FACULTY OF MECHANICAL & MANUFACTURING
ENGINEERING
SEMESTER 1 2019/2020
GROUP PROJECT
TITTLE: DESIGN A TRUSS
_______________________________________________________________
COURSE CODE
BDA20903
COURSE NAME
SOLID MECHANICS II
________________________________________________________________
GROUP MEMBERS
1. LAAVANKUMAR A/L NYANA SEGARAN (CD180077)
2. DARREN ESMITH ANAK IAN WILLIAM (CD180075)
3. ALEXIA BINTI LUKE LAITI (CD180227)
4. BABIE JANE ROGER (CD180177)
5. IRVINE KULLEH LUYOH (CD180173)
SECTION
1
LECTURER’S NAME
PROF. MADYA Ir. Ts. Dr. AL EMRAN BIN ISMAIL
SUBMISSION DATE
21/12/2019
________________________________________________________________
Executive Summary
In our assigned task, we are asked to design our own truss which is used to overcome
a trench. A trench is a type of excavation or depression in the ground that is generally deeper
than it is wide (as opposed to a wider gully, or ditch), and narrow compared with its length (as
opposed to a simple hole).
The maximum span and heights are already given which is 1.5m and 0.5m. We are to
assume that the loading is applied on the central top of the truss. In design wise, we are to
construct three design and also to take into consideration of the cost for it. Besides design,
material properties should also be taken into consideration.
The properties of different materials will also determine the strength of the bridge such
as its density, Modulus of elasticity, Poisson ratio, shear modulus, ultimate tensile strength,
minimum and maximum yield strength.
There is several software which we have survey and tried however we decided to use
http://www.federicobonfigli.com/ (Online Truss Solver). This software calculate axial forces,
the displacements of the joints, and the deformation of the elements of the structure thus also
giving a simple diagram on the deformation on our design with colour mapping and etc.
Finally at the end of completing this task, we will be selecting the best by comparing
the cost and also design of the truss.
2
Roles of team
Team Member
1. Laavankumar A/L Nyana Segaran
Roles







2. Darren Esmith Anak Ian William




3. Alexia binti Luke Laiti


Did manual and software design
calculation
Designed and proposed 3
different truss designs
Did cost estimation
Find references for report
Did appendices
Did design comparison
Compiled final report
Did software design calculation
Designed and proposed 3
different truss designs
Did executive summary
Did and finalize material and its
properties
Percentage of
Contribution
100%
100%
Did design objectives and
problem statement
Did conclusion
100%
4. Babie Jane Roger


Did manual design calculation
Did cover page
100%
5. Irvine Kulleh Luyoh

Did manual design calculation
100%
3
Design Objectives
a) Build a truss bridge that spans 1.5 and height of 0.5m.
b) To overcome a trench.
c) Measure the tensile and compressive forces in individual structural components.
d) Measure the ultimate load-carrying capacity of the truss
Problem Statement
The problem statement for this truss design is that we are required to overcome a
trench that is exist especially in drainage system. The problem that we faced is that we need to
consider the cost of our truss design to be not that costly at the same time, by using our
designated truss it can overcome the trench efficiently. On the other hand, we need to consider
the design whether it is suitable to overcome the specified problem and the workability where
it is easy to be construct so that the cost would not be spent on too much.
Besides that, we are required, also too choose the materials that is appropriate to build
the trench so that it is compatible with the location of where the trench is. Besides that, we need
to keep in mind that by building this truss besides the main reason to overcome the trench, we
may consider the truss must be long span, lightweight, reduced deflection and it can support
considerable loads.
4
Material Properties
In our task, we used the common structural steel which is A36 steel to design a truss. A36 steel
is produced in a wide variety of forms, including:

