Module 5: Theories Of Failure - VTU Updates VTU Notes
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of MaterialsModule 5: Theories of FailureObjectives:The objectives/outcomes of this lecture on “Theories of Failure” is to enablestudents for1. Recognize loading on Structural Members/Machine elements andallowable stresses.2. Comprehend the Concept of yielding and fracture.3. Comprehend Different theories of failure.4. Draw yield surfaces for failure theories.5. Apply concept of failure theories for simple designs1. Introduction:Failure indicate either fracture or permanent deformation beyond theoperational range due to yielding of a member. In the process of designing amachine element or a structural member, precautions has to be taken to avoidfailure under service conditions.When a member of a structure or a machine element is subjected to a system ofcomplex stress system, prediction of mode of failure is necessary to involve inappropriate design methodology. Theories of failure or also known as failurecriteria are developed to aid design.1.1 Stress-Strain relationships:Following Figure-1 represents stress-strain relationship for different type ofmaterials.Page 1 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of MaterialsDuctile material e.g. low carbon steelLow ductilityBrittle materialElastic – perfectly plastic materialFigure-1: Stress-Strain RelationshipBars of ductile materials subjected to tension show a linear range within whichthe materials exhibit elastic behaviour whereas for brittle materials yield zonecannot be identified. In general, various materials under similar test conditionsreveal different behaviour. The cause of failure of a ductile material need not besame as that of the brittle material.1.2Types of Failure:The two types of failure are,Yielding - This is due to excessive inelastic deformation rendering thestructural member or machine part unsuitable to perform its function. Thismostly occurs in ductile materials.Fracture - In this case, the member or component tears apart in two or moreparts. This mostly occurs in brittle materials.1.3Transformation of plane stress:Page 2 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of MaterialsFor an element subjected to biaxial state of stress the normal stress on aninclined plane is determined as,- Eq-1Similarly, on the same inclined plane the value of the shear stress is determinedas,- Eq-2The above equations (Eq-1 and Eq-2) are used to determine the condition whenthe normal stress and shear stress values are maximum/minimum bydifferentiating them with respect to θ and equating to zero. The substitution ofthe results in these equations determines maximum and minimum normal stressknown as principal stresses and maximum shear stress as indicated by thefollowing expressions (Eq-3 and Eq-4). ( ())- Eq.-3- Eq-41.4 Use of factor of safety in design:In designing a member to carry a given load without failure, usually a factor ofsafety (FS or N) is used. The purpose is to design the member in such a way thatit can carry N times the actual working load without failure. Factor of safety isdefined as Factor of Safety (FS) Ultimate Stress/Allowable Stress.2. Theories of ncipal Stress Theory (Rankine Theory)Principal Strain Theory (St. Venant’s theory)Shear Stress Theory (Tresca theory)Strain Energy Theory (Beltrami’s theory)Maximum Principal Stress Theory (Rankine theory)According to this, if one of the principal stresses σ 1 (maximum principal stress),σ2 (minimum principal stress) or σ3 exceeds the yield stress (σy ), yielding wouldPage 3 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of Materialsoccur. In a two dimensional loading situation for a ductile material wheretensile and compressive yield stress are nearly of same magnitudeσ1 σyσ2 σyYield surface for the situation is, as shown in Figure-2Figure- 2: Yield surface correspondingto maximum principal stress theoryYielding occurs when the state of stress is at the boundary of the rectangle.Consider, for example, the state of stress of a thin walled pressure vessel. Hereσ1 2σ2, σ1 being the circumferential or hoop stress and σ2 the axial stress. Asthe pressure in the vessel increases, the stress follows the dotted line. At a point(say) a, the stresses are still within the elastic limit but at b, σ 1 reaches σyalthough σ2 is still less