1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Understanding strain in stiffness matrix

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

    node1 node2

    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

    I dont understand why at node 1:


    and from equilibrium considerations at node 2:


    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:

    and if node1 is fixed then strain at node2:

    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


    User Avatar
    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


    User Avatar
    Science Advisor
    Homework Helper
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook