The strongback B C is made of an A992 steel hollow circular section with $d_{o}=60 mm$ and $d_{i}=40 mm$ . Determine the allowable maximum lifting force P without causing the strong back to buckle. F.S. =2 against buckling is desired.

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Accepted Answer

Equilibrium. The compressive force developed in the strongback can be determined by analyzing the equilibrium of joint A followed by joint B.
Joint A (Fig. a )

[latex]\stackrel{+}{ ightarrow} \Sigma F_{x}=0 ; \quad F_{A C} \cos 45^{\circ}-F_{A B} \cos 45^{\circ}=0 \quad F_{A C}=F_{A B}=F \+\uparrow \Sigma F_{y}=0 ; \quad P-2 F \sin 45^{\circ}=0 \quad F_{A B}=F_{A C}=F=0.7071 P[/latex]

Joint B (Fig. b)

[latex]\stackrel{+}{ ightarrow} \Sigma F_{x}=0 ; \quad 0.7071 P \cos 45^{\circ}-F_{B C}=0 \quad F_{B C}=0.5 P[/latex]

Section Properties. The cross-sectional area and moment of inertia are

[latex]A=\pi\left(0.03^{2}-0.02^{2} ight)=0.5\left(10^{-3} ight) \pi m ^{2} \quad I=\frac{\pi}{4}\left(0.03^{4}-0.02^{4} ight)=0.1625\left(10^{-6} ight) \pi m ^{4}[/latex]

Critical Buckling Load. Both ends can be considered as pin connections. Thus, K=1. The critical buckling load is

[latex]P_{ cr }=F_{B C}( ext { F.S. })=0.5 P(2)=P[/latex]
Applying Euler's formula,

[latex]P_{ cr }=\frac{\pi^{2} E I}{(K L)^{2}} \P=\frac{\pi^{2}\left[200\left(10^{9} ight) ight]\left[0.1625\left(10^{-6} ight) \pi ight]}{[1(4)]^{2}} \P=62.98 kN =63.0 kN[/latex]

Critical Stress. Euler's formula is valid only if [latex]\sigma_{ cr }<\sigma_{Y}[/latex].

[latex]\sigma_{ cr }=\frac{P_{ cr }}{A}=\frac{62.98\left(10^{3} ight)}{0.5\left(10^{-3} ight) \pi}=40.10 MPa <\sigma_{Y}=345 MPa[/latex]

Question 13.27: The strongback B C is made of an A992 steel hollow circular ...

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