如何设计法兰板弯矩连接

To design a bolted flange plate moment connection to AISC 360, you turn the beam end moment into a pair of flange forces and check everything that force passes through. That means slip of the flange bolts, 螺栓轴承和撕裂, 块剪, tension and compression in the flange plate, tension and flexural rupture of the beam at the holes, the weld to the column, and the column flange and web. A separate web plate carries the shear and is checked like a shear tab.

This page works through one real example from start to finish: a W24x84 beam framing into the flange of a W14x211 column, 与一个 300 急转瞬间, 一个 250 kip shear and 100 kip of axial load, 设计符合 AISC 360-16 LRFD 和 AISC 手册第 15 版. The flange plates are 1/2 in thick with eight 1 in A490 slip-critical bolts each, and the web plate is 1 in thick with ten 1 在 A490-N 螺栓中. 每个方程都用它的数字写出来, 按照与 AISC 计算表相同的顺序 SkyCiv连接设计 模块为其生成.

立即尝试我们的免费工具 购买连接设计模块

What is a flange plate moment connection?

A flange plate moment connection transfers a beam’s end moment to a column through two plates, one at each beam flange. Each plate is shop-welded to the column flange and field-bolted to the beam flange, and a separate web plate carries the shear. It is a fully restrained (阻燃剂) 矩连接 under AISC 360-16 B3.4(b).

It is also called a bolted flange plate or flange-plated moment connection. Fabricators like it because all the welding happens in the shop and the beam is only bolted in the field. 法兰板作为张力和压力偶承受力矩, the web plate is a single plate connection (a shear tab) for the shear, and the column has to take the concentrated flange forces. You can check this one directly in the 法兰盘连接计算器, and the other types are on the 瞬间连接 页.

Elevation of the example connection. The flange plates are welded to the column with 1/4 in fillets and bolted to the beam flanges. The web plate is welded with 7/16 in fillets and bolted to the beam web. 有一个 1/4 in erection gap at each flange plate, filled with shims as required.

A-A部分. Each flange plate has two lines of four bolts at a 5 1/2 在规格中, in oversized holes in the plate and standard holes in the beam flange.

How do you design a flange plate moment connection? The steps at a glance

The flange plate connection is checked first, for the flange force from the moment and axial load. The web plate is checked second, for the shear.

1. Check the detailing limits at the beam flange: 量具, 间距和边距.
2. Find the flange force and check slip of the slip-critical bolts.
3. 检查法兰板上的螺栓轴承和撕裂情况.
4. Check bolt bearing and tear-out on the beam flange.
5. 检查法兰板的块剪切是否存在三种撕裂模式.
6. Check block shear of the beam flange.
7. 检查法兰板的拉伸屈服和破裂情况.
8. Check tension rupture of the beam, 有剪切滞后.
9. Check flexural rupture of the beam at the flange bolt holes.
10. 检查法兰盘是否受压.
11. Check the detailing limit at the column face.
12. Check the fillet weld from the flange plate to the column flange.
13. Check the column flange and web for the concentrated flange force.
14. Check the detailing limits for the web plate.
15. Check bolt group shear on the web plate.
16. Check bolt bearing and tear-out on the web plate.
17. Check bolt bearing and tear-out on the beam web.
18. Combine shear, bearing and tear-out per bolt for the effective group strength.
19. Check block shear of the web plate.
20. Check shear yielding and shear rupture of the web plate.
21. Check the detailing limit for the web plate weld.
22. Check the fillet weld from the web plate to the column flange.

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.

 

 

 

部分 1: the flange plate connection

脚步 1 至 13 check the flange plates, the beam flanges, the welds to the column and the column itself, all for the flange force worked out in Step 2.

步 1: Detailing limits at the beam flange (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, and edges that are too close tear out.

最小螺栓间距为 [mathin] 2\tfrac{2}{3} d_{b} [mathin] from J3.3. The minimum edge distance for a 1 in bolt is 1 1/4 in from Table J3.4, which is the limit on the beam flange, where the holes are standard. The flange plate has oversized holes, 因此表 J3.5 添加了 1/8 在, which brings its edge distance limits to 1.375 在. The gage has its own minimum, so that the two bolt lines fit across the flange with room to install the bolts.

[math] s_{min} = 2\tfrac{2}{3}\, d_{b} = 2.667 \times (1\,in) = 2.667\,in [math]

For a minimum limit the ratio is limit over actual, and for a maximum it is actual over limit. The governing row is the plate horizontal edge distance:

[math] DCR = \frac{1.375\,in}{1.5\,in} = 0.917 [math]

DCR = 0.917 < 1.0, 行

步 2: Flange force and slip critical bolt strength (AISC 360-16 J3.8)

The moment is carried as a couple of equal and opposite forces in the two beam flanges, and the flange plate bolts have to carry that force without slipping. These bolts are slip-critical because the flange plate has oversized holes, 哪个 AISC 360-16 J3.2 allows only in slip-critical connections.

