Presentation on theme: "Normal Strain and Stress"— Presentation transcript:
1 Normal Strain and Stress Normal Strain and Stress, Stress strain diagram, Hooke’s Law
2 StrainWhen a body is subjected to load, it will deform and can be detected through the changes in length and the changes of angles between them.The deformation is measured through experiment and it is called as strain.The important of strain: it will be related to stress in the later chapter
3 Normal Strain e (epsilon) Normal strain is detected by the changes in length.e (epsilon)l’: length after deformedl: original length.Note e:dimensionlessvery small (normally is mm (=10-6 m))480(10)-6 m/m = 480 mm/m = 480 “micros” = %
4 Example 1When load P is applied, the RIGID lever arm rotates by 0.05o. Calculate the normal strain of wire BDFoundation: DL/LKnowledge required: geometrical equationRigid: no deformation on the lever
6 Example 1 When force P is applied to the rigid lever arm LBD after deformed is DB’Cosine rule can be applied hereStrain:
7 Example 1 When force P is applied to the rigid lever arm LBD after deformed is DB’Cosine rule can be applied hereStrain:
8 Example 2The force applied to the handle of the rigid lever the arm to rotate clockwise through an angle of 3o about pin A. Determine the average normal strain developed in the wire. Originally, the wire is unstretched.Discuss the approach?
10 Simple Tensile TestStrength of a material can only be determined by experimentThe test used by engineers is the tension or compression testThis test is used primarily to determine the relationship between the average normal stress and average normal strain in common engineering materials, such as metals, ceramics, polymers and composites
11 Conventional Stress–Strain Diagram Nominal or engineering stress is obtained by dividing the applied load P by the specimen’s original cross-sectional area.Nominal or engineering strain is obtained by dividing the change in the specimen’s gauge length by the specimen’s original gauge length.
13 Conventional Stress–Strain Diagram Elastic BehaviourA straight lineStress is proportional to strain, i.e., linearly elasticUpper stress limit, or proportional limit; σplIf load is removed upon reaching elastic limit, specimen will return to its original shapeYieldingMaterial deforms permanently; yielding; plastic deformationYield stress, σYOnce yield point reached, specimen continues to elongate (strain) without any increase in loadNote figure not drawn to scale, otherwise induced strains is times larger than in elastic limitMaterial is referred to as being perfectly plastic
14 Conventional Stress–Strain Diagram Strain Hardening.Ultimate stress, σuWhile specimen is elongating, its x-sectional area will decreaseDecrease in area is fairly uniform over entire gauge lengthNeckingAt ultimate stress, cross-sectional area begins to decrease in a localized region of the specimen.Specimen breaks at the fracture stress.
15 Stress–Strain Behavior of Ductile and Brittle Materials Ductile MaterialsMaterial that can subjected to large strains before it ruptures is called a ductile material.Brittle MaterialsMaterials that exhibit little or no yielding before failure are referred to as brittle materials.
16 Stress–Strain Behavior of Ductile and Brittle Materials Yield Strength0.02% strain for ductile materialStrain hardeningWhen ductile material is loaded into the plastic region and then unloaded, elastic strain is recovered.The plastic strain remains and material is subjected to a permanent set.
17 Hooke’s LawHooke’s Law defines the linear relationship between stress and strain within the elastic region.E can be used only if a material has linear–elastic behaviour.σ = stressE = modulus of elasticity or Young’s modulusε = strainE can be derived from stress and strain graph.What is it?
18 Strain Energy Modulus of Resilience When material is deformed by external loading, it will store energy internally throughout its volume.Energy is related to the strains called strain energy.Modulus of ResilienceWhen stress reaches the proportional limit, the strain-energy density is the modulus of resilience, ur:
19 ExampleThe stress–strain diagram for an aluminum alloy that is used for making aircraft parts is shown. When material is stressed to 600 MPa, find the permanent strain that remains in the specimen when load is released. Also, compute the modulus of resilience both before and after the load application.Approach to the problem:Parallel to elastic lineBoth slope is equalDistance CD can be calculated based on the slopePermanent strain: – distance CD
20 SolutionWhen the specimen is subjected to the load, the strain is approximately mm/mm.The slope of line OA is the modulus of elasticity,From triangle CBD,
21 Solution:This strain represents the amount of recovered elastic strain.The permanent strain isComputing the modulus of resilience,Note that the SI system of units is measured in joules, where 1 J = 1 Nm
22 Modulus of ToughnessModulus of toughness, ut, represents the entire area under the stress–strain diagram.It indicates the strain-energy density of the material just before it fractures.
23 ExampleThe bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. If the weight causes B to be displaced downward 0.625mm, determine the strain in wires DE and BC. Also if the wires are made of A-36 steel and have a cross-sectional area of 1.25 mm2, determine the weight W.Discuss the approach????
24 1) Calculate the displacement of D. 2) Based on displacement on D, calculate the strain and normal stress* strain in mm/mm, stress and E in MPa, F in N and length in mm
25 3) Based on normal stress at wire DE, calculate the T of wire D 4) Calculate W, based on FBD of bar DA5) Calculate normal stress of wire CB and strain of wire CBStrain can not be calculated as normal stress goes beyond yield stress (Sy = 250 MPa), elastic property is no more applied. Therefore it requires the stress and strain curve to predict the strain
26 Poisson’s Ration (nu), states that in the elastic range, the ratio of these strains is a constant since the deformations are proportional.Negative sign since longitudinal elongation (positive strain) causes lateral contraction (negative strain), and vice versa.Poisson’s ratio is dimensionless.Typical values are 1/3 or 1/4.
27 ExampleA bar made of A-36 steel has the dimensions shown. If an axial force of P is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. The material behaves elastically.
28 Discuss the approach Approach: Property A-36: E , n 1. s = P/A 2. ez = s / E3. DLz = L * ez4. ex = ey = -n ez5. DLx = L * exDLy = L * ey
29 Solution 1) The normal stress in the bar : 4) The contraction strains in both the x and y directions are2) From the table for A-36 steel, Est = 200 GPa5) The changes in the dimensions of the cross section are3) The axial elongation of the bar is therefore