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 5 kip-ft strong-axis moment.
Given Data
Column:
Column section: W8x31
Column area: 9.130 in2
Column material: A992
Base Plate:
Base plate dimensions: 15 in x 15 in
Base plate thickness: 1 in
Base plate material: A36
Concrete:
Concrete dimensions: 18 in x 18 in
Concrete thickness: 15 in
Concrete material: 3000 psi
Welds:
Weld size: 1/4 in
Filler metal classification: E70XX
Compression load transferred through welds only? NO
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Determine if Low Moment or Large Moment
The calculations for this base plate example, which is subjected to a strong-axis moment, will be based on AISC Design Guide 1, 3rd Edition. According to the design guide, base plates subjected to moment loads only are classified as the Large Moment Case. This is because no axial load is present to counteract the eccentric load acting on the connection.
For other load combinations, such as strong-axis bending with axial compression or strong-axis bending with axial tension, determining if the case is Large Moment or Low Moment depends on the actual column size, base plate size, concrete size, and the given loads.
In the free version of SkyCiv, the moment case is displayed as you enter the load values. You can access the free tool here to determine the moment case yourself. Straight from the browser, no installation required.
To learn more about Low Moment and Large Moment case, refer to this SkyCiv article: AISC DG1: Base Plates with Large Moment and Low Moment
Step-by-Step Calculations
Check #1: Weld capacity
We start the design by checking the weld capacity of the connection. We use the Blodgett equations from Design of Welded Structures by Omer W. Blodgett to determine the weld section modulus of the weld. We take the weld shape shown below as a simplification.
\(S_w = 2d_{col}b_f + \dfrac{(d_{col})^2}{3} = 2 \times 8 \text{ in} \times 8 \text{ in} + \dfrac{(8 \text{ in})^2}{3} = 149.33 \text{ in}^2\)
Next, we divide the applied bending moment by the weld section modulus to determine the weld stress per unit length. Note that in calculating the weld section modulus, we assume the critical weld stresses occur at the outermost fiber. For a weld in an I-shaped section, the critical locations are at the outer flanges. We get the weld stress as:
\(r_{uw} = \dfrac{M_z}{S_w} = \dfrac{5 \text{ kip·ft}}{149.33 \text{ in}^2} = 0.40179 \text{ kip/in}\)
Then, we calculate the capacity of the weld. For a 0.25 in weld size, the design weld throat is 0.177 in, which is obtained by multiplying the weld size by \(\dfrac{\sqrt{2}}{2}\). Using AISC 360-22 Eq. J2-4, the weld capacity per unit length is:
\(\phi r_n = \phi 0.6F_{exx}E_w k_{ds} = 0.75 \times 0.6 \times 70 \text{ ksi} \times 0.177 \text{ in} \times 1 = 5.5755 \text{ kip/in}\)
Since 0.40179 kpi < 5.5755 kpi , the column weld capacity is sufficient.
This weld check investigates the weld stress at the tension flange. Since the geometry is symmetric and the only load is a moment acting at the center, we can assume the stress at the compression flange is the same as the stress at the tension flange. Therefore, the welds at the compression flange also have enough capacity.
In this design example, however, the design inputs specify that compression is not transferred through the welds, but through bearing. Thus, in the next check, we evaluate the bearing capacity of the column.
Check #2: Calculate column bearing capacity
After evaluating the weld capacity, we make sure the column has enough capacity to bear on the base plate. This is necessary because we assume the welds do not carry the compression load.
First, we determine the compressive stress at the critical section, which is still the flange. We then add any contribution from axial compression. In this case, there is none.
\(r_u = \dfrac{N_x}{A_{col}} + \dfrac{M_z 0.5d_{col}}{I_{zz,col}} = \dfrac{0 \text{ kip}}{9.13 \text{ in}^2} + \dfrac{5 \text{ kip·ft} \times 0.5 \times 8 \text{ in}}{110 \text{ in}^4} = 2.1818 \text{ ksi}\)
Then, we use AISC 360-22 Section J7 Eq. J7-1 to determine the bearing capacity:
\(\phi r_n = \phi 1.8F_{y\_col} = 0.75 \times 1.8 \times 50 \text{ ksi} = 67.5 \text{ ksi}\)
Since 2.1818 ksi < 67.5 ksi, the column bearing capacity is sufficient.
Usually, this check is not governing.
