Basisplatten -Designbeispiel unter Verwendung von AISC 360-22 und ACI 318-19
Problemanweisung:
Determine whether the designed column-to-base plate connection is sufficient for 30 kN tension load, 3 kN Vy shear load, und 6 kN Vz shear load.
Gegebene Daten:
Spalte:
Spaltenabschnitt: W14x30
Säulenbereich: 5709.7 mm2
Säulenmaterial: A992
Grundplatte:
Grundplattenabmessungen: 12 in x 12 im
Grundplattendicke: 1/2 im
Grundplattenmaterial: A36
Fugenmörtel:
Grout Thickness: 0 mm
Beton:
Konkrete Abmessungen: 300 mmx 500 mm
Betondicke: 500 mm
Betonmaterial: 20.7 MPa
Geknackt oder ungekrönt: Geknackt
Anker:
Ankerdurchmesser: 16 mm
Effektive Einbettungslänge: 400 mm
Anchor Ending: Circular Plate
Eingebetteter Plattendurchmesser: 70 mm
Dicke eingebetteter Platten: 10 mm
Steel Material: A325N
Threads in Shear Plane: Included
Schweißnähte:
Schweißnahtgröße: 1/4 im
Füllmetallklassifizierung: E70XX
Ankerdaten (von Skyciv -Taschenrechner):
Hinweis:
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.
Schritt-für-Schritt-Berechnungen:
Prüfen #1: Berechnen Sie die Schweißkapazität
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 link for the procedure to obtain the weld demands for shear, and this link for the tension weld demands.
For this design, bleibt die weld demand at the web due to the tension load is found to be as follows, where the stress is expressed as Kraft pro Längeneinheit.
\(r_{u,\Text{Netz}} = frac{T_{u,\Text{Anker}}}{l_{\Text{eff}}} = frac{5\ \Text{kN}}{93.142\ \Text{mm}} = 0.053681\ \Text{kN / mm}\)
Außerdem, bleibt die weld stress at any part of the column section due to the shear load is determined as:
\(v_{ui} = frac{V_y}{L_{\Text{schweißen}}} = frac{3\ \Text{kN}}{1250.7\ \Text{mm}} = 0.0023987\ \Text{kN / mm}\)
\(v_{uz} = frac{V_z}{L_{\Text{schweißen}}} = frac{6\ \Text{kN}}{1250.7\ \Text{mm}} = 0.0047973\ \Text{kN / mm}\)
Since there is a combination of tension and shear loads at the Netz, we need to obtain the resultant. Expressing this as force per unit length, wir haben:
\(r_u = \sqrt{(r_{u,\Text{Netz}})^ 2 + (v_{ui})^ 2 + (v_{uz})^ 2}\)
\(r_u = \sqrt{(0.053681\ \Text{kN / mm})^ 2 + (0.0023987\ \Text{kN / mm})^ 2 + (0.0047973\ \Text{kN / mm})^ 2}\)
\(r_u = 0.053949\ \Text{kN / mm}\)
Für die Flansche, only shear stresses are present. So, the resultant is:
\(r_u = \sqrt{(v_{ui})^ 2 + (v_{uz})^ 2}\)
\(r_u = \sqrt{(0.0023987\ \Text{kN / mm})^ 2 + (0.0047973\ \Text{kN / mm})^ 2} = 0.0053636\ \Text{kN / mm}\)
Als nächstes, wir berechnen die weld capacities. For the flange, we determine the angle θ Verwendung der Vz und Vy Ladungen.
\( \theta = \tan^{-1}\!\links(\frac{v_{ui}}{v_{uz}}\richtig) = tan^{-1}\!\links(\frac{0.0023987\ \Text{kN / mm}}{0.0047973\ \Text{kN / mm}}\richtig) = 0.46365\ \Text{Arbeit} \)
Folglich, bleibt die kds factor and weld capacity are calculated using AISC 360-22 Gl. J2-5 und Gl. J2-4.