Flange force

The moment is divided by the distance between the beam flange centroids, [mathin] d_{bm} – t_{fb} [mathin]. The web plate is designed for shear only, so the axial load goes to the flange plates, half to each:

[math] P_{uF} = \frac{M_{ux}}{d_{bm} – t_{fb}} + \frac{P_{u}}{2} [math]

[math] P_{uF} = \frac{(300\,kip\text{-}ft)}{(24.1\,in) – (0.77\,in)} + \frac{(100\,kip)}{2} [math]

[math] P_{uF} = 204.31\,kip [math]

The axial tension and compression are both 100 基普, so the flange force is 204.31 kip either way: tension adds to the tension flange and compression adds to the compression flange. Every check in Part 1 uses this one value.

防滑性

在哪里 [mathin] \phi = 0.85 [mathin] is the resistance factor for oversized holes, [mathin] \mu = 0.5 [mathin] is the mean slip coefficient for Class B faying surfaces, [mathin] D_{u} = 1.13 [mathin] relates the mean installed pretension to the specified minimum, [mathin] h_{f} = 0.85 [mathin] is the filler factor, [mathin] T_{b} = 64\,kip [mathin] is the minimum pretension of a 1 in A490 bolt from Table J3.1, and there is [mathin] N_{s} = 1 [mathin] 滑移面:

[math] \phi R_{n} = \phi\, \mu\, D_{u}\, h_{f}\, T_{b}\, N_{s}\, n_{r}\, n_{c} [math]

[math] \phi R_{n} = (0.85) \times (0.5) \times (1.13) \times (0.85) \times (64\,kip) \times (1) \times (2) \times (4) [math]

[math] \phi R_{n} = 209\,kip [math]

DCR = 204.31 基普 / 209 基普= 0.978 < 1.0, 行

The filler factor of 0.85 is the J3.8 value for two or more fillers between the connected parts. The drawing calls for shims as required at the 1/4 in erection gap, so the design does not count on there being only one. 在 0.978 this is one of the two tightest checks in the connection.

步 3: 法兰板上的螺栓轴承和撕裂口 (AISC 360-16 J3.10)

Slip-critical bolts must also have enough strength as bearing bolts, in case the joint does slip. Each bolt pushes 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 end of the plate (撕掉, 1.2 螺栓的剪切强度). The bolts nearest the end of the plate have the shortest clear distance, so they are the weak ones.

Clear distance for the end bolts, 与 [mathin] d_{h} = 1.25\,in [mathin], the oversized hole dimension along the member length:

[math] l_{c1} = L_{eh,fp} – \frac{d_{h}}{2} = (1.5\,in) – \frac{(1.25\,in)}{2} = 0.875\,in [math]

Clear distance between the other bolts:

[math] l_{c2} = s_{c} – d_{h} = (3\,in) – (1.25\,in) = 1.75\,in [math]

The end bolt in each of the two lines uses [mathin] l_{c1} [mathin], and the other three in each line use [mathin] l_{c2} [mathin]:

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

[math] \phi R_{n} = (0.75) \times \left[ \min\left(1.2 \times 0.875\,in,\, 2.4 \times 1\,in\right)(0.5\,in)(65\,ksi)(2) + \min\left(1.2 \times 1.75\,in,\, 2.4 \times 1\,in\right)(0.5\,in)(65\,ksi)(2)(4 – 1) \right] [math]

[math] \phi R_{n} = 358.31\,kip [math]

DCR = 204.31 基普 / 358.31 基普= 0.570 < 1.0, 行

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

The same check applies to the beam flange, which has standard holes ([mathin] d_{h} = 1.125\,in [mathin]) and is thicker than the plate.

[math] l_{c1} = L_{eh,bf} – \frac{d_{h}}{2} = (1.5\,in) – \frac{(1.125\,in)}{2} = 0.9375\,in [math]

[math] l_{c2} = s_{c} – d_{h} = (3\,in) – (1.125\,in) = 1.875\,in [math]

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

[math] \phi R_{n} = (0.75) \times \left[ \min\left(1.2 \times 0.9375\,in,\, 2.4 \times 1\,in\right)(0.77\,in)(65\,ksi)(2) + \min\left(1.2 \times 1.875\,in,\, 2.4 \times 1\,in\right)(0.77\,in)(65\,ksi)(2)(4 – 1) \right] [math]

[math] \phi R_{n} = 591.22\,kip [math]

DCR = 204.31 基普 / 591.22 基普= 0.346 < 1.0, 行

步 5: 翼缘板块剪 (AISC 360-16 J4.3)

A block of the flange plate can tear out along the bolt lines: shear along a line of holes plus tension across the plate. With two bolt lines there are three ways for this to happen, and the weakest one governs. Each hole adds 1/16 in for damage during punching (B4.3), 和 [mathin] d_{h} = 1.25\,in [mathin] for the oversized holes in both directions.

