Exemple de conception de plaque de base utilisant AISC 360-22 et ACI 318-19
Déclaration de problème:
Determine whether the designed column-to-base plate connection is sufficient for a 20-kip tension load.
Données données:
Colonne:
Section colonne: W12x53
Zone de colonne: 15.6 in2
Matériau de colonne: A992
Plaque de base:
Dimensions de la plaque de base: 18 en x 18 in
Épaisseur de plaque de base: 3/4 in
Matériau de plaque de base: A36
Jointoyer:
Épaisseur de coulis: 1 in
Béton:
Dimensions du béton: 22 en x 22 in
Épaisseur de béton: 15 in
Matériau en béton: 4000 psi
Cracked or Uncracked: Fissuré
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
Soudures:
Taille de soudure: 1/4 in
Classification du métal de remplissage: E70XX
Anchor Data (de SkyCiv Calculator):
Definitions:
Load Path:
When a base plate is subjected to uplift (traction) les forces, 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.
Dans le Logiciel de conception de plaque de 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 de la plaque.
The assumption simplifies the base plate analysis by approximating how the uplift force spreads through the plate.
Anchor Groups:
Ce logiciel Logiciel de conception de plaque de base SkyCiv includes an intuitive feature that identifies which anchors are part of an anchor group for evaluating évasion de béton et 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. Dans le cas présent, only the tension force on the individual anchor is checked against its own effective resistance area.
Calculs étape par étape:
Vérifier #1: Calculer la capacité de soudure
Pour commencer, 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 ou 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. Dans cet exemple, 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.
\(
l_r = \frac{ré_{col} – 2t_f – 2r_{col} – s_y(n_{a,side} – 1)}{2} = frac{12.1 \, \texte{in} – 2 \fois 0.575 \, \texte{in} – 2 \fois 0.605 \, \texte{in} – 5 \, \texte{in} \fois (2 – 1)}{2} = 2.37 \, \texte{in}
\)
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.
\(
l_{eff,fin} = \min(faire, 0.5s_y) + \min(faire, l_r)
\)
\(
l_{eff,fin} = \min(2.8275 \, \texte{in}, 0.5 \fois 5 \, \texte{in}) + \min(2.8275 \, \texte{in}, 2.37 \, \texte{in}) = 4.87 \, \texte{in}
\)
Dans cet exemple, l' final effective weld length for the web anchor is taken as the effective length of the end anchor.
\(
l_{eff} = l_{eff,fin} = 4.87 \, \texte{in}
\)
Prochain, let’s calculate the load per anchor. For a given set of four (4) ancres, the load per anchor is:
\(
T_{u,ancre} = frac{N_x}{n_{a,t}} = frac{20 \, \texte{kip}}{4} = 5 \, \texte{kip}
\)
Using the calculated effective weld length, we can now determine the required force per unit length on the weld.
\(
r_u = frac{T_{u,ancre}}{l_{eff}} = frac{5 \, \texte{kip}}{4.87 \, \texte{in}} = 1.0267 \, \texte{kip / in}
\)
Maintenant, nous utiliserons 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 \(\thêta ) is taken as 90°, and the directional strength coefficient kds is calculated according to AISC 360-22 Eq. J2-5.
\(
afin que les ingénieurs puissent revoir exactement comment ces calculs sont effectués{ds} = 1.0 + 0.5(\sin(\thêta ))^{1.5} = 1 + 0.5 \fois (\sin(1.5708))^{1.5} = 1.5
\)
Ensuite, nous appliquerons AISC 360-22 Eq. J2-4 to determine the design strength of the fillet weld per unit length.
\(
\phi r_n = \phi 0.6 F_{Exx} E_{w,la toile} afin que les ingénieurs puissent revoir exactement comment ces calculs sont effectués{ds} = 0.75 \fois 0.6 \fois 70 \, \texte{KSI} \fois 0.177 \, \texte{in} \fois 1.5 = 8.3633 \, \texte{kip / in}
\)
Puisque 1.0267 kpi < 8.3633 kpi, La capacité de soudure est suffisant.
Vérifier #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 cantilever assumption.
\(
M_u = T_{u,\texte{ancre}} e = 5 \, \texte{kip} \fois 2.8275 \, \texte{in} = 14.137 \, \texte{kip} \CDOT Texte{in}
\)
Prochain, using the calculated effective weld length from the previous check as the bending width, Nous pouvons calculer le SkyCiv Foundation est un module de conception pour la conception de semelles écartées à partir des charges de superstructure of the base plate using AISC 360-22, Équation 2-1:
\(
\phi M_n = \phi F_{Y,\texte{pb}} Z_{\texte{eff}} = 0.9 \fois 36 \, \texte{KSI} \fois 0.68484 \, \texte{in}^3 = 22.189 \, \texte{kip} \CDOT Texte{in}
\)
Où,
\(
Z_{\texte{eff}} = frac{l_{\texte{eff}} (t_{\texte{pb}})^ 2}{4} = frac{4.87 \, \texte{in} \fois (0.75 \, \texte{in})^ 2}{4} = 0.68484 \, \texte{in}^ 3
\)
Puisque 14.137 poulet dans < 22.189 poulet dans, the base plate flexural yielding capacity is suffisant.