\(k_{ds} = 1.0 + 0.5(\ohne(\theta))^{1.5} = 1 + 0.5 \mal (\ohne(0.46365\ \Text{Arbeit}))^{1.5} = 1.1495\)
\(\PHI R_{n,flg} = \phi\,0.6\,F_{Exx}\,E_w\,k_{ds} = 0.75 \mal 0.6 \mal 480\ \Text{MPa} \mal 4.95\ \Text{mm} \mal 1.1495 = 1.2291\ \Text{kN / mm}\)
For the web, we calculate the angle θ using a different formula. Beachten Sie, dass Vuy is used in the formula since it represents the load parallel to the weld axis.
\( \theta = \cos^{-1}\!\links(\frac{v_{ui}}{r_u}\richtig) = \cos^{-1}\!\links(\frac{0.0023987\ \Text{kN / mm}}{0.053949\ \Text{kN / mm}}\richtig) = 1.5263\ \Text{Arbeit} \)
Verwenden von AISC 360-22 Gl. J2-5 und Gl. J2-4, bleibt die kds factor and the resulting weld capacity are determined in the same manner.
\(k_{ds} = 1.0 + 0.5(\ohne(\theta))^{1.5} = 1 + 0.5 \mal (\ohne(1.5263\ \Text{Arbeit}))^{1.5} = 1.4993\)
\(\PHI R_{n,Netz} = \phi\,0.6\,F_{Exx}\,E_w\,k_{ds} = 0.75 \mal 0.6 \mal 480\ \Text{MPa} \mal 4.95\ \Text{mm} \mal 1.4993 = 1.603\ \Text{kN / mm}\)
zuletzt, we perform base metal checks for both the column and the base plate, then obtain the governing base metal capacity.
\( \PHI R_{nbm,col} = \phi\,0.6\,F_{u,col}\,t_{col,half} = 0.75 \mal 0.6 \mal 448.2\ \Text{MPa} \mal 3.429\ \Text{mm} = 0.6916\ \Text{kN / mm} \)
\( \PHI R_{nbm,bp} = \phi\,0.6\,F_{u,bp}\,t_{bp} = 0.75 \mal 0.6 \mal 400\ \Text{MPa} \mal 12\ \Text{mm} = 2.1595\ \Text{kN / mm} \)
\( \PHI R_{nbm} = \min\big(\PHI R_{nbm,bp},\ \PHI R_{nbm,col}\big) = min(2.1595\ \Text{kN / mm},\ 0.6916\ \Text{kN / mm}) = 0.6916\ \Text{kN / mm} \)
We then compare the fillet weld capacities und base metal capacities for the weld demands at the flanges and web separately.
Schon seit 0.053949 kN / mm < 0.6916 kN / mm, Die Schweißkapazität ist ausreichend.
Prüfen #2: Berechnen Sie die Kapazität der Grundplattenflexus aufgrund der Spannungsbelastung
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.
Prüfen #3: Berechnen Sie die Ankerstange Zugkapazität
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.
Prüfen #4: Berechnen Sie die Betonausbruchkapazität in der Spannung
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.
Prüfen #5: Berechnen Sie die Ankerauszugskapazität
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.
Prüfen #6: Calculate embed plate flexural capacity
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.
Prüfen #7: Berechnen Sie die Blowout-Kapazität der Seitengesicht in der y-Richtung
Zur Berechnung der Side-Face Blowout (SFBO) Kapazität, 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 Ankergruppe.
