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CHAPTER 12
CH 12, 1PP
Problem
A direct shear test was conducted on a specimen of dry sand with a normal stress of 200 kN/m2. Failure occurred at a shear stress of 175 kN/m2. The size of the specimen tested was 75 mm × 75 mm × 30 mm (height). Determine the angle of friction, φ For a normal stress of 150 kN/m2, what shear force would be required to cause failure of the specimen?
Step-by-step solution
Show all steps
Step 1/
Given data: Size of the specimen is. Normal stress is. Shear stress at failure is.
Step 2/
a) The angle of friction is
Therefore, the angle of friction is. Step 3/ b) Given data: Normal stress is. Shear force required to cause the failure is , where is the shear stress at failure and A is the cross sectional area of the specimen. Area of the cross section is , Where, is the side of the specimen. Substituting 75 mm or 0.075 m for a in the relation, Step 4/ Shear stress at failure is
CH 12, 2PP (0)
Problem
For a dry sand specimen in a direct shear test box, the following are given:
- Angle of friction: 38°
- Size of specimen: 2 in. × 2 in. × 1.2 in. (height)
- Normal stress: 20 lb/in. Determine the shear force required to cause failure. Step-by-step solution Show all steps Step 1/ Given data: Size of the specimen is. Normal stress is. Angle of friction is. Step 2/ Shear force S required to cause the failure is , Where, is the shear stress at failure, and A is the cross sectional area of the specimen. Step 3/
Shear stress at failure is Area of the cross section is , Where, is the side of the specimen. Substituting for in the relation, Step 4/ Substituting the values in equation, The shear force, Therefore, the shear force required to cause the failure is.
force (N) 1 200 155 2 300 230 3 400 310 4 500 385 Step 2/ Theory: Drained angle of friction is Where, is the shear stress at failure. is the normal stress. Shear stress at failure is. Where, is the shear force at failure. is the area of the specimen. Normal stress is. Where, is the normal force. Compute area A of the specimen. Step 3/ Considering the test 1:
Shear stress at failure is Normal stress is Step 4/ Drained angle of friction is Therefore, the drained angle of friction is. Step 5/ Considering the test 2: Shear stress at failure is
Step 8/ Drained angle of friction is Therefore, the drained angle of friction is. Step 9/ Considering the test 4: Shear stress at failure is Normal stress is
Step 10/ Drained angle of friction is Therefore, the drained angle of friction is. Step 11/ The results are shown in the table. Test No. Shear stress at failure (N/mm^2 ) Normal stress (N/mm^2 ) Drained angle of friction 1 0.043 0.056 37. 2 0.064 0.083 37. 3 0.086 0.111 37. 4 0.107 0.139 37. Graph for shear stress at failure against normal stress.
CH 12, 4PP (2)
Problem
Repeat Problem 12.3 with the following data. Given specimen size:
- Diameter = 2 in.
- Height = 1 in. Problem 12. The following are the results of four drained, direct shear tests on a normally consolidated clay. Given:
- Size of specimen = 60 mm × 60 mm
- Height of specimen = 30 mm Draw a graph for the shear stress at failure against the normal stress, and determine the drained angle of friction from the graph.
Step-by-step solution Show all steps 100% (2 ratings) for this solution Step 1/ Calculate the area of the specimen. Here, d is the diameter of the specimen. Substitute 2 in for d. Step 2/ Consider the first test. Calculate the shear stress at failure. Here, S is the shear force at failure, and A is the area of the specimen. Substitute 37.5 lb for S , and for A. Step 3/ Calculate the normal stress at failure.
Substitute 55 lb for S , and for A. Step 6/ Calculate the normal stress at failure. Substitute 90 lb for N and for A. Step 7/ Calculate the drained angle of friction for the second test. Substitute for , and for. Therefore, the drained angle of friction for the second test is. Consider the third test. Calculate the shear stress at failure.
Substitute 70 lb for S and for A. Step 8/ Calculate the normal stress at failure. Substitute 110 lb for N , and for A. Step 9/ Calculate the drained angle of friction for the third test. Substitute for , and for. Therefore, the drained angle of friction for the third test is.
Therefore, the drained angle of friction for the fourth test is. Step 13/ The results are shown in the table: Test No. Shear stress at failure^ (lb/in
Normal stress (lb/in^2 ) Drained angle of friction 1 11.94 19.12 30. 2 17.52 28.66 31. 3 22.29 35.03 32. 4 25.48 39.81 32. Graph for shear stress at failure against normal stress.
For normally consolidated soils,. Therefore, the drained friction angle from the graph is.
