This document summarizes the analysis of a Warren truss using linear algebra and matrix operations. The author sets up an augmented matrix to model the truss, which is then partitioned and put in reduced row echelon form using MATLAB. The solutions reveal which members experience tension and compression, and the magnitude of forces. Member 1 and 6 experience tension, while members 2, 4, and 5 experience compression. Member 3 has zero force. This analysis allows the engineer to design each member appropriately based on expected stress. Linear algebra provides an efficient way to model and analyze structural systems.

1. Engineering Analysis Final Lab
Warren Truss analysis
Shawn Robinson
Due 12/10/13

2. Introduction
In engineering analysis this semester we have learned about linear algebra and its applications.
We have practiced techniques such as row operations, finding determinants, transition matrices,
and eigenvalues just to name a few. As part of the class we learned how to take the skills we
practiced and put them to use in real life applications problems, especially when it comes to
engineering applications. Using linear algebra and matrix representation of a problem can save a
lot of time when working out system of equations type problems, when the equations are linear.
In strength of materials, we learned about various material properties when structures undergo
applied loads, and how to design your structure so that they do not brake or fail. Linear algebra is
really convenient in this subject because often when solving a problem, we end up with many
linear equations with unknowns. Matrix operations let us solve these systems fairly easy. This
method can help save money and find the weakest members in a bridge or truss design, and
potentially save lives. In this lab a Warren truss will be analyzed.
Motivation
For this applications problem a Warren truss is analyzed to find the forces in each member. A
truss was chosen because it a common building structure for many different objects. Better
understanding truss analyses using an augmented matrix, is a common problem is the
engineering industry. The matrix can show the forces in each member, where then other design
parameters can be implemented to save the structure from failing. By using the forces in each
member to find the stress, you can find which member is most likely to fail. The matrix model of
the truss can be used to find the zero force members. These zero force members are represented
by the solution of a variable equal to zero. This means that under current conditions the zero
force member undergo no load or force. By knowing this information, you could then design the

3. zero force members to be cheaper when budget constraints are and issue or if building material is
limited. The designer could make the beams of the zero force members smaller by decreasing the
area, or purchase cheaper building material to reduce the cost. Another reason for choosing a
truss is that the matrix model of the truss can be modified for different situations. The Engineer
can change initial conditions of the matrix to represent the under different loads. By varying the
load amount and the load location, you can test the model matrix under different weather
conditions. Snow and rain can be represented by varying distributed loads. You can test the
model against extreme weather conditions like earthquakes, high wind, or heavy traffic. The
Warren truss was chosen because it is one of the most common repeatable truss formations,
consisting of a series of equilateral triangles with vertical members in the middle of each
triangle. The Warren truss was patented in 1848 by James Warren and Theobald Manzani. The
cross members carry both tension and compression forces depending on the load location. This
truss design is easily repeatable and usually span anywhere from 50 to 400 feet. Some well-
known Warren trusses are the Uhlerstown-Frenchtown Bridge, Washington Crossing Bridge
(below) and the Riverton Belvidere Bridge.
Mathematical Formulation
A Warren truss with 3 triangles, 13 members, 8 Joints (shown below) has an applied load of
1KN at the center of the truss. Find whether the members are in tension or compression and the
magnitude of the forces.

4. Fist the reactive forces and the support ends was found using the equation ΣFy= 0 and because
of symmetry it is found that the forces were 500N each.
Assumptions
-Each member was assumed to be in tension
- Up was positive, down was negative
-Member in Compression will yield a negative solution and Tension positive
-The augmented matrix would represent solving the truss problem by a method of sections,
where each row represents a force balance for each joint with Y, then X respectively in order
from left to right and each column represents the members.
-Because the original matrix was too large (26x13, over determined) a portioned matrix was
made representing a cut made vertically through members 6, 5, 4. Because of the symmetry of
the Warren truss, it can be assumed the new 6x6 matrix can adequately represent both side of the
truss.