than σy . Yielding will then begin at point b. This theoryof yielding has very poor agreement with experiment. However, this theory isbeing used successfully for brittle materials.2.2Maximum Principal Strain Theory (St. Venant’s Theory)According to this theory, yielding will occur when the maximum principalstrain just exceeds the strain at the tensile yield point in either simpletension or compression. If ε1 and ε2 are maximum and minimumprincipal strains corresponding to σ1 and σ2, in the limiting caseε1 (1/E)(σ1- νσ2)Page 4 of 13 σ1 σ2 Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of Materialsε2 (1/E)(σ2- νσ1) σ2 σ1 This results in,E ε1 σ1- νσ2 σ0E ε2 σ2- νσ1 σ0The boundary of a yield surface in this case is shown in Figure – 3.Figure-3:Yield surface corresponding tomaximum principal strain theory2.3Maximum Shear Stress Theory (Tresca theory)According to this theory, yielding would occur when the maximum shearstress just exceeds the shear stress at the tensile yield point. At the tensileyield point σ2 σ3 0 and thus maximum shear stress is σy /2. This gives ussix conditions for a three-dimensional stress situation:σ1- σ2 σyσ2- σ3 σyσ3- σ1 σyPage 5 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of MaterialsFigure – 4: Yield surface correspondingto maximum shear stress theoryIn a biaxial stress situation (Figure - 4) case, σ3 0 and this givesσ1σ1σ2σ1σ1σ2 σ2 σ y σ2 σy σy σy σy σyif σ1 0, σ2if σ1 0, σ2if σ2 σ1if σ1 σ2if σ1 σ2if σ2 σ1 0 0 0 0 0 0This criterion agrees well with experiment.In the case of pure shear, σ1 - σ2 k (say), σ3 0and this gives σ1- σ2 2k σyThis indicates that yield stress in pure shear is half the tensile yield stress andthis is also seen in the Mohr’s circle (Figure - 5) for pure shear.Figure – 5: Mohr’s circle forPage 6 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of Materialspure shear2.4Maximum strain energy theory (Beltrami’s theory)According to this theory failure would occur when the total strain energyabsorbed at a point per unit volume exceeds the strain energy absorbed perunit volume at the tensile yield point. This may be expressed as,(1/2)(σ1 ε1 σ2 ε2 σ3 ε3) (1/2) σy εySubstituting ε1, ε2, ε3 and εy in terms of the stresses we haveσ12 σ22 σ32 - 2 υ (σ1 σ2 σ2 σ3 σ3σ1) σy 2(σ1/ σy )2 (σ2/ σy )2 - 2ν(σ1 σ2/ σy 2) 1The above equation represents an ellipse and the yield surface is shown inF igure - 6Figure – 6: Yield surface correspondingto Maximum strain energy theory.It has been shown earlier that only distortion energy can cause yielding but inthe above expression at sufficiently high hydrostatic pressure σ1 σ2 σ3 σ(say), yielding may also occur. From the above we may write σ2(3 2ν) σy 2and if ν 0.3, at stress level lower than yield stress, yielding would occur. Thisis in contrast to the experimental as well as analytical conclusion and thetheory is not appropriate.Page 7 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of Materials2.5 Superposition of yield surfaces of different failure theories:A comparison among the different failure theories can be made by superposingthe yield surfaces as shown in figure – 7. It is clear that an immediateassessment of failure probability can be made just by plotting any experimentalin the combined yield surface. Failure of ductile materials is most accuratelygoverned by the distortion energy theory where as the maximum principal straintheory is used for brittle materials.Figure – 7: Comparison of different failure theoriesNumerical-1:A shaft is loaded by a torque of 5 KN-m. The materialhas a yield point of 350 MPa. Find the required diameter using Maximumshear stress theory. Take a factor of safety of 2.5.Torsional Shear Stress, τ 16T/πd 3, where d represents diameter of the shaftMaximum Shear Stress theory, ()Factor of Safety (FS) Ultimate Stress/Allowable StressSince σx σy 0, τmax 25.46 X 103/d3Therefore 25.46 X 103/d3 σy /(2*FS) 350*106/(2*2.5)Hence, d 71.3 mmPage 8 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of MaterialsNumerical-2:The state of stress at a point for a material is shown inthe following figure Find the factor of safety using (a) Maximum shearstress theory Take the tensile yield strength of the material as 400 MPa.From the Mohr’s circle shown below we determine,σ1 42.38MPa andσ2 -127.38MPafrom Maximum Shear Stress theory(σ1 - σ2)/2 σy /(2*FS)By