Question

Q.5 Extrusion:

A billet that is 76+X mm long with diameter = 36+Y mm is direct extruded to a diameter of 22+Y mm. The extrusion die has a die angle = (70+X)°. For the work metal, K = 600 MPa and n = 0.2+(Z/100). In the Johnson extrusion strain equation, a = 0.8 and b = 1.4.

Determine:

1. (a) extrusion ratio,

2. (b) true strain (homogeneous deformation),

3. (c) extrusion strain, and

4. (d) ram pressure and force at L = 70, 60, 50, 40, 30, 20, and 10 mm. Use of a spreadsheet calculator is

   recommended for part (d).

5. note that X=2, Y=5 , Z=2

Answer 1

A billet that is 76+X mm long with diameter = 36+Y mm is direct extruded to a diameter of 22+Y mm. The extrusion die has a die angle = (70+X)o. For the work metal, K= 600MPa and n = 0.2 + (Z/100). In the Johnson extrusion strain equation, a = 0.8 and b = 1.4. Determine (a) extrusion ratio, (b) true strain (homogeneous deformation), (c) extrusion strain, and (d) ram pressure and force at L = 70, 60, 50, 40, 30, 20, and 10 mm. Use of a spreadsheet calculator is recommended for part (d). Note that X = 2, Y = 5, Z = 2
Given Data:
Length, L = 76+2=78 mm;
Original diameter, d₀ = 36+5 = 41 mm;
Final diameter, d_f = 22+5 = 27 mm;
Die angle = 70+X = 72^o;
K = 600 MPa;
n = 0.2 + (2/100) = 0.22;
a = 0.8 and b = 1.4
Solution:
Extrusion ratio, r_x = A_o/A_f = 412/272 = 2.3056
Strain, ε = ln(r_x) = 0.8355
Extrusion strain, ε_x = a+blnr_x = 0.8 + 1.4 *(0.8355) = 1.9697
Average flow stress, Y ̅_f = Kεⁿ/(1+n) = 600*(0.8355)^0.22/(1.22) = 472.74 MPa

Pressure at different lengths, P = $\bar{\gamma {_{f}}}\times (\varepsilon _{x}+2l/d)$

l (mm)   Y_f   e_x   d (mm)   P (MPa)
70   472.74   1.9697   41   2545.390124
60   472.74   1.9697   41   2314.785246
50   472.74   1.9697   41   2084.180368
40   472.74   1.9697   41   1853.57549
30   472.74   1.9697   41   1622.970612
20   472.74   1.9697   41   1392.365734
10   472.74   1.9697   41   1161.760856