3 years ago

7

Engineering

1 answer:

Eduardwww [97]3 years ago

8 0

A. Drafters Because drafters draw the blueprints/ plans for engineering and architects

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4/5

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3 years ago

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3 × 10-4

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Answer:

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Explanation:

given data

radius of curvature = 3 × $10^{-4}$ mm

crack length = 5.5 × $10^{-2}$ mm

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to find out

maximum stress

solution

we know that maximum stress formula that is express as

$\sigma m = 2 ( \sigma o ) \sqrt{\frac{a}{\delta t}}$ ......................1

here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack

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$\sigma m = 2 ( \sigma o) \sqrt{\frac{a}{\delta t}}$

$\sigma m = 2(150) \sqrt{ \frac{\frac{5.5*10^{-2}}{2}}{3*10^{-4}}}$

σm = 2872.28 MPa

so maximum stress is 2872.28 MPa

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