Exemple de conception de plaque de base en utilisant EN 1993-1-8-2005, EN 1993-1-1-2005 et EN 1992-1-1-2004
Déclaration de problème:
Déterminez si la connexion de colonne à base de colonne conçue est suffisante pour un 50-kN tension load, 4-kN Vy shear load, et 2-kN Vz shear load.
Données données:
Colonne:
Section colonne: CHS193.7×10
Zone de colonne: 5770.0 mm²
Matériau de colonne: S460
Plaque de base:
Dimensions de la plaque de base: 300mm x 300mm
Épaisseur de plaque de base: 18mm
Matériau de plaque de base: S235
Jointoyer:
Épaisseur de coulis: 0 mm
Béton:
Dimensions du béton: 350mm x 350mm
Épaisseur de béton: 400 mm
Matériau en béton: C35/45
Craquelé ou sans crates: Fissuré
Ancres:
Diamètre d'ancrage: 16 mm
Durée d'admission efficace: 350 mm
Embedded plate diameter: 70 mm
Épaisseur de plaque intégrée: 10 mm
Anchor material: 4.8
Soudures:
Type de soudure: Fillet
Taille de soudure: 7mm
Classification du métal de remplissage: E42
Ancrer les données (de Calculateur de skyciv):
Remarques:
The purpose of this design example is to demonstrate the step-by-step calculations for capacity checks involving concurrent shear and axial loads. Some of the required checks have already been discussed in the previous design examples. Please refer to the links provided in each section.
Calculs étape par étape:
Vérifier #1: Calculer la capacité de soudure
The full tensile load is resisted by the entire weld section, tandis que le shear load components are distributed only to a portion of the total weld length. This portion is determined by projecting a 90° secteur from the center of the column to its circumference. Par conséquent, seul half of the total circumference is designed to resist the shear load.
We first compute the Longueur totale de soudure et la portion of the weld within the 90° projection.
\(L_{souder,full} = \pi d_{col} = \pi \times 193.7\ \texte{mm} = 608.53\ \texte{mm}\)
\(L_{souder} = frac{\pi d_{col}}{2} = frac{\Pi Times 193.7\ \texte{mm}}{2} = 304.26\ \texte{mm}\)
Prochain, on calcule le resultant shear load.
\(V_r = \sqrt{(V_y)^ 2 + (V_z)^ 2} = sqrt{(4\ \texte{kN})^ 2 + (2\ \texte{kN})^ 2} = 4.4721\ \texte{kN}\)
We then compute the Ordinaire et shear stresses, taking into account the assumed load distribution.
\( \sigma_{\perp} = frac{N_x}{L_{souder,full}\,a\,\sqrt{2}} = frac{40\ \texte{kN}}{608.53\ \texte{mm} \fois 4.95\ \texte{mm} \fois sqrt{2}} = 9.39\ \texte{MPa} \)
\( \votre_{\perp} = frac{N_x}{L_{souder,full}\,a\,\sqrt{2}} = frac{40\ \texte{kN}}{608.53\ \texte{mm} \fois 4.95\ \texte{mm} \fois sqrt{2}} = 9.39\ \texte{MPa} \)
\( \votre_{\parallèle} = frac{V_r}{L_{souder}\,a} = frac{4.4721\ \texte{kN}}{304.26\ \texte{mm} \fois 4.95\ \texte{mm}} = 2.9693\ \texte{MPa} \)
Après ça, on calcule le contraintes combinées utilisant EN 1993-1-8:2005 Eq. (4.1).
\(F_{w,Ed1} = sqrt{(\sigma_{\perp})^ 2 + 3\big((\votre_{\perp})^ 2 + (\votre_{\parallèle})^2\big)}\)
\(F_{w,Ed1} = sqrt{(9.39\ \texte{MPa})^ 2 + 3\big((9.39\ \texte{MPa})^ 2 + (2.9693\ \texte{MPa})^2\big)}\)
\(F_{w,Ed1} = 19.471\ \texte{MPa}\)
En même temps, Nous déterminons le stress on the base metal using the same equation.
\(F_{w,Ed2} = \sigma_{\perp} = 9.39\ \texte{MPa}\)
Prochain, on calcule le weld capacity. We first determine the ultimate tensile strength (Facteur de service humide selon NDS) de la weaker material, and then use EN 1993-1-8:2005 Eq. (4.1) to obtain the fillet weld resistance et base metal resistance.
\(f_u = \min\!\la gauche(F_{u,\texte{col}},\ F_{u,\texte{pb}},\ F_{u,w}\droite) = \min\!\la gauche(550\ \texte{MPa},\ 360\ \texte{MPa},\ 500\ \texte{MPa}\droite) = 360\ \texte{MPa}\)
\(F_{w,Rd1} = frac{f_u}{\beta_w\,(\gamma_{M2,\text{souder}})} = frac{360\ \texte{MPa}}{0.8 \fois (1.25)} = 360\ \texte{MPa}\)
\(F_{w,Rd2} = frac{0.9\,f_u}{\gamma_{M2,\text{souder}}} = frac{0.9 \fois 360\ \texte{MPa}}{1.25} = 259.2\ \texte{MPa}\)
Puisque 19.471 MPa < 360 MPa, La capacité de soudure est suffisant.
