07-Str-A5 · December 2013

Question 6 of 7: Steel beam-column AB — strength and stability

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

Notes on this paper

Paper format. 07-Str-A5 Advanced Structural Design, National Exams, December 2013 — 3 hours, closed book (handbooks and textbooks permitted, no notes). Seven questions of equal value; any five constitute a complete paper. All seven are solved here, because the set is intended as a study resource. All loads printed on the figures are unfactored.

Design data given on the paper (SI). Concrete $f'_c = 30$ MPa; structural steel $F_y = 350$ MPa; reinforcing steel $f_y = 400$ MPa. Prestressed concrete: $f'_{ci} = 35$ MPa at transfer, $f'_c = 50$ MPa, $n = 6$, $f_{pu} = 1750$ MPa, $f_{py} = 1450$ MPa, $f_{initial} = 1200$ MPa, losses $= 240$ MPa (hence $f_{pe} = 960$ MPa).

Reference texts. CSA S16 Design of Steel Structures with the CISC Handbook of Steel Construction — plate girders (Cl. 14), lateral–torsional buckling (Cl. 13.6), plastic design (Cl. 8.6, 13.7), beam-columns (Cl. 13.8), composite beams (Cl. 17), connections (Cl. 21); CSA A23.3 Design of Concrete Structures — flexure and shear (Cl. 10, 11), slenderness (Cl. 10.13–10.16), footings (Cl. 15), prestressed concrete (Cl. 18); NBCC for load combinations; C. G. Salmon, J. E. Johnson & F. A. Malhas, Steel Structures: Design and Behavior; J. G. MacGregor & J. K. Wight, Reinforced Concrete: Mechanics and Design; T. Y. Lin & N. H. Burns, Design of Prestressed Concrete Structures; L. S. Beedle, Plastic Design of Steel Frames.

Check: load factors. The paper states only that the printed loads are unfactored, and gives no dead/live split. Every question below therefore treats each printed concentrated load as specified live load and factors it by $\alpha_L = 1.5$ (NBCC principal case $1.25D + 1.5L$); member self-weight, wherever it matters, is estimated from the trial section and factored at $1.25$. A candidate assuming a different split would obtain proportionally different sizes; the method is what is examined.

Check: section properties. All steel sections selected below are quoted by their plate dimensions (flange $b \times t$, web $h \times w$) and every property — $A$, $I_x$, $I_y$, $Z$, $J$, $C_w$, $r_x$, $r_y$ — is computed from those dimensions with root fillets neglected (slightly conservative). This makes every line checkable without a handbook; a rolled shape of equal or greater properties may be substituted directly.

Check: reading of Figures 3 and 4. On the examination drawing the dimension chain under Figure 3 reads 3 m + 4 m + 7 m + 4 m, and the overall dimension reads 15 m. Reading the drawing directly resolves this: the 15 m dimension line spans B to C ($4 + 7 + 4 = 15$ m) and the 3 m is a cantilever overhang projecting beyond support B to the free end A, which carries the 80 kN load. Similarly in Figure 4 the 10.5 m is the overall width of the floor ($3 \times 3.5$ m beam spacing), not a span; the design span is the 15 m stated in the note. Both readings are self-consistent and are used throughout.

Check: support conditions in Figure 2. The frame drawing shows identical triangle-and-roller symbols at A and D, and the frame must resist the 80 kN horizontal load, so both bases are taken as pinned. The frame is then statically indeterminate to the first degree, which is the classical form of this problem for both the elastic (Questions 2, 3) and the plastic (Questions 5, 6) treatments.

The three welded I-sections selected in this paper, drawn to a common scale.

Question 6: Steel beam-column AB — strength and stability (14 + 6 = 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. The design starts from the collapse state established in Question 5, in which one column carries the plastic hinge at its top while the other is elastic. Because the lateral load reverses, column AB must be able to be either.

Given data
Quantity	Symbol	Value
Column length, pinned base	$L$	10 000 mm
Design axial compression (envelope)	$C_f$	1396.7 kN
Design moment at the top (envelope)	$M_f$	1739.8 kN·m, zero at the base
Trial section (from Question 5)	—	Flanges $340 \times 22$, web $700 \times 14$, $d = 744$ mm
Radii of gyration	$r_x$ / $r_y$	308.1 mm / 76.3 mm
Torsional constants	$J$ / $C_w$	$3.054 \times 10^6$ mm⁴ / $1.880 \times 10^{13}$ mm⁶

Find. Whether the Question 5 section satisfies the three CSA S16 Cl. 13.8 checks — cross-sectional strength, overall in-plane member strength and lateral–torsional buckling strength — and what lateral bracing that requires.

Approach. Take the axial and moment demands from the collapse state, evaluate the compressive resistance separately about each axis with the appropriate effective length, evaluate the moment resistance including lateral–torsional buckling for the braced segment, and apply the Cl. 13.8.2 interaction expression in each of its three forms.

