Geotechnical Engineering A (BSc)Coursework No 2 - Shear bo.docx
- 1. Geotechnical Engineering A (BSc) Coursework No 2 - Shear box test Level 5 Group: B Submitted to: Dr. Martin Pritchard Submission Deadline: 7th of April 2016 Name: Rui Dorey Student ID: 77149818 Contents Abstract 3 Introduction 3 Experimental procedure 4 Calculations and results 4 Discussion 11 Conclusion 11 References 11 Appendix I 11 Appendix II 11 Appendix III 11
- 2. Abstract On the shear box experiment for geotechnical module, the goal was to determine the displacement of sample of sand and also to calculated the angle of shear resistance . According to The direct shear box test BS 1377; Part 7 1990: 4 and 5 the group B has performed the experiment. In addition, the shear box test was performed on typical standard test conditions. However, the hanger on this test has suffered from four different vertical loads which were: 15, 25, 35 and 45 kg of mass. As result, on this experiment the shear stress of the four tests has increased simultaneously as the load of the test has increased, so the higher the load means the higher the shear stress and the effective normal stress. On the other hand by using the best line in order to determine the apparent cohesion on this test it was possible to see that the result of cohesion was 0 and the angle derived was: =48.76.
- 3. Introduction On the direct shear box test BS 1377; Part 7 1990: 4 and 5, the aim is to calculate and determine the volumetric displacement of a sand and the shear resistance angle. Moreover on this experiment , the shear strength of the soil in this case the sand can be defined by being the maximum shear stress that is applied at any direction. However, when the soil tested reaches the point failure means that the soil reached the maximum yield stress. Moreover, the soil shear strength is given by the frictional resistance (F) which is derived from “inter-particle forces, (N)” and in this case the pore water has no shear strength. Besides, the formula that is used to calculated the shear strength in shear box test which involves the total normal stress, apparent cohesion (C) and shear resistance angle(φ) is: tf = σntanφ BS 1377; Part 7 1990: 4 and 5. The sample used in the direct shear box is resulted by shear during the plane and it is divided by upward and downward pieces when it is applied a horizontal load on the
- 4. upper piece when the downward piece is positioned. The proving ring usually helps to apply the load and consequently the shear created by the sample normally is readable straight away and also the shear stress (t) is a result of a division of a shear by the plan area of the box, BS 1377; Part 7 1990: 4 and 5. In order to get accurate results, this test has to be repeated many times with different normal loads by using the sample with different specimens. After plotting the results, it is possible to determine the shear strength and then to get the value of the angle of shear resistance BS 1377; Part 7 1990: 4 and 5 . On the shear box test, the sample, the soil normally has a “rectangular cross-section and also is square in plan” (vickers, 1983). According to (vickers, 1983) , the shear boxes measures are 64mm square up to 254 mm square and the sample normally can be remoulded and undisturbed in case that those boxes accept “ coarse – grained soils”. However it obvious that is necessary to take care when it is tested “remoulded or coarse –grained samples” due to the fact that those samples normally “get prepared at bulk unit weight” and the relevance of the moisture content can be a problem that normally must to be taken into account (vickers, 1983). Besides, the box used on this test are “open at the top” of it and they are also a “rigid metal of construction”. In addition those boxes are “immersed in water- container” (vickers, 1983). Furthermore, the shear boxes are manufactured having two halves which one half that is the upper move in horizontal direction relatively in comparison with the other half (down half), so “the sample is sheared on horizontal plane” (vickers, 1983). Moreover the sample suffer from vertical load applied by different masses of the hanger. In addition a motor drive helps to apply horizontal shear forces into the sample in question. Besides, this “motor drive is usually multi-geared” making it possible to apply many different shear loading rates” (vickers, 1983). On the other
- 5. hand, on having ball-bearing slides mounted on the shear boxes, the free movements among both halves is facilitated (vickers, 1983). Finally, when the “proving-ring is mounted in a horizontal plane, the shear load applied on the sample in question can be recorded and then the dial gauge monitors the deformation of the proving- ring (vickers, 1983). Apparatus: NB mass of hanger = 5kg Experimental procedure · Assemble the empty shear box without the upper and loading platens. · The two halves of the box should be screwed together with the screws marked by a cross cut into the heads; the two screws marked ‘L’ should be in position but clear of the joint between the two halves. · Ensure that the apparatus is moving freely on its runners. · Fill the box with loose sand and level it off about 1mm below the top of the box. · Place the top platen, on the sand and the loading platen on the top platen. · Put the ball bearing in place and the hanger on the ball bearing. · Place a weight on the hanger. · Adjust the apparatus to take up any slack, and then zero
- 6. the proving ring dial gauge. · Remove the two screws holding the upper and lower halves of the box together.· Then screw in those marked ‘L’ until resistance is just felt: give each one a further half turn to ensure that the two halves are slightly separated so that the normal stress is being applied to the sand only, then remove them. · Switch on the motor. Record the maximum reading on the proving ring dial gauge, then switch off and slacken off the apparatus. · Dismantle the box entirely and pour all the sand back into the container. · Repeat the test three more times, increasing the mass on the hanger. · Record the results on the attached sheet. Calculations and results Sample = Sand 0.71-1.25mm Proving Ring Constant = 0.585 N/Div Mass of hanger =5 kg Area of shear box = 60mm x 60mm = 3600 mm2 Effective normal stress = Normal load / Area of shear box In order to calculate the load and shear stress values for all the vertical and horizontal displacement, it is necessary to apply the following Formulas:
- 7. Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2) The table shown below Illustrates the results of Normal load against effective stress: Normal Load (N) Effective normal stress (kN/m2) Test:1 147.15 40.875 Test:2 245.25 68.125 Test:3 343.35 95.375 Test:4 441.45 122.625 The Graph below shows the shear stress against the effective stress. Shear Stress at failure Consequently: 46.64/ 0 + 40.875 1.1410 Therefore: =48.76
- 8. First test: Horizontal Disp. (mm) Vertical Disp. (mm) Proving Ring Reading Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2)
- 12. 136.31 37.86 4.4 0.89 241 140.99 39.16 4.6 0.94 236 138.06 38.35 4.8 0.98 227 132.80 36.89 5.0 1.02 225 131.63 36.56 Second test: Horizontal Disp. (mm) Vertical Disp. (mm) Proving Ring Reading Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2)
- 17. Third test: Horizontal Disp. (mm) Vertical Disp. (mm) Proving Ring Reading Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2) 0.0 0.00 0 0.00 0.00 0.2 0.00 129 75.47 20.96 0.4 0.00 225 131.63 36.56 0.6
- 21. 73.29 5.0 1.05 446 260.91 72.48 Forth test: Horizontal Disp. (mm) Vertical Disp. (mm) Proving Ring Reading Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2) 0.0 0.00 0 0.00 0.00 0.2
- 25. 108.06 4.6 0.83 651 380.84 105.79 4.8 0.87 638 373.23 103.68 5.0 0.90 624 365.04 101.40 On the next graphs, it is highlighted the shear stress against the horizontal displacement. On this graph it is possible to see that as long the normal load increases the shear strength increases as well. The graph below shows the results of vertical displacement over the horizontal displacement. Discussion On the results shown by the shear box test , it must to be referred the results obtained on load and shear at every
- 26. simple test which were at 15,25,35,45kg of mass. In addition, those results include the effective normal stress, load and the shear stress. During this report it was also shown some line graphs comparing the four tests. Moreover, on those graphs comparing the test, it is possible to see the shear stress over the horizontal displacement which it is seen that the highest values recorded appears on the test carrying the highest weight(45kg) and it appears the other test in a respective order 35,25 and 15kg. However, on the graph that demonstrates the vertical displacement against horizontal displacement over the four test tested at the laboratory, the test with the highest weight (load) is the one with the lowest values. On the other hand , on the same graph it is possible to see that the test with 15kg and 25 kg weight were particularly almost the same. Meanwhile it is necessary to say that the on the four test the effective normal stress increases as long the weight increases and consequently the shear stress values increases simultaneously. In addition by plotting this values of effective normal stress and shear stress on a graph, it was visible the line at the failure and was possible to determine the shear resistance angle which was 48.76 degrees. Besides on this test, it was not recorded any value of apparent cohesion due to the fact that the graphs began at point 0.0 of effective stress and shear stress. However, in comparison with (vickers, 1983) results, it is possible to analyse that the trend by increasing the weights leads to increased values effective normal stress and shear stress. According to the test , it is noticeable that the results from the actual these test do not differ that much. However, the results from the actual test in comparison with the (vickers, 1983)test are slightly different, perhaps, the reason for that might be because (vickers, 1983) has used different boxes size which were at 55mm x 56mm , different proving ring constant (0.001N) and different thickness of the sample (sand) fixed at 16.04mm.
- 27. Furthermore, the reason of the growth of the load and respectively the shear stress over the time is cause the sample (sand) becomes more resistant over the time as well. On the graph of shear stress over horizontal, it possible to sees the rise of the lines on the four test until they fail. In addition, those are the point of failure of the four different test which were noticed at 147.15 , 245.25, 343.35 and 441.45 kN of load during the different tests. Conclusion To sum up the shear box device has performed with success on testing the samples of sand on the four different loads (15, 25, 35 and 45 kg of mass). As result, on the four different loads applied on the test, it was recorded the vertical displacement and proving ring reading for each load test and consequently it was calculated the load applied and shear stress values according to the readings noticed by the students ( vertical displacement and proving ring reading). In addition, those reading taken by the students were following the horizontal displacement shown on excel spread sheet given by laboratory tutor which started from 0.0 mm to 5.0 mm. however by comparing the four test (15,25,35 and 45 kg), it is possible to notice that effective stress and shear stress becomes higher once it is used a higher normal load. However the graph of effective normal stress against shear stress at the failure shows that the sample of sand (0.71 to 1.25mm) in question is good in term of quality and due to the fact that, it has and shearing resistance angle of 59.19 degrees. On the other hand, the values recorded and shown on the graph illustrating the
- 28. vertical displacement over the horizontal displacement, it possible to analyse the four different tests at same time and it has recorded almost the same values of vertical displacement until they reach 1mm of horizontal displacement, then those values recorded on 15,25,35and 45 test have increased until 1.02, 0.82,1.05 and 0.90mm respectively at the end of each test which was at 5.0 mm of horizontal displacement. References vickers, B. (1983). Shear - strength tests. In Laboratory Work in soil Mechanics . Granada Publishing.
