Base Plate Design Example using CSA S16:19 and CSA A23.3:19
Declaración del problema:
Determine whether the designed column-to-base plate connection is sufficient for a 50-kN tension load.
Datos dados:
Columna:
Sección de columna: HS324X9.5
Área de columna: 9410 mm2
Material de columna: 230GRAMO
Plato base:
Dimensiones de placa base: 500 mm x 500 mm
Espesor de la placa base: 20 mm
Material de placa base: 230GRAMO
Lechada:
Espesor de la lechada: 20 mm
Hormigón:
Dimensiones concretas: 550 mm x 550 mm
Espesor de concreto: 200 mm
Material de hormigón: 20.68 MPa
Cracked or Uncracked: Cracked
Anchors:
Anchor diameter: 19.1 mm
Effective embedment length: 130.0 mm
Hook length: 60mm
Anchor offset distance from face of column: 120.84 mm
Soldaduras:
Weld type: CJP
Clasificación de metal de relleno: E43xx
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. In the case of a circular column, the anchor tension zone includes the entire area outside the column perimeter. 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 determine the effective weld length for each anchor. El effective weld length is based on a 45° dispersion line drawn from the center of the anchor to the face of the column. If this 45° line does not intersect the column, la tangent points are used instead. Adicionalmente, if the anchors are closely spaced, the effective weld length is reduced to avoid overlap. Finalmente, the sum of all effective weld lengths must not exceed the actual weldable length available along the column circumference.
Let’s apply this to our example. Based on the given geometry, the 45° line from the anchor does not intersect the column. Como resultado, the arc length between the tangent points is used instead. This arc length must also account for any adjacent anchors, with any overlapping portions subtracted to avoid double-counting. The calculated arc length is:
\(
l_{\texto{arc}} = 254.47 \, \texto{mm}
\)
This arc length calculation is fully automated in the SkyCiv Base Plate Design Software, but it can also be performed manually using trigonometric methods. You can try the free tool from this link.
Considering the available weldable length along the column’s circumference, the final effective weld length es:
\(
l_{\texto{efecto}} = min left( l_{\texto{arc}}, \frac{\pi d_{\texto{columna}}}{n_{a,t}} \verdad) = min left( 254.47 \, \texto{mm}, \frac{\pi \times 324 \, \texto{mm}}{4} \verdad) = 254.47 \, \texto{mm}
\)
próximo, let’s calculate the load per anchor. For a given set of four (4) anclas, the load per anchor is:
\(
T_{tu,\texto{ancla}} = frac{N_X}{n_{a,t}} = frac{50 \, \texto{kN}}{4} = 12.5 \, \texto{kN}
\)
Using the calculated effective weld length, we can now compute the required force per unit length acting on the weld.
\(
v_f = \frac{T_{tu,\texto{ancla}}}{l_{\texto{efecto}}} = frac{12.5 \, \texto{kN}}{254.47 \, \texto{mm}} = 0.049122 \, \texto{kN / mm}
\)
Ahora, we refer to CSA S16:19 Cláusula 13.13.3.1 para calcular el factored resistance of the complete joint penetration (CJP) soldar. This requires the base metal resistance, expressed in force per unit length, for both the column and the base plate materials.
