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Materials processing and joining

An aluminum alloy billet 100 mm in diameter is hot extruded into a round bar 25 mm in diameter. Using the extrusion ratio R = A₀/A (the ratio of cross-sectional areas), take the ideal extrusion pressure, ignoring friction and redundant deformation, as p = Y ln R with a flow stress Y = 50 MPa. Which value is closest to p?

  1. AAbout 69 MPa
  2. BAbout 139 MPa
  3. CAbout 800 MPa
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Answer: B. About 139 MPa

Cross-sectional area is proportional to the square of the diameter, so the extrusion ratio is R = (100/25)² = 16. Then p = Y ln R = 50 × ln 16 ≈ 50 × 2.77 ≈ 139 MPa. About 69 MPa takes the logarithm of the diameter ratio of 4, and about 800 MPa uses Y × R without the logarithm. The actual extrusion pressure is considerably higher than this ideal value because of friction with the container wall and die and the redundant shear deformation inside the material. Indirect extrusion, in which the die moves and there is no sliding between billet and container, requires less pressure than direct extrusion, in which the billet slides through the container.

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