The linkage is made using two A992 steel rods, each having a circular cross section. Determine the diameter of each rod to the nearest \frac{3}{4} in. that will support a load of P=6 kip. Assume that the rods are pin connected at their ends. Use a factor of safety with respect to buckling of 1.8
Joint B :
\stackrel{+}{\rightarrow} \Sigma F_{x}=0 ; \quad F_{A B} \cos 45^{\circ}-F_{B C} \sin 30^{\circ}=0 \\F_{A B}=0.7071 F_{B C} \quad(1) \\+\uparrow \Sigma F_{y}=0 ; \quad F_{A B} \sin 45^{\circ}+F_{B C} \cos 30^{\circ}-6=0 \quad(2)Solving Eqs. (1) and (2) yields:
F_{B C}=4.392 kip \quad F_{A B}=3.106 kipFor rod A B :
P_{ cr }=3.106(1.8)=5.591 kip \\K=1.0 \quad L_{A B}=\frac{12(12)}{\cos 45^{\circ}}=203.64 in \\P_{ cr }=\frac{\pi^{2} E I}{(K L)^{2}}\\5.591=\frac{\pi^{2}(29)\left(10^{3}\right)\left(\frac{d_{A B}{4}}{64}\right)}{[(1.0)(203.64)]^{2}} \\d_{A B}=2.015 in . \quad \text { Use } d_{A B}=2 \frac{1}{8} inCheck:
\sigma_{ cr }=\frac{P_{ cr }}{A}=\frac{5.591}{\frac{\pi}{4}\left(2.125^{2}\right)}=1.58 ksi <\sigma_{Y} \quad OK
For rod B C :
Use d_{B C}=2 in.
Check:
