Ejemplo de diseño de placa base usando AISC 360-22 y ACI 318-19
Declaración del problema:
Determine whether the designed column-to-base plate connection is sufficient for a 20-kip tension load.
Datos dados:
Columna:
Sección de columna: W12x53
Área de columna: 15.6 in2
Material de columna: A992
Plato base:
Dimensiones de placa base: 18 en x 18 in
Espesor de la placa base: 3/4 in
Material de placa base: A36
Lechada:
Espesor de la lechada: 1 in
Hormigón:
Dimensiones concretas: 22 en x 22 in
Espesor de concreto: 15 in
Material de hormigón: 4000 psi
Cracked or Uncracked: Cracked
Anchors:
Anchor diameter: 3/4 in
Effective embedment length: 12 in
Embedded plate width: 3 in
Embedded plate thickness: 1/4 in
Anchor offset distance from face of column web: 2.8275 in
Soldaduras:
Tamaño de soldadura: 1/4 in
Clasificación de metal de relleno: E70XX
Anchor Data (de SkyCiv Calculator):
Definitions:
Load Path:
When a base plate is subjected to uplift (de tensión) efectivo, these forces are transferred to the anchor rods, which in turn induce bending moments in the base plate. The bending action can be visualized as cantilever bending occurring around the flanges or web of the column section, depending on where the anchors are positioned.
En el Software de diseño de placa base SkyCiv, only anchors located within the anchor tension zone are considered effective in resisting uplift. This zone typically includes areas near the column flanges or web. Anchors outside this zone do not contribute to tension resistance and are excluded from the uplift calculations.
To determine the effective area of the base plate that resists bending, a 45-degree dispersion is assumed from the centerline of each anchor rod toward the column face. This dispersion defines the effective weld length and helps establish the effective bending width del plato.
The assumption simplifies the base plate analysis by approximating how the uplift force spreads through the plate.
Anchor Groups:
El Software de diseño de placa base SkyCiv includes an intuitive feature that identifies which anchors are part of an anchor group for evaluating ruptura de concreto y concrete side-face blowout failures.
Un anchor group consists of multiple anchors with similar effective embedment depths and spacing, and are close enough that their projected resistance areas overlap. When anchors are grouped, their capacities are combined to resist the total tension force applied to the group.
Anchors that do not meet the grouping criteria are treated as single anchors. En este caso, only the tension force on the individual anchor is checked against its own effective resistance area.
Cálculos paso a paso:
Cheque #1: Calcular la capacidad de soldadura
Empezar, we need to calculate the load per anchor and the effective weld length per anchor. The effective weld length is determined by the shortest length from the 45° dispersion, constrained by the actual weld length and anchor spacing.
For this calculation, anchors are classified as either end anchors o intermediate anchors. End anchors are located at the ends of a row or column of anchors, while intermediate anchors are positioned between them. The calculation method differs for each and depends on the column geometry. En este ejemplo, there are two anchors along the web, and both are classified as end anchors.
For end anchors, the effective weld length is limited by the available distance from the anchor centerline to the column fillet. The 45° dispersion must not extend beyond this boundary.
\(
l_r = \frac{D_{columna} – 2T_F – 2r_{columna} – s_y(n_{a,lado} – 1)}{2} = frac{12.1 \, \texto{in} – 2 \veces 0.575 \, \texto{in} – 2 \veces 0.605 \, \texto{in} – 5 \, \texto{in} \veces (2 – 1)}{2} = 2.37 \, \texto{in}
\)
On the inner side, the effective length is limited by half the anchor spacing. The total effective weld length for the end anchor is the sum of the outer and inner lengths.
\(
l_{efecto,final} = \min(hacer, 0.5s_y) + \min(hacer, l_r)
\)
\(
l_{efecto,final} = \min(2.8275 \, \texto{in}, 0.5 \veces 5 \, \texto{in}) + \min(2.8275 \, \texto{in}, 2.37 \, \texto{in}) = 4.87 \, \texto{in}
\)
En este ejemplo, la final effective weld length for the web anchor is taken as the effective length of the end anchor.
