AISC Kullanarak Base Plaka Tasarım Örneği 360-22 ve ACI 318-19
Sorun Bildirimi
Tasarlanan sütun-tabaka plaka bağlantısının bir 5 kip-ft strong-axis moment.
Verilen Veriler
Sütun:
Sütun bölümü: W8x31
Sütun alanı: 9.130 içinde2
Sütun malzemesi: A992
Taban plakası:
Taban plaka boyutları: 15 x içinde 15 içinde
Taban plakası kalınlığı: 1 içinde
Taban plaka malzemesi: A36
Somut:
Somut boyutlar: 18 x içinde 18 içinde
Beton kalınlığı: 15 içinde
Beton malzeme: 3000 psi
Kaynaklar:
kaynak boyutu: 1/4 içinde
Dolgu Metal Sınıflandırması: E70XX
Sıkıştırma yükü yalnızca kaynaklar yoluyla aktarılır? HAYIR
SkyCiv Ücretsiz Aracındaki Model
Ücretsiz çevrimiçi aracımızı kullanarak yukarıdaki taban plakası tasarımını bugün modelleyin! Kayıt olmanıza gerek yok.
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 AISC Tasarım Kılavuzu 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 sütun boyutu, base plate size, concrete size, and the given yükler.
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, kurulum gerektirmez.
To learn more about Low Moment and Large Moment case, refer to this SkyCiv article: AISC DG1: Büyük Momentli ve Düşük Momentli Taban Plakaları
Adım adım hesaplamalar
Soğuk şekillendirilmiş elemanlar aşağıdakilere uygun olarak tasarlanırken #1: Weld capacity
We start the design by checking the weld capacity of the connection. We use the Blodgett equations from Kaynaklı Yapıların Tasarımı kaydeden Ömer W. Blodgett to determine the weld section modulus of the weld. We take the weld shape shown below as a simplification.
\(S_w = 2d_{col}b_f + \dfrac{(d_{col})^2}{3} = 2 \times 8 \text{ in} \times 8 \text{ in} + \dfrac{(8 \text{ in})^2}{3} = 149.33 \text{ in}^2\)
Sonraki, 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:
\(r_{uw} = \dfrac{M_z}{S_w} = \dfrac{5 \text{ kip·ft}}{149.33 \text{ in}^2} = 0.40179 \text{ kip/in}\)
Sonra, hesaplıyoruz capacity of the weld. için 0.25 in weld size, the design weld throat is 0.177 içinde, which is obtained by multiplying the weld size by \(\dfrac{\sqrt{2}}{2}\). kullanma AISC 360-22 Eşitlik. J2-4, the weld capacity per unit length is:
\(\phi r_n = \phi 0.6F_{exx}E_w k_{ds} = 0.75 \times 0.6 \times 70 \text{ ksi} \times 0.177 \text{ in} \times 1 = 5.5755 \text{ kip/in}\)
Dan beri 0.40179 KPI < 5.5755 KPI , the column weld capacity is yeterli.
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. Bu nedenle, the welds at the compression flange also have enough capacity.
Bu tasarım örneğinde, ancak, the design inputs specify that compression is not transferred through the welds, but through rulman. Böylece, in the next check, we evaluate the bearing capacity of the column.
Soğuk şekillendirilmiş elemanlar aşağıdakilere uygun olarak tasarlanırken #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.
İlk, Belirliyoruz compressive stress at the critical section, which is still the flange. We then add any contribution from axial compression. Bu durumda, there is none.
\(r_u = \dfrac{N_x}{A_{col}} + \dfrac{M_z 0.5d_{col}}{I_{zz,col}} = \dfrac{0 \text{ kip}}{9.13 \text{ in}^2} + \dfrac{5 \text{ kip·ft} \times 0.5 \times 8 \text{ in}}{110 \text{ in}^4} = 2.1818 \text{ ksi}\)
Sonra, kullanıyoruz AISC 360-22 Section J7 Eq. J7-1 belirlemek için taşıma kapasitesi:
\(\phi r_n = \phi 1.8F_{y\_col} = 0.75 \times 1.8 \times 50 \text{ ksi} = 67.5 \text{ ksi}\)
Dan beri 2.1818 ksi < 67.5 ksi, Sütun taşıma kapasitesi yeterli.
Genelde, this check is not governing.
Soğuk şekillendirilmiş elemanlar aşağıdakilere uygun olarak tasarlanırken #3: Beton Rulman Kapasitesini Hesaplayın
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 bu loaded area of the base plate, süre A2 bu 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 Bölüm 22.8.3.2.
Bunda base plate design with moment, the A1 area will be calculated using the full width of the base plate, \(B\) ve 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, hangisi muhafazakar. İçinde SkyCiv Taban Plakası yazılımı, yinelemeler 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, ve ardından, to determine the actual A1 and A2 areas.
İlk, 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, \(f_{p\_max}\), ve karşılık gelen bearing force per unit length, \(q_{max}\). Once these are determined, the corresponding Y value is determined based on Eşitlik. 4-32 of AISC DG1 3üçüncü C'yi dikkate alacak. 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.
