Engineering question ?

[Kaido]

hey guys,

wondering if any of you guys may be able to help me as i have hit a wall doing some cal for an engineering problem (mech eng).

What i am trying to do is convert Nm to MPA, and i know that is not directly possible due to Nm (force x length) and MPA (force x area).

However the date i have been given is that, i know that the value of my MPA is the max tensile strength, and the value of Nm is the apply force. I have the area of the object (72mm x 10mm)

My problem is i have forgetton the forumals

as i know if it was regarding Torque i could use Torque = T where T= ( t max * J ) / c which would give me an answer in Nm

however for tensile strength i have forgotton the forumal.

can anyone help?

[adrenalin]

Ask for you tomoz. Old man is engineer

[RK 86 WA]

actually:

http://wiki.answers.com/Q/How_do_you_convert_Mpa_to_Nm

[Kaido]

actually:

http://wiki.answers.com/Q/How_do_you_convert_Mpa_to_Nm

yeah i saw that problem is im not dealing with circular object.
my value of MPA is ruputre strength from a bend test (which is equivalent to ultimate tensile strength)

Also i may be doing things complete wrong :confused: im kind of confused atm

basiclly what i am trying to do is that i have done a test and have the loads applied to my object, i need to compare this date with some test results of materials, and pick which material will not fail under my given load.

I have work out my Shear Force and Bending Moments using SFD and BMD. however i am now stuck as i need to compare this to the Mean results from the test materials but those values are in MPA and the values i have are N and Nm.

I know that due to the type of material (brittle) and the loads (uni-axial) that i can use the Max normal stress theory (i have worked out my safety factor as beening 6).

However the MNS theory uses princple or direct stresses and im not sure how to convert my cal results into them :confused:

[duste]

σ = My/I

Meaning:
Bending Stress = (Moment * Distance from Neutral Axis to Surface) / Second Moment of Area about Neutral Axis)

From what I understand you're trying to figure out whether the applied moment will exceed the tensile strength of the material?

Given it's a rectangular cross-sectional, the second moment of area would be I=bh^3/12 where b = base width, h = height.

The neutral axis is also taken as the centre of the cross-section so y=h/2.

The moment is also the applied force your are talking about.

Alternatively you could calculate the maximum moment/force you are able to apply without exceeding the tensile strength of the material by solving for M.

EDIT: So you've got your moments from your BMDs, so just smash the largest in to the formula I mentioned above.

Also, a safety factor of 6 is fucking incredibly high, are you sure that's necessary? What is it you are working on exactly?

[Kaido]

σ = My/I

Meaning:
Bending Stress = (Moment * Distance from Neutral Axis to Surface) / Second Moment of Area about Neutral Axis)

From what I understand you're trying to figure out whether the applied moment will exceed the tensile strength of the material?

Given it's a rectangular cross-sectional, the second moment of area would be I=bh^3/12 where b = base width, h = height.

The neutral axis is also taken as the centre of the cross-section so y=h/2.

The moment is also the applied force your are talking about.

Alternatively you could calculate the maximum moment/force you are able to apply without exceeding the tensile strength of the material by solving for M.

EDIT: So you've got your moments from your BMDs, so just smash the largest in to the formula I mentioned above.

Also, a safety factor of 6 is fucking incredibly high, are you sure that's necessary? What is it you are working on exactly?

you are a champ!

I through 6 was alittle high myself, however i use the
Sf (brittle) = 2x MAX (F1,F2,F3)
I use one of the tables in my lec notes to work out the F1,2,3 values. F2 was the largest being 3 (moderatley challengine environment)

But i personally think its too big and maybe should use another table which is for recommended values for safety factors which means i would get a value of 2x2 = 4

Problem i have is the shape is a wooden peg. and for ease of cal im just assuming it to be a rec , however im not sure if that will be very accuary

[duste]

Wooden peg, eh? Mechanical Design 337?

[Kaido]

Ha, Mech Design 337?

yes ! and im completely lost for Part:A.

I have no idea if what i have done is correct or not.

