Double Angle Connection Design

Double angle connection design to AISC 360 means checking two bolt groups and the angles between them against the beam end reaction. On the beam side you check bolt shear, bearing and tear-out on the angles and the beam web, διάτμηση μπλοκ, shear yielding and rupture of the angles, and angle flexure. On the support side you repeat those checks for the outstanding legs bolted to the column. Detailing limits on spacing, Διατμητική Αντοχή Μπουλονιών, gage and connection depth apply to both sides.

This page works one real example from start to finish: a W16x26 beam framing into the flange of a W14x90 column, με α 50 kip LRFD reaction, a pair of 2L4x3-1/2×1/4 angles and three rows of 3/4 in A325-N bolts on each side, σχεδιασμένο για AISC 360-16 and the AISC Manual 15th Edition. Every equation is written out with its numbers, in the same order as the AISC calc sheet that the Σχεδιασμός σύνδεσης SkyCiv module produces for this connection.

ΔΟΚΙΜΑΣΤΕ ΤΟ ΔΩΡΕΑΝ ΕΡΓΑΛΕΙΟ ΜΑΣ ΤΩΡΑ ΑΓΟΡΑ ΣΧΕΔΙΑΣΜΟΥ ΣΥΝΔΕΣΗΣ

What is a double angle connection?

A double angle connection is a pair of steel angles, ένα σε κάθε πλευρά του ιστού της δοκού, 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 angles are detailed to flex rather than resist moment.

It is a common σύνδεση διάτμησης for beams framing into column flanges, column webs and girders. Because the beam web sits between two angles, the beam-side bolts work in double shear, and the connection is symmetric, so it does not twist the beam. The other options for the same job are the ενιαίο πιάτο (shear tab) Η σύνδεση, ο single angle connection and the shear end plate connection. You can check this example directly in the double angle connection calculator.

Elevation of the example connection. The angles are bolted to the column flange and to the beam web, με α 1/2 in clearance between the beam end and the column.

Section A-A. The long legs are bolted to the column at a 5 1/4 in gage, through horizontal short-slotted holes in the angles.

How do you design a double angle connection? The steps at a glance

1. Check the detailing limits at the beam side: απόσταση, αποστάσεις ακμών, βάθος, gage.
2. Check the beam-side bolt group in double shear.
3. Check bolt bearing and tear-out on the angles at the beam side.
4. Check bolt bearing and tear-out on the beam web.
5. Combine shear, bearing and tear-out per bolt at the beam side.
6. Check block shear of the angles at the beam side.
7. Check shear yielding and shear rupture of the angles 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 angles at the beam side.
10. Check the detailing limits at the support side.
11. Check the support-side bolt group in single shear.
12. Check bolt bearing and tear-out on the angles at the support side.
13. Check bolt bearing and tear-out on the column flange.
14. Combine shear, bearing and tear-out per bolt at the support side.
15. Check block shear of the angles at the support side.
16. Check shear yielding and shear rupture of the angles at the support side.
17. Check angle flexure at the support side using the double coped procedure.
18. Check shear and moment interaction of the angles at the support side.

Design inputs for this example

The design code is AISC 360-16 LRFD. Section properties are from the AISC Manual.

 

 

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

Before any strength check, 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 angles that are too short or too deep change how the connection rotates.

Για ένα 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 σε (Table J3.4). The maximum spacing is 24 times the thinner part, here the 1/4 in angle or the 1/4 in beam web, which gives 6 σε (J3.5). The angles have to be at least half as deep as the beam’s T dimension, so they are stiff enough to brace the beam end, and no deeper than the beam web can take. The bolt gage on the beam-side legs has to leave room to enter and tighten the bolts.

For a minimum limit the ratio is limit over actual, and for a maximum it is actual over limit. The governing row is the minimum connection depth, half the 13 5/8 in T dimension of the W16x26:

[math] DCR = \frac{d}{c} = \frac{6.8125\,in}{9\,in} = 0.75694 [math]

DCR = 0.757 < 1.0, Εντάξει

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

The beam web is sandwiched between the two angles, so every beam-side bolt is in double shear. The bolt group has to carry the full reaction through those shear planes.

