With high enough plate forces in opposite directions Bolts. How do these fail? Each pin has sheared into two pieces.

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Brittany Harper
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1 SHEAR STRENGTH In general, the shear strength of any material is the load per unit area or pressure that it can withstand before undergoing shearing failure.

2 Shearing When you Pins hear can Shear be used Failure to fasten you With high enough plate forces in probably together think two of steel Shearing plates: Pins or opposite directions Bolts. How do these fail? Each pin has sheared into two pieces.

3 The failure plane for metals will be parallel to the external shear forces. The If the internal shear shear force stress, causes τ is failure, simply then shear the shear force, stress T acting that on the results, failure plane τ f is the divided shear by strength the area, of A of the the failure material. plane: T So shear forces are those that tend τ = A to cause shear failure. Area, A

4 Since The Shear the external Force that force acts is on acting the parallel failure plane to the is failure resisted plane, by the internal strength strength of the of the material. material is thought of as its internal friction, F. This is the material s reaction to the external shear force, T.

Vector addition gives the resultant The to that and overcome tangent must thereby be of the applied cause the friction friction the to an object force, object angle, F to on of vector, is the the

5 Vector addition gives the resultant The to that and overcome tangent must thereby be of the applied cause the friction friction the to an object force, object angle, F to on of vector, is the the plane R which ratio known of where F move. acts to weight, the at W which object an angle W is rests of φ also The WRT object s the normal weight to vector, the plane. W acts known as the coefficient of friction. Friction normal problems to the failure in mechanics plane. determine the external force, T φ W W

6 SHEAR STRENGTH Consider If the and soil element is loaded of (yet soil sober): within a Soil Surface IN large SOILS soil mass: The loading of a material that undergoes shear failure is not always Soil Mass parallel to the failure plane. Soil Element Bedrock

7 σ 1 For This visual the simplicity major principal we replace stress the The load transmits stress to the distributed distribution, load designated with an equivalent σ element by inter-particle contacts. 1 due to point the load.

σ 1 σ 2 σ 3 σ 3 σ 2 σ 1 The Since The soil element we below assume squeezed the the element soil vertically is isotropic, will react will with tend the a confining to stress bulge horizontally of

8 σ 1 σ 2 σ 3 σ 3 σ 2 σ 1 The Since The soil element we below assume squeezed the the element soil vertically is isotropic, will react will with tend the a confining to stress bulge horizontally of lateral equal pressure magnitude to which will be the but the soil directed same reacts in with all upwards directions confining so and pressures it too so σis 2 = σ 23 and allowing σ 3 the us designated to other view principal it σ2 1. dimensions. directions.

9 The But friction what has force this on got this to do failure with For this to happen, a failure plane plane portion is overcome moves relative STRENGTH? develops SHEAR within FAILURE! by the the to the external soil. rest. forces and viola: Soil undergoes shear failure when one σ 1 Θ 2-D σ 1

10 1. Direct Shear angle Shear Stress of internal Test There are 3 basic laboratory at failure, friction, τtests f, φ is characterizes the 2. that pressure Triaxial can be Compression required the performed shear to strength overcome Test on soil of the the 3. samples friction Unconfined soil and to on evaluate is the one Compression surface of the its shear of Test the failure plane (a.k.a. Shear Strength). strength parameters. parameters: σ f φ Θ τ f R f

11 DIRECT SHEAR TEST Can be performed on all types of soil, moist or dry. Measures shear stress at failure on failure plane for various normal stresses. Failure plane is controlled (parallel to direction of applied load).

12 DIRECT SHEAR TEST Then This A The normal A shear prepared the forces and horizontal top a (90 a box normal failure top and a soil to has extension force base the load sample three to horizontal) are occur piston increased parts: pushed placed on a horizontal until load in opposite the is applied in sample plane the directions between to box. shears the soil. in the two: top and base: The procedure is repeated two more times using successively heavier normal loads.

13 DIRECT SHEAR TEST In This the means CV504 labs, the the failure inside plane dimensions has an of the shear box are 60 mm by 60 mm. area of 3600 mm 2. The shear force at failure (maximum) and normal load, both in Newtons are divided by this plane area to find the shear stress at failure and the normal stress in MPa. The For shear this reason, force required the soil s to shear The shear strength of the soil the strength therefore sample increases is is characterized not constant in proportion but by changes with the confining pressure. shear strength to the normal parameters: load. (c,φ).

