Base Plate Design Example using EN 1993-1-8:2005, IM 1993-1-1:2005, IM 1992-1-1:2004, und EN 1992-4:2018.
Problemanweisung:
Determine whether the designed column-to-base plate connection is sufficient for a Vy=5-kN und Vz=5-kN Querlasten.
Gegebene Daten:
Spalte:
Spaltenabschnitt: SHS 180x180x8
Säulenbereich: 5440 mm2
Säulenmaterial: S235
Grundplatte:
Grundplattenabmessungen: 350 mmx 350 mm
Grundplattendicke: 12 mm
Grundplattenmaterial: S235
Fugenmörtel:
Fugenmörtel Dicke: 6 mm
Grout material: ≥ 30 MPa
Beton:
Konkrete Abmessungen: 350 mmx 350 mm
Betondicke: 350 mm
Betonmaterial: C25/30
Geknackt oder ungekrönt: Geknackt
Anker:
Ankerdurchmesser: 12 mm
Effektive Einbettungslänge: 150 mm
Embedded plate diameter: 60 mm
Dicke eingebetteter Platten: 10 mm
Anchor material: 8.8
Other information:
- Non-countersunk anchors.
- Anchor with cut threads.
- K7 factor for anchor steel shear failure: 1.0
- Degree of Restraint of Fastener: No restraint
Schweißnähte:
Schweißtyp: Fillet Weld
Weld leg size: 8mm
Füllmetallklassifizierung: E35
Ankerdaten (von Skyciv -Taschenrechner):
Definitionen:
Lastpfad:
Mit der SkyCiv Basisplatten-Design-Software follows IM 1992-4:2018 for anchor rod design. Shear loads applied to the column are transferred to the base plate through the welds and then to the supporting concrete through the anchor rods. Friction and shear lugs are not considered in this example, as these mechanisms are not supported in the current software.
Ankergruppen:
The software includes an intuitive feature that identifies which anchors are part of an anchor group for evaluating concrete shear breakout und concrete shear pryout Fehler.
Ein Ankergruppe is defined as two or more anchors with overlapping projected resistance areas. In diesem Fall, the anchors act together, and their combined resistance is checked against the applied load on the group.
Ein single anchor is defined as an anchor whose projected resistance area does not overlap with any other. In diesem Fall, the anchor acts alone, and the applied shear force on that anchor is checked directly against its individual resistance.
This distinction allows the software to capture both group behavior and individual anchor performance when assessing shear-related failure modes.
Schritt-für-Schritt-Berechnungen:
Prüfen #1: Berechnen Sie die Schweißkapazität
We assume that the Vz shear load is resisted by the top and bottom welds, während Vy shear load is resisted exclusively by the left and right welds.
To determine the weld capacity of the top and bottom welds, we first calculate their total weld lengths.
\(
L_{w,top\&Unterseite} = 2 \links(= Abstand des Abschnitts, in dem die Scherung berücksichtigt wird, zur Fläche des nächsten Auflagers{col} – 2t_{col} – 2r_{col}\richtig)
= 2 \mal links(180 \,\Text{mm} – 2 \mal 8 \,\Text{mm} – 2 \mal 4 \,\Text{mm}\richtig)
= 312 \,\Text{mm}
\)
Als nächstes, wir berechnen die stresses in the welds.
Note that the applied Vz shear acts parallel to the weld axis, with no other forces present. This means the perpendicular stresses can be taken as zero, and only the shear stress in the parallel direction needs to be calculated.
\(
\sigma_{\Täter} = frac{N.}{(L_{w,top\&Unterseite})\,a\sqrt{2}}
= frac{0 \,\Text{kN}}{(312 \,\Text{mm}) \mal 5.657 \,\Text{mm} \mal sqrt{2}}
= 0
\)
\(
\Ihre_{\Täter} = frac{0}{(L_{w,top\&Unterseite})\,a\sqrt{2}}
= frac{0 \,\Text{kN}}{(312 \,\Text{mm}) \mal 5.657 \,\Text{mm} \mal sqrt{2}}
= 0
\)
\(
\Ihre_{\parallel} = frac{V_{mit}}{(L_{w,top\&Unterseite})\,ein}
= frac{5 \,\Text{kN}}{(312 \,\Text{mm}) \mal 5.657 \,\Text{mm}}
= 2.8329 \,\Text{MPa}
\)
Verwenden von IM 1993-1-8:2005, Gl. 4.1, the design weld stress is obtained using the directional method.
\(
F_{w,Ed1} = Quadrat{ (\sigma_{\Täter})^ 2 + 3 \links( (\Ihre_{\Täter})^ 2 + (\Ihre_{\parallel})^2 rechts) }
= Quadrat{ (0)^ 2 + 3 \mal links( (0)^ 2 + (2.8329 \,\Text{MPa})^2 rechts) }
= 4.9067 \,\Text{MPa}
\)
Zusätzlich, the design normal stress for the base metal check, pro IM 1993-1-8:2005, Gl. 4.1, is taken as zero, schon seit no normal stress is present.
\(
F_{w,Ed2} = \sigma_{\Täter} = 0
\)
Jetzt, let us assess the left and right welds. As with the top and bottom welds, we first calculate the Gesamtschweißlänge.
\(
L_{w,left\&richtig} = 2 \links(d_{col} – 2t_{col} – 2r_{col}\richtig)
= 2 \mal links(180 \,\Text{mm} – 2 \mal 8 \,\Text{mm} – 2 \mal 4 \,\Text{mm}\richtig)
= 312 \,\Text{mm}
\)
We then calculate the components of the weld stresses.
\(
\sigma_{\Täter} = frac{N.}{(L_{w,left\&richtig})\,a\sqrt{2}}
= frac{0 \,\Text{kN}}{(312 \,\Text{mm}) \mal 5.657 \,\Text{mm} \mal sqrt{2}}
= 0
\)
\(
\Ihre_{\Täter} = frac{0}{(L_{w,left\&richtig})\,a\sqrt{2}}
= frac{0 \,\Text{kN}}{(312 \,\Text{mm}) \mal 5.657 \,\Text{mm} \mal sqrt{2}}
= 0
\)
\(
\Ihre_{\parallel} = frac{V_y}{(L_{w,left\&richtig})\,ein}
= frac{5 \,\Text{kN}}{(312 \,\Text{mm}) \mal 5.657 \,\Text{mm}}
= 2.8329 \,\Text{MPa}
\)
Verwenden von IM 1993-1-8:2005, Gl. 4.1, we determine both the design weld stress and the design normal stress for the base metal check.
\(
F_{w,Ed1} = Quadrat{ \links( \sigma_{\Täter} \richtig)^ 2 + 3 \links( \links( \Ihre_{\Täter} \richtig)^ 2 + \links( \Ihre_{\parallel} \richtig)^2 rechts) }
\)
\(
F_{w,Ed1} = Quadrat{ \links( 0 \richtig)^ 2 + 3 \mal links( \links( 0 \richtig)^ 2 + \links( 2.8329 \,\Text{MPa} \richtig)^2 rechts) }
\)
\(
F_{w,Ed1} = 4.9067 \,\Text{MPa}
\)
The next step is to identify the governing weld stress between the top/bottom welds and the left/right welds. Because the weld lengths are equal and the applied loads have the same magnitude, the resulting weld stresses are equal.
\(
F_{w,Ed1} = \max(F_{w,Ed1}, \, F_{w,Ed1})
= \max(4.9067 \,\Text{MPa}, \, 4.9067 \,\Text{MPa})
= 4.9067 \,\Text{MPa}
\)
The base metal stress remains zero.
