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

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