14 DIRECT SHEAR TEST Fitting The Plotting τ axis equation a the intercept best shear of fit is Coulomb s the stress line apparent through versus failure cohesion, normal these envelope: The slope angle of this line is c of points: stress: angle the soil. τ f = c + σ ntanφ. we have an estimate of Coulomb s failure envelope of internal friction, φ of the soil. First Test Second Test Third Test Shear Stress, τ (kpa) φ τ f τ f τ f c Normal Stress, σ n (kpa)

15 TRIAXIAL COMPRESSION TEST Can be performed on all types of soil, moist or dry and can consolidate sample to in situ conditions by tracking pore water pressures. Measures vertical stress applied to soil sample and confining pressure. Shear stress on failure plane must be calculated from principal stresses.

16 TRIAXIAL COMPRESSION TEST Cylindrical The Specimens specimen specimens are is then weighed are placed prepared and in a and dimensions then plexiglas from inserted sampled measured into chamber. a latex soil. first. sleeve. The Preparation specimen varies is mounted with material between properties 2 platens (clay vs sand vs cohesive granular). diameter length

TRIAXIAL COMPRESSION TEST For Then Once an specimen assembly a the undrained the cell chamber is filled is is mounted test then with test, the is water, mounted on placed the drain the the drain air

17 TRIAXIAL COMPRESSION TEST For Then Once an specimen assembly a the undrained the cell chamber is filled is is mounted test then with test, the is water, mounted on placed the drain the the drain air pedestal valve on release Water on the valve the is valve is is forced closed and into cell cell pressure with the is increased supply opened valve compression of base open to the the and and chamber as desired well pore is locked closed. as testing value base the water for into air as the release machine. shown. collected. place. test. valve. loading ram air release valve plexiglas chamber water supply for cell (confining) pressure loading cap drainage latex sleeve or pore water specimen pressure measurement porous disc pedestal

TRIAXIAL COMPRESSION TEST Then a Major vertical Principal axial load Stress, is applied σ 1, is to the the The But The cell how effect goal pressure, is to of simulate the σ cell the pressure

18 TRIAXIAL COMPRESSION TEST Then a Major vertical Principal axial load Stress, is applied σ 1, is to the the The But The cell how effect goal pressure, is to of simulate the σ cell the pressure stresses on loading ram creating 3, is also known as combination Enter can Christian we find τ of the deviator Otto f and compressive Mohr: σ stress f from σ and 1 confining the the specimen Minor the specimen and is Principal illustrated σ in Stress. the ground. below: 3? stresses or cell the pressure: deviator stress σ : σ σ 3 σ 3 σ = Δσ+ σ 1 3 Plan View of Specimen σ 3 σ Side View of Specimen Source: commons.wikimedia.org

19 TRIAXIAL COMPRESSION TEST Herr Mohr for was any born material, in Germany on the internal shear and was and a renowned normal Civil Engineer and professor until In stresses other acting words, on ANY he plane his death on within the material, discovered MOHR S While contemplating the symmetry of caused by external stresses or loads his name, Otto started tinkering CIRCLE. with can the be determined properties of using the a circle trigonometric when he discovered transformation that... of the external stresses.

20 TRIAXIAL COMPRESSION TEST The During Ultimately, Remember then point If fit then you of one test, the tangency plot you ve circle this test plot σ 1 got and circle through of ends of a σshear 3 the Mohr s starts on when these circle Stress as shear circle! σ n one points and axis versus point failure at σoccurs 3 envelope and then and defines grows the Normal circle to the Stress? has shear right become as strength, axial tangent stress, τ f σ increases to and the normal but failure σ 3 stress, remains envelope. σconstant. f. Shear Stress, τ (kpa) τ f φ c σ 3 σ σ σ σ 1 σ 1 σ f σ 1 Normal Stress, σ n (kpa)

21 TRIAXIAL COMPRESSION TEST A As If This Geometrically, But third with one line how means test most cannot can do we that you lab find you yet be you measurements, be drawn tangent another need need one failure at to of least cell perform them envelope pressure the to two isn t ideal all three circles the bogus? from would test (one circles, a best fit is made as long as one circle in a line at help order least tangent to confirm triaxial to twice define to on all compression the three a line validity same circles) tangent material of test? the is to difficult is not out to lunch compared to the others. both. failure but at different to envelope. achieve. cell pressures. Shear Stress, τ (kpa) φ c Normal Stress, σ n (kpa)