Bolt group coefficient

The reaction acts 2 in from the beam-side bolt line, but the eccentricity is ignored here. AISC Manual page 10-9 allows this for a single vertical row of bolts with an eccentricity of no more than 3 σε, so the bolt group coefficient is simply the number of bolts:

[math] C = 3 [math]

Bolt group strength

Οπου [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} = 2 [mathin] shear planes and [mathin] h_{f} [mathin] is the filler factor from J5.2(ένα). There are no fillers, Έτσι:

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

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

[math] h_{f} = 1 [math]

[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 (2) \times (3) \times (1) [math]

[math] \phi R_{n} = 107.35\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 107.35 kip = 0.466 < 1.0, Εντάξει

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

Each beam-side bolt pushes on the edge of its hole in both angles. The angle either deforms at the hole (ρουλεμάν, capped at 2.4 Διατμητική Αντοχή Μπουλονιών) or tears a strip out to the next hole or to the bottom edge (tear-out, 1.2 Διατμητική Αντοχή Μπουλονιών). The beam pushes the bolts down, so the bottom bolt, nearest the angle’s lower edge, is the weak one.

Clear distance for the edge bolt, με [mathin] L_{ev,a} = 1.5\,in [mathin] and the standard hole dimension [mathin] d_{h} = 0.8125\,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 the inner bolts, με [mathin] s_{r} = 3\,in [mathin]:

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

Both angles bear on each bolt, so the ply thickness is [mathin] 2t_{a} = 0.5\,in [mathin], με [mathin] F_{ua} = 58\,ksi [mathin], [mathin] C = 3 [mathin] και [mathin] n_{r} = 3 [mathin]:

[math] \phi R_{n} = \phi \left[ \min\left(1.2\, l_{c1},\, 2.4\, d_{b}\right) 2t_{a} F_{ua} + \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) 2t_{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 (0.5\,in) \times (58\,ksi) + \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.5\,in) \times (58\,ksi) \times \left( (3) – 1 \right) \right] \times \left( \frac{(3)}{(3)} \right) [math]

[math] \phi R_{n} = 106.85\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 106.85 kip = 0.468 < 1.0, Εντάξει

Tear-out governs the edge bolt, because 1.2 lc1 is less than 2.4 Διπλή διάτμηση. The inner bolts are limited by bearing.

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

The same check applies to the beam web, which is a single 1/4 in ply carrying all three bolts. The beam is not coped and its top flange is well clear of the bolts, so every bolt bears like an inner bolt.

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

Οπου [mathin] t_{wb} = 0.25\,in [mathin], [mathin] F_{ub} = 65\,ksi [mathin], [mathin] C = 3 [mathin] και [mathin] \phi = 0.75 [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 (0.25\,in) \times (65\,ksi) \times (3) [math]

[math] \phi R_{n} = 65.813\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 65.813 kip = 0.760 < 1.0, Εντάξει

This is the highest strength ratio in the connection. The two angles give 0.5 in of A36 steel against the bolts, but the beam web is only 0.25 σε παχιά, so the web has about 60% of the angles’ bearing capacity even with the higher 65 ksi tensile strength.

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

Διάτμηση μπουλονιών, bearing and tear-out are not really separate checks. Each bolt carries the least of its shear strength and its bearing or tear-out strength in every ply it passes through, and the group strength is the sum of those per-bolt values.

Strength of one bolt in double shear:

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

[math] \phi r_{v} = (0.75) \times (54\,ksi) \times \frac{\pi}{4} \times \left( (0.75\,in) \right)^{2} \times (2) \times (1) = 35.785\,kip [math]

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

[math] \phi r_{n1e} = \phi\, \min\left(1.2\, l_{c1},\, 2.4\, d_{b}\right) 2t_{a} F_{ua} [math]

[math] \phi r_{n1e} = (0.75) \times \min\left(1.2 \times (1.0938\,in),\, 2.4 \times (0.75\,in)\right) \times (0.5\,in) \times (58\,ksi) = 28.547\,kip [math]

[math] \phi r_{n1i} = \phi\, \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) 2t_{a} F_{ua} [math]

[math] \phi r_{n1i} = (0.75) \times \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.5\,in) \times (58\,ksi) = 39.15\,kip [math]

