Question 1 of 7: Welded plate girder — flexure, shear and their interaction
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 1: Welded plate girder — flexure, shear and their interaction (12 + 6 + 2 = 20 marks)
Given. A two-span continuous welded plate girder, spans $L = 12$ m each, simply supported at A and B and encastré at C, carrying one unfactored 500 kN concentrated load at each mid-span, with lateral bracing at 2 m centres.
Given data
Quantity
Symbol
Value
Span, each of two
$L$
12 000 mm
Concentrated load (unfactored)
$P$
500 kN at each mid-span
Steel yield stress
$F_y$
350 MPa
Modulus of elasticity
$E$
200 000 MPa
Lateral bracing interval
$L_b$
2 000 mm
Support conditions
—
A pinned, B roller, C fixed
Find. Flange and web plate sizes such that the factored moment, factored shear and their combination are all within the resistances of CSA S16, together with the stiffening required at the supports and load points.
Figure 1 as drawn, with the factored bending-moment diagram beneath. Sagging is plotted downwards.
Approach. Analyse the two-span girder by the stiffness method for the factored loads, size the flanges from the peak factored moment using the plastic modulus, confirm the plate slendernesses keep the section Class 2, then check web shear by CSA S16 Cl. 13.4 and the moment–shear interaction by Cl. 14.6.
Factor the loads and analyse the girder. With $P_f = 1.5 \times 500 = 750$ kN at each mid-span and a trial girder self-weight of $1.386$ kN/m factored to $w_f = 1.73$ kN/m, the stiffness solution of the propped two-span beam (A pinned, B roller, C fixed) gives
$$M_{f,\text{AB}} = 1544.6\ \text{kN}\cdot\text{m (sagging)}, \qquad M_{f,\text{B}} = 1473.2\ \text{kN}\cdot\text{m (hogging)}$$
with $M_{f,\text{BC}} = 1053.6$ kN·m sagging and $M_{f,\text{C}} = 982.1$ kN·m hogging. The largest shear is at the interior support, $V_f = 508.2$ kN, and the reaction there is 934.5 kN.
Set the required plastic modulus. For a Class 1 or Class 2 section the factored moment resistance is $M_r = \phi Z F_y$ with $\phi = 0.90$, so
$$Z_{req} = \frac{M_f}{\phi F_y} = \frac{1544.6 \times 10^6}{0.90 \times 350} = 4.90 \times 10^6\ \text{mm}^3.$$
Choose plates and confirm the section class. Try flanges $250 \times 18$ mm and a web $900 \times 10$ mm, giving an overall depth $d = 936$ mm ($L/d \approx 12.8$, a normal proportion for a plate girder). The plastic modulus is
$$Z = b\,t\,(d - t) + \frac{w h^2}{4} = 250(18)(918) + \frac{10 \times 900^2}{4} = 6.156 \times 10^6\ \text{mm}^3.$$
Flange projection $b_{el}/t = 120/18 = 6.67 \le 145/\sqrt{F_y} = 7.75$, so the flanges are Class 1; the web $h/w = 90.0 \le 1700/\sqrt{F_y} = 90.9$, so the web is Class 2. The section is therefore Class 2 and the full plastic modulus may be used.
Check flexure. Lateral bracing at 2 m gives a critical elastic moment of the order of $10^4$ kN·m, far above $M_p$, so lateral–torsional buckling does not reduce the resistance and
$$M_r = \phi Z F_y = 0.90 \times 6.156 \times 10^6 \times 350 = \boxed{1939\ \text{kN}\cdot\text{m}} \; > \; 1544.6\ \text{kN}\cdot\text{m}. $$
Part (a) is satisfied with 26 % reserve, which also absorbs the difference between the trial and final self-weight.
Check shear. For an unstiffened web, $k_v = 5.34$ and $621\sqrt{k_v/F_y} = 76.7 < h/w = 90$, so the web buckles elastically and
$$F_{cri} = \frac{180\,000\,k_v}{(h/w)^2} = \frac{180\,000 \times 5.34}{90^2} = 118.7\ \text{MPa}, \qquad V_r = \phi A_w F_{cri} = 0.90(9000)(118.7) = \boxed{961\ \text{kN}}$$
against $V_f = 508.2$ kN. Part (b) is satisfied without intermediate stiffeners; only bearing stiffeners are needed.
Check the moment–shear interaction at B. The interior support carries the peak of both actions simultaneously, so apply CSA S16 Cl. 14.6:
$$0.727\frac{M_f}{M_r} + 0.455\frac{V_f}{V_r} = 0.727\frac{1473.2}{1939} + 0.455\frac{508.2}{961} = 0.552 + 0.241 = \boxed{0.79 \le 1.0}$$
so part (c) is satisfied. (Because $V_f < 0.6V_r$, the check is not in fact governing, but it is written out because two marks are assigned to it.)
Detail the stiffening. Provide bearing stiffeners at A, B and C and under each 500 kN load. At B a pair of $100 \times 12$ mm plates acting with an effective web length of $10w$ gives a bearing resistance of $\phi A F_y = 0.90(2400 + 1000)(350) = 1071$ kN against the 934.5 kN factored reaction. Weld the flanges to the web with continuous 6 mm fillet welds each side, and provide intermediate transverse stiffeners at $a/h = 1.5$ only if a thinner web is later adopted for economy.
Question 1 — final results
Item
Result
Cross-section
Welded plate girder: flanges $250 \times 18$ mm, web $900 \times 10$ mm, $d = 936$ mm (141 kg/m)