\(
A continuación se muestra un ejemplo de algunos cálculos de placa base australianos que se usan comúnmente en el diseño de placa base{r,\texto{bm}} = \phi \left( \min izquierda( F_{y,\texto{columna}} A continuación se muestra un ejemplo de algunos cálculos de placa base australianos que se usan comúnmente en el diseño de placa base{\texto{columna}}, F_{y,\texto{pb}} A continuación se muestra un ejemplo de algunos cálculos de placa base australianos que se usan comúnmente en el diseño de placa base{\texto{pb}} \verdad) \verdad)
\)
\(
A continuación se muestra un ejemplo de algunos cálculos de placa base australianos que se usan comúnmente en el diseño de placa base{r,\texto{bm}} = 0.9 \veces left( \min izquierda( 230 \, \texto{MPa} \veces 9.53 \, \texto{mm}, 230 \, \texto{MPa} \veces 20 \, \texto{mm} \verdad) \verdad) = 1.9727 \, \texto{kN / mm}
\)
Ya que 0.049122 kN / mm < 1.9727 kN / mm, 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, the moment applied to the base plate can be calculated using a viga voladiza assumption. For a circular column, the load eccentricity is determined by considering the sagitta of the welded arc, and can be calculated as follows:
\(
mi_{\texto{pipe}} = d_o + r_{\texto{columna}} \izquierda( 1 – \porque izquierda( \frac{l_{\texto{efecto}}}{2 r_{\texto{columna}}} \verdad) \verdad)
\)
\(
mi_{\texto{pipe}} = 120.84 \, \texto{mm} + 162 \, \texto{mm} \veces left( 1 – \porque izquierda( \frac{254.47 \, \texto{mm}}{2 \veces 162 \, \texto{mm}} \verdad) \verdad) = 168.29 \, \texto{mm}
\)
The induced moment is computed as:
\(
M_f = T_{tu,\texto{ancla}} mi_{\texto{pipe}} = 12.5 \, \texto{kN} \veces 168.29 \, \texto{mm} = 2103.6 \, \texto{kN} \cdot \text{mm}
\)
próximo, we will determine the bending width of the base plate. Para esto, we use the chord length corresponding to the effective weld arc.
\(
\theta_{\texto{trabajo}} = frac{l_{\texto{efecto}}}{0.5 D_{\texto{columna}}} = frac{254.47 \, \texto{mm}}{0.5 \veces 324 \, \texto{mm}} = 1.5708
\)
\(
b = d_{\texto{columna}} \izquierda( \pecado izquierda( \frac{\theta_{\texto{trabajo}}}{2} \verdad) \verdad) = 324 \, \texto{mm} \veces left( \pecado izquierda( \frac{1.5708}{2} \verdad) \verdad) = 229.1 \, \texto{mm}
\)
Finalmente, podemos calcular el factorizado flexural resistance of the base plate using CSA S16:19 Cláusula 13.5.
\(
M_r = \phi F_{y,\texto{pb}} Z_{\texto{efecto}} = 0.9 \veces 230 \, \texto{MPa} \veces 22910 \, \texto{mm}^3 = 4742.4 \, \texto{kN} \cdot \text{mm}
\)
Dónde,
\(
Z_{\texto{efecto}} = frac{b (A continuación se muestra un ejemplo de algunos cálculos de placa base australianos que se usan comúnmente en el diseño de placa base{\texto{pb}})^ 2}{4} = frac{229.1 \, \texto{mm} \veces (20 \, \texto{mm})^ 2}{4} = 22910 \, \texto{mm}^ 3
\)
Ya que 2103.6 kN-mm < 4742.4 kN-mm, the base plate flexural yielding capacity is suficiente.
Cheque #3: Calculate anchor rod tensile capacity
To evaluate the tensile capacity of the anchor rod, we refer to CSA A23.3:19 Clause D.6.1.2 and CSA S16:19 Cláusula 25.3.2.1.
primero, Determinamos el specified tensile strength of the anchor steel. This is the lowest value permitted by CSA A23.3:19 Clause D.6.1.2.
\(
F_{\texto{uta}} = min left( F_{tu,\texto{Congreso Nacional Africano}}, 1.9 F_{y,\texto{Congreso Nacional Africano}}, 860 \verdad) = min left( 400 \, \texto{MPa}, 1.9 \veces 248.2 \, \texto{MPa}, 860.00 \, \texto{MPa} \verdad) = 400 \, \texto{MPa}
\)
próximo, Determinamos el effective cross-sectional area of the anchor rod in tension using CAC Concrete Design Handbook, 3edición RD, Tabla 12.3.
