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Natalka [10]
2 years ago
9

Prove that hoop stress is twice the longitudinal stress in a cylindrical pressure vessel.

Physics
1 answer:
Iteru [2.4K]2 years ago
6 0

Answer:

Proof in explanation.

Explanation:

Consider a thin cylinder, whose thickness to diameter ration is less than 1/20, the hoop stress can be derived as follows:

Let,

L = length of cylinder

d = internal diameter of cylinder

t = thickness of wall of cylinder

P = internal pressure

σH = Hoop Stress

σL = Longitudinal Stress

Total force on half-cylinder owing to internal pressure = P x  Projected Area = P x dL    (Refer fig 9.1)

Total resisting force owing to hoop stress setup in walls = 2 σH L t

Therefore,

P d L = 2 σH L t

σH = Pd/2t      _____________ eqn (1)

Now, for longitudinal stress:

Total force on end of cylinder owing to internal pressure = P x  Projected Area = P x πd²/4

Area resisting this force = π d t        (Refer fig 9.2)

Longitudinal Stress = Force/Area

σL = (Pπd²/4)/(πdt)

σL = Pd/4t        ____________ eqn (2)

Dividing eqn (1) by eqn (2)

σH/σL = 2

<u>σH = 2 σL</u>      (Hence, Proved)

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If i = 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N ⋅ m , what are the lengt
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Answer:

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s = √[(τ)/(NIB sin θ)]

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6 0
3 years ago
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