Question 5 of 7: Square reinforced concrete section for portion A–B of member A–B–C

Question B2: Square reinforced concrete section for portion A–B of member A–B–C (20 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.

Find. The square dimension, longitudinal reinforcement and ties for the column A–B, together with the axial force, moment and shear it must carry.

[Figure not reproduced: Figure B2.1 — Figure 3 of the paper redrawn. The bases at A and E are drawn as plain hatch (built in). The real hinge at mid-span C is what makes the beam moments statically determinate. See the official exam paper.]

Approach. Exploit symmetry and the hinge to obtain the beam and column-top actions by statics alone, get the base moment from the single remaining compatibility condition, then design the square section on a strain-compatibility interaction diagram and confirm that slenderness is negligible.

1. Factor the loads. Page 1 states the loads are unfactored but Figure 3 gives no dead/live split, so a single factor of 1.5 is applied: $$\begin{aligned} & 375\ \text{kN at B and D} \\ & 900\ \text{kN at 4 m and at 12 m} \end{aligned}$$ The total factored vertical load is $2(375) + 2(900) = 2550\ \text{kN}$, so by symmetry each column carries 1275 kN.
2. Count the redundants, then find the two conditions that remove them. Two built-in bases give six reaction components against three equations of equilibrium, and the internal hinge supplies a fourth equation, leaving the frame twice redundant. But the structure and the loading are both symmetric about C, so the shear at C vanishes; and the hinge at C makes the moment there vanish too. Those two facts make the whole beam determinate.
3. Cut at the hinge and take the half beam as a free body. On the segment B–C the only actions are the 900 kN load at 4 m from B and, at C, $V = 0$ and $M = 0$. Vertical equilibrium gives the shear delivered from the column, $V_B = 900\ \text{kN}$, and moments about C give $$M_B = P_{int} \times a = 900 \times 4 = \boxed{3600\ \text{kN}\cdot\text{m}}$$ a hogging moment at the joint. Notice what this means for the rest of the beam: between the 4 m load point and the hinge there is no load, no shear and no moment at all. The beam is effectively two cantilevers off the columns joined by a zero-force link.
4. Note that the 250 kN loads do no bending work. They stand directly over the columns, so they pass straight into the columns as axial force. The axial load at the top of column A–B is therefore $$N_f = 375 + 900 = 1275\ \text{kN}$$ and the beam moment comes entirely from the interior 900 kN load.
5. Find the thrust from the one remaining compatibility condition. The column is loaded only at its ends: a moment $M = 3600\ \text{kN}\cdot\text{m}$ from the beam and a horizontal thrust $H$ from the beam axial force. By symmetry point C cannot move horizontally, and because the beam is horizontal its bending does not displace C sideways, so the condition reduces to zero horizontal displacement at the column top: $$\frac{H h^3}{3EI} + \frac{M h^2}{2EI} = 0 \;\Longrightarrow\; H = \frac{3M}{2h} = \frac{3(3600)}{2(8)} = \boxed{675\ \text{kN}}$$ Both $E$ and $I$ cancel, so the thrust is a property of the geometry alone.
6. Get the base moment, which is exactly half the top moment. Taking moments about A for the column free body, $$M_A = H h - M = \frac{3M}{2} - M = \frac{M}{2} = \boxed{1800\ \text{kN}\cdot\text{m}}$$ of opposite sense to the top. A stiffness solution that also includes the axial shortening of the beam returns 1755 kN·m, so the flexure-only closed form is 2.5 per cent conservative — which is the right side to be on, because this moment is what question B3 has to found.
7. Size the square section. The design point at the top of the column is $N_f = 1275\ \text{kN}$ with $M_f = 3600\ \text{kN}\cdot\text{m}$, an eccentricity of 2.82 m. That is far outside the section, so the member is essentially a flexural element carrying incidental compression, and the size follows from flexure. Trying square sections with the reinforcement ratio held between the Clause 10.9.1 limits of 1 and 8 per cent, a 1200 mm square with about 1.5 per cent steel is the smallest that works comfortably.
8. Build the interaction diagram and read off the resistance. With 32–30M bars ($A_{st} = 22\,400\ \text{mm}^2$, $\rho = 22\,400/1200^2 = 1.56\ \%$) arranged nine to a face, and bars at $d' = 40 + 11.3 + 15 = 66\ \text{mm}$ from each face, strain-compatibility analysis with $\varepsilon_{cu} = 0.0035$, $\alpha_1 = 0.7975$ and $\beta_1 = 0.8825$ gives $$M_r = 4545\ \text{kN}\cdot\text{m}\ \text{at}\ N_f = 1275\ \text{kN} \;\Longrightarrow\; \frac{M_f}{M_r} = \frac{3600}{4545} = \boxed{0.792}$$
9. Check the base as well as the top. The column self weight is $1.2^2 \times 8 \times 24 = 276\ \text{kN}$, so the factored axial load at the base is $1275 + 1.25(276) = 1621\ \text{kN}$. Against the smaller base moment of 1800 the utilisation is only 0.385, so the top of the column governs and one bar arrangement serves the full height.
10. Confirm slenderness is negligible — do not rely on a threshold. With $E_c = 4500\sqrt{35} = 26\,622\ \text{MPa}$, $I_g = 1200^4/12 = 1.728\times10^{11}\ \text{mm}^4$ and the Clause 10.15.3 stiffness $EI = 0.25E_c I_g$, $$P_c = \frac{\pi^2 EI}{(k l_u)^2} = \frac{\pi^2 (0.25)(26\,622)(1.728\times10^{11})}{(1.2 \times 8000)^2} = 123\,165\ \text{kN}$$ so the magnifier is $\delta = 1/(1 - N_f/P_c) = 1/(1 - 0.0104) = 1.011$. A one per cent increase is immaterial. This is worth computing explicitly rather than quoting the $kl_u/r \le 22$ rule of thumb, which the section fails at 26.7 while in fact being entirely short.
11. Design the ties. The column shear is the full frame thrust, $V_f = 675\ \text{kN}$. With at least minimum transverse steel the simplified method gives $\beta = 0.18$ and $d_v = \max(0.9(1134),\ 0.72(1200)) = 1020\ \text{mm}$: $$V_c = 0.65(0.18)\sqrt{35}\,(1200)(1020) = 848\ \text{kN} \;\Longrightarrow\; \frac{675}{848} = 0.796\ \ \checkmark$$ so no shear reinforcement is needed for strength, but the compression bars must be tied. Clause 7.6.5 caps the tie spacing at the least of $16 d_b = 478\ \text{mm}$, $48 d_{tie} = 542\ \text{mm}$ and the least column dimension, 1200 mm: use 10M ties at 300 mm, with interior legs restraining every alternate bar and closer spacing over the 1200 mm above the base and below the joint, where the moment peaks.

Final Results