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Ksenya-84 [330]
3 years ago
9

10–25. The 45° strain rosette is mounted on the surface of a shell. The following readings are obtained for each gage: ε a = −20

0(10−6), ε b = 300(10−6), and ε c = 250(10−6). Determine the in-plane principal strains.

Engineering
1 answer:
vazorg [7]3 years ago
7 0

Answer:

The answer is 380.32×10^-6

Refer below for the explanation.

Explanation:

Refer to the picture for brief explanation.

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Answer:

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Explanation:

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Making \sigma the subject then

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Since strain is given as 0.025% of the length then strain is \frac {0.025}{100}=0.00025

Now substituting E for 10\times 10^{6} psi then

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(b)

Stress, \sigma= \frac {F}{A} making A the subject then

A=\frac {F}{\sigma}

A=\frac {\pi(d_o^{2}-d_i^{2})}{4}

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We know that 2t=d_o - d_i where t is thickness

Now substituting

\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}

\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4

(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

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(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

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t=0.5(2-1.785)=0.1075 in

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