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3 years ago

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A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25

° with a plane perpendicular to the axis of the pipe. Knowing that a 66 kip axial force P is applied to the pipe, determine the normal and shearing stresses in directions respectively normal and tangential to the weld

1 answer:

Varvara68 [4.7K]3 years ago

8 0

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in

4 -in.-thick

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

= 66/ (П* (d₂²-d₁²)/4

=66/ (3.142* (12²-8²)/4

= 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

= 66* cos 25/ (П* (d₂²-d₁²)/4

=59.82/ (3.142* (12²-8²)/4

= 59.82/62.84 = 0.952kips/in²

shearing stress = tangential stress /2

= T/2 = 0.952/2 = 0.476 kips/in²

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=> $R_f = \frac{4}{5} R$

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