Stress and Strain – Problem 1 – Stress and Strain – Strength of Materials


Hello friends here in this video we are going to see a problem based on calculation of stress strain and modulus of elasticity for that here we have a question a bar 500 mm long and 22 mm in diameter is elongated by 1.2 mm and the axial pool is of 105 kilo Newton calculate the intensity of stress strain and modulus of elasticity of the bar this is the question in front of us first of all whatever is there in the question we will write that in the form of data let us get started now it is given a bar is 500 mm long and 22 mm in diameter I’ll draw the diagram for the same here I have drawn a bar or a rod whose length is 500 mm and diameter is if I look at this cross-section you’re this diameter is 22 mm so the bar having lengths and diameters is elongated by 1.2 mm this elongation means that here over this bar there would be a pull type of load that is why there is elongation so your P indicates the pull type of load on the bar and elongation is given as 1.2 mm so your elongation is denoted by Delta L that is 1.2 mm and the axial pull is of 1 0 5 kilo Newton value of P is given 1 0 5 into 10 raised to 3 Newton calculate the intensities of stress the first question is to calculate the value of stress second strain and third we have to calculate the modulus of elasticity of the bar capital e so this is the question in front of us whatever was given here we had written that in the form of data now let us try to get the solution to this problem now in the solution as the first question is stress so I will say that since stress is given by the formula is Sigma is equal to P upon a so this is the relation now when we look into this question P is given as 1 0 5 into 10 raised to 3 Newton an area we can get it from the diameter diameter is 22 mm so here I will write down this stress Sigma will be equal to P is 1 0 5 into 10 raised to 3 Newton area it is PI by 4 into diameter which is 22 square and from this I get my first answer of stress and that answer is 276 point two two new 10 per mm square this is the first answer next they are seeing us to calculate strain so after stress I’ll write down now since strain is given by the formula is strain is equal to change in length upon original length so here I’ll put the values change in length is given in the problem it is 1.2 that is the deflection and length is 500 so from this I will get strain it comes out to be 2 point 4 into 10 raise to minus 3 and strain won’t have any units it is a dimensionless quantity next after getting strain the third part is modulus of elasticity so I can say that since modulus of elasticity is given by modulus of elasticity we can get it from Hookes law in Hookes law it was modulus of elasticity was equal to stress upon strain so here I will put the values there for young’s modulus is equal to stress is two seventy six point two two our answer number one strain two point four into 10 raise to minus three so from this I will get capital e young’s modulus my answer is it comes out to be one one five zero nine 10.67 Newton per mm square so this is the answer of Young’s modulus or you can say modulus of elasticity so if I write the answers at the end here we have got stress as two seventy six point two two new 10 per mm square answer number one strain two point four into 10 raise to minus three second answer and Young’s modulus one one five zero nine one point six seven Newton per mm square so here we have completed the problem that is we have calculated whatever values that was unknown

2 Replies to “Stress and Strain – Problem 1 – Stress and Strain – Strength of Materials”

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