\(
F_{w,Ed2} = \max(F_{w,Ed2}, \, F_{w,Ed2}) = \max(0, \, 0) = 0
\)
Jetzt, we calculate the weld capacity. Zuerst, the resistance of the Filetschweißung is computed. Dann, the resistance of the base metal is determined. Mit EN 1993-1-8:2005, Gl. 4.1, the capacities are calculated as follows:
\(
F_{w,Rd1} = frac{f_u}{\beta_w \left(\Um es zu berechnen{M2,weld}\richtig)}
= frac{360 \,\Text{MPa}}{0.8 \mal (1.25)}
= 360 \,\Text{MPa}
\)
\(
F_{w,Rd2} = frac{0.9 f_u}{\Um es zu berechnen{M2,weld}}
= frac{0.9 \mal 360 \,\Text{MPa}}{1.25}
= 259.2 \,\Text{MPa}
\)
Schließlich, we compare the weld stresses with the weld capacities, and the base metal stresses with the base metal capacities.
Schon seit 4.9067 MPa < 360 MPa und 0 MPa < 259.2 MPa, the capacity of the welded connection is ausreichend.
Prüfen #2: Calculate concrete breakout capacity due to Vy shear
Following the provisions of IM 1992-4:2018, the edge perpendicular to the applied load is assessed for shear breakout failure. Only the anchors nearest to this edge are considered engaged, while the remaining anchors are assumed not to resist shear.
These edge anchors must have a concrete edge distance greater than the larger of 10·hef and 60·d, wo haben is the embedment length and d is the anchor diameter. If this condition is not met, the thickness of the base plate must be less than 0.25·hef.
If the requirements in IM 1992-4:2018, Klausel 7.2.2.5(1), are not satisfied, the SkyCiv software cannot proceed with the design checks, and the user is advised to refer to other relevant standards.
From the SkyCiv software results, the edge anchors act as einzelne Anker, since their projected areas do not overlap. Für diese Berechnung, Anchor 1 will be considered.
To calculate the portion of the Vy shear load carried by Anchor 1, the total Vy shear is distributed among the anchors nearest to the edge. This gives the perpendicular force on Anchor 1.
\(
V_{\Täter} = frac{V_y}{N_{ein,s}}
= frac{5 \,\Text{kN}}{2}
= 2.5 \,\Text{kN}
\)
Für die parallel force, it is assumed that all anchors resist the load equally. Deshalb, the parallel component of the load is calculated as:
\(
V_{\parallel} = frac{V_z}{N_{anc}}
= frac{5 \,\Text{kN}}{4}
= 1.25 \,\Text{kN}
\)
Mit der total shear load on Anchor 1 is therefore:
\(
V_{Ed} = Quadrat{ \links( V_{\Täter} \richtig)^ 2 + \links( V_{\parallel} \richtig)^ 2 }
\)
\(
V_{Ed} = Quadrat{ \links( 2.5 \,\Text{kN} \richtig)^ 2 + \links( 1.25 \,\Text{kN} \richtig)^ 2 } = 2.7951 \,\Text{kN}
\)
The first part of the capacity calculation is to determine the alpha and beta factors. Wir verwenden IM 1992-4:2018, Klausel 7.2.2.5, to set the lf dimension, und Gleichungen 7.42 und 7.43 to determine the factors.
\(
l_f = \min(h_{ef}, \, 12d_{anc})
= \min(150 \,\Text{mm}, \, 12 \mal 12 \,\Text{mm})
= 144 \,\Text{mm}
\)
\(
\alpha = 0.1 \links(\frac{l_f}{c_{1,s1}}\richtig)^{0.5}
= 0.1 \mal links(\frac{144 \,\Text{mm}}{50 \,\Text{mm}}\richtig)^{0.5}
= 0.16971
\)
\(
\beta = 0.1 \links(\frac{d_{anc}}{c_{1,s1}}\richtig)^{0.2}
= 0.1 \mal links(\frac{12 \,\Text{mm}}{50 \,\Text{mm}}\richtig)^{0.2}
= 0.07517
\)
The next step is to calculate the initial value of the characteristic resistance of the fastener. Verwenden von IM 1992-4:2018, Gleichung 7.41, the value is:
\(
V^{0}_{Rk,c} = k_9 \left( \frac{d_{anc}}{\Text{mm}} \richtig)^{\Alpha}
\links( \frac{l_f}{\Text{mm}} \richtig)^{\Beta}
\sqrt{ \frac{f_{ck}}{\Text{MPa}} }
\links( \frac{c_{1,s1}}{\Text{mm}} \richtig)^{1.5} N.