I think i may have fucked myself because i am doing 337 without completing 238 yet.

any help/input would be greatly appreciated

[duste]

Basically you work out the mean and standard deviation of the peg force data.

Multiply the mean and standard deviation by 2.5 to achieve the values required for the new design/timber to withstand.

Do the FBD/SFD/BMDs for both opening and closed conditions using these values.

Calculate bending stress from the bending moments in both conditions and calculate the standard deviation of these values also.

Calculate the mean and standard deviation of the timber rupture strength data for EACH timber.

For each of these timbers, use U_z = U_x - U_y and σ_z = sqrt(σ_x^2 + σ_y^2) then use these values in U_z = k*σ_z and solve for k. Compare this value k with the table from the failure/reliability lecture/tutorial and you'll get your failure %.

Choose the timber that has a low failure rate (not NO failure rate, because remember cost of timber increases proportionally to timber strength, so although it won't fail it will cost a fuckload).

Done.

[Kaido]

Basically you work out the mean and standard deviation of the peg force data.

Multiply the mean and standard deviation by 2.5 to achieve the values required for the new design/timber to withstand.

Do the FBD/SFD/BMDs for both opening and closed conditions using these values.

Calculate bending stress and standard deviation of this value from the bending moments in both conditions.

Calculate the mean and standard deviation of the timber rupture strength data for EACH timber.

For each of these timbers, use U_z = U_x - U_y and σ_z = sqrt(σ_x^2 + σ_y^2) then use these values in U_z = k*σ_z and solve for k. Compare this value k with the table from the failure/reliability lecture/tutorial and you'll get your failure %.

Choose the timber that has a low failure rate (not NO failure rate, because remember cost of timber increases proportionally to timber strength, so although it won't fail it will cost a fuckload).

Done.

Thankyou so much! looks like i did things completely wrong and was going off in the wrong direction! are you doing 337 atm?

For the Uz, Ux and Uy values
Uy is the max load (which i have to cal)
Ux is the strength of the part (is that 2.5x load?)

also is this correct method for the FBD for open position
http://sphotos.ak.fbcdn.net/hphotos-ak-snc4/hs803.snc4/68373_448514083575_691388575_5585767_5815235_n.jpg

Im assuming that for close position the apply force (from the class results) will be at the opposite e.g were the clamping force is? if that make sense

ps if you are at curtin let me know as i would like to buy you a beer at the tav :)

[duste]

Point C is where the spring 'leg' latches on to the peg, not where the two peg halves touch (which I called Point D, not used for opening conditions).

The force at Point A is where you apply the 2.5X, so for example if the mean of the peg force data was 12, the applied force at Point A would be 12x2.5=30N.

For closed conditions, you will use Points B, C, D with the applied force being at Point C, which is the same force (spring force) you calculated in opening conditions.

U_x is the mean rupture strength of the alternative timber.
U_y is the larger of the two mean bending stresses experienced by the peg as per your calculations.

As such, o_x is the standard deviation of the rupture strength of the alternative timber and o_y is the standard deviation of the bending stress.

Hope this helps.

[KAL SPL]

Oh boy do I feel retarded all of a sudden :p

[Fukushima]

fucking vegetarians

[duste]

Fucking drug dealers!

[Kaido]

Point C is where the spring 'leg' latches on to the peg, not where the two peg halves touch (which I called Point D, not used for opening conditions).

The force at Point A is where you apply the 2.5X, so for example if the mean of the peg force data was 12, the applied force at Point A would be 12x2.5=30N.

For closed conditions, you will use Points B, C, D with the applied force being at Point C, which is the same force (spring force) you calculated in opening conditions.

U_x is the mean rupture strength of the alternative timber.
U_y is the larger of the two mean bending stresses experienced by the peg as per your calculations.

As such, o_x is the standard deviation of the rupture strength of the alternative timber and o_y is the standard deviation of the bending stress.

Hope this helps.

Yes this Helps alot :) again sorry if i sound clueless skipping 237 wasn't the best idea i had.