Structural Shapes

Bars

Girders

Angle Iron

T iron
Below table shows the Mechanical properties of the A36 Steel:
Mechanical Properties
Values
Density
7,800 kg/𝑚3
Young’s Modulus
200 GPa (29,000,000 psi)
Poisson Ratio
0.26
Shear Modulus
75 GPa (10,900,000 psi)
Minimum Yield Strength (thickness of plate < 8 inch)
36,000 psi (250,000 kPa)
Maximum Yield Strength (thickness of plate < 8 inch)
32,000 psi (220 MPa)
Ultimate Tensile Strength
58,000-80,000psi(400-550 MPa)
Cost Estimations
Regarding the construction material and equipment to be used in this project, the table
below shows the estimated cost of the main materials used in fabricating a A36 steel truss.
Materials used to design a truss
Estimated cost
i) 14 hardwood dowels 3/8 inch in diameter
RM 20-30
and 36 inches long
ii) 11 pieces of A36 steel of 0.5m by 0.5m for
RM 40-50
the joints
Total estimated cost to build a truss is range between RM 60 – 80
5
Proposed Design
DESIGN 1
Figure 1: Pratt truss design
DESIGN 2
Figure 2: Warren truss design
6
DESIGN 3
Figure 3: Howe truss design
7
Manual Design Calculations
This section shows the calculations required for the truss design we choose which is
the Warren truss design (Design 2). The first step in the design process is to determine the
configuration of the truss. We required to consider and discuss at least one alternative design
as part of our work. For the example whose calculations are shown in this section, the
symmetric truss shown below. We assumed a load of 800N (0.8kN) is applied on the central
top of the truss.
0.8kN
0.5m
Hardwood
dowels
dowels
1.5m
Figure 4: Chosen design of truss (Warren truss – Design 2)
Calculating total force in each member:
8
(+↑) ∑Fy=0
Ay + By = 0.8kN ---(1)
(CW+) ∑MA=0
(0.8kN)(0.75m)-By(1.5m)=0
By=
(𝟎.𝟖𝒌𝑵)(𝟎.𝟕𝟓𝒎)
𝟏.𝟓𝒎
= 0.4kN ---(2)
Sub (2) into (1)
Ay = 0.8kN -By
= 0.8kN – 0.4kN
Ay = 0.4kN
By = 0.4kN
(+↑) ∑Fy=0
F4/7
0.4kN - F4/7sin63.43° = 0
4
63.43°
F4/3
𝟎.𝟒𝒌𝑵
F4/7 =𝒔𝒊𝒏𝟔𝟑.𝟒𝟑° = 0.45kN (C)
(+→) ∑Fx = 0
By
F4/7cos63.43° - F4/3 = 0
F4/3 = 0.45kNcos63.43 = 0.20kN (T)
F7/6
(+↑) ∑Fy=0
7
-F7/3sin63.43° + 0.45sin43.43° = 0
F7/3 = 0.45kN (T)
63.43°
F7/3
(+→) ∑Fx = 0
F4/7
F7/6 – F4/7cos63.43° - F7/3cos63.43° = 0
F7/6 = 0.45cos63.43° + 0.45cos63.43°
F7/6 = 0.40kN (C)
9
(+↑) ∑Fy=0
F3/6
F7/3
0.45sin63.43° - F3/6sin63.43° = 0
F3/6 = 0.45kN (C)
63.43°
F3/2
F4/3
3
(+→) ∑Fx = 0
F4/3 + F7/3cos63.43° + F3/6cos63.43° – F3/2 = 0
F3/2 = 0.20 + 0.45cos63.43° + 0.45cos63.43°
F3/2 = 0.60kN (T)
(+↑) ∑Fy=0
0.8kN
0.45sin63.43° + F6/2sin63.43° - 0.8kN = 0
F7/6
F6/5
6
63.43°
F3/6
F6/2
F6/2 = 0.45kN (C)
(+→) ∑Fx = 0
F6/5 – F7/6 – F3/6cos63.43° + F6/2cos63.43°
F6/5 = 0.40 + 0.45cos63.43° – 0.45cos63.43°
F6/5 = 0.40kN (C)
(+↑) ∑Fy=0
F2/5
F6/2
F2/5sin63.43° - 0.45sin63.43° = 0
F2/5 = 0.45kN (T)
63.43°
F2/1
2
F3/2
(+→) ∑Fx = 0
F3/2 – F6/2cos63.43° - F2/5cos63.43° - F21 = 0
F2/1 = 0.60 – 0.45cos63.43° - 0.45cos63.43°
F2/1 = 0.20kN (T)
5
(+↑) ∑Fy=0
F6/5
-F5/1sin63.43° - F2/5sin63.43° = 0
63.43°
F5/1 = -0.45kN
F5/1
F2/5
F5/1 = 0.45kN (C)
10
Final results
Members
Total Force
T or C
Ay (Joint 1)
0.4
C
By (Joint 4)
0.4
C
0.2
T
0.45
C
0.45
T
0.45
C
0.4
C
0.6
T
1/2 , 3/4
1/5 , 7/4
2/5 , 3/7
2/6 , 3/6
5/6 , 6/7
2/3
After choosing the configuration, the force in each member must be determined using
the method of joints. Most designs assume that the total external load is carried equally by two
parallel trusses that are connected by cross members. We complete the calculations by
assuming that one of the trusses carries a 800N load. The load shown in Figure 4 are based on
this assumption. After calculating the forces, the students must evaluate the potential failure
(predicted buckling load) of each member. The failure modes considered include excessive
axial stress, tensile from its joint for members in tension, and buckling of members in
compression. For the calculations presented here, the material used to build the truss (A36
steel) which has a Young’s Modulus of 200GPa.
After the force in each truss member is computed, the member loads, axial stresses,
