Ejemplo de diseño de placa base usando AISC 360-22 y ACI 318-19
Declaración del problema
Determinar si la conexión de placa de columna a base diseñada es suficiente para un 5 kip-ft strong-axis moment.
Datos dados
Columna:
Sección de columna: W8x31
Área de columna: 9.130 in2
Material de columna: A992
Plato base:
Dimensiones de placa base: 15 en x 15 in
Espesor de la placa base: 1 in
Material de placa base: A36
Hormigón:
Dimensiones concretas: 18 en x 18 in
Espesor de concreto: 15 in
Material de hormigón: 3000 psi
Soldaduras:
Tamaño de soldadura: 1/4 in
Clasificación de metal de relleno: E70XX
Carga de compresión transferida solo a través de soldaduras? 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 Guía de diseño AISC 1, 3edición RD. 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 tamaño de columna, base plate size, concrete size, and the given cargas.
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 requiere instalación.
To learn more about Low Moment and Large Moment case, refer to this SkyCiv article: AISC DG1: Placas base con momento grande y momento bajo
Cálculos paso a paso
Cheque #1: Weld capacity
We start the design by checking the weld capacity of the connection. We use the Blodgett equations from Diseño de Estructuras Soldadas por Omer W.. Blodgett to determine the weld section modulus of the weld. We take the weld shape shown below as a simplification.
\(S_w = 2d_{col}b_f + \dfrac{(d_{col})^2}{3} = 2 \times 8 \text{ in} \times 8 \text{ in} + \dfrac{(8 \text{ in})^2}{3} = 149.33 \text{ in}^2\)
próximo, 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:
\(r_{uw} = \dfrac{M_z}{S_w} = \dfrac{5 \text{ kip·ft}}{149.33 \text{ in}^2} = 0.40179 \text{ kip/in}\)
Luego, calculamos el capacity of the weld. Para 0.25 in weld size, the design weld throat is 0.177 in, which is obtained by multiplying the weld size by \(\dfrac{\sqrt{2}}{2}\). Utilizando AISC 360-22 Eq. J2-4, the weld capacity per unit length is:
\(\phi r_n = \phi 0.6F_{exx}E_w k_{ds} = 0.75 \times 0.6 \times 70 \text{ ksi} \times 0.177 \text{ in} \times 1 = 5.5755 \text{ kip/in}\)
Ya que 0.40179 KPI < 5.5755 KPI , the column weld capacity is suficiente.
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. Por lo tanto, the welds at the compression flange also have enough capacity.
En este ejemplo de diseño, sin embargo, the design inputs specify that compression is not transferred through the welds, but through Llevando. Así, in the next check, we evaluate the bearing capacity of the column.
Cheque #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.
primero, Determinamos el compressive stress at the critical section, which is still the flange. We then add any contribution from axial compression. En este caso, there is none.
\(r_u = \dfrac{N_x}{A_{col}} + \dfrac{M_z 0.5d_{col}}{I_{zz,col}} = \dfrac{0 \text{ kip}}{9.13 \text{ in}^2} + \dfrac{5 \text{ kip·ft} \times 0.5 \times 8 \text{ in}}{110 \text{ in}^4} = 2.1818 \text{ ksi}\)
Luego, usamos AISC 360-22 Section J7 Eq. J7-1 para determinar el capacidad de soporte:
\(\phi r_n = \phi 1.8F_{y\_col} = 0.75 \times 1.8 \times 50 \text{ ksi} = 67.5 \text{ ksi}\)
Ya que 2.1818 KSI < 67.5 KSI, La capacidad de carga de la columna es suficiente.
Normalmente, this check is not governing.
Cheque #3: Calcular la capacidad de carga de concreto
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 es el loaded area of the base plate, tiempo A2 es el 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 Sección de 22.8.3.2.
En esto base plate design with moment, the A1 area will be calculated using the full width of the base plate, \(B\) y el 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, que es conservador. En el Software de placa base de SkyCiv, iteraciones 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, y posteriormente, to determine the actual A1 and A2 areas.
primero, 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, \(f_{p\_max}\), y el correspondiente 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 3tercero Edición. 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.
Ahora, taking the A2 and A1 values from the iteration, usamos Eq. 4-2 of AISC Design Guide 1, 3edición RD, para determinar \(f_{p\_max}\), the design bearing strength for concrete crushing:
\(f_{p\_max} = \phi \left( \min \left( 0.85 \, (f_c’) \sqrt{\dfrac{A_2}{A_1}}, \, 1.7 \, (f_c’) \right) \right)\)
\(f_{p\_max} = 0.65 \times \left( \min \left( 0.85 \times (3 \text{ ksi}) \times \sqrt{39.941}, \, 1.7 \times (3 \text{ ksi}) \right) \right) = 3.315 \text{ ksi}\)
Adicionalmente, since this design is a Large Moment Case, the base plate has both a compression side and a tension side. En esta situación, we should allow the concrete bearing to be at its full capacity and minimize the Y value. Entonces, any remaining tensile force, based on equilibrium of forces, will be taken by the anchors at the tension side.
Given this, we set \(f_{p\_actual}\) igual a \(f_{p\_max}\).
\(f_{p\_actual} = f_{p\_max} = 3.315 \text{ ksi}\)
Ya que 3.315 ksi ≤ 3.315 KSI, La capacidad de soporte de concreto es suficiente.
Cheque #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 da \(q_{max}\)como:
\(q_{max} = f_{p\_max}B_{bp} = 3.315 \text{ ksi} \times 15 \text{ in} = 49.725 \text{ kip/in}\)
Luego, 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, usamos Eq. 4-32 of AISC DG1, 3edición RD.
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 + y – simbolos, so we get two values. De aquí, we need to determine which value makes more sense. Ya que 25.97 in is greater than the base plate length, 15 in, we will not consider this value. Entonces, usaremos 0.092925 in como el longitud del rodamiento.
\(Y = Y_{case2} = 0.092925 \text{ in}\)
próximo, we calculate the cantilever lengths. Estos son los m and n dimensions from the figure below. Esto se basa en AISC DG1 3rd Edition Section 4.3.1.
The calculated m and n cantilever values are:
\(m = \dfrac{L_{bp} – 0.95d_{col}}{2} = \dfrac{15 \text{ in} – 0.95 \times 8 \text{ in}}{2} = 3.7 \text{ in}\)
\(n = \dfrac{B_{bp} – 0.8b_f}{2} = \dfrac{15 \text{ in} – 0.8 \times 8 \text{ in}}{2} = 4.3 \text{ in}\)
Luego, Lo haremos compare the calculated Y value to the larger of m and n. Since Y is less than the larger cantilever length, usaremos AISC DG1 3rd Ed. Eq. 4-52a. This gives the required base plate thickness to be:
\(t_{p(req)} = \sqrt{\dfrac{4f_{p\_actual}Y \left( n – \dfrac{Y}{2} \right)}{\phi F_{y\_bp}}}\)
\(t_{p(req)} = \sqrt{\dfrac{4 \times 3.315 \text{ ksi} \times 0.092925 \text{ in} \times \left( 4.3 \text{ in} – \dfrac{0.092925 \text{ in}}{2} \right)}{0.9 \times 36 \text{ ksi}}} = 0.4022 \text{ in}\)
Finalmente, let’s compare it to the actual base plate thickness.
\(t_{p(actual)} = t_{bp} = 1 \text{ in}\)
Ya que 0.4022 in < 1 in, the yielding capacity of the base plate is suficiente.
Cheque #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, usamos AISC DG1 3rd Ed. Eq. 4-31.
\(T_u = q_{max}Y = 49.725 \text{ kip/in} \times 0.092925 \text{ in} = 4.6207 \text{ kip}\)
Luego, calculamos el 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. Usamos 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.
Finalmente, we determine the required base plate thickness usando Eq. 4-62a.
\(t_{p(req)} = \sqrt{\dfrac{4T_u x}{B_{bp}\phi F_{y\_bp}}} = \sqrt{\dfrac{4 \times 4.6207 \text{ kip} \times 1.749 \text{ in}}{15 \text{ in} \times 0.9 \times 36 \text{ ksi}}} = 0.25791 \text{ in}\)
Ya que 0.25791 in < 1 in, la yielding capacity of the base plate is suficiente.
Cheque #6: Comprobaciones de ancla
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 anclaje.
Resumen de diseño
El software de diseño de placa base SkyCiv puede generar automáticamente un informe de cálculo paso a paso para este ejemplo de diseño. También proporciona un resumen de los controles realizados y sus proporciones resultantes, Hacer que la información sea fácil de entender de un vistazo. A continuación se muestra una tabla de resumen de muestra, que se incluye en el informe.
Comprar software de placa base
Compre la versión completa del módulo de diseño de la placa base por sí solo sin ningún otro módulo SkyCiv. Esto le da un conjunto completo de resultados para el diseño de placa base, incluyendo informes detallados y más funcionalidad.
