Column base plates_prof_thomas_murray

COLUMN BASE PLATES
- DESIGN TO ERECTION –
Tom Murray
Emeritus Professor
Virginia Tech
Blacksburg, VA

1

Types of Column Base Plates
Compressive Axial Loads

Base Plates with Small Moments

Tensile Axial Loads

Base Plates with Large Moments

3

AISC Specification
PERTINENT SECTIONS AND TABLES
Chapter J Design of Connections
Table J3.2 Nominal Strength of
Fasteners and Threaded Parts
J8. Column Basses and Bearing on
Concrete

Chapter M Fabrication and Erection
M2.2 Thermal Cutting
M2.8 Finish of Column Bases
M4.4 Fit of Column Compression Joints
and Base Plates
5

Types of Column Base Plates
TOPICS
• Base Plate and Anchor Rod Material and Details
• Base Plates for Axial Compression
• Base Plates for Tension
• Base Plates for Axial Compression and Moment
• Shear Anchorage
• Column Erection Procedures
• Result of Poor Anchor Rod Placement
6

Base Plate and Anchor Rod
Materials and Details

7

Base Plate Material, Fabrication and Finishing
Base Plate Material
• Typical A36
• 1/8” (3mm) increments to 1-1/4 in (32 mm) then ½ in. (6 mm)
• Minimum thickness ½ in. (13 mm), typical ¾ in. (20 mm)
Fabrication and Finishing
• Plate is thermally cut
• Holes drilled or thermally cut
• Specification Section M2.2 has requirements for thermal cutting
• Finishing: Specification Section M2.8
< 2 in. (50 mm) milling not generally required
2 – 4 in.(50 – 100 mm) straightened by pressing or milling
> 4 in. (100 mm) milling required
Only mill where column bears
• Bottom surface with grout need not be milled
• Top surface not milled if complete-joint-penetration (CJP) welds
8

Base Plate Material, Fabrication and Finishing
Base Plate Welding
• Fillet welds preferred up to ¾ in. (20 mm) then groove welds
• Avoid all around symbol
• Weld for wide-flange columns subject to axial compression
Weld on one side of column flanges
to avoid rotating the section:

• HSS subject to axial compression: Weld flats only
• Shear, Tension, Moment, and Combinations
Size weld for forces not all around
No weld at fillets of wide-flange columns
Must weld at corners of HSS if tension or moment
• Shop welded preferred
• Very heavy base plates may be field welded after grouting
9

Anchor Rod Materials and Details
Anchor Rod Material

• Preferred ASTM F1554 Gr 36 except when high tension and or
moment.
10

Anchor Rod Material
• Unified Course (UNC) threads
• Standard Hex Nuts permitted but Heavy Hex nuts required for
oversize holes and with high strength anchor rods.
• Hooked anchor rods have very low pull-out strength but OK for
axial compression only.
• Headed rods or rods with nuts are preferred.

Headed Rods

Rods with Nuts
11

Anchor Rod Material
• Use of plate washers at bottom of rods can cause erection
problems and should be avoided.
• Drilled-in epoxy anchor rods may be used for light loads
• Wedge-type anchors should not be used because of potential
loosing during erection.

12

Anchor Rod Holes and Washers
• Oversize holes required because of placement tolerances.
• Recommended hole and washer sizes are in Design Guide 1
(Table 2-3) and AISC 14th Ed. Manual (Table 14-2).

• Smaller holes may be used when axial compression only.
• Plate washers required when tension and/or moment.
13

Anchor Rods Material and Details
Anchor Rod Sizing and Layout
• Recommendations
Use ¾ in. (22 mm) minimum diameter.
Use Grade 36 rod up to about 2 in. (50 mm) diameter.
Thread at least 3 in. (75 mm) beyond what is needed.
Minimum ½ in. (12 mm) distance between edge of hole
and edge of plate.
Use symmetrical pattern whenever possible.
• Provide ample clearance for nut tightening.
• Coordinate anchor bolt placement with reinforcement location.

