 I am here to discuss with you on topic tensile test on steel today. Learning outcomes of today's session, at the end of this session students will be able to explain procedure of tensile test on steel. They will be able to perform tensile test on steel in laboratory. As we know steel is an important construction material mainly used as reinforcement in the form of round bars in concrete and also at different places like roofs, frames, fermentos and so on. Now quality of this steel bars is very important mainly they have to carry tensile load. So, they are tested for tensile strength in laboratory by using relevant IS code. So, the IS codes for testing of steel are listed here. You can see IS 432 1982 is the code for mild steel and medium tensile steel bars and hard drawn steel wires for concrete reinforcement. Today we are dealing with IS this particular IS code as we are studying tensile test on mild steel bars today. So, let us see test procedure. This particular test procedure is according to IS 432 1982. Now aim of today's experiment to study tensile strength of mild steel bar specimen this is the aim. Equipments used universal testing machine universal testing machine you can see here this is the machine which is used for this particular test. Apartus used vernier caliper test specimens etc. then procedure mild steel specimen of say 20 mm diameter is say it is to be tested then diameter of this bar is measured with vernier caliper accurately remember. So, measurement of diameter of the specimen is important. Standard gauge length is marked on the specimen this is next important IS code gives standard gauge length and accordingly the gauge length should be marked and it is 5.65 under root cross section area of bar accordingly this particular gauge length is marked and bar is held in the jaws of the middle cross head and upper cross head. So, firstly it is held in the middle cross head of the UTM formally then middle cross head is taken up and the other end of the bar is fixed in the jaws of the top cross head of the UTM remember. See the specimen is prepared like this its part on two ends is having more diameter and the gauge length part is reduced in diameter. So, that we can get the results for this particular part we are testing the bar for this particular gauge length over which diameter is reduced. So, that we can confirm the failure of bar in this particular region only. Now, for taking strain readings and elongation readings extensometer is also fixed over the gauge length while testing is being performed and loading is given gradually. Gradual loading is given to the bar tensile loading and for every fixed time interval of loading say 20 kilo Newton or so extension of the bar is measured with the help of extensometer it is recorded from the extensometer the extension for each load interval. So, that we can draw stress strain graph or load versus elongation graph and graph is directly available on UTM also which we can validate the results with. Then before actual failure of the specimen takes place remember extensometer has to be taken out. So, as to avoid its damage and then yield point is noted down load at yield point is recorded then ultimate load is recorded breaking load is also recorded. These silent points we will see soon. Naking can be seen clearly before failure in the mild steel as it is a ductile material. Naking means reduction in diameter is clearly seen and before this you have to take out the extensometer and avoid the damage of the extensometer. Now, after failure of the specimen two parts are brought together and their extension in length over the gauge length and reduction in diameter at the failure point is measured. So, these are the readings taken. Now, we will discuss about these different readings in this particular part of slide. Now, reduction in diameter of the specimen is measured. Now, percent elongation of the bar is calculated. How will be the percent elongation calculated? Final gauge length minus initial gauge length divided by the final gauge length into 100 this will give us the percentage elongation of the steel bar remember. And percentage reduction in diameter how it is to be measured? Initial diameter minus final diameter divided by initial diameter into 100. So, this will give us percentage reduction in diameter. Apart from this calculate in apart from these two calculations stresses are also measured ultimate stress yield stress etcetera from the loads which you have obtained in the test. Now, there are certain specifications given by IS-432 part 1 1982 regarding the ultimate stress yield stress and percent elongation for mild steel of grade 1. This particular code gives specifications for bars less than 20 mm and bars more than 20 mm differently little different is there. When bars are less than 20 mm ultimate tensile stress should minimum tensile stress should be 410 410 Newton per mm square. Yield stress should be minimum 250 Newton per mm square and percent elongation minimum should be 23. So, these are the specifications given by IS-432. If bars are more than 20 mm up to 50 mm the ultimate tensile stress should be minimum 410 same yield stress minimum should be 240 little less 240 Newton per mm square and percent elongation is same 23 percent. Now, let us see some silent points on this stress strength curve. So, stress strength curve is very important curve which is drawn after every tensile test. And you will see that for initial portion this particular curve is straight one which shows that stress is proportional to strain up to certain limit say up to point A. So, this particular point A is known as proportional limit. Up to proportional limit the material obeys Hooke's law we know stress is proportional to strain up to proportional limit. Now, beyond proportional limit this point B is there which is elastic limit and up to elastic limit material is said to be elastic that means, if you unload the material here it will regain its original shape and size immediately. So, point B here is elastic limit. Now, once this elastic limit passes here points C and D these are the two points which are called yield points C is the upper yield point D is the lower yield point. In this region what happens load practically remains constant and strain increases continuously. So, here you can see the most part of this curve between C to D is horizontal one. So, at the same loading at the same loading the strain increases considerably. So, it is called yield stress and when higher point at higher point C this particular point is called higher yield point upper yield point and lower point C higher point C and lower point is D. So, C is the upper yield point D is the lower yield point. Then thereafter again stress increases a little till ultimate point is reached and here ultimate stress is obtained. So, maximum stress this is point E where stress taken by the specimen is maximum. And after this certain drop again occurs in the stress and breaking stress is obtained remember after ultimate stress what happens necking starts as soon as necking starts necking means reduction in diameter at the weakest point what happens load decreases a little than the ultimate and this particular breaking load is lower than the ultimate load. So, breaking stress at point F is also found out. So, here is one exercise for you students let us pause the video here a little and solve this particular problem. In a tensile test on mild steel specimen of diameter 12 mm over a gauge length of 100 mm following observations are taken yield load 28.3 kilo Newton ultimate load 47 kilo Newton find yield and ultimate stresses these are the this is the exercise. Now, here we can see the solution cross-sectional area of specimen can be found pi by 4 by d square because it is ground specimen and yield stress is equal to yield load divided by cross-sectional area. So, it is 250.35 Newton per mm square ultimate stress is ultimate load by cross-sectional area and you will get 415.78 Newton per mm square simple is the solution. So, here are two review questions for you solve them as per IS 432 1982 mild steel grade 1 bar of diameter less than 20 mm should have minimum yield strength of dash Newton per mm square pick up the correct answer and as per IS 432 mild steel of grade 1 should have minimum percent elongation of dash pick up the correct answer write your answers answers question number 1 a 250 question number 2 d 23 these are the references for today's session thank you.