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 30 kN tension load, 3 kN Vy shear load, et 6 kN Vz shear load.
Données données:
Colonne:
Section colonne: W14x30
Zone de colonne: 5709.7 mm2
Matériau de colonne: A992
Plaque de base:
Dimensions de la plaque de base: 12 en x 12 in
Épaisseur de plaque de base: 1/2 in
Matériau de plaque de base: A36
Jointoyer:
Épaisseur de coulis: 0 mm
Béton:
Dimensions du béton: 300 millimètre x 500 mm
Épaisseur de béton: 500 mm
Matériau en béton: 20.7 MPa
Craquelé ou sans crates: Fissuré
Ancres:
Diamètre d'ancrage: 16 mm
Durée d'admission efficace: 400 mm
Anchor Ending: Circular Plate
Embedded plate diameter: 70 mm
Épaisseur de plaque intégrée: 10 mm
Steel Material: A325N
Threads in Shear Plane: Included
Soudures:
Taille de soudure: 1/4 in
Classification du métal de remplissage: E70XX
Ancrer les données (de Calculateur de skyciv):
Remarque:
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
To determine the weld capacity under simultaneous loading, we first need to calculate the weld demand due to the shear load and the weld demand due to the tension load. You may refer to this lien for the procedure to obtain the weld demands for shear, and this lien for the tension weld demands.
For this design, l' weld demand at the web due to the tension load is found to be as follows, where the stress is expressed as force par unité de longueur.
\(r_{u,\texte{la toile}} = frac{T_{u,\texte{ancre}}}{l_{\texte{eff}}} = frac{5\ \texte{kN}}{93.142\ \texte{mm}} = 0.053681\ \texte{kN / mm}\)
en outre, l' weld stress at any part of the column section due to the shear load is determined as:
\(v_{ouais} = frac{V_y}{L_{\texte{souder}}} = frac{3\ \texte{kN}}{1250.7\ \texte{mm}} = 0.0023987\ \texte{kN / mm}\)
\(v_{à} = frac{V_z}{L_{\texte{souder}}} = frac{6\ \texte{kN}}{1250.7\ \texte{mm}} = 0.0047973\ \texte{kN / mm}\)
Since there is a combination of tension and shear loads at the la toile, we need to obtain the resultant. Expressing this as force per unit length, nous avons:
\(r_u = \sqrt{(r_{u,\texte{la toile}})^ 2 + (v_{ouais})^ 2 + (v_{à})^ 2}\)
\(r_u = \sqrt{(0.053681\ \texte{kN / mm})^ 2 + (0.0023987\ \texte{kN / mm})^ 2 + (0.0047973\ \texte{kN / mm})^ 2}\)
\(r_u = 0.053949\ \texte{kN / mm}\)
Pour le brides, only shear stresses are present. C'est à dire, the resultant is:
\(r_u = \sqrt{(v_{ouais})^ 2 + (v_{à})^ 2}\)
\(r_u = \sqrt{(0.0023987\ \texte{kN / mm})^ 2 + (0.0047973\ \texte{kN / mm})^ 2} = 0.0053636\ \texte{kN / mm}\)
Prochain, on calcule le weld capacities. For the flange, we determine the angle θ en utilisant le Vz et Vy de charges.
\( \theta = \tan^{-1}\!\la gauche(\frac{v_{ouais}}{v_{à}}\droite) = \tan^{-1}\!\la gauche(\frac{0.0023987\ \texte{kN / mm}}{0.0047973\ \texte{kN / mm}}\droite) = 0.46365\ \texte{travailler} \)
par conséquent, l' kds factor and weld capacity are calculated using AISC 360-22 Eq. J2-5 et Eq. J2-4.
\(afin que les ingénieurs puissent revoir exactement comment ces calculs sont effectués{ds} = 1.0 + 0.5(\sin(\thêta ))^{1.5} = 1 + 0.5 \fois (\sin(0.46365\ \texte{travailler}))^{1.5} = 1.1495\)
\(\phi r_{n,flg} = \phi\,0.6\,F_{Exx}\,E_w\,k_{ds} = 0.75 \fois 0.6 \fois 480\ \texte{MPa} \fois 4.95\ \texte{mm} \fois 1.1495 = 1.2291\ \texte{kN / mm}\)
For the web, we calculate the angle θ using a different formula. Noter que Vuy is used in the formula since it represents the load parallel to the weld axis.
