加长单板连接设计

Extended single plate connection design to AISC 360 is the shear tab procedure for a plate that reaches well past the support, usually so the beam can frame into a column web and clear the column flanges. You check the bolt group for the full eccentricity, the plate thickness for ductility, bolt bearing and tear-out on the plate and the beam web, 块剪, 剪切屈服和断裂, plate flexure and buckling, shear and flexure interaction, and the weld to the support. The long plate makes the bolt group eccentricity and the plate bending much larger than in a conventional shear tab, and that is what drives the design.

This page works one real example from start to finish: a W24x55 beam framing into the web of a W14x90 column, 与一个 100 kip LRFD reaction, through a 3/8 在x 13 在x 21 in A36 plate with fourteen 3/4 in A325-N bolts in two columns of seven. The design is to AISC 360-16 和 AISC 手册第 15 版. 每个方程都用它的数字写出来, in the same order as the AISC calc sheet that the SkyCiv连接设计 module produces for this connection.

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What is an extended single plate connection?

An extended single plate connection is a shear tab that is longer than the conventional configuration allows. The plate is shop welded to the support and field bolted to the beam web, as with any single plate, but the bolts sit far enough from the weld that the AISC Manual’s simplified Table 10-9 approach no longer applies. It is still a simple connection under AISC 360-16 B3.4, so it is detailed to rotate rather than resist moment.

The most common use is a beam framing into a column web. The beam end has to sit outside the column flanges, so the plate reaches out from the web to meet it. In this example the beam end is 7 3/4 in from the column web. A plate that long bends a lot more than a short tab, so the bolt group is designed for the full eccentricity and the plate is checked for buckling as well as yielding. It is one type of 剪力连接, alongside the double angle connection, 的 single angle connection and the shear end plate connection. You can check this example directly in the 单板连接计算器, which covers both the conventional and extended configurations.

Elevation of the example connection. The beam end sits 7 3/4 in from the column web, outside the column flanges, and the first bolt column is 1 in in from the beam end.

A-A部分. The plate has horizontal short-slotted holes and a 1/4 in fillet weld on both sides to the column web.

How do you design an extended single plate connection? The steps at a glance

1. Check the detailing limits at the beam: 螺栓间距, 边缘距离, plate depth.
2. Find the bolt group shear strength by the instantaneous center of rotation, for the full eccentricity.
3. Check plate ductility: the plate must not be stronger in bending than the bolt group.
4. Check bolt bearing and tear-out on the plate.
5. Check bolt bearing and tear-out on the beam web.
6. Combine shear, bearing and tear-out per bolt for the effective group strength.
7. Check block shear of the plate.
8. Check shear yielding and shear rupture of the plate.
9. Check plate flexure and buckling using the double coped beam procedure.
10. Check shear and flexure interaction for yielding.
11. Check shear and flexure interaction for rupture.
12. Check the detailing limit at the support.
13. Check the weld size needed to develop the plate.
14. Check the weld strength to the column web.

本示例的设计输入

设计代码为AISC 360-16 LRFD. 截面属性来自 AISC 手册.

 

 

The plate width adds up from the drawing: 7.75 in to the beam end, 1 in to the first bolt column, 3 in between columns and 1.25 in to the plate edge, 13 in in total. The plate depth is the two 1.5 in edge distances plus six 3 in spaces, 21 在.

步 1: Detailing limits at the beam side (AISC 360-16 J3.3, J3.4, J3.5)

在进行任何力量检查之前, the bolt layout has to be buildable and match the assumptions behind the strength equations. Bolts that are too close tear between the holes, edges that are too close tear out, and a plate that is too short or too deep changes how the connection rotates.

为一个 3/4 in bolt the minimum spacing is [mathin] 2\tfrac{2}{3} d_{b} = 2\,in [mathin] (J3.3) and the minimum edge distance is 1 在 (表J3.4). The plate has horizontal short slots, and its horizontal edge distance is measured from the end of the slot, so the Table J3.5 increment of 1/8 in raises its minimum to 1.125 在. The plate depth has to be at least half the beam’s T dimension, 20 3/4 in for a W24x55, so the connection is stiff enough to brace the beam end, and it must also fit within the beam web.

For a minimum limit the ratio is limit over actual, and for a maximum it is actual over limit. The governing row is the horizontal edge distance on the beam web, which sits exactly at the 1 最低限度:

[math] DCR = \frac{d}{c} = \frac{(1)}{(1)} = 1 [math]

DCR = 1.000 ≤ 1.0, 行

The plate depth is also close to its upper limit at 0.982. 一个 21 in plate with seven rows is about as deep as this beam can take.

