Question 5 of 7: B2 — Design of a slender reinforced concrete column
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 07-Str-A2 Elementary Structural Design, May 2019 sitting, 3 hours, closed book with handbooks and textbooks permitted. Seven questions of 20 marks each: Part A (A1–A3, structural steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). The candidate answers two from Part A, two from Part B and one from Part C — all seven are solved here.
Reference texts. CSA S16 Design of Steel Structures and the CISC Handbook of Steel Construction (section properties, Tables 4-4 and 7); CSA A23.3 Design of Concrete Structures with Brzev & Pao, Reinforced Concrete Design: A Practical Approach; CSA O86 Engineering Design in Wood with the Canadian Wood Council Wood Design Manual; NBCC Part 4 for load combinations.
Question 5: B2 — Design of a slender reinforced concrete column (20 marks)
Given. An 8 m interior column, pinned top and bottom, carrying axial gravity load plus a single transverse wind load at mid-height.
Given data — B2
Quantity
Symbol
Value
Unsupported height
$\ell_u$
8000 mm
End conditions
$k$
Pinned–pinned, $k=1.0$
Axial dead load
$P_D$
400 kN
Axial live load
$P_L$
300 kN
Lateral wind load at mid-height
$H$
20 kN
Concrete / steel
$f_c'$ / $f_y$
30 MPa / 400 MPa
Exposure
—
Interior (dry), 40 mm cover
Find. A square tied column section and reinforcement adequate for all governing NBCC load combinations, including slenderness effects.
Figure 5 — B2 column: pinned at both ends with a single transverse wind load at mid-height, so the first-order moment diagram is triangular with its peak at mid-height.
Approach. Compute the first-order moment, enumerate the NBCC load combinations, test slenderness against A23.3 Cl. 10.15.2, magnify the moments where required, then verify a trial section by strain compatibility on the P–M interaction surface.
Find the first-order moment. For a pinned–pinned member with a transverse point load at mid-height, the moment is triangular with its peak at the load:
$$M=\frac{H\ell_u}{4}=\frac{20(8)}{4}=40\ \text{kN}\cdot\text{m}\ \text{(unfactored)}$$
Note this is $H\ell/4$, not $H\ell/8$ — the ends are pinned, not fixed.
Enumerate the governing load combinations. From NBCC Table 4.1.3.2, the two that matter are
$$\text{Case 3: }1.25D+1.5L+0.4W\ \Rightarrow\ P_f=500+450=950\ \text{kN},\ M_{2}=0.4(40)=16\ \text{kN}\cdot\text{m}$$
$$\text{Case 4: }1.25D+1.4W+0.5L\ \Rightarrow\ P_f=500+150=650\ \text{kN},\ M_{2}=1.4(40)=56\ \text{kN}\cdot\text{m}$$
Case 3 is the heavy-axial, light-moment case and Case 4 the reverse; both must be carried through.
Adopt a trial section and check slenderness. Try a 450 mm square column with $r=0.3h=135$ mm:
$$\frac{k\ell_u}{r}=\frac{1.0(8000)}{135}=59.3$$
Because the end moments are both zero and the peak is at mid-height, $M_1/M_2=0$ and the A23.3 Cl. 10.15.2 threshold is a flat
$$34-12\left(\frac{M_1}{M_2}\right)=34\ <\ 59.3$$
so the column is slender and moment magnification is mandatory. This is the pinned-pinned trap: the threshold is at its lowest possible value precisely when the ends are least restrained.
Compute the critical buckling load. With $E_c=4500\sqrt{30}=24648$ MPa and $I_g=h^{4}/12=3.417\times10^{9}$ mm⁴, and $\beta_d$ taken as the ratio of sustained to total factored axial load,
$$EI=\frac{0.4E_cI_g}{1+\beta_d},\qquad P_c=\frac{\pi^{2}EI}{(k\ell_u)^{2}}$$
For Case 4, $\beta_d=500/650=0.769$, giving $EI=1.904\times10^{13}$ N·mm² and
$$P_c=\frac{\pi^{2}(1.904\times10^{13})}{8000^{2}}=2937\ \text{kN}$$
Magnify the Case 4 moment. With a transverse load between the supports, $C_m=1.0$:
$$\delta=\frac{C_m}{1-\dfrac{P_f}{\phi_mP_c}}=\frac{1.0}{1-\dfrac{650}{0.75(2937)}}=1.419$$
$$M_{c,4}=1.419(56)=\boxed{79.4\ \text{kN}\cdot\text{m}}$$
Magnify the Case 3 moment, respecting the minimum eccentricity. The code floor is
$$e_{\min}=15+0.03h=15+13.5=28.5\ \text{mm}\ \Rightarrow\ M_{2,\min}=950(0.0285)=27.1\ \text{kN}\cdot\text{m}$$
which exceeds the 16 kN·m from the wind, so $M_2=27.1$ kN·m is used. With $\beta_d=500/950=0.526$, $P_c=3405$ kN and $\delta=1.592$:
$$M_{c,3}=1.592(27.1)=43.1\ \text{kN}\cdot\text{m}$$
Select reinforcement and verify by strain compatibility. Adopt 8–25M ($A_{st}=4000$ mm², $\rho=1.98$ %, within the 1–8 % range of Cl. 10.9.1) in three layers at 64, 225 and 386 mm from the compression face, with 10M ties. Solving $\sum F=0$ for the neutral axis at each factored axial load and then taking moments gives
$$\text{Case 4: }P_f=650\ \text{kN}\ \Rightarrow\ M_r=310\ \text{kN}\cdot\text{m},\qquad \frac{M_{c,4}}{M_r}=\frac{79.4}{310}=\boxed{0.26}$$
$$\text{Case 3: }P_f=950\ \text{kN}\ \Rightarrow\ M_r=323\ \text{kN}\cdot\text{m},\qquad \frac{M_{c,3}}{M_r}=\frac{43.1}{323}=0.13$$
Both combinations are satisfied with a wide margin, and the wind case governs as expected.
Confirm the section is not simply an axial member. The pure-axial capacity is
$$P_{r,\max}=0.80\left[\alpha_1\phi_cf_c'(A_g-A_{st})+\phi_sf_yA_{st}\right]=3581\ \text{kN}$$
against a maximum applied 950 kN. The column is nowhere near its axial limit; what sizes it is the 1 % minimum-steel rule and the need to keep $k\ell_u/r$ within reach, not strength.
Design summary: a 450 × 450 mm tied column with 8–25M longitudinal bars and 10M ties at 300 mm centres. The utilisation of 0.26 looks generous, and it is — but the size is not arbitrary. Stepping down to a 400 mm square drops $P_c$ by 38 % (the buckling load scales with $h^{4}$), lifts the Case 4 magnifier from 1.42 to 1.90 and the eccentricity from 122 mm to 163 mm; a 350 mm square would fail outright. On a slender pinned-pinned column the section size is chosen for stiffness, and the strength check is the formality that follows.
