Question

mechanics of materials

Answer 1

youngs modulus

\[E= stress / strain \]
\[E= \frac{\sigma } {\Delta L/L}\]

Answer 2

if we change stress by \( d\sigma\) , do we get
\[E= \frac{d \sigma } {d L/L}\]
?

Answer 3

this bothers me becuse, on re-arranging
\[\frac{dL}{d \sigma}= L/E\]

Answer 4

the solution to that is exponential,
length changes exponentially when stress is increased

Answer 5

i was expecting it would be something similar to a spring,(linear)
F=kx

Answer 6

it is a spring. i don't see how you got it was exponential.\[F=\color{#c00}{\frac{YA}L} x\]

Answer 7

youngs modulus is defined as stress by strain...
lets consider unit cross section forget the A...

strain is \[\Delta L / L\]

Answer 8

and we have stress proprtional to "strain" from youngs modulus

unlike a spring, where force proportional to change in lenght

Answer 9

is it not the same case here?

Answer 10

as in, from the yongs mod formula,

if i had specimens of two different lenghths L1 and L2, I would need DIFFERENT stresses to elongate BOTH by 1 metre

Answer 11

yes of course. but the real thing is if you double the respective stresses, then the change in length would be double. all springs are different in nature.

Answer 12

doubling stress would double "strain" i.e change in length divided by original length,

the actual change in length would not double, it will be something else....
?

Answer 13

the original length is constant anyway... if a double \(cx\), you also double \(x\).

Answer 14

\(L = 10cm\)

\(\Delta L = 1cm\implies \dfrac{\Delta L}{L} = 0.1\)

\(\Delta L = 2cm\implies \dfrac{\Delta L}{L} = 0.2\)

Answer 15

yes what if you are inceasing the stress in steps?

\[\Delta L = ?\implies \dfrac{\Delta L}{1.1} \]

Answer 16

* 11.1

Answer 17

the reason why we look at "strain", that is, the percentage change in dimension, is because young's modulus is the property of a MATERIAL. we've observed that the PERCENTAGE change brought about by a given stress for all instances of a material is the same (for small stresses only, of course) no matter what the initial dimensions are

Answer 18

well doubling the stress would not double the strain in all cases..
we can conclude this looking at the graph

Answer 19

yea, i'm consider inside elastic limit

Answer 20

*considering

Answer 21

\[\frac{dL}{d \sigma}= L/E\]

This does give us an exponential relationship between \(L\) and \(\sigma\)

Answer 22

i think its boiling down to
is L a constant or a variable??

Answer 23

That eqn is telling us that the length of rod grows exponentially as the stress is increased

Answer 24

a constant for a given rod!

Answer 25

just as you would wonder, "is cross-sectional area constant or variable?"

Answer 26

so we talking for cases in which elastic limit is not reached..?

Answer 27

physics aside, I'm just interpreting the differential eqn..

Answer 28

what if i changed stress in small steps,

the new L would be the old L + the elongation

Answer 29

oh, you're talking about it like that. nice.

Answer 30

@imqwerty i only considering parts where that graph is a straight line

Answer 31

yes... if the steps are infinitely small....

Answer 32

now i fell there is something wrong with that...
because of all the spring mass problems i've solved xD

Answer 33

*feel

Answer 34

\[E= \frac{d \sigma } {d L/L}\]

this is something i made up, thinking what would happpen if i change stress in steps..

i do not know if it is correct

Answer 35

that follows directly from stress = E*strain right ?

Answer 36

yep

Answer 37

this is directly contrasting with the case of a spring,

in a spring, the actual length of the spring does not matter, only its elongation with respect to equilibrium.

however the yonugs modulus shows that its actulal length matters...

if we were going to treat 'L' as a constant,

then why define youngs mod as ratio of "stress to strain" rather than simply "stress to elongation" just like a spring?

Answer 38

i answered that before you even asked haha

the reason why we look at "strain", that is, the percentage change in dimension, is because young's modulus is the property of a MATERIAL. we've observed that the PERCENTAGE change brought about by a given stress for all instances of a material is the same (for small stresses only, of course) no matter what the initial dimensions are

Answer 39

and why do you think that in case of a spring the actual length doesn't matter?! it does! it's just that in the equation of a spring, you do not get to see the dependence on length.

Answer 40

F=kx ?

Answer 41

Yes.\[k = \frac{YA}{L}\]

Answer 42

thats the formula for ^spring constant ? i.e specifically taken in the context of springs?

Answer 43

yes, of course