Base Plate Design Example using EN 1993-1-8-2005, EN 1993-1-1-2005 and EN 1992-1-1-2004
Problem Statement:
Determine whether the designed column-to-base plate connection is sufficient for a 1500-kN compression load, 12-kN Vz shear load, and 25-kN Vy shear load.
Given Data:
Column:
Column section: HP 360×180
Column area: 23000 mm2
Column material: S275N
Base Plate:
Base plate dimensions: 750 mm x 750 mm
Base plate thickness: 25 mm
Base plate material: S235
Grout:
Grout thickness: 0 mm
Concrete:
Concrete dimensions: 750 mm x 750 mm
Concrete thickness: 380 mm
Concrete material: C20/25
Anchors:
Anchor diameter: 24 mm
Effective embedment length: 300 mm
Anchor Ending: Rectangular Plate
Embedded plate Width: 100 mm
Embedded plate thickness: 16 mm
Welds:
Weld size: 12 mm
Filler metal classification: E38
Compression load transferred through welds only? Yes
Anchor Data (from SkyCiv Calculator):
Notes:
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
In determining the weld demand, the SkyCiv calculator assumes that the Vy shear load is resisted by the web alone, the Vz shear load is resisted by the flanges alone, and the compression load is resisted by the entire section.
First, we calculate the total weld length on the section.
\(L_{\text{weld}} = 2 b_f + 2(d_{\text{col}} – 2 t_f – 2 r_{\text{col}}) + 2(b_f – t_w – 2 r_{\text{col}})\)
\(L_{\text{weld}} = 2 \times 378.8\ \text{mm} + 2 \times (362.9\ \text{mm} – 2 \times 21.1\ \text{mm} – 2 \times 15.2\ \text{mm}) + 2 \times (378.8\ \text{mm} – 21.1\ \text{mm} – 2 \times 15.2\ \text{mm})\)
\(L_{\text{weld}} = 1992.8\ \text{mm}\)
Then, we calculate the weld lengths at the flanges and the web.
\(L_{w,flg} = 2 b_f + 2(b_f – t_w – 2 r_{col}) = 2 \times 378.8\ \text{mm} + 2 \times (378.8\ \text{mm} – 21.1\ \text{mm} – 2 \times 15.2\ \text{mm}) = 1412.2\ \text{mm}\)
\(L_{w,web} = 2\,(d_{col} – 2t_f – 2r_{col}) = 2 \times (362.9\ \text{mm} – 2 \times 21.1\ \text{mm} – 2 \times 15.2\ \text{mm}) = 580.6\ \text{mm}\)
Considering the flanges first, the normal and shear stresses are calculated using EN 1993-1-8:2005 Clause 4.5.3.2.
\(\sigma_{\perp} = \frac{N_x}{L_{\text{weld}} a_{flg} \sqrt{2}} = \frac{1500\ \text{kN}}{1992.8\ \text{mm} \times 8.485\ \text{mm} \times \sqrt{2}} = 62.728\ \text{MPa}\)
\(\tau_{\perp} = \frac{N_x}{L_{\text{weld}} a_{flg} \sqrt{2}} = \frac{1500\ \text{kN}}{1992.8\ \text{mm} \times 8.485\ \text{mm} \times \sqrt{2}} = 62.728\ \text{MPa}\)
\(\eta_{\parallel} = \frac{V_z}{L_{w,flg} a_{flg}} = \frac{12\ \text{kN}}{1412.2\ \text{mm} \times 8.485\ \text{mm}} = 1.0015\ \text{MPa}\)
Using EN 1993-1-8:2005 Eq. (4.1), the design weld stress based on the directional method is then obtained.
\(F_{w,Ed1} = \sqrt{(\sigma_{\perp})^2 + 3\left((\tau_{\perp})^2 + (\eta_{\parallel})^2\right)}\)
\(F_{w,Ed1} = \sqrt{(62.728\ \text{MPa})^2 + 3 \times \left((62.728\ \text{MPa})^2 + (1.0015\ \text{MPa})^2\right)}\)
\(F_{w,Ed1} = 125.47\ \text{MPa}\)
Then, the design perpendicular stress on the base metal is determined.
\(F_{w,Ed2} = \sigma_{\perp} = 62.728\ \text{MPa}\)
For the web, we use the same formula to calculate the normal and shear stresses, which gives the corresponding design weld stress and design base metal stress.
