3.11 Recitation #11 November 25, 2003 If there’s anything you’d like covered, please let me know. Also—please let me know good times to hold a review for the final. I was thinking perhaps Thursday or Friday after classes end. Let me know if this would be appropriate. By email is fine. Today: Review vocabulary for mechanical properties of Materials. Go over stress-strain relationships, plasticity Example Problems, if time. Stress-strain diagrams The relationship between loads and deflection/stress-strain in a structure of a member can be obtained from experimental load-deflection/stress-strain curves 1LPPdL/2dL/2P/2 P/2δPPPPTTTension testBending testCompression testTorsion testShear Test The most common tests are tension test for ductile materials (steel) & compression test for brittle materials (concrete) 2Tension Test AB'CC'The upperstress levelat which thematerialbehaveselasticallyNeckingMaterial canresist more loadincreaseYieldingElasticMaterial will deformpermanently and will n to itsorginal shape upon unloading. Thedeformation that occurs is calledσfσ′uσfσyσPLσ: The upper stresslimit that strainvaries linearly withstress. Materialfollows :Stress at which aslight increase instress will result inappreciably increasin strain withoutincrease in stressNeckingStress-strain using originalarea to calculateTrue Stress-strain usingactual area to calculateterialwill return to its orginalshape if material is loadedand unloaded within thisrangeε :10 - 40 timeselastic strainyε Elastic Limit:Strain Hardening:Plastic Behaviour: NOT returplastic deformationProportional LimitHooke's LawYield Stress Ultimate stressFailure stressElastic Behaviour: MaYield strain Stress-strain diagram for ductile materials Hooke’s Law: Eσε= E is the modulus of elasticity steelE = 200 GPa concreteE = 29 GPa (21 – 29 GPa) 3Ductile Materials Materials that can be subjected to large strains before rupture Have high percent elongation −×ofoLLPercent elongation = 100L Have high percent reduction in area ofoAAPercent reduction in area = 100A−× Have capacity to absorb energy If structure made of ductile materials is overloaded, it will present large deformation before failing Some ductile materials do not exhibit a well-defined yield point, we will use offset method to define a yield strength Some ductile materials do not have linear relationship between stress and strain, we call them nonlinear materials σyσε0.002 or 0.2% offsetσε()fσε= Elastic-plastic Materials Stress-strain for structural steel will consist of elastic and perfectly plastic region. We call this kind of material elastoplastic material Analysis of structures on the basis of elastoplastic diagram is called elastoplastic analysis or plastic analysis 4Bilinear stress-strain diagram having different slopes is sometimes used to approximate the general nonlinear diagrams. This will include the strain hardening. σεyσyεσεLinearly elasticPerfectly plasticLinearly elasticStrain hardeningNonlinear Brittle Materials Materials that do not exhibit yielding before failure Some materials will show both ductile and brittle behaviours, e.g. steel with high carbon content will demonstrate brittle behaviours while steel with low carbon content will be ductile or steel subjects to low temperature will be brittle while those in the high temperature environment will be ductile Creep Deformation which increases with time under constant load (examples: rubber band; concrete bridge deck: sagging between supports due to self weight therefore the deck is constructed with an upward camber) Pδδtoδot In several situations, creep will associate with high temperature If creep becomes important, creep strength will be used in design 5Relaxation Loss of stress with time under constant strain Another manifestation of creep PrestressedwiretCreepstrengthσoσ Cyclic loading and fatigue Fracture after many cycles of loading If material is loaded into the plastic region, upon unloading elastic strain will be recovered but plastic strain remains LoadingUnloadingPermanentsetElasticrecoveryElasticregionσε 6Strain energy Energy stored internally throughout the volume of a material which is deformed by an external load FxoFoxWorkFkx Consider a linear spring having stiffness k If we apply a force , the spring will stretch Fx. The relationship between and Fx is Fkx= If we apply a force from zero to and the spring stretches to the amount of oFox, the work done is the average force magnitude times the displacement, i.e. 12ooWF=x From the conservation of energy, this work done must be equivalent to the internal work or strain energy stored within the spring when it is deformed 7dxdydzσσ If an infinitesimal element of elastic material is subjected to a normal stress σ, then tensile force on the element will be dF dxdyσ= The change in its length is dzε The work done, which equals to the strain energy stored in the element, is ()(12dU dxdy dzσε=) or 12dU dVσε= VThe total strain energy stored in a material will be VUdσε=∫ The strain energy per unit volume or the strain energy density is 12dUudVσε== If the material is linear elastic ( Eσε=, Hooke’s Law holds), the strain energy density will be 21122uEEσσσ== σεPLσPLεru If the stress σ reaches the proportional limit, the strain energy density is called the modulus of resilience u r12rPLuPLσε= 8σεtu The total strain energy density which stored in the material just before it fails is called the modulus of toughness tu Two Example Problems: 9Answer to 1. 10Answer to 2. 11More on crazing and shear deformations and zones next time. (Phenomena in amorphous polymers, as discussed yesterday)