The total tension demand of the anchor group is calculated as:
\(N_{ua} = left(\frac{N_x}{N_{ein,t}}\richtig) N_{j,G1} = left(\frac{30\ \Text{kN}}{6}\richtig) \mal 3 = 15\ \Text{kN}\)
Als nächstes, Wir bestimmen die Randabstände:
\(c_{mit,\Min.} = min(c_{\Text{links},G1},\ c_{\Text{richtig},G1}) = min(100\ \Text{mm},\ 200\ \Text{mm}) = 100\ \Text{mm}\)
\(c_{j,\Min.} = min(c_{\Text{oben},G1},\ c_{\Text{Unterseite},G1}) = min(150\ \Text{mm},\ 150\ \Text{mm}) = 150\ \Text{mm}\)
Using these edge distances, wir berechnen die anchor group capacity in accordance with ACI 318-19 Gl. (17.6.4.1).
\(N_{als} = left(\frac{1 + \dfrac{c_{j,\Min.}}{c_{mit,\Min.}}}{4} + \frac{S_{Summe,j,G1}}{6\,c_{mit,\Min.}}\richtig)\mal 13 \mal links(\frac{c_{mit,\Min.}}{1\ \Text{mm}}\richtig)\mal sqrt{\frac{EIN_{brg}}{\Text{mm}^ 2}}\ \lambda_a sqrt{\frac{f_c}{\Text{MPa}}}\mal 0.001\ \Text{kN}\)
\(N_{als} = left(\frac{1 + \dfrac{150\ \Text{mm}}{100\ \Text{mm}}}{4} + \frac{200\ \Text{mm}}{6\mal 100\ \Text{mm}}\richtig)\mal 13 \mal links(\frac{100\ \Text{mm}}{1\ \Text{mm}}\richtig)\mal sqrt{\frac{3647.4\ \Text{mm}^ 2}{1\ \Text{mm}^ 2}}\mal 1 \mal sqrt{\frac{20.68\ \Text{MPa}}{1\ \Text{MPa}}}\mal 0.001\ \Text{kN}\)
\(N_{als} = 342.16\ \Text{kN}\)
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. Jedoch, in this design example, schon seit ca₂ < 3ca₁, the SkyCiv calculator applies an additional reduction factor to account for the reduced edge capacity.
Schließlich, bleibt die design SFBO capacity ist:
\(\phi N_{als} = \phi\,N_{als} = 0.7 \mal 342.16\ \Text{kN} = 239.51\ \Text{kN}\)
Schon seit 15 kN < 239.51 kN, the SFBO capacity along the Y-direction is ausreichend.
Prüfen #8: Berechnen Sie die Blowout-Kapazität der Seitengesicht in der Z-Richtung
Following the same approach as in Prüfen #7, the total tension demand of the anchor group for the anchors closest to the Z-direction edge is:
\(N_{ua} = left(\frac{N_x}{N_{ein,t}}\richtig)N_{mit,G1} = left(\frac{30\ \Text{kN}}{6}\richtig)\mal 2 = 10\ \Text{kN}\)
Mit der Randabstände are calculated as:
\(c_{j,\Min.} = min(c_{\Text{oben},G1},\ c_{\Text{Unterseite},G1}) = min(150\ \Text{mm},\ 350\ \Text{mm}) = 150\ \Text{mm}\)
\(c_{mit,\Min.} = min(c_{\Text{links},G1},\ c_{\Text{richtig},G1}) = min(100\ \Text{mm},\ 100\ \Text{mm}) = 100\ \Text{mm}\)
Mit der nominal SFBO capacity is then determined as:
\(N_{als} = left(\frac{1 + \dfrac{c_{mit,\Min.}}{c_{j,\Min.}}}{4} + \frac{S_{Summe,mit,G1}}{6\,c_{j,\Min.}}\richtig)\mal 13 \mal links(\frac{c_{j,\Min.}}{1\ \Text{mm}}\richtig)\mal sqrt{\frac{EIN_{brg}}{\Text{mm}^ 2}}\ \lambda_a sqrt{\frac{f_c}{\Text{MPa}}}\mal 0.001\ \Text{kN}\)
\(N_{als} = left(\frac{1 + \dfrac{100\ \Text{mm}}{150\ \Text{mm}}}{4} + \frac{100\ \Text{mm}}{6\mal 150\ \Text{mm}}\richtig)\mal 13 \mal links(\frac{150\ \Text{mm}}{1\ \Text{mm}}\richtig)\mal sqrt{\frac{3647.4\ \Text{mm}^ 2}{1\ \Text{mm}^ 2}}\mal 1 \mal sqrt{\frac{20.68\ \Text{MPa}}{1\ \Text{MPa}}}\mal 0.001\ \Text{kN}\)
\(N_{als} = 282.65\ \Text{kN}\)
Since the edge distance ca₂ is still less than 3ca₁, the same modified reduction factor is applied.
