2000 0.0006 0.00030 3 10065 3000 0.001 0.00050 4 15097 4000 0.0013 0.00065 5 20130 7000 0.0023 0.00115 6 8400 0.0028 0.00140 7 8800 0.0036 8 44285 9200 0.0089 9 8800 0.1984 0.00158 10 44285 9200 0.1978 0.00461 11 9100 0.1963 0.01229 12 45795 13200 0.1924 0.03281 13 15200 0.1875 0.05980 14 76492 17000 0.1563 0.27136 15 85551 16400 0.1307 0.52037 16 82531 14800 0.1077 0.84506 17 74479 (a) Linear range (b) Offset yield (c) Complete range (c) The material is ductile since there is a large amount of deformation beyond yield. (d) The closest material to the values of and is SAE 1045 HR with = 45 kpsi, Sy, Sut, R Sy = 82 kpsi, and = 40 %. Sut R Ans. ______________________________________________________________________________ To plot true vs., the following equations are applied to the data. 2-7 Eq. (2-4) where The results are summarized in the below and plotted on the next page. The last 5 points of data are used to plot log vs log The curve fit gives = 0.2306 m log 0 = 5.1852 0 = 153.2 kpsi Ans. For 20% cold work, Eq. (2-14) and Eq. (2-17) give, = (1 = 0.1987 (1 0.2) = 0.1590 in2 A A0 W) – –
3 2000 0.0006 0.00030 10065 4 3000 0.001 0.00050 15097 5 4000 0.0013 0.00065 20130 6 7000 0.0023 0.00115 7 8400 0.0028 0.00140 8 8800 0.0036 44285 9 9200 0.0089 10 8800 0.1984 0.00158 44285 11 9200 0.1978 0.00461 12 9100 0.1963 0.01229 45795 13 13200 0.1924 0.03281 14 15200 0.1875 0.05980 76492 15 17000 0.1563 0.27136 85551 16 16400 0.1307 0.52037 82531 17 14800 0.1077 0.84506 74479 (a) Linear range (b) Offset yield (c) Complete range (c) The material is ductile since there is a large amount of deformation beyond yield. (d) The closest material to the values of Sy, Sut, and R is SAE 1045 HR with Sy = 45 kpsi, Sut = 82 kpsi, and R = 40 %. Ans. ______________________________________________________________________________ 2-7 To plot true vs., the following equations are applied to the data. Eq. (2-4) where The results are summarized in the below and plotted on the next page. The last 5 points of data are used to plot log vs log The curve fit gives m = 0.2306 log 0 = 5.1852 0 = 153.2 kpsi Ans. For 20% cold work, Eq. (2-14) and Eq. (2-17) give, A = A0 (1 – W) = 0.1987 (1 – 0.2) = 0.1590 in2