Base plaatontwerp voorbeeld met behulp van EN 1993-1-8-2005, IN 1993-1-1-2005 and EN 1992-1-1-2004
Probleemverklaring:
Bepaal of de ontworpen kolom-voetplaatverbinding voldoende is voor een 50-kN tension load, 4-kN Vy shear load, en 2-kN Vz shear load.
Gegeven gegevens:
Kolom:
Kolomgedeelte: CHS193.7×10
Kolomgebied: 5770.0 mm²
Kolommateriaal: S460
Bodemplaat:
Baseplaat afmetingen: 300mm x 300mm
Basisplaatdikte: 18mm
Basisplaatmateriaal: S235
Vocht:
Vochtdikte: 0 mm
Beton:
Concrete dimensies: 350mm x 350mm
Betonnen dikte: 400 mm
Betonnen materiaal: C35/45
Gebarsten of ongescheurd: Gebarsten
Ankers:
Ankerdiameter: 16 mm
Effectieve inbeddingslengte: 350 mm
Ingebouwde plaatdiameter: 70 mm
Ingebedde plaatdikte: 10 mm
Ankermateriaal: 4.8
Lassen:
Lastype: Fillet
Lasgrootte: 7mm
Vulmetaalclassificatie: E42
Ankergegevens (van Skyciv Calculator):
Opmerkingen:
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.
Stapsgewijze berekeningen:
Controleren #1: Lascapaciteit berekenen
The full tensile load is resisted by the entire weld section, Terwijl de shear load components are distributed only to a portion of the total weld length. This portion is determined by projecting a 90° sector from the center of the column to its circumference. Daarom, enkel en alleen half of the total circumference is designed to resist the shear load.
We first compute the Totale laslengte als de portion of the weld within the 90° projection.
\(L_{lassen,full} = \pi d_{col} = \pi \times 193.7\ \tekst{mm} = 608.53\ \tekst{mm}\)
\(L_{lassen} = frac{\pi d_{col}}{2} = frac{\pi \times 193.7\ \tekst{mm}}{2} = 304.26\ \tekst{mm}\)
De volgende, we berekenen de resultant shear load.
\(V_r = \sqrt{(V_y)^ 2 + (V_z)^ 2} = sqrt{(4\ \tekst{kN})^ 2 + (2\ \tekst{kN})^ 2} = 4.4721\ \tekst{kN}\)
We then compute the normaal en shear stresses, taking into account the assumed load distribution.
\( \sigma_{\dader} = frac{N_x}{L_{lassen,full}\,a\,\sqrt{2}} = frac{40\ \tekst{kN}}{608.53\ \tekst{mm} \keer 4.95\ \tekst{mm} \keer sqrt{2}} = 9.39\ \tekst{MPa} \)
\( \jouw_{\dader} = frac{N_x}{L_{lassen,full}\,a\,\sqrt{2}} = frac{40\ \tekst{kN}}{608.53\ \tekst{mm} \keer 4.95\ \tekst{mm} \keer sqrt{2}} = 9.39\ \tekst{MPa} \)
\( \jouw_{\parallel} = frac{V_r}{L_{lassen}\,een} = frac{4.4721\ \tekst{kN}}{304.26\ \tekst{mm} \keer 4.95\ \tekst{mm}} = 2.9693\ \tekst{MPa} \)
Daarna, we berekenen de gecombineerde spanningen gebruik makend van IN 1993-1-8:2005 Eq. (4.1).
\(F_{w,Ed1} = sqrt{(\sigma_{\dader})^ 2 + 3\big((\jouw_{\dader})^ 2 + (\jouw_{\parallel})^2\big)}\)
\(F_{w,Ed1} = sqrt{(9.39\ \tekst{MPa})^ 2 + 3\big((9.39\ \tekst{MPa})^ 2 + (2.9693\ \tekst{MPa})^2\big)}\)
\(F_{w,Ed1} = 19.471\ \tekst{MPa}\)
Tegelijkertijd, We bepalen de stress on the base metal using the same equation.
\(F_{w,Ed2} = \sigma_{\dader} = 9.39\ \tekst{MPa}\)
De volgende, we berekenen de weld capacity. We first determine the ultimate tensile strength (het programma wijst lengte toe) van de weaker material, and then use IN 1993-1-8:2005 Eq. (4.1) to obtain the fillet weld resistance en base metal resistance.
