A circular shaft has a diameter d [m] that varies with axial position x [m] according to d = 0.02(1/x²)/e*;0 < x < 3m An axial load P of 30000 N is applied to one end of the shaft whose modulus of elasticity E is 2 × 101'N/m². The axial elongation of the shaft is Ax [m] and is given Ax = (P / E) S(1 / A)dx ; A = cross-sectional area = nd²/4 Estimate Ax by 1. SIMPSON'S RULE (a → a + 2h)
A circular shaft has a diameter d [m] that varies with axial position x [m] according to d = 0.02(1/x²)/e*;0 < x < 3m An axial load P of 30000 N is applied to one end of the shaft whose modulus of elasticity E is 2 × 101'N/m². The axial elongation of the shaft is Ax [m] and is given Ax = (P / E) S(1 / A)dx ; A = cross-sectional area = nd²/4 Estimate Ax by 1. SIMPSON'S RULE (a → a + 2h)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Estimate ?? by SIMPSON’S 1/3 RULE (a → a + 2h)
![A circular shaft has a diameter d [m] that varies with axial position x [m]
according to
d = 0.02(1/x²)/e*;0<x< 3m
An axial load P of 30000 N is applied to one end of the shaft whose modulus
of elasticity E is 2 × 101N/m². The axial elongation of the shaft is 4x [m]
and is given
Ax =
(P / E) S(1 / A)dx ; A = cross-sectional area = nd?/4
Estimate Ax by
1. SIMPSON'S - RULE (a
—а+ 2h)
3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F15ff4a29-91be-4ef0-87b1-c64551fe6c6d%2F1e4af0c2-a02f-48bb-84d5-cc5943dee499%2Fbh3plh6_processed.png&w=3840&q=75)
Transcribed Image Text:A circular shaft has a diameter d [m] that varies with axial position x [m]
according to
d = 0.02(1/x²)/e*;0<x< 3m
An axial load P of 30000 N is applied to one end of the shaft whose modulus
of elasticity E is 2 × 101N/m². The axial elongation of the shaft is 4x [m]
and is given
Ax =
(P / E) S(1 / A)dx ; A = cross-sectional area = nd?/4
Estimate Ax by
1. SIMPSON'S - RULE (a
—а+ 2h)
3
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