16-Civ-A2 Elementary Structural Design · December 2018

Question 5 of 7: B2 — Uniform reinforced concrete cross-section and bar profile

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

Question 5: B2 — Uniform reinforced concrete cross-section and bar profile (8 + 6 + 6 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) a uniform rectangular cross-section, and (b) the layout of the longitudinal bars from A to C.

Approach. Assume a section, compute its self weight, analyse the determinate beam under factored loads, then check that the assumed section carries the peak sagging and hogging moments and the peak shear; finally set out the bars to follow the bending-moment diagram.

1. Trial section and factored loading. Take 400 mm x 700 mm, giving $0.40(0.70)(24) = 6.72$ kN/m of self weight. The factored loads are $$P_{f1}=P_{f2}=1.5(100)=150\ \text{kN},\qquad P_{f,C}=1.5(20)=30\ \text{kN},\qquad w_f=1.25(6.72)=8.40\ \text{kN/m}$$
2. Reactions. Taking moments about A over the full 8 m length, $$5R_B=150(2.0)+150(3.0)+30(8.0)+8.40\frac{(8.0)^{2}}{2}\quad\Longrightarrow\quad R_B=251.8\ \text{kN}$$ and $R_A = 150+150+30+8.40(8.0)-251.8 = 145.4$ kN.
3. Design moments and shear. The shear changes sign at the first point load, so the maximum sagging moment is there: $$M_{\text{sag}}=145.4(2.0)-8.40\frac{(2.0)^{2}}{2}=\boxed{274.1\ \text{kN}\cdot\text{m}}\ \text{at }x=2.0\ \text{m}$$ The overhang hogs the support: $$M_{\text{hog}}=-\left[30(3.0)+8.40\frac{(3.0)^{2}}{2}\right]=-127.8\ \text{kN}\cdot\text{m}\ \text{at B}$$ The largest shear is just left of B, $V_f = 196.6$ kN; on the overhang side it is only 55.2 kN.
4. Flexural steel for the sagging region. With 40 mm cover, 10M stirrups and 25M bars, $d = 700-40-11.3-12.6 = 636$ mm. Solving the rectangular-block equation for $M_f = 274.1$ kN·m gives $A_s = 1333$ mm². Provide 3-25M bottom (1500 mm²); then $c/d = 0.111$, deeply tension-controlled, so the section will warn by cracking and deflecting before it fails.
5. Flexural steel over the support. $M_f = 127.8$ kN·m needs only 604 mm² by analysis, but Clause 10.5.1.2 requires $$A_{s,\min}=\frac{0.2\sqrt{35}(400)(700)}{400}=828\ \text{mm}^{2}$$ Provide 3-20M top (900 mm²) running the full overhang and lapping into the span.
6. Confirm the section size. The trial 400 x 700 works at a steel ratio of $1500/(400 \times 636) = 0.59\%$ — comfortably between the minimum and the tension-controlled ceiling — and gives a span-to-depth ratio of $5000/700 = 7.1$, so deflection is not a concern. Adopt 400 mm x 700 mm.
7. Shear design. With $d_v=\max(0.9d,\,0.72h)=572$ mm, the shear at $d_v$ from the face of B is 191.8 kN, and $$V_c=0.65(0.18)\sqrt{35}(400)(572)=158.5\ \text{kN},\qquad V_s=\frac{0.85(200)(400)(572)\cot35^\circ}{300}=185.3\ \text{kN}$$ $$V_r=158.5+185.3=\boxed{343.8\ \text{kN}}\ \gg\ 191.8\ \text{kN}$$ 10M closed stirrups at 300 mm from A through B and out along the overhang; the maximum permitted spacing is $\min(0.7d_v,600)=401$ mm.
8. Bar profile along the beam — part (b). The moment diagram is sagging from A to the point of contraflexure and hogging from there to C, so:
   • Bottom: 3-25M from A to B, all three carried into the support and anchored past its face (the code requires at least one third of the positive steel to continue into a simple support). No bottom steel is needed on the overhang, but two 20M bars are carried through as stirrup-holders.
   • Top: 3-20M from the free end C, across the support at B and lapped a development length plus $d$ into the span, terminating at about 1.0 m past B where the hogging moment has decayed to zero. Top steel is essential over the whole overhang — the tension face there is the upper one.
   • Stirrups: 10M closed at 300 mm over the entire 8 m, closing around all corner bars so the cage is continuous through the support region where the moment reverses.
   The point of contraflexure sits at $x \approx 4.0$ m from A, and both the top and bottom layers must extend past it by at least $d$ = 636 mm before being cut off.

The single most important feature of the profile is that the reinforcement must change faces. Between A and the contraflexure point the tension is at the bottom; from there through B and out to C the tension is at the top. A cage detailed only for the mid-span sagging moment would leave the overhang with no tension steel at all — the classic failure of an overhanging beam.