Question

Problem 3: A consolidated-drained triaxial test is carried out on a sand specimen that is subjected to 80 kN/m2 confining pressure. The vertical deviator stress was increased slowly such that there is no built-up of pore water pressure within the specimen. The specimen failed when the addition axial stress reached 240 kN/m2. Find the friction angle of the sand. If another identical sand specimen was subjected to 150 kN/m? confining pressure, what would be the deviator stress at failure.

Answer 1

This type of problem can be solved by using simple stress relationship

Given that Confining pressure (63) = 80 kN/m2 - Deviator ar anial stress (Ed) = 240 W rhy? To find out fruction angle (0) = ? We know that the relationship between - principle stresses are given- - 5 = 63 tan tus + 2 + acutan (4st of where 0=83tod = 80t 240=320 kN/m2 - C = Cohesion and for sand czo Now putting all the values- 320 = 80 Jan445+) +o a) 3.3. darf (4540) - ten (45+) = 4 ) dan (ust!)=2 ! (45+ b ) = tun(a) = 457 = 63:43" n14 90 o

= 36.860 which y angle of shear resistance Now for second specimen - 3=150 kN/m², To calculate deviatar stress Since some specimen is used- 01-63 Jan (45+ o) + 8 = 150 dan tus+ 36.86) = ora 15004 y = 600 WN / m2 a 63 tra=600 7 od= 600-150 - Od = 450 kN/m2 which is deviator stress at failure -