8.
2.
Determine required plate thickness:
Note: Since the Mpl is expressed in units of kip-in./in.,
the plate thickness expressions can be formatted without the plate width (B) as such:
tu req =
Verification Example
LRFD
ASD
AISC Design Guide 1, 2nd Edition
4 MPlate
Base
4 M a crit Ω
u crit and Anchor Rod Design
φFy
4×11.1 kip-in.
0.90×36 ksi
= 1.17 in.
=
9.
ta req =
LRFD
ASD
e = 720 kip-in./90 kips = 8.00 in.
e = 480 kip-in./60 kips = 8.00 in.
Then, e > ekern; therefore, anchor rods are required to
resist the tensile force. The anchor rods are assumed to
be 1.5 from the plate edge.
Fy
4× 7.68 kip-in.×1.67
36 ksi
= 1.19 in.
=
Assume a 14-in. × 14-in. base plate. The effective eccentricity is
3.
Determine the length of bearing.
LRFD
ASD
3.06 ksi ×14 in.×12.5 in. ′ 2.04 ksi ×14 in.×12.5 in.
f =
2
2
= 178 kips
= 268 kips
f′=
Use plate size:
N = 19 in.
B = 19 in.
thus,
t = 14 in.
B.5.2 Example: Large Moment Base Plate Design,
Triangular Pressure Distribution Approach
Design the base plate shown in Figure B.4 for an ASD and
LRFD required strength of 60 and 90 kips, respectively, and
moments from the dead and live loads equal to 480 and 720
kip-in., respectively. The ratio of the concrete to base plate
area (A2/A1) is 4.0. Bending is about the strong axis for the
wide flange column W8×31 with d = bf = 8 in.; Fy of the
base plate and anchor rods is 36 ksi and fc′ of the concrete
is 3 ksi.
LRFD
ASD
 3.06×14 
2682 − 4 



6
268 −
× (90×5.5) + 720
A=


 3.06×14 


3
 2.04×14 
1782 − 4 



6
178 −


× (60×5.5) + 480
A=


 2.04×14 


3
= 5.27 in.
= 5.27 in.
1.
LRFD
Pu = 90 kips
M u = 720 kip-in.
φPp
A1
= 0.60(0.85)(3.0)(2)
ASD
Pa = 60 kips
M a = 480 kip-in.
Pp
ΩA1
≤ 0.60(1.7)((3.0)
φPp
A1
= 3.06 ksi
Pp
ΩA1
=
(0.85)(3.0)(2)
2.50
≤
(1.7)(3.0)
2.50
= 2.04 ksi
Figure B.4. Design example with large eccentricity.
60 / DESIGN GUIDE 1, 2ND EDITION / BASE PLATE AND ANCHOR ROD DESIGN
4.
Anchor rods are placed at a 12-in. edge distance. The
required moment strength, Mu pl or Ma pl, for a 1-in. strip
of plate due to the tension in the anchor rods is
Determine the required tensile strength of the anchor
rod.
LRFD
ASD
2.04 ksi ×5.27 in.×14 in.
3.06 ksi ×5.27 in.×14 in.
− 60 kips
− 90 kips Ta =
2
2
= 15.2 kips
= 22.8 kips
Trod = Ta / 2 = 7.660 kips
= Tu / 2 = 11.4 kips
LRFD
Tu =
Trod
5.
22.8 kips(3.2 in. −1.5 in.)
2(3.2 in. −1.5 in.)
= 11.4 in.-kips/in.
M u pl =
ASD
M a pl
15.2 kips(3.2 in. −1.5 in.)
=
2(3.2 in. −1.5 in.)
= 7.60 in.-kips/in.
Determine the required plate thickness.
The moment for this determination is to be taken at the
critical plate width. This is determined by assuming that
the load spreads at 45° to a location 0.95d of the column. The width is then taken as twice the distance from
the bolt to the critical section for each bolt, provided
that the critical section does not intersect the edge of the
plate.