14

Occupational Safety and Health Administration (OSEA)
OSEA Safety Standards for Steel Erection (2008)
• Minimum of four anchor rods.
• Base plate to resist eccentric gravity load of 300 lb (1,300 N) 18
in. (450 mm) from extrem outer face of column in each direction.
• 300 lb (1,300) represents weight of iron worker with tools.
• Exception: Post-type columns weighing less than 300 lb.
18 in.

Note: Smallest of anchor rods on
a 4 in. by 4 in. (100 mm by 100 mm)
pattern is generally sufficient.

300 lb

15

Column Base Plates
DESIGN CONSIDERATIONS
• Base Plate Bending Strength
• Concrete Crushing
• Concrete Anchorage for Tension Forces
• Shear Resistance
• Anchor Rod Tension Design
• Anchor Rod Shear Design
• Anchor Rod Embedment
• Constructability
16

Base Plates for
Axial Compression

17

Design for Axial Compression
Column Base Plate Design Limit States
• Base Plate Bending

18

Column Base Plate Bending

fpu = Ru /(BxN) = pressure due to reaction
Let m’ = max(n and m)
Mu= fpu (1) (m’2/2 < φMp
φMp = 0.9 Fy Zx = 0.9x 1 x tp2/4) Fy
Substituting:

t p , min

=

2 f pu ( m ' 2 )
0 .9 F y

19

• Lightly Loaded Base Plate
• Required concrete bearing area is less than bfdc.
d

bf

Bearing Area
20

• Lightly Loaded Base Plate
Replace m′ with l = max {m, n, λn′}, where
′
′
2 X
λ =
≤ 1 .0
1+ 1-X

1
n′ =
4

db f

2f pu l 2

t p ,min =
 4db f  Pu
0.9Fy

X=
 (d + b )2  φPp
f


Reference: Thornton, W.A., Design of Base Plates for Wide Flange
Columns – A Concatenation of Methods, AISC Engineering Journal,
21
Fourth Quarter, 1990.

HSS Sections
• Bend lines.

0.95h

0.95d

0.80d

22

Concrete Crushing
Specification Section J8. Column Bases and Bearing on Concrete
φc = 0.65
(a) On the full area of a concrete support
Pp = 0.85fc′A1
(J8-1)
(b) On less than the full area of a concrete support
(J8-2)
where Pp = 0.85fc A1 A 2 / A1 ≤ 1.7fc A1
′
′
fc’ = specified compressive strength of concrete, ksi (Mpa)
A1 = area of steel bearing on concrete, in.2
A2 = area of the portion of the supporting surface that is
geometrically similar to and concentric with the
loaded area, in.2
Note: The limit < 1.7f’c is equivalent to A2/A1 < 4
23

Design for Axial Compression -- Example

o

24”

18”

Ex.: Determine if the base plate shown is adequate.
Column: W10x33 d = 9.73 in. bf = 7.96 in.
PL 1-½ x 18 x 1’-6” A36
Concrete Pedestal:
24 in. by 24 in. (assume square)
fc′′=3.0 ksi
Pu = 250 kips
o
o

o
18”
24”

24

Concrete Crushing
A1 = 18 x 18 = 324 in2 (Plan area of base plate)
A2 = (24)(24) = 576 in2 (Plan area of concrete pedestal)
A2/A1 = 576/324 = 1.78 < 4

o

o

o

24”

= 0.65 (0.85 × 3.0 ) (324 ) 1.78
= 716 k > 250 k OK

o

18”

′
φPp = φ 0.85 f c A 1

A2
A1

18”
24”

25

Plate Bending
n = 5.82 in. m = 4.38 in.
7.96”

n ′ = (1 / 4 ) db f
= (1 / 4 ) 9 .73 x 7 .96
= 2 .20 in .
 4db f  Pu

X=
2  φP
 (d + b )
f

 p
 4x9.73x7.96  250

=
 (9.73 + 7.96 )2  716 = 0.333



4.38”
9.24”

9.73”

18”

4.38”

5.82 6.37 5.82
18”

26

Plate Bending
n = 5.82 in. m = 4.38 in.
7.96”

n ′ = (1 / 4 ) db f
= (1 / 4 ) 9 .73 x 7 .96
= 2 .20 in .
 4db f  Pu

X=
2  φP
 (d + b )
f

 p
 4x9.73x7.96  250

=
 (9.73 + 7.96 )2  716 = 0.333



4.38”
9.24”

9.73”

18”

4.38”

5.82 6.37 5.82
18”

27

Plate Bending
2 X
2 0 . 333
λ =
=
1+ 1-X
1 + 1 - 0 . 333
= 0 . 635 ≤ 1 . 0
λ n ' = 0 . 635 x 2 . 20 = 1 . 40 in .

