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Komok [63]
3 years ago
14

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
Physics
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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\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

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\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

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Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

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T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

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Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

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Substitute values from the table.

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Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

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5 0
3 years ago
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