步 2: Bolt group shear strength by the instantaneous center of rotation (AISC 360-16 J3.6)

The beam reaction reaches the plate at the bolt group, a long way from the weld, so the bolts carry a large moment as well as the shear. 瞬时旋转中心 (ICOR) method finds the point the bolt group rotates about and turns that into a coefficient C, which is how many bolts’ worth of shear strength the group actually delivers.

设计偏心率

For a conventional single plate, AISC 手册表 10-9 lets you design the bolts for a reduced eccentricity. That table does not cover the extended configuration, so the bolt group is designed for the full distance from the weld line to the centroid of the bolt group. Here that is the beam clearance, plus the edge distance to the first bolt column, plus half the column spacing:

[math] e = clr + L_{eh,bm} + \frac{s_{c}}{2} = 7.75\,in + 1\,in + \frac{3\,in}{2} = 10.25\,in [math]

Strength of one bolt

和 [mathin] \phi = 0.75 [mathin], [mathin] F_{nv} = 54\,ksi [mathin], [mathin] d_{b} = 0.75\,in [mathin] 和 [mathin] N_{s} = 1 [mathin]:

[math] \phi r_{v} = \phi\, F_{nv}\, \frac{\pi}{4} \left( d_{b} \right)^{2} N_{s}\, h_{f} = 17.892\,kip [math]

这里 [mathin] h_{f} = 1 [mathin] because there are no fillers (AISC 360-16 J5.2(一个)):

[math] h_{f} = 0.85 \lt 1 – (0.4) \times \left( (0\,in) – (0.25\,in) \right) \le 1 \;\Rightarrow\; h_{f} = 1 [math]

Locating the instantaneous center

螺栓 1 is at the bottom-left corner of the group, with the coordinates below. The IC is found by iteration, using the load-deformation curve of AISC Manual Equation (7-1) 在页面上 7-6 至 7-8. 的 使用 ICOR 进行螺栓组计算 walks through that iteration in full.

The iteration converges with the IC at:

[math] x_{ic} = -1.3636\,in, \qquad y_{ic} = 9\,in [math]

The IC sits level with the middle row and just outside the left bolt column. The bolt reactions at that location are:

That gives the two coefficients used in the rest of the design:

[math] C = 6.29 \qquad C’ = 72.195\,in [math]

Fourteen bolts deliver only 6.29 螺栓’ worth of vertical shear at a 10.25 in eccentricity, so more than half the group’s concentric strength goes into resisting the moment. C’ is the coefficient for moment alone, which the ductility check in Step 3 用途.

Bolt group strength

[math] \phi R_{n} = \phi\, F_{nv}\, \frac{\pi}{4} \left( d_{b} \right)^{2} N_{s}\, C\, h_{f} [math]

[math] \phi R_{n} = (0.75) \times (54\,ksi) \times \frac{\pi}{4} \times \left( (0.75\,in) \right)^{2} \times (1) \times (6.29) \times (1) [math]

[math] \phi R_{n} = 112.54\,kip [math]

DCR = 100 基普 / 112.54 基普= 0.889 < 1.0, 行

步 3: 板材延展性, maximum plate thickness (AISC Manual Eq. 10-3, 10-4)

A shear tab has to rotate as the beam deflects. If the plate is stronger in bending than the bolt group is in shear, the bolts fracture before the plate yields, which is a sudden, brittle failure. Capping the plate thickness makes sure the plate yields first. In the extended configuration this check is always required.

第一, the moment strength of the bolt group, 使用 [mathin] C’ = 72.195\,in [mathin] from Step 2 和 [mathin] F_{nv} = 54\,ksi [mathin]:

[math] M_{max} = \frac{F_{nv}}{0.9}\, \frac{\pi}{4}\, d_{b}^{2}\, C’ [math]

[math] M_{max} = \frac{(54\,ksi)}{0.9} \times \frac{\pi}{4} \times (0.75\,in)^{2} \times (72.195\,in) = 159.47\,kip\text{-}ft [math]

Then the thickest plate whose plastic moment does not exceed it, 与 [mathin] F_{yp} = 36\,ksi [mathin] 和 [mathin] d_{p} = 21\,in [mathin]:

[math] t_{max} = \frac{6\, M_{max}}{F_{yp}\, d_{p}^{2}} [math]

[math] t_{max} = \frac{6 \times (159.47\,kip\text{-}ft)}{(36\,ksi) \times (21\,in)^{2}} = 0.72323\,in [math]

DCR = 0.375 在 / 0.72323 在 = 0.519 < 1.0, 行

的 3/8 in plate is about half the limit. Fourteen bolts give the group a large moment strength, so there is plenty of room here.

步 4: Bolt bearing and tear-out on the single plate (AISC 360-16 J3.10)

Each bolt pushes down on the edge of its hole in the plate. The plate either deforms at the hole (轴承, capped at 2.4 螺栓的剪切强度) or tears a strip out to the next hole or to the bottom edge (撕掉, 1.2 螺栓的剪切强度). The bottom bolt in each column has the shortest clear distance, so it is the weak one.

