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 Vy=2-kip e Vz=2-kip carichi di taglio.
Dati dati:
Colonna:
Sezione colonna: HSS7X4X5/16
Area colonna: 7.59 pollici2
Materiale colonna: A36
Piastra di base:
Dimensioni della piastra di base: 12 in x 14 pollici
Spessore della piastra di base: 3/4 pollici
Materiale della piastra di base: A36
Malta:
Grout Thickness: 0.25 pollici
Calcestruzzo:
Dimensioni concrete: 12 in x 14 pollici
Spessore di cemento: 10 pollici
Materiale di cemento: 3000 psi
Crackato o non collocato: Rotto
Ancore:
Diametro dell'ancora: 1/2 pollici
Efficace lunghezza dell'incorporamento: 8 pollici
Plate washer thickness: 0.25 pollici
Plate washer connection: Welded to base plate
saldature:
Dimensione della saldatura: 1/4 pollici
Classificazione del metallo di riempimento: E70XX
Dati di ancoraggio (a partire dal Calcolatore Skyciv):
Definizioni:
Percorso di carico:
The design follows the recommendations of Guida alla progettazione AISC 1, 3Rd Edition, e ACI 318-19. 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. In questo esempio non sono considerati attrito e alette di taglio, Poiché questi meccanismi non sono supportati nell'attuale software.
Per impostazione predefinita, the applied shear load is distributed equally among all anchors, with each anchor transferring its portion of the load to the concrete support. Come alternativa, the software allows a simplified and more conservative assumption, where the entire shear load is assigned only to the anchors nearest the loaded edge. In questo caso, the shear capacity check is performed on these edge anchors alone, ensuring that potential shear failure is conservatively addressed.
Gruppi di ancoraggio:
La Software di progettazione della piastra di base Skyciv Include una caratteristica intuitiva che identifica quali ancore fanno parte di un gruppo di ancoraggio per la valutazione Breakout di taglio in cemento e Sceo di cemento Pryout fallimenti.
Un gruppo di ancoraggio è definito come due o più ancore con aree di resistenza proiettate sovrapposte. In questo caso, Le ancore agiscono insieme, e la loro resistenza combinata viene verificata rispetto al carico applicato sul gruppo.
A Ancoraggio singolo è definito come un ancoraggio la cui area di resistenza proiettata non si sovrappone a nessun altro. In questo caso, L'ancora si comporta da solo, e la forza di taglio applicata su quell'ancora viene controllata direttamente contro la sua resistenza individuale.
Questa distinzione consente al software di acquisire sia il comportamento di gruppo che le prestazioni di ancoraggio individuale quando si valutano le modalità di guasto correlate al taglio.
Calcoli passo-passo:
Dai un'occhiata #1: Calcola la capacità di saldatura
The first step is to calculate the lunghezza totale della saldatura available to resist shear. Since the base plate is welded along the perimeter of the column section, the total weld length is obtained by summing the welds on all sides.
\( L_{saldare} = 2 \sinistra( b_{col} – 2r_{col} – 2t_{col} \giusto) + 2 \sinistra( d_{col} – 2r_{col} – 2t_{col} \giusto) \)
\( L_{saldare} = 2 \volte (4\,\testo{pollici} – 2 \times 0.291\,\text{pollici} – 2 \times 0.291\,\text{pollici}) + 2 \volte (7\,\testo{pollici} – 2 \times 0.291\,\text{pollici} – 2 \times 0.291\,\text{pollici}) = 17.344\,\text{pollici} \)
Using this weld length, the applied shear forces in the y- and z-directions are divided to determine the average shear force per unit length in each direction:
\( v_{uy} = frac{V_y}{L_{saldare}} = frac{2\,\testo{kip}}{17.344\,\testo{pollici}} = 0.11531\,\text{kip/in} \)
\( v_{uz} = frac{V_Z}{L_{saldare}} = frac{2\,\testo{kip}}{17.344\,\testo{pollici}} = 0.11531\,\text{kip/in} \)
La resultant shear demand per unit length is then determined using the square root of the sum of the squares (SRSS) metodo.
