Question 7 of 8: Two-Member Frame: Cable and Bar Under Combined Yield/Buckling Checks

Question 7: Two-Member Frame: Cable and Bar Under Combined Yield/Buckling Checks (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. Wall pins at A (top) and B (bottom), $3\text{ m}$ apart. Cable $AC$ ($d=25\text{ mm}$, $\sigma_y=1{,}800\text{ MPa}$, $E=190\text{ GPa}$) runs from A to C, $4\text{ m}$ out from the wall. Bar $BC$ ($100\times100\text{ mm}$, $\sigma_y=300\text{ MPa}$, $E=200\text{ GPa}$) runs horizontally from B to C. $P=250\text{ kN}$ downward at C. FS$=1.5$ (yield), FS$=2$ (Euler buckling).

Find. Whether the cable and bar both satisfy their yield checks, and whether bar BC (in compression) satisfies the Euler buckling check — i.e. whether the structure can carry P=250 kN.

Approach. Resolve the cable geometry (a 3-4-5 triangle), solve joint C for the two member forces from equilibrium, then check each member against its allowable stress (yield, both members) and against Euler buckling (bar BC only, since it is the compression member).

1. Geometry and joint equilibrium at C. Cable length $L_{AC}=\sqrt{3^2+4^2}=5\text{ m}$. Vertical equilibrium at C: $F_{AC}\left(\dfrac{3}{5}\right)=P\Rightarrow F_{AC}=\dfrac{250}{0.6}=\boxed{416.7\text{ kN (tension)}}$. Horizontal equilibrium: $F_{BC}=F_{AC}\left(\dfrac{4}{5}\right)=\boxed{333.3\text{ kN (compression)}}$.
2. Yield check — cable AC. $A_{cable}=\tfrac{\pi}{4}(25)^2=490.9\text{ mm}^2$; $\sigma_{AC}=416{,}700/490.9=849\text{ MPa}$. Allowable $=\sigma_y/\text{FS}=1{,}800/1.5=1{,}200\text{ MPa}$. Since $849\lt1{,}200\text{ MPa}$, the cable is adequate against yielding (actual FS $=1{,}800/849=2.12\gt1.5$).
3. Yield check — bar BC. $A_{bar}=100^2=10{,}000\text{ mm}^2$; $\sigma_{BC}=333{,}300/10{,}000=33.3\text{ MPa}$, far below the allowable $300/1.5=200\text{ MPa}$ — adequate, not close to governing.
4. Buckling check — bar BC (compression member, pin-pin, $K=1$). $I_{bar}=\dfrac{100^4}{12}=8.333\times10^{6}\text{ mm}^4$; $P_{cr}=\dfrac{\pi^2EI}{(KL)^2}=\dfrac{\pi^2(200{,}000)(8.333\times10^6)}{(4{,}000)^2}=\boxed{1{,}028\text{ kN}}$. Allowable buckling load $=P_{cr}/\text{FS}=1{,}028/2=514\text{ kN}$. Since the actual compressive force $333.3\text{ kN}\lt514\text{ kN}$, bar BC is adequate against buckling too.
5. Conclusion. All three checks pass, so the structure CAN safely support $P=250\text{ kN}$; the governing (smallest) margin is the cable's yield check (FS actual $2.12$ vs. required $1.5$).