Question 1 of 7: Two-span continuous beam — lightest wide-flange section

Question A1: Two-span continuous beam — lightest wide-flange section (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 single W section that satisfies flexure (with lateral-torsional buckling over 5 m), shear and deflection for the full factored moment envelope of the two-span beam.

Approach. Solve the once-redundant beam with the three-moment equation for each patterned load case, take the moment and shear envelope, then choose the lightest section whose Class 1 or 2 lateral-torsional resistance $M_r$ over the 5 m unbraced length exceeds the demand.

1. Factor the loads. With only dead and live load present the principal combination is $1.25D + 1.5L$, and $0.9D$ is used on a span where the dead load relieves the moment being maximised. $$w_f = 1.25(15) + 1.5(10) = 33.75\ \text{kN/m}$$ $$P_f = 1.25(160) + 1.5(100) = 350\ \text{kN}$$ A trial section is needed before the self weight can be included; W530x72 is carried through below and its self weight of $0.696\ \text{kN/m}$ is factored with the dead load.
2. Solve the redundant with the three-moment equation. With $M_A = M_C = 0$ and constant $EI$, the equation for the two spans reduces to a single unknown: $$2M_B\!\left(L_1 + L_2\right) = -\frac{w L_1^{3}}{4} - \frac{3 P L_2^{2}}{8}$$ The first term is the load term of the uniformly loaded span A–B and the second that of the centrally loaded span B–C.
3. Load case 1 — full factored load on both spans (maximum hogging at B). Substituting $w = 33.75 + 1.25(0.696) = 34.62\ \text{kN/m}$, $P = 350\ \text{kN}$, $L_1 = L_2 = 5\ \text{m}$: $$2M_B(10) = -\frac{34.62(125)}{4} - \frac{3(350)(25)}{8} = -1081.9 - 3281.3$$ $$\boxed{M_B = -219.5\ \text{kN}\!\cdot\!\text{m}\ \text{(hogging)}}$$ Statics of span B–C then gives $R_C = (M_B + PL/2)/L = 133.3\ \text{kN}$ and a shear just left of the load of $V_f = 221.1\ \text{kN}$, the largest shear anywhere on the beam.
4. Load case 2 — live load on B–C only, with $0.9D$ on A–B (maximum sagging in B–C). Removing the live load from A–B and taking the dead load there at its minimum reduces the hogging restraint at B and therefore raises the sagging moment under the point load. With $w = 0.9(15) + 0.9(0.696) = 14.13\ \text{kN/m}$: $$M_B = -187.1\ \text{kN}\!\cdot\!\text{m}, \quad R_C = 139.1\ \text{kN}$$ $$\boxed{M^{+}_{BC} = R_C\left(\tfrac{L}{2}\right) = 139.1(2.5) = 345.9\ \text{kN}\!\cdot\!\text{m}}$$ This case also produces a small uplift at A of 2.1 kN, which the connection at A must be able to hold down.
5. Compute the moment-gradient factor for each unbraced segment. With lateral restraint only at the supports each 5 m span is one unbraced segment, and the quarter-point form of S16 Clause 13.6 is used because the moment reverses inside span B–C: $$\omega_2 = \frac{4M_{max}}{\sqrt{M_{max}^{2} + 4M_a^{2} + 7M_b^{2} + 4M_c^{2}}} \le 2.5$$ Span B–C returns $\omega_2 = 1.32$ (the moment is close to its peak over most of the segment, so little benefit is available); span A–B, whose moment runs from zero at A to a peak only at B, returns the ceiling value $\omega_2 = 2.5$.
6. Classify the trial section. For W530x72 taken from its nominal plate dimensions ($d = 524$, $b_f = 207$, $t_f = 10.9$, $w = 9.0\ \text{mm}$): $$\frac{b_f/2}{t_f} = \frac{103.5}{10.9} = 9.50 \le \frac{170}{\sqrt{300}} = 9.81 \ \Rightarrow\ \text{Class 2 flange}$$ $$\frac{h}{w} = \frac{502.2}{9.0} = 55.8 \le \frac{1100}{\sqrt{300}} = 63.5 \ \Rightarrow\ \text{Class 1 web}$$ The section is therefore Class 2, which develops the full plastic moment under an elastic analysis; only redistribution (plastic design) would demand Class 1. $$\phi M_p = 0.9(1.725\times10^{6})(300) = 465.8\ \text{kN}\!\cdot\!\text{m}$$
7. Check lateral-torsional buckling over the 5 m segments. Clause 13.6(a) gives the elastic critical moment $$M_u = \frac{\omega_2 \pi}{L}\sqrt{E I_y G J + \left(\frac{\pi E}{L}\right)^{2} I_y C_w}$$ With $I_y = 16.14\times10^{6}\ \text{mm}^4$, $J = 0.301\times10^{6}\ \text{mm}^4$ and $C_w = 1.063\times10^{12}\ \text{mm}^6$, span B–C returns $M_u = 486.7\ \text{kN}\!\cdot\!\text{m}$, which exceeds $0.67M_p$, so the inelastic branch applies: $$M_r = 1.15\,\phi M_p\left(1 - \frac{0.28 M_p}{M_u}\right) = 376.2\ \text{kN}\!\cdot\!\text{m}$$ Span A–B, with $\omega_2 = 2.5$, returns $M_u = 923.5$ and $M_r = 451.6\ \text{kN}\!\cdot\!\text{m}$.
8. Compare demand with resistance. $$\frac{M^{+}_{BC}}{M_{r,BC}} = \frac{345.9}{376.2} = 0.920, \quad \frac{|M_B|}{M_{r,AB}} = \frac{219.5}{451.6} = 0.486$$ Both are satisfied. The sagging check in span B–C is the binding one, and it is the unbraced length, not the plastic modulus, that sets the section: the same beam braced continuously would only need $M_f/\phi M_p = 0.74$.
9. Check shear. With $h/w = 55.8 \le 1014/\sqrt{F_y} = 58.5$ the web yields in shear before it buckles, so $F_s = 0.66F_y = 198\ \text{MPa}$: $$V_r = \phi\,A_w F_s = 0.9(524)(9.0)(198) = 840.4\ \text{kN} \gg V_f = 221.1\ \text{kN}$$ $$\boxed{\text{Adopt W530x72}, \ M_f/M_r = 0.92, \ V_f/V_r = 0.26}$$
10. Check serviceability and local effects. A live-load analysis of the same two-span beam gives a maximum deflection of about 2.1 mm in span B–C, or $L/2400$, comfortably inside the usual $L/360$. The 350 kN point load and the 346 kN reaction at B both act on a 9.0 mm web, so pairs of full-depth bearing stiffeners are specified under the point load and over support B rather than relying on a marginal Clause 14.3.2 web-crippling check.