Questions 7 7. 7- 7 7.7 0: testing techniques 445Q1 1.9 (a) The results in the table below represent shear displacement
and shear stress recorded during a direct shear test on a fracture in slate.
The shear displacement range was from^0 to^15 mm as shown in the
table below. The normal stress during the test was 0.34 MPa.
Shear displacement (mm) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
Shear stress (kPa) (^0 281 344 344 328) 281 281 297
Shear displacement (mm) 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5
Shear stress (kPa) 281 281 266 266 266 281 281 281
Shear displacement (mm) 8.0 8.5 9.0 9.5 10.0 10.5 11.0 11.5
Shear stress (kPa) 297 297 297 313 313 313 313 313
Shear displacement (mm) 12.0 12.5 13.0 13.5 14.0 14.5 15.0
Shear stress (Ea) (^313 313) 313 313 313 313 313
Use these results to determine the residual shear strength of the fracture.
(b) A series of direct shear tests was undertaken at different normal
stress values on samples from the fracture, and the peak shear stress
encountered during each test was recorded, as shown in the table below.
Normal stress(kPa) 336 648 961 1273 1586
Peakshear stress (Ea) 344 516 719 953 1156
Use these results to determine the basic friction angle, 4, and the asperity
angle, i, for the fracture. Also comment on the validity of the bi-linear
approximation for the failure locus.
Q1 1.10 The diagram below shows example results from a numerical
modelling code for predicting the elastic displacements (indicated by the