Everything in the parent note (FEM software — NASTRAN, ABAQUS (concepts and use)) is built on a handful of ideas. This page defines every symbol and word from absolute zero, in an order where each one leans only on the ones before it.
Imagine drawing a dot on a metal bracket, then pushing on the bracket. The dot slides to a new spot. The little arrow from where the dot was to where it is now is called the displacement.
Picture: the red arrow in the figure — tail at the old position, head at the new one.
Why the topic needs it: displacement is the thing we solve for. Everything else (stress, whether the bracket cracks) is computed from how points moved. When the parent writes u in Ku=F, this is it.
The bold letter u means it is not a single number but a list of numbers — one for each direction a point can move. That leads straight to the next idea.
A single point in 3D space can slide three ways: left–right (x), forward–back (y), up–down (z). Each independent way of moving is one degree of freedom.
Symbol check
The bold u is a column of all the DOF displacements of all the nodes, stacked one on top of another.
You cannot store a smooth continuous shape in a computer. So FEM covers the structure with a grid.
Picture: the smooth L-bracket on the left, the same bracket covered in red quad elements on the right. The mesh approximates the smooth shape.
Why the topic needs it: we only know how to write physics equations for a simple shape. So we solve simple shapes and glue the answers together. This gluing is called assembly: the element stiffness tables ke are stacked into one global K by adding contributions at shared nodes (Section 7 shows where ke comes from). This is literally the "finite" in Finite Element — finitely many pieces instead of infinitely smooth material.
Why quads for thin walls?
A thin bracket is basically a surface, so flat 4-node shell elements (CQUAD4) capture bending with far fewer DOF than solid bricks.
Push a soft spring and it moves a lot; push a stiff spring and it barely budges. The ratio "force needed per unit of movement" is stiffness.
Picture: the straight red line F=ku. Its slope is k. Steeper line = stiffer.
Why a matrix, not a number? A real structure has many DOF, and pushing node 1 also moves node 2 (they are connected). So we need a tableK where entry Kij says "force at DOF i caused by a unit move of DOF j." That table is the stiffness matrix, and the many-spring version of F=ku is exactly Ku=F.
A structure floating free in space has nothing to push against: give it any force and the whole thing drifts, so no unique displacement exists. Mathematically, the raw K is singular (its determinant is zero — you cannot invert it, and Ku=F has infinitely many solutions).
Why does a free-floating model give no unique u?
With nothing pinned, rigid-body drift adds any constant motion for free, so K is singular and solutions are non-unique.
Displacement tells you how points move. But to know if the metal cracks, you need what is happening inside it.
Now we need to connect the inside quantities (strain, stress) back to the nodal displacements we actually solve for. Two matrices do this, and both must be defined before use.
Why the colon in σ:δϵ (parent's virtual-work integral)? Stress and strain are 2D tables (tensors), and ":" just means "multiply matching entries and add them up" — it produces energy per volume.
The chain the parent uses is: move the nodes → get strain → get stress → check margin.ueBϵDσ
Why both E and ρ?E (via D) builds the stiffness K; ρ builds the massM (next). Stiffness resists bending; mass resists acceleration. Vibration is a fight between the two.
Pluck a structure and it wobbles at special rates. FEM finds those.
Why it matters for the parent's f1≈850 Hz: if launch shaking contains energy at 850 Hz, the bracket resonates and can break. This is the bridge to Vibration and Modal Analysis and Launch Vehicle Loads.
Convert the parent's f1
ω1=2πf1=2π(850)≈5341 rad/s.
Why do mode shapes exist only at certain frequencies?
Because ϕ=0 needs (K−ω2M) singular, i.e. det(K−ω2M)=0, which only specific ω satisfy.
There are three flavours of nonlinearity the parent relies on, and each one is just "K is no longer constant" for a different physical reason:
Material nonlinearity: past a certain stress the material stops springing back proportionally (yielding/plasticity, composite damage), so its stiffness drops as load grows.
Geometric nonlinearity: when parts bend or rotate a lot (deployable booms), the geometry itself changes enough that the stiffness depends on the current shape.
Contact nonlinearity: two surfaces touch or separate (bolted joints), and stiffness jumps the instant they come into contact.
Why iterate at all in nonlinear?
Because Fint (hence K) depends on the answer u, you cannot solve in one shot — you guess, measure the imbalance R=Fext−Fint, correct, and repeat until R≈0.