The slots run horizontally and the load is vertical, so the hole dimension across the member length is the standard [mathin] d_{h} = 0.8125\,in [mathin]. Clear distance for the edge bolt, 与 [mathin] L_{ev,pl} = 1.5\,in [mathin]:

[math] l_{c1} = L_{ev,pl} – \frac{d_{h}}{2} = (1.5\,in) – \frac{(0.8125\,in)}{2} = 1.0938\,in [math]

Clear distance between bolts, 与 [mathin] s_{r} = 3\,in [mathin]:

[math] l_{c2} = s_{r} – d_{h} = (3\,in) – (0.8125\,in) = 2.1875\,in [math]

在哪里 [mathin] t_{p} = 0.375\,in [mathin], [mathin] F_{up} = 58\,ksi [mathin], [mathin] C = 6.29 [mathin], [mathin] n_{r} = 7 [mathin] 和 [mathin] \phi = 0.75 [mathin]. The bracket is the strength of one line of seven bolts, and the factor C / n_r scales it for the eccentric group of two lines:

[math] \phi R_{n} = \phi \left[ \min\left(1.2\, l_{c1},\, 2.4\, d_{b}\right) t_{p} F_{up} + \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{p} F_{up} \left( n_{r} – 1 \right) \right] \left( \frac{C}{n_{r}} \right) [math]

[math] \phi R_{n} = (0.75) \times \left[ \min\left(1.2 \times (1.0938\,in),\, 2.4 \times (0.75\,in)\right) \times (0.375\,in) \times (58\,ksi) + \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.375\,in) \times (58\,ksi) \times \left( (7) – 1 \right) \right] \times \left( \frac{(6.29)}{(7)} \right) [math]

[math] \phi R_{n} = 177.54\,kip [math]

DCR = 100 基普 / 177.54 基普= 0.563 < 1.0, 行

Tear-out governs the edge bolt, because 1.2 lc1 is less than 2.4 D b. The inner bolts are capped by bearing.

步 5: Bolt bearing and tear-out on the beam web (AISC 360-16 J3.10)

The same check applies to the beam web. The beam is not coped and its flanges are far from the bolts, so there is no edge to tear out to, and every bolt bears like an inner bolt.

[math] l_{c2} = s_{r} – d_{h} = (3\,in) – (0.8125\,in) = 2.1875\,in [math]

在哪里 [mathin] t_{wb} = 0.395\,in [mathin], [mathin] F_{ub} = 65\,ksi [mathin] 和 [mathin] C = 6.29 [mathin]:

[math] \phi R_{n} = \phi\, \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{wb} F_{ub}\, C [math]

[math] \phi R_{n} = (0.75) \times \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.395\,in) \times (65\,ksi) \times (6.29) [math]

[math] \phi R_{n} = 218.02\,kip [math]

DCR = 100 基普 / 218.02 基普= 0.459 < 1.0, 行

的 0.395 in web is thicker than the plate and has a higher tensile strength, so it is the stronger ply in this connection.

步 6: Effective strength of the bolt group (AISC 360-16 J3.6 评论)

螺栓剪, bearing and tear-out are not really separate checks. Each bolt delivers the least of its shear strength and its bearing or tear-out strength in each ply it passes through, and the group strength is the sum of those per-bolt values, scaled for the eccentricity.

Strength of one bolt in single shear:

[math] \phi r_{v} = \phi\, F_{nv}\, \frac{\pi}{4} \left( d_{b} \right)^{2} N_{s}\, h_{f} = (0.75) \times (54\,ksi) \times \frac{\pi}{4} \times \left( (0.75\,in) \right)^{2} \times (1) \times (1) = 17.892\,kip [math]

Bearing and tear-out of one bolt on the plate, for the edge bolt and an inner bolt:

[math] \phi r_{n1e} = \phi\, \min\left(1.2\, l_{c1},\, 2.4\, d_{b}\right) t_{p} F_{up} = (0.75) \times \min\left(1.2 \times (1.0938\,in),\, 2.4 \times (0.75\,in)\right) \times (0.375\,in) \times (58\,ksi) = 21.41\,kip [math]

[math] \phi r_{n1i} = \phi\, \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{p} F_{up} = (0.75) \times \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.375\,in) \times (58\,ksi) = 29.363\,kip [math]

Bearing of one bolt on the beam web:

[math] \phi r_{n2} = \phi\, \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{wb} F_{ub} = (0.75) \times \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.395\,in) \times (65\,ksi) = 34.661\,kip [math]

Bolts are numbered from the top of each line, so bolt 7 is the edge bolt on the plate:

