Question
could you calculate 6.3 and 6.4 please?
6 A flat piece of slate with uniform thickness 25 mm is cut into the shape of a square with 100 mm long squared edges. A test is devised which allows uniform compressive oi to be applied along two edges and uniform tensile stress o2 along the other two opposite edges, as shown in Figure 2. The stresses and o2 act normally to the edges of the test specimen. The test is performed by increasing the magnitudes of and o2 simultaneously, but keeping the magnitude of always three times the magnitude of a2. 6.1 lf failure of the slate occurs when the shear stress on any plane exceeds 1 MPa, what would be the values of and 02 at the moment of failure? (4 marks) 6.2 Determine the directions of zero normal stress for the stress state at failure. 2 marks) 6.3 Would the values of 01 and at failure be changed if the rock had a tensile strength of 0.55 MPa? (3 marks) 6.4 Would the values of oi and o2 at failure be changed if a planar weakness running through the test specimen, inclined at 60° to the direction of orz, would rupture when the shear stress on it exceeds 0.75 MPa? 6 marks 60 Weakness Figure 2
Explanation / Answer
6.3 When the tensile strength of the rock is equal to0.55MPa,the value of sigma1 would be lesser than the tensile strength while the value of sigma2 will be greater than the tensile strength because the value of stress of rock lies between the two values and if the range goes beyond it,the rick would break.
Sigma1=0.50Mpa
Sigma2=0.60Mpa
6.4 The value of sigma1 and sigma2 is calculated in the orevious state and if the shear stress is said to increase to 0.75MPa, the values of sigma1 and 2 will also be changed and will be nearly equal to 0.80MPa.