\(\sigma_{\perp} = \frac{N_x}{L_{\text{weld}} a_{\text{web}} \sqrt{2}} = \frac{1500\ \text{kN}}{1992.8\ \text{mm} \times 8.485\ \text{mm} \times \sqrt{2}} = 62.728\ \text{MPa}\)
\(\tau_{\perp} = \frac{N_x}{L_{\text{weld}} a_{\text{web}} \sqrt{2}} = \frac{1500\ \text{kN}}{1992.8\ \text{mm} \times 8.485\ \text{mm} \times \sqrt{2}} = 62.728\ \text{MPa}\)
\(\tau_{\parallel} = \frac{V_y}{L_{w,\text{web}} a_{\text{web}}} = \frac{25\ \text{kN}}{580.6\ \text{mm} \times 8.485\ \text{mm}} = 5.0747\ \text{MPa}\)
\(F_{w,Ed1} = \sqrt{(\sigma_{\perp})^2 + 3\left((\tau_{\perp})^2 + (\tau_{\parallel})^2\right)}\)
\(F_{w,Ed1} = \sqrt{(62.728\ \text{MPa})^2 + 3 \times \left((62.728\ \text{MPa})^2 + (5.0747\ \text{MPa})^2\right)}\)
\(F_{w,Ed1} = 125.76\ \text{MPa}\)
\(F_{w,Ed2} = \sigma_{\perp} = 62.728\ \text{MPa}\)
We then take the governing stress between the flange and web weld groups.
\(F_{w,Ed1} = \max(F_{w,Ed1},\ F_{w,Ed1}) = \max(125.47\ \text{MPa},\ 125.76\ \text{MPa}) = 125.76\ \text{MPa}\)
\(F_{w,Ed2} = \max(F_{w,Ed2},\ F_{w,Ed2}) = \max(62.728\ \text{MPa},\ 62.728\ \text{MPa}) = 62.728\ \text{MPa}\)
Next, we calculate the weld capacity using EN 1993-1-8:2005 Eq. (4.1). The ultimate tensile strength (fu) used in this equation is the minimum value among the column, base plate, and weld metal.
\(f_u = \min(f_{u,\text{col}},\ f_{u,\text{bp}},\ f_{uw}) = \min(370\ \text{MPa},\ 360\ \text{MPa},\ 470\ \text{MPa}) = 360\ \text{MPa}\)
\(F_{w,Rd1} = \frac{f_u}{\beta_w\,(\gamma_{M2,\text{weld}})} = \frac{360\ \text{MPa}}{0.8 \times (1.25)} = 360\ \text{MPa}\)
The resistance of the base metal is also calculated using the same equation.
\(F_{w,Rd2} = \frac{0.9 f_u}{\gamma_{M2,\text{weld}}} = \frac{0.9 \times 360\ \text{MPa}}{1.25} = 259.2\ \text{MPa}\)
Finally, we compare the fillet weld resistance to the design weld stress, and the base metal resistance to the base metal stress.
Since 125.76 MPa < 360 MPa, the weld capacity is sufficient.
Check #2: Calculate concrete bearing capacity and base plate yield capacity
A design example for the concrete bearing capacity and base plate yield capacity 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 bearing capacity (Vy shear)
When shear is transferred through the anchor rods, the rods bear against the base plate. Therefore, we need to verify that the base plate has sufficient capacity to resist the bearing load at the anchor holes.
The design shear force per anchor rod is calculated as the total shear load divided by the total number of anchors.
\(F_{b,Ed} = \frac{V_y}{n_{anc}} = \frac{25\ \text{kN}}{10} = 2.5\ \text{kN}\)
Next, we determine the factors required for the bearing resistance calculation. According to EN 1993-1-8:2005 Table 3.4, we obtain the \(\alpha_d\), \(\alpha_b\), and \(k_1\) factors.
Both end and inner anchors are considered when determining the corresponding \(\alpha_d\) factors.
\(\alpha_{d,\text{end}} = \frac{l_{\text{edge},y}}{3 d_{\text{hole}}} = \frac{100\ \text{mm}}{3 \times 26\ \text{mm}} = 1.2821\)
\(\alpha_{d,\text{inner}} = \frac{s_y}{3 d_{\text{hole}}} – \frac{1}{4} = \frac{550\ \text{mm}}{3 \times 26\ \text{mm}} – \frac{1}{4} = 6.8013\)
Using the smaller \(\alpha_d\) factor, the corresponding \(\alpha_b\) factor is calculated as:
\(\alpha_b = \min\left(\alpha_{d,\text{end}},\ \alpha_{d,\text{inner}},\ \frac{F_{u,\text{anc}}}{f_{u,\text{bp}}},\ 1.0\right) = \min\left(1.2821,\ 6.8013,\ \frac{800\ \text{MPa}}{360\ \text{MPa}},\ 1\right) = 1\)
Similarly, both edge and inner bolts are considered when determining the \(k_1\) factors.
