Esempio di design della piastra di base utilizzando AISC 360-22 e ACI 318-19
Dichiarazione del problema:
Determine whether the designed column-to-base plate connection is sufficient for a 20-kip tension load.
Dati dati:
Colonna:
Sezione colonna: W12x53
Area colonna: 15.6 pollici2
Materiale colonna: A992
Piastra di base:
Dimensioni della piastra di base: 18 in x 18 pollici
Spessore della piastra di base: 3/4 pollici
Materiale della piastra di base: A36
Grout:
Grout thickness: 1 pollici
Calcestruzzo:
Dimensioni concrete: 22 in x 22 pollici
Spessore di cemento: 15 pollici
Materiale di cemento: 4000 psi
Cracked or Uncracked: Cracked
Anchors:
Anchor diameter: 3/4 pollici
Effective embedment length: 12 pollici
Embedded plate width: 3 pollici
Embedded plate thickness: 1/4 pollici
Anchor offset distance from face of column web: 2.8275 pollici
saldature:
Dimensione della saldatura: 1/4 pollici
Classificazione del metallo di riempimento: E70XX
Anchor Data (a partire dal SkyCiv Calculator):
Definitions:
Load Path:
When a base plate is subjected to uplift (trazione) forze, these forces are transferred to the anchor rods, which in turn induce bending moments in the base plate. The bending action can be visualized as cantilever bending occurring around the flanges or web of the column section, depending on where the anchors are positioned.
Nel Software di progettazione della piastra di base Skyciv, only anchors located within the anchor tension zone are considered effective in resisting uplift. This zone typically includes areas near the column flanges or web. Anchors outside this zone do not contribute to tension resistance and are excluded from the uplift calculations.
To determine the effective area of the base plate that resists bending, un carico 45-degree dispersion is assumed from the centerline of each anchor rod toward the column face. This dispersion defines the effective weld length and helps establish the effective bending width del piatto.
The assumption simplifies the base plate analysis by approximating how the uplift force spreads through the plate.
Anchor Groups:
La Software di progettazione della piastra di base Skyciv includes an intuitive feature that identifies which anchors are part of an anchor group for evaluating rottura concreta e concrete side-face blowout failures.
Un anchor group consists of multiple anchors with similar effective embedment depths and spacing, and are close enough that their projected resistance areas overlap. When anchors are grouped, their capacities are combined to resist the total tension force applied to the group.
Anchors that do not meet the grouping criteria are treated as single anchors. In questo caso, only the tension force on the individual anchor is checked against its own effective resistance area.
Calcoli passo-passo:
Dai un'occhiata #1: Calcola la capacità di saldatura
Iniziare, we need to calculate the load per anchor and the effective weld length per anchor. The effective weld length is determined by the shortest length from the 45° dispersion, constrained by the actual weld length and anchor spacing.
For this calculation, anchors are classified as either end anchors o intermediate anchors. End anchors are located at the ends of a row or column of anchors, while intermediate anchors are positioned between them. The calculation method differs for each and depends on the column geometry. In questo esempio, there are two anchors along the web, and both are classified as end anchors.
For end anchors, the effective weld length is limited by the available distance from the anchor centerline to the column fillet. The 45° dispersion must not extend beyond this boundary.
\(
l_r = \frac{d_{col} – 2t_f – 2r_{col} – s_y(n_{un carico,lato} – 1)}{2} = frac{12.1 \, \testo{pollici} – 2 \volte 0.575 \, \testo{pollici} – 2 \volte 0.605 \, \testo{pollici} – 5 \, \testo{pollici} \volte (2 – 1)}{2} = 2.37 \, \testo{pollici}
\)
On the inner side, the effective length is limited by half the anchor spacing. The total effective weld length for the end anchor is the sum of the outer and inner lengths.
\(
l_{eff,fine} = \min(Fare, 0.5s_y) + \min(Fare, l_r)
\)
\(
l_{eff,fine} = \min(2.8275 \, \testo{pollici}, 0.5 \volte 5 \, \testo{pollici}) + \min(2.8275 \, \testo{pollici}, 2.37 \, \testo{pollici}) = 4.87 \, \testo{pollici}
\)
Per questo esempio, il final effective weld length for the web anchor is taken as the effective length of the end anchor.