\(
l_{efecto} = L_{efecto,final} = 4.87 \, \texto{in}
\)
próximo, let’s calculate the load per anchor. For a given set of four (4) anclas, the load per anchor is:
\(
T_{tu,ancla} = frac{N_X}{n_{a,t}} = frac{20 \, \texto{kip}}{4} = 5 \, \texto{kip}
\)
Using the calculated effective weld length, we can now determine the required force per unit length on the weld.
\(
r_u = frac{T_{tu,ancla}}{l_{efecto}} = frac{5 \, \texto{kip}}{4.87 \, \texto{in}} = 1.0267 \, \texto{kip/in}
\)
Ahora, usaremos AISC 360-22, Chapter J2.4 to calculate the design strength of the fillet weld.
Since the applied load is purely axial tension, the angle \(\theta) is taken as 90°, and the directional strength coefficient kds is calculated according to AISC 360-22 Eq. J2-5.
\(
Suma de fuerzas de tensión de anclajes con área de cono de ruptura de concreto común{ds} = 1.0 + 0.5(\sin(\theta))^{1.5} = 1 + 0.5 \veces (\sin(1.5708))^{1.5} = 1.5
\)
Finalmente, aplicaremos AISC 360-22 Eq. J2-4 para determinar el design strength of the fillet weld per unit length.
\(
\phi r_n = \phi 0.6 F_{Exx} MI_{w,web} Suma de fuerzas de tensión de anclajes con área de cono de ruptura de concreto común{ds} = 0.75 \veces 0.6 \veces 70 \, \texto{KSI} \veces 0.177 \, \texto{in} \veces 1.5 = 8.3633 \, \texto{kip/in}
\)
Ya que 1.0267 KPI < 8.3633 KPI, La capacidad de soldadura es suficiente.
Cheque #2: Calculate base plate flexural yielding capacity due to tension load
Using the load per anchor and the offset distance from the center of the anchor to the face of the column (serving as the load eccentricity), the moment applied to the base plate can be calculated using a viga voladiza assumption.
\(
M_u = T_{tu,\texto{ancla}} e = 5 \, \texto{kip} \veces 2.8275 \, \texto{in} = 14.137 \, \texto{kip} \cdot \text{in}
\)
próximo, using the calculated effective weld length from the previous check as the bending width, podemos calcular el Calcula la capacidad de carga of the base plate using AISC 360-22, Ecuación 2-1:
\(
\phi M_n = \phi F_{y,\texto{pb}} Z_{\texto{efecto}} = 0.9 \veces 36 \, \texto{KSI} \veces 0.68484 \, \texto{in}^3 = 22.189 \, \texto{kip} \cdot \text{in}
\)
Dónde,
\(
Z_{\texto{efecto}} = frac{l_{\texto{efecto}} (A continuación se muestra un ejemplo de algunos cálculos de placa base australianos que se usan comúnmente en el diseño de placa base{\texto{pb}})^ 2}{4} = frac{4.87 \, \texto{in} \veces (0.75 \, \texto{in})^ 2}{4} = 0.68484 \, \texto{in}^ 3
\)
Ya que 14.137 pollo en < 22.189 pollo en, the base plate flexural yielding capacity is suficiente.
Cheque #3: Calculate anchor rod tensile capacity
To evaluate the tensile capacity of the anchor rod, usaremos ACI 318-19 Ecuación 17.6.1.2.
primero, Determinamos el specified tensile strength of the anchor steel. This is the lowest value permitted by ACI 318-19 Cláusula 17.6.1.2, with reference to material properties in AISC 360-22 Tabla J3.2.
\(
F_{\texto{uta}} = min left( 0.75 F_{tu,\texto{Congreso Nacional Africano}}, 1.9 F_{y,\texto{Congreso Nacional Africano}}, 125 \verdad) = min left( 0.75 \veces 120 \, \texto{KSI}, 1.9 \veces 92 \, \texto{KSI}, 125.00 \, \texto{KSI} \verdad) = 90 \, \texto{KSI}
\)
próximo, calculamos el effective cross-sectional area of the anchor rod. This is based on ACI 318-19 Commentary Clause R17.6.1.2, which accounts for thread geometry. The number of threads per inch is taken from ASME B1.1-2019 Table 1.