L型: one shear plane, tension across the bolts to one edge

[math] A_{nt} = t_{fp} \left[ (n_{r} – 2)\, s_{r} + g_{a} + L_{ev,fp} – (n_{r} – 0.5)\left( d_{h} + 0.0625\,in \right) \right] [math]

[math] A_{nt} = (0.5\,in) \times \left[ (2 – 2)(3\,in) + 5.5\,in + 3.625\,in – (2 – 0.5)(1.25\,in + 0.0625\,in) \right] = 3.5781\,in^{2} [math]

[math] A_{gv} = t_{fp} \left[ L_{eh,fp} + (n_{c} – 1)\, s_{c} \right] = (0.5\,in) \times \left[ 1.5\,in + (4 – 1)(3\,in) \right] = 5.25\,in^{2} [math]

[math] A_{nv} = t_{fp} \left[ L_{eh,fp} + (n_{c} – 1)\, s_{c} – (n_{c} – 0.5)\left( d_{h} + 0.0625\,in \right) \right] = (0.5\,in) \times \left[ 1.5\,in + (3)(3\,in) – (3.5)(1.3125\,in) \right] = 2.9531\,in^{2} [math]

和 [mathin] U_{bs} = 1 [mathin] for uniform tension:

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

[math] \phi R_{n} = (0.75) \left( 0.6 \times 65 \times 2.9531 + 1 \times 65 \times 3.5781 \right) \le (0.75) \left( 0.6 \times 50 \times 5.25 + 1 \times 65 \times 3.5781 \right) [math]

[math] \phi R_{n} = 260.81\,kip [math]

U型: two shear planes, tension between the bolt lines

[math] A_{gv} = 2\, t_{fp} \left[ L_{eh,fp} + (n_{c} – 1)\, s_{c} \right] = 2 \times (0.5\,in) \times \left[ 1.5\,in + (3)(3\,in) \right] = 10.5\,in^{2} [math]

[math] A_{nv} = 2\, t_{fp} \left[ L_{eh,fp} + (n_{c} – 1)\, s_{c} – (n_{c} – 0.5)\left( d_{h} + 0.0625\,in \right) \right] = 5.9062\,in^{2} [math]

[math] A_{nt} = t_{fp} \left[ (n_{r} – 2)\, s_{r} + g_{a} – (n_{r} – 1)\left( d_{h} + 0.0625\,in \right) \right] = (0.5\,in) \times \left[ 0 + 5.5\,in – (1)(1.3125\,in) \right] = 2.0937\,in^{2} [math]

[math] \phi R_{n} = (0.75) \left( 0.6 \times 65 \times 5.9062 + 1 \times 65 \times 2.0937 \right) \le (0.75) \left( 0.6 \times 50 \times 10.5 + 1 \times 65 \times 2.0937 \right) = 274.83\,kip [math]

2L型: two shear planes, tension from each bolt line to its edge

[math] A_{nt} = 2\, t_{fp} \left[ L_{ev,fp} – 0.5\left( d_{h} + 0.0625\,in \right) \right] = 2 \times (0.5\,in) \times \left[ 3.625\,in – 0.5 \times (1.3125\,in) \right] = 2.9687\,in^{2} [math]

[math] \phi R_{n} = (0.75) \left( 0.6 \times 65 \times 5.9062 + 1 \times 65 \times 2.9687 \right) \le (0.75) \left( 0.6 \times 50 \times 10.5 + 1 \times 65 \times 2.9687 \right) = 317.48\,kip [math]

Governing block shear

[math] \phi R_{n} = \min\left( 260.81\,kip,\, 274.83\,kip,\, 317.48\,kip \right) = 260.81\,kip [math]

DCR = 204.31 基普 / 260.81 基普= 0.783 < 1.0, 行

The L pattern governs because it has only one shear plane. 的 12.75 in plate is wider than the 9.02 in beam flange, which gives the outer strips plenty of tension area in the 2L pattern.

步 6: Block shear of the beam flange (AISC 360-16 J4.3)

The beam flange can tear out the same way. Its edges are only 1.76 in outside the bolt lines, so the critical block is the 2L pattern: the two outer strips of flange shear along the bolt lines and pull away in tension.

[math] A_{nt} = t_{fb} \left[ b_{fb} – g_{a} – (n_{r} – 1)\left( d_{h} + 0.0625\,in \right) \right] = (0.77\,in) \times \left[ 9.02\,in – 5.5\,in – (1)(1.125\,in + 0.0625\,in) \right] = 1.796\,in^{2} [math]

[math] A_{gv} = 2\, t_{fb} \left[ L_{eh,bf} + (n_{c} – 1)\, s_{c} \right] = 2 \times (0.77\,in) \times \left[ 1.5\,in + (3)(3\,in) \right] = 16.17\,in^{2} [math]

[math] A_{nv} = 2\, t_{fb} \left[ L_{eh,bf} + (n_{c} – 1)\, s_{c} – (n_{c} – 0.5)\left( d_{h} + 0.0625\,in \right) \right] = 2 \times (0.77\,in) \times \left[ 10.5\,in – (3.5)(1.1875\,in) \right] = 9.7694\,in^{2} [math]

[math] \phi R_{n} = (0.75) \left( 0.6 \times 65 \times 9.7694 + 1 \times 65 \times 1.796 \right) \le (0.75) \left( 0.6 \times 50 \times 16.17 + 1 \times 65 \times 1.796 \right) [math]

[math] \phi R_{n} = 373.31\,kip [math]

DCR = 204.31 基普 / 373.31 基普= 0.547 < 1.0, 行

步 7: 法兰板的拉伸屈服和断裂 (AISC 360-16 J4.1)

On the tension side, the flange plate is a short tension member. Its gross section must not yield, and its net section through the row of holes must not rupture.