\( r_u = \sqrt{(v_{uy})^ 2 + (v_{uz})^ 2} \)
\( r_u = \sqrt{(0.11531\,\testo{kip/in})^ 2 + (0.11531\,\testo{kip/in})^ 2} = 0.16308\,\text{kip/in} \)
Successivamente, the weld capacity is calculated using AISC 360-22 Eq. J2-4, with the directional strength coefficient taken as kds=1.0 for an HSS section. The weld capacity for a 1/4 in weld is determined as:
\( \Phi r_n = phi 0.6 F_{Exx} E_w k_{ds} = 0.75 \volte 0.6 \times 70\,\text{KSI} \times 0.177\,\text{pollici} \volte 1 = 5.5755\,\text{kip/in} \)
It is also necessary to check the base metals, both the column and the base plate, usando AISC 360-22 Eq. J4-4 to obtain the shear rupture strength. This gives:
\( \phi r_{nbm, col} = phi 0.6 F_{u\_col} t_{col} = 0.75 \volte 0.6 \times 58\,\text{KSI} \times 0.291\,\text{pollici} = 7.5951\,\text{kip/in} \)
\( \phi r_{nbm, p.p} = phi 0.6 F_{u\_bp} t_{p.p} = 0.75 \volte 0.6 \times 58\,\text{KSI} \times 0.75\,\text{pollici} = 19.575\,\text{kip/in} \)
\( \phi r_{nbm} = \min\left( \phi r_{nbm, p.p},\, \phi r_{nbm, col} \giusto) = min(19.575\,\testo{kip/in},\, 7.5951\,\testo{kip/in}) = 7.5951\,\text{kip/in} \)
Since the actual weld stress is less than both the weld metal and base metal capacities, 0.16308 KPI < 5.5755 kpi and 0.16308 KPI < 7.5951 KPI, the design weld capacity is sufficiente.
Dai un'occhiata #2: Calcola la capacità di breakout del calcestruzzo a causa di VY Shear
Perpendicular Edge Capacity:
From the layout, Ancore 1 e 4 are closest to the edge and have the shortest ca1 distance. Using these ca1 values to project the failure cones, the software identified these anchors as ancore singole, since their projected cones do not overlap. The support was also determined to be not a narrow member, so the ca1 distance is used directly without modification.
Let’s recall that the shear force is assumed to be distributed among all the anchors. Il calcolo per il Vy shear load applied to each single anchor is:
\( V_{fa\perp} = frac{V_y}{n_a} = frac{2\,\testo{kip}}{6} = 0.33333\,\text{kip} \)
Consideriamo Ancorare 1. The maximum projected area of a single anchor is calculated using ACI 318-19 Eq. 17.7.2.1.3.
\( UN_{VCo} = 4.5 (c_{a1,s1})^2 = 4.5 \volte (2\,\testo{pollici})^2 = 18\,\text{pollici}^ 2 \)
The actual projected area is then determined from the width and height of the projected failure cone.
\( B_{Vc} = min(c_{sinistra,Sforzo in alto Rinforzo},\, 1.5c_{a1,s1}) + \min(c_{giusto,Sforzo in alto Rinforzo},\, 1.5c_{a1,s1}) \)
\( B_{Vc} = min(10\,\testo{pollici},\, 1.5 \times 2\,\text{pollici}) + \min(2\,\testo{pollici},\, 1.5 \times 2\,\text{pollici}) = 5\,\text{pollici} \)
\( Per calcolarlo{Vc} = min(1.5c_{a1,s1},\, t_{conc}) = min(1.5 \times 2\,\text{pollici},\, 10\,\testo{pollici}) = 3\,\text{pollici} \)
\( UN_{Vc} = B_{Vc} Per calcolarlo{Vc} = 5\,\text{pollici} \times 3\,\text{pollici} = 15\,\text{pollici}^ 2 \)
The next step is to use Equations 17.7.2.2.1a and 17.7.2.2.1b to calculate the basic breakout strength of a single anchor. The governing capacity is taken as the lesser value.
