Question 2 of 7: Bolted beam-to-column connection and the steel tie

Question A2: Bolted beam-to-column connection and the steel tie (20 marks: 5 + 10 + 5)

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 simply supported beam bolted to a column flange at A and hung from a steel tie at B, carrying two specified point loads.

Find. A bolted shear connection at A able to carry the factored end reaction, and a steel tie at B with its end connection able to carry the factored tie force.

Approach. Take the beam as simply supported, obtain the two end reactions and factor them, then size the bolt group at A for the governing of bolt shear, bearing and block shear, and finally size the tie at B for gross-section yielding, net-section rupture and its own end bolts.

(a) Reactions and factored end forces (5 marks)

1. Specified reactions. Moments about A give $$R_B = \frac{120(1.5) + 80(3.5)}{5.0} = \frac{180 + 280}{5.0} = 92.0\ \text{kN} \qquad R_A = 200 - 92 = 108.0\ \text{kN}$$
2. Add the beam's own weight and factor. A W530 × 92 weighs $92 \times 9.81 = 0.903\ \text{kN/m}$, contributing $1.25(0.903)(5.0)/2 = 2.82\ \text{kN}$ to each end. With the applied loads taken as specified live loads, $$C_{f,A} = 1.5(108.0) + 2.82 = 164.8\ \text{kN} \qquad T_{f,B} = 1.5(92.0) + 2.82 = 140.8\ \text{kN}$$ $$\boxed{V_f = 164.8\ \text{kN at A}, \qquad T_f = 140.8\ \text{kN in the tie at B}}$$ Because the tie is vertical and the connection at A is a simple shear connection, no axial force and no end moment are transferred; the beam is genuinely simply supported, as the question states.

(b) Bolted connection at A (10 marks)

A double-angle web connection is chosen: two angles bolted through the beam web, which puts those bolts in double shear, and bolted through their outstanding legs to the column flange, where they are in single shear. The beam is shallower than the column, so no cope is needed and the beam web remains intact.

1. Shear resistance of one M20 A325 bolt (CSA S16 Clause 13.12.1.2, with $\phi_b = 0.80$, $A_b = 314\ \text{mm}^2$ and the 0.70 reduction because the threads are intercepted): $$V_r = 0.60\,\phi_b\,n\,m\,A_b\,F_u (0.70) = 0.60(0.80)(1)(1)(314)(830)(0.70) = 87.6\ \text{kN}$$ per shear plane, hence 175.2 kN per bolt in double shear through the web, and 87.6 kN per bolt in single shear at the column flange.
2. Bearing. With $\phi_{br} = 0.80$ and $F_u = 450$ MPa the bearing resistance of the 10.2 mm beam web is $$B_r = 3\,\phi_{br}\,t\,d\,F_u = 3(0.80)(10.2)(20)(450) = 220.3\ \text{kN per bolt}$$ which exceeds the double-shear value, so bolt shear governs at the web. The 24.4 mm column flange and the 9.5 mm angle legs are checked the same way and neither controls.
3. Number of bolts. Strength alone requires only $164.8/175.2 = 0.94$ bolts, so the connection is set by detailing rather than by force: a shear connection to a 533 mm deep beam should engage a reasonable proportion of the web depth, and the minimum practical group is three bolts at a 75 mm pitch with 35 mm end distances. Adopt 3 – M20 A325 bolts through the web: $$V_r = 3(175.2) = 525.6\ \text{kN} \gg 164.8\ \text{kN}$$ and the six bolts into the column flange give the same $6(87.6) = 525.6\ \text{kN}$.
4. Angles, gross shear and block shear. Take 2 – L102 × 102 × 9.5, 220 mm long. Gross shear yielding of the pair is $$V_r = 0.66\,\phi\,A_g F_y = 0.66(0.90)(2 \times 220 \times 9.5)(350)/10^3 = 869\ \text{kN}$$ Block shear (Clause 13.11) on one angle, with two 75 mm pitches, a 35 mm end distance and a 40 mm edge distance, and $U_t = 0.60$ for a single bolt line: $$T_r = \phi_u\left[U_t A_{nt} F_u + 0.6 A_{gv}\frac{F_y+F_u}{2}\right] = 0.75\left[0.6(266)(450) + 0.6(1758)(400)\right] = 370\ \text{kN}$$ per angle, 740 kN for the pair. Every alternative failure path is therefore stronger than the bolt group, so $$\boxed{V_r = 525.6\ \text{kN} > V_f = 164.8\ \text{kN} \quad (31\%\ \text{utilised})}$$
5. Why the low utilisation is the right answer. Standard framed shear connections are governed by minimum edge and end distances, by the need for erection stability, and by the requirement that a "simple" connection be flexible enough not to attract moment. Reducing to two bolts would meet the force but leaves a connection only 110 mm deep on a 533 mm beam, which is poor practice; the correct engineering statement is that the connection is detailing-governed.

(c) The steel tie at B (5 marks)

1. Gross-section yielding and net-section rupture (CSA S16 Clause 13.2). Try a flat bar 75 × 10 mm in 350W with two M20 bolts in line at each end; the hole allowance for net area is $20 + 2 + 2 = 24$ mm: $$T_r = \phi A_g F_y = 0.90(750)(350)/10^3 = 236.2\ \text{kN}$$ $$T_r = \phi_u A_{ne} F_u = 0.75\left[(75-24)(10)\right](450)/10^3 = 172.1\ \text{kN}$$ Net-section rupture governs, and $$\boxed{T_r = 172.1\ \text{kN} > T_f = 140.8\ \text{kN} \quad (82\%\ \text{utilised})}$$
2. End connection of the tie. Two M20 A325 bolts in single shear through a lapped gusset give $2(87.6) = 175.2\ \text{kN} > 140.8\ \text{kN}$, and bearing on the 10 mm bar is $3(0.80)(10)(20)(450) = 216\ \text{kN}$ per bolt. Use a 60 mm pitch and a 40 mm end distance, both above the S16 minima for a 20 mm bolt.
3. Slenderness of the tie. Clause 10.4.2.2 recommends $L/r \le 300$ for tension members so that they do not sag or vibrate. A 10 mm bar has $r = 10/\sqrt{12} = 2.89$ mm, which limits it to about 0.87 m. The tie drawn in Figure A2 is short and satisfies this, but if the hanger is longer than roughly a metre, substitute an HSS 89 × 89 × 6.4 ($A_g = 2\,115\ \text{mm}^2$, $r = 33.8$ mm, $T_r = 0.90(2\,115)(350)/10^3 = 666\ \text{kN}$), which satisfies the guidance up to about 10 m and still exceeds the required force with room to spare.