Engineering at Alberta Courses » Linear vs. Nonlinear Analysis

Nonlinear Systems of Equations: Linear vs. Nonlinear Analysis

Linear vs. Nonlinear Analysis of Structures: An Application

Most structural engineering problems in which the displacements and forces inside a structure are sought can be formulated using linear systems of equations. This is done by invoking certain assumptions that simplify the problem. However, when a structure undergoes large deformations, the problem has to be formulated using nonlinear systems of equations. The following is an example of a truss structure that aims at clarifying the difference.

Example

Two steel truss members with areas $A_1$ and $A_2$ are connected to a steel cable with an area $A_3$ as shown in the figure. All the connections are hinged connections allowing free relative rotations. Points $A$ , $C$ , and $D$ are fixed by means of hinges preventing displacements but allowing free rotations. Assuming that the horizontal displacement of point $B$ is $u_1$ to the right and the vertical displacement of point $B$ is $u_2$ downwards. Assume that Young’s modulus $E=200 GPa$ , $A_1=A_2=200 mm^2$ , and $A_3=50 mm^2$ . The force in each member is equal to $F_i=A_i\sigma_i=A_iE_i\varepsilon_i$ where $\varepsilon_i=\frac{L_i-L_{0i}}{L_{0i}}$ where $i=1,2,3$ is the member number, $L_{0i}$ is the undeformed length, $L_i$ is the deformed length, $\sigma_i$ is the longitudinal stress, $\varepsilon_i$ is the strain, and $E_i$ is Young’s modulus in member $i$ . Find the values of $u_1$ and $u_2$ at equilibrium using: 1) Linear analysis where the displacements are assumed to be small, and 2) Nonlinear analysis. Solve twice, once with $F=10,000 N$ , and another time with $F=500,000 N$ . Do the two solution strategies give the same answer?

structures1

Solution

The units adopted will be $N$ for forces, $m$ for lengths, and $N/m^2$ for $E$ . The following is the displaced shape of the structure. The middle hinge move horizontally to the right a distance $u_1$ and vertically downards a distance $u_2$ . There are two unknowns in the problem $u_1$ and $u_2$ . These can be found using two equations of equilibrium at the middle hinge. The two equations are the sum of the vertical and horizontal forces, each equal to zero.
Structures2

Linear Analysis

In linear analysis, the truss member length is assumed to only change due to displacements along its longitudinal axis. In other words, the forces in the truss member do not change if it simply rotates. In addition, the forces are assumed to be aligned with the original geometry of the structure before deformations. The strains in each member are given by: $\varepsilon_1=\frac{3-u_1-3}{3}=\frac{-u_1}{3}$ , $\varepsilon_2=\frac{3+u_1-3}{3}=\frac{u_1}{3}$ , and $\varepsilon_3=\frac{5+u_2\sin{(\theta)}-u_1\cos{(\theta)}-5}{5}=\frac{u_2\sin{(\theta)}-u_1\cos{(\theta)}}{5}$ . Therefore, the forces in each member can be calculated as follows:

$\begin{split} F_1&=E_1A_1\varepsilon_1=E\left(\frac{200}{1000000}\right)\frac{-u_1}{3}\\ F_2&=E_2A_2\varepsilon_2=E\left(\frac{200}{1000000}\right)\frac{u_1}{3}\\ F_3&=E_3A_3\varepsilon_3=E\left(\frac{50}{1000000}\right)\frac{u_2\sin{(\theta)}-u_1\cos{(\theta)}}{5} \end{split}$

where $E=200(10^9) N/m^2$ , $\sin{\theta}=\frac{4}{5}$ , and $\cos{\theta}=\frac{3}{5}$ . Assuming small deformations, equilibrium at point $B$ can be analyzed using the following forces diagram:
Structures4
The first equation is the sum of horizontal forces equals to zero:

$F_1+F_3\cos{(\theta)}-F_2=\frac{-81260000}{3}u_1+960000u_2=0$

The second equation is the sum of the vertical forces equal to zero:

$F-F_3\sin{(\theta)}=F+960000u_1-1280000 u_2=0$

These equations are linear and can be written in matrix form as follows:

$\left(\begin{matrix} \frac{-81260000}{3} & 960000\\ 960000 & -1280000\end{matrix}\right)\left(\begin{array}{c} u_1\\u_2\end{array}\right)=\left(\begin{array}{c} 0\\-F\end{array}\right)$

Setting $F=10,000 N$ yields a solution $u_1=0.28mm$ and $u_2=8.02mm$ , while setting $F=500,000$ yields $u_1=14mm$ and $u_2=401mm$ . Notice that in a linear analysis, the displacements are directly proportional to the applied load, that is, the ratio between $u_1$ at $F=10,000$ and $u_1$ at $F=500,000$ is equal to $\frac{10,000}{500,000}$ . The same applies to $u_2$ .