Schließlich, bleibt die design SFBO capacity ist:
\(\phi N_{als} = \phi\,N_{als} = 0.7 \mal 282.65\ \Text{kN} = 197.86\ \Text{kN}\)
Schon seit 10 kN < 197.86 kN, the SFBO capacity along the Z-direction ist ausreichend.
Prüfen #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.
Prüfen #10: Calculate breakout capacity (Vz 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.
Prüfen #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.
Prüfen #12: Calculate pryout capacity (Vz 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.
Prüfen #13: Berechnen Sie die Scherkapazität der Ankerstange
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.
Prüfen #14: Calculate anchor rod shear and axial capacity (AISC)
To determine the capacity of the anchor rod under combined shear and axial loads, Wir verwenden AISC 360-22 Gl. J3-3a. In this calculator, the equation is rearranged to express the result as the modified shear strength instead.
Mit der shear demand ist definiert als die shear load per anchor.
\(V_{ua} = V_{ua} = 2.5\ \Text{kN}\)
Mit der tension demand is expressed as the tensile stress in the anchor rod.
\(f_{ut} = frac{N_{ua}}{EIN_{Stange}} = frac{5\ \Text{kN}}{201.06\ \Text{mm}^ 2} = 24.868\ \Text{MPa}\)
Mit der modified shear capacity of the anchor rod is then calculated as:
\(F’_{nv} = \min\!\links(1.3\,F_{nv} – \links(\frac{F_{nv}}{\PHI F_{nt}}\richtig) f_{ut},\; F_{nv}\richtig)\)
\(F’_{nv} = \min\!\links(1.3\mal 232.69\ \Text{MPa} – \links(\frac{232.69\ \Text{MPa}}{0.75\mal 387.82\ \Text{MPa}}\richtig)\mal 24.868\ \Text{MPa},\; 232.69\ \Text{MPa}\richtig) = 232.69\ \Text{MPa}\)
We then multiply this strength by the anchor area mit AISC 360-22 Gl. J3-2.
\(\Phi R_{n,\Text{aisc}} = \phi F’_{nv} EIN_{\Text{Stange}} = 0.75 \mal 232.69\ \Text{MPa} \mal 201.06\ \Text{mm}^2 = 35.09\ \Text{kN}\)
Schon seit 2.5 kN < 35.09 kN, the anchor rod capacity is ausreichend.
Prüfen #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 auch.
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 Gl. 17.8.3.
\(ICH_{int} = frac{N_{ua}}{\phi N_n} + \frac{V_{ua}}{\phi V_n} = frac{30}{47.749} + \frac{6}{17.921} = 0.96308\)
Schon seit 0.96 < 1.2, the interaction check is ausreichend.
Entwurfszusammenfassung
Mit der Skyciv Base Plate Design Software kann automatisch einen Schritt-für-Schritt-Berechnungsbericht für dieses Entwurfsbeispiel erstellen. Es enthält auch eine Zusammenfassung der durchgeführten Schecks und deren resultierenden Verhältnisse, Die Informationen auf einen Blick leicht zu verstehen machen. Im Folgenden finden Sie eine Stichprobenzusammenfassungstabelle, Welches ist im Bericht enthalten.
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