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Traction/ Compression stress and strain exercise

  1. Aug 26, 2016 #1
    Hello everyone, could you help me with this exercise? I am stuck, and I can't find anything on the internet that solves this. Your help is very much appreciated :)

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

    Circular steel bar, clamped at extremities.
    Two parts: A(ab)=800mm² A(bc)=400mm²
    L(ab)= 400mm L(bc)=300mm
    Axial Forces= 100kN
    Young modulus E= 210GPa

    Need to find: normal stress in each bar, and the horizontal displacement at point B.

    2. Relevant equations
    Sigma (Stress) = Axial Force / Area
    Youngs modulus = Simga (stress) / Epsilon (strain)
    Epsilon (strain)= Delta L / Delta L0

    3. The attempt at a solution
    I do know how to use the formulas to calculate the normal stress, but only when I have to Axial Force of AB and BC. I don't know how to calulate it, but I know that in this instance it is F AB= 60KN and F BC= -40KN

    With this I calculate sigma ab = 60kN / 800mm² = 75MPa and sigma bc = -40kN / 400mm² = -100 MPa
    To get Delta AB I use the Young's modulus: (75MPa*0.4m)/210Gpa= 1,42857*10^(-4)m, for Delta BC I switch out the numbers and get the same amount in the opposite direction.

    Could you help me understand why in this case the force of AB is 60kN and for BC 40kN?

    Regards,
    Wolfrider
     
  2. jcsd
  3. Aug 26, 2016 #2

    SteamKing

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    Science Advisor
    Homework Helper

    Is there a diagram or picture which shows this bar?

    The bar itself is statically indeterminate, since it is fixed at both ends. In addition to the normal equations of static equilibrium, you must write additional equations describing the compatibility of displacements internal to the bar.

    See http://ocw.nthu.edu.tw/ocw/upload/8/251/Chapter_2.pdf, particularly sections 2.3 and 2.4.

    Also, see http://www.assakkaf.com/Courses/ENES220/Lectures/Lecture3 .pdf
     
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