Question

The following structural section is direct extruded from an aluminum billet in which Lo = 500 mm and Do = 110 mm. Dimensions of the cross section are given in the figure. Die angle = 90° and strength coefficient for the material is 195 MPa, and the strain hardening exponent is 0.18. Determine:

Extrusion ratio, Shape factor and Length of the extruded section if the butt remaining in the container at the end of the ram stroke is 25 mm.
Calculate the ram force required to extrude the whole material in the container (without the butt), knowing that Johnson’s equation constants are a=0.8 and b=1.2

85

Answer 1

Given that, Lo = 500 mm, Do = 110 mm, Die Angle = 90° strength coefficient for the material is = Strain Hardening Exponent=0:18 K=195 MPa, n = 0.18 195 MPa, i, e, 2 Ao r Do 4 -> (110)2 .. Extrusion ratio, pe= = 2:82 Af Af (85% 50) – (25*35) 2.25 Shape factor, Kx = 0.98 +0.02 02 (2 25 Cx- (85+50 +50 +25+25+25+25+ 35) mm = 320 mm

Extruded Section cross sectional area is = {185x50) - (25x35)} mm2 = 3375 mm2 r De 2 - 33 75 4 De=65.55 mm => Сс : rD c = 205-94 mm 2.25 320 205.94 = 1:034 Kx = 0.98 t 0.02

Let the length of the extruded Section bea mm. 4102 Х 2 xx 33 75=r (500-25) 4 rx(110)² -X475 = 1337-5 mm 4 x 3375 -> 2 = True strain, E = ln re In (2-82)=1-09 flow 195x(109,0.13 stress, ken Average Yf - In 1:18 =166.43 MPa Extrusion strain, .Ex = atbE= 0.8 + 1.2x1-04 2.048 =

For Direct Extrusion, Ram pressure is 2L p= kx Yf (Ext D. #CE 2x475 2.048 + 110 - 1.034 X 166.43x 1838.66 MPa