I was under the impression that U_x value had to be in kN, therefore mean rupture strength (e.g 100MPa for ironbark) couldn't be used??

o_y is just the the std deviation between the 2 cal results from the BMD correct?

[duste]

Well the theory itself can be used for any values and their standard deviations, but in this case we need to compare it to rupture/tensile strengths so we need a MPa value also.

No worries, good luck.

[Kaido]

Well the theory itself can be used for any values and their standard deviations, but in this case we need to compare it to rupture/tensile strengths so we need a MPa value also.

No worries, good luck.

ok got you just convert the bending stress into MPa again thanks alot for your help :)

[Wrexter]

If u need the assignment send me a pm.

Did this 3 years back.

This assignment was peanuts compared to the ass rape assignment 3 was when laurie morgan was in charge. 60 pages long of calcs, due study week.

[Fryman]

Rock climbing? Why would anybody go climb a rock? Man, things are changing. I
remember when men were "men" and women were "gals" and we called coloreds
"coloreds"...

[Kaido]

If u need the assignment send me a pm.

Did this 3 years back.

This assignment was peanuts compared to the ass rape assignment 3 was when laurie morgan was in charge. 60 pages long of calcs, due study week.

PM SENT :)

[duste]

keen2compare results also, PM sent.

[Brockas]

antilag.com - encouraging plagiarism since 2010.

[Nelos e46]

antilag.com - encouraging plagiarism since 2010.

haha,

technically its not plagiarism, since their is only one correct answer in calc(maths in general):P

[duste]

Better to help than be a pingpingpingping, sit back and watch someone fail.

No plagiarism on my behalf either - I've already done it. :)

[[PIG GTS]]

your better off getting in contact with Fish he'll help you out

[tinto]

( σ_σ )

THIS POST BROUGHT TO YOU BY AN INCOMPLETE ARTS DEGREE.

[adrenalin]

This is the email i got back from old boy

"
Kieren

Not sure exactly what he is asking. If he wants to know the axial stress then it is simply force/area. If he wants bending stress then its the moment times the distance from the neutral axis to the outermost fibre divided by the second moment of area. The second moment of area depends upon the shape.

If he wants torsional shear stress then its the torque times the radius divided by the polar moment of area. I think this is what he is after.

Nm is a torque and MPa is either a pressure or a stress. The area of 72x10 is I assume rectangular? Polar moment of inertia for a 72x10 rectangle is 2.19x10-8 m-3. So for a torque of 1 Nm, the shear stress is 1.66 MPa at the outermost point where the radius is 36.35mm. So for 10 Nm, the shear stress would be 16.6 MPa etc.

Send him this and if he needs more information, get him to explain in more detail.

Regards

Stuart Herrald
Principal/Engineering Manager
"

[upgarage]

Your dads an engineer?
hook me up with mech eng vac work :D

[duste]

Second that!

[adrenalin]

Pm with what you want to do and details and i will forward them on to him.

[Kaido]

antilag.com - encouraging plagiarism since 2010.

not plagiarism when your checking your results and method to double check they are correct :p

[Kaido]

Your dads an engineer?
hook me up with mech eng vac work :D

ian check ur pm

[Atomic_FD]

I don't even know why I'am reading this.. I'am way out of my depth.

[Fujiwara13]

Hook me up too!

[Fujiwara13]

Also, are you curtin dudes facing 25% funding cuts to your eng departments?
People are getting really anxious over here at uwa =\

[Kaido]

Hook me up too!

didn't know UWA did the same assignments?

No idea about funding cut? Our Mech department just got a new building after using a 50+ yr old building

[Fujiwara13]

;724545']didn't know UWA did the same assignments?

No idea about funding cut? Our Mech department just got a new building after using a 50+ yr old building

Ahh, more wanting hooking up with vac work. I've got a week still to do before I can graduate, but this year i didn't get anywhere.

We have totally different assignments, I've got a few mates doing mech at Curtin, doesn't sound like there are too many overlapping units.

