07-Str-A5 · May 2015

Question 6 of 7: Reinforced-concrete frame — design of member BC

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

Notes on this paper

Paper format. National Exams, May 2015 — 07-Str-A5 Advanced Structural Design. Three hours, "closed book" (textbooks and design handbooks permitted, no notes). Seven questions; any five constitute a complete paper and all are of equal value, the printed mark split being 20 marks each. All loads shown on Figures 1–4 are unfactored. All seven questions are answered here.

Design data printed on page 1 and used throughout. Concrete $f'_c = 30\ \text{MPa}$; structural steel $F_y = 350\ \text{MPa}$; reinforcing steel $f_y = 400\ \text{MPa}$; prestressed concrete $f'_{ci} = 35\ \text{MPa}$ at transfer and $f'_c = 50\ \text{MPa}$ in service, $n = 6$; $f_{pu} = 1750\ \text{MPa}$, $f_{py} = 1450\ \text{MPa}$, $f_{p,\text{initial}} = 1200\ \text{MPa}$, losses $= 240\ \text{MPa}$, hence $f_{pe} = 960\ \text{MPa}$ and the loss ratio $\eta = 0.80$.

Check: load factors. The paper says only that "all loads shown are unfactored" and gives no load classification. Every printed load is therefore treated as a specified live load and factored at 1.5; member self weight is treated as dead and factored at 1.25 (NBCC 2020 Case 2, $1.25D + 1.5L$). Resistance factors are $\phi = 0.90$ (steel), $\phi_w = 0.67$ and the extra 0.67 of CSA S16 Cl. 13.13.2.2 (welds), $\phi_{sc} = 0.80$ (shear studs), $\phi_c = 0.65$, $\phi_s = 0.85$, $\phi_p = 0.90$ (CSA A23.3). In an examination you would state exactly this and proceed.

Reference texts for 07-Str-A5.

CSA S16:19, Design of Steel Structures, with the CISC Handbook of Steel Construction (11th ed.) — Cl. 13.3 (compression), 13.5–13.6 (flexure and lateral–torsional buckling), 13.7 (bracing at plastic hinges), 13.8 (beam–columns), 13.13 (welds), 14.3–14.6 (plate girders), Cl. 17 (composite beams).
CSA A23.3:19, Design of Concrete Structures, with the CAC Concrete Design Handbook (4th ed.) — Cl. 10 (flexure and columns), Cl. 11 (shear), Cl. 18 (prestressed concrete).
Salmon, Johnson and Malhas, Steel Structures: Design and Behavior, 5th ed. — plastic analysis (Ch. 10), plate girders (Ch. 11), composite construction (Ch. 16).
MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, Canadian edition — frames, beam–columns, detailing.
Collins and Mitchell, Prestressed Concrete Structures — cable zone, transfer and service stress checks, harped profiles.
National Building Code of Canada 2020 for load combinations; CSA S6 (CHBDC) for the pedestrian-bridge serviceability and vibration criteria.

Question 6: Reinforced-concrete frame — design of member BC (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.

Quantity	Value
Frame (Figure 4)	columns $AB$, $ED$ 8 m, fixed at $A$ and $E$; beam $B$–$C$–$D$ 16 m with an internal hinge at $C$ (midspan)
Loads (specified)	500 kN at $B$, 250 kN at 4 m, 250 kN at 12 m, 500 kN at $D$
Materials	$f'_c = 30\ \text{MPa}$, $f_y = 400\ \text{MPa}$, $\alpha_1 = 0.805$, $\beta_1 = 0.895$
Trial member sizes	beam 500 × 1200 mm; columns 700 × 1000 mm

Find. The factored moment and shear envelope in $BC$, the flexural reinforcement, the stirrups, and a reinforcing arrangement for the whole member $BCD$.

Figure 4 — reinforced-concrete rigid frame with an internal hinge at midspan of the beam. Both bases are fixed.

Approach. Exploit symmetry. Structure and loading are symmetric about $C$, so the shear at $C$ is zero, and the hinge makes the moment there zero as well; the half frame is then loaded only by an axial thrust at $C$. That reduces the beam moments to pure statics — they turn out to be independent of the thrust — and only the columns need the one compatibility equation.

