16-Civ-A2 Elementary Structural Design · December 2018

Question 4 of 7: B1 — Reinforced concrete member BC of the determinate frame

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: B1 — Reinforced concrete member BC of the determinate frame (10 + 10 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. A cross-section for BC plus its flexural and shear reinforcement.

Approach. The frame has three reaction components on one rigid body, so it is determinate: solve the reactions, read the end moments of BC up each column, then design a rectangular section for the largest hogging moment and check shear.

1. Trial section and factored loads. Take BC as 450 mm wide by 900 mm deep, so its self weight is $0.45(0.90)(24) = 9.72$ kN/m. Factoring, $$P_f=1.5(200)=300\ \text{kN},\qquad H_f=1.5(80)=120\ \text{kN},\qquad w_f=1.25(9.72)=12.15\ \text{kN/m}$$
2. Reactions. Horizontal equilibrium puts the whole thrust into the pin, $A_x = H_f = 120$ kN, because the roller at D takes no horizontal force. Taking moments about A, $$8D_y=300(4)+300(8)-120(3)+12.15\frac{(8)^{2}}{2}\quad\Longrightarrow\quad D_y=453.6\ \text{kN}$$ and vertical equilibrium gives $A_y = 3(300)+12.15(8)-453.6 = 543.6$ kN.
3. End moments of the beam, read up each column. Column AB carries the 120 kN thrust over its full 6 m height with nothing else applied along it, and column CD carries the same thrust over the 3 m from the load up to C: $$M_B=H_f\,h_{AB}=120(6.0)=\boxed{720\ \text{kN}\cdot\text{m}},\qquad M_C=H_f(3.0)=360\ \text{kN}\cdot\text{m}$$ Both hog the beam (tension on the top face). Below the 80 kN load, column CD carries no shear and no moment at all, because the roller supplies no horizontal reaction.
4. Midspan moment — the beam statics check. The shear just inside B is $V = A_y - P_f = 243.6$ kN, so at midspan $$M_{\text{mid}}=-M_B+V\left(4.0\right)-\frac{w_f(4.0)^{2}}{2}=-720+974.4-97.2=157.2\ \text{kN}\cdot\text{m}\ \text{sagging}$$ Cross-check against the simple-span value: $300(8)/4 + 12.15(8)^2/8 = 697.2$ kN·m less the mean end hog $(720+360)/2 = 540$ gives 157.2 kN·m. The two routes agree.
5. Flexural steel at B (the governing section). With 40 mm cover, 10M ties and 25M bars, $d = 900-40-11.3-12.6 = 836$ mm. Using $\alpha_1 = 0.85-0.0015f_c' = 0.7975$ and $\beta_1 = 0.97-0.0025f_c' = 0.8825$, solve $$\phi_s f_y A_s\left(d-\frac{a}{2}\right)=M_f,\qquad a=\frac{\phi_s f_y A_s}{\alpha_1\phi_c f_c' b}$$ for $M_f = 720$ kN·m, which gives $A_s = 2717$ mm². Provide 6-25M top (3000 mm²), which fits in one layer across 450 mm. Then $a = 113$ mm, $c = 128$ mm and $c/d = 0.153$ — comfortably tension-controlled.
6. Steel elsewhere along the beam. At C, $M_f = 360$ kN·m needs $A_s = 1309$ mm². At midspan, $M_f = 157.2$ kN·m needs only 561 mm², so the A23.3 Clause 10.5.1.2 minimum controls: $$A_{s,\min}=\frac{0.2\sqrt{f_c'}\,b_t h}{f_y}=\frac{0.2\sqrt{35}(450)(900)}{400}=1198\ \text{mm}^{2}$$ Provide 3-25M (1500 mm²) top at C and 3-25M bottom continuous, with the six top bars at B curtailed to three once the moment falls below the three-bar capacity.
7. Shear design. The largest shear is $V_f = 243.6$ kN just inside B, reducing to 234.5 kN at the critical section $d_v$ from the face. With $d_v = \max(0.9d,\,0.72h) = 752$ mm and the general method with minimum stirrups ($\beta = 0.18$, $\theta = 35^\circ$): $$V_c=\phi_c\,\beta\sqrt{f_c'}\,b_w d_v=0.65(0.18)\sqrt{35}(450)(752)=234.4\ \text{kN}$$ $$V_s=\frac{\phi_s A_v f_y d_v\cot\theta}{s}=\frac{0.85(200)(400)(752)\cot 35^\circ}{300}=243.6\ \text{kN}$$ $$V_r=V_c+V_s=\boxed{478\ \text{kN}}\ \gg\ V_f=243.6\ \text{kN}$$ The crushing limit $0.25\phi_cf_c'b_wd_v = 1926$ kN is nowhere near, and the maximum spacing $\min(0.7d_v,600) = 527$ mm is satisfied by the 300 mm chosen. 10M closed stirrups at 300 mm throughout — minimum shear reinforcement suffices.

Notice how little of the design is driven by the 200 kN vertical loads. Because A is a pin and D a roller, the entire 80 kN horizontal load is resisted by the pin and delivered to the beam as a 720 kN·m hogging moment at B — more than four times the midspan sagging moment. A designer who analysed only the vertical loads would size the beam for about a third of the moment it actually carries.