7.96”

4.38”

9.73”

9.24”

l = max {m, n, λn′}
′
= max {4.38,5.82,1.40} = 5.82 in.

4.38”

fpu = Ru /BN = 250/(18x18)
= 0.772 ksi

18”

5.82 6.37 5.82
18”

28

Plate Bending

tp =

2f pu l 2
0.9Fy

=

2 x 0.772 x 5.82 2
0.9 x 36

7.96”

4.38”

= 1.27 in . ≤ 1.5in . OK
9.24”

9.73”

18”

4.38”

PL 1-½ x 18 x 1’-6” A36
is Adequate.

5.82 6.37 5.82
18”

29

Variation of φPn with Base Plate Thickness with Increasing Axial Force

φPn

m or n

c
c

c

BN = bfd
Required tp
30

Was the base plate too strong?

31

Base Plates for
Axial Tension

32

Design for Tensile (Uplift) Forces
Connection Limit States
• Anchor Rod Tensile Strength
• Concrete Pull-out Strength
• Concrete Anchorage Tensile Strength

33

Anchor Rod Tensile Strength
• Tensile Strength
φTn = φ FntAb
where φ = 0.75
Fnt = 0.75Fu
Fu = specified minimum tensile strength of anchor rod
= 58 ksi (300 MPa) for Gr 36
= 75 ksi (400 MPa) for Gr 55
= 125 ksi (650 MPa) for Gr 125
Notes: Ab is the nominal area of the anchor rod = πd2/4.
0.75 in 0.75Fu account for rupture (tensile) area.
See AISC 14th Ed. Manual Table 7-17 for actual areas.

34

Ex.

Determine the design strength,φTn, for the limit state
of anchor rod tension rupture. 4 - ¾ in. (20 mm)
diameter, Grade 36 anchor rods

φTn = 4 x 0.75 (0.75Fu) Ab
= 4 x 0.75 (0.75x58)(π0.752/4)
= 4 x 14.4 kips (4 x 53 kN)
= 57.6 kips (212 kN)

35

Concrete Pullout Strength
• Provisions in ACI 318-08, Section D5.3
• Design pullout strength of headed anchor
rod or anchor rod with nut:
φNp = φ ψ4Abrg8f’c
where φ = 0.70
ψ4 = 1.4 for no cracking, 1.0 if cracking exists
Abrg = net bearing area of the anchor rod head or nut, in2
f’c = specified compression strength of concrete, psi
Notes: Hooked anchor rods are not recommended with tensile
loadings.

36

Concrete Pullout Strength
• AISC Design Guide 1 Table for Strengths with Heavy Hex Nuts

Note: kN = 4.448 x kips

N/mm2 = 6.895x10-3 x psi
37

Concrete Breakout Strength
• Provisions in ACI 318-08, Section D4.2.2

Full breakout cone

Breakout cone for group anchors

Breakout cone near edge
38

Base Plate Bending Strength
• AISC Design Guide 1 recommended bend lines.
g2

g1

Alternate Critical Section is the
full width of the base plate.

39

Base Plate Bending Strength
• Other possible bend lines.

Critical
Section

Critical
Section
bf + 1 in.

