To understand the regions of a stress-strain diagram for a conventional metal an

Question

To understand the regions of a stress-strain diagram for a conventional metal and calculate percent elongation. When a uniaxial force is applied to a conventional metal, the initial response is elastic. The stress increases linearly with the strain. At a certain point, called the proportional limit, the stress begins to increase nonlinearly up to the yield point. This is typically very close to the proportional limit. After that, there is a sudden, small drop in the stress and then a constant stress yielding region. All of this occurs for very small strains (typically less than 0.005). After yielding, the material begins to strain harden. The stress increases nonlinearly with the strain until the ultimate stress is reached. The ultimate stress corresponds to the maximum load that can be sustained by the specimen. After that point, a small neck will begin to form and all the remaining deformation will occur near the neck. The engineering stress (load divided by original cross-sectional area) will decrease as the strain increases further. Finally, the fracture stress will be reached and the specimen will break. Metals are typically ductile. Two measures of ductility are percent elongation and percent reduction of area. The percent elongation is the ratio of the change in the length at fracture to the original gage length. The percent reduction of area is the ratio of the reduction in area at the neck when the specimen fractures to the original specimen area. Percent elongation = L_f - L_0/L_0 Percent reduction of area = A_0 - A_f/A_0 Large strains The figure shows the initial part of the stress-strain diagram for a ductile metal like mild steel. Identify the important points on the diagram. Drag-and-drop the appropriate labels to their respective targets. Percent elongation What is the percent elongation of the material from Part B just before failure? Express your answer as a percent.

Explanation / Answer

From the given figure failure occure at point where the stress is 400 MPa and at strain = 0.35

percentage elongation is equal to the percentage of strain at fracture = (Lf - L0)/ L0 =0.35 (strain at failure)

From this we can say percentage elongation = 0.35

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