Base Plate Design Example using CSA S16:19 and CSA A23.3:19
Problem Statement
Determine whether the designed column-to-base plate connection is sufficient for a 25 kN compression load, 5 kN Vy shear load 5 kN Vz shear load.
Given Data
Column:
Column section: HS152x152x6.4
Column area: 3610 mm2
Column material: 350W
Base Plate:
Base plate dimensions: 350 mm x 350 mm
Base plate thickness: 20 mm
Base plate material: 300W
Grout:
Grout Thickness: 0 mm
Concrete:
Concrete dimensions: 450 mm x 450 mm
Concrete thickness: 300 mm
Concrete material: 20.68 MPa
Cracked or Uncracked: Cracked
Anchors:
Anchor diameter: 12.7 mm
Effective embedment length: 250 mm
Steel Material: ASTM F1554 G36
Threads in Shear Plane: Included
Anchor Ending: Circular Plate
Welds:
Weld type: Complete Joint Penetration (CJP)
Anchor Data (from SkyCiv Calculator):
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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.
\(L_{\text{w}} = \frac{A_{\text{col}}}{t_{\text{col}}} = \frac{3610\ \text{mm}^2}{6.35\ \text{mm}} = 568.5\ \text{mm}\)
Next, we express the demand in terms of force per unit length.
\(v_{fy} = \frac{V_u}{L_w} = \frac{5\ \text{kN}}{568.5\ \text{mm}} = 0.008795\ \text{kN/mm}\)
\(v_{fz} = \frac{V_z}{L_w} = \frac{5\ \text{kN}}{568.5\ \text{mm}} = 0.008795\ \text{kN/mm}\)
The resultant load is determined as:
\(r_f = \sqrt{(v_{fy})^2 + (v_{fz})^2}\)
\(r_f = \sqrt{(0.008795\ \text{kN/mm})^2 + (0.008795\ \text{kN/mm})^2} = 0.012438\ \text{kN/mm}\)
Then, we determine the CJP weld capacity per unit length. First, we check the base metal capacities of both column and base plate using CSA S16:19 Clause 13.13.2.1(a).
\(v_{r,\text{col}} = 0.67\phi_{\text{col}}F_{u,\text{col}} = 0.67 \times 0.67 \times 6.35\ \text{mm} \times 450\ \text{MPa} = 1.2827\ \text{kN/mm}\)
\(v_{r,\text{bp}} = 0.67\phi_{\text{bp}}F_{u,\text{bp}} = 0.67 \times 0.67 \times 20\ \text{mm} \times 450\ \text{MPa} = 4.0401\ \text{kN/mm}\)
Then, we check the weld metal capacity using CSA S16:19 CLause 13.13.2.1(b).
\(v_{r,\text{weld}} = 0.67\phi_{\text{weld}}X_u = 0.67 \times 0.67 \times 6.35\ \text{mm} \times 430\ \text{MPa} = 1.2257\ \text{kN/mm}\)
We then take the minimum capacity as the governing capacity.
\(v_r = \min (v_{r,\text{bp}}, v_{r,\text{col}}, v_{r,\text{weld}}) = \min (4.0401\ \text{kN/mm}, 1.2827\ \text{kN/mm}, 1.2257\ \text{kN/mm}) = 1.2257\ \text{kN/mm}\)
Since 0.012 kN < 1.23 kN 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.
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