Bearing and tear-out 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} [math]

[math] \phi r_{n2} = (0.75) \times \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.25\,in) \times (65\,ksi) = 21.938\,kip [math]

Bolts are numbered from the top of the line, so bolt 3 is the edge bolt on the angles:

[math] \phi r_{eff,1} = \min\left( \phi r_{v},\, \phi r_{n1i},\, \phi r_{n2} \right) = \min\left( (35.785\,kip),\, (39.15\,kip),\, (21.938\,kip) \right) = 21.938\,kip [math]

[math] \phi r_{eff,2} = \min\left( \phi r_{v},\, \phi r_{n1e},\, \phi r_{n2} \right) = \min\left( (35.785\,kip),\, (28.547\,kip),\, (21.938\,kip) \right) = 21.938\,kip [math]

[math] \sum \phi r_{eff} = n_{1}\, \phi r_{eff,1} + n_{2}\, \phi r_{eff,2} = (2) \times (21.938\,kip) + (1) \times (21.938\,kip) = 65.813\,kip [math]

[math] \phi R_{n,eff} = \left( \frac{C}{n_{r}} \right) \sum \phi r_{eff} = \left( \frac{(3)}{(3)} \right) \times (65.813\,kip) = 65.813\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 65.813 kip = 0.760 < 1.0, Εντάξει

Bearing on the beam web governs all three bolts, so the effective strength equals the beam web result from Step 4. You can run the same connection in the free connection design calculator and try a heavier beam or a fourth bolt row to see how the web ratio moves.

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

A block of each angle can tear out along the bolt line: shear down the line of holes to the top of the angle, plus tension across to the angle’s free edge. With one bolt column this is an L-shaped block.

Net area in tension, με [mathin] 2t_{a} = 0.5\,in [mathin], [mathin] n_{c} = 1 [mathin], [mathin] s_{c} = 0\,in [mathin], [mathin] L_{eh,a} = 1.5\,in [mathin] και [mathin] d_{h} = 0.8125\,in [mathin]. Each hole adds 1/16 in for damage during punching (B4.3):

[math] A_{nt} = 2t_{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} = (0.5\,in) \times \left[ ((1)-1) \times (0\,in) + (1.5\,in) – ((1) – 0.5) \times ((0.8125\,in) + (0.0625\,in)) \right] = 0.53125\,in^{2} [math]

Gross and net areas in shear:

[math] A_{gv} = 2t_{a} \left[ L_{ev,a} + (n_{r}-1)\, s_{r} \right] = (0.5\,in) \times \left[ (1.5\,in) + ((3)-1) \times (3\,in) \right] = 3.75\,in^{2} [math]

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

[math] A_{nv} = (0.5\,in) \times \left[ (9\,in) – (1.5\,in) – ((3) – 0.5) \times ((0.8125\,in) + (0.0625\,in)) \right] = 2.6563\,in^{2} [math]

Block shear strength, με [mathin] U_{bs} = 1 [mathin] for uniform tension, [mathin] F_{ua} = 58\,ksi [mathin], [mathin] F_{ya} = 36\,ksi [mathin] και [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 (2.6563\,in^{2}) + (1) \times (58\,ksi) \times (0.53125\,in^{2}) \le 0.6 \times (36\,ksi) \times (3.75\,in^{2}) + (1) \times (58\,ksi) \times (0.53125\,in^{2}) \right) [math]

[math] \phi R_{n} = 83.859\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 83.859 kip = 0.596 < 1.0, Εντάξει

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.

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

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

Shear yielding (Εξ. J4-3)

[math] A_{gv} = d_{a}\, 2t_{a} = (9\,in) \times (0.5\,in) = 4.5\,in^{2} [math]

[math] \phi R_{n,sy} = \phi\, 0.6\, F_{ya} A_{gv} = (1) \times 0.6 \times (36\,ksi) \times (4.5\,in^{2}) = 97.2\,kip [math]

Shear rupture (Εξ. J4-4)

[math] A_{nv} = 2t_{a} \left[ d_{a} – n_{r}\left( d_{h} + 0.0625\,in \right) \right] = (0.5\,in) \times \left[ (9\,in) – (3) \times ((0.8125\,in) + (0.0625\,in)) \right] = 3.1875\,in^{2} [math]

