The last topic I want to look at in axial loading is thermal effects and also stresses on inclined surfaces. So firstly, thermal effects. And here is the relevant section from the reference handbook. Now, basically, what happens, of course, if we heat something up, it expands. So, for example, if I have a block of material and I increase it's temperature by delta T it expands, it gets bigger. And, in particular, it expands by the same amount, or the same fraction, in all directions, or all dimensions. So this gives rise to so called thermal strain. Epsilon T which turns out to be linearly proportional generally speaking to the temperature increase. In other words Epsilon T is equal to Alpha Delta T where Alpha is the coefficient of thermal expansion of the material. And the dimensions of Alpha are either per degree C or per degree Fahrenheit. And generally this coefficient is very small, for example, for steel Alpha is approximately 12.6 times 10 to the minus 6 Per degree C. So, it's a very small number. The expansions are very small. However, as we will see, it can result in very large stresses. So let's, for example, consider this case where I have a rod which is unconstrained and it expands. So I heat it up by an amount Delta T and it expands by an expansion lower case delta T is the thermal strain. And obviously if it's unconstrained like this, this expansion will not result in any stresses. So, the expansion from our definition Delta T is equal to epsilon T times L But epsilon T is equal to alpha delta T. Now if I compress this bar back to it's original length, then we have a strain from Hooke's law which is equal to Sigma divided by E. Stress divided by the modulus of elasticity. So equating the two, we see that the stress resulting from this thermal strain is equal to E alpha times delta T. So the easy way to understand this is, is that this stress is the stress resulting from having to compress this rod back to its original length. In other words, compressing it by and amount or equal to the thermal delta T. And this can lead to very large stresses. Now if the structure is statically determinate, then generally it will thermal strains will result in no stresses. For example in this case here where we have. Two pinned members which are pinned at B. If I heat each of these members up then this location here, B, can simply move. And there will be we know stress is resulting, however if the structure is statically indeterminate then we may or may not develop thermal stresses depending on the particular situation. For example, this situation here. This is pin connected at A and roller connected at d but it's statically indeterminate, we can't determine the forces in those members because there are too many of them. If we heat these up, if all of those members are heated by the same amount then they will all extend in Relatively the same amount. In other words, there will be no stresses set up. However, if we heated just the central members here by some amount, but didn't heat the external members then that would result in stresses. Or also, if this was a pin support here so that it was constrained and we heated up the whole structure, then it would result in thermal stresses. So generally speaking, if it's statically determinate, we won't get any thermal stresses, but if it's statically indeterminate. We may or may not develop thermal stresses. Now we will also look at conditions where we have pre-stresses or pre-strained and in that case when we change the temperature we simply add the thermal stresses. So the total stress is equal to sigma p the pre-stress. Plus the thermal stress Ea, alpha T, where the plus and minuses here refer to fact that the prestresses could be compressive or tensile and the temperature changes here can be positive or negative. So let's do an example on that we have rails of a railroad track are welded when the temperature was 60 degrees Fahrenheit. The coefficient of thermal expansion is given and the module of elasticity is 30 times 10 to the sixth pounds per square inch. If the rails are heated to 120 degrees Fahrenheit Then they try to expand and, we get a compressive stress. And the compressive stress is most nearly, which of these alternatives. So we imagine it looks like this situation, except now, we're going to compress this to reduce the length back to its original length and the total length of our rails is fixed here, so this equation applies that we previously had. Sigma, the stress, is equal Ea times E alpha delta T. E we're given as 30 x 10 to the 6th Pounds per square inch, the coefficient of thermal expansion is 6.5 time 10 to the minus 6 and delta t, the temperature change is 120 minus 60 and the answer is 11,700 pounds per square inch. And the answer is B. So indeed, you see that these, even though these expansions are very small, they can result in very large stresses. And we could also write that as 11.7 ksi, or thousand pounds per square inch. Next example we have a threaded steel rod here and the module is to be less steered to the coefficient of thermal expansion is given. And initially it's stress free so it's bolted between these two walls here but there's no stress in it. Now we're told that the allowable normal stress in the rod is 90MPa, and the maximum permissible temperature drop is most nearly which of these temperatures? In other words there's no stress in its original temperature but then we start to cool it down, the rod starts to contract, but the length is constrained by the bolts in the walls so we get some internal stress. What is this stress? We're given that the diameter is 15 millimeters. So in this case the same formula applies because there's no pre stress. Sigma is equal to E alpha Delta T. So, in this case we want to calculate the permissible temperature change. So delta T is sigma over alpha E. Substituting in the given numbers we have 90 times ten to the six, divided by twelve times ten to the minus six, et cetera. And the answer is 35.7 degrees Celsius. So the answer is C. This problem, we have a brass wire where the modular elasticity is again given, but in this case we have a pretension our force of 85 newtons, so it's already under tension. Coefficient of thermal expansion is given and we want to find the temperature change at which the wire goes slack. In other words, zero