[math] \phi r_{eff,1} = \min\left( \phi r_{v},\, \phi r_{n1i},\, \phi r_{n2} \right) = \min\left( (17.892\,kip),\, (29.363\,kip),\, (34.661\,kip) \right) = 17.892\,kip [math]

[math] \phi r_{eff,2} = \min\left( \phi r_{v},\, \phi r_{n1e},\, \phi r_{n2} \right) = \min\left( (17.892\,kip),\, (21.41\,kip),\, (34.661\,kip) \right) = 17.892\,kip [math]

Sum over one line, 与 [mathin] n_{1} = 6 [mathin] bolts at the inner value and [mathin] n_{2} = 1 [mathin] at the edge value:

[math] \sum \phi r_{eff} = n_{1}\, \phi r_{eff,1} + n_{2}\, \phi r_{eff,2} = (6) \times (17.892\,kip) + (1) \times (17.892\,kip) = 125.25\,kip [math]

[math] \phi R_{n,eff} = \left( \frac{C}{n_{r}} \right) \sum \phi r_{eff} = \left( \frac{(6.29)}{(7)} \right) \times (125.25\,kip) = 112.54\,kip [math]

DCR = 100 基普 / 112.54 基普= 0.889 < 1.0, 行

Bolt shear governs every bolt, so the effective strength equals the bolt group shear strength from Step 2. You can run the same connection in the free connection design calculator and try a larger bolt or a third bolt column to see how C moves.

步 7: Block shear of the single plate (AISC 360-16 J4.3)

A block of the plate can tear out around the bolt group: shear down the line of holes nearest the beam, plus tension across both bolt columns to the plate’s free edge. This is an L-shaped block.

Net area in tension. The tension plane runs horizontally, along the slots, so it uses the slot length [mathin] d_{h} = 1\,in [mathin], 与 [mathin] n_{c} = 2 [mathin], [mathin] s_{c} = 3\,in [mathin] 和 [mathin] L_{eh,pl} = 1.25\,in [mathin]. Each hole adds 1/16 in for damage during punching (B4.3):

[math] A_{nt} = t_{p} \left[ (n_{c}-1)\, s_{c} + L_{eh,pl} – (n_{c} – 0.5)\left( d_{h} + 0.0625\,in \right) \right] [math]

[math] A_{nt} = (0.375\,in) \times \left[ ((2)-1) \times (3\,in) + (1.25\,in) – ((2) – 0.5) \times ((1\,in) + (0.0625\,in)) \right] = 0.9961\,in^{2} [math]

Gross and net areas in shear. The shear plane runs vertically, across the slots, so the net area uses [mathin] d_{h} = 0.8125\,in [mathin]:

[math] A_{gv} = t_{p} \left[ L_{ev,pl} + (n_{r}-1)\, s_{r} \right] = (0.375\,in) \times \left[ (1.5\,in) + ((7)-1) \times (3\,in) \right] = 7.3125\,in^{2} [math]

[math] A_{nv} = t_{p} \left[ d_{p} – L_{ev,pl} – (n_{r} – 0.5)\left( d_{h} + 0.0625\,in \right) \right] [math]

[math] A_{nv} = (0.375\,in) \times \left[ (21\,in) – (1.5\,in) – ((7) – 0.5) \times ((0.8125\,in) + (0.0625\,in)) \right] = 5.1797\,in^{2} [math]

Block shear strength, 与 [mathin] F_{up} = 58\,ksi [mathin], [mathin] F_{yp} = 36\,ksi [mathin] 和 [mathin] \phi = 0.75 [mathin]. With two bolt columns the tension stress is not uniform, because the column nearest the free edge takes more of it, 所以 [mathin] U_{bs} = 0.5 [mathin]:

[math] \phi R_{n} = \phi \left( 0.6\, F_{up} A_{nv} + U_{bs} F_{up} A_{nt} \le 0.6\, F_{yp} A_{gv} + U_{bs} F_{up} A_{nt} \right) [math]

[math] \phi R_{n} = (0.75) \times \left( 0.6 \times (58\,ksi) \times (5.1797\,in^{2}) + (0.5) \times (58\,ksi) \times (0.9961\,in^{2}) \le 0.6 \times (36\,ksi) \times (7.3125\,in^{2}) + (0.5) \times (58\,ksi) \times (0.9961\,in^{2}) \right) [math]

[math] \phi R_{n} = 140.13\,kip [math]

DCR = 100 基普 / 140.13 基普= 0.714 < 1.0, 行

The shear yielding side of the equation governs. A36 has a low yield-to-tensile ratio, so the gross shear plane yields before the net shear plane ruptures.

步 8: Shear yielding and shear rupture of the single plate (AISC 360-16 J4.2)

The full reaction passes through the plate as vertical shear. The gross section must not yield, and the net section through the bolt holes must not rupture.