- 29. Appendix I Method statement A. PRIOR TO STARTING WORK 1. Each student is to wear the convenient safety boots during experiments in the laboratory. 2. Each individual is keen to review the test procedures before conducting the tests 3. Each student should be aware with the safety measures when using the experimental tools 4. Everybody must ensure what to do in case of emergency 5. The equipment needed are to be available and checked for safety and operation B. Whilst Undertaking the LABORATORY WORK 1. Ensure having dry hands before switching on or off the electrical equipment 2. Fix the two halves of the shear box with great care 3. Use the loading hangers carefully C. PRIOR TO DEPARTURE of the laboratory session 1. Check that the tools and equipment are in safe positions 2. Check that the fine equipment are returned to their places
- 30. 3. Clean the table and lab 4. Switch off the electricity source D. RESPONSIBILITIES 1. Immediate reporting of any incident to the technicians of the laboratory 2. Watching my steps in the lab and be careful not to damage or play with anything not required for the experiment 3. Keep the lab steady and quiet to allow safer work 4. Keep the place tidy and clean Appendix II Risk Assessment Form: Task – Hazards/Deficiencies Risk Current Controls WCO LIK LVL Connecting with electricity by using power tools. MIN LOW MED Workers must be wearing gloves to avoid the connection. Some of the equipment and materials may affect eyes. MIN LOW MED Workers must protect their eyes by wearing glasses. Workers may crush their fingers when putting the loads. MIN MOD
- 31. MED Workers must be careful when loading the machine and they should wear gloves. Hazard Severity (WCO) the Risk Level (LVL): MAJOR HIGH HIGH MED SERIOUS HIGH MED LOW MINOR MED LOW LOW HIGH MODRATE LOW HAZARD LIKELIHOOD Effective Normal stress vs shear shear stress at failure 0 147. 15 245.25 343.35 441.45 0 40.875 68.125 95.375 122.625 Shear Stress Vs Horizontal Disp (1)Shear Stress = Load x 10-3 / Area (kN/m2) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2000000000000002 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4000000000000004 4.5999999999999996 4.8 5 0 14.3 13.65
- 32. 13.16 25.03 34.61 40.14 43.23 45.01 45.5 45.5 45.83 46.15 46.64 44.53 43.55 42.41 41.6 41.93 35.1 38.35 37.86 39.159999999999997 38.35 36.89 36.56 (2)Shear Stress = Load x 10-3 / Area (kN/m2) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2000000000000002 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4000000000000004 4.5999999999999996 4.8 5 0 15.28 20.309999999999999 32.99 43.71 51.68 57.53 61.59 63.86 64.510000000000005 63.86 64.03 63.86 64.510000000000005 64.19 62.89 60.78 59.64 58.83 58.83 58.99 57.04 55.41 53.46 54.44 52.98 (3)Shear Stress = Load x 10-3 / Area (kN/m2) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2000000000000002 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4000000000000004 4.5999999999999996 4.8 5 0 20.96 36.56 52.81 66.14 75.56 82.06 87.26 90.68 93.28 93.28 92.14 91.16 90.84 90.03 88.56 87.91 84.34 84.01 81.58 80.599999999999994 80.44 77.84 75.239999999999995 73.290000000000006 72.48 (4)Shear Stress = Load x 10-3 / Area (kN/m2) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2000000000000002 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4000000000000004 4.5999999999999996 4.8 5 0 23.89 46.8 65.16 79.790000000000006 91.49 100.59 107.09 111.31 115.54 119.11 121.23 121.39 121.71 119.11 119.44 118.46 117.65 117.33 112.45 113.43 111.48 108.06 105.79 103.68 101.4
- 33. Vertical Disp vs Horizontal Disp (1)Vertical Disp. (mm) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2000000000000002 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4000000000000004 4.5999999999999996 4.8 5 0 0 -0.01 -0.01 -0.01 0 0.04 0.09 0.16 0.22 0.3 0.37 0.44 0.52 0.57999999999999996 0.65 0.71 0.77 0.82 0.75 0.79 0.84 0.89 0.94 0.98 1.02 (2)Vertical Disp. (mm) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2000000000000002 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4000000000000004 4.5999999999999996 4.8 5 0 0 0 0 0.02 0.01 7.0000000000000007E-2 0.11 0.17 0.22 0.28000000000000003 0.33 0.39 0.44 0.5 0.55000000000000004 0.6 0.65 0.66 0.67 0.71 0.76 0.73 0.76 0.74 0.82 (3)Vertical Disp. (mm) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2000000000000002 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4000000000000004 4.5999999999999996 4.8 5 0 0 0 0 0.01 0.04 0.08 0.12 0.18 0.24 0.31 0.38 0.44 0.5 0.56000000000000005 0.62 0.68 0.74 0.79 0.84 0.88 0.92 0.96 0.99 1.02 1.05 (4)Vertical Disp. (mm) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2000000000000002 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 4.2 4.4000000000000004 4.5999999999999996 4.8 5 0 0 -0.01 - 0.02 -0.02 0 0.02 0.05 0.09 0.14000000000000001 0.19 0.24 0.28999999999999998 0.35 0.41 0.46 0.51 0.56999999999999995 0.61 0.66 0.71 0.74 0.78 0.83 0. 87 0.9
