Base Plate Design Example using EN 1993-1-8-2005, 在 1993-1-1-2005 和EN 1992-1-1-2004
问题陈述:
Determine whether the designed column-to-base plate connection is sufficient for a 100-kN compression load.
给定数据:
柱:
列部分: HE 200 乙
列区域: 7808 毫米2
列材料: S235
底盘:
基板尺寸: 400 毫米× 400 毫米
基板厚度: 20 毫米
底板材料: S235
Grout:
Grout thickness: 20 毫米
具体:
混凝土尺寸: 450 毫米× 450 毫米
混凝土厚度: 380 毫米
混凝土材料: C20/25
焊缝:
Compression load transferred through welds only? 不
分步计算:
检查一下 #1: 计算焊接容量
Since the compression load is not transferred through welds alone, a proper contact bearing surface is required to ensure that the load is transferred via bearing. Refer to 在 1090-2:2018 条款 6.8 for contact bearing preparation.
另外, use minimum weld size specified in Eurocode.
检查一下 #2: Calculate concrete bearing capacity and base plate yield capacity
The first step is to determine the design compressive strength of the joint, which depends on the geometry of the support (具体) and the geometry of the loaded area (base plate).
We begin by calculating the alpha factor, which accounts for the diffusion of the concentrated force within the foundation.
根据 在 1992-1-1:2004, 条款 6.7, the alpha coefficient is the ratio of the loaded area to the maximum distribution area, which has a similar shape to the loaded area.
We will use the equation from 部分 6.1 of Multi-Storey Steel Buildings Part 5 通过 Arcelor Mittal, Peiner Träger, 和 Corus to calculate the alpha factor.
\(
\alpha = \min \left(
1 + \压裂{t_{\文本{conc}}}{\最高(L_{\文本{BP}}, b_{\文本{BP}})},
1 + 2 \剩下( \压裂{e_h}{L_{\文本{BP}}} \对),
1 + 2 \剩下( \压裂{e_b}{b_{\文本{BP}}} \对),
3
\对)
\)
\(
\alpha = \min \left(
1 + \压裂{380 \, \文本{毫米}}{\最高(400 \, \文本{毫米}, 400 \, \文本{毫米})},
1 + 2 \剩下( \压裂{25 \, \文本{毫米}}{400 \, \文本{毫米}} \对),
1 + 2 \剩下( \压裂{25 \, \文本{毫米}}{400 \, \文本{毫米}} \对),
3
\对)
\)
\(
\alpha = 1.125
\)
哪里,
\(
e_h = \frac{L_{\文本{conc}} – L_{\文本{BP}}}{2} = frac{450 \, \文本{毫米} – 400 \, \文本{毫米}}{2} = 25 \, \文本{毫米}
\)
\(
e_b = \frac{b_{\文本{conc}} – b_{\文本{BP}}}{2} = frac{450 \, \文本{毫米} – 400 \, \文本{毫米}}{2} = 25 \, \文本{毫米}
\)
Once the geometry is defined, we will then determine the compressive strength of the concrete using 在 1992-1-1:2004, 情商. 3.15.
\(
F_{光盘} = frac{\α_{抄送} F_{钢底板设计欧洲规范}}{\gamma_C} = frac{1 \次 20 \, \文本{兆帕}}{1.5} = 13.333 \, \文本{兆帕}
\)
下一个, we assume a value for the beta coefficient. Since grout is present, beta value can be 2/3. We will calculate the design bearing strength of the joint using the combined formulas from 在 1993-1-8:2005 情商. 6.6, 和 在 1992-1-1:2004 情商. 6.63.
\(
F_{钢底板设计欧洲规范} = \beta \alpha f_{光盘} = 0.66667 \次 1.125 \次 13.333 \, \文本{兆帕} = 10 \, \文本{兆帕}
\)
The second part involves calculating the base plate yield capacity.
Since we already have the design bearing strength of the connection, we will use this to determine the smallest cantilever distance of the base plate that experiences the full bearing load. We will refer to the SCI P358 example on page 243 和 在 1993-1-1:2005 条款 6.2.5.
\(
c = t_{\文本{BP}} \sqrt{\压裂{F_{y_{\文本{BP}}}}{3 F_{钢底板设计欧洲规范} \伽玛_{莫0}}} = 20 \, \文本{毫米} \次 sqrt{\压裂{225 \, \文本{兆帕}}{3 \次 10 \, \文本{兆帕} \次 1}} = 54.772 \, \文本{毫米}
\)
We will use this dimension to calculate the effective area of the base plate. The ‘c’ dimension we calculated may overlap or not overlap near the flange. If it overlaps, we will assume the section to be a rectangular section. If it does not overlap, we will take the shape of the column.
Without overlap
With overlap
We determined that the ‘c’ dimension does not overlap. 因此, 使用 SCI P358 pg. 243, the effective area is:
\(
A_e = 4c^2 + P_{\文本{上校}}C + 一个_{\文本{上校}} = 4 \times 54.772^2 \, \文本{毫米}^ 2 + 1182 \, \文本{毫米} \次 54.772 \, \文本{毫米} + 7808 \, \文本{毫米}可以假设为 84549 \, \文本{毫米}^ 2
\)
It is important to note that the effective area should not be less than the base plate area.
最后, 我们将使用 在 1993-1-8:2005 情商. 6.6, 和EN 1992-1-1:2004, 情商. 6.63 to calculate the design bearing resistance of the base plate connection.
\(
N_{路} = 左( \分(A_e, A_0) \对) F_{钢底板设计欧洲规范} = 左( \分(84549 \, \文本{毫米}^ 2, 160000 \, \文本{毫米}^ 2) \对) \次 10 \, \文本{兆帕} = 845.49 \, \文本{千牛}
\)
以来 845.49 千牛 > 100 千牛, the design is 充足的!
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