The truss is made from A992 steel bars, each of which has a circular cross section with a diameter of 1.5 in. Determine the maximum force P that can be applied without causing any of the members to buckle. The members are pin connected at their ends.
Members A B and B C are in compression:
Joint A :
Joint B :
\stackrel{+}{\rightarrow} \Sigma F_{x}=0 ; \quad \frac{4}{5} F_{B C}+\frac{4 P}{3}-\frac{8 P}{3}=0 \\F_{B C}=\frac{5 P}{3}Failure of rod A B :
K=1.0 \quad L=8(12)=96 in \\P_{ cr }=\frac{\pi^{2} E I}{(K L)^{2}} \\F_{A B}=\frac{4 P}{3}=\frac{\pi^{2}(29)\left(10^{3}\right)(0.2485)}{((1.0)(96))^{2}} \\P=5.79 kip \text { (controls) }Check:
P_{ cr }=F_{A B}=7.72 kip \\\sigma_{ cr }=\frac{P_{ cr }}{A}=\frac{7.72}{1.7671}=4.37 ksi <\sigma_{Y} \quad OKFailure of rod B C :
K=1.0 \quad L=5(12)=60 \text { in. } \\F_{B C}=\frac{5 P}{3}=\frac{\pi^{2}(29)\left(10^{3}\right)(0.2485)}{[(1.0)(60)]^{2}} \\P=11.9 kip