\(
UNA_{se,norte} = 215 \, \texto{mm}^ 2
\)
With these values, Solicitamos CSA A23.3:19 Eq. D.2 to compute the factored tensile resistance of the anchor rod.
\(
NORTE_{\texto{sar}} = A_{se,norte} \phi_s f_{\texto{uta}} R = 215 \, \texto{mm}^2 \times 0.85 \veces 400 \, \texto{MPa} \veces 0.8 = 58.465 \, \texto{kN}
\)
Adicionalmente, we evaluate the factored tensile resistance according to CSA S16:19 Cláusula 25.3.2.1.
\(
T_r = \phi_{ar} 0.85 UNA_{ar} F_{tu,\texto{Congreso Nacional Africano}} = 0.67 \veces 0.85 \veces 285.02 \, \texto{mm}^2 \times 400 \, \texto{MPa} = 64.912 \, \texto{kN}
\)
After comparing the two, we identify that the factored resistance calculated using CSA A23.3:19 governs in this case.
Recall the previously calculated tension load per anchor:
\(
NORTE_{fa} = frac{N_X}{n_{a,t}} = frac{50 \, \texto{kN}}{4} = 12.5 \, \texto{kN}
\)
Ya que 12.5 kN < 58.465 kN, 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 CSA A23.3:19 Clause D.6.2.3, the member does not meet the criteria for a narrow member. Por lo tanto, the given effective embedment length will be used in the calculations.
Utilizando CSA A23.3:19 Eq. D.5, calculamos el maximum projected concrete cone area para un solo ancla, based on the effective embedment length.
\(
UNA_{Recuerda} = 9 (h_{ef,s1})^2 = 9 \veces (130 \, \texto{mm})^2 = 152100 \, \texto{mm}^ 2
\)
De igual forma, we use the effective embedment length to calculate the actual projected concrete cone area of the single anchor.
\(
UNA_{Carolina del Norte} = L_{Carolina del Norte} SI_{Carolina del Norte} = 270 \, \texto{mm} \veces 270 \, \texto{mm} = 72900 \, \texto{mm}^ 2
\)
Dónde,
\(
L_{Carolina del Norte} = left( \min izquierda( C_{\texto{izquierda},s1}, 1.5 h_{ef,s1} \verdad) \verdad) + \izquierda( \min izquierda( C_{\texto{verdad},s1}, 1.5 h_{ef,s1} \verdad) \verdad)
\)
\(
L_{Carolina del Norte} = left( \min izquierda( 475 \, \texto{mm}, 1.5 \veces 130 \, \texto{mm} \verdad) \verdad) + \izquierda( \min izquierda( 75 \, \texto{mm}, 1.5 \veces 130 \, \texto{mm} \verdad) \verdad)
\)
\(
L_{Carolina del Norte} = 270 \, \texto{mm}
\)
\(
SI_{Carolina del Norte} = left( \min izquierda( C_{\texto{superior},s1}, 1.5 h_{ef,s1} \verdad) \verdad) + \izquierda( \min izquierda( C_{\texto{inferior},s1}, 1.5 h_{ef,s1} \verdad) \verdad)
\)
\(
SI_{Carolina del Norte} = left( \min izquierda( 75 \, \texto{mm}, 1.5 \veces 130 \, \texto{mm} \verdad) \verdad) + \izquierda( \min izquierda( 475 \, \texto{mm}, 1.5 \veces 130 \, \texto{mm} \verdad) \verdad)
\)
\(
SI_{Carolina del Norte} = 270 \, \texto{mm}
\)
próximo, we evaluate the factorizado basic concrete breakout resistance of a single anchor using CSA A23.3:19 Eq. D.6
\(
NORTE_{br} = k_c \phi \lambda_a \sqrt{\frac{f’_c}{\texto{MPa}}} \izquierda( \frac{h_{ef,s1}}{\texto{mm}} \verdad)^{1.5} R N
\)
\(
NORTE_{br} = 10 \veces 0.65 \veces 1 \veces sqrt{\frac{20.68 \, \texto{MPa}}{1 \, \texto{MPa}}} \veces left( \frac{130 \, \texto{mm}}{1 \, \texto{mm}} \verdad)^{1.5} \veces 1 \veces 0.001 \, \texto{kN} = 43.813 \, \texto{kN}
\)
Dónde,
- \(Suma de fuerzas de tensión de anclajes con área de cono de ruptura de concreto común{c} = 10\) para anclajes empotrados
- \(\lambda = 1.0 \) for normal-weight concrete
Ahora, we assess the effects of geometry by calculating the edge effect factor.