substitution and calculation factor of safety FS 2.356Numerical-3:A cantilever rod is loaded as shown in the followingfigure. If the tensile yield strength of the material is 300 MPa determine thePage 9 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of Materialsrod diameter using (a) Maximum principal stress theory (b) Maximumshear stress theoryAt the outset it is necessary to identify the mostly stressed element. Torsionalshear stress as well as axial normal stress is the same throughout the length ofthe rod but the bearing stress is largest at the welded end. Now among the fourcorner elements on the rod, the element A is mostly loaded as shown infollowing figureShear stress due to bending VQ/It is also developed but this is neglected due toits small value compared to the other stresses. Substituting values of T, P, F andL, the elemental stresses may be shown as in following figure.Page 10 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of MaterialsThe principal stress for the case is determined by the following equation,By Maximum Principal Stress Theory, Setting, σ1 σy we get d 26.67mmBy maximum shear stress theory by setting (σ1 – σ2)/2 σy /2, we get, d 30.63mmNumerical-4:The state of plane stress shown occurs at a critical pointof a steel machine component. As a result of several tensile tests it has beenfound that the tensile yield strength is σy 250MPa for the grade of steelused. Determine the factor of safety with respect to yield using maximumshearing stress criterion.Construction of the Mohr’s circle determinesσavg ½ (80-40) 20MPaand τm (602 252)1/2 65MPaσa 20 65 85 MPa and σb 20-65 -45 MPaThe corresponding shearing stress at yield is τ y ½ σy ½ (250) 125MPaFactor of safety, FS τ m/ τy 125/65 1.92Page 11 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of MaterialsSummary:Different types of loading and criterion for design of structuralmembers/machine parts subjected to static loading based on different failuretheories have been discussed. Development of yield surface and optimization ofdesign criterion for ductile and brittle materials were illustrated.Assignments:Assignment-1: A Force F 45,000N is necessary to rotate the shaft shownin the following figure at uniform speed. The crank shaft is made of ductile steelwhose elastic limit is 207,000 kPa, both in tension and compression. With E 207 X 106 kPa and ν 0.25, determine the diameter of the shaft usingmaximum shear stress theory, using factor of safety 2. Consider a point on theperiphery at section A for analysis (Answer, d 10.4 cm)Assignment-2: Following figure shows three elements a, b and c subjectedto different states of stress. Which one of these three, do you think will yieldfirst according to i) maximum stress theory, ii) maximum strain theory, and iii)maximum shear stress theory? Assume Poisson’s ratio ν 0.25 [Answer: i) b,ii) a, and iii) c]Page 12 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 2315CV32-Strength of MaterialsAssignment-3: Determine the diameter of a ductile steel bar if the tensileload F is 35,000N and the torsional moment T is 1800N.m. Use factor of safety 1.5. E 207*106kPa and σyp 207,000kPa. Use the maximum shear stresstheory. (Answer: d 4.1cm)Assignment-4: At a pint in a steel member, the state of stress shown inFigure. The tensile elastic limit is 413.7kPa. If the shearing stress at a point is206.85kPa, when yielding starts, what is the tensile stress σ at the pointaccording to maximum shearing stress theory? (Answer: Zero)Reference:1. Ferdinand P. Beer, E Russel Johnston Jr., John T. Dewolf and David F.Mazurek, Mechanics of Materials (SI Units), 5th Edition, Tata McGrawHill Private Limited, New Delhi2. L. S. Srinath, Advanced Mechanics of Solids, McGraw Hill, 20093. NPTEL Lecture Notes, Version 2 ME, IIT -3 lesson1.pdf)Page 13 of 13Dr. C V Srinivasa, Department of Civil EngineeringGlobal Academy of Technology, RR Nagar, Bengaluru-560098svasa@gat.ac.in, 94498 09918
VTU EDUSAT LIVE – Programme # 23 15CV32-Strength of Materials Page 1 of 13 Dr. C V Srinivasa, Department of Civil Engineering Global Academy of Technology, RR Nagar, Bengaluru-560098 svasa@gat.ac.in, 94498 09918 Module 5: Theories of Failure Objectives: The objectives/outcomes of this lecture on “Theories of Failure” is to enable students for
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