\(
l_{eff} = l_{eff,fine} = 4.87 \, \testo{pollici}
\)
Successivamente, let’s calculate the load per anchor. For a given set of four (4) ancore, the load per anchor is:
\(
T_{u,ancorare} = frac{N_x}{n_{un carico,t}} = frac{20 \, \testo{kip}}{4} = 5 \, \testo{kip}
\)
Using the calculated effective weld length, we can now determine the required force per unit length on the weld.
\(
r_u = frac{T_{u,ancorare}}{l_{eff}} = frac{5 \, \testo{kip}}{4.87 \, \testo{pollici}} = 1.0267 \, \testo{kip/in}
\)
Adesso, noi useremo AISC 360-22, Chapter J2.4 to calculate the design strength of the fillet weld.
Since the applied load is purely axial tension, the angle \(\theta) is taken as 90°, and the directional strength coefficient kds is calculated according to AISC 360-22 Eq. J2-5.
\(
Eurocodice di design con piastra di base in acciaio{ds} = 1.0 + 0.5(\senza(\theta))^{1.5} = 1 + 0.5 \volte (\senza(1.5708))^{1.5} = 1.5
\)
Infine, applicheremo AISC 360-22 Eq. J2-4 to determine the design strength of the fillet weld per unit length.
\(
\phi r_n = \phi 0.6 F_{Exx} E_{w,ragnatela} Eurocodice di design con piastra di base in acciaio{ds} = 0.75 \volte 0.6 \volte 70 \, \testo{KSI} \volte 0.177 \, \testo{pollici} \volte 1.5 = 8.3633 \, \testo{kip/in}
\)
Da 1.0267 KPI < 8.3633 KPI, La capacità di saldatura è sufficiente.
Dai un'occhiata #2: Calculate base plate flexural yielding capacity due to tension load
Using the load per anchor and the offset distance from the center of the anchor to the face of the column (serving as the load eccentricity), the moment applied to the base plate can be calculated using a a sbalzo assumption.
\(
M_u = T_{u,\testo{ancorare}} e = 5 \, \testo{kip} \volte 2.8275 \, \testo{pollici} = 14.137 \, \testo{kip} \cdot \text{pollici}
\)
Successivamente, using the calculated effective weld length from the previous check as the bending width, we can calculate the Calcola la capacità portante of the base plate using AISC 360-22, Equazione 2-1:
\(
\phi M_n = \phi F_{y,\testo{p.p}} Z_{\testo{eff}} = 0.9 \volte 36 \, \testo{KSI} \volte 0.68484 \, \testo{pollici}^3 = 22.189 \, \testo{kip} \cdot \text{pollici}
\)
Dove,
\(
Z_{\testo{eff}} = frac{l_{\testo{eff}} (t_{\testo{p.p}})^ 2}{4} = frac{4.87 \, \testo{pollici} \volte (0.75 \, \testo{pollici})^ 2}{4} = 0.68484 \, \testo{pollici}^ 3
\)
Da 14.137 pollo-in < 22.189 pollo-in, the base plate flexural yielding capacity is sufficiente.
Dai un'occhiata #3: Calculate anchor rod tensile capacity
To evaluate the tensile capacity of the anchor rod, noi useremo ACI 318-19 Equazione 17.6.1.2.
Primo, Determiniamo il specified tensile strength of the anchor steel. This is the lowest value permitted by ACI 318-19 Clausola 17.6.1.2, with reference to material properties in AISC 360-22 Tabella J3.2.
\(
f_{\testo{uta}} = min sinistra( 0.75 F_{u,\testo{anc}}, 1.9 F_{y,\testo{anc}}, 125 \giusto) = min sinistra( 0.75 \volte 120 \, \testo{KSI}, 1.9 \volte 92 \, \testo{KSI}, 125.00 \, \testo{KSI} \giusto) = 90 \, \testo{KSI}
\)
Successivamente, calcoliamo il effective cross-sectional area of the anchor rod. This is based on ACI 318-19 Commentary Clause R17.6.1.2, which accounts for thread geometry. The number of threads per inch is taken from ASME B1.1-2019 Table 1.
