07-Str-A2 · December 2014

Question 7 of 7: C1 — Bolted hanger connection in a glulam column

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2014 — 07-Str-A2 Elementary Structural Design, 3-hour duration, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel, CSA S16), Part B (B1–B3, reinforced concrete, CSA A23.3) and Part C (C1, timber, CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in total, all of equal value. All seven are solved here so the set works as a complete study resource.

Reference texts. CSA S16, Design of Steel Structures, with the CISC Handbook of Steel Construction; CSA A23.3, Design of Concrete Structures, with the CAC Concrete Design Handbook; CSA O86, Engineering Design in Wood, with the CWC Wood Design Manual; National Building Code of Canada (load combinations); Kulak & Grondin, Limit States Design in Structural Steel; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design.

Check — load factors. Note 6 on page 1 states “All loads shown are unfactored, unless otherwise specified”, but the paper never splits the figure loads into dead and live components. Throughout Parts A and B the applied figure loads are therefore treated as live load, factored by 1.5, and self-weight of concrete members as dead load, factored by 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). In Part C the paper does name the components, so 1.25D + 1.5S is used directly. If an examiner intended a different split the method is unchanged — only the numerical factor moves.

Check — figure dimensions. Every figure on this paper is a hand sketch on page 3 whose OCR text is unusable; all geometry below was read from the printed figure of the original PDF. Figure A1 is not drawn to scale horizontally (its printed 600 and 300 dimensions are authoritative, not the drawn proportions). In Figure B1 the dimension “1 m × 1 m” runs from the section centreline to the outer face, so each cell is 1 m square and the box is 2000 mm wide × 1000 mm deep; the “600 × 600” dimension line brackets the void, giving 200 mm walls and a 400 mm central web. That reading is confirmed independently by the “65 typical” cover note: 65 + 65 cover plus two 15M tie legs plus one 30M bar needs exactly 200 mm of wall.

Question 7: C1 — Bolted hanger connection in a glulam column (10 + 5 + 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. Two Douglas fir glulam beams, 130 × 304 mm, hung from opposite faces of a continuous 175 × 228 mm D.Fir-L glulam column by 8 mm steel hangers bolted through the column.

Quantity	Value
Specified dead load per beam	12 kN
Specified snow load per beam	24 kN
Column (main member) thickness	175 mm
Column face width	228 mm
Steel hanger plates	8 mm, one each side
Bolt diameter, d_F	1 in. = 25.4 mm
Relative density, D.Fir-L	G = 0.49

Find. The number of bolts required and the full connection geometry — end distance, edge distance, bolt spacing and row spacing.

Figure C1 — hanger connection elevation. Three 25.4 mm bolts in a single vertical row, one 8 mm hanger plate on each face; each beam delivers 51 kN factored.

Approach. Factor the beam reactions, note that both hangers share the same bolts so the group carries the sum in double shear, evaluate the governing European-yield-model failure mode for a steel–wood–steel joint, then fix the bolt count and lay the group out to the O86 spacing rules.