5. The matrix was then formed by finding the coefficients in front of the members for forces and
setting them equal to the resultant (which was usually 0). (See attached hand written pages for
matrix formulation)
Method and Solution
After obtaining the original matrix, the partitioned matrix representing half the truss was found
as discussed earlier. Using Mat lab, the rref command gave the solutions to the force members on
the right-hand column
>> syms x t
>> t = 45;
Original Matrix
>> x = [cosd(t) 1 0 0 0 0 0 0 0 0 0 0 0 0; 0 0 1 0 0 0 0 0 0 0 0 0 0 0; 0 -1 0 1 0 0 0 0 0 0 0 0 0 0; -cosd(t) 0 -1
0 -cosd(t) 0 0 0 0 0 0 0 0 0; -sind(t) 0 0 0 sind(t) 1 0 0 0 0 0 0 0 0; 0 0 0 0 cosd(t) 0 1 0 cosd(t) 0 0 0 0 1000;
0 0 0 -1 -sind(t) 0 0 1 sind(t) 0 0 0 0 0; 0 0 0 0 0 0 -1 0 0 0 0 0 0 0; 0 0 0 0 0 -1 0 0 0 1 0 0 0 0; 0 0 0 0 0 0 0
0 0 0 1 0 0 0; 0 0 0 0 0 0 0 -1 0 0 0 1 0 0; 0 0 0 0 0 0 0 0 -cosd(t) 0 -1 0 -cosd(t) 0; 0 0 0 0 0 0 0 0 -sind(t) -1
0 0 sind(t) 0; 0 0 0 0 0 0 0 0 0 0 0 -1 0 0]
x =
1.0e+03 *
Columns 1 through 7

6. 0.0007 0.0010 0 0 0 0 0
0 0 0.0010 0 0 0 0
0 -0.0010 0 0.0010 0 0 0
-0.0007 0 -0.0010 0 -0.0007 0 0
-0.0007 0 0 0 0.0007 0.0010 0
0 0 0 0 0.0007 0 0.0010
0 0 0 -0.0010 -0.0007 0 0
0 0 0 0 0 0 -0.0010
0 0 0 0 0 -0.0010 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
Columns 8 through 14
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0.0007 0 0 0 0 1.0000
0.0010 0.0007 0 0 0 0 0
0 0 0 0 0 0 0

7. 0 0 0.0010 0 0 0 0
0 0 0 0.0010 0 0 0
-0.0010 0 0 0 0.0010 0 0
0 -0.0007 0 -0.0010 0 -0.0007 0
0 -0.0007 -0.0010 0 0 0.0007 0
0 0 0 0 -0.0010 0 0
Partitioned matrix for symmetry
>> x = [sind(t) 0 0 0 0 0 500; cosd(t) 1 0 0 0 0 0; 0 0 1 0 0 0 0; 0 -1 0 1 0 0 0; -cosd(t) 0 -1 0 -cosd(t) 0 0; -
sind(t) 0 0 0 sind(t) 1 0]
x =
0.7071 0 0 0 0 0 500.0000
0.7071 1.0000 0 0 0 0 0
0 0 1.0000 0 0 0 0
0 -1.0000 0 1.0000 0 0 0
-0.7071 0 -1.0000 0 -0.7071 0 0
-0.7071 0 0 0 0.7071 1.0000 0
>> rref(x) ans =
1.0e+03 *
0.0010 0 0 0 0 0 0.7071
0 0.0010 0 0 0 0 -0.5000
0 0 0.0010 0 0 0 0

8. 0 0 0 0.0010 0 0 -0.5000
0 0 0 0 0.0010 0 -0.7071
0 0 0 0 0 0.0010 1.0000
Interpretation
After Matlab rref the partitioned matrix, the forces in each members is now apparent.
Solutions with positive answers represent members in tension (members 1 & 6). Solutions with
negative answers represent member in compression (members 2, 4 5). Solutions that were equal
to zero represents zero force members (member 3). The member undergoing the greatest tension
was member 6, with a load of 1KN, which is accurate because member 6 is at the top of the truss
holding more of the load. Member 5 was in the highest compression with a load of 707.1N, with
is also accurate because it is the cross member for the center of the bridge. With the forces
known for this section of the truss, it can be repeated because the other half is identical.
Conclusion
After analyzing the Warren truss, it was seen that the cross members did carry both tension and
compression, as stated by James Warren in one of his books. With this information about the
force in each member, an engineer can then appropriately design each member according to the
amount of expected stress using the equation stress=Force/Area, where the forces are the
solutions to the matrix, and the area is set by designer. After having the stress values for each
member, the engineer can select the appropriate material that is safe, stable, and cost effective.
Using Linear algebra is very useful in engineering applications because it allows for the
engineers to create a mathematical model and test it under various situations, without spending
money or very much time. It is also a quick and accurate way to get insight about other
properties of the system, like stress for trusses as shown below. It must also be noted how quick
the mat lab operations were, when it would have taken a lot of time by hand. This allows the
engineer to evaluate huge systems and structures in no time at all. This method for evaluation
structures saves time, money, and potentially lives.

9. References
Truss Structures- engr.uky.edu
Warren Truss-By Garrett Boon on January 4, 2011 - Modified March 14, 2013
SJH Engineering P.C 2012