Poisson’s Ratio – Problem 1 – Stress and Strain – Strength of Materials

Poisson’s Ratio – Problem 1 – Stress and Strain – Strength of Materials

Hello friends here in this video we will see a problem on calculation of poisons ratio here is the question whatever is given here I will read that and write it in the form of data let us get started a bar of 12 mm diameter so diameter is denoted by small D 12 mm diameter is tested on UTM UTM is universal testing machine and the following observations are noted first gauge length width is 200 mm gauge length is the original length 200 mm of the bar loaded proportional limit is 20 kilo Newton load is denoted by P 20 kilo Newton change in length that proportional limit is 0.2 mm change in length is denoted by Delta L change in diameter at proportional limit is Delta D the change in diameter 0.0025 mm calculate the value of poisons ratio this is the first question poisons ratio is denoted by mu and young’s modulus denoted by capital e so this is the question we have since the diameter is given first I will find the area of the rod in the data itself so it is PI by 4 into 12 Square and therefore area comes out to be 113 point zero 9 7 mm square so here we have to calculate mu which is called as poisons ratio so I will start with the solution part here in the solution we can start it in this way that since poisons ratio is given by mu is equal to lateral strain upon linear strain therefore mu is equal to lateral strain is denoted by e suffix L and linear strain is denoted by small E so this would be our equation number one I will highlight this and it becomes equation number one for us now since we have to find out lateral strain and linear strain so that we can calculate mu now for lateral strain it will be change in the lateral dimension and that lateral dimension is Delta D so here I will say that since lateral strain is of X L is equal to change in diameter upon the original diameter so here the change in diameter is given as 0.0025 and original diameter is 12 so from this I will first get the lateral strain and my answer is two point zero eight three into 10 raise to minus 4 now after getting lateral strain we will calculate linear strain I’ll write down also linear strain is equal to change in length upon original length change in length is given 0.2 original length is 200 mm so from this linear strain comes out to be 1 into 10 raise to minus 3 now after getting lateral strain after getting linear strain we can put the values in equation 1 to get poisons ratio so therefore put lateral strain ease of Excel and linear strain in equation number 1 so therefore mu is equal to lateral strain is 2 point 0 8 3 into 10 raise to minus 4 linear strain 1 into 10 raise to minus 3 so from this mu value comes out to be 0 point 2 0 8 this is our first answer next after getting poisons ratio the second question is to calculate modulus of elasticity now modulus of elasticity we will get it in the formula of Delta L which is deflection so I will say that now since deflection in the bar is given by Delta L is equal to PL upon a so here we want the value of e so I will shift young’s modulus that is capital e on one side and PL upon a into Delta L onto the other side putting the values P is 20 kilo Newton so it is 20 into 10 raised to 3 Newton length is given as 200 mm area we have found out in the Delta 113 point zero nine seven and change in length is zero point two so from this if I calculate I will get the answer as therefore capital e comes out to be 176 point 8 4 into 10 raised to 3 Newton per mm square this is the second answer so your in this video we have seen how to calculate the poisons ratio and young’s modulus poisons ratio our first answer it was zero point 2 0 it and the second question was young’s modulus whose value is 176 point 8 4 into 10 raised to 3 Newton per mm square once we have found out these values we can say that the problem is completed

11 comments on “Poisson’s Ratio – Problem 1 – Stress and Strain – Strength of Materials

  1. amit rajwar Post author

    sir….. in this video why did you just calculated the area of square while we are using a bar, so wont it be the area of a cylinder instead of a circle?
    please explain..

    and the way you teach is very understandable, i really like studying from you sir….you provide in depth knowledge about the subject which even most of the engineering colleges don't provide
    thank you!

  2. bhupi Post author

    Dear Sir, as you calculated the Poisson's ratio as a +ve value. Shouldn't it be a -ve value as either lateral strain or linear strain comes -ve in any case.

  3. Aerylle Tajonera Post author

    sir why this poisson's ratio answer is not in (-) . but in your previous lesson indicated that
    possion's ratio= – lateralstrain/linear strain. ??

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