# TENSILE TEST ON MILD STEEL SPECIMEN

In a tensile test of mild steel specimen, usually a round or flat bar is gradually pulled in a testing machine until it breaks. Two points, called gauge points, are marked on the central portion. The distance between these points, before the application of the load, is called gauge length of the specimen. The extensions of the gauge length and the values of the corresponding loads are required at frequent intervals. The extensions are measured by an instrument called an extensometer.

The strains corresponding to the recorded extensions are calculated by dividing the latter by the gauge length, while the stresses are calculated by dividing the loads by the original area of cross-section of the specimen. Stresses so arrived at is called nominal stress to distinguish it from actual stress which is obtained by dividing the load at a particular instant by the area of the cross-section at that instant. Actual stress is greater than nominal stress in a tensile test because the load increases, and correspondingly the area of the specimen decreases.

In figure 1, in the region BC, strain goes on increasing without the addition of any applied load. From C to D the material is ductile. There being a general reduction in the area of cross-section of the specimen. At point D, local yielding begins and a neck is formed at one point of the specimen. The stress at D is called **maximum or ultimate stress** or the tensile strength of the mild steel. Due to local necking, beyond the point of M, the stress in the material starts decreasing and the specimen breaks at the neck corresponding to stress at point F.

**Figure 1: Stress Strain Diagram**

Percentage elongation is defined as

Where is the length of the specimen at the moment of fracture, and = initial length of the specimen=gauge length.

Percentage reduction in area of the specimen is given by

Where, = original area of cross-section of the specimen, and A is the area at the neck of the fracture.

In the region upto the elastic limit, when load is removed in a tensile test, the material follows exactly the same curve back to the origin O. in this region, the material is said to be elastic. Beyond the elastic limit, between the point B and C, the material becomes perfectly plastic which means that it can deform without an increase in the applied load. The elongation of a mild steel specimen in the perfectly plastic region is typically 10 to 15 times the elongation that occurs between the onset of loading and the proportional limit.

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 its atomic and crystalline structure, resulting in increased resistance of the material to further deformation. In the stress-strain curve, CD is the strain-hardening region, beyond which necking starts.

The presence of a pronounced yield point followed by large strain is an important characteristic that is sometimes used in practical design.

**Sometimes it is not possible to locate the yield point accurately** in order to determine the yield strength of the material. For such materials the yield point stress is defined at some particular value of the permanent set. It has been observed that if load is removed in the plastic range then the unloading path line is parallel to the straight portion of the stress-strain diagram as shown in figure 2. The commonly used value of permanent set for determining the yield value of the steel is 0.2% of the maximum strain.

**Figure 2: Stress Strain curve**

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I liked it very much. If possible make all the topics available for us.

very informative…..i like it………………………makes practical very easy!

sir.i want to know the ultimate strenght of mild steel use in pressure vessels or Pressurised Air Receivers…. thanks!

when you say mild steel, it means that the steel has maximum yield strength of 250 MPa

please help me or give me some tips related to this problem ; w have a Existing MILD STEEL-Vertical Pressurized air Receiver w/ an Outside Diameter of 48in. , Height of 139in & Working pressure of 6.5 Bars.. what will be the formula to be use to find : Inside diameter, Thickness of wall , and factor of safety.

i had seen a formula's in some books, Tangential Stress = PDi/2t , Longitudinal Stress = PDi/4t, And Do = Di + 2t.. what will be the Value of Tangential Stress and Longitudinal stress to find Di & t using those formula? :'( help me.please.

good

guys as engneer its beta 2 knw much wel abt materials we use especially in designng!

yeah……..good job guys.

thanks good

The difference between mild steel and high yield steel is not clear to me

what is the percentage elongation of different sizes of mild steel bars

GREAT