Mechanical Properties of Materials.
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Mechanical Properties of Materials
The strength of a material is not only the only criterion that must be considered in designing structures. Mechanical properties such as hardness, toughness, ductility determine the selection of a material. These properties are determined by making tests on the materials and comparing the results with established standards.
Stress strain diagrams
Test results generally depend upon the size of the specimen being tested. Since it is unlikely that we will be designing a structure having parts that are the same size as the test specimen, we need to express the test results in a form that can be applied to members of any size. A simple way to achieve this objective is to convert the test results to stresses and strains.
Calculation of stress: The axial stress б in a test specimen is calculated by dividing the axial load P by the cross sectional area.
When the initial area of the specimen is used in calculation, the stress is called nominal stress(also named as engineeringstress or conventional stress)
When the actual area of the bar at the cross section where failure occurs is taken for stress calculation then this stress is called true stress.
Calculation of strain: If the initial gage length is used, then nominal strainis obtained.
Since the distance between the gages marks increases as the tensile load is applied, we can calculate the true strain (also named natural strain) at any value of the load by using the actual distance between the gage marks.
Characteristics features of a typical stress strain diagram:
Sample Description : M.S Rod
Test Name: Tensile test
Figure-1: Typical stress-strain diagram of MS Rod
The stress strain diagram of a MS Rod is shown in the Figure-1. Strains are plotted on the horizontal axis and stresses on the vertical axis. The diagrams begin with a straight line from the origin O to point A, which means that the relationship between stress and strain in this initial region is not only linear but also proportional. Beyond point A, the proportionality between stress and strain no longer exists; hence the stress at A is called the Proportional limit. The slope of the straight line from O to A is called modulus of elasticity.
With an increase in stress beyond the proportional limit, the strain begins to increase more rapidly for each increment in stress. Consequently, the stress-strain curve has a smaller and smaller slope, until, at point B, the curves becomes horizontal. Beginning at this point, considerable elongation of the test specimen occurs with no noticeable increase in the tensile force (from B to C). This phenomenon is known as yielding of the material and point B is called yield point. The corresponding stress is called yield stress.
In the region from B to C, the material becomes perfectly plastic, which means that it deforms without an increase in the applied load.
After undergoing the large strains that occur during yielding in the region BC, the steel begins to strain harden. During strain hardening, the material undergoes changes in the crystalline structure, resulting increased resistance of the material to further deformation. Elongation of the test specimen in this region requires an increase in the tensile load, and therefore stress-strain diagram has a positive slope from C to D. the load eventually reaches its maximum value and the corresponding stress at D is called ultimate stress.
Further stretching of bar is actually accompanied by a reduction in the load and fracture finally occurs at a point such as E.
When a test specimen is stretched, lateral contraction occurs. The resulting decrease in cross sectional area is too small to have a noticeable effect on the calculated value of stresses up to about point C, but beyond that point the reduction in area begins to alter the shape of the curve. In the vicinity of ultimate stress, the reduction in area of the bar becomes clearly visible and a pronounced necking of the bar occurs. If the cross sectional area at the narrow part of the neck is used to calculate the stress, the true stress-strain curve is obtained
To define a clear yield point
When a material does not have definite yield point, an arbitrary yield stress may be determined by the offset method. A straight line is drawn on the stress-strain diagram parallel to the initial linear part of the curve but offset by some standard strain, such as 0.002. The intersection of the offset line and the stress-strain curve defines the yield point.
Some important parameter
Percent elongation = (L1– L0)/L0 * 100
Where, L1 = Distance between gage marks at fracture
L0 = Original gage length
Percent reduction in area = (A0– A1)/A0 * 100
Where, A0 = Original cross sectional area
A1 = Final area
Typical stress-strain diagram for brittle materials
Figure-2: Stress-strain diagram of brittle material
Assignment – 01
The following data were obtained during a tension test of an aluminum alloy. The initial diameter of the test specimen was 0.505 inch and the gage length was 2.0 inch.
Plot the stress-strain diagram and determine the following mechanical properties:
a) proportional limit b) modulus of elasticityc) yield pointd) yield strength at 0.2% offset
e) ultimate strength f) rupture strength
Problems on Simple strain
An aluminum bar having a cross sectional area of 0.5sqinch carries the axial loads applied at the position shown in the following figure. Compute the total change in length of the bar if E = 10 x 106 psi. Assume the bar is suitably braced to prevent lateral buckling.
A bronze bar is fastened between a steel bar and an aluminum bar as shown in the figure. Axial loads are applied at the positions indicated. Find the largest value of P that will not exceed an overall deformation of 3.0mm, or the following stresses: 140 Mpa in the steel, 120 Mpa in the bronze, and 80 Mpa in the aluminum. Assume that the assembly is suitably braced to prevent buckling.
The rigid bar AB attached to two vertical rods as shown in the figure, is horizontal before the load P is applied. Determine the vertical movement of P if its magnitude is 50 KN.
A uniform concrete slab of total weight W is to be attached shown in figure, to two rods whose lower ends are on the same level. Determine the ratio of the areas of the rods so that the slab will remain level.