\(
UN_{lo so,N} = frac{\pi}{4} \sinistra( d_a – \frac{0.9743}{n_t} \giusto)^2 = \frac{\pi}{4} \volte sinistra( 0.75 \, \testo{pollici} – \frac{0.9743}{10 \, \testo{pollici}^{-1}} \giusto)^2 = 0.33446 \, \testo{pollici}^ 2
\)
With these values, we apply ACI 318-19 Equazione 17.6.1.2 to compute the design tensile strength of the anchor rod.
\(
\phi N_{per} = phi A_{lo so,N} f_{\testo{uta}} = 0.75 \volte 0.33446 \, \testo{pollici}^2 \times 90 \, \testo{KSI} = 22.576 \, \testo{kip}
\)
Recall the previously calculated tension load per anchor:
\(
N_{ua} = frac{N_x}{n_{un carico,t}} = frac{20 \, \testo{kip}}{4} = 5 \, \testo{kip}
\)
Da 5 kip < 22.576 kip, the anchor rod tensile capacity is sufficiente.
Dai un'occhiata #4: Calculate concrete breakout capacity in tension
Before calculating the breakout capacity, we must first determine whether the member qualifies as a narrow member. Secondo ACI 318-19 Clausola 17.6.2.1.2, the member meets the criteria for a narrow member. Pertanto, a modified effective embedment length must be used in the calculations.
It is determined that the modified effective embedment length, h’ef, of the anchor group is:
\(
h’_{\testo{ef}} = 5.667 \, \testo{pollici}
\)
Usando ACI 318-19 Clausola 17.6.2, calcoliamo il maximum projected concrete cone area per un'unica ancora, based on the modified effective embedment length.
\(
UN_{N_{co}} = 9 \sinistra( h’_{ef,g1} \giusto)^2 = 9 \volte sinistra( 5.6667 \, \testo{pollici} \giusto)^2 = 289 \, \testo{pollici}^ 2
\)
Allo stesso modo, we use the modified effective embedment length to calculate the actual projected concrete cone area of the anchor group.
\(
UN_{N_c} = min sinistra( n_{un carico,g1} UN_{N_{co}}, L_{N_c} B_{N_c} \giusto) = min sinistra( 4 \volte 289 \, \testo{pollici}^ 2, 22 \, \testo{pollici} \volte 22 \, \testo{pollici} \giusto) = 484 \, \testo{pollici}^ 2
\)
Dove,
\(
L_{N_c} = min sinistra( c_{\testo{sinistra},g1}, 1.5 h’_{\testo{ef},g1} \giusto)
+ \sinistra( \min \left( s_{\testo{somma},z,g1}, 3 h’_{\testo{ef},g1} \sinistra( n_{z,g1} – 1 \giusto) \giusto) \giusto)
+ \min \left( c_{\testo{giusto},g1}, 1.5 h’_{\testo{ef},g1} \giusto)
\)
\(
L_{N_c} = min sinistra( 8 \, \testo{pollici}, 1.5 \volte 5.6667 \, \testo{pollici} \giusto)
+ \sinistra( \min \left( 6 \, \testo{pollici}, 3 \volte 5.6667 \, \testo{pollici} \volte sinistra( 2 – 1 \giusto) \giusto) \giusto)
+ \min \left( 8 \, \testo{pollici}, 1.5 \volte 5.6667 \, \testo{pollici} \giusto)
\)
\(
L_{N_c} = 22 \, \testo{pollici}
\)
\(
B_{N_c} = min sinistra( c_{\testo{superiore},g1}, 1.5 h’_{\testo{ef},g1} \giusto)
+ \sinistra( \min \left( s_{\testo{somma},y,g1}, 3 h’_{\testo{ef},g1} \sinistra( n_{y,g1} – 1 \giusto) \giusto) \giusto)
+ \min \left( c_{\testo{parte inferiore},g1}, 1.5 h’_{\testo{ef},g1} \giusto)
\)
\(
B_{N_c} = min sinistra( 8.5 \, \testo{pollici}, 1.5 \volte 5.6667 \, \testo{pollici} \giusto)
+ \sinistra( \min \left( 5 \, \testo{pollici}, 3 \volte 5.6667 \, \testo{pollici} \volte sinistra( 2 – 1 \giusto) \giusto) \giusto)
+ \min \left( 8.5 \, \testo{pollici}, 1.5 \volte 5.6667 \, \testo{pollici} \giusto)
\)
\(
B_{N_c} = 22 \, \testo{pollici}
\)
Successivamente, we evaluate the basic concrete breakout strength of a single anchor using ACI 318-19 Clausola 17.6.2.2.1
\(
N_b = k_c lambda_a sqrt{\frac{f'_c}{\testo{psi}}} \sinistra( \frac{h’_{\testo{ef},g1}}{\testo{pollici}} \giusto)^{1.5} \, \testo{Trova la distribuzione delle sollecitazioni in una piastra quadrata a causa degli effetti di un foro circolare al centro sotto un carico lineare uniforme nel piano}
\)
\(
N_b = 24 \volte 1 \volte sqrt{\frac{4 \, \testo{KSI}}{0.001 \, \testo{KSI}}} \volte sinistra( \frac{5.6667 \, \testo{pollici}}{1 \, \testo{pollici}} \giusto)^{1.5} \volte 0.001 \, \testo{kip} = 20.475 \, \testo{kip}
\)
Dove,
- \(Eurocodice di design con piastra di base in acciaio{c} = 24\) per ancore gettate
- \(\lambda = 1.0 \) for normal-weight concrete
Adesso, we assess the effects of geometry by calculating the edge effect factor che per il eccentricity factor.
