6 Steel rail segments each of length 10 m are laid when the temperature is -15"C. How much minimum space should be left between the rails to allow for the exp...

Steel rails for a train track are laid in a region subject to extremes of temperature. The distance from one juncture to the next is $5.2000 \mathrm{~m},$ and the cross-sectional area of the rails is $60 . \mathrm{cm}^{2}$. If the rails touch each other without buckling at the maximum temperature, $50 .{ }^{\circ} \mathrm{C}$, how much space will there be between the rails at $-10 .{ }^{\circ} \mathrm{C} ?$

So here we have a steel rail and so part ay, they want us to find the change in length so the change in length is going to be equal to the linear expansion coefficient of steel times the original length times the change in temperature. So delta L for the steel would be equal to 1.2 times 10 to the native, fifth per degree Celsius times the original length of 12 meters, times the change in temperature, 33 degrees Celcius minus negative, nine degrees Celsius. So we find that the change in length is going to be equal to point 00605 meters. So this would be your answer for party and then for part B. They want us to find the stress so the stress is going to be equal to the Youngs. Module is times the linear expansion coefficient, times the change in temperature. So this is going to be equal. Teo the Youngs modulates of steel is two times 10 to the 11th Pascal's and then times again 1.2 times 10 to the negative fifth per degree Celsius and sometimes the change in temperature 33 plus nine this would give us 42 degree Celsius so the stress is going to be next is going to be able to negative one times 10 to the eighth. Pascal's it being negative means that the stresses compressive in nature, and that's the end of the solution. Thank you for watching.

Everyone. This is Question number 10 from Chapter 14. This problem is about train tracks being laid and were given an initial length that they're laid in 12 meter sections and initial temperature night of two degree C and then, for part, a space needed between if there just to touch at 33 degrees Celsius and B if in contact when laid. What is the stress in the tracks at 32 33 degrees C? So for part A, we have initial length. We have initial temperature, were given a final temperature, and we need how much length between them, so he essentially needed Delta l. So we should be thinking thermal expansion, which were given a crazy in the book Delta L equals l not Alfa Delta team. So we're given all these things were given l not We can find an Alfa and we're given adult a T so we can just plug these in and solve for and saw for the delta l which would be the space that is needed. So 12 meters, which is our l not are awful for steals 1.2 times 10 to the minus five and then our change in temperature is 33 minus negative to which ends up being 35. And if you plug that in, you get a Delta L of five millimeters. That's the space union between the tracks. Okay, So for part B, if they are in contact, what is the stress? So now stresses force. We should be thinking thermal stress s and were given again In the textbook, stress is equal to minus gamma Alfa Delta team plug in our numbers minus gamma for steals two times 10 to the 11th Pascal's alsa 1.2 times 10 to the minus five for steel and then delta T We just determined his 35 degrees C. And if you plug that into your calculator, you get minus 8.4 times 10 to the seventh pass cows and our minus sign here means that the force distress is compressed compressive. So you might be thinking, I thought our question for stress wass minus Alfa minus a, which is area Alfa Gamma Alpha Delta team. But stress is a pressure, not a force, and four in pressure equals force over area. So we divide this. This is we divide this a over to the other side. T enforce over area, which is pressure, which leaves us with minus alpha delta T mice Gamma Alpha Delta T Excuse me.

Mhm. This problem covers the concept of the new expansion And at the cap there are two rails each day. Uh, you didn't expand on both end violent off child up onto. So this rule will expand along this direction, violent style upon two and the other will expand by the same and the taliban. So the gap distance D. Most week we want to. The total expansion or D equals expansion will tireless L, alpha, delta T. And understood the value. So the minimum gap must be The length of the rail that is 12 m into the uh linear expansion coefficient for the steal, that is 11 in two tenders minus six, 4°C. And to the changing temperature, there is the final temperature Of 42°C- the initial temperature of -5. The responses. So the minimum gap better than the tour must be 6.2 into tenders minus three m. Or we can say the minimum get most particular into 6.2 millimeter.

So here the final length is going to be equal to the original length, multiplied by one plus the alpha. The linear expansion coefficient multiplied by the change in the temperature delta T. And we can then solve for the original length. The original length would then be equal to the final length divided by one plus alpha delta T. And so this is going to be equal to then 5.200 zero meters. This would be divided by one plus the linear expansion coefficient of 13 times 10 to the negative sixth per degree Celsius multiplied by 60.0 degrees Celsius. And we find that then the original length L. Is going to be equal to five point 19 five 195947 meters. And this would be at T equaling a negative 10 degrees Celsius. And so we can say that then the we have 4.1 millimeters between at Jason two rails. That is the end of the solution. Thank you for watching