a steel pipe of 400 mm outer diameter is fabricated


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A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° with a plane perpendicular to the axis of the pipe. Knowing that a 300-kN axial force P is applied to the pipe. Determine the normal and shearing stresses in directions respectively normal and tangential to the weld.

Answered by administrator @



Explanation:

Outer di ameter d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}

\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10

d_{1}=380 \mathrm{mm}

Given loading on the cylinder P=300 \mathrm{kN} Helix an gle of the weld form \theta=20^{\circ}

(i) Normal stress on the plane at angle \theta=20^{\circ} is

\sigma=\frac{P \cos ^{2} \theta}{A_{0}}

\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)

\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)

=12252.21 \mathrm{mm}^{2}

=12.25221 \times 10^{-9} \mathrm{m}^{2}

\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}

=-21.6 \mathrm{MPa}

(ii) Shear stress along an angle of \theta=20^{\circ} is \tau=\frac{P}{A_{0}} \cos \theta \sin \theta

=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}

=-7.86 \mathrm{MPa}


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