\)
\(
V^{0}_{Rk,c} = 1.7 \mal links( \frac{12 \,\Text{mm}}{1 \,\Text{mm}} \richtig)^{0.16971}
\mal links( \frac{144 \,\Text{mm}}{1 \,\Text{mm}} \richtig)^{0.07517}
\mal sqrt{ \frac{20 \,\Text{MPa}}{1 \,\Text{MPa}} }
\mal links( \frac{50 \,\Text{mm}}{1 \,\Text{mm}} \richtig)^{1.5}
\mal 0.001 \,\Text{kN}
\)
\(
V^{0}_{Rk,c} = 5.954 \,\Text{kN}
\)
Dann, Wir berechnen die Referenz projizierter Bereich of a single anchor, folgende IM 1992-4:2018, Gleichung 7.44.
\(
EIN_{c,V }^{0} = 4.5 \links( c_{1,s1} \richtig)^ 2
= 4.5 \mal links( 50 \,\Text{mm} \richtig)^ 2
= 11250 \,\Text{mm}^ 2
\)
Danach, Wir berechnen die Tatsächlicher projizierter Bereich of Anchor 1.
\(
B_{c,V } = \min(c_{links,s1}, \, 1.5c_{1,s1}) + \Min.(c_{richtig,s1}, \, 1.5c_{1,s1})
\)
\(
B_{c,V } = \min(300 \,\Text{mm}, \, 1.5 \mal 50 \,\Text{mm}) + \Min.(50 \,\Text{mm}, \, 1.5 \mal 50 \,\Text{mm}) = 125 \,\Text{mm}
\)
\(
Um es zu berechnen{c,V } = \min(1.5c_{1,s1}, \, t_{konz}) = \min(1.5 \mal 50 \,\Text{mm}, \, 200 \,\Text{mm}) = 75 \,\Text{mm}
\)
\(
EIN_{c,V } Die Hälfte der Wandhöhe von der Unterseite der Basis für den Fall des{c,V } B_{c,V } = 75 \,\Text{mm} \mal 125 \,\Text{mm} = 9375 \,\Text{mm}^ 2
\)
We also need to calculate the parameters for shear breakout. Wir verwenden IM 1992-4:2018, Gleichung 7.4, to get the factor that accounts for the disturbance of stress distribution, Gleichung 7.46 for the factor that accounts for the member thickness, und Gleichung 7.48 for the factor that accounts for the influence of a shear load inclined to the edge. These are calculated as follows:
\(
\Psi_{s,V } = min links( 0.7 + 0.3 \links( \frac{c_{2,s1}}{1.5c_{1,s1}} \richtig), \, 1.0 \richtig)
= min links( 0.7 + 0.3 \mal links( \frac{50 \,\Text{mm}}{1.5 \mal 50 \,\Text{mm}} \richtig), \, 1 \richtig)
= 0.9
\)
\(
\Psi_{h,V } = max links( \links( \frac{1.5c_{1,s1}}{t_{konz}} \richtig)^{0.5}, \, 1 \richtig)
= max links( \links( \frac{1.5 \mal 50 \,\Text{mm}}{200 \,\Text{mm}} \richtig)^{0.5}, \, 1 \richtig)
= 1
\)
\(
\= Abstand des Abschnitts, in dem die Scherung berücksichtigt wird, zur Fläche des nächsten Auflagers{V } = \tan^{-1} \links( \frac{V_{\parallel}}{V_{\Täter}} \richtig)
= \tan^{-1} \links( \frac{1.25 \,\Text{kN}}{2.5 \,\Text{kN}} \richtig)
= 0.46365 \,\Text{Arbeit}
\)
\(
\Psi_{\Alpha,V } = max links(
\sqrt{ \frac{1}{(\cos(\= Abstand des Abschnitts, in dem die Scherung berücksichtigt wird, zur Fläche des nächsten Auflagers{V }))^ 2 + \links( 0.5 \, (\ohne(\= Abstand des Abschnitts, in dem die Scherung berücksichtigt wird, zur Fläche des nächsten Auflagers{V })) \richtig)^ 2 } }, \, 1 \richtig)
\)
\(
\Psi_{\Alpha,V } = max links(
\sqrt{ \frac{1}{(\cos(0.46365 \,\Text{Arbeit}))^ 2 + \links( 0.5 \times \sin(0.46365 \,\Text{Arbeit}) \richtig)^ 2 } }, \, 1 \richtig)
\)
\(
\Psi_{\Alpha,V } = 1.0847
\)
One important note when determining the alpha factor is to ensure the perpendicular shear and parallel shear are identified correctly.
Schließlich, Wir berechnen die breakout resistance of the single anchor using IM 1992-4:2018, Gleichung 7.1.
\(
V_{Rk,c} = V^0_{Rk,c} \links(\frac{EIN_{c,V }}{A^0_{c,V }}\richtig)
\Psi_{s,V } \Psi_{h,V } \Psi_{ec,V } \Psi_{\Alpha,V } \Psi_{Ausbruchkegelbereich für Einzeldübel nicht durch Kanten beeinflusst,V }
\)
\(
V_{Rk,c} = 5.954 \,\Text{kN} \mal links(\frac{9375 \,\Text{mm}^ 2}{11250 \,\Text{mm}^ 2}\richtig)
\mal 0.9 \mal 1 \mal 1 \mal 1.0847 \mal 1
= 4.8435 \,\Text{kN}
\)
Applying the partial factor, the design resistance is 3.23 kN.
\(
V_{Rd,c} = frac{V_{Rk,c}}{\Um es zu berechnen{Mc}}
= frac{4.8435 \,\Text{kN}}{1.5}
= 3.229 \,\Text{kN}
\)
Schon seit 2.7951 kN < 3.229 kN, the shear breakout capacity for Vy shear is ausreichend.
Prüfen #3: Calculate concrete breakout capacity due to Vz shear
The same approach is used to determine the capacity on the edge perpendicular to the Vz shear.
Because of the symmetric design, the anchors resisting Vz shear are also identified as einzelne Anker. Let’s consider Anchor 1 again for the calculations.
Zur Berechnung der perpendicular load on Anchor 1, we divide the Vz shear by the total number of anchors nearest to the edge only. Zur Berechnung der parallel load on Anchor 1, we divide the Vy shear by the total number of anchors.
\(
V_{\Täter} = frac{V_{mit}}{N_{ein,s}}
= frac{5 \,\Text{kN}}{2}
= 2.5 \,\Text{kN}
\)
\(
V_{\parallel} = frac{V_{j}}{N_{anc}}
= frac{5 \,\Text{kN}}{4}
= 1.25 \,\Text{kN}
\)
\(
V_{Ed} = Quadrat{ \links( V_{\Täter} \richtig)^ 2 + \links( V_{\parallel} \richtig)^ 2 }
\)
\(
V_{Ed} = Quadrat{ \links( 2.5 \,\Text{kN} \richtig)^ 2 + \links( 1.25 \,\Text{kN} \richtig)^ 2 }
= 2.7951 \,\Text{kN}
\)
Using a similar approach to Check #2, the resulting breakout resistance for the edge perpendicular to Vz shear is:
\(
V_{Rd,c} = frac{V_{Rk,c}}{\Um es zu berechnen{Mc}}
= frac{4.8435 \,\Text{kN}}{1.5}
= 3.229 \,\Text{kN}
\)
Schon seit 2.7951 kN < 3.229 kN, the shear breakout capacity for Vz shear is ausreichend.
Prüfen #4: Calculate concrete pryout capacity
The calculation for the shear pryout resistance involves determining the nominal capacity of the anchors against tension breakout. The reference for tension breakout capacity is IM 1992-4:2018, Klausel 7.2.1.4. A detailed discussion of tension breakout is already covered in the SkyCiv Design Example with Tension Load and will not be repeated in this design example.
