Poisson’s Ratio – Problem 2 – Stress and Strain – Strength of Materials

Poisson’s Ratio – Problem 2 – Stress and Strain – Strength of Materials


Hello Friends here in this video we are going to see a problem on calculation of poisson’s ratio so here we have this question now whatever is given in this question I will read it and write that in the form of data first let us get started now it is given that modulus of rigidity of a material is 0.8 into 10 raised to 5 Newton per mm square modulus of rigidity is denoted by capital G next when a 6 into 6 mm square rod the area of the rod is given as it is seen here this is the width and here we have thickness so the cross-section area of the rod is a square so here B is equal to 6 mm and even T is equal to 6 mm width and thickness now area will be 6 into 6 that comes out to be 36 mm square was subjected to an axial pull of 3600 Newton this is the actual pole or we can say axial force so P is 3600 Newton it was found that the lateral dimension changed to now when the load was applied this width of 6 mm gets reduced it becomes 5.9 triple 9 1 and even the thickness is reduced it becomes 5-point Purple Line 1 mm square so this is the new area which we are getting so I will write down therefore new area is equal to five point triple 9 1 in 25.91 mm square from this area I can see that this is the change in width so it will be Delta B is equal to five point triple 9 1 mm and similarly delta T the thickness it will be five point triple line 1 mm next find Poisson ratio and modulus of elasticity so in this question I have to calculate how much is the value of MU that is Poisson ratio and how much is the modulus of elasticity so with these watch data available let us try to get the solution to this problem in the solution part as in this problem load and area are given first I will get the value of stress and before that I can even explain it with the help of a diagram in a more better manner so here is this square bar which is subjected to axial pull denoted by capital P now this axial pull it will be parallel to this length now because of the action of this axial pull the square cross-section length of this square bar will increase whereas the width and thickness will go on decreasing so if I want to draw this it will be something like this so here is the diagram that the length will increase and the other two dimensions that there is width and thickness will go on decreasing under the action of wool type of force so now first I will write down since stress induced in the square bar Sigma is equal to P upon a so therefore Sigma is equal to P is given in the problem it is 3600 Newton area it is the original area that is before deformation 36 mm square so 36 from this I will get the value of stress as 100 Newton per mm square now after getting stress the first question is we want the value of mu so in order to calculate mu which is called as Poisson ratio and it is given by lateral strain upon linear strain so to calculate mu first we should have the lateral strain and how to get that as we have Delta B and delta T we can select any one of them and find the lateral strain so here I will say that after this since lateral strain is given by suppose we are taking the width so change in width up our original width your change in width is five point triple nine one original width is six so from this I will get the value of lateral strain which comes out to be now we have to be careful here first of all this change in width it is given in the problem that it is five point nine nine one that is the changed width and if we want to calculate the how much is the change in width here this indicates the new width which I will denote it as del Delta B 1 and this will be delta T 1 now Delta B this will be 6 minus this value because the original width was 6 and now it has changed so this value comes out to be 9 into 10 raised to minus 4 divided by 6 and from this lateral strain comes out to be 1 point 5 into 10 raised to minus 4 similarly lateral strain can even be calculated as I can write down also lateral strain can even be written as change in thickness upon original thickness and therefore lateral strain becomes change in thicknesses original thickness was 6 and finally it has changed to 5 point ripple 9 1 divided by original thickness was 6 so lateral strain will be same 1 point 5 into 10 raise to minus 4 now after getting the value of lateral strain the next thing we want is linear strain so after getting this I will say that in linear strain before that I will write down poisons ratio sins and see what do we get from here since poisons ratio is given by the formula is mu is equal to lateral strain upon linear strain now if we look into this problem here they have not given us how much is the change in length they have just given the change in width and thickness the change in length is not known to us so we cannot calculate strain by using change in length upon original length formula so now here I have written that poisons ratio is lateral strain upon linear string linear strain we can calculate it by using a concept and that concept is since we know that from Hookes law Young’s modulus is equal to stress upon strain so therefore strain is equal to stress upon Young’s modulus I am going to use this formula and why I am using this formula because in this case we don’t have the value of Delta L which is the change in length we have change in width and change in thickness but change in length is not there so here what I will do therefore mu is equal to lateral strain upon in linear strain it is stress upon Young’s modulus so next this value will get multiplied on the left hand side so therefore mu into stress upon Young’s modulus is equal to lateral strain now therefore since we want to calculate mu I will give mu on one side linear strain this capital e would be multiplied divided by stress lateral strain we have calculated its value was 1.5 into 10 raise to minus 4 Young’s modulus value we do not have stress was 100 so therefore mu becomes 1.5 into 10 raise to minus 6 Young’s modulus I will keep this as equation number this would be equation one for us now after getting this equation mu and capital e that is young’s modulus both are unknown values we cannot get them by using this single equation so there is some hint given in the problem and that hint is in the form of G which is called as modulus of rigidity so we can use some relation in which we get modulus of rigidity and modulus of elasticity so here I will say that from this now since the relation between modulus of elasticity and modulus of rigidity is given by the formula is capital e is equal to 2 G 1 plus mu now in this capital e I’ll keep it as it is 2 into G the value of G is given so it is 2 into zero point 8 into 10 raised to 5 1 plus mu is 1 point 5 into 10 raised to minus 6 into capital e now I will multiply this value so therefore capital e becomes this is 160 into 10 raised to 3 and if I multiply it inside the bracket here I have this value so I am opening the bracket plus 160 into 10 raised to 3 multiplied by this term which is 1 point 5 into 10 raised to minus 6 I will get the answer as 0 point 2 for capital e now since we want to calculate capital e I will be shifting this term 0.24 e onto the left hand side so here I have capital e minus 0.2 for capital e is equal to 160 into 10 raised to 3 so therefore this becomes 0.76 capital e is equal to 160 into 10 raised to 3 so finally capital e is equal to 160 into 10 raised to 3 divided by 0.76 and hence my answer of Young’s modulus that comes out to be 210 point 5 3 into 10 raised to 3 Newton per mm square so this will be answer number one for us now after getting this answer we even want to calculate mu here so I will put this value of capital e in equation number one so therefore put capital e is equal to two hundred and ten point five three into 10 raised to 3 Newton per mm square in equation number one so here I have mu is equal to one point 5 into 10 raised to minus 6 into 2 ten point five 3 into 10 raised to 3 so this if I multiply all terms here I am getting my answer of MU as zero point three one five the second answer so here if we look into the problem they were asking us two questions value of poisons ratio and young’s modulus poisons ratio we have found out it is zero point three one five and young’s modulus is 2 10 point 5 3 into 10 raised to 3 Newton per mm square and after getting these two values we can say that the problem is completed

17 comments on “Poisson’s Ratio – Problem 2 – Stress and Strain – Strength of Materials

  1. Naveen kumar Post author

    sir appreciate your efforts but don't refer a textbook by keeping it aside and then solving the problems and rectifying mistakes and try to solve the tough questions from the text book your having .

    Reply
  2. Nanda kishore gundagani Post author

    Hello sir there is no mining subjects provided in your app and start to keep that too mining engineering nd rtu syllabus partten

    Reply
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  4. Kevin Menon Post author

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