Loads of people here are talking about how uwa's pretty much fucked, since everybody will go to curtin instead, since we're making engineering post grad after a science degree. Nothing similar going on at curtin?

[Kaido]

Ahh, more wanting hooking up with vac work. I've got a week still to do before I can graduate, but this year i didn't get anywhere.

We have totally different assignments, I've got a few mates doing mech at Curtin, doesn't sound like there are too many overlapping units.

Loads of people here are talking about how uwa's pretty much fucked, since everybody will go to curtin instead, since we're making engineering post grad after a science degree. Nothing similar going on at curtin?

not that i am aware of, it looks like curtin is starting to invest more in their engineering degrees.

how many weeks/hr do UWA need for vac work? Do you need to do a 1st aid course aswell?

[Fujiwara13]

;724557']not that i am aware of, it looks like curtin is starting to invest more in their engineering degrees.

how many weeks/hr do UWA need for vac work? Do you need to do a 1st aid course aswell?

We've got the same 400 hours business as you guys, but no first aid. Hmmm you're not third year are you? Know anybody from group D8?

[Kaido]

We've got the same 400 hours business as you guys, but no first aid. Hmmm you're not third year are you? Know anybody from group D8?

3rd atm but still got afew 2nd year units to complete. don't really know that many people as i did 2nd yr units last sem

[Kaido]

Basically you work out the mean and standard deviation of the peg force data.

Multiply the mean and standard deviation by 2.5 to achieve the values required for the new design/timber to withstand.

Do the FBD/SFD/BMDs for both opening and closed conditions using these values.

Calculate bending stress from the bending moments in both conditions and calculate the standard deviation of these values also.

Calculate the mean and standard deviation of the timber rupture strength data for EACH timber.

For each of these timbers, use U_z = U_x - U_y and σ_z = sqrt(σ_x^2 + σ_y^2) then use these values in U_z = k*σ_z and solve for k. Compare this value k with the table from the failure/reliability lecture/tutorial and you'll get your failure %.

Choose the timber that has a low failure rate (not NO failure rate, because remember cost of timber increases proportionally to timber strength, so although it won't fail it will cost a fuckload).

Done.

hey

just a quick question how do you work out the std div of the bending stress :S

[duste]

Carry the standard deviation of force throughout the calculations.

[Kaido]

Carry the standard deviation of force throughout the calculations.

sorry i dont follow?

i have work out all the stress (F_a,F_b,F_c and the bending stresses for open and close) but not sure what to do from here

[duste]

So whenever you use F_A (that was my 'opening force', the one calculated from the collation of peg force data), perform the calculation with the standard deviation of that force.

eg. If F_A were 10N and its SD was 1, you would use 10 +- 1N throughout the calculations, resulting in a +- standard deviation for each calculation thereafter, including bending stress.

(10+-1) * 2 = 20+-2 N etc etc

Follow?

[Kaido]

So whenever you use F_A (that was my 'opening force', the one calculated from the collation of peg force data), perform the calculation with the standard deviation of that force.

eg. If F_A were 10N and its SD was 1, you would use 10 +- 1N throughout the calculations, resulting in a +- standard deviation for each calculation thereafter, including bending stress.

(10+-1) * 2 = 20+-2 N etc etc

Follow?


think so

so if my M = (30.725 +- 6.975) x0.03024
0.929 +- .2109

F=30.725
Std Dev= 6.975
yeah?

[duste]

Should be half that value.

Mine was roughly 30 +- 3.8N

But yes, that's the way. It may be easier to chuck everything in an Excel spreadsheet - do one column of calculations using F, then F + sd, then F - sd. That's what I did and do for everything, makes it much much easier to recalculate things if you make an error somewhere.

[Wrexter]

email sent of assignment and excel spreadsheet i made

[Kaido]

Should be half that value.

Mine was roughly 30 +- 3.8N

But yes, that's the way. It may be easier to chuck everything in an Excel spreadsheet - do one column of calculations using F, then F + sd, then F - sd. That's what I did and do for everything, makes it much much easier to recalculate things if you make an error somewhere.

thanks dude my mean was 12.29 and std dev 2.79
so hence when i times it 2.5x i got 6.975? did i do it correctly? i only knock out 6 values

wasn't too sure if i should knock out more, i was thinking of dropping all the 20 values and 6 values

[ossie_21]

My cat's name is Mittens.

[Kaido]

email sent of assignment and excel spreadsheet i made

thanks just got it :)

[duste]

Ah okay should be fine then, I dropped more results hence why my standard deviation was lower. So long as you explain your criteria for dropping values from the collation you'll be right.

Just printed this fucker off before - 51 pages fml.

[Kaido]

Ah okay should be fine then, I dropped more results hence why my standard deviation was lower. So long as you explain your criteria for dropping values from the collation you'll be right.

Just printed this fucker off before - 51 pages fml.

ahh ok

cool cool, yeah the shaft one is fucking long as... this peg one not so much but its fucking killing! it should be the simplest of the two

[Kaido]

ok guys got another question for you. this one is relating to fluid mech.

I need to work out the variation of acceleration as a function of time, for a duct which has a constant cross section (length 1m, h= 0.2m)

Now i have the U_o=20m/s and U_1=2m/s

so i need to find out the equation for U, i am assuming it is just U=U_o +U_1 (L*H)

[duste]

I think you may have done your Acceleration differently to me, because all I needed to do for the constant-cross-section duct was to change my value of R_x.

U = as per assignment sheet

But U is a function of flow rate and cross-sectional area (as per Lecture 7 notes).

Use this, then use a_x = du/dt + u*du/dx + v*du/dy + w*du/dz to get a_x. Obviously there'll be no dy or dz in the equation so they'll cancel out. Sub in x=1 (for the exit) after this and I guarantee you'll get a different equation to your previous one.

All you have to do now for the constant cross-section duct is to change your value for R_x (radius at any point x along the duct) to 0.2 or 0.1 instead of the equation you determined earlier for the converging duct.

[Kaido]

I think you may have done your Acceleration differently to me, because all I needed to do for the constant-cross-section duct was to change my value of R_x.

U = as per assignment sheet

But U is a function of flow rate and cross-sectional area (as per Lecture 7 notes).

Use this, then use a_x = du/dt + u*du/dx + v*du/dy + w*du/dz to get a_x. Obviously there'll be no dy or dz in the equation so they'll cancel out. Sub in x=1 (for the exit) after this and I guarantee you'll get a different equation to your previous one.

All you have to do now for the constant cross-section duct is to change your value for R_x (radius at any point x along the duct) to 0.2 or 0.1 instead of the equation you determined earlier for the converging duct.

Here it what i did for the 1st section:
http://img828.imageshack.us/img828/1398/assignment.jpg

[duste]

Okay.

Find the equation for the radius at any point x, R(x). Hint: y=mx+c with x-axis being centre of duct.

Find the equation of U at inlet, which you have done: U=20+2sin(0.3t)

Find cross-sectional area at the inlet, R(0).

Find the flow rate at the inlet, m_dot_1 = A_1*U_1

Conservation of mass tells us that m_dot_1 = m_dot_2 = flow rate = Q

Find the velocity at any point x, U(x) = Q/A(x), where A(x) is found using R(x)

Use -this- velocity equation (it should have both x and t) to find the acceleration.

Substitute x=1 (ie. exit) to find the velocity at the exit.

That is for a converging duct. For a constant cross-sectional-area duct, use R(x) = const (I chose radius of inlet).

Same as before, find a(x), then x=1 to find velocity at the exit.

Plot both equations of acceleration for one cycle.

[Kaido]

cheers duste, i see what i did wrong

[Kaido]

btw is the area of the circular duct pie*r^2 ? or is it

[duste]

Yes.

[Kaido]

cheers. all fininsh now.
just like to double check my question 2.

for 1st part you find F_d

then its just a matter of working out the following
Work done, Work done (by fuel burn, e.g energy), Cost per year and then compare this values with -1% on the F_d