Question 5 of 7: Reinforced concrete design of member BC (14 + 6 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2013, 98-Civ-B2 Advanced Structural Design; 3 hours, closed book (handbooks and textbooks permitted); seven design questions, any five of which constitute a complete paper, all of equal value (20 marks each). All seven are worked below, because the set is a study resource rather than a timed sitting. Design data given on page 1 and used throughout: concrete $f^{\prime}_{c}=30\ \text{MPa}$, structural steel $F_{y}=350\ \text{MPa}$, reinforcing bar $f_{y}=400\ \text{MPa}$; for the prestressed girder $f^{\prime}_{ci}=35\ \text{MPa}$, $f^{\prime}_{c}=50\ \text{MPa}$, $n=6$, $f_{pu}=1750\ \text{MPa}$, $f_{py}=1450\ \text{MPa}$, $f_{pi}=1200\ \text{MPa}$ and losses of 240 MPa. All loads shown on the figures are unfactored.
Reference texts. CSA S16, Design of Steel Structures; CISC, Handbook of Steel Construction; Kulak & Grondin, Limit States Design in Structural Steel; CSA A23.3, Design of Concrete Structures; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition); Collins & Mitchell, Prestressed Concrete Structures; Beedle, Plastic Design of Steel Frames; National Building Code of Canada.
Check — load factor. The paper states that the loads shown are unfactored but does not separate dead from live. Every design below therefore applies a single load factor of 1.5 to the given loads, which is the NBCC live-load factor and is the conservative reading when the split is unknown. Self weight, where it is generated by a member being sized here (the reinforced concrete frame of Questions 5 and 6, and the prestressed girder of Question 7), is carried separately at 1.25. If the examiner intended a different split, only the magnitudes change — every mechanism, section classification and interaction equation is unaffected.
Question 5: Reinforced concrete design of member BC (14 + 6 marks)
Given. The frame of Figure 2 built in reinforced concrete to the Limit States Design method, all members of equal stiffness, with concrete of 30 MPa and reinforcement of 400 MPa. Member BC is the 16 m beam carrying 400 kN at mid-span, framing into pin-based columns 6 m and 8 m long, with lateral support at all joints and load points.
Given data — reinforced concrete beam BC
Quantity
Symbol
Value
Beam span, B to C
$L$
16 m (8 m each side of O)
Mid-span load
$P$
400 kN unfactored
Horizontal load at B
$H$
80 kN unfactored
Concrete strength
$f^{\prime}_{c}$
30 MPa
Reinforcement
$f_{y}$
400 MPa
Trial section
$b\times h$
600 $\times$ 1500 mm
Effective depth
$d$
1400 mm (two layers)
Find. A suitable rectangular cross-section for BC, with the amount and layout of reinforcement for both flexure and shear.
Member BC: 600 x 1500 rectangular section, 9-25M in two layers where sagging governs, 10M closed stirrups.
Approach. Choose a depth from the span, compute its self weight, run a factored elastic analysis of the whole frame including that self weight, then size the flexural steel section by section from the rectangular stress block and finish with the CSA A23.3 simplified shear method.
Choose the depth first, because it feeds the analysis. For a 16 m continuous reinforced concrete beam, A23.3 Table 9.2 requires at least $L/18.5=865\ \text{mm}$ before deflections must be computed, but strength drives a far deeper section. Take $600\times1500$; its self weight is $0.6(1.5)(24)=21.6\ \text{kN/m}$, which is $346\ \text{kN}$ over the span and therefore not something that can be left out of the analysis.
Factored elastic analysis. Solving the frame with $1.5\times$ the applied loads and $1.25\times$ the self weight gives the design moments $M_{f}=1955.7\ \text{kN}\cdot\text{m}$ sagging at O, $1906.6\ \text{kN}\cdot\text{m}$ hogging at C and only $710.0\ \text{kN}\cdot\text{m}$ hogging at B, with a maximum beam shear of $590.8\ \text{kN}$ at C. The two ends differ because the columns differ: the shorter, stiffer column at B would attract more moment were it not for the sway that the 80 kN drives, which relieves B and loads C.
Flexural steel at O and C. With $\alpha_{1}=0.85-0.0015f^{\prime}_{c}=0.805$ and $\beta_{1}=0.97-0.0025f^{\prime}_{c}=0.895$, solving $M_{r}=\phi_{s}A_{s}f_{y}(d-a/2)$ with $a=\phi_{s}A_{s}f_{y}/(\alpha_{1}\phi_{c}f^{\prime}_{c}b)$ gives $A_{s}=4353\ \text{mm}^{2}$ at O and $4237\ \text{mm}^{2}$ at C. Provide $\boxed{9\text{-}25\text{M}=4500\ \text{mm}^{2}}$ at both, in two layers.
Check ductility and minimum steel. That steel gives $a=162.4\ \text{mm}$, so $c=a/\beta_{1}=181.5\ \text{mm}$ and $c/d=0.130$: the steel strain is 0.0235, almost twelve times yield, so the section is firmly tension controlled and will warn before it fails. The minimum $A_{s}=0.2\sqrt{f^{\prime}_{c}}b_{t}h/f_{y}=2465\ \text{mm}^{2}$ is easily exceeded, and $M_{r}=2018\ \text{kN}\cdot\text{m}$ covers both critical sections.
Flexural steel at B. The hogging moment there needs only $1521\ \text{mm}^{2}$, less than the minimum, so the minimum governs and 5-25M (2500 mm$^{2}$) is carried through the top of the beam at B, giving $M_{r}=1152\ \text{kN}\cdot\text{m}$. Curtail the ninth and eighth bars from C towards mid-span only beyond the point where the reduced steel still develops the moment envelope plus the usual development length.
Shear — concrete contribution. Take the section at $d$ from the face, where $V_{f}=553\ \text{kN}$. With $d_{v}=\max(0.9d,\ 0.72h)=1260\ \text{mm}$ and minimum stirrups present, the simplified method allows $\beta=0.18$, so $$V_{c}=\phi_{c}\lambda\beta\sqrt{f^{\prime}_{c}}\,b_{w}d_{v}=0.65(0.18)\sqrt{30}(600)(1260)=485\ \text{kN}$$ leaving only 69 kN for the stirrups.
Shear — stirrups. Strength alone would allow a spacing of 1786 mm, so the detailing rules govern. The minimum area $A_{v}\ge0.06\sqrt{f^{\prime}_{c}}\,b_{w}s/f_{y}$ limits 10M double-leg stirrups to 406 mm, and the maximum spacing is $\min(0.7d_{v},600)=600\ \text{mm}$. Provide $\boxed{10\text{M}\ \text{closed stirrups at }400\ \text{mm}}$ throughout, tightened to 200 mm over the first $2d$ from each face for anchorage and confinement. The section is far from the crushing limit, $0.25\phi_{c}f^{\prime}_{c}b_{w}d_{v}=3686\ \text{kN}$.