The critical section, as shown in Figure B.5, is at 14 −
0.95(8)/2 = 3.2 in.
The required moment strength, Mu pl or Ma pl, for a 1-in.
strip of plate, determined from the bearing stress distribution in Figure B.4, is
LRFD
The required plate thickness is:
LRFD
tp =
4(12.5 in.-kips)
= 1.24 in.
0.90×36 ksi
ASD
tp =
4(8.33 in.-kips)(1.67)
= 1.24
36 ksi
Use a 14 × 14 × 1�-in. base plate.
ASD
2
M u pl =
The required moment strength due to the bearing stress
distribution is critical.
1.20 ksi ×(3.2 in.)
2
2 (3.06 ksi −1.20 ksi) × (3.2 in.) 2
+ 3
2
= 12.5 in-kips/in.
2
M a pl =
0.80 ksi ×(3.2 in.)
2
2 ( 2.04 ksi − 0.80 ksi) × (3.2 in.) 2
3
+
2
= 8.33 in-kips/in.
Figure B.5. Critical plate width for anchor bolt (tension side).
DESIGN GUIDE 1, 2ND EDITION / BASE PLATE AND ANCHOR ROD DESIGN / 61
Project: Verification Example
Engineer: Javier Encinas, PE
Descrip: Base Plate Verification
ASDIP Steel 3.2.5
STEEL BASE PLATE DESIGN
GEOMETRY
Column Section .................
W8X31
Width
Length
Column ..........
8.0
8.0
Plate ..............
14.0
14.0
Concrete Wp1 14.0 Lp1 14.0
Support
Wp2 14.0 Lp2 14.0
Rod Offset .....
5.5
5.5
Thickness of Grout ............
1.5
in
in
in
in
in
in
OK
OK
OK
OK
Page # ___
7/20/2014
www.asdipsoft.com
SERVICE LOADS (ASD)
Vertical Load P ................
60.0
Bending Moment M .........
40.0
Horizontal Load V ............
0.0
Design Eccentricity e .......
8.0
Design Eccentricity Is > L/2
MATERIALS
Plate Steel Strength Fy ....
36.0
Pier Concrete Strength f'c
3.0
kip
k-ft
kip
in
ksi
ksi
AXIALLY LOADED PLATES
Cantilever Model
Bearing Stress fp .............
0.31
Critical Section @ Long m
3.20
Critical Section @ Short n
3.80
Plate Thickness tp ..........
0.64
ksi
in
in
in
OK
Thornton Model
Bearing Strength Fp/Ω .....
2.04
Critical Section @ Int λn' .
0.81
Design Moment @ Plate ...
0.10
Plate Thickness tp ............
0.14
ksi
in
k-in/in
in
BASE PLATES WITH MOMENT
Blodgett Method
Max. Bearing Stress fp ...... 1.64
Bearing @ Critical Section
0.88
Moment @ Critical Section
7.11
Moment due to Rod Tension 4.86
Design Moment @ Plate .... 7.11
Plate Thickness tp ............. 1.15
ksi
OK
ksi
k-in/in
k-in/in
k-in/in
in
DeWolf Method
Max. Bearing Stress fp ...... 2.04
Bearing @ Critical Section
0.80
Moment @ Critical Section
8.34
Moment due to Rod Tension 3.86
Design Moment @ Plate .... 8.34
Plate Thickness tp ............. 1.24
ksi
OK
ksi
k-in/in
k-in/in
k-in/in
in
1
Project: Verification Example
Engineer: Javier Encinas, PE
Descrip: Base Plate Verification
ASDIP Steel 3.2.5
STEEL BASE PLATE DESIGN
ANCHORAGE DESIGN
Rod Material Specification .........
F1554-36
(4) Rods , fya = 36.0 ksi, futa = 58.0 ksi
Anchor Rod Size ..
1" diam. x 16.0 in emb.
