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tiny-mole [99]
3 years ago
14

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el

liptically shaped surface crack of length 0.26 mm and that has a tip radius of curvature of 0.004 mm when a stress of 1230 MPa is applied.
Engineering
1 answer:
Alex17521 [72]3 years ago
6 0

Answer:

theoretical fracture strength is 9916.58 MPa

Explanation:

given data

surface crack of length L = 0.26 mm

radius of curvature r = 0.004 mm

stress So = 1230 MPa

solution

we get here theoretical fracture strength S that is express as

S  =    S_{0} \times \sqrt{\frac{L}{r} }    .................1

here So is stress and L is length and r is radius

put here value and we get

S  =    1230 \times \sqrt{\frac{0.26}{0.004} }  

solve it we get

S  =  9916.58 MPa

so theoretical fracture strength is 9916.58 MPa

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Answer: Steel is an alloy of iron with typically a few percent of carbon to improve its strength and fracture resistance compared to iron. Many other additional elements may be present or added. Stainless steels that are corrosion and oxidation resistant need typically an additional 11% chromium.

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Write a program that uses a function called Output_Array_Info. Output_Array_Info Properties: Input Parameters: 1. A pointer to a
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Explanation:

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-----------------------------------------------------------------------------------------------------------------------------------

Program:

-----------------------------------------------------------------------------------------------------------------------------------

//header files

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using namespace std;

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3 years ago
How does a motion sensor work?
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3 0
3 years ago
A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i
Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter d_{o} = 12.8 mm

Final diameter d_{f} = 10.7

Engineering stress  \alpha _{E} = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(d_{o} )²  ; Ag = π/4(d_{f} )²

% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

= ( (d_{f} )² - (d_{o} )²) / (d_{o} )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

E_{E} = 0.431

so we substitute the value of E_{E}  into our initial equation;

True stress  \alpha _{T} = 460 ( 1 +  0.431)

True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0
2 years ago
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