16-Civ-A2 Elementary Structural Design · May 2015

Question 5 of 7: B2 — Design of beam-column AB in the determinate frame

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 5: B2 — Design of beam-column AB in the determinate frame (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
Frame geometry	A(0, 0) – B(0, 8) – C(10, 8) – D(10, 0), metres
Supports	roller at A (vertical only), hinge at D
Loads (unfactored)	400 kN at B, 350 kN at midspan of BC, 400 kN at C, 80 kN horizontal at 4 m
Materials	f'_c = 35 MPa, f_y = 400 MPa
Self-weight	ignored, as instructed

Find. Cross-sectional dimensions and longitudinal plus transverse reinforcement for member AB, which carries axial compression together with bending.

Figure B2 — determinate reinforced concrete frame: roller at A, hinge at D. Because the roller carries no horizontal force, the whole 80 kN thrust is resisted at D and column AB is unbent below the 4 m level.

Approach. Analyse the determinate frame for reactions, extract the axial force and moment diagram for AB, size a trial rectangular section, then find the steel by strain compatibility at the actual factored axial load — not by a pure-axial formula, because the eccentricity is large.

Factor the loads. Multiplying the figure loads by 1.5: 600 kN at B, 525 kN at midspan, 600 kN at C, and 120 kN horizontal at the 4 m level.
Solve the reactions. The roller at A takes vertical force only, so all horizontal equilibrium falls on the hinge at D: $$ \sum F_x = 0 \;\Rightarrow\; D_x = 120~\text{kN (acting to the left)} $$ Taking moments about D, $$ A_y = \frac{600(10) + 525(5) - 120(4)}{10} = \frac{8145}{10} = 814.5~\text{kN} $$ $$ D_y = 600 + 525 + 600 - 814.5 = 910.5~\text{kN} $$
Internal actions in AB. Below the 80 kN load the column carries no horizontal force at all, because the roller at A supplies none — so the moment is identically zero over the lower 4 m. Above that level the 120 kN shear runs up the column and the moment grows linearly to $$ M_f = 120(4.0) = \boxed{480~\text{kN}\cdot\text{m at B}} $$ The axial force is constant over the full height, since both the 600 kN head load and the 214.5 kN beam end shear are applied at B: $$ P_f = A_y = 814.5~\text{kN} \quad\text{(compression)} $$
Assess the eccentricity. $$ e = \frac{M_f}{P_f} = \frac{480\times10^6}{814.5\times10^3} = 589~\text{mm} $$ This is comparable with the whole member depth, so AB is decisively flexure-dominated. The pure-axial expression $P_{r,\max} = 0.80[\alpha_1\phi_c f'_c(A_g - A_{st}) + \phi_s f_y A_{st}]$ must not be used; the section has to be designed on its interaction diagram.
Choose a trial section. Take 500 mm wide × 700 mm deep, bending about the 700 mm dimension (the plane of the frame). With 40 mm cover, 10M ties and 30M longitudinal bars, $$ d = 700 - 40 - 11 - 15 = 634~\text{mm}, \qquad d' = 66~\text{mm} $$ Reinforce symmetrically, equal steel at each face, as is standard for a member that may see moment reversal during construction.
Strain compatibility at $P_f$. With $\varepsilon_{cu} = 0.0035$, the forces on a section with neutral axis at depth c are $$ C_c = \alpha_1\phi_c f'_c b (\beta_1 c), \quad C_s = \phi_s A_s' f_s', \quad T = \phi_s A_s f_s $$ $$ f_s' = 0.0035\,\frac{c - d'}{c}E_s \le f_y, \qquad f_s = 0.0035\,\frac{d - c}{c}E_s \le f_y $$ Solving $P_r = C_c + C_s - T = 814.5$ kN together with $M_r = 480$ kN·m about mid-depth gives $c = 115$ mm and a required $A_s = A_s' = 1215~\text{mm}^2$ per face.
Select and check the bars. Provide 3–30M at each face (2100 mm² per face, 4200 mm² total). The gross steel ratio is $$ \rho_g = \frac{4200}{500(700)} = 0.0120 $$ which satisfies the A23.3 Clause 10.9.1 and 10.9.2 limits of 1 percent minimum and 8 percent maximum. Re-solving the interaction at the provided steel, the neutral axis at $P_f = 814.5$ kN sits at $c = 120~\text{mm}$ and $$ M_r = \boxed{649~\text{kN}\cdot\text{m}} \;>\; M_f = 480~\text{kN}\cdot\text{m} $$ a flexural utilisation of 0.74 at the design axial load.
Ties. With 30M longitudinal bars and 10M ties, A23.3 Clause 7.6.5.2 limits the spacing to the least of $$ 16 d_b = 478~\text{mm}, \quad 48 d_{tie} = 542~\text{mm}, \quad b = 500~\text{mm} $$ Adopt 10M ties at 400 mm throughout, closed to 135° hooks, with the first tie 100 mm from the joint face.