Vérifier #3: Calculate anchor rod tensile capacity
To evaluate the tensile capacity of the anchor rod, nous utiliserons ACI 318-19 Équation 17.6.1.2.
Première, Nous déterminons le specified tensile strength of the anchor steel. This is the lowest value permitted by ACI 318-19 Clause 17.6.1.2, with reference to material properties in AISC 360-22 Tableau J3.2.
\(
F_{\texte{uta}} = min gauche( 0.75 F_{u,\texte{anc}}, 1.9 F_{Y,\texte{anc}}, 125 \droite) = min gauche( 0.75 \fois 120 \, \texte{KSI}, 1.9 \fois 92 \, \texte{KSI}, 125.00 \, \texte{KSI} \droite) = 90 \, \texte{KSI}
\)
Prochain, on calcule le 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.
\(
UNE_{je connais,N} = frac{\pi}{4} \la gauche( d_a – \frac{0.9743}{n_t} \droite)^2 = \frac{\pi}{4} \fois gauche( 0.75 \, \texte{in} – \frac{0.9743}{10 \, \texte{in}^{-1}} \droite)À partir de l'élévation du sol générée à partir des élévations Google 0.33446 \, \texte{in}^ 2
\)
With these values, Nous appliquons ACI 318-19 Équation 17.6.1.2 to compute the design tensile strength of the anchor rod.
\(
\phi N_{à} = phi A_{je connais,N} F_{\texte{uta}} = 0.75 \fois 0.33446 \, \texte{in}^2 \times 90 \, \texte{KSI} = 22.576 \, \texte{kip}
\)
Recall the previously calculated tension load per anchor:
\(
N_{ua} = frac{N_x}{n_{a,t}} = frac{20 \, \texte{kip}}{4} = 5 \, \texte{kip}
\)
Puisque 5 kip < 22.576 kip, the anchor rod tensile capacity is suffisant.
Vérifier #4: Calculate concrete breakout capacity in tension
Before calculating the breakout capacity, we must first determine whether the member qualifies as a narrow member. Selon ACI 318-19 Clause 17.6.2.1.2, the member meets the criteria for a narrow member. Par conséquent, 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:
\(
h’_{\texte{ef}} = 5.667 \, \texte{in}
\)
En utilisant ACI 318-19 Clause 17.6.2, on calcule le maximum projected concrete cone area pour une seule ancre, based on the modified effective embedment length.
\(
UNE_{N_{co}} = 9 \la gauche( h’_{ef,g1} \droite)À partir de l'élévation du sol générée à partir des élévations Google 9 \fois gauche( 5.6667 \, \texte{in} \droite)À partir de l'élévation du sol générée à partir des élévations Google 289 \, \texte{in}^ 2
\)
De manière similaire, we use the modified effective embedment length to calculate the actual projected concrete cone area of the anchor group.
