Zz Yy Stress & Strain: Zx Zy Yz Xy A Review
zzStress & Strain:A review zx zy yz xx yy yx yy xy xx xz zz1 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Disclaimer before beginning your problem assignment:Pick up and compare any set of textbooks on rock mechanics, soilmechanics or solid mechanics, and you will find that the discussion onMohr Circles, stress-strain analysis, matrix math, etc., either usesdifferent conventions or contains a typo that will throw your calculationsoff. Clockwise is positive, clockwise is negative, mathematical shearstrain, engineering shear strain It all seems rather confusing.But instead of becoming frustrated or condemning the proof-reader of agiven textbook (or these notes), I like to look at it as a good lesson in notrelying 100% on something, especially at the expense of your judgement.The notes that follow come from several sources and I have tried toeliminate the errors when I find them. However, when using these notesto complete your problem assignment, try to also use your judgement as towhether the answer you obtain makes sense. If not, consult a differentsource to double check to see if there was an error.On that note, if you find an error and/or a source that you wouldrecommend as having given you a clearer understanding of a particularcalculation, please let me know.2 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 4331
Understanding StressStress is not familiar: it is a tensor quantity and tensors arenot encountered in everyday life.There is a fundamental difference,difference both conceptually andmathematically, between a tensor and the more familiar quantitiesof scalars and vectors:Scalar: a quantity with magnitude only (e.g. temperature, time, mass).Vector: a quantity with magnitude and direction (e.g. force, velocity,acceleration).Tensor: a quantity with magnitude and direction, and with reference toa plane it is acting across (e.g. stress, strain, permeability).Both mathematical and engineering mistakes are easily made if thiscrucial difference is not recognized and understood.3 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433The Stress TensorThe second-order tensor which we will be examining has:-9 components of which 6 are independent;values which are point properties;values which depend on orientation relative to a set of reference axes;6 of the 9 componentspbecomingg zero at a particularporientation;three principal components;complex data reduction requirements because two or more tensorscannot, in general, be averaged by averaging the respective principalstresses.4 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 4332
Components of StressOn a real or imaginary plane through a material, there can benormal forces and shear forces. These forces create the stresstensor. The normal and shear stress components are the normaland shear forces per unit area.NormalStress ( )ShearStress ( )It should be remembered that a solid can sustain a shear force,whereas a liquid or gas cannot. A liquid or gas contains a pressure,which acts equally in all directions and hence is a scalar quantity.5 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Force and StressThe reason for this is thatit is only the force that isresolved in the first case(i.e. vector), whereas, it isboth the force and the areathat are resolved in thecase of stress (i.e. tensor).In fact, the strict definition of asecond-order tensor is a quantity thatobeys certain transformation laws as theplanes in questionpqare rotated. This iswhy the conceptualization of the stresstensor utilizes the idea of magnitude,direction and “the plane in question”.Hudson & Harrison (1997)6 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 4333
Stress as a Point PropertyBecause the acting forces will varyaccording to the orientation of A withinthe slice, it is most useful to consider thenormall stresst( N/ A) andd theth shearhstress ( S/ A) as the area A becomesvery small, eventually approaching zero.Although there are practical limitations in reducing the size ofthe area to zero, it is important to realize that the stresscomponents are defined in this way as mathematicalquantities, with the result that stress is a point property.7 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Stress Components on an Infinitesimal CubeHudson & Harrison (1997)For convenience, the shear and normal components of stress may beresolved with reference to a given set of axes, usually arectangular Cartesian x-y-z system. In this case, the body can beconsidered to be cut at three orientations corresponding to thevisiblei ibl ffaces off a cube.bTo determine all the stress components, we consider the normaland shear stresses on all three planes of this infinitesimal cube.8 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 4334
