A 500 mm wide by 3 mm thick by 2 m long aluminum 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 cross-sectional medial length of 500 mm (no stretching of the sheet due to bending). If the maximum shear stress

Question

A 500 mm wide by 3 mm thick by 2 m long aluminum 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 cross-sectional medial length of 500 mm (no stretching of the sheet due to bending). If the maximum shear stress must be limited to 75 MPa, 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 a 150 × 100 mm rectangle.

Accepted Answer

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

[latex] au_{\max }=\frac{T}{2 A_{m} t}[/latex]

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

[latex]T_{\max }=2 au_{\max } A_{m} t[/latex]

(a) Circle:

[latex]\begin{aligned}&\pi d_{m}=500 mm \quad herefore d_{m}=159.155 mm \&A_{m}=\frac{\pi}{4}(159.155 mm )^{2}=19,894.382 mm ^{2} \&T_{\max }=2 au_{\max } A_{m} t=2\left(75 N / mm ^{2} ight)\left(19,894.382 mm ^{2} ight)(3 mm )=8,952,472 N - mm =8.95 kN - m\end{aligned}[/latex]

(b) Equilateral triangle:

triangle sides are each [latex]500 mm / 3=166.667 mm[/latex]

[latex]\begin{aligned}&A_{m}=\frac{1}{2} b h=\frac{1}{2}(166.667 mm )(166.667 mm ) \sin 60^{\circ}=12,028.131 mm ^{2} \&T_{\max }=2 au_{\max } A_{m} t=2\left(75 N / mm ^{2} ight)\left(12,028.131 mm ^{2} ight)(3 mm )=5,412,659 N - mm =5.41 kN - m\end{aligned}[/latex]

(c) Square:

sides of the square are each [latex]500 mm / 4=125 mm[/latex]

[latex]\begin{aligned}&A_{m}=b h=(125 mm )(125 mm )=15,625 mm ^{2} \&T_{\max }=2 au_{\max } A_{m} t=2\left(75 N / mm ^{2} ight)\left(15,625 mm ^{2} ight)(3 mm )=7,031,250 N - mm =7.03 kN - m\end{aligned}[/latex]

(d) 150 × 100 mm rectangle:

[latex]\begin{aligned}&A_{m}=b h=(150 mm )(100 mm )=15,000 mm ^{2} \&T_{\max }=2 au_{\max } A_{m} t=2\left(75 N / mm ^{2} ight)\left(15,000 mm ^{2} ight)(3 mm )=6,750,000 N - mm =6.75 kN - m\end{aligned}[/latex]

Question 6.106: A 500 mm wide by 3 mm thick by 2 m long aluminum sheet is to...

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