Esempio di design della piastra di base utilizzando AISC 360-22 e ACI 318-19
Dichiarazione del problema
Determinare se la connessione a piastra da colonna a base progettata è sufficiente per a 5 kip-ft strong-axis moment.
Dati dati
Colonna:
Sezione colonna: W8x31
Area colonna: 9.130 pollici2
Materiale colonna: A992
Piastra di base:
Dimensioni della piastra di base: 15 in x 15 pollici
Spessore della piastra di base: 1 pollici
Materiale della piastra di base: A36
Calcestruzzo:
Dimensioni concrete: 18 in x 18 pollici
Spessore di cemento: 15 pollici
Materiale di cemento: 3000 psi
saldature:
Dimensione della saldatura: 1/4 pollici
Classificazione del metallo di riempimento: E70XX
Carico di compressione trasferito solo attraverso saldature? NO
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Determine if Low Moment or Large Moment
The calculations for this base plate example, which is subjected to a strong-axis moment, will be based on Guida alla progettazione AISC 1, 3Rd Edition. According to the design guide, base plates subjected to moment loads only are classified as the Large Moment Case. This is because no axial load is present to counteract the eccentric load acting on the connection.
For other load combinations, such as strong-axis bending with axial compression or strong-axis bending with axial tension, determining if the case is Large Moment or Low Moment depends on the actual dimensione della colonna, base plate size, concrete size, and the given carichi.
In the free version of SkyCiv, the moment case is displayed as you enter the load values. You can access the free tool here to determine the moment case yourself. Straight from the browser, nessuna installazione richiesta.
To learn more about Low Moment and Large Moment case, refer to this SkyCiv article: AISC DG1: Piastre di base con momento grande e momento basso
Calcoli passo-passo
Dai un'occhiata #1: Weld capacity
We start the design by checking the weld capacity of the connection. We use the Blodgett equations from Progettazione di strutture saldate di Omer W. Blodgett to determine the weld section modulus of the weld. We take the weld shape shown below as a simplification.
\(S_w = 2d_{col}b_f + \dfrac{(d_{col})^2}{3} = 2 \times 8 \text{ in} \times 8 \text{ in} + \dfrac{(8 \text{ in})^2}{3} = 149.33 \text{ in}^2\)
Successivamente, we divide the applied bending moment by the weld section modulus to determine the weld stress per unit length. Note that in calculating the weld section modulus, we assume the critical weld stresses occur at the outermost fiber. For a weld in an I-shaped section, the critical locations are at the outer flanges. We get the weld stress as:
\(r_{uw} = \dfrac{M_z}{S_w} = \dfrac{5 \text{ kip·ft}}{149.33 \text{ in}^2} = 0.40179 \text{ kip/in}\)
Poi, calcoliamo il capacity of the weld. Per un 0.25 in weld size, the design weld throat is 0.177 pollici, which is obtained by multiplying the weld size by \(\dfrac{\sqrt{2}}{2}\). Usando AISC 360-22 Eq. J2-4, the weld capacity per unit length is:
\(\phi r_n = \phi 0.6F_{exx}E_w k_{ds} = 0.75 \times 0.6 \times 70 \text{ ksi} \times 0.177 \text{ in} \times 1 = 5.5755 \text{ kip/in}\)
Da 0.40179 KPI < 5.5755 KPI , the column weld capacity is sufficiente.
This weld check investigates the weld stress at the tension flange. Since the geometry is symmetric and the only load is a moment acting at the center, we can assume the stress at the compression flange is the same as the stress at the tension flange. Pertanto, the welds at the compression flange also have enough capacity.
In questo esempio di progettazione, però, the design inputs specify that compression is not transferred through the welds, but through cuscinetto. così, in the next check, we evaluate the bearing capacity of the column.
Dai un'occhiata #2: Calculate column bearing capacity
After evaluating the weld capacity, we make sure the column has enough capacity to bear on the base plate. This is necessary because we assume the welds do not carry the compression load.
Primo, Determiniamo il compressive stress at the critical section, which is still the flange. We then add any contribution from axial compression. In questo caso, there is none.
\(r_u = \dfrac{N_x}{A_{col}} + \dfrac{M_z 0.5d_{col}}{I_{zz,col}} = \dfrac{0 \text{ kip}}{9.13 \text{ in}^2} + \dfrac{5 \text{ kip·ft} \times 0.5 \times 8 \text{ in}}{110 \text{ in}^4} = 2.1818 \text{ ksi}\)
Poi, noi usiamo AISC 360-22 Section J7 Eq. J7-1 per determinare il capacità portante:
\(\phi r_n = \phi 1.8F_{y\_col} = 0.75 \times 1.8 \times 50 \text{ ksi} = 67.5 \text{ ksi}\)
Da 2.1818 KSI < 67.5 KSI, La capacità del cuscinetto della colonna è sufficiente.