Tension yielding (情商. J4-1)

[math] \phi R_{n,ty} = \phi\, F_{yp}\, t_{fp}\, b_{fp} = (0.9) \times (50\,ksi) \times (0.5\,in) \times (12.75\,in) = 286.88\,kip [math]

Tension rupture (情商. J4-2)

[math] \phi R_{n,tr} = \phi\, F_{up}\, t_{fp} \left[ b_{fp} – n_{r}\left( d_{h} + 0.0625\,in \right) \right] [math]

[math] \phi R_{n,tr} = (0.75) \times (65\,ksi) \times (0.5\,in) \times \left[ 12.75\,in – (2) \times (1.25\,in + 0.0625\,in) \right] = 246.8\,kip [math]

[math] \phi R_{n} = \min\left( 286.88\,kip,\, 246.8\,kip \right) = 246.8\,kip [math]

DCR = 204.31 基普 / 246.8 基普= 0.828 < 1.0, 行

步 8: Tension rupture of the beam with shear lag (AISC 360-16 D3, J4.1)

The flange force has to get into the beam as well. Half the beam, a WT made of one flange and half the web, is treated as the tension member. The bolts connect only the flange, so the web lags behind in picking up stress, and the shear lag factor U reduces the effective net area.

From Table D3.1 case 2, 哪里 [mathin] \bar{y} = 3.0217\,in [mathin] is the distance from the flange to the centroid of the WT and l is the length of the connection, from the first bolt column to the last:

[math] U = 1 – \frac{\bar{y}}{l} = 0.66439 [math]

[math] \phi R_{n,tr} = \phi\, F_{ub}\, U \left[ 0.5\, A_{bm} – n_{r}\left( d_{h} + 0.0625\,in \right) t_{fb} \right] [math]

[math] \phi R_{n,tr} = (0.75) \times (65\,ksi) \times (0.66439) \times \left[ 0.5 \times (24.7\,in^{2}) – (2) \times (1.125\,in + 0.0625\,in) \times (0.77\,in) \right] [math]

[math] \phi R_{n,tr} = 340.78\,kip [math]

DCR = 204.31 基普 / 340.78 基普= 0.600 < 1.0, 行

步 9: Flexural rupture of the beam at the flange bolt holes (AISC 360-16 F13.1)

The bolt holes take material out of the beam’s tension flange right where the moment is largest. F13.1 checks whether those holes reduce the beam’s flexural strength, and if they do, by how much.

Gross and net area of the tension flange

[math] A_{fg} = b_{fb}\, t_{fb} = (9.02\,in) \times (0.77\,in) = 6.9454\,in^{2} [math]

[math] A_{fn} = A_{fg} – n_{r}\left( d_{h} + 0.0625\,in \right) t_{fb} = (6.9454\,in^{2}) – (2) \times (1.125\,in + 0.0625\,in) \times (0.77\,in) = 5.1166\,in^{2} [math]

Does tensile rupture apply?

[math] \frac{F_{y}}{F_{u}} = \frac{50\,ksi}{65\,ksi} = 0.76923 \le 0.8 \;\Rightarrow\; Y_{t} = 1 [math]

[math] F_{u} A_{fn} = (65\,ksi) \times (5.1166\,in^{2}) = 332.58\,kip [math]

[math] Y_{t} F_{y} A_{fg} = (1) \times (50\,ksi) \times (6.9454\,in^{2}) = 347.27\,kip [math]

手动计算可用于静定梁 [mathin] F_{u} A_{fn} \lt Y_{t} F_{y} A_{fg} [mathin], the net flange ruptures before the gross flange yields, so the limit state of tensile rupture applies.

Flexural strength at the holes (情商. F13-1)

[math] M_{n} = \frac{F_{ub} A_{fn}}{A_{fg}}\, S_{x} = \frac{(65\,ksi) \times (5.1166\,in^{2})}{(6.9454\,in^{2})} \times (196.46\,in^{3}) = 783.95\,kip\text{-}ft [math]

As a flange force, to compare with [mathin] P_{uF} [mathin]:

[math] \phi R_{n} = \frac{\phi\, M_{n}}{d_{bm} – t_{fb}} = \frac{(0.9) \times (783.95\,kip\text{-}ft)}{(24.1\,in) – (0.77\,in)} = 362.91\,kip [math]

DCR = 204.31 基普 / 362.91 基普= 0.563 < 1.0, 行

步 10: 法兰板的抗压强度 (AISC 360-16 J4.4)

On the compression side the flange plate is a short column. If it is stocky enough, it reaches its yield load without buckling, 和AISC 360-16 J4.4 lets you skip the full Chapter E column check.

The report takes the unbraced length of the plate as [mathin] L_{k,fp} = 3\,in [mathin] 与 [mathin] K = 1.2 [mathin], the recommended design value in Commentary Table C-A-7.1 for a member fixed against rotation at both ends but free to translate at one. For a rectangular plate, [mathin] r = t_{fp} / \sqrt{12} [mathin]:

[math] \frac{K L}{r} = \frac{(1.2) \times (3\,in)}{(0.5\,in) / \sqrt{12}} = 24.942 \le 25 [math]

Since KL/r is no more than 25, 情商. J4-6适用:

[math] \phi R_{n} = \phi\, F_{yp}\, t_{fp}\, b_{fp} = (0.9) \times (50\,ksi) \times (0.5\,in) \times (12.75\,in) = 286.88\,kip [math]

DCR = 204.31 基普 / 286.88 基普= 0.712 < 1.0, 行

The ratio of 24.942 is just under the limit of 25. A thinner plate or a longer unbraced length would push the plate into the Chapter E buckling equations.