Similar to Extended End-Plates

For all cases:
Define: x = distance from center of anchor bolt to critical section
w = width of critical section
Tu = force in anchor bolt
4Tu x
t p ,min =
φMn = 0.9Fywtp2/4 = Mu = Tux
0.9Fy w
40

Base Plates for Axial
Compression and Moment

41

Design for Axial Compression and Moment
Connection Limit States
• Anchor Rod Tensile Strength
• Concrete Pull-out Strength

O

• Concrete Anchorage Tensile Strength
Cases
• T = 0 (Relatively small moment)
• T > 0 (Relatively large moment)
42

Limit for T = 0
∑Fy = 0 with T=0 Base Plate is B by N in plan.
Pr = qY = fpBY
∑Mo = 0
Pre = qY2/2 – PrN/2
Then
e = N/2 – Y/2 of if fp < fp,max
with
′
′
fp ,max = 0.65x0.85fc A 2 / A1 ≤ 0.65x 1.7fc
then
ecrit = N/2 –Pr/(2Bfp,max)
Therefore

If e < ecrit, T =0 and Y = N-2e
43

For T > 0
e > ecrit and fp = fp,max
∑Fy = 0
Pr – T – fp,maxBY =0
∑Mo = 0
T(N/2+f) + Pr(N/2) – Pre – fp,maxBY2/2 = 0
Solving the resulting quadratic equation
Note: If the quantity in the
radical is less than zero, a
larger base plate is required.

Y =(f + N / 2) − (f +

N 2 2Pr (e + f )
) −
2
fp ,max B

T = fp ,max BY − T

44

Ex. 1. Determine required plate thickness and anchor rod diameter.
Given: Pu = 376 kips Mu = 940 in-kips
W12x96 A992
N = 19 in. B = 19 in. A2/A1 = 1 < 4 bf = 12.2 in. d = 12.7 in.
fc’ = 4.0 ksi Fy = 36 ksi
f = 8 in.
Solution:
′
fp ,max = 0.65x0.85fc A 2 / A1
= 0.65x0.85x4.0 1.0 = 2.21 ksi

e = Mu/Pu = 970/376 = 2.50
= 19.0/2 – 376/(2x19.0x2.21)
= 5.02 in.
e < ecrit → T = 0 and Y = N-2e = 19.0 – 2x 2.50
= 14.0 in.
45

W12x96 A992
f = 8 in.
Solution Continued:
fp = Pu/(NY)
= 376/(19.0x14.0) = 1.41 ksi
m = (N-0.95d)/2 = (19.0-0.95x12.7)/2 = 3.47 in.
m = (B-0.80bf)/2 = (19.0-0.80x12.2)/2 = 4.62 in.
tp =

2f pu n 2
0.9Fy

=

2 x1.41x 4.62 2
= 1.36 in .
0.9 x 36

Use PL 1½ x 19 x 1ft 7 in. A36
4 – ¾ in. Anchor Rods x 12 in. F1554 Gr 36
w/ heavy hex nuts at bottom.

46

W12x96 A992
f = 8 in.
Solution:
′
fp ,max = 0.65x0.85fc A 2 / A1
= 0.65x0.85x4.0 1.0 = 2.21 ksi

e = Mu/Pu = 2500/376 = 6.65
= 19.0/2 – 376/(2x19.0x2.21)
= 5.02 in.
N
2 P (e + f )
e > ecrit → T > 0 and Y =(f + N / 2) − (f + )2 − r
2

fp ,max B

47

W12x96 A992
f = 8 in.
Solution Continued:
N 2 2Pr (e + f )
Y = (f + N / 2 ) − (f + ) −
2
fp ,max B
19.0 2 2x 376(6.65 + 8.0)
) −
= (8.0 + 19.0 / 2) − (8.0 +
2
2.21x19.0
= 10.9 in.

fp = Pu/(NY)
= 376/(19.0x10.9) = 1.82 ksi

48

W12x96 A992
f = 8 in.
Solution Continued:
m = (N - 0.95d)/2 = (19.0-0.95x12.7)/2 = 3.47 in.
n = (B - 0.80bf)/2 =(19.0-0.80x12.2)/2 = 4.62 in.
tp =

2f pu m 2
0.9Fy

=

2 x1.82 x 4.62 2
= 1.54 in .
0.9 x 36

Need to check tensions side.

