Question 1 of 7: Two-span continuous beam — section for each span and a full moment connection at B

Question A1: Two-span continuous beam — section for each span and a full moment connection at B (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.

Find. The lightest adequate W section for each span, the two sections deliberately different, plus a detailed full-strength moment splice at the interior support B that accommodates the change of depth.

[Figure not reproduced: Figure A1.1 — Figure 1 of the paper redrawn: two 5 m spans, pin at A, rollers at B and C, 160 kN dead + 100 kN live at mid-span of AB, 15 kN/m dead + 10 kN/m live over BC. See the official exam paper.]

Approach. Factor the loads, solve the single redundant at B from rotational compatibility with the two spans carrying different second moments of area, envelope the pattern-loaded cases, select each section on flexure and lateral–torsional buckling, check shear and live-load deflection, then proportion the splice to develop the plastic moment of the lighter member.

1. Factor the loads. With $1.25D+1.5L$ the point load in span AB is $$P_f = 1.25(160) + 1.5(100) = 200 + 150 = 350\ \text{kN}$$ and the distributed load on span BC is $$w_f = 1.25(15) + 1.5(10) = 18.75 + 15 = 33.75\ \text{kN/m}.$$ Removing the live load leaves the dead-only values $P_f = 200\ \text{kN}$ and $w_f = 18.75\ \text{kN/m}$, which are needed for the patterned cases.
2. Write the compatibility equation at B for unequal span stiffnesses. Release the continuity at B and treat the two spans as simply supported. The end rotations at B are $P L^2/16EI_1$ from the point load and $wL^3/24EI_2$ from the UDL, and the hogging moment $M_B$ applied to both span ends removes them. Setting the relative rotation to zero, $$M_B = \frac{\dfrac{PL^2}{16 I_1} + \dfrac{wL^3}{24 I_2}}{\dfrac{L}{3 I_1} + \dfrac{L}{3 I_2}}.$$ Because the question requires the two spans to be different sections, $I_1 \ne I_2$ and this ratio cannot be cancelled — that is the point of the question.
3. Get a first estimate assuming equal stiffness. Putting $I_1 = I_2$ and full factored load on both spans, $$M_B = \frac{350(5)^2/16 + 33.75(5)^3/24}{5/3 + 5/3} = \frac{546.9 + 175.8}{3.333} = 216.8\ \text{kN}\cdot\text{m},$$ and the sagging moment at the load point in span AB is $M = P L/4 - M_B/2 = 437.5 - 108.4 = 329.1\ \text{kN}\cdot\text{m}$. These are only trial values; they will move once real sections are inserted.
4. Trial sections from the required plastic modulus. A Class 1 section needs $Z_x \ge M_f/\phi F_y$ with $\phi = 0.90$. For span AB, $Z_x \ge 329.1\times10^6/(0.9\times300) = 1.22\times10^6\ \text{mm}^3$, and for span BC, $Z_x \ge 216.8\times10^6/270 = 0.80\times10^6\ \text{mm}^3$. Try W460×68 for AB ($Z_x = 1.469\times10^6$, $I_x = 293.0\times10^6$) and W410×54 for BC ($Z_x = 1.029\times10^6$, $I_x = 183.0\times10^6\ \text{mm}^4$). Both are Class 1 in flange and web at $F_y = 300$ MPa.
5. Re-solve with the real stiffness ratio — this is where the answer changes. With $I_1/I_2 = 293.0/183.0 = 1.601$ the stiffer loaded span sheds less moment to the support: all spans fully loaded gives $M_B = 191.1\ \text{kN}\cdot\text{m}$ against the 216.8 estimated at equal stiffness, and the sagging moment in AB rises correspondingly. Patterning the live load, live on AB alone gives the worst sagging in AB and live on BC alone the worst sagging in BC.
6. Envelope the three load patterns. Tabulating $M_B$, the sagging moments and the shears either side of B for the three arrangements gives the design envelope $M_B = 191.1$, $M_{sag,AB} = 356.4$ and $M_{sag,BC} = 48.1\ \text{kN}\cdot\text{m}$, with $V_B = 213.2\ \text{kN}$ just left of B and $122.6\ \text{kN}$ just right of it. The governing moment in each span is therefore $$\boxed{\begin{aligned} M_{f,AB} &= 356.4\ \text{kN}\cdot\text{m} \\ M_{f,BC} &= 191.1\ \text{kN}\cdot\text{m}\ \text{(hogging, at B)}. \end{aligned}}$$
