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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 10-3 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 10-3 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 non-linear, 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
- Analytical Chemistry