Factored loads. The beam self weight is $0.5(1.2)(24) = 14.4\ \text{kN/m}$, factored to 18.0 kN/m; the point loads factor to 750 kN at $B$ and $D$ and 375 kN at 4 m and 12 m. Vertical equilibrium of the half frame (with $V_C = 0$) gives
$$A_y = 750 + 375 + 18.0(8) = 1269\ \text{kN}$$
Impose the hinge condition. Taking moments of all left-hand forces about $C$ and setting $M_C = 0$,
$$M_A + 8A_x = 8(1269) - 750(8) - 375(4) - 18(8)(4) = 2076\ \text{kN}\cdot\text{m}$$
The beam moment at a distance $x$ from $B$ then follows from the same free body:
$$M(x) = 2076 - 1269x + 750x + 375\langle x - 4\rangle + 9x^2$$
Notice that the 750 kN load at $B$ contributes $+750x$ and simultaneously contributes $-750x$ through $A_y$: loads standing over a column do not bend the beam at all.
The beam moment is statically determinate. Every trace of $A_x$ has cancelled, so the moment diagram in $BC$ does not depend on the redundant thrust:
$$M(0) = 2076,\quad M(4) = 144,\quad M(x) = 9(x-8)^2 \text{ for } 4 \le x \le 8, \quad M(8) = 0\ \text{kN}\cdot\text{m}$$
The whole member is in hogging — tension on top from end to end — with the maximum 2076 kN·m at the joint. Shear runs from 519 kN at $B$ to 447 kN just left of the 250 kN load, then drops to 72 kN and dies at the hinge.
The thrust, for completeness. Releasing the horizontal restraint and applying a unit force at $C$ gives $M_1 = (y - 8)$ in the column and zero in the beam, while the primary moment in the column is the constant $-M_B$. Hence
$$\delta_{11} = \int_0^8 (y-8)^2dy = \frac{512}{3}, \qquad \delta_{10} = -M_B\int_0^8 (y-8)dy = 32M_B$$ $$H = \frac{3M_B}{16} = \frac{3(2076)}{16} = 389.3\ \text{kN} \qquad\Longrightarrow\qquad M_A = M_B - 8H = -\frac{M_B}{2} = -1038\ \text{kN}\cdot\text{m}$$
The base moment is exactly half the joint moment and of opposite sign, a tidy closed-form result that follows from the primary column moment being constant.

Factored moment diagram for the half frame. The beam is entirely in hogging; the column reverses sign 2.67 m above the base.

Flexure at the critical section. Design at the face of the 1000 mm column, $x = 0.5\ \text{m}$, where $M_f = 1818.8\ \text{kN}\cdot\text{m}$. With 40 mm cover, 10M stirrups and two layers of 30M top bars, $d = 1103.8\ \text{mm}$. Solving
$$M_r = \phi_sA_sf_y\left(d - \frac{a}{2}\right),\qquad a = \frac{\phi_sA_sf_y}{\alpha_1\phi_cf'_cb}$$
for $M_r = M_f$ gives $A_s = 5423\ \text{mm}^2$. Provide 8–30M ($A_s = 5600\ \text{mm}^2$) in two layers of four:
$$a = 242.6\ \text{mm},\quad c/d = 0.246 \le \frac{700}{700+f_y} = 0.636 \quad\Longrightarrow\quad \boxed{M_r = 1871\ \text{kN}\cdot\text{m} \ \ge\ 1819\ \text{kN}\cdot\text{m}}$$
The section is comfortably tension-controlled, and $A_{s,\min} = 0.2\sqrt{f'_c}b_th/f_y = 1643\ \text{mm}^2$ is easily exceeded.
No sagging steel is required by analysis — but provide it anyway. Because loads over the columns do not bend the beam, removing either 250 kN load leaves the diagram still entirely hogging, so no pattern of the given loads produces sagging. Nevertheless provide 4–25M continuous at the bottom as structural-integrity reinforcement, lapped through the hinge region, as A23.3 Cl. 8.5 and good practice require.
Shear. At $d$ from the column face, $V_f = 490.1\ \text{kN}$. With $d_v = \max(0.9d, 0.72h) = 993.4\ \text{mm}$ and the simplified method ($\beta = 0.18$, $\theta = 35^\circ$),
$$V_c = \phi_c\beta\sqrt{f'_c}\,b_wd_v = 0.65(0.18)\sqrt{30}(500)(993.4)/10^3 = 318.3\ \text{kN}$$ $$V_s = \frac{\phi_sA_vf_yd_v\cot\theta}{s} \quad\Longrightarrow\quad \text{10M at 250 mm gives } V_r = 704\ \text{kN} \ \ge\ 490\ \text{kN}$$
Beyond 4 m the shear falls to 72 kN, but the member is 1200 mm deep so minimum stirrups are required throughout: 10M at 400 mm satisfies both $A_{v,\min} = 0.06\sqrt{f'_c}b_ws/f_y$ (which caps $s$ at 487 mm) and $s \le 0.7d_v \le 600\ \text{mm}$. The upper limit $0.25\phi_cf'_cb_wd_v = 2422\ \text{kN}$ is nowhere approached.
Detailing member $BCD$. Top steel 8–30M in two layers over each column, four of them continuous the full length and the outer four curtailed at about 5.5 m from each joint (beyond the point where $M_f$ plus the development length falls below the resistance of the remaining bars); all top bars anchored through the joint with 90° standard hooks turned down into the far column face. Bottom steel 4–25M continuous. Stirrups 10M at 250 mm for the first 4 m from each column, then 10M at 400 mm to the hinge. The hinge at $C$ is built as a keyed bearing with crossed dowels sized to transfer the 187.5 kN specified thrust and to carry no moment.