\(
UNE_{N_c} = min gauche( n_{a,g1} UNE_{N_{co}}, L_{N_c} B_{N_c} \droite) = min gauche( 4 \fois 289 \, \texte{in}^ 2, 22 \, \texte{in} \fois 22 \, \texte{in} \droite) = 484 \, \texte{in}^ 2
\)
Où,
\(
L_{N_c} = min gauche( c_{\texte{la gauche},g1}, 1.5 h’_{\texte{ef},g1} \droite)
+ \la gauche( \min gauche( s_{\texte{somme},z,g1}, 3 h’_{\texte{ef},g1} \la gauche( n_{z,g1} – 1 \droite) \droite) \droite)
+ \min gauche( c_{\texte{droite},g1}, 1.5 h’_{\texte{ef},g1} \droite)
\)
\(
L_{N_c} = min gauche( 8 \, \texte{in}, 1.5 \fois 5.6667 \, \texte{in} \droite)
+ \la gauche( \min gauche( 6 \, \texte{in}, 3 \fois 5.6667 \, \texte{in} \fois gauche( 2 – 1 \droite) \droite) \droite)
+ \min gauche( 8 \, \texte{in}, 1.5 \fois 5.6667 \, \texte{in} \droite)
\)
\(
L_{N_c} = 22 \, \texte{in}
\)
\(
B_{N_c} = min gauche( c_{\texte{Haut},g1}, 1.5 h’_{\texte{ef},g1} \droite)
+ \la gauche( \min gauche( s_{\texte{somme},Y,g1}, 3 h’_{\texte{ef},g1} \la gauche( n_{Y,g1} – 1 \droite) \droite) \droite)
+ \min gauche( c_{\texte{bas},g1}, 1.5 h’_{\texte{ef},g1} \droite)
\)
\(
B_{N_c} = min gauche( 8.5 \, \texte{in}, 1.5 \fois 5.6667 \, \texte{in} \droite)
+ \la gauche( \min gauche( 5 \, \texte{in}, 3 \fois 5.6667 \, \texte{in} \fois gauche( 2 – 1 \droite) \droite) \droite)
+ \min gauche( 8.5 \, \texte{in}, 1.5 \fois 5.6667 \, \texte{in} \droite)
\)
\(
B_{N_c} = 22 \, \texte{in}
\)
Prochain, we evaluate the basic concrete breakout strength of a single anchor using ACI 318-19 Clause 17.6.2.2.1
\(
N_b = k_c lambda_a sqrt{\frac{f'_c}{\texte{psi}}} \la gauche( \frac{h’_{\texte{ef},g1}}{\texte{in}} \droite)^{1.5} \, \texte{lbf}
\)
\(
N_b = 24 \fois 1 \fois sqrt{\frac{4 \, \texte{KSI}}{0.001 \, \texte{KSI}}} \fois gauche( \frac{5.6667 \, \texte{in}}{1 \, \texte{in}} \droite)^{1.5} \fois 0.001 \, \texte{kip} = 20.475 \, \texte{kip}
\)
Où,
- \(afin que les ingénieurs puissent revoir exactement comment ces calculs sont effectués{c} = 24\) pour ancres coulées
- \(\lambda = 1.0 \) for normal-weight concrete
Maintenant, we assess the effects of geometry by calculating the edge effect factor et la eccentricity factor.
The shortest edge distance of the anchor group is determined as:
\(
c_{a,\texte{min}} = min gauche( c_{\texte{la gauche},g1}, c_{\texte{droite},g1}, c_{\texte{Haut},g1}, c_{\texte{bas},g1} \droite)
= min gauche( 8 \, \texte{in}, 8 \, \texte{in}, 8.5 \, \texte{in}, 8.5 \, \texte{in} \droite) = 8 \, \texte{in}
\)
Selon ACI 318-19 Clause 17.6.2.4.1, the breakout edge effect factor est:
\(
\Psi_{ed,N} = min gauche( 1.0, 0.7 + 0.3 \la gauche( \frac{c_{a,\texte{min}}}{1.5 h’_{\texte{ef},g1}} \droite) \droite)
= min gauche( 1, 0.7 + 0.3 \fois gauche( \frac{8 \, \texte{in}}{1.5 \fois 5.6667 \, \texte{in}} \droite) \droite) = 0.98235
\)
Since the tension load is applied at the centroid of the anchor group, the eccentricity is zero. C'est à dire, l' eccentricity factor, also from Clause 17.6.2.4.1, est:
\(
\Psi_{ce,N} = min gauche( 1.0, \frac{1}{1 + \frac{2 et n}{3 h’_{\texte{ef},g1}}} \droite)
= min gauche( 1, \frac{1}{1 + \frac{2 \fois 0}{3 \fois 5.6667 \, \texte{in}}} \droite) = 1
\)
Aussi, both the cracking factor et la splitting factor are taken as:
\(
\Psi_{c,N} = 1
\)
\(
\Psi_{cp,N} = 1
\)
ensuite, 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:
\(
\phi N_{cbg} = \phi \left( \frac{UNE_{N_c}}{UNE_{N_{co}}} \droite) \Psi_{ce,N} \Psi_{ed,N} \Psi_{c,N} \Psi_{cp,N} N_b
\)
\(
\phi N_{cbg} = 0.7 \fois gauche( \frac{484 \, \texte{in}^ 2}{289 \, \texte{in}^ 2} \droite) \fois 1 \fois 0.98235 \fois 1 \fois 1 \fois 20.475 \, \texte{kip} = 23.58 \, \texte{kip}
\)
Ce logiciel total applied tension load on the anchor group is the product of the individual anchor load and the number of anchors:
\(
N_{ua} = gauche( \frac{N_x}{n_{a,t}} \droite) n_{a,g1} = gauche( \frac{20 \, \texte{kip}}{4} \droite) \fois 4 = 20 \, \texte{kip}
\)
Puisque 20 kips < 23.58 kips, the concrete breakout capacity is suffisant.