剪切屈服 (情商. J4-3)

[math] A_{gv} = d_{p}\, t_{p} = (21\,in) \times (0.375\,in) = 7.875\,in^{2} [math]

[math] \phi R_{n,sy} = \phi\, 0.6\, F_{yp} A_{gv} = (1) \times 0.6 \times (36\,ksi) \times (7.875\,in^{2}) = 170.1\,kip [math]

剪切断裂 (情商. J4-4)

[math] A_{nv} = t_{p} \left[ d_{p} – n_{r}\left( d_{h} + 0.0625\,in \right) \right] = (0.375\,in) \times \left[ (21\,in) – (7) \times ((0.8125\,in) + (0.0625\,in)) \right] = 5.5781\,in^{2} [math]

[math] \phi R_{n,sr} = \phi\, 0.6\, F_{up} A_{nv} = (0.75) \times 0.6 \times (58\,ksi) \times (5.5781\,in^{2}) = 145.59\,kip [math]

[math] \phi R_{n} = \min\left( \phi R_{n,sy},\, \phi R_{n,sr} \right) = \min\left( (170.1\,kip),\, (145.59\,kip) \right) = 145.59\,kip [math]

DCR = 100 基普 / 145.59 基普= 0.687 < 1.0, 行

The section is cut by the holes of one bolt column only, because the vertical section where shear is checked passes through a single line of holes.

步 9: Plate flexure using the double coped beam procedure (AISC 360-16 F11.2)

This is the check that sets an extended plate apart. The reaction acts at the bolts, 8.75 in from the weld, so the plate is a long cantilever in bending. 一个深, thin plate over that length can buckle sideways before it yields. The AISC Manual handles this with the procedure for a beam coped at both flanges, treating the plate like a beam web with the flanges cut away. The unbraced length is the distance from the weld to the first bolt column, [mathin] L_{b} = clr + L_{eh,bm} = 7.75\,in + 1\,in = 8.75\,in [mathin].

Range check

在哪里 [mathin] E = 29000\,ksi [mathin], [mathin] F_{yp} = 36\,ksi [mathin], [mathin] d_{p} = 21\,in [mathin] 和 [mathin] t_{p} = 0.375\,in [mathin]:

[math] \frac{L_{b} d}{t^{2}} = \frac{(8.75\,in) \times (21\,in)}{(0.375\,in)^{2}} = 1306.7 [math]

[math] \frac{0.08 E}{F_{y}} = \frac{0.08 \times (29000\,ksi)}{(36\,ksi)} = 64.444 \qquad \frac{1.9 E}{F_{y}} = \frac{1.9 \times (29000\,ksi)}{(36\,ksi)} = 1530.6 [math]

所以 [mathin] \frac{0.08E}{F_{y}} \lt \frac{L_{b} d}{t^{2}} \le \frac{1.9E}{F_{y}} [mathin], 和情商. F11-2 applies. 在 1306.7 the plate is close to the 1530.6 upper limit, beyond which Eq. F11-3 for elastic buckling would take over.

Section moduli of the plate

[math] Z = \frac{t_{p} \left( d_{p} \right)^{2}}{4} = \frac{(0.375\,in) \times ((21\,in))^{2}}{4} = 41.344\,in^{3} [math]

[math] S = \frac{t_{p} \left( d_{p} \right)^{2}}{6} = \frac{(0.375\,in) \times ((21\,in))^{2}}{6} = 27.563\,in^{3} [math]

Lateral-torsional buckling modification factor (AISC Manual Eq. 9-15)

With no cope on the plate, [mathin] d_{ct} = 0\,in [mathin]:

[math] C_{b} = \left[ 3 + \ln\left( \frac{L_{b}}{d_{p}} \right) \right] \left( 1 – \frac{d_{ct}}{d_{p}} \right) \ge 1.84 [math]

[math] C_{b} = \left[ 3 + \ln\left( \frac{(8.75\,in)}{(21\,in)} \right) \right] \times \left( 1 – \frac{(0\,in)}{(21\,in)} \right) \ge 1.84 [math]

[math] C_{b} = 2.1245 [math]

Flexural yielding and buckling strength

和 [mathin] \phi = 0.9 [mathin]:

[math] \phi M_{n} = \phi \left[ C_{b} \left[ 1.52 – 0.274 \left( \frac{L_{b} d}{t^{2}} \right) \left( \frac{F_{yp}}{E} \right) \right] F_{yp} S \le F_{yp} Z \right] [math]

[math] \phi M_{n} = (0.9) \times \left[ (2.1245) \times \left[ 1.52 – 0.274 \times ((1306.7)) \times \left( \frac{(36\,ksi)}{(29000\,ksi)} \right) \right] \times (36\,ksi) \times (27.563\,in^{3}) \le (36\,ksi) \times (41.344\,in^{3}) \right] [math]

