Conception de connexion à angle unique

Single angle connection design to AISC 360 means checking one angle, two bolt groups and the beam end against the beam reaction. On the beam side you check bolt shear, bearing and tear-out on the angle and the beam web, cisaillement de bloc, shear yielding and rupture of the angle, and angle flexure. If the beam is coped, you also check the coped web for block shear, cisaillement et flexion. On the support side you repeat the bolt and angle checks for the leg bolted to the girder, where the bolt group has to carry the eccentricity of the reaction. Detailing limits on spacing, distance au bord, gage, cope depth and connection depth apply throughout.

This page works one real example from start to finish: a W12x14 beam, coped at the top flange, framing into the web of a W18x35 girder with a 20 kip réaction LRFD, through a single L4x3x3/8 angle with two rows of 3/4 in A325-N bolts on each leg. The design is to AISC 360-16 et le manuel AISC 15e édition. Chaque équation est écrite avec ses chiffres, in the same order as the AISC calc sheet that the Conception de connexion SkyCiv module produces for this connection.

ESSAYEZ NOTRE OUTIL GRATUIT MAINTENANT ACHETER UN MODULE DE CONCEPTION DE CONNEXION

What is a single angle connection?

A single angle connection is one steel angle on one side of the beam web, with one leg bolted (or welded) to the web and the other leg bolted to the supporting member. It transfers the beam’s end shear and is classified as a simple connection under AISC 360-16 B3.4, so the angle is detailed to flex rather than resist moment.

It is a connexion de cisaillement used for lighter reactions, and where only one side of the beam web can be reached, such as beams framing into a girder web next to another member. Because there is only one angle, the beam-side bolts are in single shear and the load reaches the support off to one side of the beam web, so the support bolts have to carry that eccentricity. The other options for the same job are the connexion à double angle, l' plaque unique (languette de cisaillement) lien et la cisaille connexion par plaque d'extrémité. You can check this example directly in the single angle connection calculator.

Élévation de l'exemple de connexion. The beam is coped 2 7/8 in long and 1 in deep at the top to clear the girder flange, avec un 1/2 in clearance between the beam end and the girder web.

Section A-A. The long leg is bolted to the beam web, et la 3 in outstanding leg is bolted to the girder web.

How do you design a single angle connection? Les étapes en un coup d'œil

1. Check the detailing limits at the beam side: espacement, distances aux bords, profondeur, gage, cope.
2. Check the beam-side bolt group in single shear.
3. Check bolt bearing and tear-out on the angle at the beam side.
4. Vérifier le roulement des boulons et l'arrachement de l'âme de la poutre.
5. Combiner le cisaillement, bearing and tear-out per bolt at the beam side.
6. Check block shear of the angle at the beam side.
7. Check shear yielding and shear rupture of the angle at the beam side.
8. Check angle flexure at the beam side using the double coped procedure.
9. Check shear and moment interaction of the angle at the beam side.
10. Check block shear of the coped beam web.
11. Check shear yielding and shear rupture of the coped beam.
12. Check flexure of the coped beam.
13. Check the detailing limits at the support side.
14. Check the support-side bolt group with ICOR for its eccentricity.
15. Check bolt bearing and tear-out on the angle at the support side.
16. Check bolt bearing on the girder web.
17. Combiner le cisaillement, bearing and tear-out per bolt at the support side.
18. Check block shear of the angle at the support side.
19. Check shear yielding and shear rupture of the angle at the support side.
20. Check angle flexure at the support side using the double coped procedure.
21. Check shear and moment interaction of the angle at the support side.

Entrées de conception pour cet exemple

Le code de conception est AISC 360-16 LRFD. Les propriétés de la section proviennent du manuel AISC.

 

 

Étape 1: Limites de détail côté poutre (AISC 360-16 J3.3, J3.4, J3.5)

Avant tout contrôle de solidité, la disposition des boulons doit être constructible et correspondre aux hypothèses derrière les équations de résistance. Les boulons trop serrés se déchirent entre les trous, les bords trop rapprochés se déchirent, and an angle that is too short or too deep changes how the connection rotates. The cope has to be deep enough to clear the girder flange.

Pour un 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 in (Tableau J3.4). The beam-side leg has horizontal short-slotted holes, and the horizontal edge distance on the angle is measured across the end of the slot, so the Table J3.5 increment of 1/8 in for a short slot raises its minimum to 1.125 in. The horizontal edge distance on the beam web is the gage less the clearance, [mathin] 2.5\,in – 0.5\,in = 2\,in [mathin]. The angle has to be at least half as deep as the beam’s T dimension, so it is stiff enough to brace the beam end, and no deeper than the beam web can take. The bolt gage on the beam-side leg has to leave room to enter and tighten the bolts.

Pour une limite minimale, le rapport est la limite sur la valeur réelle, et pour un maximum, c'est un dépassement réel de la limite. The governing row is the minimum connection depth, half the 10 3/8 in T dimension of the W12x14:

[math] DCR = \frac{d}{c} = \frac{(5.1875)}{(6)} = 0.86458 [math]

DCR = 0.865 < 1.0, d'accord

Étape 2: Bolt group shear at the beam side (AISC 360-16 J3.6)

Each beam-side bolt passes through the angle and the beam web only, so it has one shear plane. The two bolts have to carry the full reaction between them.

Bolt group coefficient

The beam-side bolt line is 2.5 in from the angle heel, but the eccentricity is ignored here. AISC Manual page 10-116 allows this for a single vertical row of bolts with an eccentricity of no more than 3 in, so the bolt group coefficient is simply the number of bolts:

[math] C = 2 [math]

Force du groupe de boulons

Où [mathin] \phi = 0.75 [mathin], [mathin] d_{b} = 0.75\,in [mathin], [mathin] F_{nv} = 54\,ksi [mathin] for an A325-N bolt with threads included, [mathin] N_{s} = 1 [mathin] et [mathin] h_{f} = 1 [mathin] parce qu'il n'y a pas de remplissage (J5.2(a)):

[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 (2) \times (1) [math]

[math] \phi R_{n} = 35.785\,kip [math]

DCR = 20 kip / 35.785 kip = 0.559 < 1.0, d'accord

Étape 3: Bolt bearing and tear-out on the single angle at the beam side (AISC 360-16 J3.10)

Each beam-side bolt pushes on the edge of its hole in the angle. The angle either deforms at the hole (palier, plafonné à 2.4 d t Fu) or tears a strip out to the next hole or to its lower edge (arrachement, 1.2 lc t Fu). The beam pushes the bolts down on the angle, so the bottom bolt, nearest the angle’s lower edge, is the weak one.