Vérifier #5: Calculate anchor pullout capacity
The pullout capacity of an anchor is governed by the resistance at its embedded end. Pour commencer, 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, l' bearing area is calculated as:
\(
UNE_{brg} = gauche( \la gauche( b_{embed\_plate} \droite)^ 2 à droite) – UNE_{canne à pêche} = gauche( \la gauche( 3 \, \texte{in} \droite)^ 2 à droite) – 0.44179 \, \texte{in}À partir de l'élévation du sol générée à partir des élévations Google 8.5582 \, \texte{in}^ 2
\)
Où,
\(
UNE_{canne à pêche} = frac{\pi}{4} \la gauche( d_a \right)^2 = \frac{\pi}{4} \fois gauche( 0.75 \, \texte{in} \droite)À partir de l'élévation du sol générée à partir des élévations Google 0.44179 \, \texte{in}^ 2
\)
Prochain, Nous déterminons le basic anchor pullout strength utilisant ACI 318-19 Equation 17.6.3.2.2a.
\(
N_b = 8 UNE_{brg} \la gauche( F’_C Right) = 8 \fois 8.5582 \, \texte{in}^ 2 Times Left( 4 \, \texte{KSI} \droite) = 273.86 \, \texte{kip}
\)
We then apply the appropriate resistance factor and pullout cracking factor:
- Pour fissuré le béton, \(\Psi_{cp} = 1.0\)
- Pour uncracked le béton, \(\Psi_{cp} = 1.4\)
Using these, Nous calculons le design anchor pullout strength in tension par ACI 318-19 Équation 17.6.3.1.
\(
\phi N_{pn} = \phi \Psi_{c,p} N_b = 0.7 \fois 1 \fois 273.86 \, \texte{kip} = 191.7 \, \texte{kip}
\)
Recall the previously calculated tension load per anchor:
\(
N_{ua} = frac{N_x}{n_{a,t}} = frac{20 \, \texte{kip}}{4} = 5 \, \texte{kip}
\)
Puisque 5 kips < 191.7 kips, the anchor pullout capacity is suffisant.
Vérifier #6: Calculate embed plate flexural capacity
This is a supplementary check performed using the Logiciel de conception de plaques de base Skyciv to verify that the embedded plate has sufficient flexural capacity and will not yield under the applied pullout loads.
Première, 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.
\(
b’ = frac{b_{embed\_plate} – d_a}{2} = frac{3 \, \texte{in} – 0.75 \, \texte{in}}{2} = 1.125 \, \texte{in}
\)
Prochain, on calcule le moments de flexion induced by the uniform bearing pressure. This pressure represents the force transferred from the anchor pullout action onto the embedded plate.
\(
m_f = \frac{\la gauche( \frac{T_a}{UNE_{brg}} \droite) \la gauche( b’ \droite)^ 2}{2} = frac{\la gauche( \frac{5 \, \texte{kip}}{8.5582 \, \texte{in}^ 2} \droite) \fois gauche( 1.125 \, \texte{in} \droite)^ 2}{2} = 0.36971 \, \texte{kip}
\)
Ensuite, using the calculated moment and given material properties, we will determine the minimum required plate thickness to resist flexural yielding.
\(
t_{min} = sqrt{\frac{4 m_f}{\Phi f_{y\_ep}}} = sqrt{\frac{4 \fois 0.36971 \, \texte{kip}}{0.9 \fois 36 \, \texte{KSI}}} = 0.21364 \, \texte{in}
\)
Recall actual embedded plate thickness:
\(
t_{actual} = t_{embed\_plate} = 0.25 \, \texte{in}
\)
Puisque 0.21364 in < 0.25 in, the embedded plate flexural capacity is suffisant.
Vérifier #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 Clause 17.6.4 are not met. Par conséquent, side-face blowout failure along the Y-direction will not occur.
Vérifier #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 Clause 17.6.4 are not met. Par conséquent, side-face blowout failure along the Z-direction will not occur.
Résumé de la conception
Ce logiciel Logiciel de conception de plaques de base Skyciv can automatically generate a step-by-step calculation report for this design example. Il fournit également un résumé des contrôles effectués et de leurs ratios résultants, rendre les informations faciles à comprendre en un coup d'œil. Vous trouverez ci-dessous un échantillon de tableau de résumé, qui est inclus dans le rapport.
Rapport d'échantillon de skyciv
Cliquez ici Pour télécharger un exemple de rapport.
Logiciel d'achat de plaques de base
Achetez la version complète du module de conception de la plaque de base seul sans aucun autre module Skyviv. Cela vous donne un ensemble complet de résultats pour la conception de la plaque de base, y compris des rapports détaillés et plus de fonctionnalités.