Concrete Is Cracked at Service Load Level
Tension Analysis (kip)
Total Tension Force N ..........
15.4 kip
Tension Force per Rod Ni ....
7.7 kip
Anchor Reinf: Use 2 Bars #5 per Rod
Failure Mode
Ω
Nn N / Nn/Ω
Steel Strength Nsa
2.00
35.1
0.44
Rebars Strength Nrg
2.00
74.4
0.41
Conc. Breakout Ncbg
2.00
N.A.
N.A.
Pullout Strength Npn
2.00
36.0
0.43
Side Blowout Nsbg
2.00
N.A.
N.A.
N / Nn/Ω Tension Design Ratio ....
0.44 OK
Page # ___
7/20/2014
www.asdipsoft.com
SUMMARY OF RESULTS
Design Moment @ Plate ...
8.3 k-in/in
Plate Thickness tp ............
1.24 in
Max. Bearing Stress fp .....
2.04 ksi
Bearing Strength Fp/Ω ......
2.04 ksi
fp / Fp/Ω Design Ratio ..............
1.00 OK
DESIGN IS DUCTILE
Shear Analysis (kip)
Shear Taken by Anchor Rods only
Total Shear Force V ...........
0.0 kip
Shear Force per Rod Vi ......
0.0 kip
All Anchor Rods Are Effective
No Reinforcing Bars Provided
Failure Mode
Ω
Vn V / Vn/Ω
Steel Strength Vsa
2.31
16.9
0.00
Rebars Strength Vrg
2.31
N.A.
N.A.
Conc. Breakout Vcbg
2.14
11.9
0.00
Conc. Pryout Vcpg
2.14
52.8
0.00
V / Vn/Ω Shear Design Ratio ......
0.00 OK
Tension-Shear Interaction
Combined Stress Ratio ...........
0.25
OK
DESIGN CODES
Steel design .............
AISC 360-10 (14th Ed.)
Base plate design ....
AISC Design Series # 1
Anchorage design ...
ACI 318-11 Appendix D
2
Project: Verification Example
Engineer: Javier Encinas, PE
Descrip: Base Plate Verification
ASDIP Steel 3.2.5
STEEL BASE PLATE DESIGN
Tension Breakout
Page # ___
7/20/2014
www.asdipsoft.com
Shear Breakout
3
Project: Verification Example
Engineer: Javier Encinas, PE
Descrip: Base Plate Verification
STEEL BASE PLATE DESIGN
ASDIP Steel 3.2.5
Column Section .................
Width
Length
Column ..........
in
Plate ..............
in
OK
Wp1
Lp1
in
OK
Wp2
Lp2
in
OK
Rod Offset .....
in
OK
Thickness of Grout ............
in
Concrete
Support
Bearing stress
Page # ___
7/20/2014
www.asdipsoft.com
Vertical Load P ................
kip
Bending Moment M .........
k-ft
Horizontal Load V ............
kip
Design Eccentricity e .......
8.0
in
Design Eccentricity Is > L/2
Plate Steel Strength Fy ....
