6 X 440 kips) = 968 kips
ASD
Pa= D + L = 220 kips + 440 kips = 660 kips
LRFD
ASD
Determine B and N
LRFD
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ASD
B and N turned out to be the same by LRFD and ASD.
LRFD determine thickness
Pp= (0.6)(0.85)(3 ksi)(24 in X 28 in) = 1028 kips > Pu= 968 kips so its OK
l=maximum of m, n, =7.493 in
Now we have come up with a plate that is 21/4 X 24 X 2-4 that is on top of a
concrete pedestal of the same size, 24 X 2-4.
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ASD determine thickness
Since B and N are the same (24 x 28) and m and n are the same (7.4925 in and
7.00 in). Therefore l will be the same (7.493 inches).
So, in this case it made no difference which you use, LRFD or ASD. Again, you dont
have to use both methods of design. I am only doing this for illustration of how it is
done by both methods. It would be a good idea to check LRFD by using ASD. You
should come up with the same or close to the same answers.
Now let us look at Case II where . This is the case where the concrete
pedestal is much larger than the base plate. This will yield the smallest base plateof all three cases. When A2=A1, you get the largest base plate for all three cases.
Example 2
Design a base plate of A36 steel (Fy=36 ksi) for a W12 X 65 column (Fy=50 ksi)
that supports a dead load of 150 kips and a live load of 300 kips. The concrete has
a compressive strength fc=3 ksi and the pedestal or footing is 9 ft X 9 ft.
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As you can see, d=12.1 in and bf=12.0 in
LRFD
Pu = 1.2(D)+1.6(L) = (1.2 X 150 kips)+(1.6 X 300 kips) = 660 kips
ASD
Pa = D+L = 150 kips + 300 kips = 450 kips
Since
Determine required area, B and N
LRFD
Since d is about equal to bf than B and N should be about equal.
Now we have a square base plate that is 16 inches by 1 foot 4 inches.
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ASD
Again, since d and bfare about equal than B and N should be equal.
Again B and N are equal for LRFD and ASD. Now we check to see if A2>4A1.
A2= (12 in/ft X 9 ft)2= 11,664 in2
4A1= (4)(16 in)2= 1,024 in2
A2 is much greater than 4A1.
LRFD
Pp=0.85fcA1(2)=(0.6)(0.85)(3 ksi)(256 in2)(2)=783 kips>Pu=660 kips so OK
Round up to the nearest 1/4", use 11/2
Now we have come up with a plate, 11/2 X 16 X 1-4 on top of a concrete
pedestal or footing that is 9 ft X 9 ft.
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ASD
B and N are the same (16 X 16) so m and n are the same (2.25 in and 3.20 in).
Since should be 1.0 than will be the same (3.01 inches).
We came up with the same plate 11/2 X 16 X 1-4 on top of a concrete pedestal
or footing that is 9 ft X 9 ft. So in this example, it didnt matter if you use LRFD or
ASD, you come up with the same size plate.
Now, let us look at Case III where A1
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ASD
Pa = D+L = 220 kips + 440 kips = 660 kips
Try A1= BN = (22 in X 26 in) = 572 in2
Since A2 is 3 inches larger on each side of the base plate, than
A2 = (28 in X 32 in) = 896 in2
LRFD
Now A1 = BN = (22 in X 24 in) = 528 in2
Than A2 = (28 in X 30 in) = 840 in2
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Now we have come up with a plate 21
/4 X 22 X 2-0on a concrete pedestalthat is 28 X 30.
ASD
If B and N are the same (22 in X 24 in) = 528 in2 Than A2= (28 in X 30 in) = 840
in2
Since B an N are the same than m, n, n are the same and the maximum is 6.00 in.
Use 2 1/4 inch
Again we came up with the same plate (2 1/4 in X 22 in X 24 in) by LRFD and ASD
and the same pedestal (28 in X 30 in). If Pp was not greater than Pu than you
would need to increase the size if A1 and if Pp/ was not greater than Pa than youincrease A1.
Beam Bearing Plates
Now we will look at beam bearing plates. These are plates placed on top of concrete
or masonry walls and between the wall and the steel beam.
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The dimension B=2k+2n. The dimension k is in the steel shapes table. Use the k des
value in the table. Beam bearing plates are mentioned in section 14 of the Steel
Construction Manual. It is very rare the no bearing plate will be required. In that
case, B=bf. The length of bearing, N, may be determined by the available wall
thickness, clearance requirements or by the minimum required based on local web
yielding or web crippling. N, cannot be greater than the wall thickness. N, should be
greater than or equal to 4 inches. The dimensions B and N should be rounded up to
the nearest inch. The thickness, t, should be in increments of1/8 inch up to 1.25
inches and increments of inch thereafter.
Now, n = (B/2)-k
LRFD ASD
Rn>=Ru Rn/>=Ra
There are three limit states that we need to check. They are all in section J of the
specification. In section J8, we use the same equations for base plates to determine
the bearing strength on concrete.
N+2.5k
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Now we use equation J8-1 when N is the same as the wall thickness and we use
equation J8-2 when N is less than the wall thickness.