\(k_{1,\text{edge}} = \min\left(2.8\left(\frac{l_{\text{edge},z}}{d_{\text{hole}}}\right) – 1.7,\ 1.4\left(\frac{s_z}{d_{\text{hole}}}\right) – 1.7,\ 2.5\right)\)
\(k_{1,\text{edge}} = \min\left(2.8 \times \frac{75\ \text{mm}}{26\ \text{mm}} – 1.7,\ 1.4 \times \frac{150\ \text{mm}}{26\ \text{mm}} – 1.7,\ 2.5\right) = 2.5\)
\(k_{1,\text{inner}} = \min\left(1.4\left(\frac{s_z}{d_{\text{hole}}}\right) – 1.7,\ 2.5\right) = \min\left(1.4 \times \frac{150\ \text{mm}}{26\ \text{mm}} – 1.7,\ 2.5\right) = 2.5\)
The governing \(k_1\) factor, corresponding to the smaller value, is:
\(k_1 = \min(k_{1,\text{edge}},\ k_{1,\text{inner}}) = \min(2.5,\ 2.5) = 2.5\)
Finally, we calculate the bearing resistance using the equation from EN 1993-1-8:2005 Table 3.4.
\(F_{b,Rd} = \frac{k_1 \alpha_b f_{u\_bp} d_{anc} t_{bp}}{\gamma_{M2,anchor}} \frac{2.5 \times 1 \times 360 \text{ MPa} \times 24 \text{ mm} \times 25 \text{ mm}}{1.25} = 432 \text{ kN} \)
Since 2.5 kN < 432 kN, the base plate bearing capacity is sufficient.
Check #4: Calculate base plate bearing capacity (Vz shear)
The calculation for the bearing capacity under Vz shear follows the same procedure as that for Vy shear, but considering the geometry along the Vz shear axis.
The anchor demand due to Vz shear is:
\(F_{b,Ed} = \frac{V_z}{n_{anc}} = \frac{12\ \text{kN}}{10} = 1.2\ \text{kN}\)
Using EN 1993-1-8:2005 Table 3.4, the factors are determined as follows:
\( \alpha_{d,\text{end}} = \frac{l_{\text{edge},z}}{3 d_{\text{hole}}} = \frac{75\ \text{mm}}{3 \times 26\ \text{mm}} = 0.96154 \)
\( \alpha_{d,\text{inner}} = \frac{s_z}{3 d_{\text{hole}}} – \frac{1}{4} = \frac{150\ \text{mm}}{3 \times 26\ \text{mm}} – \frac{1}{4} = 1.6731 \)
\( \alpha_b = \min\!\left(\alpha_{d,\text{end}},\ \alpha_{d,\text{inner}},\ \frac{F_{u,\text{anc}}}{f_{u,\text{bp}}},\ 1.0\right) = \min\!\left(0.96154,\ 1.6731,\ \frac{800\ \text{MPa}}{360\ \text{MPa}},\ 1\right) = 0.96154 \)
\(k_{1,\text{edge}} = \min\!\left(2.8\left(\frac{l_{\text{edge},y}}{d_{\text{hole}}}\right) – 1.7,\ 1.4\left(\frac{s_y}{d_{\text{hole}}}\right) – 1.7,\ 2.5\right)\)
\(k_{1,\text{edge}} = \min\!\left(2.8 \times \left(\frac{100\ \text{mm}}{26\ \text{mm}}\right) – 1.7,\ 1.4 \times \left(\frac{550\ \text{mm}}{26\ \text{mm}}\right) – 1.7,\ 2.5\right) = 2.5\)
\(k_{1,\text{inner}} = \min\!\left(1.4\left(\frac{s_y}{d_{\text{hole}}}\right) – 1.7,\ 2.5\right) = \min\!\left(1.4 \times \left(\frac{550\ \text{mm}}{26\ \text{mm}}\right) – 1.7,\ 2.5\right) = 2.5\)
\(k_1 = \min\!\left(k_{1,\text{edge}},\ k_{1,\text{inner}}\right) = \min(2.5,\ 2.5) = 2.5\)
Finally, the design bearing resistance of the base plate is:
\(F_{b,Rd} = \frac{k_1 \alpha_b f_{u,bp} d_{anc} t_{bp}}{\gamma_{M2,\text{anchor}}} = \frac{2.5 \times 0.96154 \times 360\ \text{MPa} \times 24\ \text{mm} \times 25\ \text{mm}}{1.25} = 415.38\ \text{kN}\)
Since 1.2 kN < 415 kN, the base plate bearing capacity is sufficient.
Check #5: Calculate concrete breakout capacity (Vy shear)
A design example for the concrete breakout capacity is already discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Check #6: Calculate concrete breakout capacity (Vz shear)
A design example for the concrete breakout capacity is already discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Check #7: Calculate concrete pryout capacity
A design example for the capacity of the concrete against shear pryout force is already discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Check #8: Calculate anchor rod shear capacity
A design example for the anchor rod shear capacity 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
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