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goblinko [34]
3 years ago
6

Q9. A cylindrical specimen of a metal alloy 54.8 mm long and 10.8 mm in diameter is stressed in tension. A true stress of 365 MP

a causes the specimen to plastically elongate to a length of 61.8 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 54.8 mm to a length of 64.7 mm. (10 points)
Engineering
1 answer:
wolverine [178]3 years ago
7 0

Answer:

σ = 391.2 MPa

Explanation:

The relation between true stress and true strain is given as:

σ = k εⁿ

where,

σ = true stress = 365 MPa

k = constant

ε = true strain = Change in Length/Original Length

ε = (61.8 - 54.8)/54.8 = 0.128

n = strain hardening exponent = 0.2

Therefore,

365 MPa = K (0.128)^0.2

K = 365 MPa/(0.128)^0.2

k = 550.62 MPa

Now, we have the following data:

σ = true stress = ?

k = constant = 550.62 MPa

ε = true strain = Change in Length/Original Length

ε = (64.7 - 54.8)/54.8 = 0.181

n = strain hardening exponent = 0.2

Therefore,

σ = (550.62 MPa)(0.181)^0.2

<u>σ = 391.2 MPa</u>

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Mirror Radius (mm) Bending Failure Stress (MPa)

.603                                         225

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( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

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n ( mean value ) =  ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

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