\( V_{b1} = 7 \sinistra( \frac{\min(l_e,\, 8d_a)}{d_a} \giusto)^{0.2} \sqrt{\frac{d_a}{\testo{pollici}}} \lambda_a sqrt{\frac{f'_c}{\testo{psi}}} \sinistra( \frac{c_{a1,s1}}{\testo{pollici}} \giusto)^{1.5} \,\testo{Trova la distribuzione delle sollecitazioni in una piastra quadrata a causa degli effetti di un foro circolare al centro sotto un carico lineare uniforme nel piano} \)
\( V_{b1} = 7 \volte sinistra( \frac{\min(8\,\testo{pollici},\, 8 \times 0.5\,\text{pollici})}{0.5\,\testo{pollici}} \giusto)^{0.2} \volte sqrt{\frac{0.5\,\testo{pollici}}{1\,\testo{pollici}}} \volte 1 \volte sqrt{\frac{3\,\testo{KSI}}{0.001\,\testo{KSI}}} \volte sinistra( \frac{2\,\testo{pollici}}{1\,\testo{pollici}} \giusto)^{1.5} \times 0.001\,\text{kip} \)
\( V_{b1} = 1.1623\,\text{kip} \)
\( V_{b2} = 9 \lambda_a sqrt{\frac{f'_c}{\testo{psi}}} \sinistra( \frac{c_{a1,s1}}{\testo{pollici}} \giusto)^{1.5} \,\testo{Trova la distribuzione delle sollecitazioni in una piastra quadrata a causa degli effetti di un foro circolare al centro sotto un carico lineare uniforme nel piano} \)
\( V_{b2} = 9 \volte 1 \volte sqrt{\frac{3\,\testo{KSI}}{0.001\,\testo{KSI}}} \volte sinistra( \frac{2\,\testo{pollici}}{1\,\testo{pollici}} \giusto)^{1.5} \times 0.001\,\text{kip} = 1.3943\,\text{kip} \)
\( V_b = \min(V_{b1},\, V_{b2}) = min(1.1623\,\testo{kip},\, 1.3943\,\testo{kip}) = 1.1623\,\text{kip} \)
Successivamente, il breakout capacity parameters are determined. La breakout edge effect factor is calculated according to ACI 318-19 Clausola 17.7.2.4, che per il thickness factor is calculated according to Clausola 17.7.2.6.1.
\( \Psi_{ed,V } = \min\left(1.0,\, 0.7 + 0.3 \sinistra( \frac{c_{a2,s1}}{1.5c_{a1,s1}} \giusto) \giusto) = \min\left(1,\, 0.7 + 0.3 \volte sinistra( \frac{2\,\testo{pollici}}{1.5 \times 2\,\text{pollici}} \giusto) \giusto) = 0.9 \)
\( \Psi_{h,V } = \max\left( \sqrt{ \frac{1.5c_{a1,s1}}{t_{conc}} },\, 1.0 \giusto) = \max\left( \sqrt{ \frac{1.5 \times 2\,\text{pollici}}{10\,\testo{pollici}} },\, 1 \giusto) = 1 \)
Infine, ACI 318-19 Clausola 17.7.2.1(un carico) is used to determine the concrete breakout capacity of a single anchor in shear. The calculated capacity for Vy shear in the perpendicular direction is 0.69 kips.
\( \phi V_{cb\perp} = phi a sinistra( \frac{UN_{Vc}}{UN_{VCo}} \giusto) \Psi_{ed,V } \Psi_{c,V } \Psi_{h,V } V_b \)
\( \phi V_{cb\perp} = 0.65 \volte sinistra( \frac{15\,\testo{pollici}^ 2}{18\,\testo{pollici}^ 2} \giusto) \volte 0.86 \volte 1 \volte 1 \times 1.1623\,\text{kip} = 0.56661\,\text{kip} \)
The calculated capacity for Vy shear nel perpendicular direction is 0.56 kips.
Parallel Edge Capacity:
Failure along the edge parallel to the load is also possible in this scenario, so the concrete breakout capacity for the parallel edge must be determined. The anchors or anchor group considered are those aligned with the parallel edge. conseguentemente, il ca1 edge distance is measured from the anchor to the edge along the Z-direction. Based on the figure below, the failure cone projections overlap; perciò, the anchors are treated as a group.
Astuccio 1:
Astuccio 2:
We refer to ACI 318-19 Figura. R17.7.2.1b for the different cases used when evaluating anchor groups. In this base plate design, welded plate washers are specifically used. Pertanto, solo Astuccio 2 is checked.
The required load for the anchor group in Case 2 is taken as the carico di taglio totale.
\( V_{fa\parallel,case2} = V_y = 2\,\text{kip} \)
In calculating the capacity for the Case 2 failure, the anchors considered are the rear anchors. Di conseguenza, the ca1 edge distance is measured from the rear anchor group to the failure edge.
With this ca1 distance and edge orientation, it must be verified whether the support qualifies as a narrow member. Following ACI 318-19 Clausola 17.7.2.1.2, the SkyCiv Base Plate software identified the support as narrow. Pertanto, il modified ca1 distance si usa, which is calculated to be 6.667 pollici.
The same steps as in the perpendicular case are followed: calculating the projected failure areas, il basic single-anchor breakout strength, che per il breakout parameters. The calculated values for each step are shown below.
