Large Displacements of 2D Truss Structures

01

Overview and assumptions

Consider a planar structure assembled from straight, prismatic truss elements. Each element has two nodes and carries axial force only.

The global nodal-displacement and nodal-force vectors are denoted by $\boldsymbol{u}$ and $\boldsymbol{F}$.

Notation

$\boldsymbol{u}$

Nodal-displacement vector in the global reference frame

$\boldsymbol{F}$

Nodal-force vector in the global reference frame

$U$

Elastic deformation energy (internal energy)

$V$

Potential of the applied nodal forces

$\sigma,\ \varepsilon$

Axial stress and axial strain, respectively

$E$

Young's modulus, with $\sigma=E\varepsilon$

$\mathcal{V}_{el}=AL$

Element volume; $A$ is the cross-sectional area and $L$ is the initial length

02

Element kinematics

For initial nodal coordinates$(x_1,y_1)$ and$(x_2,y_2)$, the initial length is:

With element displacement vector$\boldsymbol{u}_{el}=\begin{pmatrix}u_1&v_1&u_2&v_2\end{pmatrix}^{T}$, the deformed length becomes:

The engineering axial strain, constant along the element, is:

Here, $x_1,x_2,y_1,y_2$ are the nodal coordinates, and$u_1,u_2,v_1,v_2$ are the corresponding nodal displacements.

03

Total potential energy

The total potential energy of the structure is the sum of the elastic deformation energy and the potential of the applied nodal forces. Here the applied loads are assumed to be conservative, displacement-independent nodal forces:

The summation extends over the deformation energy of all truss elements. Because the strain is constant along each element, the structural energy becomes:

Since the strain is constant within a truss element of cross-sectional area $A$,

The minimum-total-potential-energy principle requires

Substitution of the element energy gives the nonlinear system in the nodal displacements:

04

Element force and tangent stiffness

At truss-element level, the deformation energy is:

In general, the element internal-force vector and tangent stiffness matrix are obtained as derivatives of the element energy:

For the complete structure, assembly gives:

The structural tangent matrix is the Hessian of the internal energy:

05

Newton–Raphson solution

The nonlinear algebraic system$\boldsymbol{\Psi}=\boldsymbol{0}$ is solved iteratively. At iteration$i$,

A displacement-based convergence measure is:

where $n$ is the number of active equations.

06

MATLAB implementation

The following symbolic MATLAB program computes$\boldsymbol{f}_{el}$ and$\boldsymbol{k}_{el}$ as derivatives of $U_{el}$.

% k_sym
clear
syms x1 x2 u1 u2 y1 y2 v1 v2 L EA real
U=EA/2/L*(sqrt((x2-x1+u2-u1)^2+(y2-y1+v2-v1)^2)-L)^2;
fel=[diff(U,u1); diff(U,v1); diff(U,u2); diff(U,v2)];
fel=simplify(fel)
kel=[diff(fel,u1) diff(fel,v1) diff(fel,u2) diff(fel,v2)];
kel=simplify(kel)

The resulting expressions are copied into thestiff.m subroutine. The nonlinear analysis is started by running main.m, which calls:

• gen.m — generates the numerical input data;
• stiff.m — assembles and solves the nonlinear system iteratively;
• strain_stress.m — computes strains and stresses in all truss elements.

The Newton–Raphson update in stiff.m is implemented as

dS=-K\F;
S=S+dS;
err=sqrt(dS'*dS/neq);

Exact strain derivatives and the compact approximation

Introduce the current element-axis components and two compact quantities:

Direct differentiation of$\varepsilon=L_f/L-1$ gives the exact gradient and Hessian:

For very small axial deformation, one may set $L_f\approx L$. The additional approximation$\partial^2\varepsilon/\partial\boldsymbol{u}_{el}^2\approx(1/L^2)\boldsymbol{G}$ omits the second term of the exact Hessian:

This is a computational approximation, not the exact second derivative obtained merely by substituting$L_f=L$. The supplied deriv.m uses the exact expressions, while retaining their compact form.

07

Numerical examples

Example 1 — vertical loading

The complete numerical input data are given ingen.m. A vertical load$V=100\ \mathrm{N}$ is applied at the free end. The deformed configurations and load–displacement curves show the strongly nonlinear geometric response of the structure.

Example 2 — combined horizontal and vertical loading

A horizontal load $H=200\ \mathrm{N}$ is added to the vertical load$V=100\ \mathrm{N}$. The MATLAB and ANSYS displacement results are identical to the displayed precision.

08

Extension to a nonlinear material

Example 3 — nonlinear material response

In Example 3, some element strains exceed 10%, so the small-strain material interpretation is no longer appropriate. In the formulation below,$\varepsilon$ is engineering strain and$\sigma$ is its work-conjugate nominal stress. If the relation is supplied as a nonlinear function$\sigma=f(\varepsilon)$, the elastic strain-energy density is the area under the stress–strain curve. Therefore,

The function $\sigma(\varepsilon)$ is assumed to be known experimentally. Differentiation gives the consistent internal-force vector:

With the material tangent$E_t=d\sigma/d\varepsilon$, the consistent element tangent is:

The tangent $\partial\sigma/\partial\varepsilon$ is computed from the same shape-preserving interpolation used for the tabulated stress–strain relation. The program uses this material tangent together with the exact strain Hessian, so the assembled matrix is consistent with the internal-force vector.

09

Linear limiting case

When both displacement and strain are small, retaining only first-order terms in the deformation expression gives:

This follows from$\sqrt{1+a}\approx1+a/2$ for $a\ll1$. The resulting problem is linear, with loads proportional to displacements. The strain derivatives are:

The linear element stiffness matrix is therefore:

With $c=\cos\theta$ and $s=\sin\theta$, this becomes:

10

MATLAB program files

Each archive contains the MATLAB source files and supporting data for the corresponding implementation.