Check — slenderness. With $r \approx 0.3h = 210$ mm and no lateral bracing anywhere in the frame, an effective length factor of k = 2.0 gives $k l_u/r = 2(8000)/210 = 76$, far above the A23.3 Clause 10.15.2 threshold. The member is genuinely slender and a rigorous design would magnify the moment for second-order effects. The 175 kN·m of flexural reserve computed above (649 against 480) corresponds to a permissible magnifier of 1.35, which comfortably brackets the magnification a P-delta analysis of this frame would produce; if a stiffer answer is wanted, deepening the section to 500 × 800 raises M_r without changing the bar count.

Quantity	Value
Reactions	A_y = 814.5 kN, D_x = 120 kN, D_y = 910.5 kN
Factored axial force, P_f	814.5 kN (compression, constant)
Factored moment, M_f	480 kN·m at B (zero below the 80 kN load)
Eccentricity, e	589 mm — flexure-dominated
Section	500 mm × 700 mm
Longitudinal steel	6–30M (3 per face), ρ_g = 1.20 percent
Required steel per face	1215 mm² (2100 mm² provided)
Neutral-axis depth at P_f	120 mm
Moment resistance at P_f	649 kN·m (utilisation 0.74)
Ties	10M at 400 mm, closed, 135° hooks

Topic. Design of a reinforced concrete beam-column by strain compatibility.

Key relations. $P_r = \alpha_1\phi_c f'_c b\beta_1 c + \phi_s A_s'f_s' - \phi_s A_s f_s$; $M_r = C_c(h/2 - a/2) + C_s(h/2 - d') + T(d - h/2)$; $f_s = 0.0035 E_s (d-c)/c \le f_y$; $e = M_f/P_f$.

Why this works. A member carrying both thrust and moment has no single strength — it has an interaction envelope, and the eccentricity tells you which part of that envelope you are working in. Small eccentricity means the section is essentially a column and fails by concrete crushing with all the steel in compression; large eccentricity means it is essentially a beam with a helpful axial load and fails by steel yielding. Here e = 589 mm against a 700 mm depth places the section firmly on the tension-controlled branch, where the axial load actually increases the moment capacity by delaying the tension steel's yield. Solving the two equilibrium equations simultaneously for the neutral-axis depth and the steel area is the only way to land on the correct point of that envelope. Note also how the roller at A shapes the whole problem: it makes the lower half of the column a pure axial strut and pushes all the bending into the upper half.

Common pitfall. Designing this member with the maximum axial-load formula, which assumes concentric loading and would suggest an enormous capacity that the section cannot deliver at e = 589 mm. A second error is to forget that the moment in AB is zero below the horizontal load, and to detail the heavy reinforcement over the full height when only the upper region needs it — although in practice the bars are run through for constructability.

Source. CSA A23.3, Design of Concrete Structures, Clauses 10.1, 10.9, 10.10 and 10.15; CAC Concrete Design Handbook, Chapter 3 (column interaction diagrams); MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Chapter 11.

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