16-Civ-A2 Elementary Structural Design · May 2015

Question 1 of 7: A1 — Design of the bolted connection at B

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

Notes on this paper

Paper format. National Examinations, May 2015 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, closed book (textbooks and design handbooks permitted; one approved Casio or Sharp calculator). Seven questions in three parts: Part A (A1–A3, structural 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, 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”, but the paper never splits the figure loads into dead and live components. Every applied figure load in Parts A and B is therefore treated as live load and factored by 1.5; concrete self-weight, where it is not expressly excluded, would be dead load at 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). Part C names its components, so 1.25D + 1.5L is applied directly there. If an examiner intended a different split, the method below is unchanged — only the numerical factor moves.

Check — figure dimensions. All four figures are hand sketches on page 3; every dimension below was read from the printed figures. Figure A2 reads 80 – 300 – 80 across the bottom plate (overall 460 mm wide) with the 300 mm dimension spanning the two web centrelines, and 250 mm overall depth. Figure B1 reads 3.0 m top, 2.5 m bottom, 2.0 m deep, walls “300 constant” and cover “70 typical”. Figure B2 reads 5 m + 5 m span, 4 m + 4 m column height, roller at A and hinge at D.

Question 1: A1 — Design of the bolted connection at B (10 + 10 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
Beam BC	W360×79, span 3.0 + 3.0 = 6.0 m, roller at C
Column AB	W460×106, fixed base at A, 8.0 m high
Point load on BC	300 kN at midspan (unfactored)
Steel	G40.21 350W: F_y = 350 MPa, F_u = 450 MPa
Bolts	ASTM A325, F_u = 825 MPa

Find. A bolted connection at B that safely transfers the factored end reaction of beam BC into the column, with every limit state (bolt shear, bearing, angle shear, block shear) shown to be satisfied.

[Figure not reproduced. See the official exam paper.]

Approach. Establish first that the joint at B must be a shear (pin) connection for the frame to be determinate as the question states, then take the beam end reaction as the only force on the connection and check bolt shear, bearing, gross shear on the angles and block shear in turn.

Confirm the joint is a shear connection. The frame has a fixed base at A (three reaction components) and a roller at C (one), so 4 reactions act on a single rigid body. A rigid joint at B would leave the structure indeterminate to the first degree, contradicting the word determinate in the question. One release — a moment release at B — restores determinacy: $$ n = r - 3 - (\text{releases}) = 4 - 3 - 1 = 0 $$ The bolted joint therefore carries shear only, and beam BC behaves as a simple span with a pin at B and a roller at C.
Factor the load and take the beam end reaction. The 300 kN is unfactored (page-1 note 6) and is treated as live load, so $$ P_f = 1.5 \times 300 = 450~\text{kN} $$ For a simply supported span with the load at midspan the reactions are equal, giving the design shear on the connection $$ V_f = \frac{P_f}{2} = \boxed{225~\text{kN}} $$
Choose the connection type and bolt size. Adopt a standard double-angle framing connection: two L102×102×9.5 angles, one each side of the beam web, bolted to the web with M20 A325 bolts in double shear and to the column flange with M20 A325 bolts in single shear. Assume threads intercept the shear planes (the conservative and usual assumption for short grips), for which S16 Clause 13.12.1.2(c) reduces the shear resistance by 0.70: $$ V_r = 0.60\,\phi_b\, n\, m\, A_b\, F_u \times 0.70 $$ with $\phi_b = 0.80$, $A_b = 314~\text{mm}^2$ and $F_u = 825$ MPa. Per bolt per shear plane, $$ V_r = 0.60 (0.80)(314)(825)(0.70) = 87.0~\text{kN} $$
Size the beam-web bolt group. Each web bolt has two shear planes. Using three bolts at 75 mm pitch, $$ V_r = 3 \times 2 \times 87.0 = 522~\text{kN} \; > \; 225~\text{kN} $$ The utilisation is $225/522 = 0.43$. Two bolts (348 kN) would also be sufficient in strength, but three are adopted so the angle leg is deep enough to give the joint the rotational flexibility a simple connection needs while still developing the reaction with reserve.
Check the column-side bolts. The outstanding legs bear on the column flange with three bolts per angle in single shear, six in total: $$ V_r = 6 \times 1 \times 87.0 = 522~\text{kN} \; > \; 225~\text{kN} $$
Check bearing (S16 13.12.1.2a). Bearing is governed by the thinnest connected ply, the W360×79 web at $t = 9.4$ mm: $$ B_r = 3\,\phi_{br}\, t\, d\, F_u = 3(0.80)(9.4)(20)(450) = 203~\text{kN per bolt} $$ Three bolts give 609 kN. On the paired angles $t = 2 \times 9.5 = 19$ mm gives 410 kN per bolt, and the 20.6 mm column flange is thicker still. Bearing does not govern.
Check gross shear on the angles (S16 13.4.2). With three bolts at 75 mm pitch and 35 mm end distances the angle length is 230 mm: $$ V_r = 0.66\,\phi\, A_g\, F_y = 0.66(0.90)(2 \times 230 \times 9.5)(350) = 908~\text{kN} $$
Check block shear in the beam web (S16 13.11). Tear-out of the bolt line towards the beam end, with a 50 mm end distance and 22 mm holes, gives $A_{gv} = 9.4(2 \times 75 + 35) = 1739~\text{mm}^2$ and $A_{nt} = 9.4(50 - 11) = 367~\text{mm}^2$. For a coped web with a single bolt line $U_t = 0.30$: $$ T_r + V_r = \phi_u\left[U_t A_{nt} F_u + 0.60 A_{gv}\frac{F_y + F_u}{2}\right] $$ $$ = 0.75\left[0.30(367)(450) + 0.60(1739)(400)\right] = 350~\text{kN} $$ Block shear is the lowest of the four resistances and therefore governs, but it still exceeds the 225 kN demand by 56 percent.
Verify the detailing rules. The 75 mm pitch exceeds the minimum $2.7d = 54$ mm, the 35 mm edge distance meets the 26 mm minimum for a sheared edge on an M20 bolt, and the 230 mm angle length leaves the connection well clear of the W360×79 fillet region.