[math] \phi M_{n} = 111.63\,kip\text{-}ft [math]

The plastic moment cap governs: 111.63 kip-ft is [mathin] \phi F_{yp} Z [mathin]. With C_b above 2, the buckling expression comes out well above the cap, so even at L_b d / t² = 1306.7 the plate reaches its full plastic moment. The equivalent shear capacity at the bolts is:

[math] \phi V_{n,y} = \frac{\phi M_{n}}{L_{b}} = \frac{(111.63\,kip\text{-}ft)}{(8.75\,in)} = 153.09\,kip [math]

Flexural rupture (AISC Manual Part 15 桌子 15-3, 部分 9 情商. 9-4)

The net plastic section modulus removes the holes of one bolt column, 与 [mathin] d_{h} = 0.8125\,in [mathin], [mathin] s_{r} = 3\,in [mathin] 和 [mathin] n_{r} = 7 [mathin]. With an odd number of rows the last term applies:

[math] Z_{net} = \frac{t_{p} \left( d_{p} \right)^{2}}{4} – t_{p} \left( d_{h} + 0.0625\,in \right) s_{r} \left\lfloor \frac{n_{r}^{2}}{4} \right\rfloor – \left[ n_{r}\ \text{odd} \right] \frac{t_{p} \left( d_{h} + 0.0625\,in \right)^{2}}{4} [math]

[math] Z_{net} = \frac{(0.375\,in) \times ((21\,in))^{2}}{4} – (0.375\,in) \times ((0.8125\,in) + (0.0625\,in)) \times (3\,in) \times \left\lfloor \frac{(7)^{2}}{4} \right\rfloor – \left[ (7)\ \text{odd} \right] \times \frac{(0.375\,in) \times ((0.8125\,in) + (0.0625\,in))^{2}}{4} = 29.459\,in^{3} [math]

[math] \phi M_{n} = \phi\, F_{up} Z_{net} = (0.75) \times (58\,ksi) \times (29.459\,in^{3}) = 106.79\,kip\text{-}ft [math]

[math] \phi V_{n,r} = \frac{\phi M_{n}}{L_{b}} = \frac{(106.79\,kip\text{-}ft)}{(8.75\,in)} = 146.46\,kip [math]

Governing flexural capacity

[math] \phi V_{n} = \min\left( \phi V_{n,y},\, \phi V_{n,r} \right) = \min\left( (153.09\,kip),\, (146.46\,kip) \right) = 146.46\,kip [math]

DCR = 100 基普 / 146.46 基普= 0.683 < 1.0, 行

Flexural rupture through the holes governs, by a small margin over yielding and buckling.

步 10: Shear and flexure interaction of the single plate, 屈服 (AISC Manual Part 10)

The plate carries shear and bending at the same section at the same time, so the two have to be checked together. The Manual uses a circular interaction. This one combines the flexural yielding strength from Step 9 with the shear yielding strength from Step 8.

[math] I_{nt} = \left( \frac{V_{u}}{\phi V_{n}} \right)^{2} + \left( \frac{V_{u}}{\phi R_{n,sy}} \right)^{2} [math]

[math] I_{nt} = \left( \frac{(100\,kip)}{(153.09\,kip)} \right)^{2} + \left( \frac{(100\,kip)}{(170.1\,kip)} \right)^{2} = 0.7723 [math]

DCR = 0.7723 / 1.0 = 0.772 < 1.0, 行

步 11: Shear and flexure interaction of the single plate, 破裂 (AISC Manual Part 10)

The same interaction, with the flexural rupture strength from Step 9 and the shear rupture strength from Step 8. Both act on the net section through the holes.

[math] I_{nt} = \left( \frac{V_{u}}{\phi V_{n}} \right)^{2} + \left( \frac{V_{u}}{\phi R_{n,sr}} \right)^{2} [math]

[math] I_{nt} = \left( \frac{(100\,kip)}{(146.46\,kip)} \right)^{2} + \left( \frac{(100\,kip)}{(145.59\,kip)} \right)^{2} = 0.938 [math]

DCR = 0.938 / 1.0 = 0.938 < 1.0, 行

This is the highest strength ratio in the connection. Each of the two rupture checks on its own is under 0.69, but together they come close to the limit. That is typical of a long plate, where the moment at the bolt line is large compared with the shear.

步 12: Detailing limit at the support side

The fillet weld sits in the gap between the column web and the end of the beam, so the weld cannot be larger than that clearance. With the beam held 7.75 in off the column web, this check has a lot of room.

[math] DCR = \frac{d}{c} = \frac{(0.25)}{(7.75)} = 0.032258 [math]

DCR = 0.032 < 1.0, 行

步 13: Weld required to develop the single plate (AISC Manual Part 10)

The ductility check in Step 3 makes the plate yield first. That only works if the weld does not fail before the plate does. Two fillet welds, each at least 5/8 of the plate thickness, develop the strength of the plate.

