Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2016 — 98-Civ-A2 Elementary Structural Design, 3 hours, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). The candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in all, of equal value. All seven are solved here, because the set is a study resource. Page 1 states that all loads shown are unfactored, so the load combination is applied by the candidate: the point and distributed loads drawn on the figures are treated as live (1.5) and self-weight as dead (1.25), per NBCC 2020 Table 4.1.3.2 case 2 (1.25D + 1.5L).
Reference texts.
CSA S16:19, Design of Steel Structures (Clauses 13.3, 13.8, 13.13) with the CISC Handbook of Steel Construction, 11th ed., Part 6 section tables.
CSA A23.3:19, Design of Concrete Structures (Clauses 10 and 11), with Brzev & Pao, Reinforced Concrete Design: A Practical Approach, 3rd ed.
CSA O86:19, Engineering Design in Wood, with the Canadian Wood Council Wood Design Manual, 2020.
Kassimali, Structural Analysis, 6th ed. (SI), Chapters 3, 5 and 12 for the determinate/indeterminate analysis feeding each design.
Check — figure readings. Page 3 of 3 of this paper carries all six figures as a single hand-drawn composite. Each panel was read directly from the printed figure; every dimension quoted below is taken from it. One genuine inconsistency was found and is flagged in Question 5: Figure B2 carries two conflicting height dimensions (7 m against the column, 9 m at the right-hand margin) for the same distance between A and the beam.
Given. A frame with a pinned base at A, a 7 m column AB, a rigid corner at B and an 8 m beam BC on a roller at C. A horizontal load of 100 kN acts at B and a vertical load of 400 kN at midspan of BC, both unfactored. fc' = 35 MPa, fy = 400 MPa.
Find. Rectangular sections and the flexural and shear reinforcement for both beam BC and column AB, with the bar layout.
Figure B2 — the determinate frame: two reactions at the pin A and one at the roller C.
Check — conflicting height dimension. Figure B2 dimensions the height from A to the beam twice: 7 m on the dimension line drawn against the column, and 9 m on the right-hand margin. Both dimension lines span the same two extension lines, so one of them is a drafting error on the original. The 7 m value is adopted here because its extension lines terminate on the A support and on joint B themselves. Under the 9 m reading the joint moment would rise from 1050 to 1350 kN·m and the maximum beam moment from 1905 to 2055 kN·m, which would require the beam to deepen from 1500 mm to about 1600 mm; the design method is unchanged. Per NOTE 1 on page 1 this assumption would be submitted with the answer paper.
Approach. The frame has three reactions on one rigid body, so it is statically determinate: take moments about A for the roller reaction, recognise that the entire joint moment at B is delivered up the column by the horizontal load, then design the beam for its peak sagging moment and the column as a flexure-dominated member by strain compatibility.
Factor the loads and add self-weight. Trial a 500 × 1500 beam:
$$H_f=1.5(100)=150\ \text{kN},\qquad P_f=1.5(400)=600\ \text{kN}$$
$$w_{sw}=0.500(1.500)(24)=18.0\ \text{kN/m}\ \Rightarrow\ 1.25(18.0)=22.5\ \text{kN/m}$$
Reactions. Taking moments about A, with the roller at C 8 m away and the horizontal load acting 7 m above A,
$$C_y=\frac{600(4)+150(7)+22.5(8)^2/2}{8}=521.2\ \text{kN}$$
$$A_y=600+22.5(8)-521.2=258.8\ \text{kN},\qquad A_x=150\ \text{kN}$$
Moment at the joint B. The base is pinned, so the column moment is zero at A and rises linearly to
$$M_B=H_f\,h=150(7)=\boxed{1050\ \text{kN}\cdot\text{m}}$$
Checking the same quantity from the beam side, \(C_y(8)-600(4)-22.5(8)^2/2=1050\) kN·m — the two free bodies agree, which confirms the reactions.
