Solving for Force and Restraint in Materials with Temperature Changes

[Dell]

how do i go about solving this

im not sure I've learned this material yet but i have learned defrmations and deflections,
i need to somehow find the force applied to the bars due to their exopansion and the restraint.

what is \alpha exactly? the increase in length for each degree F??

 

[mgb_phys]

\alpha[/tex] exactly? the increase in length for each degree F??

Yes it's the fraction increase in length with temperature.

 

[Dell]

line the strain per degree?
can i use superposition here, find the strain on each material annd then add them? how will i take into account the frame if the total deflection comes more than 0.02in

 

[mgb_phys]

Yes, you assume the base doesn't change an the two parts each change in length by whatever factor.
Then you have two bars that are under a certain strain (they want to be longer than they are) you then use Young's modulus to fin the stress and so the force

 

[Dell]

so then can i say that
εxx=α*T
εyy=0
εzz=0

for aluminium
εxx=α*T=2.56*10^-3

for bronze
εxx=α*T=2.02*10^-3

now i have all the strains for both materials
from a table i get the poisson ratios
aluminuim->0.33
bronze->0.34

(aluminium) σxx= 37.93*10³Pa
(Bronze) σxx= 59.1*10³Panow what i tried here was Fx=σxx*A but that doesn't come right, how do i tak the interaction between the 2 materials into account?? i think that may be where I am going wrong?

 

[mgb_phys]

so calculate how long each bar would expand to if it was free.
then if this is more than the 0.2" extra space - you have two bars with a total length of X compressed into 27.2" from the youngs moduluses work out how much each bar would be compressed ad there fore the stress and strain.
Note the bars are different materials so the stronger one will squash the weaker one

 

[Dell]

for Bronze
εxx=α*T=0.00202
A=2.5
E=15*10⁶
v=0.34

according to hookes law

σxx=\frac{E(1-v)εxx}{(1+v)(1-2v)}

=\frac{15*10⁶(1-0.34)*0.00202}{(1+0.34)(1-2*0.34)}=46637.12687

Fx=σxx/A=46637.12687/2.5
Fx=18654.85075

for aluminium
εxx=α*T=0.00256
A=3
E=10*10⁶
v=0.33

according to hookes law

σxx=\frac{E(1-v)εxx}{(1+v)(1-2v)}

=\frac{10*10⁶(1-0.33)*0.00256}{(1+0.33)(1-2*0.33)}=37930.11942

Fx=σxx/A=37930.11942/3
Fx=12643.37314Fx_Bronze-Fx_Al= 6011.47

 

[Dell]

another way i looked at it was through the deflection, now since i only have force on the x axis

delta=εxx*L
=σxx*L/E
=(F*L)/(A*E)

now i know that the total deflection is 0.02

0.02=(F*15)/(3*10^7)+(F*12)/(2.5*15*10^6)

F=24390.2439

i just can't get anywhere near the correct answer, obviously i am doing something fundamentally wrong

 

[mgb_phys]

Bronze a=10.1e-6, original length = 12"

Al a=12.8e-6 orig = 15"

so after 200f,
Al = 200*10.1e-6 * 15 = 0.0303 extra, so 15.03 total
Br = 200*12.8e-6 *12 = 0.0307 extra, so 12.03 total

total extra length = 0.061 an we only have 0.02 so we have 0.041 extra compression providing the stress

The force in both bars must be the same - otherwise they would move!
So we have an unknown expansion of each but a known total expansion.
We know the total strain of the two bars
From the strain and E (and A) we can find the stress and so the force.

Should just be a couple of simultaneous equations

 

[Dell]

εtot=(0.02) / 27
=7.407407e-4

σ=ε*E
F=ε*E*A

F=al*10e6*3
F=br*15e6*2.5
al*15+br*12=0.02

after solving i get
F=24390.2439
al=8.13e-4
br=5.65e-4still no good

i can see where this might be wrong, nowhere here do i take into account the amount that each material expands.
but i have no idea how to fix it

 

[Dell]

i think i found a mistake, in the 3rd equation

F=al*10e6*3
F=br*15e6*2.5
al*15+br*12=0.041

F=50 000
al=1/600
br=1/750delta=al*L

=15/600=0.025still not the right answers

 

[Dell]

i think i might finally have it (again)

the strain i found in my last post was the strain that the force f applied, but i cannot find the deflection from this

what i think i need to do is find some kind of effective strain which is the difference between the strain had there been no restriction and the strain i just found ??

 

[Dell]

got it! thanks for all the help