Stress Tensor ConventionsThus, we arrive at 9 stress components comprised of 3 normal and6 shear components.Hudson & Harrison (1997)The standard convention for denoting these componentsis that the first subscript refers to the plane on whichthe stress component actsacts, and the second subscriptdenotes the direction in which it acts.For normal stresses, compression is positive. For shearstresses, positive stresses act in positive directions onnegative faces (a negative face is one in which the outwardnormal to the face points in the negative direction).9 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Stress Components on a CubeGeotechnical Engineering-Right-hand systems-Compression positive-Tension negativeHudson & Harrison (1997)10 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 4335
Symmetry in the Stress MatrixAlthough we arrive at 9 stress components inthe stress matrix, we can assume that thebody is in equilibrium. By inspecting theequilibrium of forces at a point in terms ofthese 9 stress components, we can see thatfor there to be a resultant moment of zero,then the shear stresses opposite from oneanother must be equal in magnitude.Thus, by considering momentequilibriumilib iaroundd theh x, y andd zaxes, we find that: xy yx yz zy11 of 79 xz zxErik Eberhardt – UBC Geological EngineeringEOSC 433Symmetry in the Stress MatrixIf we consider the stress matrixagain, we find that it issymmetrical about the leadingdiagonal.It is clear then that the state of stress at a point is definedcompletely by six independent components (3 normal and 3 shear).Remembering back now, it can be noted that a scalar quantity canbe completely specified by 1 value and a vector by 3 values, but atensor requires 6 values.Whatever method is used to specify the stress state,there must be 6 independent pieces of information!!12 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 4336
Principal StressesThe actual values of the 6 stress components inthe stress matrix for a given body subjected toloading will depend on the orientation of thecube in the body itself.If we rotate the cube, it should be possibleto find the directions in which the normalstress components take on maximum andminimum values. It is found that in thesedirections the shear components on all facesof the cube become zero!symmetryThe principal stresses are defined as thosenormal components of stress that act onplanes that have shear stress componentswith zero magnitude!13 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #1Q. Add the following 2-D stressstates, and find the principalstresses and directions of theresultant stress state.A.Hint: Solve the problem graphically using a Mohr’s circle plot.Hudson & Harrison (1997)14 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 4337
Example #1 (Solution)Q. Add the following 2-D stressstates, and find the principalstresses and directions of theresultant stress state.A. Step 1: Draw xy and lm axes for the first stress state, andthen plot the corresponding Mohr circle.15 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #1 (Solution)Q. Add the following 2-D stressstates, and find the principalstresses and directions of theresultant stress state.A. The stresses transformed to the xy axes are then:MPa16 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 4338
Example #1 (Solution)Q. Add the following 2-D stressstates, and find the principalstresses and directions of theresultant stress state.A. Step 2: Draw xy and lm axes for the second stress state,and then plot the corresponding Mohr circle.17 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #1 (Solution)Q. Add the following 2-D stressstates, and find the principalstresses and directions of theresultant stress state.A. The stresses transformed to the xy axes are then:MPa18 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 4339