49

W12x96 A992
f = 8 in.
Solution Continued:
T = fp,max BY – Pu
= 2.21x19.0x10.9 – 376 = 80.4 kips
Try 2 – Anchor Rods each Side
Required plate thickness:
X = f – 0.95d/2 = 8.0 -0.95x12.7/2
= 1.96 in.
x
w = bf + 1.0 = 12.2 + 1.0 = 13.0 in.
f
4T x
4 x 80 .4 x1.96
t p ,min =
=
= 1.22 in .
0.9Fy w
0.9 x 36 x13 .0
Use PL 1½ x 19 x 1ft 7 in. A36 50

W12x96 A992
f = 8 in.
Solution Continued:
Trod = 80.4/2 = 40.2 kips per anchor rod
Try 1 -1/4 in. diameter
φTn = 0.75 (0.75Fu) Ab
= 0.75 (0.75x58)(π1.252/4)
= 40.0 kips ≈ 40.2 kips Say OK
Check Pullout:
From AISC Design Guide 1, Table 3.2, φTn = 50.2 kips OK
51

Shear Anchorage
of Base Plates

52

Shear Anchorage of Base Plates
Transferring Shear Forces (ACI 318-08 and ACI 349-06)
• Friction between base plate and grout or concrete surface.
φVn = φμPu < φ0.2fc’Ac or φ800Ac
where φ = 0.75
μ = coefficient of friction = 0.4
Pu = factored compressive axial force
Ac = plan area of base plate, BN
• Bearing of the column and base plate and/or shear lug against
concrete.

φPu,brg = 0.55fc’Abrg
where Abrg = vertical bearing area
53

Shear Anchorage of Base Plates
Transferring Shear Forces (ACI 318-08 and ACI 349-06)
• Shear strength of the anchor rods.
Not recommended because of oversize holes in base plate.
• Hairpins and Tie Rods

54

Column Erection
Procedures

55

Column Erection Procedures
Anchor rods are positioned and placed in the concrete before it
cures. Plastic caps are placed over the anchor bolts to protect
laborers.

56

Concrete block-outs may be made so that the base plate and
anchor bolts are recessed below floor level for safety and
aesthetics.

57

Slotted holes are allow for difference in standard field tolerances
between concrete and steel.

58

The column is leveled by adjusting the anchor bolt nuts below the
plate. Grout will be placed under and around the base plate to
transfer axial forces to the concrete below.

59

Grout will be placed under and around the base plate to transfer
axial forces to the concrete below. At least it should be.

60

Result of Poor
Anchor Rod Placement

The following slides are courtesy of Fisher &
Kloiber- AISC Field Fixes presentation.
62

Anchor rods in the wrong location.

63


64


65

Anchor rods too short.

66

Anchor rods too long.

67

Shop rework fo column and base plate.

68

Anchor rods

69

Poor Anchor Rod Placement
Anchor bolts run over by a crane.

70

Poor Anchor Rod Placement
Anchor bolts run over by something.

71

Our next IMCA Manual, due to come out shortly, will state in the Code
of Standard Practice, that in addition to the requirement that the
contractor responsible for placing the anchors in site, present a drawing
showing the as-placed location of the anchors, that the supplier of the
structure provide the necessary templates for this purpose. They are to be
so designed as to insure the correct position, both vertically (including
the length of the anchor extending above the concrete surface) and
horizontally, show the location of the building axis relative to the anchor
group and leave permanent marks of the axis in the concrete surface. The
price of the templates is recommended to be the average price of the steel
structure. (We save so much on the cost of erection that we would be glad to
give the templates away free, but don't tell my clients.)
We have been using this method over the past 5 years in all
of our jobs. We are delighted in not having had a single problem in placing
the columns, the structures are all self-plumbing, the bolts all easily
inserted and the time for erection greatly shortened. On a job we are now
doing, from the date our bid was accepted, it took 7 weeks to design,
fabricate, erect the steel structure and install 20,000 sq. ft. of decking,
with shear anchors hand welded, for a three story school building, working
one 11 hr shift both at the shop and the site. A second similar 30,000
sq. ft. building will be finished this week, 8 weeks after the first one.

72

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