Generalmente, this check is not governing.
Dai un'occhiata #3: Calcolare la capacità portante del calcestruzzo
To calculate the concrete bearing capacity, we need two areas. See the figure below as a guide.
- A1 – Base plate loaded area
- A2 – Area of the concrete support with the same shape as A1
A1 è il loaded area of the base plate, while A2 è il same loaded area projected at a 2H:1V slope into the concrete support. More details on these areas can be found in ACI 318-19 Sezione 22.8.3.2.
In questo base plate design with moment, the A1 area will be calculated using the full width of the base plate, \(B\) che per il actual bearing length, \(Y\). The actual bearing length is measured from the edge of the base plate at the compression side towards the center of the plate.
\( A_1 = B \times Y \)
For most engineers, calculating A1 and A2 requires a lot of work, so they simply assume A1 = A2, che è conservatore. Nel Software della piastra base SkyCiv, iterazioni are performed to determine the actual A1 values and to maximize the capacity of the base plate.
The table below shows the iteration steps from SkyCiv to determine the bearing length Y, e successivamente, to determine the actual A1 and A2 areas.
Primo, an initial Y is assumed, and the A2 and A1 values are calculated based on the initial Y. These are then used to calculate the maximum bearing strength of the concrete, \(f_{p\_max}\), e il corrispondente bearing force per unit length, \(q_{max}\). Once these are determined, the corresponding Y value is determined based on Eq. 4-32 of AISC DG1 3rd Edizione. If the calculated Y does not match the initial Y, another iteration step is performed. A more detailed step-by-step calculation of the last iteration is shown in the next section.
Adesso, taking the A2 and A1 values from the iteration, noi usiamo Eq. 4-2 of AISC Design Guide 1, 3Rd Edition, to determine \(f_{p\_max}\), the design bearing strength for concrete crushing:
\(f_{p\_max} = \phi \left( \min \left( 0.85 \, (f_c’) \sqrt{\dfrac{A_2}{A_1}}, \, 1.7 \, (f_c’) \right) \right)\)
\(f_{p\_max} = 0.65 \times \left( \min \left( 0.85 \times (3 \text{ ksi}) \times \sqrt{39.941}, \, 1.7 \times (3 \text{ ksi}) \right) \right) = 3.315 \text{ ksi}\)
Inoltre, since this design is a Large Moment Case, the base plate has both a compression side and a tension side. In questa situazione, we should allow the concrete bearing to be at its full capacity and minimize the Y value. Quindi, any remaining tensile force, based on equilibrium of forces, will be taken by the anchors at the tension side.
Given this, we set \(f_{p\_actual}\) uguale a \(f_{p\_max}\).
\(f_{p\_actual} = f_{p\_max} = 3.315 \text{ ksi}\)
Da 3.315 ksi ≤ 3.315 KSI, La capacità del cuscinetto in cemento è sufficiente.
Dai un'occhiata #4: Calculate base plate yielding capacity at the bearing interface
The next step is to determine if the base plate has enough capacity for the given load. As the moment load pushes the base plate against the concrete, the base plate has a tendency to yield due to flexure. We call this check the base plate yielding check on the bearing interface. There is also a separate yielding due to the tension load, and it will be covered in the next design check.
To check the base plate yielding, we need to determine the \(q_{max}\), which is the compression force per unit length. AISC DG1 3rd Ed. Eq. 4-37 dà \(q_{max}\)come:
\(q_{max} = f_{p\_max}B_{bp} = 3.315 \text{ ksi} \times 15 \text{ in} = 49.725 \text{ kip/in}\)
Poi, we will calculate the Y-dimension. Although the Y-dimension is already calculated in the prior check, this section will show how we use the equations from AISC DG1. For this Large Moment Case, noi usiamo Eq. 4-32 of AISC DG1, 3Rd Edition.
For the plus case:
\(Y_{case1} = \left( f + \dfrac{L_{bp}}{2} \right) + \sqrt{\left( \left( f + \dfrac{L_{bp}}{2} \right)^2 \right) – \dfrac{2M_z}{q_{max}}}\)
\(Y_{case1} = \left( 5.5315 \text{ in} + \dfrac{15 \text{ in}}{2} \right) + \sqrt{\left( \left( 5.5315 \text{ in} + \dfrac{15 \text{ in}}{2} \right)^2 \right) – \dfrac{2 \times 5 \text{ kip·ft}}{49.725 \text{ kip/in}}}\)
\(Y_{case1} = 25.97 \text{ in}\)
For the minus case:
\(Y_{case2} = \left( f + \dfrac{L_{bp}}{2} \right) – \sqrt{\left( \left( f + \dfrac{L_{bp}}{2} \right)^2 \right) – \dfrac{2M_z}{q_{max}}}\)
\(Y_{case2} = \left( 5.5315 \text{ in} + \dfrac{15 \text{ in}}{2} \right) – \sqrt{\left( \left( 5.5315 \text{ in} + \dfrac{15 \text{ in}}{2} \right)^2 \right) – \dfrac{2 \times 5 \text{ kip·ft}}{49.725 \text{ kip/in}}}\)
\(Y_{case2} = 0.092925 \text{ in}\)
Note that this equation has + e – simboli, so we get two values. Da qui, we need to determine which value makes more sense. Da 25.97 pollici is greater than the base plate length, 15 pollici, we will not consider this value. Quindi, noi useremo 0.092925 pollici as the lunghezza del cuscinetto.