[math] \phi R_{n,sr} = \phi\, 0.6\, F_{ua} A_{nv} = (0.75) \times 0.6 \times (58\,ksi) \times (3.1875\,in^{2}) = 83.194\,kip [math]

[math] \phi R_{n} = \min\left( \phi R_{n,sy},\, \phi R_{n,sr} \right) = \min\left( (97.2\,kip),\, (83.194\,kip) \right) = 83.194\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 83.194 kip = 0.601 < 1.0, Εντάξει

Βήμα 8: Flexure of the double angle at the beam side, double coped procedure (AISC 360-16 F11.2)

The reaction is carried from the beam-side bolt line to the heel of the angle, so each angle leg acts as a short cantilever plate in bending. Ενα βαθύ, thin leg can buckle sideways before it yields. The AISC Manual handles this with the procedure for a beam coped at both flanges, treating each leg like a beam web with the flanges cut away. The unbraced length is the beam-side gage, [mathin] L_{b} = 2\,in [mathin].

Range check

Οπου [mathin] E = 29000\,ksi [mathin], [mathin] F_{ya} = 36\,ksi [mathin], [mathin] d_{a} = 9\,in [mathin] και [mathin] t_{a} = 0.25\,in [mathin]:

[math] \frac{L_{b} d}{t^{2}} = \frac{(2\,in) \times (9\,in)}{(0.25\,in)^{2}} = 288 [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]

Έτσι [mathin] \frac{0.08E}{F_{y}} \lt \frac{L_{b} d}{t^{2}} \le \frac{1.9E}{F_{y}} [mathin], και Εξ. F11-2 applies.

Section moduli of one angle leg

[math] Z = \frac{t_{a} \left( d_{a} \right)^{2}}{4} = \frac{(0.25\,in) \times ((9\,in))^{2}}{4} = 5.0625\,in^{3} [math]

[math] S = \frac{t_{a} \left( d_{a} \right)^{2}}{6} = \frac{(0.25\,in) \times ((9\,in))^{2}}{6} = 3.375\,in^{3} [math]

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

With no cope on the angles, [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\,in)}{(9\,in)} \right) \right] \times \left( 1 – \frac{(0\,in)}{(9\,in)} \right) \ge 1.84 [math]

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

Flexural strength

Με [mathin] \phi = 0.9 [mathin], and a factor of 2 for the two angles:

[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] 2 [math]

[math] \phi M_{n} = (0.9) \times \left[ (1.84) \times \left[ 1.52 – 0.274 \times ((288)) \times \left( \frac{(36\,ksi)}{(29000\,ksi)} \right) \right] \times (36\,ksi) \times (3.375\,in^{3}) \le (36\,ksi) \times (5.0625\,in^{3}) \right] \times 2 [math]

[math] \phi M_{n} = 27.337\,kip\text{-}ft \;(328.04\,kip\text{-}in) [math]

The plastic moment cap [mathin] F_{ya} Z [mathin] governs, so the legs are not prone to buckling at this depth and thickness.

Equivalent shear capacity

[math] \phi V_{n} = \frac{\phi M_{n}}{L_{b}} = \frac{(27.337\,kip\text{-}ft)}{(2\,in)} = 164.03\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 164.03 kip = 0.305 < 1.0, Εντάξει

Βήμα 9: Shear and moment interaction of the double angle at the beam side (AISC Manual Eq. 10-5)

The angle legs carry 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 between the shear yielding strength and the flexural strength.

The moment at the angle is the reaction times the unbraced length, με [mathin] V_{r} = 50\,kip [mathin] και [mathin] L_{b} = 2\,in [mathin]:

[math] M_{r} = V_{r}\, L_{b} = (50\,kip) \times (2\,in) = 8.3333\,kip\text{-}ft [math]

Με [mathin] \phi V_{n} = 97.2\,kip [mathin] from shear yielding in Step 7 και [mathin] \phi M_{n} = 27.337\,kip\text{-}ft [mathin] from Step 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{(50\,kip)}{(97.2\,kip)} \right)^{2} + \left( \frac{(8.3333\,kip\text{-}ft)}{(27.337\,kip\text{-}ft)} \right)^{2} = 0.35753 [math]

DCR = 0.35753 / 1.0 = 0.358 < 1.0, Εντάξει

Βήμα 10: Detailing limits at the support side (AISC 360-16 J3.3, J3.4, J3.5)

The outstanding legs are bolted to the column flange, so the same spacing and edge rules apply, plus two gage limits. The bolts must sit far enough out from the beam web to be entered and tightened, and they must stay far enough in from the edge of the column flange.

