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Vertical deflection, stress, and strain Help!

  1. Feb 22, 2010 #1
    The shear modulus of aluminum is 2.4 × 1010 N/m2. An aluminum nail of radius
    7.5 × 10–4 m projects 0.035 m horizontally outward from a wall. A man hangs a wet raincoat of weight 25.5 N from the end of the nail.
    Assuming the wall holds its end of the nail, what is the vertical deflection of the other end of the nail?

    What is the stress for this situation?

    What is the strain?





    Can someone work this out? I am super stuck! Thanks in advance
     
  2. jcsd
  3. Feb 22, 2010 #2

    PhanthomJay

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    I'm assuming you are looking for shear stress and shear deflection for this problem, and not the bending stress and deflection due to bending. Have you not been given a formula for determining deflections and stresses? You'll have to try to work it out first.
     
  4. Feb 22, 2010 #3
    So I did some more digging, this is what i got, is it correct?

    E= 7.0 x 10^10 N/m^2 (youngs modulus of aluminum)
    A= 1.77 x 10 ^-6 N/m^2

    vertical deflection: deltaY= (F/A)(L/G) (25N/(1.77 x 10 ^-6 N/m^2))/(0.035m/(2.4×10^10 N/m^2)) = 2.06 x10^-5 m

    stress= F/A = 25N/(1.77 x 10 ^-6 N/m^2)= 1.4 x 10^7 N/m^2

    E = stress/strain ---> strain = stress/E
    (1.4 x 10^7 N/m^2)/(7.0 x 10^10 N/m^2)= 2.0x10^-4 = strain

    ?? good?
     
  5. Feb 22, 2010 #4

    PhanthomJay

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    I am not sure of the question. It asks for vertical deflection, stress, and strain, but it gives only a shear modulus, G. Shear deformations and strains are a function of G, whereas bending deformations and strains are a function of E, the elastic modulus. I don't know why the problem does not give an E value...you shouldn't have to look it up. Unfortunately, I never use the metric system in engineering, so i don't have a good feel for the numbers..too many darn decimals in SI, so i don't have the will to crank out the numbers to see if bending deformations are more significant than shear deformations for this case, or vice versa. In any case, don't confuse shear stresses and shear strain with bending stresses and bending strains. Where the shear deformation can be calculated (approximated)as you have stated (delta = FL/AG), the bending deformation is FL^3/3EI. And while the the shear stress is F/A, the bending stress is Mc/I, and where the the shear strain is delta_shear/L ,the bending strain is bending stress/E. So where are we now, i don't know, i've even managed to confuse myself :frown:. Please restate the problem as written.
     
  6. Feb 23, 2010 #5
    I wasn't sure what to due for calculating the strain, so i used the eq. tht included E....
    the question is word for word:

    The shear modulus of aluminum is 2.4 × 1010 N/m2. An aluminum nail of radius
    7.5 × 10–4 m projects 0.035 m horizontally outward from a wall. A man hangs a wet raincoat of weight 25.5 N from the end of the nail.
    Assuming the wall holds its end of the nail, what is the vertical deflection of the other end of the nail?

    What is the stress for this situation?

    What is the strain?
     
  7. Feb 23, 2010 #6
    Is anyone certain of how to calculate the strain?
     
  8. Feb 23, 2010 #7

    PhanthomJay

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    This problem is poorly worded. They give you the shear modulus, G, but not the elastic modulus, E. An then they ask you to calculate deflection, stress, and strain, when they shuuld have specified whether they were looking for deflection due to shear or bending , and shear stress vs. bending stresses, and shear strain vs. bending strain, or both. I cranked out some real rough numbers, and it looks like shear stresses and deflections pale in comparison to bending stresses and deflection at the free end as caused by bending moments. But since they gave you G and not E, i guess they are asking for shear strains and stresses and deflections, which are extremely small. Have you studied these equations? They are usually not covered as well as the bending stresses and strains and deflections.
     
  9. Feb 23, 2010 #8

    PhanthomJay

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    see comments above
     
  10. Feb 23, 2010 #9
    yes, I too have been looking into it,


    The above strain is wrong, as you stated with E value my book states shear strain to be:

    strain= ΔL= (1/G)(F/A)

    =(1/2.4×10^10 N/m^2)(25.5N/(1.77 x 10 ^-6 N/m^2) =2.10 x 10^-5 m
    so I am going with this for my solution for strain.
     
  11. Feb 23, 2010 #10

    PhanthomJay

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    Correct answer, wrong terminology for shear strain. If you are trying to find shear strain, you must use G, not E. Shear strain = shear stress/G, or , alternatively, shear strain = shear deformation/L , or shear deformation = (FL/AG)/L = F/AG, same result. Do not call shear strain delta L; delta L has units of length;strain has no dimension (it is a dimensionless quantity).
     
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