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Do determine the required thickness, use the following equations:
LRFD
ASD
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For a beam bearing plate, use only equations J10-5a or J10-5b.
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Example 4
Design a beam bearing plate for a W18 X 71 beam (Fy=36 ksi) that sit on a
reinforced concrete wall (fc=3 ksi) that is 8 inches thick. The bearing plate material
is A36 (Fy=36 ksi). The end reaction are dead load of 30 kips and the live load is 50
kips.
As you can see, d=18.5 in, bf=7.64 in, tw=0.495 in, tf=0.810 in and kdes=1.21 in
LRFD
Ru= 1.2D+1.6L = (1.2 X 30 kips)+(1.6 X 50 kips) = 116 kips
Check B= 10> bf= 7.64so this is OK
A1= BN = 10 X 8= 80 in2
Web local yielding
Rn = (2.5k+N)Fywtw = ((2.5 x 1.21)+8)(36 ksi)(0.495) = 196 kips
Rn = 1.00(196 kips) = 196 kips>Ru = 116 kips so this is OK
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Web crippling
N/d = 8/18.5= 0.432 > 0.2 so use equation J10-5b
Rn=0.75(222 kips)=166 kips > Ru= 116 kips so this is OK
Use a beam bearing plate of11/4 in x 8 in x 10 in
ASD
Ra= D+L = 30 kips+50 kips = 80 kips
N=8 inches
A1= BN = 10 x 8 = 80 in2
Web local yielding
Rn= (2.5k+N)Fywtw= ((2.5 x 1.21 in)+8 in)(36 ksi)(0.495 in) = 196 kips
Rn/ = 196 kips/1.50 = 131 kips > Ra = 80 kips so it is OK
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Web crippling
N/d = 8/18.5 = 0.432 > 0.2 so use equation J10-5b
Rn/ = 222 kips/2.00 = 111 kips > Ra= 80 kips so this is OK
So we came up with the same size plate, 1 1/4 in x 8 in x 10 in. Again it didntmatter if you use LRFD or ASD; you come up with the same size plate. You dont
have to use both methods but it is a good check. This problem can be done in
Excel. I dont see how Engineers get by without Excel.
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BEAM BEARING PLATE
DESIGN
INPUT LRFD
Ru= 116 kips
Beam= W18 x 71required
A1= 75.8 in
d= 18.5 in B= 9.48 in Round 10 in
bf= 7.64 in A1= 80 in
tw= 0.495 in Web Local Yielding
tf= 0.810 in Rn= 196.5 kips
k= 1.21 in Rn= 196.5 kips Ru= 116 kips OK
fy= 36 ksi Web Crippling
N/d= 0.432 > 0.2
Concrete Rn= 221.7 kips
Wall= 8.0 in Rn= 166.3 kips Ru= 116 kips OK
fc'= 3.0 ksi n= 3.79 in
Thickness
Loads t= 1.13 in
Dead= 30 kips
Live= 50 kips Plate 1 1/4" x 8" x 10"
Example 5
A W21 x 62 beam (Fy=50 ksi) is resting on a concrete wall, fc=3 ksi. The dead load
is 26 kips and the live load is 43 kips. Do we even need a bearing plate?
As you can see from the shapes table, d=21 in, bf=8.24 in, tw=0.400 in, tf=0.615 in
and kdes=1.12 in
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LRFD
Ru= 1.2D+1.6L = (1.2 x 26 kips)+(1.6 x 43 kips) = 100 kips
Now if N=thickness of the wall, which is unknown and B = b f= 8.24 in than we need
to determine the thickness of the wall. The thickness of the bearing plate t = tf=
0.615 inches.
Now, as long as the wall is 13 inches thick, we dont need a bearing plate.
ASD
Ra= D+L = 26 kips + 43 kips = 69 kips
Again n=3.00 in
Based on ASD, the wall would need to be 14 inches thick to not use a bearing plate.
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Example 6
A W24 x 68 beam (Fy=50 ksi) is resting on a 8 inch concrete wall, fc=3 ksi. N will
equal the thickness of the wall. The bearing plate material is A36 steel, Fy=36 ksi.
The dead load is 30 kips and the live load is 60 kips. Size the bearing plate.
As you can see from the shapes table, d=23.7 in, bf=8.97 in, tw=0.415 in, tf=0.585in and kdes=1.09 in. This can be done in Excel.
BEAM BEARING PLATE
DESIGN
INPUT LRFD
Ru= 132 kips
Beam= W24 x 68
required
A1= 86.3 in
d= 23.7 in B= 10.78 in Round 11 in
bf= 8.97 in A1= 88 in
tw= 0.415 in Web Local Yielding
tf= 0.585 in Rn= 222.5 kips
k= 1.09 in Rn= 222.5 kips Ru= 132 kips OK
fy= 50 ksi Web Crippling
N/d= 0.338 > 0.2
Concrete Rn= 236.4 kips
Wall= 8.0 in Rn= 177.3 kips Ru= 132 kips OK
fc'= 3.0 ksi n= 4.41 in
Thickness
Loads t= 1.34 in
Dead= 30 kips
Live= 60 kips Plate 1 1/2" x 8" x 11"