Question 5 of 7: Reinforced-Concrete Design of Beam AB

Question 5: Reinforced-Concrete Design of Beam AB (10 + 5 + 5 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 same L-frame geometry and loading as Question 3, now to be built in reinforced concrete, so the analysis must be redone with concrete member stiffnesses and the beam self weight included.

Find. (a) a rectangular beam section for AB with its flexural and shear reinforcement, and (b) the long-term mid-span deflection of that beam.

Approach. Re-analyse the frame elastically with cracked-section stiffnesses, because the concrete members have a quite different stiffness ratio from the steel ones and they carry their own weight. Design the three critical sections for the resulting moments, check shear at the support, then compute the immediate deflection with an effective moment of inertia and multiply the sustained part by the creep factor.

1. Part (a) — set up the concrete analysis. A 450 mm by 900 mm section gives $I_{g} = 450\times900^{3}/12 = 2.734\times10^{10}$ mm$^{4}$ and a self weight of $0.45\times0.90\times24 = 9.72$ kN/m, factored to 12.15 kN/m. A23.3 Cl 10.14.1.2 requires reduced stiffnesses for the analysis of a frame: $0.35I_{g}$ for the flexural member and $0.70I_{g}$ for the column. With $E_{c} = 3300\sqrt{30} + 6900 = 24\,975$ MPa this halves the beam stiffness relative to the column, which pushes moment towards the built-in end A.
2. Read off the design actions. The elastic analysis of the frame under the factored loads and the factored self weight returns $$M_{A} = -686.0,\quad M_{\text{mid}} = +594.1,\quad M_{B} = -420.1\ \text{kN.m}$$ with $V_{A} = 344.3$ kN and $V_{B} = 277.9$ kN (which sum to the total load of 622.2 kN, as they must). The column takes 1028 kN axially at B, rising to 1149 kN at the base once its own weight is added.
3. Fix the effective depth and the flexural relation. With 40 mm cover, 10M stirrups and 25M bars in one layer, $d = 900 - 40 - 11.3 - 12.6 = 836$ mm. The resistance of a singly reinforced rectangular section is $$M_{r} = \phi_{s}A_{s}f_{y}\left(d - \frac{a}{2}\right),\qquad a = \frac{\phi_{s}A_{s}f_{y}}{\alpha_{1}\phi_{c}f'_{c}b}$$ with $\alpha_{1}\phi_{c}f'_{c}b = 0.805\times0.65\times30\times450 = 7064$ N/mm.
4. Design the three critical sections. Solving the quadratic at each: $$M_{A} = 686\ \Rightarrow\ A_{s} = 2609\ \text{mm}^{2}\ \rightarrow\ \boxed{6\text{-}25\text{M top at A}}\ (3000\ \text{mm}^{2},\ M_{r} = 779\ \text{kN.m})$$ $$M_{\text{mid}} = 594\ \Rightarrow\ A_{s} = 2234\ \text{mm}^{2}\ \rightarrow\ 5\text{-}25\text{M bottom}\ (2500\ \text{mm}^{2},\ M_{r} = 660\ \text{kN.m})$$ $$M_{B} = 420\ \Rightarrow\ A_{s} = 1547\ \text{mm}^{2}\ \rightarrow\ 4\text{-}25\text{M top at B}\ (2000\ \text{mm}^{2},\ M_{r} = 536\ \text{kN.m})$$ All exceed the Cl 10.5.1.2 minimum of $0.2\sqrt{f'_{c}}b_{t}h/f_{y} = 1109$ mm$^{2}$, and the deepest stress block gives $c/d = 0.193$, comfortably tension controlled.
5. Check that the section is not over-reinforced. The largest steel ratio is $\rho = 3000/(450\times836) = 0.0080$, which puts the neutral axis at $c = 161$ mm and the steel strain at $(836-161)/161\times0.0035 = 0.0147$, seven times yield. The section will warn before it fails, which is the intent of the ultimate strength method.
6. Design the shear reinforcement. Take $d_{v} = \max(0.9d,\ 0.72h) = 752$ mm, so the design shear at $d_{v}$ from A is $344.3 - 12.15\times0.752 = 335.2$ kN. With minimum stirrups present the simplified method gives $\beta = 0.18$ and $\theta = 35^{\circ}$: $$V_{c} = \phi_{c}\lambda\beta\sqrt{f'_{c}}b_{w}d_{v} = 0.65(0.18)\sqrt{30}(450)(752)\times10^{-3} = 217.0\ \text{kN}$$ leaving $V_{s} = 118.2$ kN for the stirrups. Two-leg 10M ties ($A_{v} = 200$ mm$^{2}$) would allow 618 mm spacing on strength, but Cl 11.3.8.1 caps it at $0.7d_{v} = 527$ mm, so use $$\boxed{10\text{M closed stirrups at 400 mm centres throughout}}$$ giving $V_{r} = 217.0 + 183.0 = 400$ kN and a utilisation of 0.84.
7. Part (b) — find the cracked and effective stiffnesses. With $n = E_{s}/E_{c} = 8.01$ and 2500 mm$^{2}$ at mid-span, the cracked neutral axis solves $b(kd)^{2}/2 = nA_{s}(d - kd)$ to give $kd = 232$ mm, hence $$I_{cr} = \frac{b(kd)^{3}}{3} + nA_{s}(d - kd)^{2} = 9.18\times10^{9}\ \text{mm}^{4}$$ The cracking moment is $M_{cr} = f_{r}I_{g}/y_{t}$ with $f_{r} = 0.6\lambda\sqrt{f'_{c}} = 3.29$ MPa, giving 199.6 kN.m, and the service mid-span moment from the same frame analysis at unfactored loads is 413.7 kN.m. Branson's expression then gives $$I_{e} = I_{cr} + (I_{g} - I_{cr})\left(\frac{M_{cr}}{M_{a}}\right)^{3} = 1.122\times10^{10}\ \text{mm}^{4} = 0.41I_{g}$$
8. Compute the immediate deflection. Re-running the frame at service load with $E_{c}I_{e}$ on the beam gives a mid-span deflection of 4.83 mm downwards. The built-in end at A and the partial restraint from the column at B between them remove most of the simple-span deflection, which for the same load on a simply supported beam would be about six times larger.
9. Apply the creep multiplier. A23.3 Cl 9.8.2.5 gives the additional long-term deflection as $\zeta = s/(1 + 50\rho')$ times the immediate deflection of the sustained load. With $s = 2.0$ for five years or more and compression steel $A'_{s} = 1000$ mm$^{2}$ ($\rho' = 0.00266$), $\zeta = 2.0/1.133 = 1.765$, so $$\Delta_{LT} = \Delta_{i}(1 + \zeta) = 4.83\times2.765 = \boxed{13.4\ \text{mm}} = L/599$$ which is inside the $L/480$ limit of Table 9.3 for members supporting non-sensitive construction.