\(
UNA_{se,norte} = frac{\pi}{4} \izquierda( d_a – \frac{0.9743}{n_t} \verdad)^2 = \frac{\pi}{4} \veces left( 0.75 \, \texto{in} – \frac{0.9743}{10 \, \texto{in}^{-1}} \verdad)^2 = 0.33446 \, \texto{in}^ 2
\)
With these values, Solicitamos ACI 318-19 Ecuación 17.6.1.2 to compute the design tensile strength of the anchor rod.
\(
\phi N_{a} = phi A_{se,norte} F_{\texto{uta}} = 0.75 \veces 0.33446 \, \texto{in}^2 \times 90 \, \texto{KSI} = 22.576 \, \texto{kip}
\)
Recall the previously calculated tension load per anchor:
\(
NORTE_{ua} = frac{N_X}{n_{a,t}} = frac{20 \, \texto{kip}}{4} = 5 \, \texto{kip}
\)
Ya que 5 kip < 22.576 kip, the anchor rod tensile capacity is suficiente.
Cheque #4: Calculate concrete breakout capacity in tension
Before calculating the breakout capacity, we must first determine whether the member qualifies as a narrow member. De acuerdo a ACI 318-19 Cláusula 17.6.2.1.2, the member meets the criteria for a narrow member. Por lo tanto, a modified effective embedment length must be used in the calculations.
It is determined that the modified effective embedment length, h’ef, of the anchor group is:
\(
h’_{\texto{ef}} = 5.667 \, \texto{in}
\)
Utilizando ACI 318-19 Cláusula 17.6.2, calculamos el maximum projected concrete cone area para un solo ancla, based on the modified effective embedment length.
\(
UNA_{NORTE_{co}} = 9 \izquierda( h’_{ef,g1} \verdad)^2 = 9 \veces left( 5.6667 \, \texto{in} \verdad)^2 = 289 \, \texto{in}^ 2
\)
De igual forma, we use the modified effective embedment length to calculate the actual projected concrete cone area of the anchor group.
\(
UNA_{N_c} = min left( n_{a,g1} UNA_{NORTE_{co}}, L_{N_c} SI_{N_c} \verdad) = min left( 4 \veces 289 \, \texto{in}^ 2, 22 \, \texto{in} \veces 22 \, \texto{in} \verdad) = 484 \, \texto{in}^ 2
\)
Dónde,
\(
L_{N_c} = min left( C_{\texto{izquierda},g1}, 1.5 h’_{\texto{ef},g1} \verdad)
+ \izquierda( \min izquierda( s_{\texto{suma},z,g1}, 3 h’_{\texto{ef},g1} \izquierda( n_{z,g1} – 1 \verdad) \verdad) \verdad)
+ \min izquierda( C_{\texto{verdad},g1}, 1.5 h’_{\texto{ef},g1} \verdad)
\)
\(
L_{N_c} = min left( 8 \, \texto{in}, 1.5 \veces 5.6667 \, \texto{in} \verdad)
+ \izquierda( \min izquierda( 6 \, \texto{in}, 3 \veces 5.6667 \, \texto{in} \veces left( 2 – 1 \verdad) \verdad) \verdad)
+ \min izquierda( 8 \, \texto{in}, 1.5 \veces 5.6667 \, \texto{in} \verdad)
\)
\(
L_{N_c} = 22 \, \texto{in}
\)
\(
SI_{N_c} = min left( C_{\texto{superior},g1}, 1.5 h’_{\texto{ef},g1} \verdad)
+ \izquierda( \min izquierda( s_{\texto{suma},y,g1}, 3 h’_{\texto{ef},g1} \izquierda( n_{y,g1} – 1 \verdad) \verdad) \verdad)
+ \min izquierda( C_{\texto{inferior},g1}, 1.5 h’_{\texto{ef},g1} \verdad)
\)
\(
SI_{N_c} = min left( 8.5 \, \texto{in}, 1.5 \veces 5.6667 \, \texto{in} \verdad)
+ \izquierda( \min izquierda( 5 \, \texto{in}, 3 \veces 5.6667 \, \texto{in} \veces left( 2 – 1 \verdad) \verdad) \verdad)
+ \min izquierda( 8.5 \, \texto{in}, 1.5 \veces 5.6667 \, \texto{in} \verdad)
\)
\(
SI_{N_c} = 22 \, \texto{in}
\)
próximo, we evaluate the basic concrete breakout strength of a single anchor using ACI 318-19 Cláusula 17.6.2.2.1
\(
N_b = k_c lambda_a sqrt{\frac{f’_c}{\texto{psi}}} \izquierda( \frac{h’_{\texto{ef},g1}}{\texto{in}} \verdad)^{1.5} \, \texto{lbf}
\)
\(
N_b = 24 \veces 1 \veces sqrt{\frac{4 \, \texto{KSI}}{0.001 \, \texto{KSI}}} \veces left( \frac{5.6667 \, \texto{in}}{1 \, \texto{in}} \verdad)^{1.5} \veces 0.001 \, \texto{kip} = 20.475 \, \texto{kip}
\)
Dónde,
- \(Suma de fuerzas de tensión de anclajes con área de cono de ruptura de concreto común{c} = 24\) para anclajes empotrados
- \(\lambda = 1.0 \) for normal-weight concrete
Ahora, we assess the effects of geometry by calculating the edge effect factor y el eccentricity factor.