Example #1 (Solution)Q. Add the following 2-D stressstates, and find the principalstresses and directions of theresultant stress state.A. Step 3: Adding the two xy stress states givesMPa19 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #1 (Solution)Q. Add the following 2-D stressstates, and find the principalstresses and directions of theresultant stress state.A. Step 4: Plotting the Mohr circle for the combined stressstate and reading off the principal stresses and the principaldirections gives the required values 1 37.4 MPa 2 2727.66 MPawith 1 being rotated 19ºclockwise from the xdirection20 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43310
Example #2Q. A stress state has been measured where: 1 15 MPa, plunging 35º towards 085º 2 10 MPa, plunging 43º towards 217º 3 8 MPa,MP plungingli 27º towardstd 335ºFind the 3-D stress tensor in the right-handed x-y-zcoordinate system with x horizontal to the east, yhorizontal to the north and z vertically upwards.A.Perhaps before proceeding with this problem it would help toreview some matrix math.21 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Stress Transformation – Step 1The matrix equation to conduct stress transformation is as follows: where the stress components are assumed known in the x-y-zcoordinate system and are required in another coordinate system lm-n inclined with respect to the first.The term lx is the direction cosine of the angle between the x-axisand l-axis. Physically, it is the projection of a unit vector parallelto l on to the x-axis, with the other terms similarly defined.22 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43311
Stress Transformation – Step 2Expanding this matrix equation inorder to obtain expressions forthe normal component of stress inthe l-direction and shear on thel-face in the m-direction gives:The other four necessary equations (i.e.(i e for y, z, yz and zx) are found using cyclicpermutation of the subscripts in the equationsabove.23 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Stress Transformation – Step 3It is generally most convenient to refer to the orientation of a plane onwhich the components of stress are required using dip direction/dip anglenotation ( , ). The dip direction is measured clockwise bearing fromNorth and the dip angle is measured downwards from the horizontal plane.If we use a right-handed coordinate system with x north, y east andz down, and take n as the normal to the desired plane, then:And the rotationmatrix becomes:Note that right-handed systems are always used for mathematical work. Thereare two obvious choices for a right-handed system of axes: x East, y North andz up; or x North, y East and z down. There are advantages to both, and sobeing adept with both is important.24 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43312
Example #2 (Solution)Q. A stress state has been measured where: 1 15 MPa, plunging 35º towards 085º 2 10 MPa, plunging 43º towards 217º 3 8 MPa, plunging 27º towards 335ºFind the 3-D stress tensor in the right-handed xyz co-ordinatesystem with x, horizontal to the east; y, horizontal to the north;and z, vertically upwards.A. Step 1: The stress transformation equations are given bywhich can be written as lmn R xyzRT25 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #2 (Solution)Q. A stress state has been measured where: 1 15 MPa, plunging 35º towards 085º 2 10 MPa, plunging 43º towards 217º 3 8 MPa, plunging 2727º towards 335335ºFind the 3-D stress tensor (where x east, y north, z up).The equation lmn R xyzRT means that, if we know theA. stresses relative to the xyz axes (i.e. xyz) and theorientation of the lmn axes relative to the xyz axes (i.e. R),we can then compute the stresses relative to the lmn axes(i.e. lmn).However, in this problem we have been given the principalstresses, which is a stress state relative to some lmn system(i.e. lmn), where the lmn axes correspond to the principaldirections.26 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43313
Example #2 (Solution)Q. A stress state has been measured where: 1 15 MPa, plunging 35º towards 085º 2 10 MPa, plunging 43º towards 217º 3 8 MPa, plunging 2727º towards 335335ºFind the 3-D stress tensor (where x east, y north, z up).A. As we know the principal directions relative to the xyz axes,we are able to compute R. Thus, we need to evaluate lxyz,and we do this using the inverse of the stress transformationequations: xyz RT lmnRNotice that since the rotation matrix is orthogonal, we do notneed to use the inverse of R, i.e. R-1, and thus R-1 RT.27 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #2 (Solution)Q. A stress state has been measured where: 1 15 MPa, plunging 35º towards 085º 2 10 MPa, plunging 43º towards 217º 3 8 MPa, plunging 2727º towards 335335ºFind the 3-D stress tensor.A. Step 2: With the given data forthe principal directions the matrix R is computed as:28 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43314
