Question

A cylindrical rod of brass originally 10 mm in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 380 MPa and a ductility of at least 15 %EL are desired. Furthermore, the final diameter must be 7.5 mm. Explain how this may be accomplished. Use the graphs given in previous question.

Answer 1

Step-by-step explanation:

From the information given:

original diameter
`d_o` = 10 mm

final diameter
`d_f =` 7.5 mm

Cold work tensile strength of brass = 380 MPa

Recall that;

`\text {The percentage CW }= (\pi ((d_o)/(2))^2 - \pi((d_f)/(2))^2 )/(\pi((d_o)/(2))^2) * 100`

`\implies (\pi ((10)/(2))^2 - \pi((7.5)/(2))^2 )/(\pi((10)/(2))^2) * 100`

`\implies43.87\% \ CW`

→ At 43.87% CW, Brass has a tensile strength of around 550 MPa, which is greater than 380 MPa.

→ At 43.87% CW, the ductility is less than 5% EL, As a result, the conditions aren't met.

To achieve 15% EL, 28% CW is allowed at most

i.e

The lower bound cold work = 15%

The upper cold work = 28%

The average =
`(15+28)/(2)` = 21.5 CW

Now, after the first drawing, let the final diameter be
`d_o^'`; Then:

`4.5\% \ CW = (\pi ((d_o^')/(2))^2 - ((7.5)/(2))^2)/(\pi ((d_o^')/(2))^2)* 100`

By solving:

`d_o^'} = 8.46 mm`

To meet all of the criteria raised by the question, we must first draw a wire with a diameter of 8.46 mm and then 21.5 percent CW on it.