Question 6 of 7: Reinforced-concrete design of member BC for flexure and shear

Question 6: Reinforced-concrete design of member BC for flexure and shear (15 + 5 = 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. The portal of Figure 4: a 16 m beam BD on 8 m columns fixed at A and E, with an internal hinge at C, the mid-point of the beam. Loading is symmetric — 500 kN at each of B and D and 250 kN at 4 m and 12 m.

Find. The flexural and shear reinforcement for member BC, and a reinforcing arrangement for the whole beam BCD.

Approach. Exploit symmetry to reduce the frame to a half-frame whose only redundant is the horizontal thrust at C, analyse it, then design the beam section at the joint for flexure and check shear by the CSA A23.3 simplified method.

1. Part (a) — exploit the symmetry. Both the structure and the loading are symmetric about C. At a point on the axis of symmetry the vertical shear must vanish, and the hinge already forces the moment to vanish, so the only action crossing C is a horizontal thrust. The frame therefore reduces to a half-frame A–B–C, fixed at A and restrained horizontally at C, with one redundant.
2. Factored loads on the half-frame. $$P_B = 1.5(500) = 750\ \text{kN},\qquad P_{4\text{m}} = 1.5(250) = 375\ \text{kN}$$ $$w_{beam} = w_{col} = 1.25(23.04) = 28.8\ \text{kN/m}$$
3. The half-beam is a cantilever. Because neither moment nor vertical shear crosses C, all of the beam load must be carried back to joint B. Taking moments about B for B–C, $$M_B = 375(4) + 28.8(8)(4) = 1500 + 921.6 = \boxed{2421.6\ \text{kN}\cdot\text{m (hogging)}}$$ with the shear at the face of the joint $$V_B = 375 + 28.8(8) = 605.4\ \text{kN}$$ A stiffness solution of the half-frame reproduces the same joint moment and returns a horizontal thrust of 454 kN at C, a base moment $M_A = 1210.8$ kN·m and a column axial load of 1585.8 kN.
Factored bending moments. With no moment and no shear transferred at C, each half of the beam cantilevers from its own joint and the whole beam moment is concentrated at B and D.
4. Trial section and effective depth. A moment of this size on a 16 m beam calls for a deep member; take 600 × 1600 mm. This is also the section adopted for the columns, so the paper’s instruction to assume the same stiffness for all members is satisfied exactly rather than approximately. With two layers of 30M bars and 40 mm cover, $d = 1500$ mm.
5. Flexural reinforcement. Estimating the lever arm at $0.9d$, $$A_s \approx \frac{M_f}{\phi_sf_y(0.9d)} = \frac{2421.6\times10^6}{0.85(400)(1350)} = 5276\ \text{mm}^2$$ Provide 8 – 30M ($A_s = 5600\ \text{mm}^2$) in the top face. Checking exactly, with $\alpha_1 = 0.805$ and $\beta_1 = 0.895$, $$a = \frac{\phi_sA_sf_y}{\alpha_1\phi_cf'_cb} = \frac{0.85(5600)(400)}{0.805(0.65)(30)(600)} = 202\ \text{mm},\qquad c = 226\ \text{mm}$$ $$M_r = \phi_sA_sf_y\left(d-\frac{a}{2}\right) = \boxed{2664\ \text{kN}\cdot\text{m}} \ge 2421.6$$ a ratio of 0.91. The strain check $c/d = 0.151$ is far below the 0.5 balanced-section limit, so the section is comfortably tension-controlled, and $A_{s,min} = 0.2\sqrt{f'_c}b_th/f_y = 2629\ \text{mm}^2$ is easily exceeded.
6. Part (b) — shear. With $d_v = \max(0.9d,\ 0.72h) = 1350$ mm and the simplified method ($\beta = 0.18$, $\theta = 35^{\circ}$) for a member carrying at least minimum stirrups, $$V_c = \phi_c\lambda\beta\sqrt{f'_c}\,b_wd_v = 0.65(0.18)\sqrt{30}(600)(1350)/10^3 = 519\ \text{kN}$$ $$V_s = V_f - V_c = 605.4 - 519 = 86.4\ \text{kN}$$ The stirrup requirement is nominal, so minimum transverse reinforcement governs: $$s_{max} = \frac{A_vf_y}{0.06\sqrt{f'_c}b_w} = \frac{400(400)}{0.06\sqrt{30}(600)} = 811\ \text{mm}, \quad\text{but Cl 11.3.8 caps } s \text{ at } 600\ \text{mm}$$ Use 15M double-leg stirrups at 300 mm over the first 3 m from B, where the moment gradient is steepest and the top steel needs the anchorage, and at 500 mm over the remainder. The upper limit $V_{r,max} = 0.25\phi_cf'_cb_wd_v = 3949$ kN is nowhere approached, so web crushing is not an issue.
7. Reinforcing details for member BCD. The moment diagram falls from 2422 kN·m at each joint to zero at the hinge, so the top steel is curtailed in stages: all 8 – 30M continue 1.5 m past the face of the column, four are cut at about 3 m (where $M_f = 735$ kN·m, matched by 3 bars) and three continue through to C to hold the stirrups and to carry the small hogging moment that any load asymmetry would produce. Three 30M bars run continuously in the bottom face for the same reason and to resist handling and temperature effects. At joint B every top bar is anchored with a standard 90° hook turned down into the column, and the joint core is confined with 15M ties at 150 mm. At C the two halves are detailed as a true hinge: the section is locally narrowed to a throat about one third of the beam width, the longitudinal steel is stopped and replaced by crossing dowels, so that the 454 kN thrust is transmitted while rotation stays free.
Reinforcement at the two critical beam sections and in the column — all members 600 × 1600 mm, so the equal-stiffness assumption is exact.
8. Note on the thrust. The beam also carries the 454 kN horizontal thrust as axial compression, which over a gross area of 960 000 mm² is only 0.47 MPa. It is beneficial (it slightly raises the flexural resistance) and is neglected in the design, but the same thrust must be delivered to the foundations, so the base connections and the ground beam between A and E have to be detailed for it.