Question 8 of 8: Deflection by virtual work (tie-rod frame)

Question 8: Deflection by virtual work (tie-rod frame) (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. Two-force tie rod $D$–$B$ (length 6.25 m, $AE=4.5\times10^4$ kN); bent beam $A$–$B$–$C$ ($EI=2.4\times10^5$ kN·m², inextensible); 60 kN down at C. Find. horizontal deflection $\delta_{C,h}$.

Approach. The structure is determinate. Compute the real tie force and beam moments; apply a unit horizontal load at C for the virtual forces; then $\displaystyle \delta_{C,h}=\int\frac{Mm}{EI}\,ds+\frac{N n L}{AE}$.

1. Real forces. Treating $A$–$B$–$C$ as a free body with the tie force $T$ along $B$→$D$ (unit $(-0.8,0.6)$): $\sum M_A: 3T-60(9)=0\Rightarrow T=\boxed{180\text{ kN (tension)}}$. Reactions $A_x=144$, $A_y=-48$ kN. Real beam moments: $M_B=-240$ kN·m, $M_C=0$.
2. Virtual system. Unit horizontal load at C gives virtual tie force $n=1$ and virtual moments $m(x)$ from $A_x'=-0.2$, $A_y'=-0.6$.
3. Tie-rod term. $\dfrac{NnL}{AE}=\dfrac{180(1)(6.25)}{4.5\times10^4}=0.025\text{ m}=\boxed{25\text{ mm}}$.
4. Bending term. Integrating $Mm/EI$ over A–B and B–C gives $5+5 = 10$ mm.
5. Total. $\delta_{C,h}=25+10=\boxed{35\text{ mm}}$ (horizontal, in the direction of the applied unit load).

The axial flexibility of the slender tie rod dominates the response: it supplies 25 of the 35 mm, while the two stiff, inextensible beams bend only enough to add 10 mm. This is typical of tie-braced systems — the serviceability deflection is governed by the rod's $AE$, so increasing the rod area (not the beam depth) is the efficient way to stiffen point C. Had the rod been treated as rigid, the predicted deflection would be under-estimated by more than two-thirds.