The shortest edge distance of the anchor group is determined as:
\(
C_{a,\texto{min}} = min left( C_{\texto{izquierda},s1}, C_{\texto{verdad},s1}, C_{\texto{superior},s1}, C_{\texto{inferior},s1} \verdad) = min left( 475 \, \texto{mm}, 75 \, \texto{mm}, 75 \, \texto{mm}, 475 \, \texto{mm} \verdad) = 75 \, \texto{mm}
\)
De acuerdo a CSA A23.3:19 Eq. D.10 and D.11, the breakout edge effect factor es:
\(
\Psi_{ed,norte} = min left( 1.0, 0.7 + 0.3 \izquierda( \frac{C_{a,\texto{min}}}{1.5 h_{ef,s1}} \verdad) \verdad) = min left( 1, 0.7 + 0.3 \veces left( \frac{75 \, \texto{mm}}{1.5 \veces 130 \, \texto{mm}} \verdad) \verdad) = 0.81538
\)
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 factorizado concrete breakout resistance of the single anchor:
\(
NORTE_{cbr} = left( \frac{UNA_{Carolina del Norte}}{UNA_{Recuerda}} \verdad) \Psi_{ed,norte} \Psi_{c,norte} \Psi_{cp,norte} NORTE_{br} = left( \frac{72900 \, \texto{mm}^ 2}{152100 \, \texto{mm}^ 2} \verdad) \veces 0.81538 \veces 1 \veces 1 \veces 43.813 \, \texto{kN} = 17.122 \, \texto{kN}
\)
Recall the previously calculated tension load per anchor:
\(
NORTE_{fa} = frac{N_X}{n_{a,s}} = frac{50 \, \texto{kN}}{4} = 12.5 \, \texto{kN}
\)
Ya que 12.5 kN < 17.122 kN the concrete breakout capacity is suficiente.
This concrete breakout calculation is based on Anchor ID #1. The same capacity will apply to the other anchors due to the symmetric design.
Cheque #5: Calculate anchor pullout capacity
The pullout capacity of an anchor is governed by the resistance at its embedded end. For hooked anchors, it is dependent on its hook length.
We compute the factored basic anchor pullout resistance por CSA A23.3:19 Eq. D.17.
\(
NORTE_{pr} = \Psi_{c,pag} 0.9 \fi (f’_c) e_h d_a R = 1 \veces 0.9 \veces 0.65 \veces (20.68 \, \texto{MPa}) \veces 60 \, \texto{mm} \veces 19.05 \, \texto{mm} \veces 1 = 13.828 \, \texto{kN}
\)
Recall the previously calculated tension load per anchor:
\(
NORTE_{fa} = frac{N_X}{n_{a,t}} = frac{50 \, \texto{kN}}{4} = 12.5 \, \texto{kN}
\)
Ya que 12.5 kN < 13.828 kN, the anchor pullout capacity is suficiente.
Cheque #6: Calculate side-face blowout capacity in Y-direction
This calculation is not applicable for hooked anchors.
Cheque #7: Calculate side-face blowout capacity in Z-direction
This calculation is not applicable for hooked anchors.
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
Sample report will be added soon.
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