\( UN_{VCo} = 4.5 (c_{‘a1,g2})^2 = 4.5 \volte (6.6667\,\testo{pollici})^2 = 200\,\text{pollici}^ 2 \)
\( UN_{Vc} = B_{Vc} Per calcolarlo{Vc} = 14\,\text{pollici} \times 10\,\text{pollici} = 140\,\text{pollici}^ 2 \)
\( V_{b1} = 7.0733\,\text{kip} \)
\( V_{b2} = 8.4853\,\text{kip} \)
\( V_b = \min(V_{b1},\, V_{b2}) = min(7.0733\,\testo{kip},\, 8.4853\,\testo{kip}) = 7.0733\,\text{kip} \)
\( \Psi_{ed,V } = 1.0 \)
\( \Psi_{h,V } = 1.0 \)
The equation for the parallel edge capacity differs from the perpendicular edge capacity. ACI 318-19 Clausola 17.7.2.1(c) is applied, where the breakout equation is multiplied by 2.
\( \phi V_{cbg\parallel} = 2 \phi \left( \frac{UN_{Vc}}{UN_{VCo}} \giusto) \Psi_{ed,V } \Psi_{c,V } \Psi_{h,V } V_b \)
\( \phi V_{cbg\parallel} = 2 \volte 0.65 \volte sinistra( \frac{140\,\testo{pollici}^ 2}{200\testo{pollici}^ 2} \giusto) \volte 1 \volte 1 \volte 1 \times 7.0733\,\text{kip} = 6.4367\,\text{kip} \)
The calculated capacity for Vy shear nel parallelo direction is 6.43 kips.
We now assess the perpendicular and parallel failures separately.
- For the perpendicular edge failure, da 0.33 kip < 0.56 kip, the design concrete shear breakout capacity is sufficiente.
- For the parallel edge failure, da 2 kip < 6.43 kip, the design concrete shear breakout capacity is sufficiente.
Dai un'occhiata #3: Calcola la capacità di breakout del calcestruzzo a causa del taglio VZ
The base plate is also subjected to Vz shear, so the failure edges perpendicular and parallel to the Vz shear must be checked. Using the same approach, the perpendicular and parallel capacities are calculated as 2.45 kips e 1.26 kips, rispettivamente.
Perpendicular Edge:
Parallel Edge:
These capacities are then compared to the required strengths.
- For the perpendicular edge failure, da 2 kip < 2.45 kip, the concrete shear breakout capacity is sufficiente.
- For the parallel edge failure, da 0.33 kip < 1.26 kip, the concrete shear breakout capacity is sufficiente.
Dai un'occhiata #4: Calcola la capacità del cemento.
La concrete cone for pryout failure is the same cone used in the tensile breakout check. To calculate the shear pryout capacity, the nominal tensile breakout strength of the single anchors or anchor group must first be determined. The detailed calculations for the tensile breakout check are already covered in the SkyCiv Design Examples for Tension Load.
It is important to note that the anchor group determination for shear pryout is different from that for shear breakout. Pertanto, the anchors in the design must still be checked to determine whether they act avere un gruppo or as ancore singole against the shear pryout failure. The classification of the support as a narrow section must also be verified and should follow the same conditions used for tension breakout.
From the SkyCiv calculations, il nominal tensile breakout strength del gruppo di ancoraggio è 12.772 kips. With a pryout factor of kcp=2, the design pryout capacity is:
\( \phi V_{cpp} = \phi k_{cp} N_{cbg} = 0.65 \volte 2 \volte 12.772 \,\testo{kip} = 16.604\,\text{kip} \)
The required strength is the resultant of the applied shear loads. Since all anchors belong to a single group, the total resultant shear is assigned to the group.
\( V_{Fare} = sqrt{(V_y)^ 2 + (V_Z)^ 2} = sqrt{(2\,\testo{kip})^ 2 + (2\,\testo{kip})^ 2} = 2.8284\,\text{kip} \)
\( V_{Fare} = sinistra( \frac{V_{Fare}}{n_a} \giusto) N_{un carico,G1} = sinistra( \frac{2.8284\,\testo{kip}}{6} \giusto) \volte 6 = 2.8284\,\text{kip} \)
Since the total shear load is less than anchor group capacity, 2.82 kips < 18.976 kips, the design pryout capacity is sufficiente.
Dai un'occhiata #5: Calcola la capacità di taglio dell'asta di ancoraggio
Recall that in this design example, shear is distributed to all anchors. The total shear load per anchor is therefore the resultant of its share of the Vy load and its share of the Vz load.
