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One shaft two diameters angle of deformation

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Compute the angle of twist of the free end relative to the fixed end of the steel bar:
    200 N*m, 80 x 10^9 GPa (shear modulus of elasticity)

    (Length 1: 1.2 m, dia of .040 m on left, length of .4 m dia of .020 m, on right)


    2. Relevant equations

    angle = (torque)(length) / (shear modulus of elasticity)(J of shaft)

    3. The attempt at a solution

    (200 n*m)(1.2 m) / (80 x 10^9)(25.1 x 10^8) = 1st section of shaft
    (200 n*m)(0.4 m) / (80 x 10^9)(1.5 x 10^8) = 2nd section of shaft

    = 0.012 + 0.0064 = 0.018 rad

    Book answer: 0.0756 rad

    Thanks!
     
  2. jcsd
  3. Feb 21, 2012 #2
    no idea?
     
  4. Feb 21, 2012 #3

    nvn

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    togo: Exponent or units on G (shear modulus of elasticity) listed wrong in section 1 of post 1, but listed correctly in section 3. Exponent on J listed wrong in post 1. J values rounded or truncated too much. Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits. Unit symbol for newtons is uppercase N, not lowercase n. Lowercase n means nano. Numbers less than 1 must always have a zero before the decimal point. E.g., 0.4 m, not .4 m. Try again.
     
    Last edited: Feb 21, 2012
  5. Feb 22, 2012 #4
    rounded or truncated too much. Did you run the calculation yourself? I've been using about 4 sig figs throughout the book with accurate results up to this point. G is correct afaik.

    ps.
    .040^4 = 25.1 x 10^-8 according to my calculator with no additional figures
    .020^4 = 16 x 10^-8 also, no additional figures
     
    Last edited: Feb 22, 2012
  6. Feb 23, 2012 #5

    nvn

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    togo: That is not the correct formula for J. Check your book to find the relevant equation for J. Try again.
     
  7. Feb 23, 2012 #6
    J = (pi x d^4) / 32
     
  8. Feb 23, 2012 #7

    nvn

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    Right.
     
  9. Feb 24, 2012 #8
    that's the formula I used, thanks

    answer is still wrong
     
  10. Feb 24, 2012 #9

    nvn

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    togo: Try it again, correcting the mistakes explained in post 3. Show your work.
     
  11. Feb 25, 2012 #10
    .040^4 = 25.1 x 10^-8
    .020^4 = 16 x 10^-8

    J1 = .04 x 10^-8 x (pi/32) = 2.513 x 10^-7
    J2 = .02 x 10^-8 x (pi/32) = 1.571 x 10^-8

    (200 n*m)(1.2 m) / (80 x 10^9)(2.513 x 10^-7) = 1st section of shaft = 0.0119 rad
    (200 n*m)(0.4 m) / (80 x 10^9)(1.571 x 10^-8) = 2nd section of shaft = 0.0636 rad

    =0.0756 rad
     
  12. Mar 1, 2012 #11

    nvn

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    togo: The formula you listed for J1 and J2 in post 10 is currently wrong, and does not match the right-hand side. Also, review the information in post 3.

    (1) The unit symbol for newton is uppercase N, not lowercase n. Lowercase n means nano.

    (2) Numbers less than 1 must always have a zero before the decimal point. E.g., 0.04, not .04. See the international standard (ISO 31-0). Or see any credible text book.​
     
    Last edited: Mar 1, 2012
  13. Mar 2, 2012 #12
    thanks for that nvn. please don't respond to my help threads anymore

    I will be seeing an instructor about this one

    you will note that the correct numbers are in the final formula and you've neglected getting to the heart of the matter for half a month now
     
    Last edited: Mar 2, 2012
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