SKM754C-69-17102513271_0009.jpg - ME 467 NAME Exam 1 Fall 2017 20 An aluminum product is made through indirect extrusion The aluminum starts out as a

SKM754C-69-17102513271_0009.jpg - ME 467 NAME Exam 1 Fall...

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Unformatted text preview: ME 467 NAME: Exam 1 Fall 2017 20. An aluminum product is made through indirect extrusion. The aluminum starts out as a round bar, with Do = 50 mm and L = 80mm, with Ao = 1963.50 mm. The product final 12 pt cross-section is as shown in the figure below (in mm), an equilateral triangle. The process results in Y = 321.73 MPa with an extrusion strain &x = 2.33. a) Determine the maximum extrusion pressure required, in MPa. b) Determine the maximum force required, in N. PAX = Kx Yf Ex T1 ( 5 0 mm ) 2 S 4 - 1963, 495 me 4 42_7/3 4 ( 30 mm ) = 389. 71 1 mm 2 30 OR 30 mm "ha sin 60- h= 25, 981 mm - Af = 2 (1/5 )h b= 389.71l mm Kx = 0. 98 + 6. 2 ( -7 2, 25 TT DC Af = - DC - (Af ) (4 ) - 22.275 mm TT Ce = TTDC = TT ( 22, 275 mm) = 69, 9804 man CX = 3 ( 30 mm ) = 90 mmm 2. 25 K x = 0. 48+6, 02 (69,98mm) 2 , 2> - . 98+0. 02 ( 1. 286 ) - 1.015 Pmax = Kx If E x = ( 1, 015 ) ( 3 21, 73 MPz ) ( 2. 33 ) - pmay = 761. 042 MP d b ) F = pmax As = ( 761 , 04 2 m A . ) ( 19 63 , 495 mm ? ) F = 1,494, 302. 707 N B Page 9 of 13...
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