\( v_{Fare,y} = frac{V_y}{n_a} = frac{2\,\testo{kip}}{6} = 0.33333\,\text{kip} \)
\( v_{Fare,z} = frac{V_Z}{n_a} = frac{2\,\testo{kip}}{6} = 0.33333\,\text{kip} \)
\( V_{Fare} = sqrt{(v_{Fare,y})^ 2 + (v_{Fare,z})^ 2} \)
\( V_{Fare} = sqrt{(0.33333\,\testo{kip})^ 2 + (0.33333\,\testo{kip})^ 2} = 0.4714\,\text{kip} \)
Questo dà il shear stress on the anchor rod come:
\( f_v = \frac{V_{Fare}}{UN_{asta}} = frac{0.4714\,\testo{kip}}{0.19635\,\testo{pollici}^ 2} = 2.4008\,\text{KSI} \)
Because a plate washer is present, un eccentric shear load is induced in the anchor rod. The eccentricity is taken as half of the distance measured from the top of the concrete support to the center of the plate washer, accounting for the thickness of the base plate. Fare riferimento a Guida alla progettazione AISC 1, 3rd Edition Section 4.3.3.
\( e = 0.5 \sinistra( \frac{t_{pw}}{2} + t_{p.p} \giusto) = 0.5 \volte sinistra( \frac{0.25\,\testo{pollici}}{2} + 0.75\,\testo{pollici} \giusto) = 0.4375\,\text{pollici} \)
The moment from the eccentric shear is then expressed as an axial stress in the anchor rod. Using the section modulus, the axial stress due to this moment is calculated as:
\( Z_{asta} = frac{\pi}{32} (d_a)^3 = \frac{\pi}{32} \volte (0.5\,\testo{pollici})^3 = 0.012272\,\text{pollici}^ 3 \)
\( f_t = \frac{V_{Fare} e}{Z_{asta}} = frac{0.4714\,\testo{kip} \times 0.4375\,\text{pollici}}{0.012272\,\testo{pollici}^ 3} = 16.806\,\text{KSI} \)
ACI Anchor Rod Shear Capacity:
Following ACI 318-19 Clausola 17.7.1, the design strength is then determined. A 0.8 reduction factor is applied due to the presence of grout pads. The design capacity is therefore:
\( \phi V_{per,Qui} = 0.8 \phi 0.6 UN_{lo so,v} f_{uta} = 0.8 \volte 0.65 \volte 0.6 \times 0.1419\text{pollici}^2 \times 90\text{KSI} = 3.9845\text{kip} \)
Come alternativa, il SkyCiv Base Plate software allows the 0.8 simplification to be disabled, and use the actual grout pad thickness in the calculations. In questo caso, the total eccentricity includes the grout pad, and the combined shear and axial strength is determined in accordance with AISC provisions.
AISC Anchor Rod Shear Capacity:
Primo, il nominal shear and tensile stresses are determined for an A325 rod.
\( F_{nv} = 0.45 F_{u,anc} = 0.45 \volte 120\ \testo{KSI} = 54\ \testo{KSI} \)
\( F_{nt} = 0.75 F_{u,anc} = 0.75 \volte 120\ \testo{KSI} = 90\ \testo{KSI} \)
The AISC method uses AISC 360-22 Eq. J3-3a, which may be expressed to include the effects of axial stress. This is carried out as follows.
\( F’_{nv} = min sinistra( 1.3 F_{nv} – \sinistra( \frac{F_{nv}}{\Phi f_{nt}} \giusto) f_t,\; F_{nv} \giusto) \)
\( F’_{nv} = min sinistra( 1.3 \volte 54\ \testo{KSI} – \sinistra( \frac{54\ \testo{KSI}}{0.75 \volte 90\ \testo{KSI}} \giusto) \volte 16.806\ \testo{KSI},\; 54\ \testo{KSI} \giusto) = 54\ \testo{KSI} \)
The design shear capacity from the AISC method is then calculated as:
\( \Phi R_{n,\mathrm{aisc}} = \phi F’_{nv} UN_{asta} = 0.75 \volte 54\ \testo{KSI} \volte 0.19635\ \testo{pollici}^2 = 7.9522\)
To ensure both methods are covered, the governing capacity is taken as the lesser of the two, che è 3.98 kip.
\( \phi V_n = \min \left( \phi V_{per,Qui},\; \Phi R_{n,\mathrm{aisc}} \giusto) = min (3.9845\ \testo{kip},\; 7.9522\ \testo{kip}) = 3.9845\ \testo{kip} \)
Since the shear load per anchor rod is less than the governing anchor rod capacity in shear, 0.47 kip < 3.98 kip, the design anchor rod shear capacity is sufficiente.
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