\(f_u = \min\!\links(F_{u,\tekst{col}},\ f_{u,\tekst{bp}},\ f_{u,w}\Rechtsaf) = \min\!\links(550\ \tekst{MPa},\ 360\ \tekst{MPa},\ 500\ \tekst{MPa}\Rechtsaf) = 360\ \tekst{MPa}\)
\(F_{w,Rd1} = frac{f_u}{\beta_w\,(\gamma_{M2,\text{lassen}})} = frac{360\ \tekst{MPa}}{0.8 \keer (1.25)} = 360\ \tekst{MPa}\)
\(F_{w,Rd2} = frac{0.9\,f_u}{\gamma_{M2,\text{lassen}}} = frac{0.9 \keer 360\ \tekst{MPa}}{1.25} = 259.2\ \tekst{MPa}\)
Sinds 19.471 MPa < 360 MPa, De lascapaciteit is voldoende.
Controleren #2: Bereken de buigcapaciteit van de basisplaat als gevolg van spanningsbelasting
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.
Controleren #3: Bereken de betonuitbraakcapaciteit in spanning
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.
Controleren #4: Bereken het pull -outcapaciteit van het anker
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.
Controleren #5: Bereken de side-face blowoutcapaciteit in Y-richting
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.
Controleren #6: Bereken side-face blowoutcapaciteit in Z-richting
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.
Controleren #7: Calculate base plate bearing capacity at anchor holes (Vy afschuiving)
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.
Controleren #8: Calculate base plate bearing capacity at anchor holes (Vz-afschuiving)
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.
Controleren #9: Calculate concrete breakout capacity (Vy afschuiving)
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.
Controleren #10: Calculate concrete breakout capacity (Vz-afschuiving)
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.
Controleren #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.
Controleren #12: Bereken de afschuifcapaciteit van de ankerstang
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. Echter, in dit voorbeeld, the shear acts without a lever arm. Daarom, 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, Raadpleeg deze link.
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. Jij kan try out the free tool vandaag.
Controleren #13: Calculate anchor steel interaction check
We gebruiken IN 1992-4:2018 Tafel 7.3 Eq. (7.54) om de 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 is:
\(IK_{int} = links(\frac{N_{Ed}}{N_{Rd,s}}\Rechtsaf)^ 2 + \links(\frac{V_{Ed}}{V_{Rd,s}}\Rechtsaf)^2)
\(IK_{int} = links(\frac{10\ \tekst{kN}}{49.22\ \tekst{kN}}\Rechtsaf)^ 2 + \links(\frac{1.118\ \tekst{kN}}{38.604\ \tekst{kN}}\Rechtsaf)➔⡔ Koop generieke tadalafil 0.042117\)
Sinds 0.042 < 1.0, the anchor rod steel failure interaction check is voldoende.
Controleren #14: Calculate concrete failure interaction check
de obstructie onder het dakoppervlak interaction check is required for concrete failures under simultaneous shear and tensile loading. Voor deze, we gebruiken IN 1992-4:2018 Tafel 7.3 Eq. (7.55) en Eq. (7.56).
Here are the resulting ratios for all tensile checks.
Here are the resulting ratios for all shear checks.
Eerste, we check using Eq. (7.55) and compare the result to the maximum interaction limit of 1.0.
\(IK_{\tekst{case1}} = links(\links(\frac{N_{Ed}}{N_{Rd}}\Rechtsaf)^{1.5}\Rechtsaf) + \links(\links(\frac{V_{Ed}}{V_{Rd}}\Rechtsaf)^{1.5}\Rechtsaf)\)
\(IK_{\tekst{case1}} = links(\links(\frac{40}{45.106}\Rechtsaf)^{1.5}\Rechtsaf) + \links(\links(\frac{4.1231}{14.296}\Rechtsaf)^{1.5}\Rechtsaf) = 0.99\)
De volgende, we check using Eq. (7.56) and compare the result to the maximum interaction limit of 1.2.
\(IK_{\tekst{geval2}} = frac{N_{Ed}}{N_{Rd}} + \frac{V_{Ed}}{V_{Rd}} = frac{40}{45.106} + \frac{4.1231}{14.296} = 1.1752\)
Sinds 0.99 < 1.0 en 1.175 < 1.2, de concrete failure interaction check is voldoende.
Ontwerp Samenvatting
De Skyciv Base Plate Design Software Kan automatisch een stapsgewijze berekeningsrapport genereren voor dit ontwerpvoorbeeld. Het biedt ook een samenvatting van de uitgevoerde controles en hun resulterende verhoudingen, De informatie in één oogopslag gemakkelijk te begrijpen maken. Hieronder is een sample samenvattende tabel, die is opgenomen in het rapport.
Skyciv Sample Report
Klik hier om een voorbeeldrapport te downloaden.
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