\( \theta = \cos^{-1}\!\la gauche(\frac{v_{ouais}}{r_u}\droite) = \cos^{-1}\!\la gauche(\frac{0.0023987\ \texte{kN / mm}}{0.053949\ \texte{kN / mm}}\droite) = 1.5263\ \texte{travailler} \)
En utilisant AISC 360-22 Eq. J2-5 et Eq. J2-4, l' kds factor and the resulting weld capacity are determined in the same manner.
\(afin que les ingénieurs puissent revoir exactement comment ces calculs sont effectués{ds} = 1.0 + 0.5(\sin(\thêta ))^{1.5} = 1 + 0.5 \fois (\sin(1.5263\ \texte{travailler}))^{1.5} = 1.4993\)
\(\phi r_{n,la toile} = \phi\,0.6\,F_{Exx}\,E_w\,k_{ds} = 0.75 \fois 0.6 \fois 480\ \texte{MPa} \fois 4.95\ \texte{mm} \fois 1.4993 = 1.603\ \texte{kN / mm}\)
dernièrement, we perform base metal checks for both the column and the base plate, then obtain the governing base metal capacity.
\( \phi r_{Nbm,col} = \phi\,0.6\,F_{u,col}\,t_{col,half} = 0.75 \fois 0.6 \fois 448.2\ \texte{MPa} \fois 3.429\ \texte{mm} = 0.6916\ \texte{kN / mm} \)
\( \phi r_{Nbm,pb} = \phi\,0.6\,F_{u,pb}\,t_{pb} = 0.75 \fois 0.6 \fois 400\ \texte{MPa} \fois 12\ \texte{mm} = 2.1595\ \texte{kN / mm} \)
\( \phi r_{Nbm} = \min\big(\phi r_{Nbm,pb},\ \phi r_{Nbm,col}\big) = min(2.1595\ \texte{kN / mm},\ 0.6916\ \texte{kN / mm}) = 0.6916\ \texte{kN / mm} \)
We then compare the fillet weld capacities et base metal capacities for the weld demands at the flanges and web separately.
Puisque 0.053949 kN / mm < 0.6916 kN / mm, 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 traction de la tige d'ancrage
A design example for the anchor rod tensile capacity is already discussed in the Base Plate Design Example for Tension. Please refer to this link for the step-by-step calculation. Please refer to this link for the step-by-step calculation.
Vérifier #4: Calculer la capacité de rupture du béton en tension
A design example for the capacity of the concrete in tension breakout is already discussed in the Base Plate Design Example for Tension. Please refer to this link for the step-by-step calculation. Please refer to this link for the step-by-step calculation.
Vérifier #5: Calculer la capacité d'arrachement de l'ancre
A design example for the anchor pull out capacity is already discussed in the Base Plate Design Example for Tension. Please refer to this link for the step-by-step calculation. Please refer to this link for the step-by-step calculation.
Vérifier #6: Calculer la capacité de flexion de la plaque intégrée
A design example for the supplementary check on the embedded 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 #7: Calculer la capacité d'éruption de la face latérale dans la direction en y
Pour calculer le Side-Face Blowout (SFBO) capacité, we first determine the total tension force on the anchors closest to the edge. For this check, we will evaluate the capacity of the edge along the Y-direction.
Since the failure cone projections of the SFBO along the Y-direction overlap, the anchors are treated as an groupe d'ancrage.
The total tension demand of the anchor group is calculated as:
\(N_{faire} = gauche(\frac{N_x}{n_{a,t}}\droite) n_{Y,G1} = gauche(\frac{30\ \texte{kN}}{6}\droite) \fois 3 = 15\ \texte{kN}\)
Prochain, Nous déterminons le distances aux bords:
\(c_{z,\min} = min(c_{\texte{la gauche},G1},\ c_{\texte{droite},G1}) = min(100\ \texte{mm},\ 200\ \texte{mm}) = 100\ \texte{mm}\)
\(c_{Y,\min} = min(c_{\texte{Haut},G1},\ c_{\texte{bas},G1}) = min(150\ \texte{mm},\ 150\ \texte{mm}) = 150\ \texte{mm}\)
Using these edge distances, on calcule le anchor group capacity in accordance with ACI 318-19 Eq. (17.6.4.1).
\(N_{comme} = gauche(\frac{1 + \dfrac{c_{Y,\min}}{c_{z,\min}}}{4} + \frac{s_{somme,Y,G1}}{6\,c_{z,\min}}\droite)\fois 13 \fois gauche(\frac{c_{z,\min}}{1\ \texte{mm}}\droite)\fois sqrt{\frac{UNE_{brg}}{\texte{mm}^ 2}}\ \lambda_a sqrt{\frac{f_c}{\texte{MPa}}}\fois 0.001\ \texte{kN}\)
\(N_{comme} = gauche(\frac{1 + \dfrac{150\ \texte{mm}}{100\ \texte{mm}}}{4} + \frac{200\ \texte{mm}}{6\fois 100\ \texte{mm}}\droite)\fois 13 \fois gauche(\frac{100\ \texte{mm}}{1\ \texte{mm}}\droite)\fois sqrt{\frac{3647.4\ \texte{mm}^ 2}{1\ \texte{mm}^ 2}}\fois 1 \fois sqrt{\frac{20.68\ \texte{MPa}}{1\ \texte{MPa}}}\fois 0.001\ \texte{kN}\)
\(N_{comme} = 342.16\ \texte{kN}\)
In the original equation, a reduction factor is applied when the anchor spacing is less than 6ca₁, assuming the headed anchors have sufficient edge distance. Par contre, in this design example, puisque ca₂ < 3ca₁, the SkyCiv calculator applies an additional reduction factor to account for the reduced edge capacity.
Ensuite, l' design SFBO capacity est:
\(\phi N_{comme} = \phi\,N_{comme} = 0.7 \fois 342.16\ \texte{kN} = 239.51\ \texte{kN}\)
Puisque 15 kN < 239.51 kN, the SFBO capacity along the Y-direction is suffisant.
Vérifier #8: Calculer la capacité d'éruption de la face latérale dans la direction z
Following the same approach as in Vérifier #7, the total tension demand of the anchor group for the anchors closest to the Z-direction edge is:
\(N_{faire} = gauche(\frac{N_x}{n_{a,t}}\droite)n_{z,G1} = gauche(\frac{30\ \texte{kN}}{6}\droite)\fois 2 = 10\ \texte{kN}\)
Ce logiciel distances aux bords are calculated as:
\(c_{Y,\min} = min(c_{\texte{Haut},G1},\ c_{\texte{bas},G1}) = min(150\ \texte{mm},\ 350\ \texte{mm}) = 150\ \texte{mm}\)
\(c_{z,\min} = min(c_{\texte{la gauche},G1},\ c_{\texte{droite},G1}) = min(100\ \texte{mm},\ 100\ \texte{mm}) = 100\ \texte{mm}\)
Ce logiciel nominal SFBO capacity is then determined as:
\(N_{comme} = gauche(\frac{1 + \dfrac{c_{z,\min}}{c_{Y,\min}}}{4} + \frac{s_{somme,z,G1}}{6\,c_{Y,\min}}\droite)\fois 13 \fois gauche(\frac{c_{Y,\min}}{1\ \texte{mm}}\droite)\fois sqrt{\frac{UNE_{brg}}{\texte{mm}^ 2}}\ \lambda_a sqrt{\frac{f_c}{\texte{MPa}}}\fois 0.001\ \texte{kN}\)
\(N_{comme} = gauche(\frac{1 + \dfrac{100\ \texte{mm}}{150\ \texte{mm}}}{4} + \frac{100\ \texte{mm}}{6\fois 150\ \texte{mm}}\droite)\fois 13 \fois gauche(\frac{150\ \texte{mm}}{1\ \texte{mm}}\droite)\fois sqrt{\frac{3647.4\ \texte{mm}^ 2}{1\ \texte{mm}^ 2}}\fois 1 \fois sqrt{\frac{20.68\ \texte{MPa}}{1\ \texte{MPa}}}\fois 0.001\ \texte{kN}\)
\(N_{comme} = 282.65\ \texte{kN}\)
Since the edge distance ca₂ is still less than 3ca₁, the same modified reduction factor is applied.
