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Poisson again - totally stumped as compressive column?

  1. Aug 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Column supports a mass on its' top. So force is downwards.
    Column properties:
    Do = 50mm (outer dia)
    Di = 40mm (inner dia)
    E = 250 GNm^-2 (modulus of elasticity)
    V = 0.33 (Poissons ratio)



    2. Relevant equations

    Poissons ratio = Transverse strain = - εt / εl

    Transverse strain εt = (Reduction in Diameter - Original Diameter) / Original Diameter

    Axial strain = Transverse strain / V

    Δd = εtd

    ε = εt / -V

    ε = σ / E

    σ = F/A


    3. The attempt at a solution

    Those that saw my recent post will realise I've worked out how to calculate axial strain / transverse strain etc.
    But that was where strain was 'pulling' a member and thereby reducing its' diameter and increasing its' length.

    This problem is the opposite! There are also some additional technical questions related to strain gauges and wheatstone bridges. But I'll discuss this with my tutor Monday when he is available...

    In the mean time I want to crack the first part of the problem.

    The question:
    Sketch a graph of expected strain against applied load, for the range 0 to 500kg. Make load the horizontal axis and strain the vertical.
    The column is fixed vertically and the mass is placed on top.

    So intuitively the diameter will increase and the length will decrease.

    My first question is do I apply the same formula as before in my last post? Or will being a column effect the required equations?

    My previous workings:

    Transverse strain εt = (Reduction in Diameter - Original Diameter) / Original Diameter
    (0.3999 - 0.4000) / 0.4000 = -0.00025 Transverse strain (εt)

    Axial strain = Transverse strain / V
    -0.00025 / 0.3 = .00083333333 Axial strain

    I confirmed these by dividing Transverse strain by Axial strain = -0.3

    Since decrease in diameter = transverse strain x Original Diameter then:
    Δd = εtd = -0.00025 x .40 = 0.1mm which is correct.

    As εt = -Vε Then:
    -0.00025 = -0.3 x ε

    ε = εt / -V
    ε = 0.00083333333

    ε = σ / E
    σ is unknown ∴
    σ = ε x E
    σ = 16666666 Pa (stress value)

    σ = F / A

    Force = σ x A (Area)

    -----------------------------------------------------------------------------------------------------------

    So far I have done the following:

    Calculated area by A = ∏/4 [(0.05m^2) - (0.04m^2)]
    A = 7.0685834 x10^-4 m^2
    E = 250 x10^9 Nm^-2
    l = 0.2m
    Force = mass x gravity

    I calculated force at 50kg intervals to 500kg
    So F = 490.5N @ 50kg
    to F = 4905N @ 500kg

    I then calculated stress as σ = F/A , using the 10 values of force I derived.
    So σ = 693915.6 Pa @ 490.5N
    to σ = 6939155.6 Pa @ 4905N

    So I believe I have the load (force) worked out to plot already correctly?

    And can use ε = σ / E to calculate strain for the vertical axis?

    Many many thanks,

    Lloyd
     
    Last edited: Aug 1, 2013
  2. jcsd
  3. Aug 4, 2013 #2
    How to calculate strain against an applied load on top of a column

    1. The problem statement, all variables and given/known data
    This question: (I also posted in Engineering but had no replies :-(, but have since done more work, so hopefully will get a nudge?).

    Tubular column supports a mass on top (column is vertical). X is the horizontal, Y is the vertical.
    Outside diameter = 50mm
    Inside diameter = 40mm
    Modulus (youngs) = 250 GN m^-2
    Poison's = 0.33
    Mass varies from 0 - 500kg.
    Column length = 200mm
    A strain gauge is applied to the column and measures strain in the horizontal plane.

    Ignore the weight of the column for all calculations.

    Q1: Sketch a graph of the expected strain against the applied load, for the range 0 - 500kg. Make the load the horizontal axis on the graph and the train the vertical axis.


    2. Relevant equations

    Area = ∏/4 [(Do^2) - (Di^2)]
    F = mg
    σ = F/A
    ε = σ/E


    3. The attempt at a solution

    So using the above equations, and from the question, I don't think the strain gauge position is relevant (yet - there are more questions after). I "think" the question is saying calculate strain downwards? On this basis I have done the following:

    Area = ∏/4 [(Do^2) - (Di^2)] = 7.0685834x10^-4 m^2 (I've ignored the length for area, as most examples for strain do this in the examples in book but am unsure if I'm right to omit the length?)

    Calculated F = mg for values 50kg to 500kg in ten increments giving 490.5N to 4905N

    Calculated σ = F/A for above F values giving 693915.55Pa to 6939155.59Pa

    Calculated ε = σ/E for above values giving 0.00000277566 to 0.00002775662

    Plotted a graph with x axis Load(kg) 0 to 500kg and y axis strain (ε x10^-6) 0 to 28

    My graph looks good as a straight diagonal line with strain increasing from 0 to 27.75662 x10^-6 at 500Kg.

    Does that seem right?

    Thank you!
     
  4. Aug 4, 2013 #3

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    LDC1972: The answers to your four questions in post 1 are, yes, no, yes, yes. The answer to your last question in post 2 is, yes.

    (1) By the way, always leave a space between a numeric value and its following unit symbol. E.g., 0.2 m, not 0.2m. See the international standard for writing units (ISO 31-0).

    (2) Two unit symbols multiplied together must be separated by, e.g., an asterisk or a space. E.g., N*m, not Nm.
     
    Last edited: Aug 4, 2013
  5. Aug 5, 2013 #4
    Thank you for the reply.

    I do apologise for the messy post and double posting(!) - I've been "told off"!

    Thanks for the reply, I'm going to keep at this all week, it's a long question in the whole part.....
     
  6. Aug 6, 2013 #5
    Further help / confirmation needed please?

    Question 1: calculate change in column length:

    I have done the change in ten increments of 50 kg through to 500 Kg. But to keep the post short, for ultimate length at 500 kg / 4905 N = Δl = εlo (change in length = strain value at 4905 N x original length) =
    27.75662x10^-6 m x 0.2 m = 0.005551324mm
    Lo - Δl = 199.9944487 mm

    Question 2: Here's where I'm getting confused! Question is increase in diameter? Now, once I have these values I can start to also get strain in the horizontal plane as well, which leads me nicely to sorting out the most appropriate positioning of the strain gauge.

    What I have done so far:

    Increase in diameter at maximum load (500 Kg / 4905 N):
    εl = Δl / lo (compressive strain = change in length / original length)
    εl = 0.00000555132 m / 0.2 m = 0.0000277566

    εt = Δd/ do (expansion strain = change in diameter / original diameter)
    Δd = unknown, therefore:
    We do know poisson's = 0.33, therefore:
    v = εt / εl transpose to
    εt = v x εl = 0.33 x 0.0000277566
    εt = 0.00000915967 (expansion strain).

    Δd = εtd = 0.00000915967 m x 0.05 m = 0.00045798 mm (expansion strain at 500kg load of outer diameter of column).

    If the above is right then I can proceed with the question change in inner diameter in same manner? And having those values calculate strain in the horizontal plane?

    Thank you.....
     
  7. Aug 7, 2013 #6
    I've just realised that these values should be negative as stress and strain are are compressive in the vertical plane?
     
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