Beam Bearing Plates

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 and 1/4” inch increments thereafter.

The design bearing strength, φcPp, and the allowable bearing strength, Pp/Ωc for

column bases and bearing on concrete are found in J8 of the specification.

Note:

Geometry:

As you can see:

B = 2n + 0.8bf and 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.

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

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

ASD

B and N turned out to be the same by LRFD and ASD.

LRFD determine thickness

φPp= (0.6)(0.85)(3 ksi)(24 in X 28 in) = 1028 kips > Pu= 968 kips so it’s 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”.

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 don’t

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 plate

of 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.

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.

ASD

Again, since d and bf are 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

φPp=φ0.85fc’A1(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.

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 didn’t matter if you use LRFD or

ASD, you come up with the same size plate.

Now, let us look at Case III where A1<A2<4A1.

Example 3

Design a base plate for a W12 X 152 column (Fy=50 ksi) that supports a dead load

of 220 kips and a live load of 440 kips. The plate material is A36 steel (Fy=36 ksi).

It will sit on a concrete pedestal whose compressive strength (fc’) is 3 ksi and is

3 inches wider on each side of the base plate.

As you can see, d=13.7 in and bf=12.5 in.

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

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

Now we have come up with a plate 21/4” X 22” X 2’-0” on a concrete pedestal

that 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 φPp was not greater than Pu than you

would need to increase the size if A1 and if Pp/Ω was not greater than Pa than you

increase 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.

The dimension B=2k+2n. The dimension k is in the steel shapes table. Use the kdes

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 of 1/8 inch up to 1.25

inches and increments of ¼ inch thereafter.

Now, n = (B/2)-k

LRFD ASD

φRn>=Ru Rn/Ω>=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

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.

LRFD

ASD

The next two limit states are found in section J10 of the specification. They are web

local yielding and web crippling but use only the part that applies to the reactions

on a beam. If we have a concentrated load on top of a beam like another beam, we

would use the other part for a bearing plate between the two beams.

We would use only equation J10-3 for a beam bearing plates on concrete.

Do determine the required thickness, use the following equations:

LRFD

ASD

For a beam bearing plate, use only equations J10-5a or J10-5b.

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= 7.64” so 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

φRn = 1.00(196 kips) = 196 kips>Ru = 116 kips so this is OK

Web crippling

N/d = 8”/18.5” = 0.432 > 0.2 so use equation J10-5b

φRn=0.75(222 kips)=166 kips > Ru= 116 kips so this is OK

Use a beam bearing plate of 11/4 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/Ω = 196 kips/1.50 = 131 kips > Ra = 80 kips so it is OK

Web crippling

N/d = 8”/18.5” = 0.432 > 0.2 so use equation J10-5b

Rn/Ω = 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 didn’t

matter if you use LRFD or ASD; you come up with the same size plate. You don’t

have to use both methods but it is a good check. This problem can be done in

Excel. I don’t see how Engineers get by without Excel.

BEAM BEARING PLATE DESIGN

INPUT

LRFD

Ru= 116 kips

Beam= W18 x 71

required A1= 75.8 in²

d= 18.5 in B= 9.48 in Round 10 in bf= 7.64 in A1= 80 in²

tw= 0.495 in Web Local Yielding

tf= 0.810 in Rn= 196.5 kips

k= 1.21 in φRn= 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 φRn= 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

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 = bf= 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 don’t 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.

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=0.585

in 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 in²

d= 23.7 in B= 10.78 in Round 11 in bf= 8.97 in A1= 88 in²

tw= 0.415 in Web Local Yielding

tf= 0.585 in Rn= 222.5 kips

k= 1.09 in φRn= 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 φRn= 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"