The following is the Mathematica code to formulate and solve the above system of equations using the “Solve” function.

View Mathematica Code

Clear[F, Ee];
Ee = 200 (10^9);
eps1 = -u1/3;
A1 = A2 = 200/1000000;
A3 = 50/1000000;
eps2 = u1/3;
Cth = 3/5;
Sth = 4/5;
eps3 = (-u1*Cth + u2*Sth)/5;
F1 = Ee*A1*eps1;
F2 = Ee*A2*eps2;
F3 = Ee*A3*eps3;
Eq1 = Expand[(F1 + F3*Cth - F2)]
Eq2 = Expand[F - F3*Sth]
sol1 = Solve[{Eq1 == 0.0, Eq2 == 0.0 /. F -> 10000}]
sol2 = Solve[{Eq1 == 0.0, Eq2 == 0.0 /. F -> 500000}]

View Python Code

import sympy as sp
sp.init_printing(use_latex=True)
u1, u2, F = sp.symbols('u1 u2 F')
Ee = 200*(10**9)
eps1 = -u1/3
A1 = A2 = 200/1000000
A3 = 50/1000000
eps2 = u1/3
Cth = 3/5
Sth = 4/5
eps3 = (-u1*Cth + u2*Sth)/5
F1 = Ee*A1*eps1
F2 = Ee*A2*eps2
F3 = Ee*A3*eps3
Eq1 = F1 + F3*Cth - F2
Eq2 = F - F3*Sth
display("Eq1: ",Eq1)
display("Eq2: ",Eq2)
# Using Sympy's built-in linear solve function
display("sol1: ",sp.linsolve((Eq1, Eq2.subs({'F':10000})),(u1, u2)))
display("sol2: ",sp.linsolve((Eq1, Eq2.subs({'F':500000})),(u1, u2)))

Nonlinear Analysis

In a nonlinear analysis, the analysis takes into consideration the displaced position of the structure without any simplified assumptions. Looking at the first truss member, the displaced structure is shown in the following figure.
Structures5
For this member, the original length $L_{01}=3$ , the final length $L_1=\sqrt{u_2^2+(3-u_1)^2}$ , the strain $\varepsilon_1=\frac{L_1-L_{01}}{L_{01}}$ and the angle of inclination $\theta_1$ is such that $\sin{(\theta_1)}=\frac{u_2}{L_1}$ and $\cos{(\theta_1)}=\frac{3-u_1}{L_1}$ .

For the second truss member, the displaced structure is shown in the following figure.
Structures6
For this member, the original length $L_{02}=3$ , the final length $L_2=\sqrt{u_2^2+(3+u_1)^2}$ , the strain $\varepsilon_2=\frac{L_2-L_{02}}{L_{02}}$ and the angle of inclination $\theta_2$ is such that $\sin{(\theta_2)}=\frac{u_2}{L_2}$ and $\cos{(\theta_2)}=\frac{3+u_1}{L_2}$ .

For the third truss member, the displaced structure is shown in the following figure.
structures9
For this member, the original length $L_{03}=5$ , the final length $L_3=\sqrt{(4+u_2)^2+(3-u_1)^2}$ , the strain $\varepsilon_3=\frac{L_3-L_{03}}{L_{03}}$ and the angle of inclination $\theta_3$ is such that $\sin{(\theta_3)}=\frac{4+u_2}{L_3}$ and $\cos{(\theta_3)}=\frac{3-u_1}{L_3}$ .

Finally, the equations of equilibrium can be written taking into consideration the inclination of each member:
Structures8
The first equation is the sum of horizontal forces equals to zero:

$F_1\cos{(\theta_1)}+F_3\cos{(\theta_3)}-F_2\cos{(\theta_2)}=0$

The second equation is the sum of the vertical forces equal to zero:

$F-F_1\sin{(\theta_1)}-F_2\sin{(\theta_2)}-F_3\sin{(\theta_3)}=0$

The following is a screenshot of the Mathematica output of the two equations:
Structures10

Setting $F=10,000 N$ yields a solution $u_1=0.28mm$ and $u_2=8.02mm$ , while setting $F=500,000 N$ yields $u_1=11.6mm$ and $u_2=340mm$ . For small deformations under small applied loads, the linear analysis and nonlinear analysis produce exactly the same results. In this situation, the rotations of the members are very small such that the original shape and the deformed shape can be assumed to be the same. However, under large loads, the linear analysis is no longer valid. The two solutions at $F=500,000 N$ are far from each other with the nonlinear analysis solution giving less displacement. In this particular situation, as the structure deforms, the nonlinear effects cause the structure to be more rigid and therefore exhibit less deformations. Cable structures with large loads and large deformations in the cables need to be analyzed using a nonlinear analysis rather than a linear analysis.

The following is the Mathematica code to formulate and solve the above nonlinear system of equations using the “FindRoot” function with an initial guess of $u_1=0$ , and $u_2=0$ .