From the SkyCiv software calculations, the nominal capacity of the section for tension breakout is 44.61 kN.
We then use IM 1992-4:2018, Equation 7.39a, to obtain the design characteristic resistance. Verwenden von k8 = 2, the capacity is 59.48 kN.
\(
V_{Rd,cp} = frac{k_8 N_{cbg}}{\gamma_C}
= frac{2 \mal 44.608 \,\Text{kN}}{1.5}
= 59.478 \,\Text{kN}
\)
In the shear pryout check, all anchors are effective in resisting the full shear load. From the image generated by the SkyCiv software, all failure cone projections overlap with each other, making the anchors act as an Ankergruppe.
Deshalb, the required resistance of the anchor group is the total resultant shear load of 7.07 kN.
\(
V_{res} = Quadrat{(V_y)^ 2 + (V_z)^ 2}
= Quadrat{(5 \,\Text{kN})^ 2 + (5 \,\Text{kN})^ 2}
= 7.0711 \,\Text{kN}
\)
\(
V_{Ed} = left(\frac{V_{res}}{N_{anc}}\richtig) N_{ein,G1}
= left(\frac{7.0711 \,\Text{kN}}{4}\richtig) \mal 4
= 7.0711 \,\Text{kN}
\)
Schon seit 7.0711 kN < 59.478 kN, the shear pryout capacity is ausreichend.
Prüfen #5: Calculate anchor rod shear capacity
The calculation of the anchor rod shear capacity depends on whether the shear load is applied with a moment arm. Um dies zu bestimmen, Wir beziehen uns auf IM 1992-4:2018, Klausel 6.2.2.3, where the thickness and material of the grout, the number of fasteners in the design, the spacing of the fasteners, and other factors are checked.
Mit der Skyciv Base Plate Design Software performs all the necessary checks to determine whether the shear load acts with or without a lever arm. For this design example, it is determined that the shear load is nicht applied with a lever arm. Deshalb, Wir verwenden IM 1992-4:2018, Klausel 7.2.2.3.1, for the capacity equations.
We begin by calculating the characteristic resistance of the steel fastener using IM 1992-4:2018, Gleichung 7.34.
\(
V^0_{Rk,s} = k_6 A_s f_{u,anc}
= 0.5 \mal 113.1 \,\Text{mm}^2 mal 800 \,\Text{MPa}
= 45.239 \,\Text{kN}
\)
Als nächstes, we apply the factor for the ductility of the single anchor or the anchor group, taking k7 = 1.
\(
V_{Rk,s} = k_7 V^{0}_{Rk,s}
= 1 \mal 45.239 \,\Text{kN}
= 45.239 \,\Text{kN}
\)
We then obtain the partial factor for steel shear failure mit IM 1992-4:2018, Tabelle 4.1. For an anchor with 8.8 des Materials , the resulting partial factor is:
\(
\Um es zu berechnen{MS,Schub-}
= max links( 1.0 \links( \frac{F_{u,anc}}{F_{j,anc}} \richtig), \, 1.25 \richtig)
= max links( 1 \mal frac{800 \,\Text{MPa}}{640 \,\Text{MPa}}, \, 1.25 \richtig)
= 1.25
\)
Applying this factor to the characteristic resistance, the design resistance is 36.19 kN.
\(
V_{Rd,s} = frac{V_{Rk,s}}{\Um es zu berechnen{MS,Schub-}}
= frac{45.239 \,\Text{kN}}{1.25}
= 36.191 \,\Text{kN}
\)
Mit der required shear resistance per anchor rod is the resultant shear load divided by the total number of anchor rods, which calculates to 1.77 kN.
\(
V_{Ed} = frac{\sqrt{ (V_y)^ 2 + (V_z)^ 2 }}{N_{anc}}
\)
\(
V_{Ed} = frac{\sqrt{ (5 \,\Text{kN})^ 2 + (5 \,\Text{kN})^ 2 }}{4}
= 1.7678 \,\Text{kN}
\)
Schon seit 1.7678 kN < 36.191 kN, the anchor rod steel shear capacity 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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