- 34. 1 Geotechnical Engineering: Stresses in soil – shear stress Dr Martin Pritchard Contents Shear strength Failure criterion for soils Mohr-Coulomb – failure criterion for soils Shear strength tests Stress path – failure envelope Tutorial questions Learning outcome: Assess, from first principles, the importance of the relevance of shear strength parameters. Specific text: http://environment.uwe.ac.uk/geocal/SoilMech/stresses/stresses. htm http://environment.uwe.ac.uk/geocal/glossary/GLOSS_A.HTM
- 35. Shear strength of soils Defined as the maximum shear stress that can be applied to that soil in any direction When this maximum has been reached the soils yields & is regarded to have failed The pore water has no shear strength Shear surface Relative displacement of soil mass F F N N There are two components of shear strength; these are friction (f) and cohesion (c). NB pore water has no shear strength. The values of c and f are known as the shear strength
- 36. parameters. In 1773 Coulomb developed an expression for these parameters, which defines a straight-line failure envelope. Coulomb’s equation: where: c = apparent cohesion f = angle of shearing resistance sn = normal stress on failure plane The first scientific study of soil mechanics was undertaken by French physicist Charles-Augustin de Coulomb, who published a theory of earth pressure in 1773. Coulomb’s work and a theory of earth masses published by Scottish engineer William Rankine in 1857 are still primary tools used to quantify earth stresses. These theories have been amended in the 20th century to take into account the influence of cohesion, a more recently discovered property of soils that causes them to behave somewhat differently under stress than Rankine and Coulomb predicted. He formulated the Coulomb’s law, which deals with the electrostatic interaction between electrically charged particles. The coulomb, SI unit of electric charge, was named after him. Born in Angoulême, France to a wealthy family, Charles- Augustin de Coulomb was the son of Henri Coulomb, an inspector of the Royal Fields in Montpellier. The family soon moved to Paris, where Coulomb studied mathematics at the famous Collège des Quatre-Nations. A few years later in 1759, he was enrolled at the military school of Mézières. He graduated from Ecole du Génie at Mézières in 1761. Coulomb worked in the West Indies as a military engineer for
- 37. almost nine years. When he came back to France, he was quite ill. During the French Revolution, Coulomb lived in his estate at Blois, where he mostly carried out scientific research. He was made an inspector of public instruction in 1802. 5 tf = c + σn tan f Shear stress (tf) kN/m2 Normal stress (σn) kN/m2 f c Coulomb’s equation (1773) sn tan f Cohesionless soils Shear stress (tf) kN/m2 Normal stress (sn) kN/m2 f tf = tan f e.g. sand & gravels (only friction)
- 38. Shear stress (tf) kN/m2 Normal stress (sn) kN/m2 Frictionless soils In terms of total stress, any saturated soil, which is not allowed to drain, will exhibit no, or very little frictional resistance, e.g. clays c f = 0 undrained strength envelope tf = c e.g. undrained saturated clays & silts (only cohesion) tf = c + sn tan f Shear stress (tf) kN/m2 Normal stress (sn) kN/m2 f c Partially saturated mixtures of cohesive & frictional material e.g. clayey sands, silty sands (both friction & cohesion) Mohr-Coulomb failure criterion for soils The force acting within a soil at a given point can be resolved into three principal stresses. These stresses act at right angles to each other and are termed:
- 39. Major principal stress (σ1) Intermediate principal stress (σ2) Minor principal stress (σ3) Definition: Principal stress is the normal stress acting on a principal plane A Principal plane is the plane on which no shear stress will occur If a soil sample is subject to an all-round pressure (sh or in a triaxial test s3) and the vertical pressure (sv or in a triaxial test s1 = s3+ DS) then the shear and normal stresses on any plane within the soil can be represented by points on the circumference of a circle, i.e. using Mohr circle construction Test conditions Foundations built on a saturated silty clay bearing capacity required the soil must be able to take the load before the uw has had time to dissipate soil will become stronger with time as uw dissipate & effective stress increases
- 40. settlement (dissipation of uw) will occur which may itself cause problems critical stage from a strength point of view is immediately after the load is applied Long term stability of a cut slope or retaining wall effective stress parameters obtained from a drained or undrained test with the measurement of uw Type of test depends on information required e.g. Undrained test Drained test or undrained with measurement of uw Effective stress Shear stress (tf) kN/m2 Normal stress (sn) kN/m2 Shear strength of a soil can be expressed in terms of effective stress tf’ = c’ + sntan f’ f f’ c c’