Check #3: Calculate concrete bearing capacity
To calculate the concrete bearing capacity, we need two areas. See the figure below as a guide.
- A1 – Base plate loaded area
- A2 – Area of the concrete support with the same shape as A1
A1 is the loaded area of the base plate, while A2 is the same loaded area projected at a 2H:1V slope into the concrete support. More details on these areas can be found in ACI 318-19 Section 22.8.3.2.
In this base plate design with moment, the A1 area will be calculated using the full width of the base plate, \(B\) and the actual bearing length, \(Y\). The actual bearing length is measured from the edge of the base plate at the compression side towards the center of the plate.
\( A_1 = B \times Y \)
For most engineers, calculating A1 and A2 requires a lot of work, so they simply assume A1 = A2, which is conservative. In the SkyCiv Base Plate software, iterations are performed to determine the actual A1 values and to maximize the capacity of the base plate.
The table below shows the iteration steps from SkyCiv to determine the bearing length Y, and subsequently, to determine the actual A1 and A2 areas.
First, an initial Y is assumed, and the A2 and A1 values are calculated based on the initial Y. These are then used to calculate the maximum bearing strength of the concrete, \(f_{p\_max}\), and the corresponding bearing force per unit length, \(q_{max}\). Once these are determined, the corresponding Y value is determined based on Eq. 4-32 of AISC DG1 3rd Edition. If the calculated Y does not match the initial Y, another iteration step is performed. A more detailed step-by-step calculation of the last iteration is shown in the next section.
Now, taking the A2 and A1 values from the iteration, we use Eq. 4-2 of AISC Design Guide 1, 3rd Edition, to determine \(f_{p\_max}\), the design bearing strength for concrete crushing:
\(f_{p\_max} = \phi \left( \min \left( 0.85 \, (f_c’) \sqrt{\dfrac{A_2}{A_1}}, \, 1.7 \, (f_c’) \right) \right)\)
\(f_{p\_max} = 0.65 \times \left( \min \left( 0.85 \times (3 \text{ ksi}) \times \sqrt{39.941}, \, 1.7 \times (3 \text{ ksi}) \right) \right) = 3.315 \text{ ksi}\)
Additionally, since this design is a Large Moment Case, the base plate has both a compression side and a tension side. In this situation, we should allow the concrete bearing to be at its full capacity and minimize the Y value. So, any remaining tensile force, based on equilibrium of forces, will be taken by the anchors at the tension side.
Given this, we set \(f_{p\_actual}\) equal to \(f_{p\_max}\).
\(f_{p\_actual} = f_{p\_max} = 3.315 \text{ ksi}\)
Since 3.315 ksi ≤ 3.315 ksi, the concrete bearing capacity is sufficient.
Check #4: Calculate base plate yielding capacity at the bearing interface
The next step is to determine if the base plate has enough capacity for the given load. As the moment load pushes the base plate against the concrete, the base plate has a tendency to yield due to flexure. We call this check the base plate yielding check on the bearing interface. There is also a separate yielding due to the tension load, and it will be covered in the next design check.
To check the base plate yielding, we need to determine the \(q_{max}\), which is the compression force per unit length. AISC DG1 3rd Ed. Eq. 4-37 gives \(q_{max}\)as:
\(q_{max} = f_{p\_max}B_{bp} = 3.315 \text{ ksi} \times 15 \text{ in} = 49.725 \text{ kip/in}\)
Then, we will calculate the Y-dimension. Although the Y-dimension is already calculated in the prior check, this section will show how we use the equations from AISC DG1. For this Large Moment Case, we use Eq. 4-32 of AISC DG1, 3rd Edition.