Factored load. With dead and snow named explicitly, the NBCC principal combination applies: $$P_f = 1.25D + 1.5S = 1.25(12) + 1.5(24) = \boxed{51\ \text{kN per beam}}$$ Because a bolt passes right through plate–column–plate, both hangers are carried by the same fasteners: $$P_{f,\text{group}} = 2(51) = \boxed{102\ \text{kN}}$$
Embedment strengths. The load runs down the column, i.e. parallel to the grain, so $$f_1 = 50\,G\,(1-0.01d_F) = 50(0.49)\left[1-0.01(25.4)\right]=18.28\ \text{MPa}$$ For the steel side plates, $f_2 = 3f_u = 3(450) = 1350$ MPa.
Evaluate the yield modes. For one bolt in double shear through a 175 mm main member with 8 mm side plates:
• wood crushes over the full main-member thickness: $f_1d_Ft_m = 18.28(25.4)(175) = 81.2$ kN
• both steel plates crush: $2f_2d_Ft_s = 2(1350)(25.4)(8) = 549$ kN
• the bolt forms a plastic hinge at each shear plane: $2\sqrt{2}\,d_F^{\,2}\sqrt{f_yf_1/3} = 79.3$ kN, taking $f_y = 310$ MPa for an ASTM A307 bolt
The last of these governs, so $n_u = 79.3$ kN per bolt. With $t_m/d_F = 6.9$ the bolt is slender relative to the timber, which is exactly why bending of the fastener, not crushing of the wood, controls.
Factored resistance per bolt. With $\phi = 0.8$ and, for a dry, untreated, standard-term (snow) load case, $K_D = K_{SF} = K_T = 1.0$: $$N_r = \phi\,n_u\,K_DK_{SF}K_T = 0.8(79.3) = \boxed{63.4\ \text{kN per bolt}}$$
Number of bolts. $$n = \frac{102}{63.4} = 1.61$$ Two bolts would satisfy strength, but a two-bolt hanger offers no redundancy and little rotational restraint, so adopt three 25.4 mm bolts in a single vertical row. Applying the row factor $J_R = 0.95$ for three bolts in a row: $$N_{r,\text{group}} = 3(63.4)(0.95)=181\ \text{kN} \; > \; 102\ \text{kN} \quad\checkmark$$
Connection geometry (O86 Table 12.4.6). All distances are multiples of the 25.4 mm bolt diameter: $$\text{end distance} = 7d_F = 178\ \text{mm} \qquad \text{edge distance} = 1.5d_F = 38\ \text{mm}$$ $$\text{spacing along the row} = 4d_F = 102\ \text{mm} \qquad \text{row spacing} = 3d_F = 76\ \text{mm}$$
Lay the bolts out. Place the single row on the centreline of the 228 mm face, giving 114 mm to each edge — three times the 38 mm minimum. Space the three bolts at 110 mm vertically (above the 102 mm minimum), so the group is 220 mm deep. The column is continuous past the connection in both directions, so the 178 mm end-distance requirement is satisfied automatically; it governs only if the hanger is placed near a column end, which should be avoided.
Check the steel hanger. Each plate carries 51 kN across a net section on the 228 mm face with one 27.4 mm hole: $$A_n = (228-27.4)(8) = 1605\ \text{mm}^2$$ $$T_r = 0.85\phi\,A_nF_u = 0.85(0.9)(1605)(450)=487\ \text{kN} \; \gg \; 51\ \text{kN} \quad\checkmark$$ The 8 mm plate is governed by detailing and bearing, not by tension — provide at least $1.5d_F = 38$ mm of plate beyond each hole and check bolt bearing on the plate at the hanger fabrication stage.

Check — yield-mode expressions. The three modes evaluated above are the Johansen yield equations that CSA O86 Clause 12.4.4 adopts, written for a steel–wood–steel joint. Clause numbering and the exact coefficient set differ slightly between O86 editions; the governing mode and the resulting bolt count are unaffected, but the numerical resistance should be confirmed against the edition in force. The connection has ample reserve (181 kN against 102 kN demand), so a 10–15 % shift in n_u would not change the answer.

Quantity	Value
Factored load per beam	51 kN
Factored load on the bolt group	102 kN
Wood embedment strength, f₁	18.28 MPa (parallel to grain)
Governing yield mode	bolt plastic hinges, n_u = 79.3 kN
Resistance per bolt, N_r	63.4 kN
Bolts required / provided	1.61 → 3 × 25.4 mm, one row
Group resistance	181 kN > 102 ✓
End distance	178 mm (7d)
Edge distance	38 mm minimum; 114 mm provided
Bolt spacing along the row	102 mm minimum; 110 mm provided
Row spacing (if two rows used)	76 mm (3d)
Hanger plate	8 mm, T_r = 487 kN ✓

Topic: Laterally loaded bolted timber connections — the European yield model and O86 geometry rules.

Key relations: $f_1 = 50G(1-0.01d_F)$ parallel to grain and $22G(1-0.01d_F)$ perpendicular; yield modes comparing wood crushing $f_1d_Ft_m$, plate crushing $2f_2d_Ft_s$ and fastener hinging; $N_r=\phi n_u n_F J_R K_DK_{SF}K_T$ with $\phi = 0.8$; minimum geometry $7d_F$ end, $1.5d_F$ edge, $4d_F$ in-row spacing, $3d_F$ between rows.

Why this works: A bolt in wood does not fail like a bolt in steel. Because timber is soft, the fastener embeds into it, and the joint’s capacity depends on a competition between the wood crushing under bearing and the bolt itself bending into a plastic hinge. Johansen’s model enumerates every kinematically admissible combination of those two mechanisms and takes the smallest — a genuine lower-bound plasticity argument. Slender bolts in thick members hinge (as here, with $t_m/d_F = 6.9$); stubby bolts in thin members crush the wood instead. The geometry rules exist for a different reason entirely: they prevent brittle splitting and row tear-out, which the yield model does not describe at all, so they must be satisfied independently of any strength calculation.

Common pitfall: Designing each hanger for its own 51 kN and forgetting that a single bolt passes through both plates and therefore carries 102 kN in double shear — that halves the required bolt count. Also common is using the perpendicular-to-grain embedment strength (which is less than half the parallel value) when the load actually runs down the column axis, and treating the geometry minimums as advisory: a bolt group that satisfies strength but sits 60 mm from a loaded end will split the member long before the bolts yield.

Source: CSA O86, Engineering Design in Wood, Clause 12.4 and Table 12.4.6; Canadian Wood Council, Wood Design Manual, Connections chapter.

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