ksi
Pier Concrete Strength f'c
ksi
60.0 / (14.0 * 14.0) = 0.3 ksi
Bearing strength
= 0.85 * 3.0 *
= 5.1 ksi
Under-strength factor Ω = 2.50
Bearing strength ratio =
ACI 9.3.2.4
=
0.3
2.0 / 2.50
= 0.15
< 1.0 OK
Critical section m =
0.5 * (14.0 - 0.95 *8.0) = 3.2 in
Critical section n =
0.5 * (14.0 - 0.80 *8.0) = 3.8 in
[
=
+
=
Controlling section
-
ACI 10.14.1
4 * 8.0 * 8.0
(8.0 + 8.0)²
] * 0.15 = 0.15
AISC-DG#1 3.1.2
AISC-DG#1 3.1.2
= 0.40
= 2.0 in
Max (3.2, 3.8, 0.40 * 2.0) = 3.8 in
Plate moment
0.3 * 3.8² / 2 = 2.2 k-in/in
Plate thickness
= 3.8 *
= 0.64 in
AISC-DG#1 3.1.2
1
Project: Verification Example
Engineer: Javier Encinas, PE
Descrip: Base Plate Verification
STEEL BASE PLATE DESIGN
ASDIP Steel 3.2.5
Eccentricity
40.0 * 12 / 60.0 = 8.0 in
Page # ___
7/20/2014
www.asdipsoft.com
> L / 6 = 14.0 / 6 = 2.3in
Bearing length
Y = 1.5 * (14.0 / 2 + 5.5) - 0.5 *
Max bearing stress
[
+
]-
= 5.3 in
AISC-DG#1 B.4.2
2.0 ksi
Tension
2.0 * 5.3 * 14.0 / 2 - 60.0 = 15.4 kip
Bearing at critical section
2.0 * (1 - 3.2 / 5.3) = 0.8 ksi
Moment due to bearing
AISC-DG#1 B.4.2
AISC-DG#1 3.1.2
Mb = 0.8 * 3.2² / 2 + (2.0 - 0.8) * 3.2² / 3 = 8.3 k-in/in
Moment due to tension
Mt = 7.7 * [3.2 - (14.0 / 2 - 5.5)] / [2 * (3.2 - (14.0 / 2 - 5.5))] = 3.9 k-in/in
Plate thickness
=
= 1.24 in
2
Project: Verification Example
Engineer: Javier Encinas, PE
Descrip: Base Plate Verification
STEEL BASE PLATE DESIGN
ASDIP Steel 3.2.5
Page # ___
7/20/2014
www.asdipsoft.com
Rod Material Specification ...... F1554-36 , Use (4) Rods , fya = 36.0 ksi, futa = 58.0 ksi
Anchor Rod Size .... 1" diam. x 16.0 in emb. ,
Ase = 0.61 in² , Abrg = 1.50 in²
ACI D.5
Total tension force N = 15.4 kip ,
# of tension rods = 2 , Tension force per rod Ni = 7.7 kip
- Steel strength of anchors in tension
Steel strength
ACI D.5.1
0.606 * 58.0 = 35.1 kip
ACI Eq. (D-2)
Under-strength factor Ω = 2.00
Steel strength ratio =
ACI D.4.3
=
7.7
= 0.44 < 1.0 OK
35.1 / 2.00
ACI D.4.1.1
- Concrete breakout strength of anchors in tension
ACI D.5.2
Anchor reinforcement: Use 2 bars #5 per rod
Bar strength
0.31 * 2 * 2 * 60 = 74.4 kip
Under-strength factor Ω = 2.00
Bar strength ratio =
ACI D.5.2.9
=
15.4
= 0.41 < 1.0 OK
74.4 / 2.00
Effective embedment
ACI D.4.1.1
19.50 / 1.5 = 13.00 in
ACI D.5.2.3
Anchor group area
Anc = (19.5 + 8.5) * (8.5 + 11.0 + 8.5) = 784.0 in²
Single anchor area
Single anchor strength
Eq. (D-5)
= 24
Eccentricity factor
= 61.6 kip
1.00 (No eccentric load)
Edge effects factor
Cracking factor
ACI D.5.2.1
9 * (13.0)² = 1521.0 in²
= 0.7 + 0.3
1.00
Eq. (D-6)
ACI D.5.2.4
8.5
= 0.83
1.5 * 13.0
(Cracked concrete at service load level)
ACI D.5.2.5
ACI D.5.2.6
Breakout strength
784.0
1.00 * 0.83 * 1.00 * 61.6 = 26.4 kip
1521.0
Eq. (D-4)
Under-strength factor Ω = 2.00
