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Unformatted text preview: Q2.4 Solution (a) The normal stress on the sample is given by the hanger load (kg) 9.81 (N/kg) the sample area, 0.06m 0.06m = 3.6 103 m 2 , 1000 to convert from Pa to kPa. The shear force on the sample is given by 1.1 (the number of proving ring dial divisions  the number of divisions at zero load), i.e. 1.1 (n  91). To convert this to the shear stress, it is necessary to divide the shear force by the area of the sample, 0.06m 0.06m = 3.6 103 m 2 , and divide by 1000 to convert from Pa to kPa. Hanger load, kg Normal stress, kPa Peak shear load, N Peak shear stress, kPa 3 8.175 40.7 11.31 10 27.25 78.1 21.69 20 54.5 130.9 36.36 These data are plotted on a graph of against ' in Figure Q2.4. The peak strength failure envelope is highly nonlinear, with ' peak = 55 at ' 8 kPa, falling to ' peak = 34 at ' 55 kPa...
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This note was uploaded on 12/28/2011 for the course CHM 4302 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk
 Analytical Chemistry

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