- 41. Measurement of shear strength parameters: Shear box test: This is the simplest form of laboratory shear strength test and is often referred to as the direct shear test as it relates the shear stress at failure directly to the normal stress, thus the failure envelope may be plotted directly from the results. Shear box test (BS 1377; Part 7; 1990: 4 & 5) Triaxial Compression Test (BS 1377; Part 7 & 8: 1990) Shear Vane Test (In-situ test) Shear box test: Typical results from a shear box test Shear Stress (t) kN/m2 Horizontal Displacement (mm)
- 42. sn = 20 kN/m2 sn= 40 kN/m2 sn= 80 kN/m2 Loose Dense tf (20kN/m2) tf (40kN/m2) tf (80kN/m2) tf (20kN/m2) tf (40kN/m2) tf (80kN/m2) Normal Stress (sn) kN/m2 Shear Stress (t) kN/m2 20 40 80 Dense Loose Medium Medium Ultimate or residual shear strength f’
- 43. Volumetric Displacement Horizontal Displacement (mm) Vertical Displacement (mm) Compression – volume decrease Dilation – volume increase sn=20 kN/m2 sn=40 kN/m2 sn=80 kN/m2 At the maximum shearing resistance (tf) = the rate of maximum
- 44. volume change (dilation) A thin rupture zone of the soil at critical density is produced Typical results Example 4 (shear box) The following results were recorded during a shear box test on a cohesive soil: If the specimen size was 60 mm x 60 mm, plot the failure envelope and determine the apparent cohesion and angle of shearing resistance. Area of shear box = 60 x 60 = 3600 mm2 or 3600/(1000x1000) = 3.6x10-3 m2
- 45. = 20.3 kN/m2 = 30.3 kN/m2 cohesive soil therefore obtain a ‘c’ value C = 25 kN/m2 ϕ = 14o Example 5 (shear box): Two specimens of sand were tested in a shear box at the same constant normal stress of 200 kN/m2. Specimen A was prepared in a loose state, and specimen B was compacted into a dense state. Plot the stress/strain and vertical displacement curves for the two tests. Determine the angle of shearing resistance corresponding to the loose and dense states. Loose sand: τ rises slowly to reach a value of 88 kN/m2. Dense sand: τ rises steeply to reach a peak value of 125 kN/m2, before falling slowly to level off at the same approx. ultimate
- 46. value of 88 kN/m2. 88 kN/m2 125 kN/m2 The horiz. disp. vs vertical disp. plot shows that: As the loose specimen is sheared, a reduction in volume takes place. As the dense specimen is sheared, after an initial amount of small compression, volume expansion (dilation) takes place. Shear stress (kN/m2) - D 0.0 50.0 100.0 150.0 200.0 250.0 300.0 350.0 400.0 450.0 0.0 67.0 111.0 124.0 120.0 103.0 94.0 88.0 88.0 87.0 Shear stress (kN/m2) - L 0.0 50.0 100.0 150.0 200.0 250.0 300.0 350.0 400.0 450.0 0.0 35.0 57.0 72.0 79.0 84.0 84.0 88.0 89.0 90.0 Vert. Disp (10-2 mm) - D 0.0 50.0 100.0 150.0 200.0 250.0 300.0 350.0 400.0 450.0 0.0 -3.0 3.0 11.0 18.0 23.0 26.0 27.0 28.0 28.0 Vert. Disp (10-2 mm) - L 0.0 50.0 100.0 150.0 200.0 250.0 300.0 350.0 400.0 450.0 0.0 - 6.0 -9.0 -11.0 -13.0 -14.0 -14.0 -14.0 - 14.0 -14.0 Horiz. disp (mm) Shear stress (kN/m2) Loose sand, ϕ = 24o Dense sand, ϕ= 32o Dense 0.0 200.0 0.0 125.0 Loose 0.0 200.0 0.0 88.0 Normal stress (kN/m2) Shear stress (kN/m2)
- 47. Triaxial Compression Test (BS 1377; Part 7; 1990) The apparatus consists of a cell, which is filled with water under pressure; the specimen is loaded vertically, via a proving ring to measure load. The vertical load on the specimen is increased until failure occurs, the vertical strain being recorded at the same time using a dial gauge. The test is repeated on different specimens from the same soil, using different values of cell pressure. Triaxial Test Equipment Triaxial Test Cell Test Sample Stresses on specimen in triaxial cell Cell Pressure Deviator Stress =P/A s1=s3+P/A s1 = Major principal stress s3 = Minor principal stress P/A = (s1-s3) = Deviator Stress (DS) The deviator stress is the load on the specimen, P, divided by the cross sectional area of the specimen. However, as the
- 48. sample is compressed during the test, the cross sectional area will increase. Therefore, in calculating the deviator stress an allowance for the change in area must be considered. For the calculation of deviator stress, it is assumed that the volume of the specimen remains constant and that the sample will deform as a cylinder. where P = vertical load, which is measured by a proving ring (kN) A = Area calculated using the following method Brittle failure (shear) Plastic failure (barrelling) failure at 20% strain Strain (e) % Deviator Stress (kN/m2) Typical stress/strain graphs
