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

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


Hello Friends here in this video we will see a problem based on temperature stress and strain here we have a question I’ll read the question and write the data accordingly a rod 10 meter long length of the rod is given 10 meters so it is 10 into 10 raised to 3 mm at 10 degree Celsius is heated to 70 degree Celsius so 10 degree Celsius is the initial temperature T 1 and 70 degree Celsius it is the final temperature next if the free expansion is prevented find the magnitude and nature of stress induced that is we have to calculate the magnitude of stress and its nature take capital e young’s modulus is given to point 1 into 10 raised to 5 Newton per mm square and alpha value 12 into 10 raise to minus 6 per degree celsius so this is the question we have now based on this let us try to get the solution I will read it once again that since it is given a rod is 10 meter long and it is heated so first I will draw the diagram to explain that suppose here we have a rod whose length is capital L now this rod it is made up of metal and then it is heated here I am drawing a heating member which provides heat to this rod now when this metal rod is heated at first the temperature of the metal rod was 10 degree Celsius and after heating the temperature has become 70 degree Celsius next as the rod is made up of metal then because of the heat there are chances of this rod to expand and this expansion of the rod is called as free expansion denoted by Delta L now if we are heating such a rod and then we are allowing it to expand freely based on its free expansion value then there will be no stress developed in this rod but if we try to stop this free expansion completely or partially suppose in this example here we have a wall which is trying to stop the free expansion of the metal rod so when the free expansion is prevented at that time there will be temperature stress developed in this metal rod so first thing to start in the problem I will find this free expansion value that is since the rod has been heated from 10 degree Celsius to 70 degree Celsius therefore there will be increase in length of the rod so increase in length is equal to free expansion of the rod therefore free expansion it is denoted by Delta L and it is equal to alpha into delta T into L so I will go on putting the values Delta L is equal to alpha it is 12 into 10 raise to minus 6 delta T since the temperature difference is T 2 minus T 1 so it is 70 minus 10 original length is 10 meter so in mm it is 10 into 10 raised to 3 so from this here I will get the value I’ll write it onto the next page the free expansion comes out to be 7 point 2 mm so the metal rod I am just drawing the diagram to explain this in a more effective manner fearfully freely expand and this value is Delta L now if the free expansion of this metal rod it is given that in the problem that the free expansion is prevented so when we are preventing the free expansion it means we are not allowing this metal rod to expand to 7.2 mm because of the change in temperature it means suppose the free expansion is prevented by having a wall here so because of this wall this metal rod won’t be able to expand by delta L and hence I can say that hence if the free expansion is prevented then thermal stresses and strains will be developed in the rod and as we know that when the thermal stresses and strains are developed we need to find the value so first I will say that therefore thermal strain it is equal to alpha into delta T therefore that is equal to alpha is 12 into 10 raise to minus 6 and the temperatures range is 70 minus 10 so strain the answer which I have it is 7 point 2 into 10 raise to minus 4 so this much is the thermal strain next after getting the thermal strain I can find the thermal stress so therefore thermal stress Sigma is equal to thermal strain into Young’s modulus this formula has come from Hookes law so therefore thermal stress is equal to thermal strain was 7.2 into 10 raise to minus 4 into Young’s modulus given in the problem two point 1 into 10 raised to 5 so from this the stress I have it is 150 1.2 Newton per mm square so this is my answer and in the problem they are also saying to calculate the nature of stress so as we can see that when the metal rod was freely expanding stress-induced was 0 but if we are attaching a wall here it means there is a force acting from the wall towards the metal rod and that force is compressive so I can say that therefore nature of stress will be compressive because we are avoiding the free expansion and this is the second answer so in this video we have seen the problem on temperature stress and strain

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