The shortest edge distance of the anchor group is determined as:
\(
c_{un carico,\testo{min}} = min sinistra( c_{\testo{sinistra},g1}, c_{\testo{giusto},g1}, c_{\testo{superiore},g1}, c_{\testo{parte inferiore},g1} \giusto)
= min sinistra( 8 \, \testo{pollici}, 8 \, \testo{pollici}, 8.5 \, \testo{pollici}, 8.5 \, \testo{pollici} \giusto) = 8 \, \testo{pollici}
\)
Secondo ACI 318-19 Clausola 17.6.2.4.1, the breakout edge effect factor è:
\(
\Psi_{ed,N} = min sinistra( 1.0, 0.7 + 0.3 \sinistra( \frac{c_{un carico,\testo{min}}}{1.5 h’_{\testo{ef},g1}} \giusto) \giusto)
= min sinistra( 1, 0.7 + 0.3 \volte sinistra( \frac{8 \, \testo{pollici}}{1.5 \volte 5.6667 \, \testo{pollici}} \giusto) \giusto) = 0.98235
\)
Since the tension load is applied at the centroid of the anchor group, the eccentricity is zero. così, il eccentricity factor, also from Clause 17.6.2.4.1, è:
\(
\Psi_{ec,N} = min sinistra( 1.0, \frac{1}{1 + \frac{2 e’_N}{3 h’_{\testo{ef},g1}}} \giusto)
= min sinistra( 1, \frac{1}{1 + \frac{2 \volte 0}{3 \volte 5.6667 \, \testo{pollici}}} \giusto) = 1
\)
Inoltre, both the cracking factor che per il splitting factor are taken as:
\(
\Psi_{c,N} = 1
\)
\(
\Psi_{cp,N} = 1
\)
Poi, we combine all these factors and use ACI 318-19 Eq. 17.6.2.1b to evaluate the concrete breakout strength of the anchor group:
\(
\phi N_{cbg} = \phi \left( \frac{UN_{N_c}}{UN_{N_{co}}} \giusto) \Psi_{ec,N} \Psi_{ed,N} \Psi_{c,N} \Psi_{cp,N} N_b
\)
\(
\phi N_{cbg} = 0.7 \volte sinistra( \frac{484 \, \testo{pollici}^ 2}{289 \, \testo{pollici}^ 2} \giusto) \volte 1 \volte 0.98235 \volte 1 \volte 1 \volte 20.475 \, \testo{kip} = 23.58 \, \testo{kip}
\)
La total applied tension load on the anchor group is the product of the individual anchor load and the number of anchors:
\(
N_{ua} = sinistra( \frac{N_x}{n_{un carico,t}} \giusto) n_{un carico,g1} = sinistra( \frac{20 \, \testo{kip}}{4} \giusto) \volte 4 = 20 \, \testo{kip}
\)
Da 20 kips < 23.58 kips, the concrete breakout capacity is sufficiente.
Dai un'occhiata #5: Calculate anchor pullout capacity
The pullout capacity of an anchor is governed by the resistance at its embedded end. Iniziare, we calculate the bearing area of the embedded plate, which is the net area after subtracting the area occupied by the anchor rod.