Vérifier #2: Calculer la capacité de rendement en flexion de la plaque de base due à la charge de tension
A design example for the base plate flexural yielding capacity is already discussed in the Base Plate Design Example for Tension. Please refer to this link for the step-by-step calculation.
Vérifier #3: Calculer la capacité de rupture du béton en tension
A design example for the capacity of the concrete in breakout due to tension load is already discussed in the Base Plate Design Example for Tension. Please refer to this link for the step-by-step calculation.
Vérifier #4: Calculer la capacité d'arrachement de l'ancre
A design example for the anchor pullout capacity is already discussed in the Base Plate Design Example for Tension. Please refer to this link for the step-by-step calculation.
Vérifier #5: Calculer la capacité d'éruption de la face latérale dans la direction en y
A design example for the side-face blowout capacity in Y-direction is already discussed in the Base Plate Design Example for Tension. Please refer to this link for the step-by-step calculation.
Vérifier #6: Calculer la capacité d'éruption de la face latérale dans la direction z
A design example for the side-face blowout capacity in Z-direction is already discussed in the Base Plate Design Example for Tension. Please refer to this link for the step-by-step calculation.
Vérifier #7: Calculate base plate bearing capacity at anchor holes (Vy Shear)
A design example for the base plate bearing capacity in the anchor holes for Vy shear is already discussed in the Base Plate Design Example for Compression and Shear. Please refer to this link for the step-by-step calculation.
Vérifier #8: Calculate base plate bearing capacity at anchor holes (Cisaillement vz)
A design example for the base plate bearing capacity in the anchor holes for Vz shear is already discussed in the Base Plate Design Example for Compression and Shear. Please refer to this link for the step-by-step calculation.
Vérifier #9: Calculate concrete breakout capacity (Vy Shear)
A design example for the concrete capacity in breakout failure due to Vy shear is already discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Vérifier #10: Calculate concrete breakout capacity (Cisaillement vz)
A design example for the concrete capacity in breakout failure due to Vz shear is already discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Vérifier #11: Calculate pryout capacity
A design example for the concrete pryout capacity is already discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Vérifier #12: Calculer la capacité de cisaillement de la tige d'ancrage
The effect of the tension load on the anchor rod capacity is considered in this check if the shear force acts with a lever arm. Par contre, dans cet exemple, the shear acts without a lever arm. Par conséquent, the interaction between shear and tensile stresses on the anchor rod will be evaluated separately in the interaction check.
For the step-by-step calculation of the shear capacity without a lever arm, veuillez vous référer à ce lien.
The SkyCiv Base Plate Design software can perform all the necessary checks to determine whether the shear load acts with or without a lever arm. Tu peux try out the free tool aujourd'hui.
Vérifier #13: Calculate anchor steel interaction check
Nous utilisons EN 1992-4:2018 Le tableau 7.3 Eq. (7.54) pour évaluer le interaction between the shear and tensile stresses on the anchor rod. By substituting the tensile stress and capacity as well as the shear stress and capacity into the equation, the resulting interaction value est:
\(JE_{int} = gauche(\frac{N_{Ed}}{N_{Rd,s}}\droite)^ 2 + \la gauche(\frac{V_{Ed}}{V_{Rd,s}}\droite)^2)
\(JE_{int} = gauche(\frac{10\ \texte{kN}}{49.22\ \texte{kN}}\droite)^ 2 + \la gauche(\frac{1.118\ \texte{kN}}{38.604\ \texte{kN}}\droite)À partir de l'élévation du sol générée à partir des élévations Google 0.042117\)
Puisque 0.042 < 1.0, the anchor rod steel failure interaction check is suffisant.
Vérifier #14: Calculate concrete failure interaction check
SkyCiv automatise également les calculs de charge de neige au sol avec quelques paramètres interaction check is required for concrete failures under simultaneous shear and tensile loading. Pour ça, nous utilisons EN 1992-4:2018 Le tableau 7.3 Eq. (7.55) et Eq. (7.56).
Here are the resulting ratios for all tensile checks.
Here are the resulting ratios for all shear checks.
Première, we check using Eq. (7.55) and compare the result to the maximum interaction limit of 1.0.
\(JE_{\texte{case1}} = gauche(\la gauche(\frac{N_{Ed}}{N_{Rd}}\droite)^{1.5}\droite) + \la gauche(\la gauche(\frac{V_{Ed}}{V_{Rd}}\droite)^{1.5}\droite)\)
\(JE_{\texte{case1}} = gauche(\la gauche(\frac{40}{45.106}\droite)^{1.5}\droite) + \la gauche(\la gauche(\frac{4.1231}{14.296}\droite)^{1.5}\droite) = 0.99\)
Prochain, we check using Eq. (7.56) and compare the result to the maximum interaction limit of 1.2.
\(JE_{\texte{cas2}} = frac{N_{Ed}}{N_{Rd}} + \frac{V_{Ed}}{V_{Rd}} = frac{40}{45.106} + \frac{4.1231}{14.296} = 1.1752\)
Puisque 0.99 < 1.0 et 1.175 < 1.2, l' concrete failure interaction check est suffisant.
Résumé de la conception
Ce logiciel Logiciel de conception de plaques de base Skyciv peut générer automatiquement un rapport de calcul étape par étape pour cet exemple de conception. 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.
