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Understanding strain in stiffness matrix

  1. May 30, 2009 #1
    1. The problem statement, all variables and given/known data

    node1 node2
    O---------------------------------------O

    a rod with 2 nodes:
    x1=axial force at node1
    x2=axial force at node2
    u1,axial deflection at node1
    u2,axial deflection at node2
    l=elements length
    E=young's modulus
    A=cross sectional area of rod element
    e=strain

    I dont understand why at node 1:

    e=(u1-u2)/l

    and from equilibrium considerations at node 2:

    e=(u2-u1)/l

    2. Relevant equations

    3. The attempt at a solution

    From what i understand, if one end is fixed say node2 then strain at node 1 will be:
    u1/l

    and if node1 is fixed then strain at node2:
    u2/l

    at node 1, if i assume that the direction of force/displacement from node1 to node2 is positive isnt that supposed to be negative because of compression?

    and i dont understand how u1-u2 or u2-u1 works? Can somebody explain it to me and give me an example?
     
  2. jcsd
  3. May 30, 2009 #2

    djeitnstine

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    Gold Member

    Hello xpp8 welcome to PF

    strain is defined as deflection per unit length, i.e. [tex]\epsilon = \frac{\delta}{L_0}[/tex] where [tex]\delta = L' - L_0[/tex] i.e. the change in length of the member due to the axial force. where [tex] L_0[/tex] is the original length and [tex]L'[/tex] is the new length.

    [tex]\epsilon [/tex] is important because a large deflection for a small length can easily allow the member the exceed the elastic limit or [tex]\sigma_{y}[/tex] and become plastic or permanently deformed.

    Try it...stretch a small rubber band -of length [tex] a [/tex] fixed at one end for some large [tex]\delta [/tex] just before it pops, it can probably remain elastic. then stretch an even smaller rubber band say of length [tex]\frac{1}{5}a [/tex] do you think the smaller of the two can withstand the same [tex]\delta [/tex]? no

    This is characterized by strain [tex]\epsilon = \frac{\delta}{L_0}[/tex]
     
  4. May 30, 2009 #3
    Hi djeitnstine,

    How do i relate (L'-Lo)/Lo with e=(u1-u2)/l ?

    if in equilibrium(tension/compression with X1=-X2, E and A are constant) U1 and U2 will be the same in magnitude but different in direction, what about in non-equilibrium(X1 not equal to -X2 ) state where the direction and magnitude might be different. How do i visualize such elastic behaviour?
     
  5. May 30, 2009 #4

    PhanthomJay

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    Gold Member

    Seems like a strange sign convention. Let's assume node 1 is fixed, and node 2 is free to move, and the member is in tension and in equilibrium. Then the displacement u1 of node 1 is 0, the displacement of node 2 is u2 to the right, and the strain at either node node is u2/l. If node 2 is fixed and node 1 is free, then u2 is 0 and the displacement of node 1 is u1 to the left, and the strain at either node is u1/l. If both nodes 1 and 2 are both free to move, lets say equally from the mid point, then u1 = u2, and the displacement of node 1 is u1 to the left, the displacemnt of node 2 is u2 to the right, and the strain is (u1 + u2)/l at either node. In all cases, the strain is the same. Your formula seems to choose a sign convention that calls strain positive if the deformation is to the right, and negative if it is to the left. It's all in the plus and minus sign.
     
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