The angles have horizontal short-slotted holes here. 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 from 1 σε 1.125 σε. The edge distance on the column flange is what is left of the 14.5 in flange outside the 5.25 in gage:

[math] L_{eh,sup} = \frac{b_{fs} – g}{2} = \frac{(14.5\,in) – (5.25\,in)}{2} = 4.625\,in [math]

The governing row is the minimum bolt gage at the support:

[math] DCR = \frac{d}{c} = \frac{(4.375)}{(5.25)} = 0.83333 [math]

DCR = 0.833 < 1.0, Εντάξει

Βήμα 11: Bolt group shear at the support side (AISC 360-16 J3.6)

Each support-side bolt passes through one angle leg and the column flange, so it has one shear plane. There are six of them, three per angle, and they share the reaction equally because the eccentricity at the support is zero.

Οπου [mathin] d_{b} = 0.75\,in [mathin], [mathin] N_{s} = 1 [mathin], [mathin] N_{V} = 6 [mathin] bolts in shear, [mathin] h_{f} = 1 [mathin] with no fillers (J5.2(ένα)) και [mathin] \phi = 0.75 [mathin]:

[math] \phi R_{n} = \phi\, F_{nv}\, \frac{\pi}{4} \left( d_{b} \right)^{2} N_{s}\, N_{V}\, 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) \times (1) [math]

[math] \phi R_{n} = 107.35\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 107.35 kip = 0.466 < 1.0, Εντάξει

The result matches Step 2. Three bolts in double shear and six bolts in single shear have the same number of shear planes.

Βήμα 12: Bolt bearing and tear-out on the double angle at the support side (AISC 360-16 J3.10)

On the support side the angles are pushed down while the column flange 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. Each angle is a single 1/4 in ply.

The holes are short slots running horizontally, so the dimension across the slot in the direction of the load is the standard [mathin] d_{h} = 0.8125\,in [mathin]:

[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]

Οπου [mathin] t_{a} = 0.25\,in [mathin], [mathin] F_{ua} = 58\,ksi [mathin], [mathin] N_{V} = 6 [mathin] και [mathin] n_{r} = 3 [mathin]. One line of bolts is calculated and then scaled by [mathin] N_{V}/n_{r} [mathin] for the two angles:

[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{N_{V}}{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 (0.25\,in) \times (58\,ksi) + \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.25\,in) \times (58\,ksi) \times \left( (3) – 1 \right) \right] \times \left( \frac{(6)}{(3)} \right) [math]

[math] \phi R_{n} = 106.85\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 106.85 kip = 0.468 < 1.0, Εντάξει

This equals the beam-side result in Step 3. Two lines of single 1/4 in plies give the same total bearing thickness as one line through both angles.

Βήμα 13: Bolt bearing and tear-out on the column flange (AISC 360-16 J3.10)

The bolts also bear on the column flange. The column 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]

Οπου [mathin] t_{fs} = 0.71\,in [mathin], [mathin] F_{us} = 65\,ksi [mathin] και [mathin] N_{V} = 6 [mathin]:

[math] \phi R_{n} = \phi\, \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{fs} F_{us}\, N_{V} [math]

[math] \phi R_{n} = (0.75) \times \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.71\,in) \times (65\,ksi) \times (6) [math]

[math] \phi R_{n} = 373.81\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 373.81 kip = 0.134 < 1.0, Εντάξει

Βήμα 14: Effective strength of the bolt group at the support side (AISC 360-16 J3.6 Commentary)

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.

