07-Str-A2 · May 2016

Question 3 of 7: A3 — Steel beam ABC of the determinate frame

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

Notes on this paper

Paper format. National Examinations — May 2016, 07-Str-A2 Elementary Structural Design. Three-hour closed-book examination (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, structural steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3), Part C (C1, timber to 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, because the set is a study resource rather than an examination script. Page 1 states that all loads shown are unfactored, so the load factors are applied within each solution.

Reference texts.

CSA S16:19, Design of Steel Structures — Clauses 11 (class of section), 13.3 (compressive resistance), 13.5–13.6 (bending), 13.8 (axial compression and bending), 13.13 (welds).
CISC, Handbook of Steel Construction, 12th ed. — section property tables, weld tables.
CSA A23.3:19, Design of Concrete Structures — Clauses 10 (flexure and axial load), 11 (shear), 7.6 (ties).
CSA O86:19, Engineering Design in Wood, with the Canadian Wood Council Wood Design Manual 2020 — Clause 7 (glued-laminated timber).
NBCC 2020 — Table 4.1.3.2 load combinations.

Check — assumptions common to the whole paper. (i) Every load drawn in Figures A1, A2, A3, B3 and quoted in C1 is a specified (unfactored) load; unless a question names the load type, the point loads are treated as live (factor 1.5) and concrete/steel self-weight as dead (factor 1.25), per NBCC Table 4.1.3.2 case 2. (ii) Steel is CSA G40.21 350W, so F_y = 350 MPa, E = 200 000 MPa. (iii) Question B3 states f'_c = 35 MPa and f_y = 400 MPa; because B1 and B2 quote no material strengths, the same pair is adopted for them and the assumption is stated in each solution. (iv) Section properties are recomputed from the nominal plate dimensions of each rolled shape rather than read off a table, so every number below is reproducible; they agree with the CISC Handbook to better than 1 %.

Question 3: A3 — Steel beam ABC of the determinate frame (8 + 12)

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 ABC	horizontal, 8.0 m from A to C in four 2.0 m bays
Support at A	roller (vertical reaction only)
Internal hinge at B	4.0 m from A
Column CD	8.0 m, rigidly framed to the beam at C, fixed at D
Vertical point loads (unfactored)	250 kN at 2 m, at 6 m and at C (8 m)
Horizontal load on the column (unfactored)	60 kN, applied 4.0 m below C
Steel	G40.21 350W, F_y = 350 MPa

Find. The bending-moment and shear envelope in beam ABC under factored loads, and the lightest rolled W section that satisfies flexure, shear and class requirements.

[Figure not reproduced: Figure A3 (redrawn). Roller at A, internal hinge at B, rigid corner at C, fixed base at D — four reaction components less three equilibrium equations less one condition equation gives a determinate frame. See the official exam paper.]

Approach. Confirm determinacy, use the zero-moment condition at the hinge to release the left span, work rightwards to the rigid corner at C, then select a W section from the plastic moment requirement and verify class, shear and lateral support.