[math] W_{duc} = \frac{5}{8}\, t_{p} = \frac{5}{8} \times (0.375\,in) = 0.23438\,in [math]

DCR = 0.23438 在 / 0.25 在 = 0.938 < 1.0, 行

The nearest standard size above 0.234 在是 1/4 在, so this ratio is set by rounding up, not by the load.

步 14: Fillet weld capacity to the column web (AISC 360-16 J2.4)

最后的, the welds have to carry the reaction itself. Their strength depends on size, length and load direction. The base metal on either side of the weld, here the plate and the column web, can also limit how much of the weld size counts.

Base metal limit on weld size (AISC Manual Eq. 9-2)

This is the largest weld size, in sixteenths of an inch, that the thinner connected part can support. The plate is welded on both sides, so half its thickness is available to each weld. 和 [mathin] t_{p} = 0.375\,in [mathin], [mathin] F_{up} = 58\,ksi [mathin], [mathin] t_{ws} = 0.44\,in [mathin] 和 [mathin] F_{us} = 65\,ksi [mathin]:

[math] D_{max} = \frac{\frac{t_{p}}{2} F_{up} \le t_{ws} F_{us}}{3.09\,kip/in \times \left( F_{EXX} / 70\,ksi \right)} [math]

[math] D_{max} = \frac{(0.1875\,in) \times (58\,ksi) \le (0.44\,in) \times (65\,ksi)}{(3.09\,kip/in) \times (1)} = 3.5194 [math]

So only 3.52/16 在, 关于 0.220 在, 的 1/4 in weld can be counted. 的 3/8 in A36 plate cannot develop a full 1/4 in weld on both faces, and the plate rather than the 0.44 in column web sets the limit.

Effective length

Two lines of weld, each reduced by 2W at the ends:

[math] L_{w} = 2 \left[ d_{p} – 2W \right] = 2 \times \left[ (21\,in) – 2 \times (0.25\,in) \right] = 41\,in [math]

焊接强度

The load runs parallel to the welds, 所以 [mathin] \theta = 0 [mathin] 并且没有定向增加. 和 [mathin] F_{EXX} = 70\,ksi [mathin] 和 [mathin] \phi = 0.75 [mathin]:

[math] \phi R_{n} = \phi\, 0.6\, F_{EXX} \frac{\sqrt{2}}{2} \left( W \le \frac{D_{max}}{16} \right) \left[ 1 + 0.5 \left( \sin\theta \right)^{1.5} \right] L_{w} [math]

[math] \phi R_{n} = (0.75) \times 0.6 \times (70\,ksi) \times \frac{\sqrt{2}}{2} \times \left( (0.25\,in) \le \frac{(3.5194)}{(16)} \right) \times \left[ 1 + 0.5 \times (\sin(0))^{1.5} \right] \times (41\,in) [math]

[math] \phi R_{n} = 200.88\,kip [math]

DCR = 100 基普 / 200.88 基普= 0.498 < 1.0, 行

Summary of the extended single plate connection design checks

The connection passes every check. 最高比例, 1.000, is a detailing limit: the bolts are exactly the minimum 1 in from the beam end, 和 21 in plate is at 0.982 of its maximum depth. Of the checks that compare load with capacity, shear and flexure interaction for rupture governs at 0.938, followed by bolt group shear at 0.889. Both come from the 10.25 in eccentricity, which cuts the 14 螺栓到 6.29 effective bolts and puts a large moment into the plate at the bolt line. The weld-to-develop ratio of 0.938 is a sizing rule, not a load check.

如果检查失败要更改什么

• Bolt group shear or effective strength: use a larger bolt or a stronger grade, such as A325-X (threads excluded) or A490. Adding a bolt row raises C, but here the plate is already near its maximum depth, so a third bolt column is the other option.
• Interaction or plate flexure: 使用较厚的板, as long as the ductility check in Step 3 still passes. Reducing the beam clearance shortens L_b and the eccentricity, which helps every plate and bolt check.
• 延展性 (t_max): use a thinner plate, or add bolts so the group’s moment strength rises. A thicker plate makes this check worse.
• Tear-out on the plate: increase the vertical edge distance below the bottom bolt row.
• 块剪, shear yielding or shear rupture of the plate: use a thicker or deeper plate. A longer horizontal edge distance also adds to the block shear tension area.
• 焊接: the base metal limit caps the useful weld size at about 0.220 in for a 3/8 A36板, so a bigger weld adds nothing. A thicker or higher-strength plate raises that cap.
• A different connection: if the plate cannot be made to work, 一个 double angle connection or a shear end plate with the beam coped to fit between the column flanges avoids the long cantilever altogether.

经常问的问题

传统单板连接和扩展单板连接有什么区别?

