6463 aluminium alloy

Transcription

For this Screencast we are going to discuss material selection specifically we're going to discuss how we'll use elastic properties of materials to choose candidates for our dessign problem If we take a look at our problem statement we have a cylindrical rod that is 380 millimeters long. Its gonna to have a diameter of 10 millimeters We anticipate that it's gonna be subjected to a tensile load. The rod is not to experience plastic defamation and is not to have an elongation more than 0.9 millimeters and the applied load is going to be 24,500 Newton's we have 4medals metals or alloys that are listed and we need to take a look at which of these is going to be a candidate So to go ahead and start let's mark down our known values. We know force is equal to 24,500 Newton's the diameter of the rod is going to be ten millimeters making our radius 5 millimeters. We know the initial length is gonna be 380 millimeters and we know that our elongation delta l is going to be no more than 0.9 millimeters The equations that we will need for this problem include engineering stress where Sigma equals F/A_o. Here F is going to be our applied force and A_o is going to be our cross-sectional area we'll also be using engineering strain where epsilon equals delta "l" of "l_o". Here delta "l" is our change in length or our elongation and "l_o" is going to be our original length of the rod lastly we are going to be using Hooke's law which allows us to relate our modulus of elasticity also known as stiffness which is "e" to the engineering stress and strain The first step to solving this problem is to determine the applied stress that the rod is going to end up seeing with our applied load. We can do this using our engineering stress equation so Sigma equaling F/A_o. Our force was given as 24,500 Newton's and our radius was given as five millimeters Now keep in mind we want to make sure that our final units are in Newton meters^2 or MPa So we need to go ahead and convert our five millimeters into meters so our final answer is 3.11*10^8 Newton meters which is equal to 311 MPa Just as a reminder one MPa is equal to 1*10^6 Newton meters^2. So our 311 MPa is the applied stress that we're going to end up seeing on this rod. What this means is that we need to have a material that can stand the applied stress without plastic deformation taking place. So if we looked back up at our table that includes all the material properties and specifically take a look at it the column with yield stress we'll notice that there's actually only two materials that have the yield stress that are going to be greater than the applied stress. One is the brass alloy at 345 MPa and the other one is the steel alloy at 450 MPa. So at this point we can go ahead and eliminate aluminum as well as copper as potential material for this problem So the next step we need to go ahead and do is check brass and steel for the elongation requirement using Hooke's law and engineering strain we can go ahead and find the maximum elongation for both brass and steel so using Hooke's law where Sigma is equal to "e" time epsilon we can solve for epsilon and we also know that epsilon is equal to delta "l" over "l_o". As a reminder delta "l" over "l_o" comes directly from the engineering strain equation we can now solve for delta "l" and this now gives us an equation to be able to determine what the maximum elbongated is for each material. So we can now evaluate both brass and steel. So we can go ahead and solve for delta "l" for brass where we'll have 311 MPa over 100,000 MPa times 380 millimeters One thing to note is if we go back up and take a look at our table our values for Youngs modulus are given in GPa So has to be able to convert that to MPa. That is why we have 100,000 MPa for this problem So is we go ahead and calculate this we find that our finally answers gonna be 1.18 millimeters for brass and now we need to go ahead and calculate the elongation for steel. So going through the same process we'll move down and go ahead and calculate delta "l" for steel. Using the properties for steal we have 311 MPa over 207,000 MPa So that is Young's modulus for steel times our 380 millimeters. So the final elongation for steel would be 0.57 millimeters What these values represent is the elongation we can expect if we end up using brass or we end up using steel So in this case we know that delta "l" max can only be 0.9 millimeters so brass is going to elongate 1.18 millimeters while steel is going to elongate 0.57 millimeters So for this problem steel is the only material that will end up working and if we go back and double check our steel is gonna satisfy both our elongation requirment as well as our requirement that we ended up needing for stress.