Question 2 of 7: A2 — Bolted beam-to-column connection
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2015 — 07-Str-A2 Elementary Structural Design, 3-hour closed-book paper (textbooks and design handbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). The candidate answers two from Part A, two from Part B and the one question in Part C — five solutions, all of equal value. All seven are solved here, because the set is a study resource rather than an exam script.
CISC, Handbook of Steel Construction — section property tables for W-shapes and hollow structural sections.
CSA A23.3, Design of Concrete Structures — Clauses 10 (flexure and axial), 11 (shear), 10.15 (slender columns).
CSA O86, Engineering Design in Wood — Clause 6 (modification factors), Clause 7 (sawn lumber, bending and shear).
Canadian Wood Council, Wood Design Manual — specified strengths for Beam and Stringer grades.
Check — load factoring. Note 6 on page 1 states “all loads shown are unfactored”, so every load taken from a figure is a specified load and is factored here as 1.25D + 1.5L (NBCC Table 4.1.3.2, Case 2), treating the drawn point loads as live and member self-weight as dead. Where a question states a load directly in its text without a dead/live split (B3), that load is taken as already factored; this is stated again at that question.
Beam W530×109, G40.21M 350W (web $w = 11.6$ mm, mass 109 kg/m)
$F_y = 350$ MPa, $F_u = 450$ MPa
Column W610×241, G40.21M 350W (flange 31.0 mm)
bolting surface for the connection
Beam layout (Figure A2)
bolted to the column at A, tie rod at 4 m, free end B at 6 m
Specified point loads
140 kN at 2 m from A; 100 kN at B
Beam self-weight
1.069 kN/m (dead)
Find. The factored reaction the beam delivers to the column, and a complete bolted shear connection sized for it.
[Figure not reproduced: Figure A2 (redrawn): the connection at A and the tie rod at 4 m are the two supports; the 2 m length beyond the rod is an overhang. See the official exam paper.]
Approach. Treat the beam as determinate on two supports — the connection at A and the tie rod at 4 m — take moments about the rod to isolate the reaction at A, factor it, then size bolts for shear, bearing and block shear.
Specified reaction at the connection. Taking moments about the tie rod at $x = 4$ m, with the 100 kN tip load working to lift the beam at A,
$$R_A=\frac{140(4-2)-100(6-4)}{4}=\frac{280-200}{4}=20.0\ \text{kN}$$
The beam self-weight, $w = 1.069$ kN/m over 6 m with its resultant at $x = 3$ m, contributes a further
$$R_{A,D}=\frac{1.069(6)(4-3)}{4}=1.60\ \text{kN}$$
Factored reaction. Applying 1.25D + 1.5L, with the drawn point loads taken as live,
$$V_f=1.5(20.0)+1.25(1.60)=\boxed{32.0\ \text{kN}}$$
For interest, vertical equilibrium puts $T_f = 336$ kN in the tie rod — the rod, not the connection, carries the bulk of the load, because the 100 kN tip load levers most of the 140 kN load onto the rod. The connection is therefore a light one, and the design will be governed by detailing minima rather than by strength.
Choose the connection type. Adopt a flexible double-angle (clip-angle) shear connection: 2 L102×102×9.5 bolted to the beam web and to the column flange with M20 ASTM F3125 Grade A325 bolts in a bearing-type joint, threads intercepted by a shear plane. This detail transmits shear while allowing the beam end to rotate, which is what a “simply supported” idealisation demands.
Bolt shear resistance. With $A_b = 314\ \text{mm}^{2}$, $F_u = 825$ MPa, $\phi_b = 0.80$ and the 0.70 reduction for threads in the shear plane (S16 Clause 13.12.1.2),
$$V_r=0.70\left(0.60\,\phi_b\,n\,m\,A_b\,F_u\right)$$
per bolt this is 87.1 kN in single shear and 174.2 kN in double shear.
Bolt group. Use two M20 bolts through the beam web (double shear, one plane per angle) and four M20 bolts to the column flange (two per angle, single shear), at 70 mm pitch with 40 mm end distance. Then
$$V_{r,\text{web}}=2(174.2)=348\ \text{kN},\qquad V_{r,\text{flange}}=4(87.1)=348\ \text{kN}$$
Both are more than ten times $V_f = 32.0$ kN, confirming that two bolts — the S16 practical minimum for a shear connection — are sufficient.
Bearing. With $B_r=3\phi_{br}tdF_u$ and $\phi_{br} = 0.80$, bearing on the 11.6 mm beam web gives 250.6 kN per bolt and bearing on the paired 9.5 mm angle legs gives more still; the 31.0 mm column flange is never critical. Bearing does not govern.
Block shear. Using S16 Clause 13.11 with $\phi_u = 0.75$, $U_t = 1.0$ and 24 mm holes,
$$T_r=\phi_u\left[U_tA_{nt}F_u+0.6A_{gv}\frac{F_y+F_u}{2}\right]$$
which yields 261 kN per angle (522 kN for the pair) and 359 kN for the beam-web tear-out path. The gross shear resistance of the two angle legs is 299 kN. Every rupture mode clears $V_f$ by an order of magnitude.
Detailing checks. Minimum pitch is $2.7d = 54$ mm (70 mm used); minimum end and edge distance for a 20 mm bolt at a sheared edge is 32 mm (40 mm used); maximum edge distance is $12t$ or 150 mm. The angles are cut 200 mm long, leaving ample clearance below the beam flange for rotation. The governing statement is therefore that the connection is minimum-size controlled.
Check. The tie rod is treated as a rigid vertical support; if the rod is long and elastic enough to stretch appreciably, the beam becomes propped rather than simply supported and $R_A$ rises. Even a rod force redistribution of 100 % would leave the two-bolt connection adequate, so the conclusion is insensitive to this assumption.
