A cylindrical specimen of cold-worked steel has a Brinell hardness of 250.(a) Estimate its ductility in percent elongation.(b) If the specimen remained cylindrical during deformation and its original radius was 5 mm (0.20 in.), determine its radius after deformation.

A) Ductility = 11% EL

B) Radius after deformation = 4.27 mm

Explanation:

A) From equations in steel test,

Tensile Strength (Ts) = 3.45 x HB

Where HB is brinell hardness;

Thus, Ts = 3.45 x 250 = 862MPa

From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.

Also, from image 2,at CW of 27%,

Ductility is approximately, 11% EL

B) Now we know that formula for %CW is;

%CW = (Ao - Ad)/(Ao)

Where Ao is area with initial radius and Ad is area deformation.

Thus;

%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100

%CW = [1 - (rd)²/(ro)²]

1 - (%CW/100) = (rd)²/(ro)²

So;

(rd)²[1 - (%CW/100)] = (ro)²

So putting the values as gotten initially ;

(ro)² = 5²([1 - (27/100)]

(ro)² = 25 - 6.75

(ro) ² = 18.25

ro = √18.25

So ro = 4.27 mm

answer:

the answer is d

explanation:

answer:

a

explanation: