Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 18 mm and 11 mm, respectively. The second specimen, with an initial radius of 12 mm, must have the same deformed hardness as the first specimen; compute the second specimen's radius, in mm, after deformation.

For the deform two cylinder specimen the answer is 62.65% CW,The second specimen is 7.33 mm

Explanation:

Solution

Now,

Let calculate the percentage of the cold work to be carried out to deform the two cylinder specimens to have the same hardness.

% CW = A₀ - Ad /A₀ * 100

=πr²₀ - πr²d /πr²₀ * 100

so,

r²₀ - r²d/ r²₀  * 100

= 18² - 11²/18² * 100

= 324 -121/324 * 100

= 62.65% CW

Now,

We calculate for the deformed radius of the second specimen for the same direction

so,

r₂ = r₀√ 1 - %CW /100

= 12 √ 1 - 6265/ 100

= 12 * 0.6111

r₂ = 7.33 mm

explanation:

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answer:

explanation:

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