Example #2 (Solution)Q. A stress state has been measured where: 1 15 MPa, plunging 35º towards 085º 2 10 MPa, plunging 43º towards 217º 3 8 MPa, plunging 2727º towards 335335ºFind the 3-D stress tensor (where x east, y north, z up).A. Step 3: Since the matrix lmn is given by the 3-D stress tensor, xyz RT lmnR solves as:29 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #2 (Solution)Q. A stress state has been measured where: 1 15 MPa, plunging 35º towards 085º 2 10 MPa, plunging 43º towards 217º 3 8 MPa, plunging 2727º towards 335335ºFind the 3-D stress tensor (where x east, y north, z up).A. or: if orientation and matrix values are not rounded, amore accurate answer may be obtained the 3-D stress tensor,tensor xyz RT lmnR then solvesmore exactly as:30 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43315
Example #3Q. For our previously given rock mass with the stress state: 1 15 MPa, plunging 35º towards 085º 2 10 MPa, plunging 43º towards 217º 3 8 MPa, plunging 2727º towards 335335º a fault has been mapped with an orientation of 295º/50º.Determine the stress components in a local coordinate systemaligned with the fault. Assume for this question that thepresence of the fault does not affect the stress field.A. Hint: Here we use the same methodology to find the 3-Dstress tensor in an lmn coordinate system where the n-axiscoincides with the normal to the fault and the l-axiscoincides with the strike of the fault.31 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #3 (Solution)Q. Determine the stress components in a local coordinate systemaligned with the fault. 1 2 3 fault15 MPa, plunging 35º towards 085º10 MPa, plunging 43º towards 217º8 MPa, plunging 27º towards 335ºorientation 295º/50º.A. Step 1: We therefore need to determine lmn, where lmn aregiven by the orientation of the fault. With the l-axis parallelto the strike of the plane and the n-axis normal to theplane, the m-axis becomes the dip. The tend and plunge ofeach axes are then as followsfollows:32 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43316
Example #3 (Solution)Q. Determine the stress components in a local coordinate systemaligned with the fault. 1 2 3 fault15 MPa, plunging 35º towards 085º10 MPa, plunging 43º towards 217º8 MPa, plunging 27º towards 335ºorientation 295º/50º.A. Step 2: The matrix R, then computes as:33 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #3 (Solution)Q. Determine the stress components in a local coordinate systemaligned with the fault. 1 2 3 fault15 MPa, plunging 35º towards 085º10 MPa, plunging 43º towards 217º8 MPa, plunging 27º towards 335ºorientation 295º/50º.A. Step 3: From Q #2, the matrix xyz is:Step 4: As a result, thematrix lmn R xyzRT, thesolution to which is:34 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43317
Example #3 (Solution)Q. Determine the stress components in a local coordinate systemaligned with the fault. 1 2 3 fault15 MPa, plunging 35º towards 085º10 MPa, plunging 43º towards 217º8 MPa, plunging 27º towards 335ºorientation 295º/50º.A. or: if orientation and matrix values are not rounded, amore accurate answer may be obtained. the 3-D stresstensor, lmn R xyzRT then solves more exactly as:35 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #4Q. Six components of stress are measured at a point: xx 14.0 MPa yy 34.8 MPa zz 16.116 1 MPMPa xy -0.6 MPa yz 6.0 MPa xz -2.12 1 MPaMPDetermine the principal stresses and their direction cosines.A.Before proceeding with this problem, we must define theinvariants of stress.36 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43318
Stress Invariants – Step 1When the stress tensor is expressed with reference to sets ofaxes oriented in different directions, the components of the tensorchange. However, certain functions of the components do notchange.hTheseThare knownkas stresstiinvariants,i t expressedd as I1, I2,I3, where:The expression for the first invariant,invariant I1, indicates that for agiven stress state, whatever its orientation, the values of thethree normal stresses will add up to the same value I1.37 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Stress Invariants – Step 2When the principal stresses have to be calculated from thecomponents of the stress tensor, a cubic equation can be used forfinding the three values 1, 2, 3:Because the values of the principal stresses must be independent ofthe choice of axes, the coefficients I1, I2, I3 must be invariantwith respect to the orientation of the axes. It can also be notedfrom the first invariant that:I1 xx yy zz 1 2 338 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43319
