Question 3 of 7: Beam ABD on a shear connection and a steel tie

Question A3: Beam ABD on a shear connection and a steel tie (8 + 12 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. Beam A–B–D: A–B=3 m, B–D=1 m; a vertical steel tie from B up to a fixed anchor C; a 200 kN point load at the free tip D; G40.21 350W steel (F_y=350 MPa) assumed throughout. Self-weight ignored per the question.

Find. (a) An adequate rolled section for beam ABD; (b) the required size of tie BC.

Approach. The bolted shear connection at A transmits vertical shear only (no moment); the tie BC can only pull. With just the 200 kN tip load, solve statics for the tie tension and end reaction, obtain the V and M envelope for the beam, then size a rolled W-shape and the tie for the factored forces.

1. Reactions (unfactored). Taking moments about A (tie force R_B up at x=3 m, 200 kN down at x=4 m): $$\Sigma M_A=0:\ R_B(3)=200(4)\Rightarrow R_B=\boxed{266.7\text{ kN (tension in tie)}}$$ $$\Sigma F_y=0:\ R_A = 200-266.7=\boxed{-66.7\text{ kN (connection resists 66.7 kN uplift)}}$$
2. Beam shear and moment. Between A and B, V is constant at −66.7 kN and M runs linearly from 0 to $$-66.7(3)=-200\text{ kN}\cdot\text{m}$$ (hogging) at B; past B, V jumps to $$-66.7+266.7=200\text{ kN}$$ and M returns linearly to zero at the free tip D. The whole beam is hogging (top fibre tension) with no sagging region; the governing values are $$M_{max}=200\text{ kN}\cdot\text{m},\ V_{max}=200\text{ kN}$$ at B. Factoring by 1.5: $$M_f=\boxed{300\text{ kN}\cdot\text{m}},\quad V_f=\boxed{300\text{ kN}}$$
3. (a) Select beam ABD. Required plastic modulus: $$Z_x \ge \frac{M_f}{\phi F_y}=\frac{300\times10^6}{0.9(350)}=952\times10^3\text{ mm}^3$$ Try W410×60 (Z_x=1200×10³ mm³, d=407 mm, t_w=7.7 mm, per CISC Handbook): $$M_r=\phi Z_xF_y=0.9(1200\times10^3)(350)=\boxed{378\text{ kN}\cdot\text{m}} \ge 300\ \checkmark$$ $$V_r=\phi(0.66F_y)(dt_w)=0.9(0.66)(350)(407\times7.7)=\boxed{652\text{ kN}} \ge 300\ \checkmark$$
4. (b) Design tie BC. Factored tension: $$T_f = 266.7(1.5)=\boxed{400\text{ kN}}$$ Gross-yield sizing (CSA S16 Cl. 13.2): $$A_{req}=\frac{T_f}{\phi A_gF_y/A_g}=\frac{400\times10^3}{0.9(350)}=\boxed{1270\text{ mm}^2}$$ giving a minimum round-bar diameter of 40.2 mm. Select a 45 mm ∅ round bar: $$A=\pi(45)^2/4=1590\text{ mm}^2,\qquad T_r=\phi A F_y=0.9(1590)(350)=\boxed{501\text{ kN}} \ge 400\ \checkmark$$ A threaded end reduces the net area at the root (≈75% of gross for a standard UNC thread on this diameter); even so, $$T_{ru}=\phi_uA_nF_u=0.75(0.75\times1590)(450)=403\text{ kN}$$, essentially matching T_f – upsize the threaded ends (e.g. forged/upset ends) or step up to 50 mm ∅ if the ends are simply cut threads.