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 listed above as radius 1 and radius 2. The second specimen, has the initial radius listed, must have the same deformed hardness as the first specimen; compute the second specimen's radius, in mm, after deformation.

The second specimen's radius, in mm, after deformation = 9mm

Explanation:

The radii of both cylindrical specimens are missing.

I'll assume values for them.

Let r1 = initial radius of first cylinder = 20mm

Let r2 = deformed radius of the first cylinder = 15mm

Let r = Initial radius of second cylinder = 16mm.

Now, we can proceed

For the two cylinders to have the same deformed hardness,

They must be deformed by the same percentage.

To get the deformed radius of the second cylindrical specimen, we'll need to calculate the percentage deformation of specimen 1.

This is given by

(Ao - Ad) /Ao

Where Ao = Original Area = πr²

Ao = πr1²

Ao = π20²

Ao = 400π

Ad = Deformed Area = πr²

Ad = πr2²

Ad = π15²

Ad = 225π

So, %deformation = %d =(400π - 225π)/400π

= 175π/400π

= 0.4375

%d = 43.75%

Now, we'll solve for the deformed radius of the second cylindrical specimen.

Note that, the two specimens have the same % deformation (to have the same hardness).

So, we have

rd - ro = 1 - %d

Where rd = deformed radius

ro = original radius = 16

rd/ro = 1 - %d becomes

rd / 16 = 1 - 43.75%

rd / 16 = 56.25%

rd = 16 * 56.25%

rd = 9mm

answer: 500 wph.

explanation: