Bolted Beam-to-Column Connection and Steel Tie: 16-Civ-A2 Elementary Structural Design, December 2013, Q2

Question A2: Bolted Beam-to-Column Connection and Steel Tie (5 + 10 + 5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 5.0 m beam W530×92 of G40.21M 350W spans from a bolted connection at the column face (point A) to a vertical steel tie at point B. From Figure A2 the specified point loads are 120 kN at 1.5 m from A and 80 kN at 3.5 m from A; the remaining 1.5 m carries no load. Beam web thickness $w = 10.2$ mm, flange thickness 15.6 mm, depth 533 mm. The column is a W610×195 with a 24.4 mm flange.

Find. A bolted shear connection at A and a steel tie at B, both sized to the factored reactions.

Figure A2 — beam W530×92 bolted to a W610×195 column at A and hung from a vertical steel tie at B. Loads shown are specified (unfactored).

Approach. Take moments about A to split the load between the connection and the tie, factor the reactions, then size the bolt group at A for shear and bearing and the tie at B for gross yielding and net-section rupture.

Specified reactions. The beam is simply supported between A and the tie at B, so $$R_B = \frac{120(1.5) + 80(3.5)}{5.0} = \frac{460}{5.0} = 92.0 \text{ kN}, \qquad R_A = 200 - 92 = 108.0 \text{ kN}$$ The vertical tie has no horizontal component, so it carries pure tension equal to $R_B$.
Factored actions. Treating the figure loads as live load, $$V_f = 1.5(108.0) = \boxed{162 \text{ kN at A}}, \qquad T_f = 1.5(92.0) = \boxed{138 \text{ kN in the tie}}$$
Choose the connection type and bolt. A double-angle (clip-angle) shear connection is used: two L90×90×10 angles of grade 300W, bolted to the beam web with the bolts in double shear and to the column flange with the bolts in single shear. Bolts are M20 ASTM A325 ($F_u = 830$ MPa, $A_b = 314$ mm²) in punched holes 22 mm diameter, with threads intercepted by a shear plane.
Bolt shear resistance. For threads intercepted, S16 applies a 0.70 reduction to the 0.60 shear factor: $$V_r = 0.70(0.60)\phi_b n A_b F_u = 0.70(0.60)(0.80)(2)(314)(830) = 175 \text{ kN per bolt}$$ Three bolts through the web therefore give $3(175) = 526$ kN, comfortably above $V_f = 162$ kN. On the column flange each bolt has one shear plane, $V_r = 87.6$ kN, and six bolts (three per angle leg) provide 526 kN.
Bearing on the beam web. The thin 10.2 mm web usually governs a shear connection, so it must be checked explicitly: $$B_r = 3\phi_{br} t d F_u = 3(0.80)(10.2)(20)(450) = 220 \text{ kN per bolt}$$ Three bolts give 661 kN > 162 kN. Bearing is therefore not critical, and with 35 mm end distance and 75 mm pitch the S16 minimum edge and spacing rules (2.7$d$ = 54 mm pitch, 26 mm edge for a sheared edge) are satisfied.
Angle capacity. With three bolts at 75 mm pitch and 35 mm end distances the angles are 295 mm long. Their combined gross shear resistance is $$V_r = 0.66 \phi A_{gv} F_y = 0.66(0.90)(2 \times 295 \times 10)(300) = 1051 \text{ kN}$$ which is an order of magnitude above the demand, confirming that the 10 mm angle thickness is set by fabrication practice rather than strength. Adopt: 2 — L90×90×10 × 295 mm long, with 3 M20 A325 bolts to the beam web and 6 M20 A325 bolts to the column flange.
Design the tie at B. A flat bar 90 × 12 mm of 300W steel is proposed, connected at each end by two M20 bolts in 22 mm holes. Gross-section yielding governs first: $$T_r = \phi A_g F_y = 0.90(90 \times 12)(300) = 292 \text{ kN}$$ and net-section rupture, with the full width connected so no shear-lag reduction applies, $$T_r = 0.85\phi_u A_{ne} F_u = 0.85(0.75)\big[(90-22)(12)\big](450) = 234 \text{ kN}$$ Both exceed $T_f = 138$ kN, so $$\boxed{\text{tie} = 90 \times 12 \text{ mm flat bar, 300W, } T_r = 234 \text{ kN}}$$ Because the member is in tension only, no slenderness check is required; a nominal $L/r \le 300$ limit for sag is met by the 12 mm thickness over the short hanger length.

Item	Design	Demand	Resistance
Reaction at A	—	162 kN	—
Web bolts (double shear)	3 — M20 A325	162 kN	526 kN
Bearing on 10.2 mm web	3 bolts	162 kN	661 kN
Column-flange bolts (single shear)	6 — M20 A325	162 kN	526 kN
Clip angles	2 — L90×90×10 × 295	162 kN	1051 kN
Steel tie at B	90 × 12 flat bar, 300W	138 kN	234 kN

Topic. Simple (shear) beam-to-column connections and tension-member design to CSA S16.

Key relations. Bolt shear $V_r = 0.60 \phi_b n A_b F_u$ (×0.70 when threads are intercepted); bearing $B_r = 3\phi_{br} t d F_u$; block-shear and gross-shear checks on the connecting elements; tension member $T_r = \min\{\phi A_g F_y,\; 0.85 \phi_u A_{ne} F_u\}$.

Why this works. A clip-angle connection is designed as a pure shear (pinned) joint: the angles are flexible enough out of plane that the beam end can rotate, so the connection is required to carry only the vertical reaction, which is exactly what the simply supported analysis assumes. The load path is checked in sequence — bolt shear, plate bearing, and the connected material itself — because failure of any link ends the connection. For the tension tie, two limit states compete: yielding across the whole gross area, which is ductile and therefore given the more generous factor, and fracture across the reduced net area at the holes, which is brittle and consequently carries both a lower $\phi_u$ and the additional 0.85 factor.

Common pitfall. Forgetting the 0.70 threads-intercepted reduction and reporting a bolt capacity 40 % too high; and checking bearing on the connecting angles instead of the beam web, when it is the thinner 10.2 mm web that governs. A third trap is factoring the moment rather than the reaction — the connection sees the end shear only.

Source. Kulak & Grondin, Limit States Design in Structural Steel, Ch. 4 and Ch. 9; CISC Handbook of Steel Construction, CSA S16 Cl. 13.2 (tension), 13.10 (fasteners), 13.12 (bearing).

Question 2 of 7: Bolted Beam-to-Column Connection and Steel Tie

Question A2: Bolted Beam-to-Column Connection and Steel Tie (5 + 10 + 5 marks)