Base Plate Design Example using AISC 360-22 and ACI 318-19
Problem Statement:
Determine whether the designed column-to-base plate connection is sufficient for a 25 kips compression load, 3 kips Vy shear load 1 kip Vz shear load.
Given Data:
Column:
Column section: HSS6x0.312
Column area: 5.220 in2
Column material: A36
Base Plate:
Base plate dimensions: 12 in x 12 in
Base plate thickness: 1/2 in
Base plate material: A36
Grout:
Grout Thickness: 3/4 in
Concrete:
Concrete dimensions: 13 in x 13 in
Concrete thickness: 8 in
Concrete material: 3000 psi
Cracked or Uncracked: Cracked
Anchors:
Anchor diameter: 1/2 in
Effective embedment length: 5 in
Steel Material: A325N
Threads in Shear Plane: Included
Anchor Ending: Rectangular Plate
Welds:
Weld size: 1/4 in
Filler metal classification: E70XX
Transfer compression load via welds: Yes
Anchor Data (from SkyCiv Calculator):
Note:
The purpose of this design example is to demonstrate the step-by-step calculations for capacity checks involving concurrent shear and axial loads. Some of the required checks have already been discussed in the previous design examples. Please refer to the links provided in each section.
Step-by-Step Calculations:
Check #1: Calculate weld capacity
Given that the column compression load is transferred via welds, we need to consider the resultant load of the compression and shear loads in determining the strength of the welds.
To evaluate the weld capacity, we first determine the total weld length based on the column dimensions.
\(L_{\text{weld}} = \pi d_{\text{col}} = \pi \times 6\ \text{in} = 18.85\ \text{in}\)
Next, we express the demand in terms of force per unit length.
\(c_u = \frac{N_x}{L_{\text{weld}}} = \frac{25\ \text{kip}}{18.85\ \text{in}} = 1.3263\ \text{kip/in}\)
\(v_{uy} = \frac{V_y}{L_{\text{weld}}} = \frac{3\ \text{kip}}{18.85\ \text{in}} = 0.15915\ \text{kip/in}\)
\(v_{uz} = \frac{V_z}{L_{\text{weld}}} = \frac{1\ \text{kip}}{18.85\ \text{in}} = 0.053052\ \text{kip/in}\)
The resultant load is determined as:
\(r_u = \sqrt{(c_u)^2 + (v_{uy})^2 + (v_{uz})^2}\)
\(r_u = \sqrt{(1.3263\ \text{kip/in})^2 + (0.15915\ \text{kip/in})^2 + (0.053052\ \text{kip/in})^2}\)
\(r_u = 1.3369\ \text{kip/in}\)
Then, we determine the fillet weld capacity per unit length using AISC 360-22 Eq. J2-4. Note that for HSS sections, kds is always equal to 1.0.
\(k_{ds} = 1.0 + 0.5\big(\sin(\theta)\big)^{1.5} = 1 + 0.5 \times \big(\sin(0)\big)^{1.5} = 1\)
\(\phi r_n = \phi \, 0.6 F_{exx} E_w k_{ds} = 0.75 \times 0.6 \times 70\ \text{ksi} \times 0.177\ \text{in} \times 1 = 5.5755\ \text{kip/in}\)
The next capacity to check is the base metal capacity of the connecting elements. This is also expressed as force per unit length. We use AISC 360-22 Eq. J4-4 for both the column and base plate capacities.
\( \phi r_{nbm,col} = \phi\,0.6\,F_{u,col}\,t_{col} = 0.75 \times 0.6 \times 58\ \text{ksi} \times 0.291\ \text{in} = 7.5951\ \text{kip/in} \)
\( \phi r_{nbm,bp} = \phi\,0.6\,F_{u,bp}\,t_{bp} = 0.75 \times 0.6 \times 58\ \text{ksi} \times 0.5\ \text{in} = 13.05\ \text{kip/in} \)
We then take the minimum capacity as the governing base metal capacity.
\(\phi r_{nbm} = \min\big(\phi r_{nbm,bp},\ \phi r_{nbm,col}\big) = \min(13.05\ \text{kip/in},\ 7.5951\ \text{kip/in}) = 7.5951\ \text{kip/in}\)
Finally, we compare both the fillet weld capacity and the base metal capacity against the weld demand.
Since 1.3369 kip/in < 5.5755 kip/in and 1.3369 kip/in < 7.5951 kip/in the weld capacity is sufficient.
Check #2: Calculate bearing capacity of column
A design example for the bearing capacity of the column is already discussed in the Base Plate Design Example for Compression. Please refer to this link for the step-by-step calculation.
Check #3: Calculate base plate flexural yielding capacity due to compression load
A design example for the base plate flexural yielding capacity is already discussed in the Base Plate Design Example for Compression. Please refer to this link for the step-by-step calculation.
Check #4: Concrete bearing capacity
A design example for the concrete bearing capacity is already discussed in the Base Plate Design Example for Compression. Please refer to this link for the step-by-step calculation.
Check #5: Concrete breakout capacity (Vy Shear)
A design example for the concrete breakout capacity due to Vy shear is ready discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Check #6: Concrete breakout capacity (Vz Shear)
A design example for the concrete breakout capacity due to Vz shear is already discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Check #7: Concrete pryout capacity
A design example for the capacity of the concrete section against pryout is already discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Check #8: Anchor Rod Shear Capacity
A design example for the shear capacity of the anchor rod is already discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Design Summary
The SkyCiv Base Plate Design software can automatically generate a step-by-step calculation report for this design example. It also provides a summary of the checks performed and their resulting ratios, making the information easy to understand at a glance. Below is a sample summary table, which is included in the report.
SkyCiv Sample Report
Click here to download a sample report.
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