- 49. The triaxial compression tests are commonly undertaken on undrained specimens, where a rubber membrane seals the specimen within the cell. These results are in terms of total stress. However, it is possible to measure the pore water pressures during the shearing stage, allowing the effective stress parameters to be recorded. Another method of assessing the effective stress of a sample is to apply the load at a very slow rate and allow the sample to drain. Undrained Consolidated undrained with uw measurement Consolidated drained The various triaxial compression tests area described in detail in the following parts of BS 1377, 1990; 24hrs
- 50. 1). Undrained test*: Drainage is prevented throughout the test, so that no dissipation of pore pressure is possible. Parameters obtained: cu and fu Typical site problem: immediate bearing capacity of foundations in saturated clay. 2). Consolidated — undrained test*: Free drainage is allowed for (usually) 24 hours under cell pressure only to allow the specimen to consolidate or to become saturated. Drainage is then prevented and pore-pressure readings taken during the application of axial load (i.e. shearing stage). Parameters obtained: c’ and f’ (i.e. referred to effective stress) and ccu and fcu (i.e. referred to total stress) Typical site problem: sudden change in load, after an initial stable period, e.g. rapid drawdown of water behind a dam; or where effective stress analysis is required, e.g. slope stability. 3). Consolidated - drained test: Free drainage is allowed during a consolidation stage and drainage maintained during the axial loading (which is carried out at a slow rate) so that no increase in pore pressure occurs. Parameters obtained: c’d and f’d Typical site problem: long-term slope stability Shear stress (tf) kN/m2 normal stress (sn) kN/m2
- 51. Undrained shear strength of saturated clays In terms of total stress, any saturated soil, which is not allowed to drain, will exhibit no, or very little frictional resistance e.g. clays cu f = 0 undrained strength envelope t = cu Having tested three specimens from the same sample at three different cell pressures, the shear strength parameters may be assessed using a construction known as a Mohr circle diagram. This diagram may be explained by way of the following example. Example 6 (triaxial test) The stress/strain graphs for three specimens taken from a single sample of silty clay are shown below. Calculate the shear strength parameters for the clay. Strain (e) % Deviator Stress (kN/m2) 672 573 425
- 52. 425 573 672 425+100 = 525 573+200 = 773 672+300 = 972 A B C Mohr circle diagram 425 573 672 425+100 = 525 573+200 = 773 672+300 = 972 Shear Stress (τ) kN/m2 Cell Pressure kN/m2 400 300 200 100 0 0 200 400 600 800 1000
- 53. s1=525 s3=100 Sample A Sample B s1=773 s3=200 s1=972 s3=300 Sample C σ1-σ3 Diameter of Circle Radius of Circle = { cu = 130 kN/m2 f = 16o Shear vane test (in-situ test)
- 54. The vane test is a simple in-situ test suitable for use on saturated clay at the bottom of a trial pit or borehole on site: The vane is driven or pushed into the soil and a measured torque (T) applied to it until it rotates. The failure surface is the curved surface plus the flat ends of the ‘cylinder’ of soils whose diameter and height are that of the vane. The torque required to cause failure in a saturated clay is: c where: T = Torque c = Cohesion d = Diameter h = height of vane t f = c + s n tan f Shear stress (t f
- 55. ) kN/m 2 Normal stress (s n ) kN/m 2 f c s n tan f Author/ Editor Craig, RE. Title Craig’s Soil Mechanics Edition 7th Publication Year 2004 Publisher Taylor & Francis ISBN 9780415561266 Recommended reading ✓ Author/ Editor Whitlow, R. Title Basic Soil Mechanics Edition 4th Publication Year 2000 Publisher Prentice Hall ISBN 9780582381094 Recommended reading ✓
- 56. Author/ Editor Vickers, B. Title Laboratory Works in Soil Mechanics (Scanned copy of Chapter 4: ‘Shear-strength Tests’ is available to students on X-stream) Edition 2nd Publication Year 1983 Publisher Granada Recommended reading ü Author/ Editor Vickers, B. Title Laboratory Works in Soil Mechanics (Scanned copy of Chapter 4: ‘Shear-strengthTests’ is available to students on X-stream) Edition 2 nd Publication Year 1983 Publisher Granada Recommended reading ü Author/ Editor Vickers, B. Title Laboratory Works in Soil Mechanics (Scanned copy of Chapter 4: ‘Shear-strength Tests’ is available to students on X-stream) Edition 2nd
- 57. Publication Year 1983 Publisher Granada Recommended reading Author/ Editor Craig, RE. Title Craig’s Soil Mechanics Edition 7th Publication Year 2004 Publisher Taylor & Francis ISBN 9780415561266 Recommended reading ✓ Author/ Editor Whitlow, R. Title Basic Soil Mechanics Edition 4th Publication Year 2000 Publisher Prentice Hall ISBN 9780582381094