Reinforcing arrangement for member BCD. Top steel is continuous over the whole member because the beam never goes into sagging.

Result	Value
Factored column axial load	1269 kN
Beam moment at the joint $B$ / at 4 m / at the hinge	2076 / 144 / 0 kN·m (all hogging)
Design moment at the column face	1818.8 kN·m
Frame thrust $H = 3M_B/16$ / base moment	389.3 kN / −1038 kN·m
Beam section	500 × 1200 mm, $d = 1103.8$ mm
Flexural steel	$A_s$ required 5423 mm² → 8–30M top ($M_r = 1871$ kN·m, 0.972)
Bottom steel	4–25M continuous (integrity; not required by analysis)
Shear at $d$ from the face / $V_c$	490.1 kN / 318.3 kN
Stirrups	10M at 250 mm for 4 m from each column, then 10M at 400 mm

Topic. Limit-states design of a beam in a fixed-base portal containing an internal hinge, and the simplification that symmetry buys.

Key relations. At an axis of symmetry under symmetric loading, shear $= 0$; an internal hinge adds $M = 0$; force method $\delta_{10} + X\delta_{11} = 0$ with $\delta_{ij} = \int M_iM_j\,ds/EI$; $M_r = \phi_sA_sf_y(d - a/2)$ with $a = \phi_sA_sf_y/(\alpha_1\phi_cf'_cb)$; simplified shear $V_c = \phi_c\beta\sqrt{f'_c}b_wd_v$ with $\beta = 0.18$, $V_s = \phi_sA_vf_yd_v\cot\theta/s$.

Why this works. The frame has six reaction components and only four equations (three of statics plus the hinge), so it is twice indeterminate; symmetry disposes of one redundant and the hinge condition of another, leaving a single unknown thrust. What makes the question tractable is that the unit-load moment diagram for that thrust is zero everywhere in the beam, so the beam's moments are fixed by statics alone and the compatibility integral is needed only for the column. That also explains the clean result $M_{\text{base}} = -M_{\text{top}}/2$: with a constant primary moment down the column, the integrals reduce to $H = 3M_B/16$ regardless of the load pattern.

Common pitfall. Assuming the beam must sag somewhere and detailing a conventional bottom-steel cage at midspan. It never sags: the hinge kills the moment at $C$ and the 250 kN loads are close enough to the columns that the diagram decays monotonically. The second pitfall is including the 500 kN column loads in the beam moment calculation; they pass straight into the columns and cancel out of $M(x)$ exactly.

Source. CSA A23.3:19 Cl. 10 (flexure) and Cl. 11.3 (simplified shear method); MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, frames and detailing chapters; Hibbeler, Structural Analysis, force method and symmetric structures.

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Quantity	Value
Frame (Figure 4)	columns \(AB\), \(ED\) 8 m, fixed at \(A\) and \(E\); beam \(B\)–\(C\)–\(D\) 16 m with an internal hinge at \(C\) (midspan)
Loads (specified)	500 kN at \(B\), 250 kN at 4 m, 250 kN at 12 m, 500 kN at \(D\)
Materials	\(f'_c = 30\ \text{MPa}\), \(f_y = 400\ \text{MPa}\), \(\alpha_1 = 0.805\), \(\beta_1 = 0.895\)
Trial member sizes	beam 500 × 1200 mm; columns 700 × 1000 mm

Result	Value
Factored column axial load	1269 kN
Beam moment at the joint \(B\) / at 4 m / at the hinge	2076 / 144 / 0 kN·m (all hogging)
Design moment at the column face	1818.8 kN·m
Frame thrust \(H = 3M_B/16\) / base moment	389.3 kN / −1038 kN·m
Beam section	500 × 1200 mm, \(d = 1103.8\) mm
Flexural steel	\(A_s\) required 5423 mm² → 8–30M top (\(M_r = 1871\) kN·m, 0.972)
Bottom steel	4–25M continuous (integrity; not required by analysis)
Shear at \(d\) from the face / \(V_c\)	490.1 kN / 318.3 kN
Stirrups	10M at 250 mm for 4 m from each column, then 10M at 400 mm