Assign4persadcUTT Chvorinov's RuleQuestion.pptx
7
The sketch above shows two designs of tie-down hardware made for road-hauled freight. They are both produced from cast steel, and are listed as “7050Ring”, and “8040Rectangle”. In both cases, the general cross-section is approximately circular with a 10mm diameter. The 7050Ring has a 70mm outside diameter, and a 50mm inside diameter. The ring is a torus. Note #2 on Page 2 gives the volume and surface area of a torus. The 8040Rectangle outer rectangle is 80mm x 40mm, and the inside is 60mm x 20mm. You may assume that the 8040Rectangle can be decomposed into two long cylindrical pieces- each 55mm long, and two short cylindrical pieces – each15mm long. The remaining four curved corners can be represented by the ring shown in Figure 3(c). This ring can also be treated as a torus with a 25mm outside diameter and a 5mm inside diameter, as depicted in Figure 3(d). Both designs are to be produced in metal molds of similar design. Solidification time in the mold is predicted using Chvorinov's Rule with a measured mould constant of 1.0 s/mm 2 . Calculate the solidification times for the “7050Ring” and for the “8040Rectangle”. NOTE #1: The mould constant C m (s /mm 2 ) in Chvorinov's Rule for a casting of Volume V (mm 3 ) and Surface Area A (mm 2 ) is related to the time for the casting to solidify t TS (s) by C m = t TS / (V/A) 2 . 3 (a) 7050Ring 3 (b) 8040Rectangle 3(c) Elements of 8040Rect. 3(d) A torus 25mmOD & 5mmID This ring can also be treated as a torus with a 25mm outside diameter and a 5mm inside diameter as overlaid on the rule in (d).
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Transcript of Assign4persadcUTT Chvorinov's RuleQuestion.pptx
The sketch above shows two designs of tie-down hardware made for road-hauled freight. They are both produced from cast steel, and are listed as “7050Ring”, and “8040Rectangle”. In both cases, the general cross-section is approximately circular with a 10mm diameter. The 7050Ring has a 70mm outside diameter, and a 50mm inside diameter. The ring is a torus.
Note #2 on Page 2 gives the volume and surface area of a torus.
The 8040Rectangle outer rectangle is 80mm x 40mm, and the inside is 60mm x 20mm. You may assume that the 8040Rectangle can be decomposed into two long cylindrical pieces- each 55mm long, and two short cylindrical pieces – each15mm long. The remaining four curved corners can be represented by the ring shown in Figure 3(c). This ring can also be treated as a torus with a 25mm outside diameter and a 5mm inside diameter, as depicted in Figure 3(d).
Both designs are to be produced in metal molds of similar design. Solidification time in the mold is predicted using Chvorinov's Rule with a measured mould constant of 1.0 s/mm2.
Calculate the solidification times for the “7050Ring” and for the “8040Rectangle”.
NOTE #1: The mould constant Cm (s /mm2) in Chvorinov's Rule for a casting of Volume V (mm3) and Surface Area A (mm2) is related to the time for the casting to solidify tTS(s) by Cm = tTS / (V/A)2.
3 (a) 7050Ring 3 (b) 8040Rectangle 3(c) Elements of 8040Rect.
3(d) A torus 25mmOD & 5mmID
This ring can also be treated as a torus with a 25mm outside diameter and a 5mm inside diameter as overlaid on the rule in (d).
NOTE #1: The mould constant Cm (s /mm2) in Chvorinov's Rule for a casting of Volume V (mm3) and Surface Area A (mm2) is related to the time for the casting to solidify tTS(s) by Cm = tTS / (V/A)2.
NOTE #2: The Volume V (mm3) and Surface Area A (mm2) formulae for a ring torus are as given below. The sketch is of a cut along a diameter of such a ring torus.
The next four pages reproduce the problems that we solved in class.
We calculated Solidification Times….. Use n = 2 unless you are given a different value.
