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While calculating hoop's stress we consider projected area. Infact in bearing stress from pins in a truss something similar is applicable.Can you explain why we are taking something like this instead of the curved area?

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We cut the vessel in half as shown below.

It is the vertical projection considered because the pressure inside the cylindrical vessel due to fluid acts perpendicular to the surface of the cylindrical vessel, such that differential force is perpendicular at any point. Assuming that we take the horizontal component and vertical component of each of the differential forces, all horizontal component of the differential force is actually countered on the other side of the vessel, therefore giving an equilibrium in horizontal direction.

The hoop now only considers vertical component of the force, therefore, whatever configuration of the vessel, we need to take only the vertical component, thus we only need the area of the vessel perpendicular to the force being considered.

As pointed in a comment, this analysis is similar to bearing pins in truss member in a way because in analysis of pins, we take the diameter of the pin as equal to the diameter of the hole, thus the pressure in the pin is EQUAL in the half of the pin, as the other side is unloaded. The bearing area is then calculated as the projected area because of what is explained above. In reality, this is not what happens because we tend to increase the hole diameter for ease in installation. Pressure vessels on the other hand is analyzed similar to this because fluid will tend to pressurized all directions with the same value. Thus, bearing area projection is similar in these two cases.

Pressure acts on side of the vessel

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  • $\begingroup$ The question asked for the stresses due to a pin, not an internal pressure which is equal in all directions. $\endgroup$
    – Solar Mike
    Aug 19, 2017 at 10:48
  • $\begingroup$ Can you explain why we are taking something like this instead of the curved area? .. I answered this because he must be thinking to the bearing stress or axial stress in truss members. When he asked about the curved area, is there anything that he means other than the projected area of the inside of the vessel? I believe the curved area statement is similar on how we derive the stressed area when we compute bearing stress in pins, isn't it? $\endgroup$
    – Jem Eripol
    Aug 19, 2017 at 10:51
  • $\begingroup$ The Op mentions a "pin" ( see "Infact in bearing stress from pins") while you talk about pressure... A pin will not load a bearing the same way as a pressure vessel is loaded. $\endgroup$
    – Solar Mike
    Aug 19, 2017 at 10:56
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    $\begingroup$ i beg to disagree. How we compute stressed area in bearing pins is similar on how we compute projected area in pressure vessels. OP is talking about the similarities of how bearing area is to the projected area in pressure vessels. $\endgroup$
    – Jem Eripol
    Aug 19, 2017 at 10:57
  • $\begingroup$ Then take a crankshaft out of an old engine and show that the pressure load has been equal around the perimeter... Then I will believe you, until then .... $\endgroup$
    – Solar Mike
    Aug 19, 2017 at 10:58
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Because some of the curved surface is not being loaded.

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  • $\begingroup$ Explain elaborately with picture $\endgroup$ Aug 14, 2017 at 9:52
  • $\begingroup$ Look at the diagrams you have already and think about how the curved surface supports the load : one extreme is force perpendicular to surface and the other : force parallel to surface $\endgroup$
    – Solar Mike
    Aug 14, 2017 at 9:57
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    $\begingroup$ @RajorshiKoyal "Explain elaborately with picture" - sorry, but this isn't a "write a free textbook just for you" site - especially for people who don't even know the magic work "please" :-) $\endgroup$
    – alephzero
    Aug 15, 2017 at 19:40

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