ACI D.4.3
Breakout strength ratio =
=
15.4
= 1.17 > 1.0 NG
26.4 / 2.00
Bar strength ratio controls (0.41 < 1.17)
ACI D.5.2.9
- Concrete pullout strength of anchors in tension
Single anchor strength
Cracking factor
ACI D.5.3
8 * 1.50 * 3.0 = 36.0 kip
1.00
(Cracked concrete at service load level)
Pullout strength
1.00 * 36.0 = 36.0 kip
ACI Eq. (D-14)
ACI D.5.3.6
ACI Eq. (D-13)
Under-strength factor Ω = 2.00
Pullout strength ratio =
ACI D.4.1.1
ACI D.4.3
=
7.7
= 0.43
36.0 / 2.00
< 1.0 OK
ACI D.4.1.1
3
Project: Verification Example
Engineer: Javier Encinas, PE
Descrip: Base Plate Verification
STEEL BASE PLATE DESIGN
ASDIP Steel 3.2.5
- Concrete side-face blowout strength of anchors in tension
Page # ___
7/20/2014
www.asdipsoft.com
ACI D.5.4
Side-face blowout Nsbg = N.A. (Embed < 2.5 Ca₁ , 16.0 < 2.5 * 8.5 = 21.3)
ACI D.5.4.1
Tension Design Ratio =
ACI D.4.1.1
= 0.44 < 1.0 OK
ACI D.5
Shear resisted by Anchor Rods only
Total shear force V = 0.0 kip ,
(anchor rods are welded to the base plate)
Shear per rod Vi = 0.0 kip , (all anchor rods are effective)
- Steel strength of anchor rods in shear
Steel strength
0.6 * 0.61 * 58.0 * 0.80 = 16.9 kip
Under-strength factor Ω = 2.31
Steel strength ratio =
ACI D.6.1.2
ACI D.4.3
=
0.0
= 0.00
16.9 / 2.31
< 1.0 OK
- Concrete breakout strength of anchors in shear
ACI D.4.1.1
ACI D.5.2
No Reinforcing bars provided
Anchor group area
Avc = (1.5 * 8.00) * (8.50 + 11.00 + 8.50) = 336.0 in²
Single anchor area
ACI D.6.2.1
4.5 * (8.00)² = 288.0 in²
Eq. (D-32)
Single anchor strength
Vb =
Vb = 11.2 kip
Eccentricity factor
1.00 (No eccentric load)
Edge effects factor
Cracking factor
= 0.7 + 0.3
1.00
Eq. (D-33)
ACI D.6.2.5
8.50
= 0.91
1.5 * 8.0
(Cracked concrete at service load level)
Thickness factor
ACI D.6.2.6
ACI D.6.2.7
ACI D.6.2.8
Breakout strength
336.0
288.0
1.00 * 0.91 * 1.00 * 1.00 * 11.2 = 11.9 kip
Under-strength factor Ω = 2.14
Breakout strength ratio =
Eq. (D-31)
ACI D.4.3
=
0.0
= 0.00 < 1.0 OK
11.9 / 2.14
Breakout strength ratio controls (0.00 < 0.00)
ACI D.4.1.1
ACI D.6.2.9
4
Project: Verification Example
Engineer: Javier Encinas, PE
Descrip: Base Plate Verification
STEEL BASE PLATE DESIGN
ASDIP Steel 3.2.5
Page # ___
7/20/2014
www.asdipsoft.com
- Concrete pryout strength of anchors in shear
Pryout strength
2.0 * 11.9 = 52.8 kip
ACI D.6.3.1
Under-strength factor Ω = 2.14
ACI D.4.3
Pryout strength ratio =
Shear Design Ratio =
=
0.0
= 0.00
52.8 / 2.14
< 1.0 OK
= 0.00 < 1.0 OK
ACI D.4.1.1
ACI D.4.1.1
Combined Stress Ratio =
Combined Stress Ratio =
+
= 0.25
< 1.0 OK
ACI RD.7
Anchorage design is ductile
Steel design ...................
AISC 360-10 (14th Ed.)
Base plate design ..........
AISC Design Series # 1
Anchorage design .........
ACI 318-11 Appendix D
5