For the plus case:
\(Y_{case1} = \left( f + \dfrac{L_{bp}}{2} \right) + \sqrt{\left( \left( f + \dfrac{L_{bp}}{2} \right)^2 \right) – \dfrac{2M_z}{q_{max}}}\)
\(Y_{case1} = \left( 5.5315 \text{ in} + \dfrac{15 \text{ in}}{2} \right) + \sqrt{\left( \left( 5.5315 \text{ in} + \dfrac{15 \text{ in}}{2} \right)^2 \right) – \dfrac{2 \times 5 \text{ kip·ft}}{49.725 \text{ kip/in}}}\)
\(Y_{case1} = 25.97 \text{ in}\)
For the minus case:
\(Y_{case2} = \left( f + \dfrac{L_{bp}}{2} \right) – \sqrt{\left( \left( f + \dfrac{L_{bp}}{2} \right)^2 \right) – \dfrac{2M_z}{q_{max}}}\)
\(Y_{case2} = \left( 5.5315 \text{ in} + \dfrac{15 \text{ in}}{2} \right) – \sqrt{\left( \left( 5.5315 \text{ in} + \dfrac{15 \text{ in}}{2} \right)^2 \right) – \dfrac{2 \times 5 \text{ kip·ft}}{49.725 \text{ kip/in}}}\)
\(Y_{case2} = 0.092925 \text{ in}\)
Note that this equation has + and – symbols, so we get two values. From here, we need to determine which value makes more sense. Since 25.97 in is greater than the base plate length, 15 in, we will not consider this value. So, we will use 0.092925 in as the bearing length.
\(Y = Y_{case2} = 0.092925 \text{ in}\)
Next, we calculate the cantilever lengths. These are the m and n dimensions from the figure below. This is based on AISC DG1 3rd Edition Section 4.3.1.
The calculated m and n cantilever values are:
\(m = \dfrac{L_{bp} – 0.95d_{col}}{2} = \dfrac{15 \text{ in} – 0.95 \times 8 \text{ in}}{2} = 3.7 \text{ in}\)
\(n = \dfrac{B_{bp} – 0.8b_f}{2} = \dfrac{15 \text{ in} – 0.8 \times 8 \text{ in}}{2} = 4.3 \text{ in}\)
Then, we will compare the calculated Y value to the larger of m and n. Since Y is less than the larger cantilever length, we will use AISC DG1 3rd Ed. Eq. 4-52a. This gives the required base plate thickness to be:
\(t_{p(req)} = \sqrt{\dfrac{4f_{p\_actual}Y \left( n – \dfrac{Y}{2} \right)}{\phi F_{y\_bp}}}\)
\(t_{p(req)} = \sqrt{\dfrac{4 \times 3.315 \text{ ksi} \times 0.092925 \text{ in} \times \left( 4.3 \text{ in} – \dfrac{0.092925 \text{ in}}{2} \right)}{0.9 \times 36 \text{ ksi}}} = 0.4022 \text{ in}\)
Finally, let’s compare it to the actual base plate thickness.
\(t_{p(actual)} = t_{bp} = 1 \text{ in}\)
Since 0.4022 in < 1 in, the yielding capacity of the base plate is sufficient.
Check #5: Calculate base plate yielding capacity at the tension interface
The required tensile force on the anchors at the tension side is equal to the resulting compression force on the compression side. This is because the base plate is only loaded with moment loads, and so the resulting couple forces are equal in magnitude.
To calculate the required strength of anchors, we use AISC DG1 3rd Ed. Eq. 4-31.
\(T_u = q_{max}Y = 49.725 \text{ kip/in} \times 0.092925 \text{ in} = 4.6207 \text{ kip}\)
Then, we calculate the eccentricity of this tensile force from the column support. This is taken from the center of the anchor rod to the center of the column flange. We use Eq. 4-61 of the design guide to determine the eccentricity. In the equation, f is the distance of the anchor rod from the column centerline.
\(x = \left( f – \dfrac{d_{col}}{2} \right) + \dfrac{t_f}{2} = \left( 5.5315 \text{ in} – \dfrac{8 \text{ in}}{2} \right) + \dfrac{0.435 \text{ in}}{2} = 1.749 \text{ in}\)
Note that in this analysis, we take the whole base plate width as the section to resist the flexure, as we closely follow the AISC DG1 3rd Ed. guidelines. Other conservative bending widths could also be used.
Finally, we determine the required base plate thickness using Eq. 4-62a.
\(t_{p(req)} = \sqrt{\dfrac{4T_u x}{B_{bp}\phi F_{y\_bp}}} = \sqrt{\dfrac{4 \times 4.6207 \text{ kip} \times 1.749 \text{ in}}{15 \text{ in} \times 0.9 \times 36 \text{ ksi}}} = 0.25791 \text{ in}\)
Since 0.25791 in < 1 in, the yielding capacity of the base plate is sufficient.
Check #6: Anchor Checks
The rest of the anchor design checks are not shown in this example. Please refer to other SkyCiv Base Plate Design Examples that covers capacity of anchorage.
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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