Establish the design actions. From the collapse statics of Question 5, the beam delivers 312.9 kN to column AB and 472.9 kN to column CD, and each column carries the 900 kN joint load plus 23.8 kN of self-weight. The envelope for a reversible lateral load is therefore $$C_f = 1396.7\ \text{kN} \quad\text{with}\quad M_f = M_p = 1739.8\ \text{kN}\cdot\text{m at the top, } 0 \text{ at the pinned base.}$$
Set the bracing. The examination note gives lateral support only at loads and joints, which for this column means A and B alone. At an unbraced weak-axis length of 10 m, $KL/r_y = 131$, the critical elastic moment falls to $M_u = 1314$ kN·m — below $0.67M_p = 1669$ kN·m, so the elastic branch applies and $M_r = \phi M_u = 1183$ kN·m, well under the 1740 kN·m the member must deliver. The interaction then reads $0.63 + 1.25 = 1.88$, a clear failure. Question 5 has already established that Cl. 13.7 requires bracing within 5452 mm of the column hinge in any case, so brace each column at its third points, $L_y = 3333$ mm. This is the decisive design decision in this question.
Cross-sectional strength (Cl. 13.8.2 a). With $C_r = \phi AF_y = 7799$ kN and $M_r = \phi ZF_y = 2241$ kN·m, and $U_{1x}$ taken as 1.0, $$\frac{C_f}{C_r} + 0.85\frac{U_{1x}M_{fx}}{M_{rx}} = \frac{1396.7}{7799} + 0.85\frac{1739.8}{2241} = 0.179 + 0.660 = \boxed{0.84 \le 1.0}.$$
Overall in-plane member strength (Cl. 13.8.2 b). Bending is about the strong axis with $K = 1.0$ in the plane, so $KL/r_x = 10\,000/308.1 = 32.5$ and $\lambda_x = 0.43$, giving $C_r = 7237$ kN from the S16 column curve ($n = 1.34$). Then $$\frac{1396.7}{7237} + 0.85\frac{1739.8}{2241} = 0.193 + 0.660 = 0.85 \le 1.0.$$
Lateral–torsional buckling strength (Cl. 13.8.2 c). Out of plane, $KL/r_y = 3333/76.3 = 43.7$ and $\lambda_y = 0.58$, giving $C_r = 6668$ kN. For the braced segment the critical elastic moment is $$M_u = \frac{\omega_2\pi}{L}\sqrt{EI_yGJ + \left(\frac{\pi E}{L}\right)^{2}I_yC_w} = 9575\ \text{kN}\cdot\text{m},$$ far above $0.67M_p = 1669$ kN·m, so the inelastic branch applies and returns the full $M_r = \phi M_p = 2241$ kN·m. The interaction gives $$\frac{1396.7}{6668} + 0.85\frac{1739.8}{2241} = 0.209 + 0.660 = \boxed{0.87 \le 1.0}.$$
Confirm the local stability. Class 1 has already been demonstrated in Question 5 for this axial load ($h/w = 50.0$ against a limit of 51.9, flange $b_{el}/t = 7.41$ against 7.75), which is a requirement rather than an option here: a plastic hinge must be able to rotate, and only a Class 1 section can do so without local buckling.
Check the base and the frame stability. The pinned base transmits 174.0 kN of horizontal shear and 1396.7 kN of axial load and no moment, so provide a base plate with four anchor rods placed on the bending axis (so as not to create unintended fixity), a shear lug or a tie into the floor slab for the horizontal force, and a levelling detail that does not restrain rotation. Frame sway stability is provided by the frame action itself; with the members Class 1 and the hinge bracing in place, the mechanism assumed in Question 5 can actually form.
Conclude. The Question 5 section satisfies all three interaction checks at 0.84–0.87 with the third-point bracing in place, so it is retained for column AB. Its governing check is lateral–torsional buckling, and the margin would vanish entirely if the column were braced only at its ends.

Question 6 — final results
Check	Resistances	Interaction
(a) Cross-sectional strength	$C_r = 7799$ kN, $M_r = 2241$ kN·m	0.84 ✓
(b) In-plane member strength	$C_r = 7237$ kN ($KL/r_x = 32.5$)	0.85 ✓
(c) Lateral–torsional buckling	$C_r = 6668$ kN ($KL/r_y = 43.7$), $M_u = 9575$ kN·m	0.87 ✓
Section adopted	Welded I: flanges $340 \times 22$, web $700 \times 14$, $d = 744$ mm, Class 1
Bracing required	Lateral support at the column third points (3.33 m)
Base	True pin: 4 anchor rods on the bending axis, shear lug for 174 kN

Topic. Beam-column design to CSA S16 Cl. 13.8 — the three interaction checks and what each one is protecting against.

Key relations. $C_f/C_r + 0.85U_{1x}M_{fx}/M_{rx} \le 1.0$ evaluated three ways; $C_r = \phi AF_y(1 + \lambda^{2n})^{-1/n}$ with $n = 1.34$ and $\lambda = (KL/r)\sqrt{F_y/\pi^2E}$; $M_u$ from the standard lateral–torsional buckling expression; $M_r = 1.15\phi M_p(1 - 0.28M_p/M_u) \le \phi M_p$ when $M_u > 0.67M_p$.

Why this works. The three checks are not repetition. The first asks whether the cross-section itself can carry the actions; the second asks whether the member can buckle in the plane of bending; the third asks whether it can buckle sideways and twist. They use the same interaction form but different resistances, and different members are caught by different checks. Here the third governs, and it governs because $r_y$ is small — a deep welded I optimised for moment is inherently weak out of plane. Bracing is far cheaper than steel: dropping the unbraced length from 10 m to 3.33 m raises $C_r$ from below the applied load to 6668 kN and lifts $M_r$ to its full plastic value, without adding a kilogram to the member.

Common pitfall. Using the elastic frame moment of 1846 kN·m here. Once the frame is plastically designed, the elastic peak is not the design action — redistribution has been explicitly accepted, and the member need only deliver $M_p$. The converse pitfall is more dangerous: designing the member for $M_p$ but bracing it as if it were elastic, so the hinge cannot actually form and the assumed mechanism never develops.

Source. CSA S16 Cl. 13.3, 13.6, 13.7 and 13.8; Salmon, Johnson & Malhas, Ch. 12 (beam-columns).

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