Question 3 of 7: Welded corner connection at B, and the beam-column AB
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 07-Str-A5 Advanced Structural Design, National Examinations, May 2013 — three hours, "closed book" with handbooks and textbooks permitted and one approved Casio or Sharp calculator. Seven questions are printed; any five constitute a complete paper and all questions are of equal value (20 marks each, split 12 + 6 + 2 or 12 + 8 or 14 + 6 or 15 + 5 as printed on page 1). All seven questions are solved here, because this set is a study resource rather than an examination script.
Design data printed on page 1 (used throughout). Design in SI. Concrete $f'_c=30$ MPa; structural steel $F_y=350$ MPa; rebar $f_y=400$ MPa. Prestressed concrete: $f_{ci}=35$ MPa at transfer, $f'_c=50$ MPa, $n=6$, $f_{pu}=1750$ MPa, $f_{py}=1450$ MPa, $f_{p,initial}=1200$ MPa, losses in prestress $=240$ MPa. All loads shown on the figures are unfactored.
Reference texts. CSA S16 Design of Steel Structures and the CISC Handbook of Steel Construction — plate girders (Cl. 14), plastic design (Cl. 8.6), beam-columns (Cl. 13.8), composite beams (Cl. 17) and connections (Cl. 21); CSA A23.3 Design of Concrete Structures — flexure, shear, columns (Cl. 10), footings (Cl. 15), joints (Cl. 21.7) and prestressed concrete (Cl. 18); NBCC for the 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 "all loads shown are unfactored" and gives no dead/live split for the concentrated loads. Every question below therefore treats the printed concentrated loads as specified live load and factors them by $\alpha_L=1.5$ (NBCC principal load case $1.25D+1.5L$); self-weight, where it is significant (the plate girder, the concrete beam, the composite slab, the prestressed girder), is estimated from the trial section and factored at $1.25$. A candidate who assumed a different split would obtain proportionally different member sizes; the method is what is being examined.
Check: section properties. Rolled-shape properties (area, $I$, $S$, $Z$, $r$) are quoted from the CISC Handbook of Steel Construction, which candidates are permitted to bring into this examination. They are stated explicitly wherever they are used, so every subsequent line can be checked against them.
Question 3: Welded corner connection at B, and the beam-column AB (14 + 6 = 20 marks)
Given. The frame of Figure 2 with W610 × 155 in both members ($d=611$ mm, $b_f=324$ mm, $t_f=19.0$ mm, $w=12.7$ mm, $A=19\,800$ mm$^2$, $Z_x=4750\times10^{3}$ mm$^3$, $I_y=108\times10^{6}$ mm$^4$, $r_x=255$ mm, $r_y=73.9$ mm), $M_p=1472$ kN·m from Question 2, and the collapse reactions $A_x=64$ kN, $A_y=982$ kN. Electrode E49xx ($X_u=490$ MPa), $F_y=350$ MPa.
Find. (a) A welded knee detail at B able to develop the full plastic moment of the members, including whatever panel-zone reinforcement is needed; (b) whether W610 × 155 satisfies CSA S16 Cl. 13.8 as the beam-column AB.
(a) The welded corner connection at joint B
Approach. In plastic design a connection is not proportioned for the moment that happens to reach it — it must be able to deliver $M_p$ with enough rotation capacity for the mechanism to form elsewhere. So the knee is designed for $M_f=M_p=1472$ kN·m, the flange forces are carried by complete-joint-penetration groove welds, and the panel zone is checked and stiffened.
Flange force to be transferred. Idealising the moment as a flange couple over the lever arm $d-t_f$,
$$T_f=\frac{M_p}{d-t_f}=\frac{1472\times10^{6}}{611-19}=\boxed{2486\ \text{kN}}$$
This force must pass from the beam flange, across the corner, into the column flange.
Flange welds. Use complete-joint-penetration (CJP) groove welds, made from the outside with backing bars, at both the top and the bottom flange of the beam where it meets the column. A CJP weld in tension is matched to the base metal, so no weld calculation is required beyond specifying matching electrodes (E49xx for 350W) and full inspection; the strength check is the flange itself. A square-cut, fully welded knee is preferred here over a bolted end plate because the mechanism relies on the joint rotating plastically without slip.
Panel-zone shear. The knee panel, $d_b\times d_c=611\times611$ mm, must carry the flange force less the shear brought up by the column:
$$V_{f,panel}=T_f-V_{col}=2486-64=2422\ \text{kN}$$
Its own resistance (S16 Cl. 13.4.1.1, an unstiffened web in shear) is
$$V_r=\phi(0.55F_y)wd_c=0.9(0.55)(350)(12.7)(611)=1344\ \text{kN}$$
which is only 55% of what is required. The panel must be reinforced.
Diagonal stiffener. Add a diagonal stiffener across the knee, from the inside corner to the outside corner. It resists the shortfall as a strut/tie at 45°:
$$F_{dia}=\left(V_{f,panel}-V_r\right)\sqrt{2}=(2422-1344)\sqrt{2}=1525\ \text{kN}$$
$$A_{req}=\frac{F_{dia}}{\phi F_y}=\frac{1525\times10^{3}}{0.9(350)}=\boxed{4840\ \text{mm}^{2}}$$
Provide a pair of plates $130\times20$ mm, one each side of the web ($2\times2600=5200$ mm$^2$), groove-welded into the corners of the panel. Their width/thickness ratio $130/20=6.5$ is inside the Class 1 limit $145/\sqrt{F_y}=7.75$, so the strut cannot buckle locally before it yields.
Continuity plates. Provide horizontal stiffener pairs in line with each beam flange, thickness at least $t_f=19$ mm (use 20 mm), fitted between the column flanges and welded all round. They carry the 2486 kN flange force into the panel and prevent column-flange bending and web crippling. As a check on the alternative of leaving them out, the unstiffened web resistances at a concentrated flange force — web crippling and web yielding — are both well below 2486 kN for a 12.7 mm web, so the plates are mandatory rather than optional.
Web welds. The beam web is joined to the column with a pair of 10 mm fillet welds over the clear web depth $d-2t_f=573$ mm:
$$V_r=2(0.67\phi_w A_wX_u)=2\left[0.67(0.67)(0.707\times10\times573)(490)\right]\times10^{-3}=1782\ \text{kN}$$
which comfortably exceeds the beam shear at B. Alternatively a CJP weld may be specified across the whole section for simplicity of inspection.
Welded knee at joint B: CJP groove welds at both beam flanges, continuity plates in line with them, and a diagonal stiffener pair carrying the panel-shear shortfall.
(b) Is W610 × 155 adequate as the beam-column AB?
Actions on AB. From the collapse statics of Question 2, member AB carries $C_f=A_y=982$ kN of axial compression with $M_{f}=384$ kN·m at B and zero at the pinned base A. Its length is 6 m and the frame is braced (lateral support at all joints and load points), so $K=1.0$.
Compressive resistance about both axes. Lateral support is provided only at the joints, so $KL=6000$ mm applies about the strong and the weak axis. With
$$\lambda=\frac{KL}{r}\sqrt{\frac{F_y}{\pi^{2}E}},\qquad C_r=\phi AF_y\left(1+\lambda^{2n}\right)^{-1/n},\ n=1.34$$
the strong axis gives $\lambda_x=(6000/255)(0.01332)=0.313$ and $C_{rx}=\boxed{6037\ \text{kN}}$, while the weak axis gives $\lambda_y=(6000/73.9)(0.01332)=1.081$ and $C_{ry}=3425$ kN. The weak axis is by far the more slender, and it is the one that enters the third of the three checks below.
Moment resistance. The moment varies linearly from 0 to 384 kN·m, so $\kappa=0$ and $\omega_2=1.75$. With $J=1.89\times10^{6}$ mm$^4$ and $C_w=I_y(d-t_f)^{2}/4=9.46\times10^{12}$ mm$^6$,
$$M_u=\frac{\omega_2\pi}{L}\sqrt{EI_yGJ+\left(\frac{\pi E}{L}\right)^{2}I_yC_w}=3470\ \text{kN}\cdot\text{m}$$
Since $M_u>0.67M_p=1114$ kN·m, $M_r=1.15\phi M_p\left(1-0.28M_p/M_u\right)=1490$ kN·m, capped at $\phi M_p=1496$ kN·m; take $M_r=1490$ kN·m.
Amplification factor. $C_e=\pi^{2}EA/(KL/r_x)^{2}=70\,595$ kN, and for a member with no transverse load and $\kappa=0$, $\omega_1=0.6$, so $U_{1x}=\omega_1/(1-C_f/C_e)=0.60$, taken as 1.0 for the strength checks as Cl. 13.8.2 requires.
Cl. 13.8.2 interaction — all three checks.
$$\text{cross-section strength:}\quad \frac{C_f}{\phi AF_y}+\frac{0.85U_{1x}M_{fx}}{M_{rx}}=\frac{982}{6237}+\frac{0.85(384)}{1496}=0.376$$
$$\text{overall member strength:}\quad \frac{C_f}{C_{rx}}+\frac{0.85U_{1x}M_{fx}}{M_{rx}}=\frac{982}{6037}+\frac{0.85(384)}{1496}=0.381$$
$$\text{lateral-torsional buckling:}\quad \frac{C_f}{C_{ry}}+\frac{0.85U_{1x}M_{fx}}{M_{rx}}=\frac{982}{3425}+\frac{0.85(384)}{1490}=\boxed{0.506}\le 1.0$$
The third check uses the weak-axis $C_{ry}$ and the laterally unsupported $M_r$, so it governs — but at barely half the available capacity. W610 × 155 is more than adequate for AB.
Interpret the margin. The section was sized by the beam requirement $M_p=1472$ kN·m and the column happens to be lightly loaded, which is exactly what the constant-$M_p$ instruction of Question 2 produces. If economy mattered, AB could be a much lighter section — but then the frame would no longer have uniform $M_p$ and the mechanism analysis of Question 2 would have to be redone with member-dependent hinge capacities.