- 60. Effective stress Total stress obtained from laboratory or in-situ tests. It is essential that the conditions of the tests are reported: 2. Drained or Undrained with pore pressure measured = where uf = pore pressure at failure A subscript ‘d’ can also be used to further denote that the parameters have been obtained under fully drained conditions (as in the ‘triaxial consolidated drained test’), e.g.: Material f ' peak f ' ult Dense well graded SAND or angular GRAVEL5535 Medium desne uniform SAND4032 Dense slightly clayey SILT4732 Sandy silty CLAY3530 Shaley CLAY3535
- 61. Silty CLAY (London Clay)2115 Sheet1Materialf' peakf' ultDense well graded SAND or angular GRAVEL5535Medium desne uniform SAND4032Dense slightly clayey SILT4732Sandy silty CLAY3530Shaley CLAY3535Silty CLAY (London Clay)2115 Sheet2 Sheet3 Normal load (N)73191309427545 Shear load at failure (N)109139170197227 ( ) Load stress Area s = ( ) 3 3 7310 ' 3.610 n normalstress s - - ´ = ´ ( ) 3 3 10910
- 62. 3.610 f Shearstress t - - ´ = ´ Normal stress (kN/m 2 )20.3 Shear stress at failure (kN/m 2 )30.3 Sheet1Normal load (N)73191309427545Shear load at failure (N)109139170197227Normal stress (kN/m2)20.3Shear stress at failure (kN/m2)30.3 Sheet2 Sheet3 Sheet1Normal load (N)73191309427545Shear load at failure (N)109139170197227 Sheet2 Sheet3 0 10 20 30 40 50 60 70 050100150200 Normal stress (kN/m 2 )
- 63. Shear stress at failure (kN/m 2 ) Normal stress (kN/m 2 )20.353.185.8118.6151.4 Shear stress at failure (kN/m 2 )30.338.647.254.763.1 Chart120.353.185.8118.6151.4 Normal stress (kN/m2) Shear stress at failure (kN/m2) 30.3 38.6 47.2 54.7 63.1 Sheet1Normal load (N)73191309427545Shear load at failure (N)10913917019722720.353.185.8118.6151.430.338.647.254.76 3.1 Sheet1 Normal stress (kN/m2) Shear stress at failure (kN/m2) Sheet2 Sheet3 Sheet1Normal load (N)73191309427545Shear load at failure (N)109139170197227Normal stress (kN/m2)20.353.185.8118.6151.4Shear stress at failure (kN/m2)30.338.647.254.763.1 Sheet2 Sheet3 Loose state
- 64. Horiz. Disp (10-‐2 mm) -‐ L 0 50 100 150 200 250 300 350 400 450 Vert. Disp (10-‐2 mm) -‐ L 0 -‐6 -‐9 -‐11 -‐13 -‐14 -‐14 -‐14 -‐14 -‐14 Shear stress (kN/m2) -‐ L 0 35 57 72 79 84 84 88 89 90 Dense state Horiz. Disp (10-‐2 mm) -‐ D 0 50 100 150 200 250 300 350 400 450 Vert. Disp (10-‐2 mm) -‐ D 0 -‐3 3 11 18 23 26 27 28 28 Shear stress (kN/m2)
- 65. -‐ D 0 67 111 124 120 103 94 88 88 87 Loose state Horiz. Disp (10 -2 mm) - L050100150200250300350400450 Vert.Disp (10 -2 mm) - L0-6-9-11-13-14-14-14-14-14 Shear stress (kN/m 2 ) - L0355772798484888990 Dense state Horiz. Disp (10 -2 mm) - D050100150200250300350400450 Vert.Disp (10 -2 mm) - D0-3311182326272828 Shear stress (kN/m 2 ) - D06711112412010394888887 ( ) 100% o X Strain L e =´ ( ) 13 P
- 66. Deviatorstress A ss æö =- ç÷ èø ( ) ()) oooo VolumeVALALALX ===- ( ) ( ) 1 oo o VA orAorA LX e == -- Test TypePart No Para. 1Undrained triaxial compression test78 & 9 2Consolidated undrained triaxail compression test with pore water pressure measurement87 3Consolidated drained triaxail compression test 88 Sheet1Test TypePart NoPara.1Undrained triaxial compression test78 & 92Consolidated undrained triaxail compression test with pore water pressure measurement873Consolidated drained triaxail compression test88 Sheet2
- 67. Sheet3 Sample Cell Pressure ( s 3) A100 B200 C300 A100 B200 C300 Cell Pressure ( s 3 ) Deviator Stress ( s 1 - s 3 )=P/A Sample Major principal Stress s 1 13 P A ss =+ Sheet1SampleCell Pressure (s3)A100B200C300 Sheet2 Sheet3 Sheet1SampleDeviator Stress (s1-s3)=P/ACell Pressure (s3)Major principal Stress s1A100B200C300 Sheet2
- 68. Sheet3 A100 B200 C300 Cell Pressure ( s 3 ) Deviator Stress ( s 1 - s 3 )=P/A Sample Major principal Stress s 1 13 2 ss - æö ç÷ èø Sheet1SampleDeviator Stress (s1-s3)=P/ACell Pressure (s3)Major principal Stress s1A100B200C300 Sheet2 Sheet3 ( ) ( ) 2 /3
- 69. 2 d TorqueTchd p =+ Test 1 Normal Load (kg) =…………15…………….Normal Load (N) =……147.15… Effective normal stress (kN/m2) =……40.88………. Horizontal Disp. (mm) Vertical Disp. (mm) Proving Ring Reading Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2)
- 73. 217 126.9 35.3 4.4 1.15 213 124.6 34.6 4.6 1.18 207 121.1 33.6 4.8 1.22 199 116.4 32.3 5.0 1.24 191 111.7 31.0 Test 2 Normal Load (kg) =…………25…………….Normal Load (N) =………245.25………
- 74. Effective normal stress (kN/m2)…68.13…. Horizontal Disp. (mm) Vertical Disp. (mm) Proving Ring Reading Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2) 0.0 0.00 0 0 0 0.2 0.00 105 61.4 17.1 0.4 0.00 205 119.9 33.3
- 78. 204.8 56.9 5.0 1.15 339 198.3 55.1 Test 3 Normal Load (kg) =…………35……………. Normal Load (N) =…………343.35……. Effective normal stress (kN/m2) =……95. 38..……. Horizontal Disp. (mm) Vertical Disp. (mm) Proving Ring Reading Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2)
- 82. 4.2 1.01 479 280.2 77.8 4.4 1.05 442 258.6 71.8 4.6 1.08 443 259.2 72.0 4.8 1.12 448 262.1 72.8 5.0 1.15 447 261.5 72.6 Test 4 Normal Load (kg) =………45…………….Normal Load (N) =………441.45…….