Stress Invariants – Step 3Each principal stress is related to a principal stress axis, whosedirection cosines can be obtained, for example for 1, through aset of simultaneous, homogeneous equations in x1, y1 , z1, basedon the dot product theorem of vector analysis:Where:Substituting for x1, y1 , z1in the dot product relationfor any unit vector gives:Brady & Brown (1993)39 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Stress Invariants – Step 4Proceeding in a similar way, the vectors of direction cosines for theintermediate and minor principal stresses axes, i.e. ( x2, y2 , z2)and ( x3, y3 , z3) are obtained by repeating the calculations butsubstituting 2 and 3.Where:Where:40 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 43320
Stress Invariants – Step 5The procedure for calculating the principal stresses and theorientations of the principal stress axes is simply the determinationof the eigenvalues of the stress matrix, and the eigenvector foreach eigenvalue. Thus, some simple checks can be performed toassess the correctness of the solution:Invariance of the sum of thenormal stresses requires that:The condition of orthogonality requires thateach of the three dot products of the vectorsof the direction cosines must vanish:41 of 79Erik Eberhardt – UBC Geological EngineeringEOSC 433Example #4 (Solution)Q. Six components of stress are measured
Pick up and compare any set of textbooks on rock mechanics, soil mechanics or solid mechanics, and you will find that the discussion on Mohr Circles, stress-strain analysis, matrix math, etc., either uses different conventions or contains a typo that will throw your calculations off. Clockwise is positive, clockwise is negative, mathematical shear
1. Stress-Strain Data 10 2. Mohr Coulomb Strength Criteria and 11 Stress Paths 3. Effect of Different Stress Paths 13 4. Stress-Strain Data for Different Stress 1, Paths and the Hyperbolic Stress-Strain Relationship 5. Water Content versus Log Stress 16 6. Review 17 B. CIU Tests 18 1. Stress-Strain Data 18 2.
The incremental creep strain, especially the comparatively large transient creep strain in concrete and some steels [8] is included in the stress-strain relationship. The current total mechanical strain, E i , is obtained from the total mechanical strain at last time step, E i - 1, and the incremental stress related strain, A E i :
For measuring the strain in three different directions strain rosettes are used. Strain rosettes are three strain gages positioned in a rosette-like layout. Therefore by measuring three linearly independent strain in three direction, the components of the
Strain is the amount of deformation of a body due to an applied force. More specifically, strain (e) is defined as the fractional change in length, as shown in Figure 1. Figure 1. Definition of Strain Strain can be positive (tensile) or negative (compressive). Although dimensionless, strain
Given: a strain gage subjected to in-plane strains referenced to the a-t coordinate system: εa, εt, γat Define three types of strain gage strain sensitivities Sa axial strain sensitivity ( R/R)/ εa St transverse strain sensitivity ( R/R)/ εt Ss shear strain sensitivity ( R/R)/ γat
2.5 State of Strain at a Point 73 2.6 Engineering Materials 80 2.7 Stress–Strain Diagrams 82 2.8 Elastic versus Plastic Behavior 86 2.9 Hooke’s Law and Poisson’s Ratio 88 2.10 Generalized Hooke’s Law 91 2.11 Hooke’s Law for Orthotropic Materials 94 2.12 Measurement of Strain: Strain Rosette 97 2.13 Strain Energy 101 2.14 Strain Energy in Common Structural Members 104
3.6.2 Symptom: Strain gage fails to give a reading 15 4 GENERAL CONSIDERATIONS 16 4.1 Conversion of reading to strain changes 16 4.2 Stress strain relationship 16 4.3 Positioning of weldable strain gages 16 4.3.1 Pressure shaft 17 4.3.2 Steel pile or strut 17 4.3.3 Wye section 19 4.3.4 I-beams 20 4.3.5 Strain gages mounted on flanges 21
CHAPTER 4 RESULT AND DISCUSSION 25 4.1 Introduction 25 4.2 Mechanical Properties 25 4.2.1 Tensile Test 25 4.2.2 Engineering Stress and Engineering Strain 31 4.2.3 True Stress and True Strain 33 4.2.4 True Stress Strain and Engineering Stress Strain 35 4