Ensuite, l' design SFBO capacity est:
\(\phi N_{comme} = \phi\,N_{comme} = 0.7 \fois 282.65\ \texte{kN} = 197.86\ \texte{kN}\)
Puisque 10 kN < 197.86 kN, the SFBO capacity along the Z-direction est suffisant.
Vérifier #9: Calculate breakout capacity (Vy Shear)
A design example for the concrete breakout capacity in 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 breakout capacity (Cisaillement vz)
A design example for the concrete breakout capacity in 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 #11: Calculate pryout capacity (Vy Shear)
A design example for the capacity of the concrete against pryout 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 #12: Calculate pryout capacity (Cisaillement vz)
A design example for the capacity of the concrete against pryout 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 #13: Calculer la capacité de cisaillement de la tige d'ancrage
A design example for the anchor rod shear capacity is already discussed in the Base Plate Design Example for Shear. Please refer to this link for the step-by-step calculation.
Vérifier #14: Calculate anchor rod shear and axial capacity (AISC)
To determine the capacity of the anchor rod under combined shear and axial loads, nous utilisons AISC 360-22 Eq. J3-3A. In this calculator, the equation is rearranged to express the result as the modified shear strength instead.
Ce logiciel shear demand est défini comme le shear load per anchor.
\(V_{faire} = V_{faire} = 2.5\ \texte{kN}\)
Ce logiciel tension demand is expressed as the tensile stress in the anchor rod.
\(F_{ut} = frac{N_{faire}}{UNE_{canne à pêche}} = frac{5\ \texte{kN}}{201.06\ \texte{mm}^ 2} = 24.868\ \texte{MPa}\)
Ce logiciel modified shear capacity of the anchor rod is then calculated as:
\(F’_{nv} = \min\!\la gauche(1.3\,F_{nv} – \la gauche(\frac{F_{nv}}{\Phi f_{NT}}\droite) F_{ut},\; F_{nv}\droite)\)
\(F’_{nv} = \min\!\la gauche(1.3\fois 232.69\ \texte{MPa} – \la gauche(\frac{232.69\ \texte{MPa}}{0.75\fois 387.82\ \texte{MPa}}\droite)\fois 24.868\ \texte{MPa},\; 232.69\ \texte{MPa}\droite) = 232.69\ \texte{MPa}\)
We then multiply this strength by the anchor area utilisant AISC 360-22 Eq. J3-2.
\(\phi R_{n,\texte{AISC}} = \phi F’_{nv} UNE_{\texte{canne à pêche}} = 0.75 \fois 232.69\ \texte{MPa} \fois 201.06\ \texte{mm}À partir de l'élévation du sol générée à partir des élévations Google 35.09\ \texte{kN}\)
Puisque 2.5 kN < 35.09 kN, the anchor rod capacity is suffisant.
Vérifier #15: Calculate interaction checks (ACI)
When checking the anchor rod capacity under combined shear and tension loads using ACI, a different approach is applied. For completeness, we also perform the ACI interaction checks in this calculation, which include other concrete interaction checks ainsi que.
Here are the resulting ratios for all ACI tension checks:
And here are the resulting ratios for all ACI shear checks:
We get the check with the largest ratio and compare it to the maximum interaction ratio using ACI 318-19 Eq. 17.8.3.
\(JE_{int} = frac{N_{faire}}{\phi N_n} + \frac{V_{faire}}{\phi V_n} = frac{30}{47.749} + \frac{6}{17.921} = 0.96308\)
Puisque 0.96 < 1.2, the interaction check is 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.
