The objective of this experiment is to investigate the behavior of steel specimen under a tensile test and to determine it's properties.

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Lee Rose
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1 Objective: The objective of this experiment is to investigate the behavior of steel specimen under a tensile test and to determine it's properties. Introduction: Mechanical testing plays an important role in evaluating fundamental properties of engineering materials as well as in developing new materials and in controlling the quality of materials for use in design and construction. If a material is to be used as part of an engineering structure that will be subjected to a load, it is important to know that the material is strong enough and rigid enough to withstand the loads that it will experience in service. As a result engineers have developed a number of experimental techniques for mechanical testing of engineering materials subjected to tension, compression, bending or torsion loading. The most common type of test used to measure the mechanical properties of a material is the Tension Test. Theory: Some the terminology and equations follow. Engineering Strain: l i l o l l o l o where, e is the engineering strain l o is initial gauge length, and l i is the instantaneous gauge length. Engineering Stress: F Applied Force A o Initial Cross - Sectional Area. li True Strain: t ln ln1. lo F Applied Force True Stress: t 1. A Instantane ous Cross - Sectional Area i Strain Hardening Coefficient: The letter n in the equation t K n t or log( t ) log(k) n log( t ). Elongation= lf lo 1% lo Ductility (Area Reduction) = Ao Af 1% Ao εf Toughness= Area under the true stress strain curve= σ dε Elastic specific energy per unit volume = 1 σ ε 2 1

between two fixtures in UTM which clamp it.

2 Apparatus and materials: -UTM: universal testing machine. -steel specimen. Procedures: 1) The steel specimen which has known dimensions, like length and crosssectional area placed between two fixtures in UTM which clamp it. 2) Then we begin applying Load to the material gripped at one end while the other end is fixed. We keep increasing the load while at the same time measuring the change in length of the sample. 2

3 True Stress (KPa) Stress (KPa) Results and Discussion: The engineering stress-engineering strain curve of the tested material. 12 Stress Strain curve Strain (*1^3) The true stress-true strain curve of the tested material True stress strain curve True strain 3

log True Stress The logarithmic relation between true stress and true strain. Log true stress vs. log true strain 3.2213 +.218x = y 3.5 3 2.5 2 1.5 1.5-3.5-3 -2.5-2 -1.5-1 -.5 log True strain Log K=3.

4 log True Stress The logarithmic relation between true stress and true strain. Log true stress vs. log true strain x = y log True strain Log K= K= n=.218 Difference between true and engineering stress strain curves. It can be noted that the true stress and strain are practically indistinguishable from the engineering stress and strain at small deformations, as shown below in Figure. Also, the strain becomes large and the cross-sectional area of the specimen decreases, the true stress can be much larger than the engineering stress. 4

5 Stress (KPa) Mechanical properties for the tested specimens from curves Elastic Limit Stress Strain curve UTS Fracture 6 Yield Stress 4 2 Proportional Limit Strain (*1^3) a. Modulus of elasticity = Slop until proportional Limit Pa b. Proportional limit (4.741, ) c. Elastic limit (26.95, ) d. Yield strength KPa e. Elastic specific energy per unit volume = 1 2 σ ε = = 1 2 *67252*26.95 = Pa, 1 J/m 3 f. Ultimate tensile strength KPa g. Fracture stress KPa εf h. Toughness=Area under the true stress strain curve= σ dε= K ε n dε ; K = , n =.218 i. Elongation = lf lo lo Toughness = KPa 1% = % = % 5

6 j. Ductility = Ao Af Ao 1% = % = % Given L o =1mm and d o =7.5mm A o = π 4 do2 = mm 2 L f =L o +ΔL L f =L o + εf L o L f = *1= mm A o L o = A f L f *1=A f * A f = mm 2 Questions Q1. Consider the engineering stress-strain curve, the curve after necking goes down and the engineering stress decreases with the increase of engineering strain. Explain why? Solution: This happens because the stress value is calculated using original cross-sectional area. After Ultimate point, necking takes place i.e. there is a big reduction in crosssectional area, but for stress calculation, original cross section is taken into account. That s why there is a dip in the curve, which is wrong. This thing is not evident in true stress-strain curve. Stress value is calculated using instantaneous cross-sectional area and hence there is no dip. Q2. Discuss the effects of; temperature; strain rate; and deformation rate on the shape of the true stress- true strain curve Solution: Increasing temperature of material mechanical behavior e.g yield stress, flow stress, plastic instability, fracture stress reduce. On the other hand, increasing strain rate will cause increasing in mechanical behavior. Decreasing deformation rate will produce accurate data of stress-strain curve. 6

7 Q3. The true strain can be expressed as: ε=ln (1/ (1-r)) where r is the percentage reduction of area. Prove this relation. Solution: r = ((A -A)/A (*1% ε =ln (l/l ) =ln (A /A) =ln (1/ (A/A )) =ln (1/ ((A -A +A)/A )) =ln (1/ (1-(A -A)/A )) ε =ln (1/ (1-r)) In a tensile test, why doesn't necking occur at the center always? The reasons behind this phenomenon can be summarized as follow: Shape of the sample and especially, distance of the area under analysis from fixtures and type of cross section. The amount and location of defects of the specific sample. The material under analysis and it's plastic state prior to testing. The forces distribution on the specimen if symmetric or not, in this case the specimen has gripped at one end. If the material is homogeneous and there are no particular concentrations of defects, then the centerline, being at the center of the area with smallest cross section, is at the middle of the volume with the highest resolved shear stress and that will, eventually during the trial, reach the Critical resolved shear stress and begin necking. 7

8 Conclusions: As concluded, tensile test provides results about the Mechanical properties such as Yield strength and UTS which has important role in design and construction. Beside that the relation between stress and strain can be obtained after the test specimen has broken. The original cross section and length are also compared to the final cross section and length to obtain reduction in area and elongation. 8

ME 2570 MECHANICS OF MATERIALS

ME 2570 MECHANICS OF MATERIALS Chapter III. Mechanical Properties of Materials 1 Tension and Compression Test The strength of a material depends on its ability to sustain a load without undue deformation