Prove that hoop stress is twice the longitudinal stress in a cylindrical pressure vessel.
1 answer:
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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Answer:
a = 9.86 m/s²
Explanation:
given,
distance between the centers of wheel = 156 cm
center of mass of motorcycle including rider = 77.5 cm
horizontal acceleration of motor cycle = ?
now,
The moment created by the wheels must equal the moment created by gravity.
take moment about wheel as it touches the ground, here we will take horizontal distance between them.
then, take the moment around the center of mass. Since the force on the ground from the wheels is horizontal, we need the vertical distance.
now equating both the moment
m g d = F h
d is the horizontal distance
h is the vertical distance
m g d = m a h
term of mass get eliminated
g d = a h
so,
a = 9.86 m/s²