1
$\begingroup$

The state problem of a structure at equilibrium is generally expressed as:

$K(x) ~u = F(x)$

where $K$ is the stiffness matrix, $u$ is the displacement vector, $F$ is the force vector, and $x$ is the design variable (could be cross-sectional areas for example).

What I don't understand is that how does this equation express equilibrium when $F$ and $u$ aren't zero? I have this confusion because I'm linking it to Newton's second law of motion, where if the sum of forces is equal to zero, then we have equilibrium. If such linking makes sense..?

$\endgroup$
3
  • 1
    $\begingroup$ All it's saying is that forces are cancelled out by the structure (that deforms in response to the force). If it didn't cancel, then you would have acceleration, and that means you have change in velocity over time. Change over time is not equilibrium. $\endgroup$
    – Abel
    Commented Mar 2 at 22:55
  • $\begingroup$ @Abel thank you! But, we can't think of acceleration and displacement together cancelling the force? $\endgroup$
    – user134613
    Commented Mar 2 at 23:57
  • $\begingroup$ You can have an inertial frame at a constant velocity that is at 'equilibrium' but not an accelerating frame (which we sometimes use in free body diagrams of accelerating objects). The key is this equilibrium business, and it ties to Newton's realization that an object in constant velocity motion could stay in motion. $\endgroup$
    – Abel
    Commented Mar 3 at 0:07

2 Answers 2

0
$\begingroup$

You are right, assuming the structure is subject to $$ F_0cos(\omega t+ \phi) \ $$ and we disregard the transient part of the motion of the structure when the vibration has settled into steady state. It is subject to harmonic accelaration $ \ddot\omega$.

And the simple damped harmonic equation is of the form

$$ m \ \ddot x +C\dot x+Kx=F_0*cos(\omega t+ \phi) $$

$$x(t)=A cos(\omega t+\phi)$$ However if the structure is over damped anf Forcing function is constant at $Equilibrium $ indicates the acceleration is converged to zero and the equation you have is valid.

$K(x) ~u = F(x)$

$\endgroup$
0
$\begingroup$

You are having confusion at a fundamental level, (and yes you are aware of it as you also mention Newton's law, which is the beginning of everything in structural engineering).

In the equation you mention, F=uK there is no net F that would violate Newton's law, and would make your worries right. In the equation you mention, all action forces are already countered by reaction forces. This equation simply tells us how the structure deforms under a given force and stiffness that is all. It is simply a force deformation relationship.

The equations that answers your concern are simply the ones below, which is the starting point of structural engineering and the most fundamental equation for a civil engineer (which is based on Newton's Law):

For a structure to be in equilibrium, Sum of forces in x direction = 0 Sum of forces in y direction = 0 Sum of moments around z axis = 0 (This is thinking in 2D. If you think in 3D, there would also be sum of forces in z = 0, sum of moments around x and y = 0, which would make 6 equations)

To better visualize, forget about the structure. Think of a spring. How do you calculate its deformation of a simple linear spring under a given force? You do it by the equation: F=kx, where F is the force, k is the linear stiffness of spring and x is the linear deforation. With this we know force deformation relationship of the spring. Does it mean there is net force here that violates Newton's Law? No.

Similarly the equation you gave is simply the force deformation relationship of a structure. Nothing violates Newton's Law there. K is simply the global stiffness matrix, which is generated during structural analysis, after considering all member arrangements, stiffnesses and connection conditions in 3D. For a dynamic analysis it is a little more complicated where, in addition to deformation, also acceleration and velocity also are taken into account, where they are multiplied by viscosity and acceleration respectively and the equation also becomes time dependent. Ultimately the structure obeys this relationship while deforming.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.