Question

The following structural section is direct extruded from an aluminum billet in which Lo = 500 mm and Do = 100 mm. Dimensions of the cross section are given in the figure. Die angle = 90° and strength coefficient for the material is 190 MPa, and the strain hardening exponent is 0.17. Determine Extrusion ratio, Shape factor and Length of the extruded section if the butt.1 remaining in the container at the end of the ram stroke is 30 mm Calculate the ram force required to extrude the whole material in the container.2 (without the butt), knowing that Johnson's equation constants are a=0.8 and b=1.2 -50 25 25 - 25

Answer 1

solution: Given: le=500 mm bo=100 mm K = 190mpg n =0:17 XT002 I Extrusion ratio = Ao AF 75X50 – 25X25 r= 2.5136 b_shape factor 2.25 Ks = 0.98 +0.02 Cx Сс C2 → perimeter of extruded cross sections Cc- perimeter of a circle with some area Cro 50+75+50+25+25+25+25+25 = 300mm Areq = 75750-25X25 = II Do? 4 Da = 63.0 7 Cu = TI DE 198.179 225 Ks = 0.98 +0.02 300 1981

= 1030 a length of entrided portion Applying mass conservation *1002(500 – 20) = (75X50 - 25x250xL 4 L = 1181-32 mm :: L = 1.181m a Ram force If=ken 1th - ex at bear 0.8+1.2 ln (2.5136) 31.906 ..haug - Keh -0.17 19ox 1906 1 +0:17 181.21 mpa. lth +(6x+ HQL 181.21 (1-906 +2.500 Do 2157.48 N/mm2

.. Ram force = Px Ao 300 = 2157.48 XLX4002 north = 1.69 MN