The holes are short slots running horizontally and the load is vertical, so the hole dimension in the direction of the load is the standard [mathin] d_{h} = 0.8125\,in [mathin]. Distance libre pour le boulon de bord, avec [mathin] L_{ev,a} = 1.5\,in [mathin]:

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

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

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

Où [mathin] t_{a} = 0.375\,in [mathin], [mathin] F_{ua} = 58\,ksi [mathin], [mathin] C = 2 [mathin] et [mathin] n_{r} = 2 [mathin]:

[math] \phi R_{n} = \phi \left[ \min\left(1.2\, l_{c1},\, 2.4\, d_{b}\right) t_{a} F_{ua} + \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{a} F_{ua} \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( (2) – 1 \right) \right] \times \left( \frac{(2)}{(2)} \right) [math]

[math] \phi R_{n} = 50.773\,kip [math]

DCR = 20 kip / 50.773 kip = 0.394 < 1.0, d'accord

Tear-out governs the edge bolt, because 1.2 lc1 is less than 2.4 db. The other bolt is limited by bearing.

Étape 4: Appui de boulon et arrachement sur l'âme de la poutre (AISC 360-16 J3.10)

Le même contrôle s'applique à l'âme de la poutre. Here the web moves down against bolts held by the angle, so the bolts bear on the top of the web holes, and the edge bolt is the top one, 3.575 in below the cut edge of the cope.

Clear distance for the top bolt, avec [mathin] L_{ev,bm} = 3.575\,in [mathin] et [mathin] d_{h} = 0.8125\,in [mathin]:

[math] l_{c1} = L_{ev,bm} – \frac{d_{h}}{2} = (3.575\,in) – \frac{(0.8125\,in)}{2} = 3.1688\,in [math]

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

Où [mathin] t_{wb} = 0.2\,in [mathin], [mathin] F_{ub} = 65\,ksi [mathin], [mathin] C = 2 [mathin] et [mathin] n_{r} = 2 [mathin]:

[math] \phi R_{n} = \phi \left[ \min\left(1.2\, l_{c1},\, 2.4\, d_{b}\right) t_{wb} F_{ub} + \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{wb} F_{ub} \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 (3.1688\,in),\, 2.4 \times (0.75\,in)\right) \times (0.2\,in) \times (65\,ksi) + \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.2\,in) \times (65\,ksi) \times \left( (2) – 1 \right) \right] \times \left( \frac{(2)}{(2)} \right) [math]

[math] \phi R_{n} = 35.1\,kip [math]

DCR = 20 kip / 35.1 kip = 0.570 < 1.0, d'accord

Both bolts are capped by bearing at 2.4 db, because the long edge distance below the cope keeps tear-out out of play. Ce logiciel 0.2 dans le Web est plus mince que le 3/8 in angle, so it is the weaker ply on the beam side.

Étape 5: Effective strength of the bolt group at the beam side (AISC 360-16 J3.6 Commentaire)

Cisaille à boulon, le roulement et l'arrachement ne sont pas vraiment des contrôles séparés. Each bolt carries the least of its shear strength and its bearing or tear-out strength in every ply it passes through, et la force du groupe est la somme de ces valeurs par boulon.

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 angle, for the edge bolt and the inner bolt:

[math] \phi r_{n1e} = \phi\, \min\left(1.2\, l_{c1},\, 2.4\, d_{b}\right) t_{a} F_{ua} = (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_{a} F_{ua} = (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 and tear-out of one bolt on the beam web, for the edge bolt and the inner bolt:

[math] \phi r_{n2e} = \phi\, \min\left(1.2\, l_{c1},\, 2.4\, d_{b}\right) t_{wb} F_{ub} = (0.75) \times \min\left(1.2 \times (3.1688\,in),\, 2.4 \times (0.75\,in)\right) \times (0.2\,in) \times (65\,ksi) = 17.55\,kip [math]

[math] \phi r_{n2i} = \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 (0.2\,in) \times (65\,ksi) = 17.55\,kip [math]

Bolts are numbered from the top of the line. Verrouiller 1 is the edge bolt on the beam web and an inner bolt on the angle; verrouiller 2 is the edge bolt on the angle:

[math] \phi r_{eff,1} = \min\left( \phi r_{v},\, \phi r_{n1i},\, \phi r_{n2e} \right) = \min\left( (17.892\,kip),\, (29.363\,kip),\, (17.55\,kip) \right) = 17.55\,kip [math]

[math] \phi r_{eff,2} = \min\left( \phi r_{v},\, \phi r_{n1e},\, \phi r_{n2i} \right) = \min\left( (17.892\,kip),\, (21.41\,kip),\, (17.55\,kip) \right) = 17.55\,kip [math]

[math] \sum \phi r_{eff} = n_{1}\, \phi r_{eff,1} + n_{2}\, \phi r_{eff,2} = (1) \times (17.55\,kip) + (1) \times (17.55\,kip) = 35.1\,kip [math]

[math] \phi R_{n,eff} = \left( \frac{C}{n_{r}} \right) \sum \phi r_{eff} = \left( \frac{(2)}{(2)} \right) \times (35.1\,kip) = 35.1\,kip [math]

DCR = 20 kip / 35.1 kip = 0.570 < 1.0, d'accord

Bearing on the beam web governs both bolts, though bolt shear is only 2% plus haut. Vous pouvez exécuter la même connexion dans le calculateur de conception de connexion gratuit and try a third bolt row to see how the ratio moves.

Étape 6: Block shear of the single angle at the beam side (AISC 360-16 J4.3)

A block of the angle can tear out along the bolt line: shear down the line of holes, plus tension across to the angle’s free edge. Avec une colonne de boulons, c'est un bloc en forme de L.

Net area in tension. The tension plane runs horizontally, along the slots, so it uses the slot length [mathin] d_{h} = 1\,in [mathin], avec [mathin] t_{a} = 0.375\,in [mathin], [mathin] n_{c} = 1 [mathin], [mathin] s_{c} = 0\,in [mathin] et [mathin] L_{eh,a} = 1.5\,in [mathin]. Chaque trou ajoute 1/16 en cas de dommages lors du poinçonnage (B4.3):

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

[math] A_{nt} = (0.375\,in) \times \left[ ((1)-1) \times (0\,in) + (1.5\,in) – ((1) – 0.5) \times ((1\,in) + (0.0625\,in)) \right] = 0.36328\,in^{2} [math]

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

[math] A_{gv} = t_{a} \left[ L_{ev,a} + (n_{r}-1)\, s_{r} \right] = (0.375\,in) \times \left[ (1.5\,in) + ((2)-1) \times (3\,in) \right] = 1.6875\,in^{2} [math]

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

[math] A_{nv} = (0.375\,in) \times \left[ (6\,in) – (1.5\,in) – ((2) – 0.5) \times ((0.8125\,in) + (0.0625\,in)) \right] = 1.1953\,in^{2} [math]

Résistance au cisaillement des blocs, avec [mathin] U_{bs} = 1 [mathin] pour une tension uniforme, [mathin] F_{ua} = 58\,ksi [mathin], [mathin] F_{ya} = 36\,ksi [mathin] et [mathin] \phi = 0.75 [mathin]:

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

[math] \phi R_{n} = (0.75) \times \left( 0.6 \times (58\,ksi) \times (1.1953\,in^{2}) + (1) \times (58\,ksi) \times (0.36328\,in^{2}) \le 0.6 \times (36\,ksi) \times (1.6875\,in^{2}) + (1) \times (58\,ksi) \times (0.36328\,in^{2}) \right) [math]

[math] \phi R_{n} = 43.14\,kip [math]

DCR = 20 kip / 43.14 kip = 0.464 < 1.0, d'accord

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.

Étape 7: Shear yielding and shear rupture of the single angle at the beam side (AISC 360-16 J4.2)

The full reaction passes down through the angle leg as vertical shear. La section brute ne doit pas céder, et la section du filet à travers les trous de boulons ne doit pas se rompre.

Cédant au cisaillement (Eq. J4-3)

[math] A_{gv} = d_{a}\, t_{a} = (6\,in) \times (0.375\,in) = 2.25\,in^{2} [math]

[math] \phi R_{n,sy} = \phi\, 0.6\, F_{ya} A_{gv} = (1) \times 0.6 \times (36\,ksi) \times (2.25\,in^{2}) = 48.6\,kip [math]

Rupture par cisaillement (Eq. J4-4)

[math] A_{nv} = t_{a} \left[ d_{a} – n_{r}\left( d_{h} + 0.0625\,in \right) \right] = (0.375\,in) \times \left[ (6\,in) – (2) \times ((0.8125\,in) + (0.0625\,in)) \right] = 1.5938\,in^{2} [math]

[math] \phi R_{n,sr} = \phi\, 0.6\, F_{ua} A_{nv} = (0.75) \times 0.6 \times (58\,ksi) \times (1.5938\,in^{2}) = 41.597\,kip [math]

[math] \phi R_{n} = \min\left( \phi R_{n,sy},\, \phi R_{n,sr} \right) = \min\left( (48.6\,kip),\, (41.597\,kip) \right) = 41.597\,kip [math]

DCR = 20 kip / 41.597 kip = 0.481 < 1.0, d'accord

Étape 8: Flexure of the single angle at the beam side, procédure à double face (AISC 360-16 F11.2)

The reaction is carried from the beam-side bolt line to the heel of the angle, so the leg acts as a short cantilever plate in bending. Un profond, thin leg can buckle sideways before it yields. The AISC Manual handles this with the procedure for a beam coped at both flanges, treating the leg like a beam web with the flanges cut away. The unbraced length is the beam-side gage, [mathin] L_{b} = 2.5\,in [mathin]. The leg is also checked for flexural rupture through the bolt holes.

Range check

Où [mathin] E = 29000\,ksi [mathin], [mathin] F_{ya} = 36\,ksi [mathin], [mathin] d_{a} = 6\,in [mathin] et [mathin] t_{a} = 0.375\,in [mathin]:

[math] \frac{L_{b} d}{t^{2}} = \frac{(2.5\,in) \times (6\,in)}{(0.375\,in)^{2}} = 106.67 [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]

Donc [mathin] \frac{0.08E}{F_{y}} \lt \frac{L_{b} d}{t^{2}} \le \frac{1.9E}{F_{y}} [mathin], et éq. F11-2 s'applique.

Section moduli of the angle leg

[math] Z = \frac{t_{a} \left( d_{a} \right)^{2}}{4} = \frac{(0.375\,in) \times ((6\,in))^{2}}{4} = 3.375\,in^{3} [math]

[math] S = \frac{t_{a} \left( d_{a} \right)^{2}}{6} = \frac{(0.375\,in) \times ((6\,in))^{2}}{6} = 2.25\,in^{3} [math]

Facteur de modification du déversement (Éq. manuel AISC. 9-15)

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

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

[math] C_{b} = \left[ 3 + \ln\left( \frac{(2.5\,in)}{(6\,in)} \right) \right] \times \left( 1 – \frac{(0\,in)}{(6\,in)} \right) \ge 1.84 [math]

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

Flexural yielding and buckling strength

Avec [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_{ya}}{E} \right) \right] F_{ya} S \le F_{ya} Z \right] [math]

[math] \phi M_{n} = (0.9) \times \left[ (2.1245) \times \left[ 1.52 – 0.274 \times ((106.67)) \times \left( \frac{(36\,ksi)}{(29000\,ksi)} \right) \right] \times (36\,ksi) \times (2.25\,in^{3}) \le (36\,ksi) \times (3.375\,in^{3}) \right] [math]

[math] \phi M_{n} = 9.1125\,kip\text{-}ft \;(109.35\,kip\text{-}in) [math]

The plastic moment cap [mathin] F_{ya} Z [mathin] governs, so the leg is not prone to buckling at this depth and thickness. The equivalent shear capacity is:

[math] \phi V_{n,y} = \frac{\phi M_{n}}{L_{b}} = \frac{(9.1125\,kip\text{-}ft)}{(2.5\,in)} = 43.74\,kip [math]

Flexural rupture (Partie du manuel AISC 15 Le tableau 15-3, Partie 9 Eq. 9-4)

The net plastic section modulus removes the bolt holes, avec [mathin] d_{h} = 0.8125\,in [mathin] et [mathin] s_{r} = 3\,in [mathin]. With an even number of rows the last term is zero:

[math] Z_{net} = \frac{t_{a} \left( d_{a} \right)^{2}}{4} – t_{a} \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_{a} \left( d_{h} + 0.0625\,in \right)^{2}}{4} [math]

[math] Z_{net} = \frac{(0.375\,in) \times ((6\,in))^{2}}{4} – (0.375\,in) \times ((0.8125\,in) + (0.0625\,in)) \times (3\,in) \times \left\lfloor \frac{(2)^{2}}{4} \right\rfloor – \left[ (2)\ \text{odd} \right] \times \frac{(0.375\,in) \times ((0.8125\,in) + (0.0625\,in))^{2}}{4} = 2.3906\,in^{3} [math]

[math] \phi M_{n} = \phi\, F_{ua} Z_{net} = (0.75) \times (58\,ksi) \times (2.3906\,in^{3}) = 8.666\,kip\text{-}ft [math]

[math] \phi V_{n,r} = \frac{\phi M_{n}}{L_{b}} = \frac{(8.666\,kip\text{-}ft)}{(2.5\,in)} = 41.597\,kip [math]

Governing flexural capacity

[math] \phi V_{n} = \min\left( \phi V_{n,y},\, \phi V_{n,r} \right) = \min\left( (43.74\,kip),\, (41.597\,kip) \right) = 41.597\,kip [math]

DCR = 20 kip / 41.597 kip = 0.481 < 1.0, d'accord

Flexural rupture governs. It happens to give the same 41.597 kip as shear rupture in Step 7.

Étape 9: Shear and moment interaction of the single angle at the beam side (Éq. manuel AISC. 10-5)

The angle leg carries shear and bending at the same section at the same time, donc les deux doivent être vérifiés ensemble. The Manual uses a circular interaction between the shear yielding strength and the flexural strength.

The moment at the angle is the reaction times the unbraced length, avec [mathin] V_{r} = 20\,kip [mathin] et [mathin] L_{b} = 2.5\,in [mathin]:

[math] M_{r} = V_{r}\, L_{b} = (20\,kip) \times (2.5\,in) = 4.1667\,kip\text{-}ft [math]

Avec [mathin] \phi V_{n} = 48.6\,kip [mathin] from shear yielding in Step 7 et [mathin] \phi M_{n} = 9.1125\,kip\text{-}ft [mathin] de l'étape 8:

[math] I_{nt} = \left( \frac{V_{r}}{\phi V_{n}} \right)^{2} + \left( \frac{M_{r}}{\phi M_{n}} \right)^{2} [math]

[math] I_{nt} = \left( \frac{(20\,kip)}{(48.6\,kip)} \right)^{2} + \left( \frac{(4.1667\,kip\text{-}ft)}{(9.1125\,kip\text{-}ft)} \right)^{2} = 0.37843 [math]

DCR = 0.37843 / 1.0 = 0.378 < 1.0, d'accord

Étape 10: Block shear of the coped beam web (AISC 360-16 J4.3)

The top cope removes the beam’s top flange at the connection, so a block of the web can tear out along the bolt line: shear up the line of holes to the cut edge of the cope, plus tension across to the beam end.

Net area in tension, avec [mathin] t_{wb} = 0.2\,in [mathin], [mathin] L_{eh,bm} = 2\,in [mathin] and the standard hole [mathin] d_{h} = 0.8125\,in [mathin]:

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

[math] A_{nt} = (0.2\,in) \times \left[ ((1)-1) \times (0\,in) + (2\,in) – ((1) – 0.5) \times ((0.8125\,in) + (0.0625\,in)) \right] = 0.3125\,in^{2} [math]

Gross area in shear, from the cope edge down to the bottom bolt, avec [mathin] L_{ev,bm} = 3.575\,in [mathin]:

[math] A_{gv} = t_{wb} \left[ L_{ev,bm} + (n_{r}-1)\, s_{r} \right] = (0.2\,in) \times \left[ (3.575\,in) + ((2)-1) \times (3\,in) \right] = 1.315\,in^{2} [math]

Net area in shear. The report writes it with a depth term [mathin] d_{coped} = 10.15\,in [mathin]; the bracket [mathin] d_{coped} – L_{ev,bm} = 6.575\,in [mathin] is the same shear plane length as in the gross area:

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

[math] A_{nv} = (0.2\,in) \times \left[ (10.15\,in) – (3.575\,in) – ((2) – 0.5) \times ((0.8125\,in) + (0.0625\,in)) \right] = 1.0525\,in^{2} [math]

Avec [mathin] F_{ub} = 65\,ksi [mathin], [mathin] F_{yb} = 50\,ksi [mathin], [mathin] U_{bs} = 1 [mathin] et [mathin] \phi = 0.75 [mathin]:

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

[math] \phi R_{n} = (0.75) \times \left( 0.6 \times (65\,ksi) \times (1.0525\,in^{2}) + (1) \times (65\,ksi) \times (0.3125\,in^{2}) \le 0.6 \times (50\,ksi) \times (1.315\,in^{2}) + (1) \times (65\,ksi) \times (0.3125\,in^{2}) \right) [math]

[math] \phi R_{n} = 44.822\,kip [math]

DCR = 20 kip / 44.822 kip = 0.446 < 1.0, d'accord

Étape 11: Shear yielding and shear rupture of the coped beam (AISC 360-16 J4.2)

At the cope, the full reaction passes through the web that is left below the cut, without the top flange. That reduced web is checked for shear yielding on its gross area and shear rupture through the bolt holes.

The depth of web remaining is the beam depth less the cope depth, [mathin] d_{coped} = 11.9\,in – 1\,in = 10.9\,in [mathin].

Cédant au cisaillement (Eq. J4-3)

[math] A_{gv} = d_{coped}\, t_{wb} = (10.9\,in) \times (0.2\,in) = 2.18\,in^{2} [math]

[math] \phi R_{n,sy} = \phi\, 0.6\, F_{yb} A_{gv} = (1) \times 0.6 \times (50\,ksi) \times (2.18\,in^{2}) = 65.4\,kip [math]

Rupture par cisaillement (Eq. J4-4)

[math] A_{nv} = t_{wb} \left[ d_{coped} – n_{r}\left( d_{h} + 0.0625\,in \right) \right] = (0.2\,in) \times \left[ (10.9\,in) – (2) \times ((0.8125\,in) + (0.0625\,in)) \right] = 1.83\,in^{2} [math]

[math] \phi R_{n,sr} = \phi\, 0.6\, F_{ub} A_{nv} = (0.75) \times 0.6 \times (65\,ksi) \times (1.83\,in^{2}) = 53.528\,kip [math]

[math] \phi R_{n} = \min\left( (65.4\,kip),\, (53.528\,kip) \right) = 53.528\,kip [math]

DCR = 20 kip / 53.528 kip = 0.374 < 1.0, d'accord

Étape 12: Flexure of the coped beam (Partie du manuel AISC 9, Eq. 9-7)

The reaction acts at the beam end, so the coped section at the inside face of the cope carries a moment equal to the reaction times its distance from the bolts. Without its top flange the web there is a tee with an unbraced top edge in compression, which can buckle locally. The Manual’s coped beam procedure checks the web slenderness against a limit that depends on the cope geometry.

Entrées: [mathin] F_{yb} = 50\,ksi [mathin], [mathin] D_{ct} = 1\,in [mathin], [mathin] c = L_{ct} = 2.875\,in [mathin], no bottom cope, [mathin] \phi = 0.9 [mathin]. The section moduli of the tee remaining at the cope are [mathin] S_{x} = 5.6132\,in^{3} [mathin] et [mathin] Z_{x} = 9.5686\,in^{3} [mathin].

Web slenderness (Eq. 9-11)

[math] \lambda = \frac{h_{o}}{t_{wb}} = \frac{(10.9\,in)}{(0.2\,in)} = 54.5 [math]

Buckling adjustment factor (Eq. 9-14a) and plate buckling coefficient (Eq. 9-13a)

Puisque [mathin] c \le d_{bm} [mathin] et [mathin] c \le h_{o} [mathin]:

[math] f = 2 \left( \frac{c}{d_{bm}} \right) = 2 \times \left( \frac{(2.875\,in)}{(11.9\,in)} \right) = 0.48319 [math]

[math] k = 2.2 \left( \frac{h_{o}}{c} \right)^{1.65} = 2.2 \times \left( \frac{(10.9\,in)}{(2.875\,in)} \right)^{1.65} = 19.835 [math]

Limiting slenderness for a compact web (Eq. 9-12)

[math] \lambda_{p} = 0.475 \sqrt{\frac{\left( f k \ge 1.61 \right) E}{F_{yb}}} [math]

[math] \lambda_{p} = 0.475 \times \sqrt{\frac{\left( (0.48319) \times (19.835) \ge 1.61 \right) \times (29000\,ksi)}{(50\,ksi)}} = 35.414 [math]

Flexural strength (Eq. 9-7)

Puisque [mathin] \lambda_{p} \lt \lambda \le 2\lambda_{p} [mathin], the web is noncompact and the strength is interpolated between the plastic and elastic moments:

[math] \phi M_{n} = \phi \left[ F_{yb} Z_{x} – \left( F_{yb} Z_{x} – F_{yb} S_{x} \right) \left( \frac{\lambda}{\lambda_{p}} – 1 \right) \right] [math]

[math] \phi M_{n} = (0.9) \times \left[ (50\,ksi) \times (9.5686\,in^{3}) – \left( (50\,ksi) \times (9.5686\,in^{3}) – (50\,ksi) \times (5.6132\,in^{3}) \right) \times \left( \frac{(54.5)}{(35.414)} – 1 \right) \right] [math]

[math] \phi M_{n} = 27.889\,kip\text{-}ft [math]

Equivalent shear capacity

The lever arm is the cope eccentricity, the cope length plus the beam end clearance, [mathin] e_{c} = 2.875\,in + 0.5\,in = 3.375\,in [mathin]:

[math] \phi V_{n} = \frac{\phi M_{n}}{e_{c}} = \frac{(27.889\,kip\text{-}ft)}{(3.375\,in)} = 99.16\,kip [math]

DCR = 20 kip / 99.16 kip = 0.202 < 1.0, d'accord

Étape 13: Detailing limits at the support side (AISC 360-16 J3.3, J3.4, J3.5)

The outstanding leg is bolted to the girder web, so the same spacing and edge rules apply, plus a gage limit that keeps the bolts far enough out from the beam web to be entered and tightened. The angle holes here are standard, so the horizontal edge distance minimum is the plain 1 dans le tableau J3.4. The actual edge distance is what is left of the 3 in leg outside the 1.75 in gage, 1.25 in.

The governing row is the minimum bolt gage:

[math] DCR = \frac{d}{c} = \frac{(1.5)}{(1.75)} = 0.85714 [math]

DCR = 0.857 < 1.0, d'accord

Étape 14: Bolt group shear at the support side by the instantaneous center of rotation (AISC 360-16 J3.6)

This is where a single angle differs most from a double angle. The beam web sits to one side of the support bolt line, so the reaction reaches the girder bolts 1.85 in off their line, and the two bolts carry a moment as well as the shear. Le centre de rotation instantané (ICOR) La méthode trouve le point autour duquel le groupe de boulons tourne et le transforme en un coefficient C, the number of bolts’ valeur de la résistance au cisaillement que le groupe offre réellement.

Excentricité de conception

The eccentricity is the support gage plus half the beam web, measured to the beam web centreline:

[math] e = g + \frac{t_{wb}}{2} = 1.75\,in + \frac{0.2\,in}{2} = 1.85\,in [math]

Résistance d'un boulon

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

Localisation du centre instantané

Verrouiller 1 est au bas de la ligne, avec les coordonnées ci-dessous. Le CI est trouvé par itération, en utilisant la courbe charge-déformation de l'équation manuelle AISC (7-1) sur les pages 7-6 à 7-8. Ce logiciel calcul du groupe de boulons à l'aide d'ICOR parcourt cette itération dans son intégralité.

L'itération converge avec le IC à:

[math] x_{ic} = -1.2162\,in, \qquad y_{ic} = 1.5\,in [math]

Les réactions du boulon à cet endroit sont:

That gives the bolt group coefficient:

[math] C = 1.24 [math]

Force du groupe de boulons

Avec [mathin] N_{s} = 1 [mathin], [mathin] h_{f} = 1 [mathin] (J5.2(a)) et [mathin] \phi = 0.75 [mathin]:

[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 (1.24) \times (1) [math]

[math] \phi R_{n} = 22.187\,kip [math]

DCR = 20 kip / 22.187 kip = 0.901 < 1.0, d'accord

This is the governing check in the connection. With only two bolts 3 in apart, a 1.85 in eccentricity cuts the group from 2 boulons’ worth of strength to 1.24, a 38% perte.

Étape 15: Bolt bearing and tear-out on the single angle at the support side (AISC 360-16 J3.10)

On the support side the angle is pushed down while the girder web holds the bolts in place, so the bolts bear upward on the angle holes. The top bolt, nearest the angle’s upper edge, is the edge bolt here. The group strength is scaled by C/n to account for the eccentricity.

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

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

Où [mathin] t_{a} = 0.375\,in [mathin], [mathin] F_{ua} = 58\,ksi [mathin], [mathin] C = 1.24 [mathin] et [mathin] n_{r} = 2 [mathin]:

[math] \phi R_{n} = \phi \left[ \min\left(1.2\, l_{c1},\, 2.4\, d_{b}\right) t_{a} F_{ua} + \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{a} F_{ua} \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( (2) – 1 \right) \right] \times \left( \frac{(1.24)}{(2)} \right) [math]

[math] \phi R_{n} = 31.479\,kip [math]

DCR = 20 kip / 31.479 kip = 0.635 < 1.0, d'accord

Étape 16: Bolt bearing on the girder web (AISC 360-16 J3.10)

The bolts also bear on the girder web. The girder runs past the connection in both directions, 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]

Où [mathin] t_{ws} = 0.3\,in [mathin], [mathin] F_{us} = 65\,ksi [mathin] et [mathin] C = 1.24 [mathin]:

[math] \phi R_{n} = \phi\, \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{ws} F_{us}\, 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 (0.3\,in) \times (65\,ksi) \times (1.24) [math]

[math] \phi R_{n} = 32.643\,kip [math]

DCR = 20 kip / 32.643 kip = 0.613 < 1.0, d'accord

Étape 17: Effective strength of the bolt group at the support side (AISC 360-16 J3.6 Commentaire)

As at the beam side, each bolt delivers the least of its shear strength and its bearing or tear-out strength in each ply, and the group strength is the sum, scaled here by C/n for the eccentricity.

[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 angle, for the edge bolt and the inner bolt:

[math] \phi r_{n1e} = (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} = (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 girder web:

[math] \phi r_{n2} = \phi\, \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{ws} F_{us} = (0.75) \times \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.3\,in) \times (65\,ksi) = 26.325\,kip [math]

Bolts are numbered from the top of the line, alors boulon 1 is the edge bolt on the support side:

[math] \phi r_{eff,1} = \min\left( \phi r_{v},\, \phi r_{n1e},\, \phi r_{n2} \right) = \min\left( (17.892\,kip),\, (21.41\,kip),\, (26.325\,kip) \right) = 17.892\,kip [math]

[math] \phi r_{eff,2} = \min\left( \phi r_{v},\, \phi r_{n1i},\, \phi r_{n2} \right) = \min\left( (17.892\,kip),\, (29.363\,kip),\, (26.325\,kip) \right) = 17.892\,kip [math]

[math] \sum \phi r_{eff} = n_{1}\, \phi r_{eff,1} + n_{2}\, \phi r_{eff,2} = (1) \times (17.892\,kip) + (1) \times (17.892\,kip) = 35.785\,kip [math]

[math] \phi R_{n,eff} = \left( \frac{C}{n_{r}} \right) \sum \phi r_{eff} = \left( \frac{(1.24)}{(2)} \right) \times (35.785\,kip) = 22.187\,kip [math]

DCR = 20 kip / 22.187 kip = 0.901 < 1.0, d'accord

Bolt shear governs both bolts, so the effective strength equals the bolt group shear strength from Step 14.

Étape 18: Block shear of the single angle at the support side (AISC 360-16 J4.3)

The same L-shaped block can tear out of the outstanding leg. The holes here are standard, so both the tension and shear planes use [mathin] d_{h} = 0.8125\,in [mathin], and the tension leg is the 1.25 in horizontal edge distance.

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

[math] A_{nt} = (0.375\,in) \times \left[ ((1)-1) \times (0\,in) + (1.25\,in) – ((1) – 0.5) \times ((0.8125\,in) + (0.0625\,in)) \right] = 0.30469\,in^{2} [math]

[math] A_{gv} = t_{a} \left[ L_{ev,a} + (n_{r}-1)\, s_{r} \right] = (0.375\,in) \times \left[ (1.5\,in) + ((2)-1) \times (3\,in) \right] = 1.6875\,in^{2} [math]

[math] A_{nv} = (0.375\,in) \times \left[ (6\,in) – (1.5\,in) – ((2) – 0.5) \times ((0.8125\,in) + (0.0625\,in)) \right] = 1.1953\,in^{2} [math]

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

[math] \phi R_{n} = (0.75) \times \left( 0.6 \times (58\,ksi) \times (1.1953\,in^{2}) + (1) \times (58\,ksi) \times (0.30469\,in^{2}) \le 0.6 \times (36\,ksi) \times (1.6875\,in^{2}) + (1) \times (58\,ksi) \times (0.30469\,in^{2}) \right) [math]

[math] \phi R_{n} = 40.591\,kip [math]

DCR = 20 kip / 40.591 kip = 0.493 < 1.0, d'accord

Étape 19: Shear yielding and shear rupture of the single angle at the support side (AISC 360-16 J4.2)

The outstanding leg carries the reaction into the girder as vertical shear, so it gets the same gross and net section checks as the beam-side leg. Both legs are 6 in long with two standard-width holes across the load, so the numbers match Step 7.

[math] \phi R_{n,sy} = \phi\, 0.6\, F_{ya} A_{gv} = (1) \times 0.6 \times (36\,ksi) \times (2.25\,in^{2}) = 48.6\,kip [math]

[math] \phi R_{n,sr} = \phi\, 0.6\, F_{ua} A_{nv} = (0.75) \times 0.6 \times (58\,ksi) \times (1.5938\,in^{2}) = 41.597\,kip [math]

[math] \phi R_{n} = \min\left( (48.6\,kip),\, (41.597\,kip) \right) = 41.597\,kip [math]

DCR = 20 kip / 41.597 kip = 0.481 < 1.0, d'accord

Étape 20: Flexure of the single angle at the support side, procédure à double face (AISC 360-16 F11.2)

The outstanding leg bends between the support bolt line and the beam web. The lever arm is the support eccentricity from Step 14, [mathin] L_{b} = 1.85\,in [mathin].

Range check

[math] \frac{L_{b} d}{t^{2}} = \frac{(1.85\,in) \times (6\,in)}{(0.375\,in)^{2}} = 78.933 [math]

The limits are the same as Step 8, 64.444 et 1530.6, so Eq. F11-2 s'applique. The section moduli are also the same: [mathin] Z = 3.375\,in^{3} [mathin] et [mathin] S = 2.25\,in^{3} [mathin].

Facteur de modification du déversement (Éq. manuel AISC. 9-15)

[math] C_{b} = \left[ 3 + \ln\left( \frac{(1.85\,in)}{(6\,in)} \right) \right] \times \left( 1 – \frac{(0\,in)}{(6\,in)} \right) \ge 1.84 [math]

[math] C_{b} = 1.84 [math]

Flexural yielding and buckling strength

[math] \phi M_{n} = (0.9) \times \left[ (1.84) \times \left[ 1.52 – 0.274 \times ((78.933)) \times \left( \frac{(36\,ksi)}{(29000\,ksi)} \right) \right] \times (36\,ksi) \times (2.25\,in^{3}) \le (36\,ksi) \times (3.375\,in^{3}) \right] [math]

[math] \phi M_{n} = 9.1125\,kip\text{-}ft [math]

[math] \phi V_{n,y} = \frac{\phi M_{n}}{L_{b}} = \frac{(9.1125\,kip\text{-}ft)}{(1.85\,in)} = 59.108\,kip [math]

Flexural rupture

The net plastic section modulus is the same as Step 8, [mathin] Z_{net} = 2.3906\,in^{3} [mathin]:

[math] \phi M_{n} = \phi\, F_{ua} Z_{net} = (0.75) \times (58\,ksi) \times (2.3906\,in^{3}) = 8.666\,kip\text{-}ft [math]

[math] \phi V_{n,r} = \frac{\phi M_{n}}{L_{b}} = \frac{(8.666\,kip\text{-}ft)}{(1.85\,in)} = 56.212\,kip [math]

Governing flexural capacity

[math] \phi V_{n} = \min\left( (59.108\,kip),\, (56.212\,kip) \right) = 56.212\,kip [math]

DCR = 20 kip / 56.212 kip = 0.356 < 1.0, d'accord

The moment capacities match the beam side. Only the lever arm is shorter, 1.85 en contre 2.5 in, so the equivalent shear capacity is higher.

Étape 21: Shear and moment interaction of the single angle at the support side (Éq. manuel AISC. 10-5)

The support side uses the same circular interaction as Step 9, with the shorter 1.85 in lever arm.

[math] M_{r} = V_{r}\, L_{b} = (20\,kip) \times (1.85\,in) = 3.0833\,kip\text{-}ft [math]

[math] I_{nt} = \left( \frac{V_{r}}{\phi V_{n}} \right)^{2} + \left( \frac{M_{r}}{\phi M_{n}} \right)^{2} [math]

[math] I_{nt} = \left( \frac{(20\,kip)}{(48.6\,kip)} \right)^{2} + \left( \frac{(3.0833\,kip\text{-}ft)}{(9.1125\,kip\text{-}ft)} \right)^{2} = 0.28384 [math]

DCR = 0.28384 / 1.0 = 0.284 < 1.0, d'accord

Summary of the single angle connection design checks

La connexion passe tous les contrôles. The highest ratio, 0.901, is bolt shear at the girder, where the two bolts carry the reaction at a 1.85 in eccentricity and deliver only 1.24 boulons’ worth of strength. The effective strength check at the support gives the same 0.901, because bolt shear is the lowest value for both bolts. The next highest are the two detailing limits, 0.865 for the minimum angle depth and 0.857 for the support gage. On the beam side, bearing on the 0.2 in web governs at 0.570, and the coped beam has plenty of reserve.

Que changer si une vérification échoue

• Bolt shear or effective strength at the support: ajouter une rangée de boulons, which means a deeper angle and a higher C, ou utilisez un boulon plus gros. A325-X (threads excluded) or A490 bolts raise the shear stress. Reducing the support gage shortens the eccentricity, within the minimum gage limit.
• Appui sur l'âme de la poutre: add a bolt row or use a larger bolt, which raises the 2.4 d cap on bearing.
• Tear-out on the angle: increase the vertical edge distance at the top or bottom of the angle, or use a thicker angle.
• Cisaillement de bloc, shear rupture or shear yielding of the angle: use a thicker or deeper angle. A longer horizontal edge distance also raises the block shear tension area.
• Angle flexure or interaction: reduce the gage so the lever arm L_b is shorter, or use a thicker angle.
• Coped beam block shear, shear or flexure: make the cope shorter or shallower, move the bolts further from the cope, or add a bolt row. A deeper beam or a heavier web helps all three.
• A double angle instead: if the support bolts cannot be made to work, a connexion à double angle removes the eccentricity at the support and doubles the beam-side shear planes.

Questions fréquentes

Why is the eccentricity ignored at the beam but not at the girder?

AISC Manual page 10-116 lets you neglect the eccentricity of a single vertical row of bolts when it is no more than 3 in, alors le 2.5 in eccentricity at the beam-side bolts is ignored and C equals the bolt count, 2. At the girder, the bolt line sits 1.85 in to one side of the beam web, and the report designs those bolts by the instantaneous center of rotation method, which gives C = 1.24.

Why does the support bolt group govern a single angle connection?

There is only one line of bolts at the support, each in single shear, and the reaction arrives off to one side of that line. The eccentricity makes the bolts resist a moment as well as the shear, which reduces the group to 1.24 boulons’ worth of strength in this example. A double angle avoids this because the two angles sit symmetrically about the beam web.

Why is the beam coped, and what does the cope change?

The beam’s top flange is at the same level as the girder’s top flange, so it is cut back 2 7/8 in long and 1 in deep to clear the girder flange. Without its top flange the beam end is weaker, so the coped web gets its own checks for block shear to the cut edge, shear on the reduced depth, and local buckling of the web under the moment at the cope.

Why use short-slotted holes in the leg on the beam web?

Horizontal short slots in the beam-side leg give the erector fit-up tolerance along the beam, which helps when the beam length or the girder position is slightly off. Because the slots are horizontal and the load is vertical, bearing and tear-out use the standard hole dimension. The slot length only appears in the block shear tension area and in the edge distance increment from Table J3.5.

Why does the edge bolt switch between the beam web and the angle?

At the beam side the beam web moves down against the bolts, so the bolts bear on the top of the web holes and the top bolt, nearest the cope, is the edge bolt in the web. The angle resists that load, so the bolts bear on the bottom of its holes and the bottom bolt is its edge bolt. At the girder the angle is pushed down against bolts held by the girder, so the top bolt is the edge bolt on the angle.

When should you use a single angle instead of a double angle?

A single angle suits lighter reactions and places where only one side of the beam web can be reached, for example when another beam frames into the opposite side of the girder at the same point. It uses less material and fewer bolts, but its eccentricity at the support costs bolt capacity, and the beam-side bolts are in single shear instead of double shear.

Cet exemple suit-il la 15e édition de l'AISC ou le manuel de la 16e édition de l'AISC?

Il suit l'AISC 360-16 et le manuel AISC 15e édition, qui est écrit autour de cette spécification. Le manuel AISC 16e édition s'associe au dernier 360-22 spécification. The limit states for a single angle connection are the same in both, mais les références aux clauses et aux pages se déplacent, alors vérifiez-les par rapport à l'édition utilisée par votre projet.

Comment cela changerait-il pour AISC ASD au lieu de LRFD?

Les états limites et les résistances nominales sont identiques. Avec AISC ASD, vous comparez la réaction au niveau de service à la résistance nominale divisée par un facteur de sécurité au lieu de la multiplier par un facteur de résistance.. Un facteur de résistance de 0.75 devient un facteur de sécurité de 2.00, shear yielding’s 1.00 devient 1.50, and flexure’s 0.90 devient 1.67. Les combinaisons de charges changent également, donc la demande est plus faible.

Design a single angle connection for free

Chaque numéro sur cette page provient d'une seule solution. Pick the single angle configuration, entrez votre faisceau, soutien, angle, boulons, cope and reaction, et voyez chacune de ces vérifications dans votre navigateur. Il n'y a rien à installer.

CONCEVOIR UNE CONNEXION GRATUITEMENT

Le module SkyCiv Connection Design est le logiciel de conception de connexion derrière cet exemple. Il vous donne le rapport de calcul complet avec chaque équation et vos chiffres, comme les étapes ci-dessus, plus toutes les capacités et DCR, selon AISC 360-16 (LRFD et TSA) et AS 4100:2020. Ce logiciel calculateur de boulons covers the bolt group checks on their own.

ACHETER UN MODULE DE CONCEPTION DE CONNEXION

Références

• AISC 360-16, Spécification pour les bâtiments en acier de construction: B3.4, B4.3, F11.2 (Eq. F11-2), J3.3, J3.4 (Tableau J3.4), J3.5 (Tableau J3.5), J3.6 (Eq. J3-1) et commentaire, J3.10 (Eq. J3-6a, J3-6c), J4.2 (Eq. J4-3, J4-4), J4.3 (Eq. J4-5), J5.2(a).
• Manuel de construction en acier AISC, 15e édition: Partie 7, Eq. 7-1 (des pages 7-6 à 7-8); Partie 9, Eq. 9-4, 9-7, 9-11, 9-12, 9-13a, 9-14un et 9-15; Partie 10, de SkiCiv 10-116 et éq. 10-5 (de SkiCiv 10-90); Partie 15, Le tableau 15-3.
• Calcul du groupe de boulons à l'aide d'ICOR
• Conception d'assemblages en acier utilisant AISC 360-16
• Boulons dans la conception de connexion SkyCiv