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 35227 8400 0.0028 0.00140 7 42272 8800 0.0036 0.00180 8 44285 9200 0.0089 0.00445 9 46298 8800 0.1984 10 44285 9200 0.1978 0.00461 11 46298 9100 0.1963 0.01229 12 45795 0.1924 0.03281 13 66428 0.1875 0.05980 14 76492 17000 0.1563 0.27136 15 85551 16400 0.1307 16 82531 14800 0.1077 0.84506 17 (a) Linear range (b) Offset yield (c) Complete (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 table below and plotted on the next page. The last 5 points of data are used to plot log vs log The 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 35227 7 8400 0.0028 0.00140 42272 8 8800 0.0036 0.00180 44285 9 9200 0.0089 0.00445 46298 10 8800 0.1984 44285 11 9200 0.1978 0.00461 46298 12 9100 0.1963 0.01229 45795 13 0.1924 0.03281 66428 14 0.1875 0.05980 76492 15 17000 0.1563 0.27136 85551 16 16400 0.1307 82531 17 14800 0.1077 0.84506 (a) Linear range (b) Offset yield (c) Complete (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 table below and plotted on the next page. The last 5 points of data are used to plot log vs log The 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
Solution-Manual-for-Shigleys-Mechanical-Engineering-Design-10th-Edition-by-Budynas.docx