Calling all engineers.

J

JJ_

New member
Hi guys wondering if you can help. Im in my third year going on to my fourth at University.



I have however two small problems that I would like some help with. Two probably simple question although I just can't get my head round where to start. Give me a buffer and some SSR and im good, give me some euler's or pressurised cylinders and im stuck! lol.



Ok first question goes like this.



Thin steel cylinder, closed ends, 180mm diameter and 10mm thick. Yield stress of 250MN/m2.



Determine the internal pressure. Then if the internal pressure is 0.75p then fine (using tresca) tensile focre to cause yielding and the applied bending moment to cause yielding.



I have a hunch I should use mohr's circle but im not sure what the units correspond to.



Next and final question.......



A steel bush is to be shrunk on to a steel shaft so that the internal diameter is extended to 0.512mm above its original size. The insied and outside diameters of the bush are 203mm and 305mm.



Find the normal pressure between the bush and shaft

Fine the hoop stress at the inner and outer surfaces of the line.



E=207 GN/m2 and V=0.28



I know I have to find the pressure using lame's but how do I start with this if I have E and not the pressure.



Thanks guys, hopefully I can help you sometime!
 
I re-formatted my brain's hard drive at the exact moment they handed me my Mech. Eng. degree. Your post just brings up bad memories for me :lol :cry:
 
Blah, mechanical engineers :grinno:



I'm not required to start thinking for another month but if I catch any of my ME friend's online I'll ask them if they know.
 
JJ_ said:
Hi guys wondering if you can help. Im in my third year going on to my fourth at University.



I have however two small problems that I would like some help with. Two probably simple question although I just can't get my head round where to start. Give me a buffer and some SSR and im good, give me some euler's or pressurised cylinders and im stuck! lol.



Ok first question goes like this.



Thin steel cylinder, closed ends, 180mm diameter and 10mm thick. Yield stress of 250MN/m2.
Click to expand...



No loading? Just geometry and material properties?



Determine the internal pressure.
Click to expand...



Presure = Force / Area. Without any force, the pressure is zero. I think you need to reread the problem, or I'm in left field. It's been a hard week. I'm probably in left field.



Then if the internal pressure is 0.75p then fine (using tresca) tensile focre to cause yielding
Click to expand...



Whoa! Tresca failure criterion = Failure model based on the maximum shearing stress required to initiate yielding.



OK, IF you had a tensile force, you could use Mohres circle to find the shear forces, but again, since you have no loading information...



and the applied bending moment to cause yielding.
Click to expand...



Well, that you could know. Try starting with a good statics book, and look up some formulas for beams. Your beam is a cantilevered cylinder. You want to solve for the load where the maximum stress = yield strength of the material.



I have a hunch I should use mohr's circle but im not sure what the units correspond to.
Click to expand...



Understanding Mohr's circle can be challenging. Read as much as you can about it. Shigley and Timashenko were good back in my college days, but that was too many revisions ago. Bruhn also has a fantastic book for aircraft structures, but the methods are applicable to other structures. There are also tutorials available on the internet - google "Mohr's Circle" and you'll be faced with lots of info. Take a look at this:



http://www.roymech.co.uk/Useful_Tables/Mechanics/Mohrs_circle.html



as an example.



Next and final question.......



A steel bush is to be shrunk on to a steel shaft so that the internal diameter is extended to 0.512mm above its original size. The insied and outside diameters of the bush are 203mm and 305mm.



Find the normal pressure between the bush and shaft

Fine the hoop stress at the inner and outer surfaces of the line.



E=207 GN/m2 and V=0.28



I know I have to find the pressure using lame's but how do I start with this if I have E and not the pressure.
Click to expand...



I could be wrong, but I think you're making things too difficult for yourself. Maybe because I learned engineering 20 years ago when Finite Element Models were run in batch mode on mainframes, and all you got for output were arrays of numbers, and RPN calculators where some of my best friends, but...



The stress and strain are related by the modulus of elasticity, right? You know how much strain you have by the forced change in geometry. So the radial and hoop stresses are related to the radial expansion and increases in perimeter of the inner and outer diameters.



OK, so maybe the shaft will compress a bit. So you'll do some iterations, and within a handfull of calculations you'll be within the number of significant digits for the calculation (i.e., you have 2 and 3 digit numbers, so I bet you iterate maybe twice at most to get a good 3 digit answer, and then another just to verify that those 3 digits won't change as you expend computation power to get a better number - this is what the old folks of my generation had in mind when they did all these calculations on slide rules). Realistically, my experience is that with a bushing that thin compared to the shaft, the shaft isn't going to be doing a lot of shrinking. As a practical matter, it would be ignored, but as a practical matter, your professors are theoretical, and not practical. :)



Does that give you enough to go on? I'd hate to spell it ALL out and spoil your fun, and this smells too much like work to me.



Hope this helps! Standard disclaimers apply: YMMV - Objects in mirrors may be closer than they appear.



:bestwish:
 
OK, rereading your post, I mis-spoke. With a bushing 51mm thick over a shaft that's 203mm in diameter, compression of the shaft will be a factor (assuming similar material - copper bushing over titanium shaft would be... a golf club! :)). Still a handfull of calculations, at most, if you reiterate.



Sorry. Like I said, it's been a hard week.



OK, for your beam calculation, take a look at this:



http://www.engineersedge.com/beam_bending/beam_bending9.htm



Now, I could say that the section modulus "is left to the student as an excercise." But for a cylinder, it's pretty easy.



Hint: Section modulus is going to be related to the moment of inertia.



Hint^2: Moment of Inertia of a cylinder can be taken as the moment of inertia of a cylinder based on the OD, minus the moment of inertia of a cylinder based on the ID.



See also:

http://darkwing.uoregon.edu/~struct/courseware/461/461_lectures/461_lecture38/461_lecture38.html

http://feaservices.com/fund_struct/fund_struct.html



or, say, Mark's Standard Handbook for Mechanical Engineers.
 
They only made us materials guys take 2 mechanics courses, still learned to hate Mohr and his ******* circle. I think Arved is right on the first one (and the second one), you forgot to give us a load. The second one sounds like a real pain if you have to include the steel shaft compressing, but it sounds like the problem will let you consider the shaft to be uncompressible. Good luck!
 
Ben Z. said:
They only made us materials guys take 2 mechanics courses, still learned to hate Mohr and his ******* circle. I think Arved is right on the first one (and the second one), you forgot to give us a load. The second one sounds like a real pain if you have to include the steel shaft compressing, but it sounds like the problem will let you consider the shaft to be uncompressible. Good luck!
Click to expand...



OK, for a Materials Engineer... Picture a tensile failure in, say, a bar. The 45-degree cut that results (either a big \, V, or microsopically, the jagged surface)? That's due to a shear failure. Visual example of Mohr's circle!



Same with a (wire) cable failure (as in a throttle cable). You get these 45-degree cones at the fracture surface.



Since steel, for instance, is ~1/2 as strong in shear as it is in tensile, the failures will, at some level of observation, appear to be shear failures.



At least, that's how the Materials Engineers at work tell me it works.



All the best,
 
Thanks guys for all your help. I managed to get through most of it last night. Its just an exam paper that was given out last year, Im trying to guess what questions will come up.



Arved what do you work as now? As i feel I may be headed into your area of work. As my course involves this stuff for the most part, along with some CAD for good measure. lol!
 
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