Every limit state is satisfied with the governing resistance 350 kN against a demand of 225 kN, so the connection below is adopted.

Item	Value
Factored beam end reaction, V_f	225 kN
Connection	2 – L102×102×9.5 angles, 230 mm long
Beam web bolts	3 – M20 A325, double shear, 75 mm pitch
Column flange bolts	6 – M20 A325 (3 per angle), single shear
Bolt shear resistance	522 kN
Bearing on the W360×79 web	203 kN per bolt (609 kN total)
Gross shear on the angle pair	908 kN
Block shear in the beam web	350 kN (governs)
Governing utilisation	225 / 350 = 0.64

Topic. Simple (shear) beam-to-column connections in limit states steel design.

Key relations. Bolt shear $V_r = 0.60\phi_b n m A_b F_u$ (× 0.70 when threads intercept the shear plane); bearing $B_r = 3\phi_{br} t d F_u$; gross shear $V_r = 0.66\phi A_g F_y$; block shear $T_r + V_r = \phi_u[U_t A_{nt}F_u + 0.60 A_{gv}(F_y+F_u)/2]$.

Why this works. The word determinate in the question is not decoration — it is the statement that fixes the connection type. Counting reactions and releases shows that the joint at B must release moment, which collapses a potentially messy frame problem into a simple-span beam whose end reaction is half the point load. Once the design force is a pure vertical shear, the connection becomes a sequence of independent limit-state checks on the same force, and the design is complete when the smallest of them exceeds the demand. Double-angle connections are the natural choice because the thin outstanding legs deform enough to relieve any end moment that does try to build up, keeping the real behaviour consistent with the pinned idealisation used in the analysis.

Common pitfall. Assuming a rigid joint at B and analysing an indeterminate frame — this both contradicts the question and produces a connection moment the bolts were never sized for. A second common error is to use the full unreduced bolt shear (124 kN per plane) by assuming threads are excluded from the shear plane, which is only valid if the grip is detailed to guarantee it.

Source. CSA S16, Design of Steel Structures, Clauses 13.4, 13.11 and 13.12; CISC Handbook of Steel Construction, Part 3 (bolted connections); Kulak & Grondin, Limit States Design in Structural Steel, Chapter 9.

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