Peak beam moment. Working back from C, \(M(x)=C_yx-11.25x^2\) until the point load is reached at x = 4 m:
$$M_{max}=521.2(4)-11.25(4)^2=\boxed{1905\ \text{kN}\cdot\text{m}}$$
It is worth pausing on the structure of this answer. Remove the 100 kN horizontal load and \(A_x\) vanishes, the joint moment goes to zero and BC becomes an ordinary simple span. Everything that makes this a frame problem rather than a beam problem comes from that single horizontal force, and it enters twice: once as the 1050 kN·m hogging moment at B, and once through the extra 131 kN it adds to the roller reaction.
Beam flexural steel. With two layers of 30M bars, \(d=1500-(40+11.3+15.0+27.5)=1406\) mm. For 1905 kN·m the requirement is 4225 mm²; provide 8–30M = 5600 mm² in two layers of four:
$$a=\frac{0.85(5600)(400)}{0.7975(0.65)(35)(500)}=210\ \text{mm},\qquad
M_r=0.85(5600)(400)\left(1406-105\right)=\boxed{2477\ \text{kN}\cdot\text{m}}$$
At the joint the beam hogs 1050 kN·m; 4–30M (2800 mm²) top steel gives \(M_r=1290\) kN·m. Minimum steel is 2218 mm², satisfied at both faces. Clear bar spacing in a 500 mm width is 92 mm, comfortably above the 42 mm minimum.
Beam shear. \(d_v=\max(0.9d,0.72h)=1266\) mm; at that distance from C the shear is \(521.2-22.5(1.266)=492.7\) kN. With 10M stirrups at 250 mm,
$$V_c=0.65(0.18)\sqrt{35}(500)(1266)=438\ \text{kN},\qquad V_s=\frac{0.85(200)(400)(1266)\cot35^\circ}{250}=492\ \text{kN}$$
$$V_r=\boxed{930\ \text{kN}}\ >\ 492.7\ \text{kN}$$
Column AB. Adopt 500 × 1200 with the 1200 mm dimension in the plane of the frame. The axial load is small — \(N_f=258.8+1.25(0.5)(1.2)(24)(7)=384.8\) kN — against a moment of 1050 kN·m, so
$$e=\frac{M_f}{N_f}=\frac{1050\times10^{3}}{384.8}=2729\ \text{mm}\ \gg\ h$$
This is a flexure-dominated member: the pure-axial expression \(P_{r,max}=0.80[\alpha_1\phi_cf_c'(A_g-A_{st})+\phi_sf_yA_{st}]\) is meaningless here and must not be used. Solving the strain-compatibility equations at \(N_f=384.8\) kN with 5–30M each face (3500 mm²) puts the neutral axis at c = 103 mm and gives
$$M_r=\boxed{1494\ \text{kN}\cdot\text{m}}\ >\ 1050\ \text{kN}\cdot\text{m}\qquad(\text{utilisation }0.70)$$
The steel ratio is \(\rho=7000/600\,000=1.17\ \%\), which satisfies the 1 % minimum of Clause 10.9.1 — and it is that minimum, not strength, that fixes the bar count.
Column shear and ties. The column shear is the full 150 kN over its whole height. With \(d_v=1020\) mm, \(V_c=0.65(0.18)\sqrt{35}(500)(1020)=353\) kN > 150 kN, so only nominal ties are needed: 10M ties at 350 mm (below the 16-bar-diameter limit of 478 mm and the 48-tie-diameter limit of 542 mm).
Reinforcement layout. Beam BC: 8–30M bottom in two layers of four, curtailed to four bars beyond the point of contraflexure; 4–30M top running continuously from the joint into the span and anchored into the column with a standard hook; 10M closed stirrups at 250 mm throughout. Column AB: 5–30M at each 500 mm face (ten bars in total), continuous from the pin at A into the joint at B, with the outer face bars — the tension face under the 100 kN load — fully anchored into the beam; 10M ties at 350 mm.
Check — column slenderness. The roller at C provides no lateral restraint, so the frame sways and the column effective length factor is at least 2.0 (pinned base). Then \(kl_u/r=2(7000)/(0.3\times1200)=38.9\), which exceeds the 22 threshold of A23.3 Clause 10.15.2 for a member in single curvature, so a rigorous design would apply the moment magnifier of Clause 10.15.3. Because the section is only 70 % utilised the magnified moment still fits; the reserve should nonetheless be confirmed against the actual sway stiffness of the completed building.
