Question

hapter 20 Saved Required information A specimen of steel having an initial diameter of 0.503 in was tested in tension using a gauge length of 2 in. The following data were obtained for the elastic and plastic states: For the elastic state Pi(lb) Alin) 1000 2000 3000 4000 0.0004 0.0006 0.001 0.0013 0.0023 0.0028 0.0036 0.0089 8400 8800 9200 For the plastic state Pi(lb) 8800 9200 9100 13200 15200 17000 16400 14800 A lin2 0.1984 0.1978 0.1963 0.1924 0.1875 0.1563 0.1307 0.1077 III LIIT NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.

Answer 1

9200/AO 8400/Ao Stress, Ib/in2 0.0005 0 .002 0.003 0.0035 0.001 0.0015 0.0014 0.0025 Strain, in/in 0.0040 0045 0.005 wa Aol. 1978-1

17000/AO 17000/AO Sut' 14800/AO 9200/AO Stress, Ib/in2 3 0.9 0.5 0.6 0.7 Ao/. 1077-1 Åol.1978-1 ^o/. 1563-1 Strain, inlin

For area 1, ut; = }(0.0014) (salt) UT, = 3 (0.0014) (651952) = 29.5904 For area 2, uz; = ((0.1976 - 1) - 0.0014) (200) UT, = (6.1997 - 1) - 0.0014) (1997) = 148.8379 For area 3, UT, = ((0.16 - 1) - (0.1678 - 1)) (170) UT, = ((6.1887 - 1) - 06.1996 - 1)) (4.7054) ut, = 22819.7957

For area 4, ur, = ((6.107 - 1) - (0.1878 - 1)) (17000) ut, = (06.1987 - 1) - 06.1996 - 1)) (10.704) ut, = 71900.4688 For area 5, ur; = ș ((1.1868 - 1) - (0.1976 - 1)) ( 1970 - 1900) ur. = į (C6.1987 - 1) - 06.1987 - 1)) (9.709 - 2017) ut, = 5235.1296 For area 6, usa = ((0.10 - 1) - (CSR - 2)) ( 1970 19.) ut, = } (6.1987 - 1) - 06.1962 - 1)) (4.7969 UT, = 3175.8083

The modulus of toughness is ut = (ut, + ut, + ut, + ur.) - (ut: + ut ) UT = (29.5904 + 148.8379 +22819.7957 + 71900.4688) - (5235.1296 + 3175.8083) in: up = 86487.755 in )

Ut= 86.487 X 10³ in.lb/in³