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BabaBlast [244]
3 years ago
7

In a tensile test on a steel specimen, true strain is 0.171 at a stress of 263.8 MPa. When true stress is 346.2 MPa, true strain

is 0.226. Determine strain hardening exponent, n, in the flow curve for the plastic region of this steel.
Engineering
1 answer:
swat323 years ago
6 0

Answer:n=0.973

Explanation:

Given

When True strain\left ( \epsilon _T_1\right )=0.171

at \sigma _1=263.8 MPa

When True stress\left ( \sigma _2\right )=346.2 MPa

true strain \left ( \epsilon _T_2\right )=0.226

We know

\sigma =k\epsilon ^n

where \sigma=True stress

\epsilon=true strain

n=strain hardening exponent

k=constant

Substituting value

263.8=k\left ( 0.171\right )^n------1

346.2=k\left ( 0.226\right )^n-----2

Divide 1 & 2 to get

\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n

1.312=\left ( 1.3216\right )^n

Taking Log both side

ln\left ( 1.312\right )=nln\left ( 1.3216\right )

n=0.973

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Answer:

a)Δs = 834 mm

b)V=1122 mm/s

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Explanation:

Given that

s = 15t^3 - 3t\ mm

a)

When t= 2 s

s = 15t^3 - 3t\ mm

s = 15\times 2^3 - 3\times 2\ mm

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At t= 4 s

s = 15t^3 - 3t\ mm

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So the displacement between 2 s to 4 s

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Δs = 834 mm

b)

We know that velocity V

V=\dfrac{ds}{dt}

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At t=  5 s

V=45t^2-3

V=45\times 5^2-3

V=1122 mm/s

We know that acceleration a

a=\dfrac{d^2s}{dt^2}

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3 years ago
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Ksju [112]

Answer:

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Answer:

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