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Stress / strain - pt2

  1. Feb 6, 2010 #1
    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.
     
    Last edited: Feb 6, 2010
  2. jcsd
  3. Feb 6, 2010 #2

    tiny-tim

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    Hi lemon! :smile:

    (have a delta: ∆ :wink:)
    (oooh … correct homework template, but without the clutter i like it! :biggrin:)

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

    what am i missing? :redface:
     
  4. Feb 6, 2010 #3
    Hay tT:
    Ahhh!! Silly me :redface:

    strain = 130/25000 = 0.0052
     
  5. Feb 6, 2010 #4
    ™ … 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
     
  6. Feb 6, 2010 #5

    tiny-tim

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    Too much stress and strain! :biggrin:
     
  7. Feb 7, 2010 #6
    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?
     
  8. Feb 7, 2010 #7

    tiny-tim

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    Hi lemon ! :smile:

    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). :wink:
     
  9. Feb 8, 2010 #8
    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
     
  10. Feb 14, 2010 #9
    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?
     
  11. Feb 15, 2010 #10
    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) Ep=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
     
  12. Feb 16, 2010 #11

    tiny-tim

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    Looks ok! :smile:

    (except the decimal point in the cm :wink:)
     
  13. Feb 16, 2010 #12
    ooopppss!!!:redface:
    c) Δx=εxL. 1.5x10-4x2.0m=3x10-4m, or 0.0003m
    or 0.03cmo:)
     
  14. Feb 16, 2010 #13
    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
     
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