Question

L Bar stock of initial diameter - 90 mm is drawn with a draft - 15 mm. The draw die has an entrance angle = 18", and the coefficient of friction at the work-die interface=0.08. The metal behaves as a perfectly plastic material with yield stress = 105 MPa. Determine (a) area reduction, (b) draw stress, (c) draw force required for the operation, and (d) power to perform the operation if exit velocity- 1.0 m/min.

Answer 1

(a) Area Reduction:

AR = 0.5

(b) Draw Stress:

Draw Stress = 210 MPa

(c) Draw Force:

F = 1.963 kN (approximately)

(d) Power:

P = 1.963 kW (approximately)

Step-by-step explanation:

(a) Area Reduction:

The area reduction (AR) can be calculated using the equation:

AR = (Initial Area - Final Area) / Initial Area

Given:

Initial diameter (D1) = 90 mm

Draft (d) = 15 mm

First, we need to calculate the final diameter (D2):

D2 = D1 - 2d = 90 mm - 2 * 15 mm = 60 mm

The initial area (A1) can be calculated using the formula for the area of a circle:

A1 = π * (D1/2)^2

The final area (A2) can be calculated similarly:

A2 = π * (D2/2)^2

Now, we can calculate the area reduction:

AR = (A1 - A2) / A1

(b) Draw Stress:

The draw stress can be calculated using the equation:

Draw Stress = (Yield Stress) / (1 - AR)

Given:

Yield stress (σ) = 105 MPa

(c) Draw Force:

The draw force (F) can be calculated using the equation:

F = Draw Stress * A1

(d) Power:

The power (P) can be calculated using the equation:

P = F * V

Given:

Exit velocity (V) = 1.0 m/min

Let's calculate the values:

(a) Area Reduction:

AR = (A1 - A2) / A1

A1 = π * (D1/2)^2

A2 = π * (D2/2)^2

(b) Draw Stress:

Draw Stress = (Yield Stress) / (1 - AR)

(c) Draw Force:

F = Draw Stress * A1

(d) Power:

P = F * V

Now, we can substitute the given values and calculate the results.