The shortest edge distance of the anchor group is determined as:
\(
C_{a,\texto{min}} = min left( C_{\texto{izquierda},g1}, C_{\texto{verdad},g1}, C_{\texto{superior},g1}, C_{\texto{inferior},g1} \verdad)
= min left( 8 \, \texto{in}, 8 \, \texto{in}, 8.5 \, \texto{in}, 8.5 \, \texto{in} \verdad) = 8 \, \texto{in}
\)
De acuerdo a ACI 318-19 Cláusula 17.6.2.4.1, the breakout edge effect factor es:
\(
\Psi_{ed,norte} = min left( 1.0, 0.7 + 0.3 \izquierda( \frac{C_{a,\texto{min}}}{1.5 h’_{\texto{ef},g1}} \verdad) \verdad)
= min left( 1, 0.7 + 0.3 \veces left( \frac{8 \, \texto{in}}{1.5 \veces 5.6667 \, \texto{in}} \verdad) \verdad) = 0.98235
\)
Since the tension load is applied at the centroid of the anchor group, the eccentricity is zero. Así, la eccentricity factor, also from Clause 17.6.2.4.1, es:
\(
\Psi_{CE,norte} = min left( 1.0, \frac{1}{1 + \frac{2 y N}{3 h’_{\texto{ef},g1}}} \verdad)
= min left( 1, \frac{1}{1 + \frac{2 \veces 0}{3 \veces 5.6667 \, \texto{in}}} \verdad) = 1
\)
Adicionalmente, both the cracking factor y el splitting factor are taken as:
\(
\Psi_{c,norte} = 1
\)
\(
\Psi_{cp,norte} = 1
\)
Luego, we combine all these factors and use ACI 318-19 Eq. 17.6.2.1b to evaluate the concrete breakout strength of the anchor group:
\(
\phi N_{cbg} = \phi \left( \frac{UNA_{N_c}}{UNA_{NORTE_{co}}} \verdad) \Psi_{CE,norte} \Psi_{ed,norte} \Psi_{c,norte} \Psi_{cp,norte} Nótese bien
\)
\(
\phi N_{cbg} = 0.7 \veces left( \frac{484 \, \texto{in}^ 2}{289 \, \texto{in}^ 2} \verdad) \veces 1 \veces 0.98235 \veces 1 \veces 1 \veces 20.475 \, \texto{kip} = 23.58 \, \texto{kip}
\)
El total applied tension load on the anchor group is the product of the individual anchor load and the number of anchors:
\(
NORTE_{ua} = left( \frac{N_X}{n_{a,t}} \verdad) n_{a,g1} = left( \frac{20 \, \texto{kip}}{4} \verdad) \veces 4 = 20 \, \texto{kip}
\)
Ya que 20 kips < 23.58 kips , the concrete breakout capacity is suficiente.
Cheque #5: Calculate anchor pullout capacity
The pullout capacity of an anchor is governed by the resistance at its embedded end. Empezar, we calculate the bearing area of the embedded plate, which is the net area after subtracting the area occupied by the anchor rod.
