Materials Science And Engineering
Materials Science And Engineering
10th Edition
ISBN: 9781119405498
Author: Callister, William D., Jr, RETHWISCH, David G., Jr., 1940- Author.
Publisher: Wiley,
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Chapter 10, Problem 25QAP

(a)

To determine

To sketch: The continuous cooling curve to yield fine pearlite and proeutectoid cementite.

(b)

To determine

To sketch: The continuous cooling curve to yield martensite.

(c)

To determine

To sketch: The continuous cooling curve to yield martensite and proeutectoid cementite.

(d)

To determine

To sketch: The continuous cooling curve to yield coarsepearlite and proeutectoid cementite.

(e)

To determine

To sketch: The continuous cooling curve to yield martensite, fine pearlite, and proeutectoid cementite.

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Can you rewrite the solution because it is unclear? AM (+) = 8(1+0.5 cos 1000kt +0.5 ros 2000 thts) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. J4000 t j11000rt $14+) = 45 jqooort +4e + e + e j 12000rt. 12000 kt + e +e +e Le jsoort -; goon t te +e Dcw> = 885(W- 100007) + 8 IS (W-10000) - USB
Can you rewrite the solution because it is unclear? Q2 AM ①(+) = 8 (1+0.5 cos 1000πt +0.5 ros 2000kt) $4+) = 45 = *cos 10000 πt. 8 cos wat + 4 cosat + 4 cos Wat coswet. j1000016 +4e -j10000πt j11000Rt j gooort -j 9000 πt + e +e j sooort te +e J11000 t + e te j 12000rt. -J12000 kt + с = 8th S(W- 100007) + 8 IS (W-10000) <&(w) = USB -5-5 -4-5-4 b) Pc 2² = 64 PSB = 42 + 4 2 Pt Pc+ PSB = y = Pe c) Puss = PLSB = = 32 4² = 8 w 32+ 8 = × 100% = 140 (1)³×2×2 31 = 20% x 2 = 3w 302 USB 4.5 5 5.6 6 ms Ac = 4 mi = 0.5 mz Ac = 4 ५ M2 = =0.5
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