7. The stiffness ratio disqualifies the obvious lighter section. W460×60 has $\phi M_p = 0.9(300)(1.266\times10^6) = 341.8\ \text{kN}\cdot\text{m}$, which comfortably exceeds the equal-stiffness estimate of 329.1 but is less than the true 356.4. A candidate who never re-ran the analysis with the real $I_1/I_2$ would select an inadequate section by about 4 per cent.
8. Check span AB in flexure including lateral–torsional buckling. Lateral support is taken at A, at the load point and at B, so the unbraced length is 2.5 m. With $\omega_2$ from the Clause 13.6 quadratic form the moment gradient over each half span is steep enough that $M_r$ reaches the full $\phi M_p = 0.9(300)(1.469\times10^6) = 396.7\ \text{kN}\cdot\text{m}$. Hence $$\frac{M_f}{M_r} = \frac{356.4}{396.7} = 0.898 \le 1.0 \quad \checkmark$$
9. Check span BC, where the 5 m unbraced length does bite. Over span BC the compression flange near B is the bottom flange, unbraced over the whole 5 m. With $\omega_2 = 1.746$ for the moment diagram running from 191.1 hogging at B to zero at C, Clause 13.6(a) gives $M_r = 237.3\ \text{kN}\cdot\text{m}$, below $\phi M_p = 277.8$, so buckling governs: $$\frac{M_f}{M_r} = \frac{191.1}{237.3} = 0.805 \le 1.0 \quad \checkmark$$
10. Check shear (Clause 13.4.1.1). Both webs are stocky, so $F_s = 0.66F_y$ and $V_r = \phi A_w F_s$ gives 744 kN for W460×68 and 539 kN for W410×54. The ratios are $213.2/744 = 0.286$ and $122.6/539 = 0.228$ — shear is nowhere near critical, which is normal for a rolled beam at these spans.
11. Check live-load deflection. Superposing the simply supported case and the continuity moment, live load on AB alone gives $\Delta = PL^3/48EI_1 - M_B L^2/16EI_1 = 4.44 - 0.96 = 3.48\ \text{mm}$, and live on BC alone gives $5wL^4/384EI_2 - M_B L^2/16EI_2 = 2.22 - 0.82 = 1.40\ \text{mm}$. Against the usual floor criterion $L/360 = 13.9\ \text{mm}$ both are comfortable, at $L/1435$ and $L/3565$ respectively.
12. Size the moment splice at B. A full moment connection must develop the member rather than merely the applied moment, so design for the greater of $M_f = 191.1$ and the $\phi M_p = 277.8\ \text{kN}\cdot\text{m}$ of the lighter W410×54: $M_{con} = 277.8\ \text{kN}\cdot\text{m}$. Resolving this into a flange couple over the lever arm $d - t_f = 403 - 10.9 = 392.1\ \text{mm}$, $$T_f = \frac{M_{con}}{d - t_f} = \frac{277.8\times10^6}{392.1} = \boxed{708.5\ \text{kN}}$$ in each flange plate.
13. Proportion the plates, welds and web splice. A 200 × 20 mm 300W plate gives $T_r = \phi A F_y = 0.9(200)(20)(300) = 1080\ \text{kN}$, a ratio of 0.656 — the surplus is deliberate, because the plate must also accommodate the bolt or weld pattern. A 10 mm fillet weld with E49XX electrodes carries $V_r = 0.67\phi_w(0.707D)X_u = 1.555\ \text{kN/mm}$, so $708.5/1.555 = 456\ \text{mm}$ of weld is needed, i.e. 228 mm each side of each plate. The web splice carries the 213 kN shear: three M22 A325 bolts in double shear provide $3 \times 353 = 1060\ \text{kN}$, an ample margin that keeps the plate compact.
14. Resolve the change of depth. The two beams differ by $459 - 403 = 56\ \text{mm}$. Align the top flanges so the floor above stays flat and let the whole difference appear at the soffit: the top flange plate runs straight across, and the bottom plate steps through a taper not steeper than 1:2.5 so the flange force turns without a stress concentration. Stiffen the deeper web on both sides at the step, in line with the shallower beam soffit, to carry the vertical component of the turned flange force.

Final Results