- 83. Effective normal stress (kN/m2) =……122.63…………. Horizontal Disp. (mm) Vertical Disp. (mm) Proving Ring Reading Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2) 0.0 0.00 0 0 0 0.2 0.00 159 93.0 25.8 0.4 0.00 273 159.7 44.4 0.6
- 87. 102.7 5.0 1.09 610 356.9 99.1 LEEDS BECKETT UNIVERSITY CIVIL ENGINEERING GEOTECHNICAL ENGINEERING A (BSc) Laboratory Experiment: The direct shear box test BS 1377; Part 7 1990: 4 and 5 Object of Experiment: To determine the angle of shear resistance and volumetric displacement of a sand using the direct shear box test. Theory: The shear strength of a soil may be defined as the maximum shear stress that can be applied to that soil in any direction. When this maximum has been reached the soil yields and is regarded to have failed. It should be
- 88. appreciated that the shear strength of a soil is derived from the frictional resistance, F, generated from inter-particle forces, N; the pore water has no shear strength. The shear strength is a function of the total normal stress to that plane, and is given by: tanf ncτ σ φ= + In the direct shear box the sample is caused to shear along the plane dividing the upper and lower pieces by applying a horizontal load to the upper piece while the lower piece is held in position. The load is generally applied via a proving ring, hence the load causing the sample to shear can be read directly and the shear stress, t, is the load causing shear divided by the plan area of the box. The test is repeated several times on different specimens of the same sample using different normal loads. The results are then be plotted, to give the shear strength envelope, form which a value of φ may be obtained. Apparatus:
- 89. NB mass of hanger = 5kg The apparatus, as shown above, comprises a square box construction in two separate pieces, an upper piece and a lower piece. The vertical normal load is applied directly through the upper pressure plate and it is divided by the plan area of the box to give the normal stress σ. The volumetric behavior of the soil is also determined during the test by measuringthe amount of horizontal displacement and vertical displacement using dial gauges. Method: Assemble the empty shear box as shown, without the upper and loading platens. The two halves of the box should be screwed together with the screws marked by a crosscut into the heads; the two screws marked ‘L’ should be in position but clear of the joint between the two halves. Ensure that the apparatus is moving freely on its runners. Fill the box with loose sand and level it off about 1mm below the top of the box. Place the top platen, on the sand and the loading platen on the top platen. Put the ball bearing in place and the hanger on the ball bearing. Place a weight on the hanger. Adjust the apparatus to take up any slack, and then zero the proving ring dial gauge.
- 90. REMOVE THE TWO SCREWS HOLDING THE UPPER ANDLOWER HALVES OF THE BOX TOGETHER, then screw in those marked ‘L’ until resistance is just felt: give each one a further half turn to ensure that the two halves are slightly separated so that the normal stress is being applied to the sand only,then remove them. Switch on the motor. Record the maximum reading on the proving ring dial gauge, then switch off and slacken off the apparatus. Dismantle the box entirely and pour all the sand back into the container. Repeat the test threemore times, increasingthe mass on the hanger. Record the results on the attached sheet. Results and Calculations: • Calculate the values of effective normal stress (σ’) for each loading condition. • Plot τ against horizontal movementfor each loading condition. • Determine values for shear stress at failure (τf) • Plot a graph of τf against σ’. Draw the straight line of best fit through the origin and
- 91. the points, and calculate the value of the angle of shearing resistance of the soil from the co-ordinates of any convenient pointon the line. • Plot vertical displacement against horizontal displacement. Conclusion: • Comment on the value of φ for the sand you tested. • Explain the main points to the vertical vs horizontal displacement plot. • Comment on the use of this test and your findings. Indicative Reading: Manual of Soil Laboratory Testing Vol.1 – K.H. Head Basic Soil Mechanics – Whitlow, R. Laboratory Work in Soil Mechanics — B. Vickers (Chapter 4: Shear-strength test is available to download from My-Beckett). Name……………...…….……..………….. Group/Date………………………………. Mass of hanger = 5kg Sample =………………………………….. Proving Ring Constant = ……………... Area of shear box = ………………….. Test 1 Test 2 Normal Load (kg) =……………………….Normal Load (N)
- 92. =…………………. Normal Load (kg) =……………………….Normal Load (N) =…………………. Effective normal stress (kN/m2) =………………………..……………. Effective normal stress (kN/m2) =………………………..……………. 0.0 0.0 0.2 0.2 0.4 0.4 0.6 0.6 0.8 0.8 1.0 1.0 1.2 1.2 1.4 1.4 1.6 1.6 1.8 1.8 2.0 2.0 2.2 2.2 2.4 2.4 2.6 2.6 2.8 2.8 3.0 3.0 3.2 3.2 3.4 3.4 3.6 3.6 3.8 3.8 4.0 4.0 4.2 4.2 4.4 4.4 4.6 4.6 4.8 4.8 5.0 5.0 Proving Ring Reading
- 93. Shear Stress = Load x 10-3 / Area (kN/m2) Horizontal Disp. (mm) Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2) Vertical Disp. (mm) Horizontal Disp. (mm) Proving Ring Reading Vertical Disp. (mm) Load = Proving Ring Reading x Constant (N) Name……………...…….……..…………..
- 94. Group/Date………………………………. Mass of hanger = 5kg Sample =………………………………….. Proving Ring Constant = ……………... Area of shear box = ………………….. Test 3 Test 4 Normal Load (kg) =……………………….Normal Load (N) =…………………. Normal Load (kg) =……………………….Normal Load (N) =…………………. Effective normal stress (kN/m2) =………………………..……………. Effective normal stress (kN/m2) =………………………..……………. 0.0 0.0 0.2 0.2 0.4 0.4 0.6 0.6 0.8 0.8 1.0 1.0 1.2 1.2 1.4 1.4 1.6 1.6 1.8 1.8 2.0 2.0 2.2 2.2 2.4 2.4 2.6 2.6 2.8 2.8 3.0 3.0 3.2 3.2 3.4 3.4 3.6 3.6 3.8 3.8 4.0 4.0 4.2 4.2 4.4 4.4 4.6 4.6
- 95. 4.8 4.8 5.0 5.0 Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2) Vertical Disp. (mm) Vertical Disp. (mm) Horizontal Disp. (mm) Horizontal Disp. (mm) Proving Ring Reading Load = Proving Ring Reading x Constant (N) Shear Stress = Load x 10-3 / Area (kN/m2)
- 96. Proving Ring Reading 2b - BSc Shear box lab handout2b Lab sheet for shear box