07-Str-A2 · December 2018

Question 3 of 7: A3 — Stub beam connection and steel tie

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

Notes on this paper

Paper format. National Examinations, December 2018 — 07-Str-A2 Elementary Structural Design. Three hours, CLOSED BOOK (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3), Part C (C1, timber to CSA O86). A candidate submits five solutions — two from Part A, two from Part B and the one question in Part C — all of equal value. All seven questions are solved below, because this set is a study resource rather than an exam script. Page 1 states that all loads shown are unfactored, so every load case is factored here (1.5 on the specified live/imposed loads, 1.25 on self weight) before any resistance is compared against it.

Reference texts.

CSA S16:19, Design of Steel Structures — and CISC, Handbook of Steel Construction, 11th ed. (section tables, Table 2 class limits, Clause 13.8 interaction, Clause 13.5 effective widths).
CSA A23.3:19, Design of Concrete Structures — and Cement Association of Canada, Concrete Design Handbook, 4th ed.
CSA O86:19, Engineering Design in Wood — and Canadian Wood Council, Wood Design Manual (glulam selection tables).
Kulak & Grondin, Limit States Design in Structural Steel, 11th ed., Chs. 4, 5, 8, 9.
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition), Chs. 4, 6, 11.
National Building Code of Canada, Table 4.1.3.2 (load combinations).

Check — assumptions declared under Note 1 of the paper. (i) In A2 the single dimension h is the wall thickness of the whole fabricated panel (top plate and both stems), which is how the figure dimensions it. (ii) In A3 the bolted joint at the column is a simple (shear-only) connection and the steel tie is a two-force member, which is what makes the stub beam determinate. (iii) In B1/B3 the left column A–B is taken as 6 m high, level with the roller at D, as Figure B1 draws it; the beam self weight is included, the column self weight is not (it does not change any design action at C). (iv) In B2 and C1 the beam is drawn with a tapered soffit, but part (a) asks for a uniform cross-section, so a prismatic member is designed.

Question 3: A3 — Stub beam connection and steel tie (12 + 8 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.

Quantity	Value
Stub beam	W360 x 79, $d$ = 354 mm, web $w$ = 9.4 mm, $F_y$ = 350 MPa, $F_u$ = 450 MPa
Column	W610 x 140, flange $t$ = 22.2 mm
Point load (unfactored)	100 kN at 1.5 m from the column face
Vertical steel tie	at 3.0 m from the column face (the beam end)
Beam self weight	79 kg/m = 0.775 kN/m

Find. The bolted beam-to-column connection and the cross-section of the steel tie, both to CSA S16.

Figure A3 — the stub beam is bolted to the column flange at one end and hung from a vertical steel tie at the other, with the 100 kN load at midspan.

Approach. The bolted joint is a simple shear connection and the tie is a two-force member, so the stub beam is a determinate simply supported span of 3.0 m. Find the two end reactions under factored load, then size a double-angle bolted connection for the reaction and a plate tie for the hanger force, checking every limit state of each.

