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Stress and Strain of Aluminum - Lab Report Example

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The report "Stress and Strain of Aluminum" focuses on the laboratory analysis of the stress and strain of the structure of aluminum molecules. From the graph, the stress of a material decreases gradually with an increase in strain, this means the material becomes nonlinear as stress is increased…
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Stress and Strain of Aluminum
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stress and strain Case Study November 02, Memos Prof. November 02 a) the graph of stress (Pa x1010) against strain From the graph stress of a material decreases gradually with increase in strain this means the material becomes non linear as stress is increased. This may be associated with rearrangement of molecules with the structure of aluminum. As strain is increased atoms move to new positions to create new equilibrium in the material. This is because of their nature in mobility and dislocation. The graph is linear to a certain point but it deviates when the limit is reached. A fracture may be experienced if a strain continues beyond the proportionate limit. At zero the graph is starting to form linearity, however it reaches at 100 when it starts to decrease which can be associated with the proportionate limit. b) The graph of stress against strain reduced in a range just larger than the original portion. 2. a) Straine) is the fractional length change of a stretched material, while stress (σe) is the force per unit area of the stretched material. Therefore, deformation is a change in the size or shape of the object. Strain=  Stress =  and has SI units which are the same as those of pressure N/m2 or Pa . Where A is the initial cross-sectional area, Lo is the initial gauge length , and L is the change in gauge length. According to Hooke’s law, the deformation is proportional to the deforming forces as long as they are not too large. F= k L where k is constant and it depends on the length and cross sectional area of the object. So Hooke’s law written in stress will be  = And length change is ( L) is proportional to the magnitude of the deforming forces, Y depends on the inherent stiffness of the material from which the object is composed. k = Y , therefore, Y is the constant of proportionality called Young’s modulus which will be given by the slope of the stress-strain curve. Young’s modulus or elastic modulus has the same units as those of stress (Pa or N/M2) and can be thought of as the inherent stiffness of a material because it measures the resistance of the material to elongation or compression. So, materials that stretch easily and are flexible such as rubber have low Young’s modulus. While materials that are stiff such as steel have high Young’s modulus; it takes a lager stress to produce the same strain. From data young’s modulus is calculated as change in y-axis divided by change in x-axis Y (slope) = = = 2.117610 Young’s modulus (E) from the data is 2.117610Pa b) Yield stress is the stress which is required to deform the material it is at that point when a permanent deformation takes place. It is usually at 0.2%; in this case of aluminum yield stress begins at 0.4%. At the point there is intersection between strain and yield stress and strain is called off-set stress. As strain is increased, many materials eventually deviate from this linear proportionality, the point of departure being termed the proportional limit. This nonlinearity is usually associated with stress-induced “plastic” flow in the specimen. Here the material is undergoing a rearrangement of its internal molecular or microscopic structure, in which atoms are being moved to new equilibrium positions. This plasticity requires a mechanism for molecular mobility, which in crystalline materials can arise from dislocation motion. Materials lacking this mobility, for instance by having internal microstructures that block dislocation motion, are usually brittle rather than ductile. The stress-strain curve for brittle materials are typically linear over their full range of strain, eventually terminating in fracture without appreciable plastic flow. c) Ultimate stress/ strength is the maximum stress that can be withstood without breaking. It is the stress which is called true stress it is calculated as  = σu - σ0.2 The stress at the ultimate strain is calculated as shown below σt= σu (l+e) where σt= 0.2, e=11918.55 σt= σu (l+e) Therefore, σu =  =1.678 x10 -5 Elasticity is the property of complete and immediate recovery from an imposed displacement on release of the load, and the elastic limit is the value of stress at which the material experiences a permanent residual strain that is not lost on unloading. The residual strain induced by a given stress can be determined by drawing an unloading line from the highest point reached on the see curve at that stress back to the strain axis, drawn with a slope equal to that of the initial elastic loading line. This is done because the material unloads elastically, there being no force driving the molecular structure back to its original position. A closely related term is the yield stress, denoted σY in these modules; this is the stress needed to induce plastic deformation in the specimen. Since it is often midcult to pinpoint the exact stress at which plastic deformation begins, the yield stress is often taken to be the stress needed to induce a specimen amount of permanent strain, typically 0.2%. The construction used to find this “offset yield stress” is shown in Fig. 2, in which a line of slope E is drawn from the strain axis at e = 0.2%; this is the unloading line that would result in the specified permanent strain. The stress at the point of intersection with the σe − e curve is the offset yield stress. the figure above shows the engineering stress-strain curve for copper with an enlarged scale, now showing strains from zero up to specimen fracture. Here it appears that the rate of strainhardening2 diminishes up to a point labeled UTS, for Ultimate Tensile Strength. Beyond that point, the material appears to strain soften, so that each increment of additional strain requires a smaller stress. 3). the Young’s modules in the hand book is 70 x 10 9Pa while the yield strength for aluminum is 15-20 MPa and ultimate stress is 40-50 MPa. From this figures it is clear that the figures, which were obtained, are different and did not tally at all the reasons being due to error in carrying out experiment such as measuring the length or improper taking of the reading or the nature of the equipments that were used. These equipments may have had problems therefore that may have been the reason why the data did not match. The ultimate strength and the breaking point that is the yielding stress are close together in the case of aluminum. If this strengths are exceeded a fracture will occur abruptly. An expression has been derived for the complete stress-strain curve for aluminum. The expression involves the ultimate tensile strength and strain. It has been shown to produce stress-strain curves which are in good agreement with tests over the full range of strains up to the ultimate tensile strain. Expressions are also derived for the ultimate tensile strength and strain in terms of the Ramberg-Osgood parameters. The expressions are generally in reasonable agreement with experimental data. The maximum deviations observed are of the order of 20% for the ultimate tensile strength and 40% for the ultimate tensile strain. The percentage difference between young’s modulus is = - 230.56% The yield strength for this is = 5.96-7 If the tensile or compressive stress exceeds the proportional limit, the strain is no longer proportional to the stress. The solid returns to its original length when the stress is removed as long as the stress does not exceed the elastic limit. If the stress exceeds the elastic limit, the material is permanently deformed. For lager stress the material fractures when the stress reaches the breaking point. A ductile material continues to stretch beyond its ultimate tensile strength without breaking; the stress then decreases from the ultimate strength. For brittle substance, the ultimate strength and the breaking point are close together. For compression, the stresses are roughly proportional right up to the breaking point, whereas tension there is a marked deviation. 4). Using Mohr to calculate failure the following formula is adopted. - =  =  = Where σ1 is maximum stress is 43.6 10 and σ3 is minimum stress 0.110 . At 90 o Mohr circle will have the following σ =0.310 =  =  == 21.7510 0.31021.7510- 21.7510-0.310 21.7510-0.310 = 21.7210 References Boyer, H.F., Atlas of Stress-Strain Curves, ASM International, Metals Park, Ohio, 1987. Courtney, T.H., Mechanical Behavior of Materials, McGraw-Hill, New York, 1990. Hayden, H.W., W.G. Moffatt and J. Wulff, The Structure and Properties of Materials: Vol. III Mechanical Behavior, Wiley, New York, 1965. Read More
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