22 TRIAXIAL COMPRESSION TEST Once then So Instead Remember The the for we Centre radius have each of doing the of the test, deviator the this shear the Mohr s graphically, shear strength stress, Circle, strength, σ R parameters, we C is = is half can σthen: 1 τ- use f σ and 3, trigonometry to which φ and normal is the c defining stress, diameter find equations the σ f of failure can or: for τ the be Mohr s found. envelope, Circle. f and σ f using the angle of the failure plane, Θ and the values of σ 1 and σ 3 φshear Stress, τ (kpa) c σ1 σ3 R= specimen failure plane C 2 σ 3 Θ R σ1 σ3 C= + 2 R σ 1 Θ Normal Stress, σ n (kpa)

23 TRIAXIAL COMPRESSION TEST s ABC To EBC follow = & 90 BCF the so are ACB trig both we isosceles = label 90 - φ the and & EBF vertices: DBC is 90. = φ EFB = 90 DCB Θ & BCF = 180 =180 2Θ = 2(90- - φ Θ) = 2Θ A φshear Stress, τ (kpa) τ f c Rearranging: θ= E Θ σ 3 B D σ f φ 2Θ 45 o φ 2 C σ 1 F σ1 σ 2 σ C= R= σ 3 Θ Normal Stress, σ n (kpa)

24 TRIAXIAL COMPRESSION TEST Also In So DBC, in knowing DBC, side side BD φ you is DC the can = Rcos(180-2Θ) same find Θ as and τ f. using σ Θ and the σ 1 & σ 3 values for each trial, f = C Rcos(180-2Θ) ( o ) or 1 τ = Rsin 2θ = ( σc + Rcos(2Θ) 1 σ )sin2θ f 3 τ f and σ f can be found 2 for each trial. 1 τ = ( σ σ ) 1 3 sin2θ (Eqn. 4.3) f 2 A φshear Stress, τ (kpa) τ f c σ f = 1 2 ( σ + σ ) + ( σ σ ) cos2θ (Eqn. 4.4) E 1 Θ σ 3 3 B D σ f φ 2Θ 3 C σ 1 F o φ θ= 45 2 σ1 σ = 2 1+ σ C= 2 R 3 σ 3 Θ Normal Stress, σ n (kpa)

25 TRIAXIAL COMPRESSION TEST And, As What happens when the pore water is not Therefore, Typically, the external pressure deviator increases, stress the failure internal is The One the pore normal final allowed fairly apparent the water constant stress, word failure to pressure drain cohesion, σon envelope for (UNDRAINED (acting each c different will is in be typically a TEST)? the opposite horizonal line and φ cell same u for each trial and f for nomenclature All stress symbols each trial equal to direction to the external) pressure. increases u = used will then the 0. shear in All stress symbols be σ to match (and trivialize) strength, 3 + cused in DRAINED UNDRAINED u the τtests are not tests are usually primed σ f effect. 1,σ 3,σ f primed σ (The and τ f radii indicating 1,σ 3,σ f and τ all that f indicating the same) they are in that they are in terms of TOTAL terms of EFFECTIVE STRESS STRESS and the shear strength and the shear strength parameters are denoted (φ parameters are denoted (φ,c ). u,c u ) Shear Stress, τ (kpa) c u = τ f φ u 0 σ f σ f σ f Normal Stress, σ n (kpa)

26 UNCONFINED COMPRESSION TEST Is performed mainly on cylindrical, moist clay specimens sampled from bore holes. Measures vertical stress applied to soil sample with no confining pressure. Shear stress on failure plane is determined similarly to undrained triaxial compression test.

27 UNCONFINED COMPRESSION TEST Instead The If The the Mohr s Because This axial of calling circle load it continues starts the deviator 0 and grow stress, increases until σ, failure it is The If q steadily called occurs u a point is does q u analogous σdoes 3 maximize as either of = 0 tangency not and maximize q uf before the is before 15% strain unconfined is the reached when triaxial the compressive then specimen s of the circle the 15% diameter circle becoming strain, compression the stress, q shear and of then test. q u. strength the tangent maximum to is reached qfailure u value or 15% envelope is used strain. u at as when 15% failure circle, q uf. strain envelope the shear shear is used defines strength, failure to occurs. define the the shear τ f and normal unconfined strength, stress τcompressive f and at failure, normal strength stress, σ f are both σ f of. estimated to be half of q the specimen, q uf. uf Shear Stress, τ (kpa) c = τ f q u q u q u σ f q u q u q u q uf Normal Stress, σ n (kpa)

Soil strength. the strength depends on the applied stress. water pressures are required

Soil strength. the strength depends on the applied stress. water pressures are required Soil Strength Soil strength u Soils are essentially frictional materials the strength depends on the applied stress u Strength is controlled by effective stresses water pressures are required u Soil strength