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: LRFD
tu req =
4 M u crit φFy
4×11.1 kip-in. 0.90×36 ksi = 1.17 in. =
9.
Assume a 14-in. × 14-in. base plate. The effective eccentricity is LRFD
ASD
e = 720 kip-in./90 kips = 8.00 in.
e = 480 kip-in./60 kips = 8.00 in.
ASD
ta req =
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.
4 M a crit Ω Fy
4× 7.68 kip-in.×1.67 36 ksi = 1.19 in. =
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