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Question 6.105: A 24 in. wide by 0.100 in. thick by 100 in. long steel sheet...

A 24 in. wide by 0.100 in. thick by 100 in. long steel sheet is to be formed into a hollow section by bending through 360° and welding (i.e., butt-welding) the long edges together. Assume a crosssectional medial length of 24 in. (no stretching of the sheet due to bending). If the maximum shear stress must be limited to 12 ksi, determine the maximum torque that can be carried by the hollow section if
(a) the shape of the section is a circle.
(b) the shape of the section is an equilateral triangle.
(c) the shape of the section is a square.
(d) the shape of the section is an 8 × 4 in. rectangle.

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The maximum shear stress for a thin-walled section is given by Eq. (6.25)

\tau_{\max }=\frac{T}{2 A_{m} t}

and thus, the maximum torque that can be carried by the hollow section is

T_{\max }=2 \tau_{\max } A_{m} t

(a) Circle: 

\begin{aligned}&\pi d_{m}=24 \text { in. } \quad \therefore d_{m}=7.639437 \text { in. } \\&A_{m}=\frac{\pi}{4}(7.639437 \text { in. })^{2}=45.8366  in .^{2} \\&T_{\max }=2 \tau_{\max } A_{m} t=2(12  ksi )\left(45.8366  in. ^{2}\right)(0.100  in .)=110.0 \text { kip-in. }\end{aligned}

(b) Equilateral triangle: 

triangle sides are each 24  in. / 3=8  in.

\begin{aligned}&A_{m}=\frac{1}{2} b h=\frac{1}{2}(8  in .)(8  in .) \sin 60^{\circ}=27.7128  in.^{2} \\&T_{\max }=2 \tau_{\max } A_{m} t=2(12  ksi )\left(27.7128  in. ^{2}\right)(0.100  in .)=66.5  kip – in .\end{aligned}

(c) Square: 

sides of the square are each 24  in. / 4=6  in.

\begin{aligned}&A_{m}=b h=(6 \text { in. })(6 \text { in. })=36 \text { in. }^{2} \\&T_{\max }=2 \tau_{\max } A_{m} t=2(12  ksi )\left(36  in .^{2}\right)(0.100  in .)=86.4 \text { kip-in. }\end{aligned}

(d) 8 × 4 in. rectangle: 

\begin{aligned}&A_{m}=b h=(8  in. )(4  in .)=32  in. ^{2} \\&T_{\max }=2 \tau_{\max } A_{m} t=2(12  ksi )\left(32  in.{ }^{2}\right)(0.100  in .)=76.8  kip – in .\end{aligned}

 

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