Analytical Chem Soil HW Solutions 19

# Analytical Chem Soil HW Solutions 19 - Q2.4 Solution (a)...

This preview shows page 1. Sign up to view the full content.

19 Determination of peak strengths Q2.4 The following results were obtained from a shearbox test on a 60 mm × 60 mm sample of dry sand of unit weight 18 kN/m 3 . Reading on proving ring deflexion dial gauge (divisions) Zero force 91 Peak shear force for a hanger load of 3kg 128 Peak shear force for a hanger load of 10kg 162 Peak shear force for a hanger load of 20kg 210 One division on the proving ring dial gauge corresponds to a force of 1.1N across the proving ring. (a) Plot the data on a graph of shear stress against normal effective stress, and sketch the peak strength failure envelope. (b) What is the peak resistance to shear on a horizontal plane at a depth of 3 m below the top of a dry embankment made from this soil? (c) A model of the embankment is constructed from the same sand at a scale of 1:10. What is the peak resistance to shear on a horizontal plane at a depth of 300mm below the top of the model? (d) Would you expect the model to behave in the same way as the real embankment?
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 12/28/2011 for the course CHM 4302 taught by Professor Stuartchalk during the Fall '11 term at UNF.

Ask a homework question - tutors are online