tension and zero force, zero stress. In the wire. And which of these answers is the correct one. So, first I'll calculate the cross sectional area. Pi by four d squared. The diameter is two millimeters, so the cross sectional area is 3.14 times 10 to the minus six meters squared. Here is our formula sigma is equal to with the plus and minuses in there the pre-stress plus the thermal stress, so in this case we know that it's initially under tension, so sigma P is going to be positive. So, Delta T is also going to be positive in this case, so we want this to be equal to zero. So firstly the prestressed sigma P is the pretense force T divided by the cross sectional area. Is equal to 2.71 times ten to the seventh newtons per square meter. That's the pretension. So substituting it in we have sigma is equal to Sigma P minus EA Alpha Delta T, which we want to be equal to zero. So therefore delta T, the temperature change is sigma P over a E is equal to, substituting in the numbers, is equal to 12.6 degrees Fahrenheit, degrees Celsius, sorry, which is a positive number so the answer is B. In other words, as we would intuitively expect we have to heat up this rod to expand it. So there's no residual force in it and no residual stress or strain. Next I want to look at stresses on inclined surfaces. So far for actually loaded members, we've only looked at normal stresses. In other words, if I take a plane through the Cross section here. The force on that and the distresses are normal. And the magnitude of the normal stress is the force divided by the cross sectional area. But what if I make an angled cut across here? Such as this angle here? Now, if I draw a free body diagram of the left hand side here, I get something like this. And in side view, like this. And now we see that, for this to be an equilibrium, we must have normal and tangential components of force. So we have a normal component of force and stress here. And also a tangential component of force and stress in this direction. And to analyze this, if I have a plane of arbitrary angle fatal like this, then this is a relatively simple exercise in geometric geometry and force balances. We find that the tangential, the normal component of force, sigma theta, or the normal stress sigma theta, is (sigma x)/2 Into 1 plus co-sign 2 theta, and the tangential stress tell theta is equal to minus sigma X over 2, sign theta where sigma X is the normal or the actual stress P over A. And if we plot out this relationship we get something like this for planes of different angle, where sigma theta here, this curve, is the normal stress and this curve is tal theta the tangential stress. So, the stresses vary shown here. And the maximum value of the normal stress, sigma max, is equal to the actual stress, sigma x. And this occurs on a plane which is in this direction, in other words, normal to the applied force. And the maximum value of the shear stress is half the actual stress sigma x over 2 and this occurs on planes which are oriented at 45 degrees. Either that direction or that direction to the actual force. And in between they vary as shown here. So for example, if I look at two elements here and Element A, which is who's faces are perpendicular and parallel to the main axis and another element, B, whose faces are plus minus 45 degrees to the main axis. The force distributions look like this. First for element A we have only actual stresses or normal stresses, rather, sigma x which are equal to the main actual Force and no other normal stresses or sheer stresses. Whereas the element which is oriented at 45 degrees looks like this. This is subject to normal stresses of sigma x over 2. Sigma X over two. And these are all in tension and shear stresses which are equal to, also equal to sigma X over two, which are in the directions as show. So the magnitudes of the stresses are all equal for that particular case. But for arbitrary angles we have relationships which are as shown here. So let me do a numerical example on that. We have a rectangular steel bar carrying a tensile load P. The allowable stresses in tension and shear are 14,500 psi. And 7,100 PSI respectively. The maximum permissible value of the actual IP is most nearly which of there alternatives. Now in this case, we don't know which one of these is governing Whether it's the normal stress or the tangential stress, so we have to calculate both of them. So, firstly let's compute the principal actual stress, sigma x, force divided by area. Or, rearranging to get the force, which is what we're looking for here, is equal to sigma x times the cross-sectional area. The maximum normal stress, sigma max, is equal to sigma x. Therefore for that case, the actual load, P, is sigma max times A and the maximum actual stress we're given is 14,500 pounds per square inch, so the load is 43,500 pounds. The maximum sheer stress, tell max, is equal to sigma X over two, so in that case the actual load is P times Two times Tau max times A, or two times 7,100 times two times one point five, is 42,600 pounds. So, the governing one out of those two is the lower one which is 42,600 pounds. So, the answer is B. The final example, we have a brick here which is being compressed in a testing machine by some force P. The ultimate sheer stress is twelve thousand, twelve hundred pounds per square inch, and the ultimate compressive stress is 3600 pounds per square inch The force P required to the brick is most nearly which of these alternatives? Now in this case we know that as usual for the maximum compressive stress is equal to sigma X and the maximum shear stress is sigma X over two. But in this case we can see that the maximum shear stress here, 1200, is less than half of the maximum normal stress. Therefore we know right away that it's going to be the shear stress which is going to govern this problem. And we don't need to calculate them separately. So in that case we know that taw max is equal to the sigma x over 2 or P over 2A. So rearranging the force is equal to 2 times the cross-sectional area times maximum shear stress is equal to 2 times 4 times 2.5. And the ultimate or the allowable shear stress is 1,200 pounds per square inch, so calculating the answer is 24,000 pounds and the answer is B. And this concludes our discussion of thermal effects and stresses on inclined planes.
