How Do Material Properties Affect Wire Deformation?

[lemon]

1. A copper wire has length 200cm and radius 0.40mm. A mass of 1.00kg is suspended from the end and allowed to come to rest. The Young modulus of copper is 130Pa. Calculate:
a) the stress
b) the strain
c) the extension
d) the elastic energy stored
Repeat the calculations for the case of a carbon fibre of Young modulus 7500GPa.
2. stress=F/A
strain=deltax/L
E=stress/strain
let gravity=-10m/s
let deltax = ^x

3.
a) F = 1.0kg x 10m/s = 10N
Stress = 10/π0.0004²=25000Pa, 25kPa

b) Now I'm already stuck because I don't have ^x to calculate the strain.

 

[tiny-tim]

Hi lemon!

(have a delta: ∆ )

1. A copper wire has length 200cm and radius 0.40mm. A mass of 1.00kg is suspended from the end and allowed to come to rest. The Young modulus of copper is 130Pa. Calculate:
a) the stress
b) the strain
…
2. stress=F/A
strain=deltax/L
E=stress/strain
…
Stress = 10/π0.0004²=25000Pa, 25kPa

b) Now I'm already stuck because I don't have ^x to calculate the strain.

(oooh … correct homework template, but without the clutter … i like it! )

I don't understand … you have the stress, and you have E = stress/strain …

what am i missing? ​

 

[lemon]

Hay tT:
Ahhh! Silly me

strain = 130/25000 = 0.0052

 

[lemon]

™ … for copying-and-pasting … ™
π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô
see also Redbelly98's "useful stuff" list

Thank you

 

[tiny-tim]

Ahhh! Silly me

Too much stress and strain!

 

[lemon]

ok. So if I haven't made anymore silly mistakes:

For the Copper wire:
F=10N
Stress=25kPa
Strain=0.0052
Δx=0.0052x2.0m = 0.0104m
Ep=1/2x10x0.0104 = 0.052J

For the Carbon fibre:
F=10N
Stress=25kPa
Strain=25/(7500x10⁹) = 3.3x10^-9
Δx=(3.3x10^-9)x2.0=6.6x10^-9
Ep=1/2x10x(6.6x10^-9)=33x10^-9J

How'd I do?

 

[tiny-tim]

Hi lemon !

Fine, except that for the copper, you seem to have used E = strain/stress, and I think if the original length is given in cm, you should give the extension in cm also (well, maybe only for the copper).

 

[lemon]

tiny-tim Hi lemon !

Fine, except that for the copper, you seem to have used E = strain/stress

I haven't calculated for Young's modulus in this question tT. Not sure I understand.


and I think if the original length is given in cm, you should give the extension in cm also (well, maybe only for the copper).

Copper: Δx=0.0052x2.0m = 0.0104m = 1.04cm
Carbon fibre: Δx=(3.3x10-9)x2.0=6.6x10-9m

 

[lemon]

errr! I have just recalculated - not sure if my calculator has a problem suddenly but when I calculate the stress I get:
10N/∏x0.0004²=19894367.89

What's going on here?

 

[lemon]

ok I have to say sorry sorry sorry to everyone.
did some bum calculations there.

I have hopefully recalculated correctly.

a) F=1.0kg x 10m/s = 10N
δ=10/∏0.0004²=19894367.89 or 20x10⁶ (2s.f.) or 20.0MPa

b) E=δ/ε. ε=δ/E, Young's modulus
20.0MPa/130GPa=1.5x10^-4 (2s.f.)

c) Δx=εxL. 1.5x10^-4x2.0m=3x10^-4m, or 0.0003cm

d) E_p=1/2FΔx=1/2(10)(3x10^-4m)=1.5x10^-3J

Would somebody be good enough to see if I have finally got my head straight please

 

[tiny-tim]

Looks ok!

(except the decimal point in the cm )

 

[lemon]

ooopppss!
c) Δx=εxL. 1.5x10-4x2.0m=3x10-4m, or 0.0003m
or 0.03cm

 

[lemon]

Repeat the calculations for the case of a carbon fibre of Young modulus 7500GPa.

F=10N
δ=20.0MPa
E=7500GPa

E=20.0MPa/7500GPa=2.67x10^-6

Δx=(2.67x10^-6)x(3x10^-4)=8.01x10^-10m
0.801nm, or 80.1ncm