# Understanding strain in stiffness matrix

1. May 30, 2009

### xp8pbp

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. May 30, 2009

### djeitnstine

Hello xpp8 welcome to PF

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

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

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

This is characterized by strain $$\epsilon = \frac{\delta}{L_0}$$

3. May 30, 2009

### xp8pbp

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?

4. May 30, 2009

### PhanthomJay

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.