Shear strength of one bolt, from the group strength in Step 11:

[math] \phi r_{v} = \frac{\phi R_{n}}{N_{V}} = \frac{(107.35\,kip)}{(6)} = 17.892\,kip [math]

Bearing and tear-out of one bolt on one angle, 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_{a} F_{ua} = (0.75) \times \min\left(1.2 \times (1.0938\,in),\, 2.4 \times (0.75\,in)\right) \times (0.25\,in) \times (58\,ksi) = 14.273\,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 (0.25\,in) \times (58\,ksi) = 19.575\,kip [math]

Bearing of one bolt on the column flange:

[math] \phi r_{n2} = \phi\, \min\left(1.2\, l_{c2},\, 2.4\, d_{b}\right) t_{fs} F_{us} = (0.75) \times \min\left(1.2 \times (2.1875\,in),\, 2.4 \times (0.75\,in)\right) \times (0.71\,in) \times (65\,ksi) = 62.303\,kip [math]

Bolts are numbered from the top of the line, so bolt 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),\, (14.273\,kip),\, (62.303\,kip) \right) = 14.273\,kip [math]

[math] \phi r_{eff,2} = \min\left( \phi r_{v},\, \phi r_{n1i},\, \phi r_{n2} \right) = \min\left( (17.892\,kip),\, (19.575\,kip),\, (62.303\,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 (14.273\,kip) + (2) \times (17.892\,kip) = 50.058\,kip [math]

[math] \phi R_{n,eff} = \left( \frac{N_{V}}{n_{r}} \right) \sum \phi r_{eff} = \left( \frac{(6)}{(3)} \right) \times (50.058\,kip) = 100.12\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 100.12 kip = 0.499 < 1.0, Εντάξει

Here the inner bolts are limited by AISC bolt shear and the top bolt by tear-out of the angle. The thick column flange never governs.

Βήμα 15: Block shear of the double angle at the support side (AISC 360-16 J4.3)

The same L-shaped block can tear out of each outstanding leg. The difference from Step 6 is the hole size: the tension plane runs horizontally, along the slots, so the net tension area uses the slot length [mathin] d_{h} = 1\,in [mathin]. The shear planes run vertically, across the slots, so they use [mathin] d_{h} = 0.8125\,in [mathin].

[math] A_{nt} = 2t_{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} = (0.5\,in) \times \left[ ((1)-1) \times (0\,in) + (1.5\,in) – ((1) – 0.5) \times ((1\,in) + (0.0625\,in)) \right] = 0.48438\,in^{2} [math]

[math] A_{gv} = 2t_{a} \left[ L_{ev,a} + (n_{r}-1)\, s_{r} \right] = (0.5\,in) \times \left[ (1.5\,in) + ((3)-1) \times (3\,in) \right] = 3.75\,in^{2} [math]

[math] A_{nv} = (0.5\,in) \times \left[ (9\,in) – (1.5\,in) – ((3) – 0.5) \times ((0.8125\,in) + (0.0625\,in)) \right] = 2.6563\,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 (2.6563\,in^{2}) + (1) \times (58\,ksi) \times (0.48438\,in^{2}) \le 0.6 \times (36\,ksi) \times (3.75\,in^{2}) + (1) \times (58\,ksi) \times (0.48438\,in^{2}) \right) [math]

[math] \phi R_{n} = 81.82\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 81.82 kip = 0.611 < 1.0, Εντάξει

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

The outstanding legs carry the reaction into the column as vertical shear, so they get the same gross and net section checks as the beam-side legs.

Shear yielding (Εξ. J4-3)

[math] A_{gv} = d_{a}\, 2t_{a} = (9\,in) \times (0.5\,in) = 4.5\,in^{2} [math]

[math] \phi R_{n,sy} = \phi\, 0.6\, F_{ya} A_{gv} = (1) \times 0.6 \times (36\,ksi) \times (4.5\,in^{2}) = 97.2\,kip [math]

Shear rupture (Εξ. J4-4)

[math] A_{nv} = 2t_{a} \left[ d_{a} – n_{r}\left( d_{h} + 0.0625\,in \right) \right] = (0.5\,in) \times \left[ (9\,in) – (3) \times ((0.8125\,in) + (0.0625\,in)) \right] = 3.1875\,in^{2} [math]

[math] \phi R_{n,sr} = \phi\, 0.6\, F_{ua} A_{nv} = (0.75) \times 0.6 \times (58\,ksi) \times (3.1875\,in^{2}) = 83.194\,kip [math]

[math] \phi R_{n} = \min\left( (97.2\,kip),\, (83.194\,kip) \right) = 83.194\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 83.194 kip = 0.601 < 1.0, Εντάξει

Βήμα 17: Flexure of the double angle at the support side, double coped procedure (AISC 360-16 F11.2)

The outstanding legs bend between the support bolt line and the beam web. The lever arm is half of what is left of the 5.25 in gage after the 0.25 in beam web:

[math] L_{b} = \frac{g – t_{wb}}{2} = \frac{(5.25\,in) – (0.25\,in)}{2} = 2.5\,in [math]

Range check

[math] \frac{L_{b} d}{t^{2}} = \frac{(2.5\,in) \times (9\,in)}{(0.25\,in)^{2}} = 360 [math]

The limits are the same as Step 8, 64.444 και 1530.6, so Eq. F11-2 applies. The section moduli are also the same: [mathin] Z = 5.0625\,in^{3} [mathin] και [mathin] S = 3.375\,in^{3} [mathin].

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

[math] C_{b} = \left[ 3 + \ln\left( \frac{(2.5\,in)}{(9\,in)} \right) \right] \times \left( 1 – \frac{(0\,in)}{(9\,in)} \right) \ge 1.84 [math]

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

Flexural strength

[math] \phi M_{n} = (0.9) \times \left[ (1.84) \times \left[ 1.52 – 0.274 \times ((360)) \times \left( \frac{(36\,ksi)}{(29000\,ksi)} \right) \right] \times (36\,ksi) \times (3.375\,in^{3}) \le (36\,ksi) \times (5.0625\,in^{3}) \right] \times 2 [math]

[math] \phi M_{n} = 27.337\,kip\text{-}ft \;(328.04\,kip\text{-}in) [math]

The plastic cap governs again, so the flexural strength is the same as at the beam side. Only the lever arm changes.

Equivalent shear capacity

[math] \phi V_{n} = \frac{\phi M_{n}}{L_{b}} = \frac{(27.337\,kip\text{-}ft)}{(2.5\,in)} = 131.22\,kip [math]

DCR = 50 δέρμα μόσχου ακατέργαστου / 131.22 kip = 0.381 < 1.0, Εντάξει

Βήμα 18: Shear and moment interaction of the double angle at the support side (AISC Manual Eq. 10-5)

The support side uses the same circular interaction as Step 9, but the lever arm is 2.5 in instead of 2 σε, so the moment is larger.

[math] M_{r} = V_{r}\, L_{b} = (50\,kip) \times (2.5\,in) = 10.417\,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{(50\,kip)}{(97.2\,kip)} \right)^{2} + \left( \frac{(10.417\,kip\text{-}ft)}{(27.337\,kip\text{-}ft)} \right)^{2} = 0.4098 [math]

DCR = 0.4098 / 1.0 = 0.410 < 1.0, Εντάξει

Summary of the double angle connection design checks

The connection passes every check. The highest ratio, 0.833, is a detailing limit rather than a strength check: ο 5 1/4 in support gage against its 4.375 in minimum. Of the checks that compare load with capacity, bolt bearing on the 1/4 in beam web governs at 0.760, which is also why the effective strength of the beam-side bolt group is 0.760. The angles themselves have plenty of reserve, with block shear at the support side the highest of them at 0.611.

What to change if a check fails

• Bearing on the beam web or beam-side effective strength: add a bolt row, which means deeper angles, or pick a beam with a thicker web. A larger bolt raises the 2.4 d cap on bearing.
• Διάτμηση μπουλονιών: add a bolt row or use a larger bolt. A325-X (threads excluded) or A490 bolts raise the shear stress.
• Tear-out on the angles: increase the vertical edge distance at the top or bottom of the angles, or use thicker angles.
• Διάτμηση μπλοκ, shear rupture or shear yielding of the angles: use thicker or deeper angles. 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 thicker angles.
• Support gage or edge distance: adjust the gage within the column flange width, or switch the leg orientation so the wider leg sits where the gage needs room.

Συχνές ερωτήσεις

Why is the 2 in eccentricity ignored on the beam side?

AISC Manual page 10-9 lets you neglect the eccentricity of a single vertical row of bolts when it is no more than 3 σε. Here the beam-side bolts are 2 in from the angle heel, so the bolt group coefficient is taken as the full bolt count, 3, and every bolt carries an equal share of the reaction. On the support side the bolts are symmetric about the beam web, so the eccentricity there is zero.

Why are the bolts in double shear at the beam and single shear at the column?

At the beam, each bolt passes through one angle, the beam web and the other angle, so it is cut on two planes. At the column, each bolt passes through one angle leg and the column flange only, so it is cut on one plane. Because there are six support bolts against three beam bolts, both sides end up with six shear planes and the same 107.35 kip bolt shear strength.

Why does the edge bolt switch from the bottom to the top between the two sides?

The beam pushes its bolts down on the angles, so the angles tear out toward their bottom edge at the beam side. At the column the angles are pushed down while the column flange holds its bolts, so the bolts bear upward on the angle holes and tear-out runs toward the top edge. The clear distance is the same, but it applies to bolt 3 on one side and bolt 1 on the other.

Why use short-slotted holes in the outstanding legs?

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

Why is a double coped beam procedure used to check an angle?

Each angle leg is a thin, deep plate bending over a short span between the bolt line and its support, which is the same situation as a beam web with both flanges coped away. The AISC Manual procedure for double coped beams covers lateral-torsional buckling of that plate. In this example the legs are short enough that the plastic moment cap governs, so buckling does not reduce the strength.

Which check usually governs a double angle connection?

Two angles put two plies against every beam-side bolt, while the beam web is a single ply. So when the beam web is thin, bearing on the web tends to control before the angles do. In this example bearing on the 0.25 in W16x26 web governs at 0.760, ahead of block shear of the angles at 0.611.

Does this example follow the AISC 15th edition or the AISC 16th edition Manual?

It follows AISC 360-16 and the AISC Manual 15th Edition, which is written around that Specification. The AISC 16th Edition Manual pairs with the later 360-22 Προσδιορισμός. The limit states for a double angle connection are the same in both, but clause and page references move, so check them against the edition your project uses.

How would this change for AISC ASD instead of LRFD?

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. A resistance factor of 0.75 becomes a safety factor of 2.00, shear yielding’s 1.00 becomes 1.50, and flexure’s 0.90 becomes 1.67. The load combinations change too, so the demand is lower.

Design a double angle connection for free

Every number on this page comes from one solve. Pick the double angle configuration, enter your beam, υποστήριξη, γωνίες, bolts and reaction, and see each of these checks in your browser. Δεν υπάρχει τίποτα για εγκατάσταση.

ΣΧΕΔΙΑΣΤΕ ΜΙΑ ΣΥΝΔΕΣΗ ΔΩΡΕΑΝ

The SkyCiv Connection Design module is the connection design software behind this example. It gives you the full calculation report with every equation and your numbers in it, like the steps above, plus every capacity and DCR, για AISC 360-16 (LRFD and ASD) και ως 4100:2020. ο αριθμομηχανή μπουλονιών covers the bolt group checks on their own.

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βιβλιογραφικές αναφορές

• AISC 360-16, Προδιαγραφές για δομικά κτίρια από χάλυβα: B3.4, B4.3, F11.2 (Εξ. F11-2), J3.3, J3.4 (Table J3.4), J3.5 (Table J3.5), J3.6 (Εξ. J3-1) and Commentary, J3.10 (Εξ. J3-6a, J3-6c), J4.2 (Εξ. J4-3, J4-4), J4.3 (Εξ. J4-5), J5.2(ένα).
• Εγχειρίδιο AISC Steel Construction, 15ου Έκδοση: Μέρος 9, Εξ. 9-15 (σελίδα 9-9); Μέρος 10, σελίδα 10-9 και Εξ. 10-5 (σελίδα 10-90).
• Υπολογισμός ομάδας μπουλονιών με χρήση ICOR
• Design of steel connections using AISC 360-16
• Μπουλόνια in SkyCiv Connection Design