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A thin walled spherical shell is subjected to an internal pressure. If the radius of the shell is increased by 1% and the thickness is reduced by 1%, with the internal pressure remaining the same, the percentage change in the circumferential (hoop) stress is
- a)0
- b)1
- c)1.08
- d)2.02
Correct answer is 'D'. Can you explain this answer?
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Verified Answer
A thin walled spherical shell is subjected to an internal pressure. If...
Hoop stress for a thin spherical shell (σh) = Pr/2t
By applying logarithm on both sides, we get
log(σh) = log(P/2) + log(r) − log(t)
Differentiating the above equation, f(t) = sin t and it is given that dr/r = 0.01 and dt/t = 0.01
Up on substituting we get, dσh/σh = 0.02
∴ Percentage increase will be 2%
Most Upvoted Answer
A thin walled spherical shell is subjected to an internal pressure. If...
Solution:
Given data:
Increase in radius = 1%
Decrease in thickness = 1%
Internal pressure = constant
We know that the circumferential stress (hoop stress) in a thin-walled spherical shell is given by:
σθ = Pr/t
where σθ is the circumferential stress, P is the internal pressure, r is the radius, and t is the thickness of the shell.
Let us assume that the initial radius and thickness of the shell are r1 and t1 respectively. Therefore, the initial circumferential stress can be given as:
σθ1 = P*r1/t1
After increasing the radius by 1% and decreasing the thickness by 1%, the new radius and thickness become:
r2 = r1 + 0.01r1 = 1.01r1
t2 = t1 - 0.01t1 = 0.99t1
Therefore, the new circumferential stress can be given as:
σθ2 = P*r2/t2
Substituting the values of r2 and t2, we get:
σθ2 = P*(1.01r1)/(0.99t1)
To find the percentage change in circumferential stress, we can use the formula:
% change = (|σθ2-σθ1|/σθ1)*100
Substituting the values of σθ1 and σθ2, we get:
% change = (|P*(1.01r1)/(0.99t1) - P*r1/t1|/(P*r1/t1))*100
Simplifying the above equation, we get:
% change = (|1.01/0.99 - 1|)*100
% change = 2.02%
Therefore, the percentage change in the circumferential stress is 2.02%, which is closest to option (d).
Given data:
Increase in radius = 1%
Decrease in thickness = 1%
Internal pressure = constant
We know that the circumferential stress (hoop stress) in a thin-walled spherical shell is given by:
σθ = Pr/t
where σθ is the circumferential stress, P is the internal pressure, r is the radius, and t is the thickness of the shell.
Let us assume that the initial radius and thickness of the shell are r1 and t1 respectively. Therefore, the initial circumferential stress can be given as:
σθ1 = P*r1/t1
After increasing the radius by 1% and decreasing the thickness by 1%, the new radius and thickness become:
r2 = r1 + 0.01r1 = 1.01r1
t2 = t1 - 0.01t1 = 0.99t1
Therefore, the new circumferential stress can be given as:
σθ2 = P*r2/t2
Substituting the values of r2 and t2, we get:
σθ2 = P*(1.01r1)/(0.99t1)
To find the percentage change in circumferential stress, we can use the formula:
% change = (|σθ2-σθ1|/σθ1)*100
Substituting the values of σθ1 and σθ2, we get:
% change = (|P*(1.01r1)/(0.99t1) - P*r1/t1|/(P*r1/t1))*100
Simplifying the above equation, we get:
% change = (|1.01/0.99 - 1|)*100
% change = 2.02%
Therefore, the percentage change in the circumferential stress is 2.02%, which is closest to option (d).
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A thin walled spherical shell is subjected to an internal pressure. If...
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A thin walled spherical shell is subjected to an internal pressure. If the radius of the shell is increased by 1% and the thickness is reduced by 1%, with the internal pressure remaining the same, the percentage change in the circumferential (hoop) stress isa)0b)1c)1.08d)2.02Correct answer is 'D'. Can you explain this answer?
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