Question

please make sure it is correct

Calculate the percentage of each phases of 1040 steel at 500 °C present using lever rule WF=0.02, WE=0.76, Wc=6.7 Composition Cara 10 15 20 1600 538- 1493 1400 L 2500 13940 yu 1200 11470 2.14 4.30 Austite 2000 Temperature (°C) 1000 Temperature (°F) 9120 Fec 800 .. 1500 7270 0.76 0.022 "o, Ferrite 600 FC Dementie 1000 5 6.70 F 3 Composition witc) OI Ferrite 6%, Cementite 94% Ferrite 94%, Pearlite 6% Ferrite 6%, Pearlite 94% o IV Ferrite 94%, Cementite 6%

Answer 1

for gien 1040 steel. low alloy steel 0.400) 800 727 176 d. 600 b 500 1 6.67 400 0.40 5.76 ->CZ Fe in 1040 Steel at given problem we only know compositiom at 727°C Wc = 6.67 WE 0:76). WE=0.022). 500°C But we are asked for we don't know the value at point a 10%) chart = 0.400% Love carbon nog 1040 steel fom leuch mule Ferrite 6,7 -0.4 =.943 6 6.7 ~ 0.02 94.3%. applying pear lite 1-943 (Q+ Fe30) Cementite) =o.os7 5.7 % Ferritet ñ 6% Termite 94.3% Ansa II Pearlite 267.6

if you take value between any 0 -0.22 500°c as aboue solution won't give exact answer be callse did not use exact c% at soo°C at a take (by observing the dogram) we o.oos a = ue d 0.941 Fernte =Wao 6.7 -0.4 6.7 -0.005 e = 94.1). And Pearlite = 1- Wa 1 -0.94] 05 099 socha 0.005 0.4 = I' = 5.99 %. Aus

Please understand in first solution i have taken value which was given in question .

In second solution i have taken value of point a at 500^oC by observing the diagram as exact value of point a is not given in question .this will give more appropriate answer.

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