Confirm the structure is determinate. The frame has four reaction components (one at the roller A, three at the fixed base D) on a single rigid body, with one internal hinge supplying one condition equation: $$\text{DSI} = r - 3 - c = 4 - 3 - 1 = 0$$ so the frame can be solved by statics alone — which is exactly what the word “determinate” in the question is telling us.
Factor the loads and add the beam’s own weight. Anticipating a section of roughly 174 kg/m (confirmed in Step 6): $$P_{f} = 1.5(250) = 375\ \text{kN}, \quad H_{f} = 1.5(60) = 90\ \text{kN}, \quad w_{f} = 1.25(1.707) = 2.13\ \text{kN/m}$$
Release the left span at the hinge. Taking moments about B for the free body A–B, where the roller at A supplies only a vertical force: $$\sum M_{B} = 0:\quad A_{y}(4.0) - 375(2.0) - 2.13(4.0)(2.0) = 0 \quad\Longrightarrow\quad A_{y} = \boxed{191.8\ \text{kN}}$$ The sagging peak sits under the first point load: $$M_{\max}^{+} = 191.8(2.0) - \frac{2.13(2.0)^{2}}{2} = 379.3\ \text{kN}\cdot\text{m}$$ and the moment returns to zero at B, as the hinge demands.
Carry the hinge force across into span BC. Vertical equilibrium of A–B gives the force the right-hand segment must supply: $$V_{B} = 375 + 2.13(4.0) - 191.8 = 191.8\ \text{kN}\ \text{downward on BC}$$ Because A is a roller, no horizontal force can travel along the beam, so the beam carries zero axial force — a point worth noting before Question 5, where the same fact controls the column.
Hogging moment at the rigid corner C. Taking moments about C for the free body B–C: $$M_{C} = 191.8(4.0) + 375(2.0) + 2.13(4.0)(2.0) = \boxed{1534\ \text{kN}\cdot\text{m}}$$ $$V_{C} = 191.8 + 375 + 375 + 2.13(4.0) = 950.3\ \text{kN}$$ The hogging moment at C is four times the sagging peak, so the corner governs the beam design outright.
Select the section. Requiring φZ_xF_y ≥ 1534 kN·m: $$Z_{x,\text{req}} = \frac{1534\times10^{6}}{0.9(350)} = 4.87\times10^{6}\ \text{mm}^{3}$$ Try W610 × 174 (d = 616, b = 325, t = 21.6, w = 14.0 mm): $$Z_{x} = 2bt\left(\frac{d-t}{2}\right)+\frac{w(d-2t)^{2}}{4} = 5.32\times10^{6}\ \text{mm}^{3}$$ $$M_{r} = 0.9\left(5.32\times10^{6}\right)(350) = \boxed{1676\ \text{kN}\cdot\text{m}} \;>\; 1534\ \text{kN}\cdot\text{m}$$ Utilisation 0.92. The next lighter section in the series, W610 × 155, gives M_r = 1476 kN·m — 4 % short — so W610 × 174 is the economical choice.
Class of section. Against the Class 1 limits 145/√F_y = 7.75 and 1100/√F_y = 58.8: $$\begin{aligned} \frac{b}{2t} &= \frac{162.5}{21.6} = 7.52\ \checkmark \\ \frac{h}{w} &= \frac{572.8}{14.0} = 40.9\ \checkmark \end{aligned}$$ Class 1 confirmed, so φZ_xF_y is legitimately available and the plastic moment used in Step 6 stands.
Shear at the corner. With h/w = 40.9 below 1014/√F_y = 54.2, $$V_{r} = \phi\left(0.66F_{y}\right)dw = 0.9(0.66)(350)(616)(14.0) = 1793\ \text{kN} \;>\; 950\ \text{kN}\ \checkmark$$ Combined moment-and-shear interaction (Clause 14.6) need not be checked because V_f/V_r = 0.53 is below the 0.60 threshold.

Check — lateral support. The hogging region near C puts the bottom flange in compression, where a floor deck offers no restraint. The design above assumes the beam is laterally braced at the load points and at C, giving an unbraced length of 2.0 m; at that spacing L < L_u and the full plastic moment is available. If the bottom flange were unbraced over the whole 4 m of span BC, M_r would fall by roughly 10 % and W610 × 217 would be required instead. Show the bottom-flange bracing on the drawings.

Result	Value
Degree of static indeterminacy	0 — determinate
Reaction at roller A (unfactored / factored)	125 kN / 191.8 kN
Hogging moment at C (unfactored / factored)	1000 kN·m / 1534 kN·m
Sagging peak at 2 m (factored)	379.3 kN·m
Shear at C (factored)	950.3 kN
Base moment at D (unfactored)	1240 kN·m
Required plastic modulus	4.87×10⁶ mm³
Beam ABC	W610 × 174, Class 1, M_r = 1676 kN·m (utilisation 0.92)
Shear resistance	1793 kN (utilisation 0.53)

Topic. Statics of a determinate frame containing an internal hinge, and plastic-section design of the resulting beam.

Key relations. DSI = r − 3 − c for a single-body plane frame; zero moment at an internal hinge; M_r = φZ_xF_y for a Class 1 or 2 section (S16 Clause 13.5); V_r = φ(0.66F_y)dw for a stocky web (Clause 13.4.1.1); Class 1 limits b/2t ≤ 145/√F_y and h/w ≤ 1100/√F_y.

Why this works. An internal hinge converts a moment-carrying joint into a shear-only joint, which supplies the fourth equation the four reactions need. In practice this means the frame can be dismantled: everything to the left of the hinge is a simple span that can be solved on its own, and the shear it hands over at the hinge becomes a known load on the right-hand segment. Because the roller at A cannot deliver horizontal force, the beam is free of axial load and the whole horizontal thrust from the 60 kN load travels to the fixed base — a detail that leaves the beam a pure flexural member and lets it be designed on φZF_y alone. The hogging moment at C is large because the segment BC is effectively a 4 m cantilever hanging off the hinge and loaded by both the hinge shear and two point loads.

Common pitfall. Treating the beam as a simply-supported member spanning A to C and designing for the sagging moment, which misses the hogging moment at C by a factor of four. The second is forgetting that the bottom flange is in compression near C and quietly assuming full lateral support — the reason the assumption is flagged explicitly above.

Source. CSA S16:19 Clauses 11.2, 13.4, 13.5 and 13.6; CISC Handbook of Steel Construction, 12th ed., Part 5 (beam selection tables and moment resistances).

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