For a rectangular embedded plate, il bearing area is calculated as:
\(
UN_{brg} = sinistra( \sinistra( b_{embed\_plate} \giusto)^2 \right) – UN_{asta} = sinistra( \sinistra( 3 \, \testo{pollici} \giusto)^2 \right) – 0.44179 \, \testo{pollici}^2 = 8.5582 \, \testo{pollici}^ 2
\)
Dove,
\(
UN_{asta} = frac{\pi}{4} \sinistra( d_a \right)^2 = \frac{\pi}{4} \volte sinistra( 0.75 \, \testo{pollici} \giusto)^2 = 0.44179 \, \testo{pollici}^ 2
\)
Successivamente, Determiniamo il basic anchor pullout strength usando ACI 318-19 Equation 17.6.3.2.2a.
\(
N_b = 8 UN_{brg} \sinistra( f’_c \right) = 8 \volte 8.5582 \, \testo{pollici}^2 \times \left( 4 \, \testo{KSI} \giusto) = 273.86 \, \testo{kip}
\)
We then apply the appropriate resistance factor and pullout cracking factor:
- Per incrinato calcestruzzo, \(\Psi_{cp} = 1.0\)
- Per uncracked calcestruzzo, \(\Psi_{cp} = 1.4\)
Using these, calcoliamo il design anchor pullout strength in tension per ACI 318-19 Equazione 17.6.3.1.
\(
\phi N_{pn} = \phi \Psi_{c,p} N_b = 0.7 \volte 1 \volte 273.86 \, \testo{kip} = 191.7 \, \testo{kip}
\)
Recall the previously calculated tension load per anchor:
\(
N_{ua} = frac{N_x}{n_{un carico,t}} = frac{20 \, \testo{kip}}{4} = 5 \, \testo{kip}
\)
Da 5 kips < 191.7 kips, the anchor pullout capacity is sufficiente.
Dai un'occhiata #6: Calculate embed plate flexural capacity
This is a supplementary check performed using the Software di progettazione della piastra di base Skyciv to verify that the embedded plate has sufficient flexural capacity and will not yield under the applied pullout loads.
Primo, we determine the length of the free (unsupported) end of the embedded plate, measured from the edge of the support to the face of the rod.
\(
b’ = frac{b_{embed\_plate} – d_a}{2} = frac{3 \, \testo{pollici} – 0.75 \, \testo{pollici}}{2} = 1.125 \, \testo{pollici}
\)
Successivamente, calcoliamo il momento flettente induced by the uniform bearing pressure. This pressure represents the force transferred from the anchor pullout action onto the embedded plate.
\(
m_f = \frac{\sinistra( \frac{T_a}{UN_{brg}} \giusto) \sinistra( b’ \giusto)^ 2}{2} = frac{\sinistra( \frac{5 \, \testo{kip}}{8.5582 \, \testo{pollici}^ 2} \giusto) \volte sinistra( 1.125 \, \testo{pollici} \giusto)^ 2}{2} = 0.36971 \, \testo{kip}
\)
Infine, using the calculated moment and given material properties, we will determine the minimum required plate thickness to resist flexural yielding.
\(
t_{min} = sqrt{\frac{4 m_f}{\phi F_{y\_ep}}} = sqrt{\frac{4 \volte 0.36971 \, \testo{kip}}{0.9 \volte 36 \, \testo{KSI}}} = 0.21364 \, \testo{pollici}
\)
Recall actual embedded plate thickness:
\(
t_{actual} = t_{embed\_plate} = 0.25 \, \testo{pollici}
\)
Da 0.21364 pollici < 0.25 pollici, the embedded plate flexural capacity is sufficiente.
Dai un'occhiata #7: Calculate side-face blowout capacity in Y-direction
This calculation is not applicable for this example, as the conditions specified in ACI 318-19 Clausola 17.6.4 are not met. Pertanto, side-face blowout failure along the Y-direction will not occur.
Dai un'occhiata #8: Calculate side-face blowout capacity in Z-direction
This calculation is not applicable for this example, as the conditions specified in ACI 318-19 Clausola 17.6.4 are not met. Pertanto, side-face blowout failure along the Z-direction will not occur.
Riepilogo del progetto
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