Simple maths question

[Punx0r]

Ok, I suck at maths.

It's a simple question:

If the tensile strain in an object is measured at 0.0002 and the dia. after loading is 484.3mm, what was the original dia.?

The answer is also simple and intuitive: divide the first number by the second to give the change in dimension.

But assuming you didn't intuitively know that, how would you work it out? Assuming you're just given the formula for strain (strain = change in dim. / orig. dimension)?

I know it's simple, but I just can't rearrange that simple formula to give the answer. I've spent an hour covering a page in jibberish trying to transpose the original formula to the one that actually gives the answer.

[618ireland]

It's not that simple really

We know two things

1) Orig dim - change in dim = 484.3mm
2) 0.0002 = change in dim. / orig. dimension........... so change in dim. = 0.0002 * orig. dimension

So let's look at 1) for a second...
Orig dim = 484.3mm + change in dim (call this line Rover)

then into line Rover throw in the value for change in dim. derived from 2)
...Orig dim = 484.3mm + (0.0002 * orig. dimension)

rearrange .......Orig dim - 484.3mm = (0.0002 * orig. dimension)
and rearrange again ..........Orig dim - 484.3mm - (0.0002 * orig. dimension) = 0
and rearrange again..........Orig dim - (0.0002 * orig. dimension) = 484.3mm

common denominator of orig dim on the rhs of that equation............
so ..........Orig dim (1-0.0002) = 484.3mm
so .......Orig dim = 484.3mm / (1-0.0002) = 484.3mm / 0.9998 = 484.397mm to 3 dec pl

[618ireland]

.............

If the tensile strain in an object is measured at 0.0002 and the dia. after loading is 484.3mm, what was the original dia.?

The answer is also simple and intuitive: divide the first number by the second to give the change in dimension.

But assuming you didn't intuitively know that, how would you work it out? Assuming you're just given the formula for strain (strain = change in dim. / orig. dimension)?

I know it's simple..............

The reason it's not simple is that neither the actual change in diameter or the original is known

[radddogg]

[Punx0r]

Aha! A man who obviously did better than me at school

I don't think it's a biggy, but the given answer is 484.203mm. I think in line 1) you wanted a plus rather than minus? I assume the working still stands though

If it helps any I was using:

O = original dimension
C = Change in dimension
S = Strain
T = C+O

I also have a question about shear modulus if you're not up to much

[618ireland]

.........

I don't think it's a biggy, but the given answer is 484.203mm. I think in line 1) you wanted a plus rather than minus? I assume the working still stands though ..........

To be honest I dunno about that, if the answer is 484.202mm than the thingy got bigger after the strain was experienced, I threw in a minus rather than a plus as I presumed the tensile strain experienced was a pulling apart, thus the original diameter should be greater than the diameter after loading, seems as it was 484.3mm after loading I'd be reckoning I'm right and the book is wrong

[Drew]

the dimension would increase after strain. i think the confusion comes from halfway through 618's first post, must have been thinking dimension for most of it, but then gets into diameter by the second post. a diameter would decrease after strain but not uniformly over a length.

rearranging furmula is easy enough. you just need to counteract whats on one side of the equals sign with the other.

given a = b/c

as c is under the divide sign you need to times it on the other side to counteract. this becomes c X a = b

Drew

[Chris_TI]

???

[618ireland]

Apologies Drew is quite right. I fell into the diameter muck as dia is mentioned in the first post .

[Dave]

If that's a simple maths question, what's a difficult one??!!!

Cheers
Dave

[t'mill]

Contender for the dullest thread in internet history

[escortpop]

Contender for the dullest thread in internet history

[E_T_V]

Sommat to do with poissons ratio I think.

[Punx0r]

A good point has been raised: why would something experiencing tensile strain get fatter?

The full question:

"The tensile strain in a component was measured using a strain gauge at 0.0002. Determine the original diameter of the component if the diameter after loading was measured at 484.3mm. Ans: 484.203mm"

I'm just lacking decent technique on the rearranging formula.

I've got the way you mentioned drew: a = b/c, rearrange for b:

ac = bc/c -> ac = b

It occured to me yesterday though, it seems sometimes you must divide the divided side. Arrange for c:

a/b = b/(b/c) -> a/b = c

I don't know if that's right, but it seems to work.

Anyway, I mentioned shear modulus:

"The maximum shear stress in a shaft is 45MN/m^2. If G for the shaft material is 79 GN/mm^2, determine the shear strain. Ans: 0.57^-3 radians)"

I thought that was a missprint - that it should have been 79 GN/m^2. But you must use mm to the get the answer, and the answer itself seems sensible?

Contender for the dullest thread in internet history

After the threads on The X-Factor and Big Brother?

[vinny19791]

What the hell is this logger

[Beaker]

I've got the way you mentioned drew: a = b/c, rearrange for b:

ac = bc/c -> ac = b

It occured to me yesterday though, it seems sometimes you must divide the divided side. Arrange for c:

a/b = b/(b/c) -> a/b = c

To me it seems you are over thinking it.

a=b/c ....Another way to think about it assume the equation is a always a numerator/denominator. To rearrange - anything on the bottom right, moves to the top left. Anything on the bottom left, moves to the top right.

[Drew]

I've got the way you mentioned drew: a = b/c, rearrange for b:

ac = bc/c -> ac = b

It occured to me yesterday though, it seems sometimes you must divide the divided side. Arrange for c:

a/b = b/(b/c) -> a/b = c

I don't know if that's right, but it seems to work.

it doesnt. lets say:
a = 0.5
b = 5
c = 10
so a = b/c ~ 0.5 = 5/10

if you wanted to know what c was on its own id first move it over the equals sign
c X a = b

then move a across by dividing the other side, leaving;
c = b/a ~ 10 = 5/0.5

Drew

[Punx0r]

That's definitely a better way. I was just doing equal things to both sides of the equation and trying to cancel things out. Then fudged the check

Definitely need to reducate myself a bit here...

Edit: just noticed I put it the wrong way around. a/b = c should have been b/a = c - typo...

What the hell is this logger

Secondary school maths.

[Beaker]

What the hell is this logger

Excitement! No that is not sarcasm I love maths.

[Punx0r]

I hate maths. Boo!

[Null_Byte]

I have just skim read this so do excuse the silly question, but isn't there is a vital point missing in your original formula?

strain = change in dim. / orig. dimension

simply rearranging the formula to give orig. dim will not work, as the final measurement is not the change in the measurement.

for example:

If we take S as strain, O as your original measurement and F as your final measurement the formula would look more like:

Code: Select all

S = O - F
    ____
      O

you can then work this through fairly easily

Code: Select all

S = O - F 
    _   _
    O   O

Code: Select all

S = 1 - F/O

You can go two ways now, but I'll do it this way for the sake of simplicity

Code: Select all

F + S =1
_
O

Code: Select all

F = 1 - S
_
O

Multiply both sides by O

Code: Select all

F = O - SO

Factorise

Code: Select all

F = O(1-S)

Find O

Code: Select all

 F     =  O
___
1-S

Substituting your values would give 484.39mm


Probably simpler ways to do it, but it has been a long while since I did this sort of thing.

[Din]

Secondary school maths.

I dont remember doing this sort of thing in maths...

There again i got a D in GCSE maths sooo

[Punx0r]

We did I promptly forgot most of it though... Searching the web for help on these sorts of things often brings up learning packages for schoolkids. The other day someone asked me to multiply/divide two fractions and I just couldn't remember how. In fact, I looked it up, got the hand of it, and right now I can't remember it

Jon, I think what you've got there is what I was looking for, I think.

You're right, there is a variable missing in the original formula. That's why I was throwing in T (your F) as well (total dim. original + change).

In your first step, don't you want s = F - O, rather than O - F? To give change in dim.?

[Null_Byte]

In your first step, don't you want s = F - O, rather than O - F? To give change in dim.?

No, I don't think so, because your final value is going to be smaller than your original value if it is stretched.

[618ireland]

Jon got the same answer as me

[Punx0r]

Ah yes, I forgot about that issue. So it seems the question is wrong, then? It should have said compressive, not tensile, strain?

[618ireland]

I reckon the reference to diameter should be length in the question.

[paul888]

I Hate maths.

[Dave_ZS]

wow, that's just fried me noggin

welcome to : helpwithsocalledsimplemathsquestionstech.net

[t'mill]

Contender for the dullest thread in internet history

After the threads on The X-Factor and Big Brother? [/quote]

Bring back the egg thread

[bjrespect]

Contender for the dullest thread in internet history

After the threads on The X-Factor and Big Brother?

Bring back the egg thread [/quote]

now that was a eggselent thread

[peteT16]

Contender for the dullest thread in internet history

After the threads on The X-Factor and Big Brother?

Bring back the egg thread

now that was a eggselent thread [/quote]

That's a terrible yolk

[bjrespect]

your yolking not this again

i thort these sort of threads were eggstinct

[t'mill]

I'm glad it wasn't laid to rest. It was a smashing thread

[bjrespect]

i think somebody may have poached the thread