\(Y = Y_{case2} = 0.092925 \text{ in}\)
Successivamente, we calculate the cantilever lengths. Queste sono le m and n dimensions from the figure below. Questo si basa su AISC DG1 3rd Edition Section 4.3.1.
The calculated m and n cantilever values are:
\(m = \dfrac{L_{bp} – 0.95d_{col}}{2} = \dfrac{15 \text{ in} – 0.95 \times 8 \text{ in}}{2} = 3.7 \text{ in}\)
\(n = \dfrac{B_{bp} – 0.8b_f}{2} = \dfrac{15 \text{ in} – 0.8 \times 8 \text{ in}}{2} = 4.3 \text{ in}\)
Poi, noi compare the calculated Y value to the larger of m and n. Since Y is less than the larger cantilever length, noi useremo AISC DG1 3rd Ed. Eq. 4-52un carico. This gives the required base plate thickness to be:
\(t_{p(req)} = \sqrt{\dfrac{4f_{p\_actual}Y \left( n – \dfrac{Y}{2} \right)}{\phi F_{y\_bp}}}\)
\(t_{p(req)} = \sqrt{\dfrac{4 \times 3.315 \text{ ksi} \times 0.092925 \text{ in} \times \left( 4.3 \text{ in} – \dfrac{0.092925 \text{ in}}{2} \right)}{0.9 \times 36 \text{ ksi}}} = 0.4022 \text{ in}\)
Infine, let’s compare it to the actual base plate thickness.
\(t_{p(actual)} = t_{bp} = 1 \text{ in}\)
Da 0.4022 pollici < 1 pollici, the yielding capacity of the base plate is sufficiente.
Dai un'occhiata #5: Calculate base plate yielding capacity at the tension interface
The required tensile force on the anchors at the tension side is equal to the resulting compression force on the compression side. This is because the base plate is only loaded with moment loads, and so the resulting couple forces are equal in magnitude.
To calculate the required strength of anchors, noi usiamo AISC DG1 3rd Ed. Eq. 4-31.
\(T_u = q_{max}Y = 49.725 \text{ kip/in} \times 0.092925 \text{ in} = 4.6207 \text{ kip}\)
Poi, calcoliamo il eccentricity of this tensile force from the column support. This is taken from the center of the anchor rod to the center of the column flange. Usiamo Eq. 4-61 of the design guide to determine the eccentricity. In the equation, f is the distance of the anchor rod from the column centerline.
\(x = \left( f – \dfrac{d_{col}}{2} \right) + \dfrac{t_f}{2} = \left( 5.5315 \text{ in} – \dfrac{8 \text{ in}}{2} \right) + \dfrac{0.435 \text{ in}}{2} = 1.749 \text{ in}\)
Note that in this analysis, we take the whole base plate width as the section to resist the flexure, as we closely follow the AISC DG1 3rd Ed. guidelines. Other conservative bending widths could also be used.
Infine, we determine the required base plate thickness usando Eq. 4-62un carico.
\(t_{p(req)} = \sqrt{\dfrac{4T_u x}{B_{bp}\phi F_{y\_bp}}} = \sqrt{\dfrac{4 \times 4.6207 \text{ kip} \times 1.749 \text{ in}}{15 \text{ in} \times 0.9 \times 36 \text{ ksi}}} = 0.25791 \text{ in}\)
Da 0.25791 pollici < 1 pollici, il yielding capacity of the base plate is sufficiente.
Dai un'occhiata #6: Controlli di ancoraggio
The rest of the anchor design checks are not shown in this example. Please refer to other SkyCiv Base Plate Design Examples that covers capacity of ancoraggio.
Riepilogo del progetto
Il software di progettazione della piastra di base Skyciv può generare automaticamente un rapporto di calcolo passo-passo per questo esempio di progettazione. Fornisce inoltre un riepilogo dei controlli eseguiti e dei loro rapporti risultanti, rendere le informazioni facili da capire a colpo d'occhio. Di seguito è riportata una tabella di riepilogo del campione, che è incluso nel rapporto.
Acquista software di base
Acquista da solo la versione completa del modulo di progettazione della piastra di base senza altri moduli SkyCiv. Questo ti dà un set completo di risultati per la progettazione della piastra di base, tra cui report dettagliati e più funzionalità.