The conventional configuration in AISC Manual Part 10 covers a single vertical row of 2 至 12 螺栓, with the bolt line no more than 3 1/2 in from the weld and limits on edge distance and plate thickness. Within those limits you can use Table 10-9, with its reduced bolt eccentricity and no plate buckling check. Anything outside them, such as this plate with two bolt columns 8.75 在和 11.75 in from the weld, is the extended configuration. The bolt group is then designed for the full eccentricity, the plate thickness must pass the ductility check, and the plate is checked for buckling.

Why is the bolt eccentricity 10.25 在?

In the extended configuration the bolt group is designed for the distance from the weld line to the centroid of the bolts. That is the 7.75 in beam clearance, plus the 1 in edge distance to the first bolt column, plus half of the 3 in spacing between the two columns, 10.25 in in total.

Why does a beam framing into a column web need an extended plate?

The column flanges are in the way. A W14x90 flange is 14.5 在宽, so it projects about 7 in from the web on each side, and the beam end has to sit outside that line. The plate is welded to the column web and reaches out past the flange tips to the beam, which is why this example has a 7.75 in clearance and a 13 in wide plate. The other option is to cope the beam flanges so the beam fits between the column flanges, which lets you use a shorter, conventional plate.

Why is the block shear uniformity factor 0.5?

With two bolt columns, the tension across the block is not shared evenly. The bolt column closer to the free edge of the plate takes more of it. AISC 360-16 J4.3 and its Commentary use U_bs = 0.5 for non-uniform tension stress, and the AISC Manual applies that to single plates with more than one bolt column.

Why does only part of the 1/4 in weld count?

手动均衡器. 9-2 limits the weld size to what the base metal can carry. The plate is 3/8 在厚, A36, and welded on both faces, so each weld can only be backed by half of it. That gives 3.52 sixteenths, or about 0.220 在, of usable weld size. 的 1/4 in weld is still needed to meet the 5/8 t_p rule in Step 13, but its strength is calculated on the smaller size.

Why does shear and flexure interaction govern?

In a conventional tab the moment at the bolt line is small, so interaction rarely matters. Here the bolts are 8.75 in from the weld, so the moment is large. The plate’s flexural rupture strength (146.46 kip as equivalent shear) and its shear rupture strength (145.59 基普) are almost equal. Squaring and adding the two ratios gives 0.938, even though each one on its own is below 0.69.

此示例是否遵循 AISC 第 15 版或 AISC 第 16 版手册?

它遵循 AISC 360-16 和 AISC 手册第 15 版, 这是围绕该规范编写的. AISC 第 16 版手册与后续版本配对 360-22 规范. The limit states for an extended single plate connection are the same in both, but clause and page references move, 因此请根据您的项目使用的版本检查它们.

对于 AISC ASD 而不是 LRFD 来说，这会如何变化?

The limit states and nominal strengths are identical. With AISC ASD you compare the service-level reaction against the nominal strength divided by a safety factor instead of multiplying by a resistance factor. 阻力系数为 0.75 成为安全系数 2.00, shear yielding’s 1.00 变成 1.50, and flexure’s 0.90 变成 1.67. 荷载组合也发生变化, so the demand is lower.

Design an extended single plate connection for free

此页上的每个数字都来自一个解决方案. Pick the single plate configuration, 输入你的光束, 柱子, 盘子, bolts and reaction, 并在浏览器中查看每项检查. 没有什么可安装的.

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SkyCiv Connection Design 模块是本示例背后的连接设计软件. 它为您提供完整的计算报告，其中包含每个方程和您的数字, 就像上面的步骤一样, 加上每个容量和 DCR, 对于AISC 360-16 (LRFD 和 ASD) 并作为 4100:2020. 的 螺栓计算器 和 焊接计算器 cover the bolt group and weld checks on their own.

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参考资料

• AISC 360-16, 钢结构建筑规范: B3.4, B4.3, F11.2 (情商. F11-2), J2.4 (情商. J2-4), J3.3, J3.4 (表J3.4), J3.5 (表 J3.5), J3.6 (情商. J3-1) 及评论, J3.10 (情商. J3-6a, J3-6c), J4.2 (情商. J4-3, J4-4), J4.3 (情商. J4-5), J5.2(一个).
• AISC 钢结构手册, 15第版: 部分 7, 首先介绍你自己和你申请的职位 7-6 至 7-8 (情商. 7-1); 部分 9, 情商. 9-2, 9-4 和 9-15; 部分 10, 桌子 10-9, 页 10-87 and page 10-89 (情商. 10-3, 10-4); 部分 15, 桌子 15-3.
• 使用 ICOR 计算螺栓组
• 使用 AISC 进行钢连接设计 360-16
• 螺栓 和 焊缝 SkyCiv 连接设计