Question 6 of 7: B3 — Design of Column BC in a Determinate Concrete Frame
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2013 — 98-Civ-A2 Elementary Structural Design, 3-hour duration, CLOSED BOOK (handbooks and textbooks permitted, no notes). Seven questions are printed: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete) and 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 total, all of equal value. All seven are solved here, because the set is a study resource. All loads shown on the figures are unfactored; NBCC load combination 1.25D + 1.5L is applied throughout. Solutions are to CSA-S16 (steel), CAN/CSA-A23.3 (concrete) and CSA-O86 (timber), as the paper directs.
CISC, Handbook of Steel Construction — section-property tables for C-shapes, W-shapes and hollow structural sections.
CSA A23.3, Design of Concrete Structures — Clauses 10 (flexure and axial load) and 11 (shear, general and simplified methods).
Brzev & Pao, Reinforced Concrete Design: A Practical Approach — Chapters 3–6 and 9.
CSA O86, Engineering Design in Wood — Clause 6 (sawn lumber), Tables 6.3.1B and 6.4.5.
Canadian Wood Council, Wood Design Manual — beam selection and modification factors.
Check — assumptions carried through this paper. (1) The 60 kN, 50 kN, 40 kN, 200 kN, 250 kN and 300 kN forces on Figures A1–B3 are specified live loads, so a load factor of 1.5 is used; concrete self-weight is factored at 1.25. (2) Steel members are assumed continuously laterally supported unless an unbraced length is printed — no unbraced lengths appear on this paper. (3) In Figure B2 the "65 typical" dimension is taken as the distance from the tension face to the centroid of the bottom bars, and the 15M stirrups are assumed at 300 mm on centre in each stem (the figure gives no spacing). (4) CSA O86 Table 6.3.1B Beam-and-Stringer specified strengths for D.Fir-L No.1 are used in C1.
Question 6: B3 — Design of Column BC in a Determinate Concrete Frame (10 + 10 marks)
Given. L-shaped frame: roller support at A, horizontal member AB of 6.0 m (dimensioned 3 m + 3 m), rigid corner at B, column BC 5.0 m long descending to a pin at C. Unfactored live loads: 250 kN at 3.0 m from A on the beam, and 300 kN applied directly at joint B. $f_c' = 35$ MPa, $f_y = 400$ MPa; the column is short (no slenderness amplification).
Find. A square cross-section and its longitudinal and tie reinforcement for member BC.
Figure B3 — determinate frame; the roller at A and the pin at C give three reaction components, and the 250 kN load is placed so that the beam moment at B vanishes.
Approach. Solve the frame statically, establish the axial force and bending moment delivered to the column head, then size a short tied column from the A23.3 axial-resistance expression, choose bars within the 1–8 % limits and detail the ties.
Check determinacy. The roller at A supplies one reaction component and the pin at C supplies two, three in all, matching the three equations of planar statics for a single rigid body — the frame is statically determinate, as the question states.
Reactions. Taking moments about C, only the roller reaction $A_y$ and the 250 kN beam load have a lever arm (the 300 kN load acts directly over the column line):
$$\sum M_C=0:\quad A_y(6.0)=250(6.0-3.0)\ \Rightarrow\ A_y=125\ \text{kN}\uparrow$$
$$\sum F_y=0:\quad C_y=250+300-125=425\ \text{kN}\uparrow,\qquad \sum F_x=0:\quad C_x=0$$
Actions delivered to the column head. The bending moment in the beam immediately to the left of B is
$$M_B=A_y(6.0)-250(3.0)=750-750=0$$
Because joint B is monolithic, the column-top moment equals the beam-end moment, which is zero; and with $C_x = 0$ there is no shear in the column either. The member is therefore in pure axial compression:
$$\boxed{N=425\ \text{kN compression},\quad M=0\ \text{throughout BC}}$$
This is not a coincidence of the arithmetic — the 250 kN load sits at mid-span of AB, which is precisely the position that makes the simple-beam moment at B vanish.
Factored axial load. Both loads are specified live loads:
$$P_f=1.5(425)=638\ \text{kN}$$
Trial section. Try a 300 × 300 mm square column. The gross area is $A_g = 90\,000$ mm². Choose 4–20M longitudinal bars, $A_{st} = 4(300) = 1\,200$ mm², giving
$$\rho_t=\frac{A_{st}}{A_g}=\frac{1\,200}{90\,000}=1.33\ \%$$
which sits inside the A23.3 limits of 1 % minimum and 8 % maximum for a tied column.
Axial resistance. A23.3 Clause 10.10.4 caps the resistance of a tied column at 80 % of the concentric value, which is how the code accounts for unavoidable accidental eccentricity:
$$P_{r,max}=0.80\left[\alpha_1\phi_cf_c'(A_g-A_{st})+\phi_sf_yA_{st}\right]$$
$$P_{r,max}=0.80\left[0.7975(0.65)(35)(88\,800)+0.85(400)(1\,200)\right]=0.80\left[1\,611+408\right]\ \text{kN}$$
$$\boxed{P_{r,max}=1\,615\ \text{kN}\;>\;P_f=638\ \text{kN}\quad\checkmark}$$
The utilisation is only 0.39, but 300 mm is already at the practical minimum for a 5 m cast-in-place column carrying a beam, so the section is governed by buildability and by the 1 % minimum steel rule rather than by strength.
Ties. Use 10M ties. The spacing must not exceed the least of 16 longitudinal bar diameters, 48 tie diameters, or the least column dimension:
$$s\le\min\left[16(19.5),\;48(11.3),\;300\right]=\min(312,\;542,\;300)=300\ \text{mm}$$
$$\boxed{\text{10M ties at 300 mm on centre}}$$
A single closed square tie engages all four corner bars, satisfying the requirement that every corner bar be laterally supported by a tie corner bent to no more than 135°.
Detailing at the ends. Because joint B is monolithic, the column bars are carried into the beam and lapped or anchored for their full development length so that the joint can transmit any incidental moment. At the pinned base C the bars are dowelled into the footing but the connection is detailed without moment continuity — a small central dowel group and no confining collar — so that the idealised pin assumed in the analysis is actually realised on site. Tie spacing is halved through the depth of the beam-column joint.
Check: the analysis result $M_B = 0$ depends on reading support A as a roller (one vertical reaction) and the 250 kN load as acting at mid-span of AB. If A were instead a pin, the frame would be indeterminate and the column would attract moment; the question's own statement that the frame is determinate confirms the roller reading.