Factored loads. The 100 kN is a specified imposed load and the beam self weight is dead load: $$P_f=1.5(100)=150\ \text{kN},\qquad w_f=1.25(0.775)=0.969\ \text{kN/m}$$
Reactions on the 3.0 m determinate span. The load sits at midspan, so the point load splits evenly and the self weight splits evenly: $$R_f=T_f=\frac{150}{2}+\frac{0.969(3.0)}{2}=75.0+1.45=\boxed{76.5\ \text{kN}}$$ The bolted connection therefore transfers 76.5 kN of vertical shear (and no moment), and the tie carries 76.5 kN of tension.
Choose the connection type. Use a pair of angles 2L90 x 90 x 8 bolted to the beam web with M20 ASTM A325 bolts in double shear, and bolted to the column flange with M20 bolts in single shear. Because the beam is 354 mm deep, two bolts at 75 mm pitch fit comfortably within the web depth and give the joint the rotational flexibility a simple connection needs.
Bolt shear at the beam web. With threads intercepted by a shear plane (Clause 13.12.1.2, the 0.70 multiplier) and two shear planes per bolt, $$V_r=0.60\,\phi_b\,n\,m\,A_b\,F_u=0.60(0.80)(1)(2)(314)(830)(0.70)=175.1\ \text{kN per bolt}$$ Two bolts give 350.3 kN against the 76.5 kN required.
Bearing on the beam web. Clause 13.12.1.2(a) with $\phi_{br}=0.80$ and the 9.4 mm web: $$B_r=3\phi_{br}\,t\,d\,F_u=3(0.80)(9.4)(20)(450)=203.0\ \text{kN per bolt}$$ so bolt shear, at 175.1 kN, remains the weaker of the two. Bearing on the paired 8 mm angles (16 mm of combined thickness) gives 345.6 kN per bolt and does not govern.
Angle-to-column bolts. Four M20 bolts (two per angle leg) act in single shear at $0.60(0.80)(314)(830)(0.70)=87.6$ kN each, a group resistance of 350.3 kN. The W610 x 140 flange is 22.2 mm thick, so bearing and flange bending are not concerns for a load of this size.
Connection verdict. The governing resistance is $$V_{r,\text{group}}=2(175.1)=\boxed{350.3\ \text{kN}}\quad\text{against}\quad V_f=76.5\ \text{kN}$$ a utilisation of 0.22. Two bolts is the practical minimum for a shear connection, so the joint is minimum-size-governed rather than load-governed.
Design the steel tie. Try a 60 x 12 mm plate with a single M20 bolt (22 mm hole) at each end. Gross-section yielding (Clause 13.2(a)(i)): $$T_r=\phi A_g F_y=0.90(60)(12)(350)=226.8\ \text{kN}$$ Net-section fracture (Clause 13.2(a)(iii)) with a shear-lag factor $U=0.85$ appropriate to a plate connected through its full width at a single bolt line: $$T_r=0.85\,\phi_u A_{ne}F_u=0.85(0.75)\left[(60-22)(12)\right](450)=130.8\ \text{kN}$$ Fracture governs, so $T_r = 130.8$ kN against $T_f = 76.5$ kN — a utilisation of 0.58.
Slenderness of the tie. For a 12 mm plate $r_{\min}=t/\sqrt{12}=3.46$ mm; Clause 10.4.2.2 recommends $L/r \le 300$ for tension members, so the unbraced tie length must not exceed $300(3.46)=1.04$ m. A longer tie should be a round bar or a small HSS instead of a flat plate.

Both components are governed by minimum practical sizes rather than by the applied load, which is the normal outcome for a 100 kN stub bracket in heavy framing. The one result that must not be skipped is the net-section check on the tie: it is 42 % below the gross-yield value, so a designer who stops at $\phi A_gF_y$ overstates the tie by nearly a factor of two.

Result	Value
Factored point load $P_f$ / self weight $w_f$	150 kN / 0.969 kN/m
Connection shear $V_f$ = tie tension $T_f$	76.5 kN
Connection adopted	2L90 x 90 x 8 with 2-M20 A325 bolts (double shear) to the web, 4-M20 to the column flange
Bolt double-shear resistance / bearing on the 9.4 mm web	175.1 kN / 203.0 kN per bolt
Connection group resistance $V_r$ / utilisation	350.3 kN / 0.22
Tie adopted	60 x 12 mm plate, M20 bolt each end
Tie gross yield / net fracture (governs)	226.8 kN / 130.8 kN
Tie utilisation	0.58

Topic. Simple shear connections and tension members: CSA S16 Clauses 13.2, 13.12 and 13.10.

Key relations. Bolt shear $V_r = 0.60\phi_b n m A_b F_u$ (times 0.70 when threads are intercepted); bearing $B_r = 3\phi_{br} t d F_u$; tension member $T_r = \min\left[\phi A_gF_y,\ 0.85\phi_u A_{ne}F_u\right]$; simple-span reactions from statics.

Why this works. Declaring the joint a simple connection and the tie a two-force member is not a convenience — it is what makes the stub beam statically determinate, and it is the assumption the detailing must then honour. A pair of thin angles bolted only to the web is flexible enough to rotate as the beam deflects and therefore genuinely delivers the shear-only reaction the analysis assumed; a full-depth welded moment joint would not, and the beam would then attract an end moment nobody designed for. Choosing the analysis model and choosing the detail are one decision, not two.

Common pitfall. Sizing a tension member on gross yielding alone. Once a hole is punched, net-section fracture is a separate limit state with a different resistance factor ($\phi_u = 0.75$ instead of $\phi = 0.90$) and it very often governs, as it does here.

Source. CSA S16:19, Clauses 13.2, 13.10, 13.12.1; CISC Handbook of Steel Construction, Part 3 (bolted connections); Kulak & Grondin, Chs. 3 and 9.

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