16-Civ-A2 Elementary Structural Design · December 2017
Question 3 of 7: A3 — Beam ABD and its steel tie
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2017 — 16-Civ-A2 Elementary Structural Design. Three hours, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete) and Part C (C1, timber). The candidate answers two from Part A, two from Part B and the single Part C question — five solutions of equal value. All seven are solved here, because the set is a study resource. Page 1 states that all loads shown are unfactored, so every question below applies its own load combination.
Check: figure page. All five figures are hand-drawn on page 3 of the paper. Every dimension used below is cross-checked against the labelled dimension chains (for example 200 overhang + 3 × 200 stems + 2 × 600 voids + 200 overhang = 2200 in Figure B3). Where the exam leaves a quantity undimensioned (the tie length in Figure A3, member thicknesses in Figure B1) the assumption is stated in the question concerned, as NOTE 1 on page 1 invites.
Question 3: A3 — Beam ABD and its steel tie (8 + 12 = 20 marks)
Find. (a) a rolled section for beam ABD, and (b) a section and end connection for the vertical tie BC.
Figure A3 - beam ABD: bolted shear connection at A to the W530x138 column, vertical steel tie BC at B (3 m from A) and a 200 kN unfactored point load at the free tip D, 1 m beyond B.
Approach. The beam is determinate — a vertical reaction at the bolted seat A and the tie force at B — so resolve the statics first, then select the beam for the hogging moment over B and check shear and lateral-torsional buckling; finally size the tie as a tension member for gross yielding and net-section fracture.
Factored load. Page 1 states the loads are unfactored, and the 200 kN at D is a live load applied to a structure whose self weight the note tells us to ignore, so$$P_f = 1.5(200) = 300\ \text{kN}$$
Reactions. Taking moments about A, the tie at B (3 m from A) must balance the load at D (4 m from A):$$T_f = \frac{300(4.0)}{3.0}=\boxed{400\ \text{kN (tension)}}$$ Vertical equilibrium then gives $R_A = 300 - 400 = -100$ kN. The negative sign is the physically important result: the connection at A is in uplift, holding the beam down with 100 kN, because the tie is inboard of the load. A seat angle alone would not do — the bolts must be designed for tension-side shear reversal.
Design actions on the beam. Between A and B there is no load, so the moment varies linearly from zero at the pinned seat to the value produced by the overhang; over BD the free body is simply the 300 kN at 1 m:$$M_f = 300(1.0) = 300\ \text{kN}\cdot\text{m}\ \text{(hogging at B)},\qquad V_f = 300\ \text{kN}$$ Because the moment is hogging everywhere, the bottom flange is the compression flange along the whole member.
Select the beam. A Class 1 or 2 section with full lateral support needs $Z_x \ge M_f/(\phi F_y) = 300\times10^{6}/(0.9\times350) = 952\times10^{3}$ mm3. Try W410 × 60 ($d$ = 407, $b$ = 178, $t$ = 12.8, $w$ = 7.7 mm, $Z_x = 1190\times10^{3}$ mm3, $I_y = 12.0\times10^{6}$ mm4). Classification: flange $b/2t = 6.95 < 7.75$ and web $h/w = 49.5 < 58.8$, so the section is Class 1, and$$M_r=\phi Z_xF_y = 0.90(1190\times10^{3})(350)=\boxed{375\ \text{kN}\cdot\text{m}} \; > \; 300\ \text{kN}\cdot\text{m}$$
Check lateral-torsional buckling of segment AB. The compression flange is braced at the seat A and at the tie connection B, so $L = 3000$ mm, and the moment runs linearly from 0 to $M_f$ giving $\omega_2 = 1.75$. With $J = 307\times10^{3}$ mm4 and $C_w = I_y(d-t)^2/4$,$$M_u=\frac{\omega_2\pi}{L}\sqrt{EI_yGJ+\left(\frac{\pi E}{L}\right)^2I_yC_w}=1007\ \text{kN}\cdot\text{m}$$ This far exceeds $0.67M_p = 279$ kN·m, so S16 13.6(a) applies and returns a value capped at the full $\phi M_p$ — lateral-torsional buckling does not reduce the resistance. The utilisation is $300/375 = 0.80$.
Check shear. The web is stocky ($h/w = 49.5 < 1014/\sqrt{F_y} = 54.2$), so it yields in shear before it buckles and $F_s = 0.66F_y$:$$V_r=\phi A_w(0.66F_y)=0.90(407)(7.7)(0.66\times350)/10^3=651\ \text{kN}\;>\;300\ \text{kN}$$ Shear is comfortable, as expected for a member selected on moment. Answer (a): W410 × 60 in 350W steel.
Design the tie BC (part b). The tie carries $T_f = 400$ kN in pure tension. Two limit states apply. Gross-section yielding needs $A_g \ge T_f/(\phi F_y) = 400\times10^{3}/(0.9\times350) = 1270$ mm2; net-section fracture, S16 13.2(a)(iii), needs $A_{ne} \ge T_f/(0.85\phi_uF_u) = 400\times10^{3}/(0.85\times0.75\times450) = 1394$ mm2. Select HSS 102 × 102 × 6.4 (class C, $A_g = 2320$ mm2), connected by slotting the tube over a gusset and welding along the four faces so the shear-lag factor is 0.90:$$T_r=\min\left[0.90(2320)(350),\;0.85(0.75)(0.90\times2320)(450)\right]/10^3=\boxed{599\ \text{kN}}$$ giving a utilisation of $400/599 = 0.67$. Net-section fracture governs, which is typical for a welded HSS tie.
Complete the load path at A. The 100 kN hold-down must pass through the bolted shear connection into the W530 × 138 column web. With M20 A325 bolts in single shear through threads, $V_r = 0.60\phi_b n m A_b F_u = 0.60(0.80)(1)(314)(380)/10^3 = 57.3$ kN per bolt, so two M20 bolts suffice on strength. Detail a minimum of three bolts in a standard double-angle or shear-tab connection for stability during erection and to satisfy the minimum-edge-distance and pitch rules.
Check: the tie length is not dimensioned. Figure A3 shows C directly above B against a hatched support but gives no height, so the tie has been designed purely as a tension member. That is safe for any length — a tie carries no compression under this loading — but if the load could ever reverse (uplift on the walkway, erection condition) the member would need a compressive check and the length would then be required.