步 11: Detailing limit at the column face

The flange plate weld sits in the gap between the column flange and the end of the beam, so the weld cannot be larger than the beam clearance.

[math] DCR = \frac{W}{clr} = \frac{0.25\,in}{0.5\,in} = 0.5 [math]

DCR = 0.500 < 1.0, 行

步 12: Fillet weld capacity, flange plate to column flange (AISC 360-16 J2.4)

The whole flange force passes through the fillet welds that join the flange plate to the column. Their fillet weld capacity depends on size, length and load direction. The base metal on either side of the weld, here the plate and the column flange, can also limit how much of the weld size counts.

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

[math] D_{max} = \frac{ \min\left( \frac{t_{fp}}{2} F_{up},\; t_{fs} F_{us} \right) }{3.09\,kip/in \times \left( F_{EXX} / 70\,ksi \right)} = \frac{ \min\left( (0.25\,in)(65\,ksi),\; (1.56\,in)(65\,ksi) \right) }{(3.09\,kip/in)(1)} = 5.2589 [math]

So up to 5.26/16 in of weld size counts. 的 1/4 在 (4/16 在) weld is below that, so its full size is used.

Effective length

The weld runs across the plate width on both faces of the plate, each line reduced by 2W at the ends. The plate is narrower than the column flange, so the plate width sets the length:

[math] L_{w} = 2 \left[ \min\left( b_{fp},\, b_{fs} \right) – 2W \right] = 2 \times \left[ \min\left( 12.75\,in,\, 15.8\,in \right) – 2 \times (0.25\,in) \right] = 24.5\,in [math]

焊接强度

The flange force acts perpendicular to the weld lines, 所以 [mathin] \theta = 90^{\circ} [mathin] and the directional factor [mathin] 1 + 0.5 \sin^{1.5}\theta [mathin] 是 1.5:

[math] \phi R_{n} = \phi\, 0.6\, F_{EXX} \frac{\sqrt{2}}{2} \min\left( W,\, \frac{D_{max}}{16} \right) (1.5)\, L_{w} [math]

[math] \phi R_{n} = (0.75) \times 0.6 \times (70\,ksi) \times \frac{\sqrt{2}}{2} \times (0.25\,in) \times (1.5) \times (24.5\,in) [math]

[math] \phi R_{n} = 204.64\,kip [math]

DCR = 204.31 基普 / 204.64 基普= 0.998 < 1.0, 行

This is the governing check of the whole connection. 的 1/4 in weld carries the flange force with almost nothing to spare, and only because the 1.5 directional factor applies. 没有它, the weld would need to be about 3/8 在.

步 13: Column flange and web under the flange force (AISC 360-16 J10)

Each flange plate delivers its force to the column as a concentrated load. The column flange can bend, the web can yield, cripple or buckle under it, and the panel zone has to carry the moment as shear. If any of these fail, the column needs stiffeners or a doubler plate, or a heavier section.

Flange local bending (情商. J10-1)

[math] \phi R_{n,flb} = \phi\, 6.25\, F_{ys}\, t_{fs}^{2} = (0.9) \times 6.25 \times (50\,ksi) \times (1.56\,in)^{2} = 684.45\,kip [math]

Web compression buckling (情商. J10-8)

The report uses the column’s T dimension, [mathin] T_{sup} = 10\,in [mathin], for the web depth in this equation, 与 [mathin] Q_{f} = 1 [mathin]:

[math] \phi R_{n,wcb} = \phi \left( \frac{24\, t_{ws}^{3} \sqrt{E\, F_{ys}}}{T_{sup}} \right) Q_{f} = (0.9) \times \left( \frac{24 \times (0.98\,in)^{3} \times \sqrt{(29000\,ksi) \times (50\,ksi)}}{(10\,in)} \right) \times (1) = 2448\,kip [math]

Strictly, J10.5 applies when compressive forces act on both column flanges at the same level, as when beams frame in from both sides. Checking it for a one-sided connection is conservative, and here it is nowhere near governing.

Web local crippling (情商. J10-4)

The length of bearing [mathin] l_{b} = 0.5\,in [mathin] is the flange plate thickness:

[math] \phi R_{n,wlc} = \phi\, 0.8\, t_{ws}^{2} \left[ 1 + 3 \left( \frac{l_{b}}{d_{sup}} \right) \left( \frac{t_{ws}}{t_{fs}} \right)^{1.5} \right] \sqrt{\frac{E\, F_{ys}\, t_{fs}}{t_{ws}}}\, Q_{f} [math]

[math] \phi R_{n,wlc} = (0.75) \times 0.8 \times (0.98\,in)^{2} \times \left[ 1 + 3 \times \left( \frac{0.5\,in}{15.7\,in} \right) \times \left( \frac{0.98\,in}{1.56\,in} \right)^{1.5} \right] \times \sqrt{\frac{(29000\,ksi) \times (50\,ksi) \times (1.56\,in)}{(0.98\,in)}} \times (1) = 917.11\,kip [math]

