Question

7.D7 A cylindrical brass rod having a minimum tensile strength of 450 MPa (65,000 psi), a ductility of at least 13%EL, and a final diameter of 12.7 mm (0.50 in.) is desired. Some brass stock of diameter 19.0 mm (0.75 in.) that has been cold worked 35% is available. Describe the procedure you would follow to obtain this material. Assume that brass experiences cracking at 65%CW.

Answer 1

Answer: CW = 59%

Step-by-step explanation:

Given Data:

Do = ?

D1 = 12.7mm

Average = 28% CW

%CW = π ([Do/2]^2 - π [ D1/2]^2 ) / π [ Do/2 ]^2 *100

Input data into the formula

28 = π ([Do/2]^2 - π [ 12.7/2]^2 ) / π [ Do/2 ]^2 *100

Do = 15.1mm

When CW = 35%

35 = π ([Do/2]^2 - π [ 35/2]^2 ) / π [ Do/2 ]^2 *100

= 23.6mm

Now we calculate the total CW needed.

CW= π ([Do/2]^2 - π [ D1/2]^2 ) / π [ Do/2 ]^2 *100

When Do = 23.6mm

D1 = 15.1mm

%CW = π ([23.6/2]^2 - π [ 15.1/2]^2 ) / π [ 23.6/2 ]^2 *100

CW = 59%