Yeah I can do, I currently have a sample that's 20x40mm.
The treatment is for homonegisation of the material so should be the same, over time I am expecting to see an increase in delta phases and gamma prime and a decrease in laves phase.
I have tried XRD but for some reason only showed the...
Afternoon all,
Hopefully somebody can help me, I'm doing my final year project and it's looking at the effect of heat treatment on in17, when I run an XRD scan I found that I all the phases sort of hid behind the matrix and so can't really make them out.
So I've been looking at using the SEM...
Good morning,
I'm trying to workout the time for an element to diffuse at set distance in microns.
I have the distance, the diffusion coefficient, just unsure which equation I actually use.
X= sqrt DT or the other one x= sqrt 2DT.
I can't seem to figure out when you use one and not the...
No, we have just performed a tensile and 3point bend test on a
(0/90/0/90/0/90/0/90) 8layers
(0/45/-45/90/90/-45/45/0) 8layers,
the question i need to answer is how the stacking sequence effects the properties. I've tried reading laminate theory but just don't seem to understand. Am i right in...
Evening all,
im hoping somebody can help or point me in the right direction.
We have tested these two laminates, one tensile and other three point bend test. The (0/90) has a higher uts, elastic modules, and flexure strength.
(0/+45/-45/90) has a higher flexural modulus. Both roughly same...
Hi again thanks for your update.
x(t)= c1sin(5√5t)+c2sin(5√5t)+1/109cos(4t)
x(t)= 0 c1= -1/109
x'(t)= (-4sin(4t)/109)- c1(5√5sin(5√5t))/(2√t)+c2(5√5tcos(5√5t))/(2√t)st
just checking this is ok before i sub c1 and -0.5 into x'(t) to get c2. solution will then be
x(t) with values of c1 and...
Ok can you see if this is correct.
y"+0y'+25=0
so roots are r+/-sqrt25
r=0+5i
r=0-5i
into
y(t)= Acos4t+Bsin4ty(t)=0
y'(t)=0
0+0+125A=1cos(4*0)
A=1/125
x(t)=Acos4t+Bsin4t+(1/125)
x'(t) = 4Acos(4t)−4Bsin(4t)
sub initial value into it
x(0)=0
X'(0)=-0.5
X(t)= A+(1/125)
A= -(1/125)
x'(t)=...
Thank you. I'm guessing the second part. M(d^2x/dt^2)=0.2cos(4t)-kx.I'm guessing you rearrange to get d^2x/dt^2+k/mx=0.2cos(4t)/m then do the same process?
If I had worked it out correctly I would have got the same. Is that the answer to the question? I think I'm going to have to do some more reading. Do you do the same process for every first order ode?
So I'm guessing I do the step I did become
Subbing x(0)=0.8 x'(0)=0
So therefore
0.8=c1+c2+ 299/625
-c1+0.3216=c2
X'(0)=0= (-25/2)c1+(-10)c2
Sub them into I get.
=1.2864e^(-25/2)t-0.9684e^-10t
Thank you for helping I'm literally learning as you post.
Hi Mark.
Thanks for you reply. I don't really now what to do. I'm guessing it goes it the form.
y"+22.5y+125=0? Dont know what to do with the g+(kl) /m side of the equation
Regards
James.
Afternoon All
I have a math question I don't actually have a clue what to do. Can some help me out.
A mass M is suspended vertically by a damped spring of length L and stiffness k such that the distance x between the centre of the mass and the top of the springis given by
M (d^2 x)/(dt^2...