For a rectangular embedded plate, la bearing area is calculated as:
\(
UNA_{brg} = left( \izquierda( B_{embed\_plate} \verdad)^2 Derecha) – UNA_{vara} = left( \izquierda( 3 \, \texto{in} \verdad)^2 Derecha) – 0.44179 \, \texto{in}^2 = 8.5582 \, \texto{in}^ 2
\)
Dónde,
\(
UNA_{vara} = frac{\pi}{4} \izquierda( d_a \right)^2 = \frac{\pi}{4} \veces left( 0.75 \, \texto{in} \verdad)^2 = 0.44179 \, \texto{in}^ 2
\)
próximo, Determinamos el basic anchor pullout strength usando ACI 318-19 Equation 17.6.3.2.2a.
\(
N_b = 8 UNA_{brg} \izquierda( F’_C Derecha) = 8 \veces 8.5582 \, \texto{in}^2 \times \left( 4 \, \texto{KSI} \verdad) = 273.86 \, \texto{kip}
\)
We then apply the appropriate resistance factor and pullout cracking factor:
- Por cracked hormigón, \(\Psi_{cp} = 1.0\)
- Por uncracked hormigón, \(\Psi_{cp} = 1.4\)
Using these, Calculamos el design anchor pullout strength in tension por ACI 318-19 Ecuación 17.6.3.1.
\(
\phi N_{pn} = \phi \Psi_{c,pag} N_b = 0.7 \veces 1 \veces 273.86 \, \texto{kip} = 191.7 \, \texto{kip}
\)
Recall the previously calculated tension load per anchor:
\(
NORTE_{ua} = frac{N_X}{n_{a,t}} = frac{20 \, \texto{kip}}{4} = 5 \, \texto{kip}
\)
Ya que 5 kips < 191.7 kips , the anchor pullout capacity is suficiente.
Cheque #6: Calculate embed plate flexural capacity
This is a supplementary check performed using the Software de diseño de placa base de SkyCiv to verify that the embedded plate has sufficient flexural capacity and will not yield under the applied pullout loads.
primero, we determine the length of the free (unsupported) end of the embedded plate, measured from the edge of the support to the face of the rod.
\(
b’ = frac{B_{embed\_plate} – d_a}{2} = frac{3 \, \texto{in} – 0.75 \, \texto{in}}{2} = 1.125 \, \texto{in}
\)
próximo, calculamos el momento flector induced by the uniform bearing pressure. This pressure represents the force transferred from the anchor pullout action onto the embedded plate.
\(
m_f = \frac{\izquierda( \frac{T_a}{UNA_{brg}} \verdad) \izquierda( b’ \verdad)^ 2}{2} = frac{\izquierda( \frac{5 \, \texto{kip}}{8.5582 \, \texto{in}^ 2} \verdad) \veces left( 1.125 \, \texto{in} \verdad)^ 2}{2} = 0.36971 \, \texto{kip}
\)
Finalmente, using the calculated moment and given material properties, we will determine the minimum required plate thickness to resist flexural yielding.
\(
A continuación se muestra un ejemplo de algunos cálculos de placa base australianos que se usan comúnmente en el diseño de placa base{min} = sqrt{\frac{4 m_f}{\phi F_{y\_ep}}} = sqrt{\frac{4 \veces 0.36971 \, \texto{kip}}{0.9 \veces 36 \, \texto{KSI}}} = 0.21364 \, \texto{in}
\)
Recall actual embedded plate thickness:
\(
A continuación se muestra un ejemplo de algunos cálculos de placa base australianos que se usan comúnmente en el diseño de placa base{actual} = t_{embed\_plate} = 0.25 \, \texto{in}
\)
Ya que 0.21364 in < 0.25 in, the embedded plate flexural capacity is suficiente.
Cheque #7: Calculate side-face blowout capacity in Y-direction
This calculation is not applicable for this example, as the conditions specified in ACI 318-19 Cláusula 17.6.4 are not met. Por lo tanto, side-face blowout failure along the Y-direction will not occur.
Cheque #8: Calculate side-face blowout capacity in Z-direction
This calculation is not applicable for this example, as the conditions specified in ACI 318-19 Cláusula 17.6.4 are not met. Por lo tanto, side-face blowout failure along the Z-direction will not occur.
Resumen de diseño
El Software de diseño de placa base de SkyCiv can automatically generate a step-by-step calculation report for this design example. 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.
Informe de muestra de SkyCiv
haga clic aquí Para descargar un informe de muestra.
Comprar software de placa base
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