Şimdi, taking the A2 and A1 values from the iteration, kullanıyoruz Eşitlik. 4-2 of AISC Design Guide 1, 3Rd Edition, to determine \(f_{p\_max}\), the design bearing strength for concrete crushing:
\(f_{p\_max} = \phi \left( \min \left( 0.85 \, (f_c’) \sqrt{\dfrac{A_2}{A_1}}, \, 1.7 \, (f_c’) \right) \right)\)
\(f_{p\_max} = 0.65 \times \left( \min \left( 0.85 \times (3 \text{ ksi}) \times \sqrt{39.941}, \, 1.7 \times (3 \text{ ksi}) \right) \right) = 3.315 \text{ ksi}\)
bunlara ek olarak, since this design is a Large Moment Case, the base plate has both a compression side and a tension side. Bu durumda, we should allow the concrete bearing to be at its full capacity and minimize the Y value. Yani, any remaining tensile force, based on equilibrium of forces, will be taken by the anchors at the tension side.
Given this, we set \(f_{p\_actual}\) eşittir \(f_{p\_max}\).
\(f_{p\_actual} = f_{p\_max} = 3.315 \text{ ksi}\)
Dan beri 3.315 ksi ≤ 3.315 ksi, Beton rulman kapasitesi yeterli.
Soğuk şekillendirilmiş elemanlar aşağıdakilere uygun olarak tasarlanırken #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. Eşitlik. 4-37 verir \(q_{max}\)gibi:
\(q_{max} = f_{p\_max}B_{bp} = 3.315 \text{ ksi} \times 15 \text{ in} = 49.725 \text{ kip/in}\)
Sonra, 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, kullanıyoruz Eşitlik. 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 + ve – semboller, so we get two values. Buradan, we need to determine which value makes more sense. Dan beri 25.97 içinde is greater than the base plate length, 15 içinde, we will not consider this value. Yani, kullanacağız 0.092925 içinde olarak yatak uzunluğu.
\(Y = Y_{case2} = 0.092925 \text{ in}\)
Sonraki, we calculate the cantilever lengths. Bunlar m and n dimensions from the figure below. Bu temeldir AISC DG1 3rd Edition Section 4.3.1.
The calculated m and n cantilever values are:
\(m = \dfrac{L_{bp} – 0.95d_{col}}{2} = \dfrac{15 \text{ in} – 0.95 \times 8 \text{ in}}{2} = 3.7 \text{ in}\)
\(n = \dfrac{B_{bp} – 0.8b_f}{2} = \dfrac{15 \text{ in} – 0.8 \times 8 \text{ in}}{2} = 4.3 \text{ in}\)
Sonra, güneş panelinin maruz kalma kategorisi rüzgara karşı 45 compare the calculated Y value to the larger of m and n. Since Y is less than the larger cantilever length, kullanacağız AISC DG1 3rd Ed. Eşitlik. 4-52a. This gives the required base plate thickness to be:
\(t_{p(req)} = \sqrt{\dfrac{4f_{p\_actual}Y \left( n – \dfrac{Y}{2} \right)}{\phi F_{y\_bp}}}\)
\(t_{p(req)} = \sqrt{\dfrac{4 \times 3.315 \text{ ksi} \times 0.092925 \text{ in} \times \left( 4.3 \text{ in} – \dfrac{0.092925 \text{ in}}{2} \right)}{0.9 \times 36 \text{ ksi}}} = 0.4022 \text{ in}\)
En sonunda, let’s compare it to the actual base plate thickness.
\(t_{p(actual)} = t_{bp} = 1 \text{ in}\)
Dan beri 0.4022 içinde < 1 içinde, the yielding capacity of the base plate is yeterli.
Soğuk şekillendirilmiş elemanlar aşağıdakilere uygun olarak tasarlanırken #5: Calculate base plate yielding capacity at the tension interface
The required tensile force on the anchors at the tension side Viagra Satın Almak İçin Telefon Numarası Viagra Satın Alın 🔹 En İyi Online İlaç 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, kullanıyoruz AISC DG1 3rd Ed. Eşitlik. 4-31.
\(T_u = q_{max}Y = 49.725 \text{ kip/in} \times 0.092925 \text{ in} = 4.6207 \text{ kip}\)
Sonra, hesaplıyoruz 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. Kullanıyoruz Eşitlik. 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.
En sonunda, we determine the required base plate thickness kullanma Eşitlik. 4-62a.
\(t_{p(req)} = \sqrt{\dfrac{4T_u x}{B_{bp}\phi F_{y\_bp}}} = \sqrt{\dfrac{4 \times 4.6207 \text{ kip} \times 1.749 \text{ in}}{15 \text{ in} \times 0.9 \times 36 \text{ ksi}}} = 0.25791 \text{ in}\)
Dan beri 0.25791 içinde < 1 içinde, NS yielding capacity of the base plate is yeterli.
Soğuk şekillendirilmiş elemanlar aşağıdakilere uygun olarak tasarlanırken #6: Çapa Kontrolleri
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 demirleme.
Tasarım Özeti
SkyCiv Base Plaka Tasarım Yazılımı, bu tasarım örneği için otomatik olarak adım adım hesaplama raporu oluşturabilir. Ayrıca gerçekleştirilen kontrollerin ve bunların sonuç oranlarının bir özetini sağlar, Bir bakışta bilginin anlaşılmasını kolaylaştırmak. Aşağıda bir örnek özet tablosu var, rapora dahildir.
Base Plaka Yazılımı Satın Alın
Base Plaka Tasarım Modülünün tam sürümünü başka bir SkyCiv modül olmadan kendi başına satın alın. Bu size taban plakası tasarımı için tam bir dizi sonuç verir, ayrıntılı raporlar ve daha fazla işlevsellik dahil.
