Project: Base Plate 5
Billing Reference: 2200.025
IES Employee, IES
File: Example Report.vcbp
Monday, January 07, 2013
Base Plate Analysis
Analysis Classification (AISC Design Guide #1)
Load Set: Load Set 1
Load Combination: 1.4D
Support Strength, f'c = 4 Ksi
Loaded Area, A1 = 484 in^2
Support Area, A2 =The largest area contained on the
support that is geometrically similar to and concentric
with the loaded area.
A2 = 484 in^2
(ACI 10.17.1)
= 2.21 Ksi
Pu = 140 K (Compression is positive)
Mun = 0 K-ftĺH% 0 in
Mub = 116.67 K-ftĺH1 -10. in
Edge Lengths:
B = 22 in
N = 22 in^2
Design Guide #1 sign convention
Loads and eccentricities shown in positive direction
(Equations 3.3.7)
N direction:
= 9.5603 in vs. eN = -10. in
B direction:
= 9.5603 in vs. eB = 0 in
Classification:
Axial Compression with a Large Moment in the strong direction.
Page 1 of 3
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Project: Base Plate 5
Billing Reference: 2200.025
IES Employee, IES
File: Example Report.vcbp
Monday, January 07, 2013
Base Plate w/ Large Strong-Axis Moment (AISC Design Guide #1, Section 3.4)
Load Set: Load Set 1
Load Combination: 1.4D
Base Plate Parameters:
B = 22 in, N = 22 in, f = 9.5 in, Pu = 140 K, eN = 10. in
For large moments, fpu = f fpn (See Support Bearing Capacity)
fpu = 2.21 Ksi
Check Base Plate Dimensions:
420.25 in^2 > 112.3 in^2
Base Plate area is adequate
Base plate with large moment
(Fig. 3.4.1 AISC Design Guide #1)
Calculate Bearing Length, Y:
= 2.9515 in
Base Plate Bending Demand: Tension Interface
= 2.21 Ksi * 22 in * 2.9515 in - 140 K = 3.501 K
Moment arm for tension anchor group, x = 3.4675 in
w = Effective width for tension anchor group, where the load distributes at 45 degree angles
w = 16.87 in
= 0.71959 ft-K/ft
Base Plate Bending Demand: Bearing Interface
Column Depth, d = 12.7 in
Column Width, b = 12.2 in
Depth Reduction Factor, a = 0.95 (sect 3.1.3)
Width Reduction Factor, g = 0.95 (sect 3.1.3)
= 4.9675 in
= 6.12 in
Mu(n) calculation is a crude attempt at including some 2-way
bending in the analysis. The approach is based on the note at
the end of section 3.4.2.
Figure shows bending lines of column and
illustrates base plate under strong axis bending.
Effective width shown by dashed lines.
= 2.21 Ksi * 6.12 in^2.0 / 2.0 = 41.387 ft-K/ft
Since Y < m
= 2.21 Ksi * 2.9515 in * (4.9675 in - 2.9515 in / 2.0) = 22.776 ft-K/ft
Mu(Bearing) = 41.387 ft-K/ft
Controlling Mu = max( 0.71959 ft-K/ft , 41.387 ft-K/ft) = 41.387 ft-K/ft
Page 2 of 3
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Project: Base Plate 5
Billing Reference: 2200.025
IES Employee, IES
File: Example Report.vcbp
Monday, January 07, 2013
Base Plate Design
Support Bearing Capacity (ACI 318-08 10.14.1)
Load Set: Load Set 1
Load Combination: 1.4D
fpu = 2.21 Ksi
= 2.21 Ksi
f = 0.65
Support Strength, f'c = 4 Ksi
Loaded Area, A1 = 484 in^2
Support Area, A2 = 484.01 in^2
(The largest area contained on the support that is geometrically similar to and concentric with
the loaded area)
Unity = fpu / (ffpn) = 1.000
Plate Bending (AISC 360-10 F11)
Load Set: Load Set 1
Load Combination: 1.4D
Mu = 3.4489 K-ft
(Eqn F11-1)
For rectangular sections, 1.6 * My will not control
f = 0.90 , Fy = 36 Ksi, t = 2.5 in, w = width = 1 in
fMn = 4.2188 K-ft
Unity = Mu / (f * Mn) = 0.818
Base Plate Detailing (AISC 360-10 J3)
Messages:
Bolt spacing is adequate.
Bolt edge distances are adequate.
Page 3 of 3
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