Web local yielding (情商. J10-2)

[math] \phi R_{n,wly} = \phi\, F_{ys}\, t_{ws} \left( 5\, k_{des} + l_{b} \right) = (1) \times (50\,ksi) \times (0.98\,in) \times \left( 5 \times 2.16\,in + 0.5\,in \right) = 553.7\,kip [math]

Panel zone web shear (情商. J10-9)

[math] \phi R_{n,wpz} = \phi\, 0.6\, F_{ys}\, d_{sup}\, t_{ws} = (0.9) \times 0.6 \times (50\,ksi) \times (15.7\,in) \times (0.98\,in) = 415.42\,kip [math]

Governing column strength

[math] \phi R_{n} = \min\left( 684.45,\, 2448,\, 917.11,\, 553.7,\, 415.42 \right) = 415.42\,kip [math]

DCR = 204.31 基普 / 415.42 基普= 0.492 < 1.0, 行

The W14x211 column has a 1.56 in flange and a 0.98 in web, so it needs no stiffeners or doubler plate for this flange force. Panel zone shear governs the column checks.

部分 2: the web plate connection

脚步 14 至 22 check the web plate, a single plate connection that carries the 250 kip shear. Because the flange plates make the joint fully restrained, the web plate does not have to rotate like a simple shear tab, and the checks that exist only to keep a shear tab flexible are not part of this design.

步 14: Detailing limits for the web plate (AISC 360-16 J3.3, J3.4, J3.5)

The web plate bolt layout has to meet the same spacing and edge distance rules, and the plate has to be deep enough to stabilise the beam web while still fitting between the beam flanges.

The plate has horizontal short-slotted holes. The horizontal edge distance is measured along the slot, 因此表 J3.5 添加了 1/8 in and the limit becomes 1.375 在. The vertical edge distance is measured across the slot and stays at 1.25 在.

[math] DCR = \frac{1.375\,in}{1.5\,in} = 0.917 [math]

DCR = 0.917 < 1.0, 行

步 15: Bolt group shear on the web plate (AISC 360-16 J3.6)

The ten web bolts share the 250 kip shear. In a simple shear tab, the eccentricity of the reaction would reduce the group to fewer than ten bolts’ 力量的价值. Here the report uses a bolt group coefficient C = 10, the full bolt count, because the end moment is taken by the flange plates and the web bolts carry the shear directly.

在哪里 [mathin] \phi = 0.75 [mathin], [mathin] F_{nv} = 68\,ksi [mathin] for an A490-N bolt, [mathin] N_{s} = 1 [mathin] shear plane, 和 [mathin] h_{f} = 1 [mathin] because there are no fillers (J5.2(一个)):

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

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

[math] \phi R_{n} = (0.75) \times (68\,ksi) \times \frac{\pi}{4} (1\,in)^{2} \times (1) \times (10) \times (1) [math]

[math] \phi R_{n} = 400.55\,kip [math]

DCR = 250 基普 / 400.55 基普= 0.624 < 1.0, 行

For a bolt group where eccentricity does reduce C, 的 使用 ICOR 进行螺栓组计算 shows how the coefficient is found.

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

Each bolt pushes down on its hole in the web plate. The bottom bolt in each line has the short clear distance to the plate edge, so it is the weak one. The slots run horizontally and the load is vertical, so the hole dimension across the slot, [mathin] d_{h} = 1.125\,in [mathin], 用来.

[math] l_{c1} = L_{ev,pl} – \frac{d_{h}}{2} = (1.5\,in) – \frac{(1.125\,in)}{2} = 0.9375\,in [math]

[math] l_{c2} = s_{r} – d_{h} = (3\,in) – (1.125\,in) = 1.875\,in [math]

One line of bolts is worked out and scaled by [mathin] C / n_{r} = 10 / 5 [mathin] for the 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] \frac{C}{n_{r}} [math]

[math] \phi R_{n} = (0.75) \times \left[ \min\left(1.125,\, 2.4\right)(1)(65) + \min\left(2.25,\, 2.4\right)(1)(65)(5 – 1) \right] \times \frac{10}{5} [math]

[math] \phi R_{n} = 987.19\,kip [math]

DCR = 250 基普 / 987.19 基普= 0.253 < 1.0, 行

步 17: 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) – (1.125\,in) = 1.875\,in [math]

[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 1.875\,in,\, 2.4 \times 1\,in\right) \times (0.47\,in) \times (65\,ksi) \times (10) [math]

[math] \phi R_{n} = 515.53\,kip [math]

DCR = 250 基普 / 515.53 基普= 0.485 < 1.0, 行

步 18: Effective strength of the web bolt group (AISC 360-16 J3.6 评论)

In practice 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.

[math] \phi r_{v} = \phi\, F_{nv}\, \frac{\pi}{4} d_{b}^{2}\, N_{s}\, h_{f} = (0.75) \times (68\,ksi) \times \frac{\pi}{4} (1\,in)^{2} \times (1) \times (1) = 40.055\,kip [math]

[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(1.125\,in,\, 2.4\,in\right) \times (1\,in) \times (65\,ksi) = 54.844\,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(2.25\,in,\, 2.4\,in\right) \times (1\,in) \times (65\,ksi) = 109.69\,kip [math]

[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(2.25\,in,\, 2.4\,in\right) \times (0.47\,in) \times (65\,ksi) = 51.553\,kip [math]

Bolts are numbered from the top of the line, so bolt 5 is the edge bolt on the plate:

[math] \sum \phi r_{eff} = n_{1}\, \phi r_{eff,1} + n_{2}\, \phi r_{eff,2} = (4) \times (40.055\,kip) + (1) \times (40.055\,kip) = 200.28\,kip [math]

[math] \phi R_{n,eff} = \frac{C}{n_{r}} \sum \phi r_{eff} = \frac{10}{5} \times (200.28\,kip) = 400.55\,kip [math]

DCR = 250 基普 / 400.55 基普= 0.624 < 1.0, 行

Every bolt is limited by AISC bolt shear, not by bolt bearing. 的 1 in plate and the 0.47 in web are thick enough that the bolts give out first, so the effective strength equals the bolt group shear strength from Step 15. You can run the same connection in the free connection design calculator and try a thinner web plate to see when bearing starts to govern.

步 19: Block shear of the web plate (AISC 360-16 J4.3)

A block of the web plate can tear out along the bolts: shear down the outer line of holes plus tension across the two bolt columns to the plate’s free edge.

The net tension area runs along the slots, so it uses the slot length, [mathin] d_{h} = 1.3125\,in [mathin]:

[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] = (1\,in) \times \left[ (1)(3\,in) + 1.5\,in – (1.5)(1.3125\,in + 0.0625\,in) \right] = 2.4375\,in^{2} [math]

Gross and net shear areas, 与 [mathin] d_{h} = 1.125\,in [mathin] across the slots:

[math] A_{gv} = t_{p} \left[ L_{ev,pl} + (n_{r} – 1)\, s_{r} \right] = (1\,in) \times \left[ 1.5\,in + (4)(3\,in) \right] = 13.5\,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] = (1\,in) \times \left[ 15\,in – 1.5\,in – (4.5)(1.1875\,in) \right] = 8.1563\,in^{2} [math]

With two bolt columns, the column nearer the edge carries more of the tension than the other, so the stress across the tension area is not uniform and [mathin] U_{bs} = 0.5 [mathin]:

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

[math] \phi R_{n} = (0.75) \left( 0.6 \times 65 \times 8.1563 + 0.5 \times 65 \times 2.4375 \right) \le (0.75) \left( 0.6 \times 50 \times 13.5 + 0.5 \times 65 \times 2.4375 \right) [math]

[math] \phi R_{n} = 297.98\,kip [math]

DCR = 250 基普 / 297.98 基普= 0.839 < 1.0, 行

步 20: Shear yielding and shear rupture of the web plate (AISC 360-16 J4.2)

The full shear passes through the web plate. 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} = (15\,in) \times (1\,in) = 15\,in^{2} [math]

[math] \phi R_{n,sy} = \phi\, 0.6\, F_{yp} A_{gv} = (1.00) \times 0.6 \times (50\,ksi) \times (15\,in^{2}) = 450\,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] = (1\,in) \times \left[ 15\,in – (5) \times (1.125\,in + 0.0625\,in) \right] = 9.0625\,in^{2} [math]

[math] \phi R_{n,sr} = \phi\, 0.6\, F_{up} A_{nv} = (0.75) \times 0.6 \times (65\,ksi) \times (9.0625\,in^{2}) = 265.08\,kip [math]

[math] \phi R_{n} = \min\left( 450\,kip,\, 265.08\,kip \right) = 265.08\,kip [math]

DCR = 250 基普 / 265.08 基普= 0.943 < 1.0, 行

Shear rupture governs the web plate. Five holes take almost 40% out of the shear area of the 15 在盘子里.

步 21: Detailing limit for the web plate weld

Like the flange plate weld, the web plate weld has to fit in the gap between the column flange and the end of the beam.

[math] DCR = \frac{W}{clr} = \frac{0.438\,in}{0.5\,in} = 0.876 [math]

DCR = 0.876 < 1.0, 行

步 22: Fillet weld capacity, web plate to column flange (AISC 360-16 J2.4)

最后的, the two fillet welds on the web plate have to carry the shear into the column flange.

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

The plate is welded on both sides, so half its thickness is available to each weld:

[math] D_{max} = \frac{ \min\left( \frac{t_{p}}{2} F_{up},\; t_{fs} F_{us} \right) }{3.09\,kip/in \times \left( F_{EXX} / 70\,ksi \right)} = \frac{ \min\left( (0.5\,in)(65\,ksi),\; (1.56\,in)(65\,ksi) \right) }{(3.09\,kip/in)(1)} = 10.518 [math]

的 7/16 in weld is well below 10.5/16 在, so its full size counts.

Effective length

[math] L_{w} = 2 \left( d_{p} – 2W \right) = 2 \times \left[ 15\,in – 2 \times (0.4375\,in) \right] = 28.25\,in [math]

焊接强度

The shear runs parallel to the welds, 所以 [mathin] \theta = 0 [mathin] 并且没有定向增加:

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

[math] \phi R_{n} = (0.75) \times 0.6 \times (70\,ksi) \times \frac{\sqrt{2}}{2} \times (0.4375\,in) \times \left[ 1 + 0.5 \sin^{1.5}(0) \right] \times (28.25\,in) [math]

[math] \phi R_{n} = 275.29\,kip [math]

DCR = 250 基普 / 275.29 基普= 0.908 < 1.0, 行

设计检查总结

Flange plate connection

Web plate connection

The connection passes every check. On the flange side, the weld from the flange plate to the column governs at 0.998, and slip of the slip-critical bolts is close behind at 0.978. Both are checked against the same 204.31 kip flange force, so this flange plate moment connection has almost no spare capacity at the flange plates. On the web side, shear rupture of the web plate governs at 0.943. The column needs no stiffeners, and its panel zone is the weakest column check at 0.492.

如果检查失败要更改什么

• Slip of the flange bolts: add a column of bolts, or use larger bolts for a higher pretension. Fewer fillers would raise the filler factor from 0.85 toward 1.0, but the shim stack is set by the erection gap.
• Weld to the column: increase the fillet size, as long as it still fits within the beam clearance, or use a wider plate for a longer weld.
• 法兰盘张力, block shear or bearing: use a thicker or wider plate, or a longer edge distance at the end of the plate.
• 法兰盘受压: keep KL/r at or below 25 with a thicker plate or a shorter unbraced length. 多于 25 the plate has to be checked as a column under Chapter E.
• Beam tension or flexural rupture at the holes: these depend on the beam flange and the hole layout, so use a heavier beam or fewer holes across the flange.
• Column limit states: add transverse stiffeners or a doubler plate, or use a heavier column if that is preferred.
• 腹板剪切破裂或块剪切: make the plate deeper with another bolt row, or thicker.
• Web plate weld: increase the weld size within the beam clearance, or make the plate deeper.

经常问的问题

Why are the flange plate bolts slip-critical?

The flange plates have oversized holes, which give the erector tolerance to fit the beam between the plates. AISC 360-16 J3.2 仅在滑动关键连接中允许过大的孔, so the hole type decides the bolt type. That in turn means a prepared faying surface, here Class B with a slip coefficient of 0.5. Fabricators normally set these choices before the project starts.

How is the flange force calculated from the moment?

The moment is divided by the distance between the centroids of the two beam flanges, the beam depth less one flange thickness. That gives a tension force in one flange and an equal compression force in the other. Half of the axial load is added to each flange plate, because the web plate is designed for shear only. In this example the flange force is 204.31 基普.

Why is there an erection gap and shim plate at the flange plate?

The gap is detailed on purpose. Beam depths vary within mill tolerance, and the gap lets the beam drop between the top and bottom flange plates during erection. Shims then fill whatever gap is left. Because the number of shims is not known in advance, the slip check uses the AISC 360-16 filler factor for two or more fillers, 0.85, so the design holds however many shims go in.

Why is the web plate designed differently from a simple shear tab?

A simple shear tab has to rotate, so it gets a maximum plate thickness, a weld sized to develop the plate, and a bolt group reduced for eccentricity. In a flange plate moment connection the flange plates make the joint fully restrained and take the moment. The web plate only carries the shear, so this example uses all ten bolts at full strength and has no ductility checks.

Does the column need stiffeners?

Not in this example. The W14x211 column passes flange local bending, 网络本地产量, 网络瘫痪, web compression buckling and panel zone shear, with the panel zone governing at a ratio of 0.492. If any of those fail, transverse stiffeners or a doubler plate are added, or a heavier column is used instead if that is preferred.

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. 条款编号和一些规定, 包括滑动关键螺栓的填充系数, 新版本中发生了变化, 因此，请根据您的项目使用的版本检查参考文献.

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 forces 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, 和 0.90 变成 1.67. 荷载组合也发生变化, so the flange force is lower.

Design a flange plate moment connection for free

此页上的每个数字都来自一个解决方案. Pick the flange plate configuration, 输入你的光束, 柱子, 螺栓, 片刻, 剪切和轴向载荷, 并在浏览器中查看每项检查. 没有什么可安装的.

免费设计连接

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.

购买连接设计模块

参考资料

• AISC 360-16, 钢结构建筑规范: B3.4, B4.3, D3 (表D3.1), F13.1 (情商. F13-1), J2.4 (情商. J2-4), J3.2, J3.3, J3.4 (表J3.4), J3.5 (表 J3.5), J3.6 (情商. J3-1) 及评论, J3.8 (情商. J3-4, Table J3.1), J3.10 (情商. J3-6a, J3-6c), J4.1 (情商. J4-1, J4-2), J4.2 (情商. J4-3, J4-4), J4.3 (情商. J4-5), J4.4 (情商. J4-6), J5.2(一个), J10.1 (情商. J10-1), J10.2 (情商. J10-2), J10.3 (情商. J10-4), J10.5 (情商. J10-8), J10.6 (情商. J10-9); Commentary Table C-A-7.1.
• AISC 钢结构手册, 15第版: 部分 1 截面属性; 部分 9, 情商. 9-2; 部分 10, 有效的焊接长度.
• 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