The strongback is made of an A992 steel hollow circular section with the outer diameter of $d_{o}=60 mm$ . If it is designed to withstand the lifting force of P=60 kN, determine the minimum required wall thickness of the strong back so that it will not buckle. Use F.S. =2 against buckling.

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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 60-2 F \sin 45^{\circ}=0 \quad F_{A B}=F_{A C}=F=42.43 kN ( T )[/latex]

Joint B (Fig. b)

[latex]\stackrel{+}{ ightarrow} \Sigma F_{x}=0 ; \quad 42.43 \cos 45^{\circ}-F_{B C}=0 \quad F_{B C}=30 kN ( C )[/latex]

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

[latex]A=\frac{\pi}{4}\left(0.06^{2}-d_{i}^{2} ight) \quad I=\frac{\pi}{4}\left[0.03^{4}-\left(\frac{d_{i}}{2} ight)^{4} ight]=\frac{\pi}{64}\left(0.06^{4}-d_{i}^{4} ight)[/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}( FS. )=30(2)=60 kN[/latex]
Applying Euler's formula,

[latex]P_{ cr }=\frac{\pi^{2} E I}{(K L)^{2}} \60\left(10^{3} ight)=\frac{\pi^{2}\left[200\left(10^{9} ight) ight]\left[\frac{\pi}{64}\left(0.06^{4}-d_{i}^{4} ight] ight.}{[1(4)]^{2}} \d_{i}=0.04180 m =41.80 mm[/latex]

Thus, [latex]t=\frac{d_{a}-d_{i}}{2}=\frac{60-41.80}{2}=9.10 mm[/latex]

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

[latex]\sigma_{ cr }=\frac{P_{ cr }}{A}=\frac{60\left(10^{3} ight)}{\frac{\pi}{4}\left(0.06^{2}-0.04180^{2} ight)}=41.23 MPa <\sigma_{Y}=345 MPa[/latex]

Question 13.28: The strongback is made of an A992 steel hollow circular sect...

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