and buckling loads are calculated for the truss and are summarized in a single table, as shown
in Table 1. Students are required to submit their results in this format, as it makes checking the
results more convenient for the grader and forces the students to be complete in their analysis.
Next, students are required to complete a second table to help them arrive at the
appropriate predicted failure load. The factor of safety corresponding to an applied load of
800N on a single truss is computed for each member, and the most critical member is selected
as the member with the lowest factor of safety. Table 2 summarizes these calculations.
11
1) Finding Moment of Inertia
2) Finding buckling load
K = 1 (Pinned ends)
L = 0.5m
E = 𝟐𝟎𝟎𝒙𝟏𝟎𝟗 (A36 steel)
3) Axial Stress
Axial stress, 𝝈𝒙 =
𝑻𝒐𝒕𝒂𝒍 𝒎𝒆𝒎𝒃𝒆𝒓 𝒐𝒇 𝒇𝒐𝒓𝒄𝒆 (𝒌𝑵)
𝑴𝒆𝒎𝒃𝒆𝒓 𝒐𝒇 𝒂𝒓𝒆𝒂 (𝒎𝒎𝟒 )
Table 1 – Summary of forces, stresses, and buckling loads on members.
Members
Member
length (m)
Member
Area
(mm^2)
Member I
(mm^4)
Type of Member
1/2 , 3/4
1/5 , 7/4
2/5 , 3/7
2/6 , 3/6
5/6 , 6/7
2/3
0.5
0.5
0.5
0.5
0.5
0.5
100
100
100
100
100
100
208.33
208.33
208.33
208.33
208.33
208.33
single flat member
single flat member
single flat member
single flat member
single flat member
single flat member
Total
Axial Stress Buckling
Member of T or C
(MPa)
Load (kN)
Force (kN)
0.2
-0.45
0.45
-0.45
-0.4
0.6
T
C
T
C
C
T
2
4.5
4.5
4.5
4
6
N/A
1.64
N/A
1.64
1.64
N/A
12
Table 2 – Evaluation of factors of safety for a load of 800N. applied to one truss.
T or
Members
C
1/2 , 3/4
1/5 , 7/4
2/5 , 3/7
2/6 , 3/6
5/6 , 6/7
2/3
T
C
T
C
C
T
Total Member
of Force (kN)
0.2
-0.45
0.45
-0.45
-0.4
0.6
Member
Axial Stress
(MPa)
2
4.5
4.5
4.5
4
6
Tensile
FS
Load to buckle
a member (kN)
Buckling
Load FS
4
N/A
1.78
N/A
N/A
1.33
N/A
1.64
N/A
1.64
1.64
N/A
N/A
3.64
N/A
3.64
4.1
N/A
Force to cause tensile = 800N
Tensile FS = Force to cause tensile/total member force
Buckling load FS = Load to buckle member/total member force
After completing Table 2, the predicted failure load is calculated as 800 times the
smallest factor of safety in Table 2 times the number of parallel trusses. The smallest factor of
safety occurs in members 2/3 is approximately 1.33. Thus, for this design which employs no
parallel trusses, the predicted failure load is 1 064 N. (800 x 1.33 x 1). The expected failure
mode for the truss is buckling in member 2/3.
13
Software Design Calculation
DESIGN 2 (WARREN TRUSS DESIGN)
Software calculation prove manual calculation for force in member (axial force) is correct.
Online software analysis http://www.federicobonfigli.com/EN/TrussSolver.aspx
14
Finding buckling load:
Software calculation prove manual calculation for buckling load is correct which is 1640 N or
1.64kN.
Online software analysis:
https://www.efunda.com/formulae/solid_mechanics/columns/calc_column_critical_load.cfm#calc
15
Design Comparison
Description
Materials
Warren Truss
Howe Truss
Pratt Truss
Is a series of
Has diagonal
Has diagonal
isosceles triangles
members that slant
members that slant
or equilateral
away from the
down toward the
triangles.
middle.
middle.
Heavier steel or
Steel is not
iron is required for
economical for the
the triangles to
members that
handle the
handle compressive
compressive force.
force.
The downward
beam attaches to
Shape
The diagonal
vertical beam to
beams make a “v”
make a sideways
along the entire
“z”, in the middle
structure.
of the structure, an
upside down “v” is
formed.
Easy of
Very easy and
Construction
quick to build.
Cost
Cost effective
Thinner and lighter
steel or iron is used
for the diagonal
members.
The downward
beam attaches to
vertical beam to
make a sideways
“z”, in the middle
of the structure, a
“v” is formed.
Longer built time,
Longer build time,
more complex
more complex
construction.
construction.
High cost
High cost
Table 3: Comparisons between Warren Truss, Howe Truss and Pratt Truss
16
Conclusion
In conclusion, we used A36 metal, which is mild carbon steel for our truss design to
overcome the trench because it have carbon content of less than 0.3%. This material is easy to
be machined, welded, and formed making it extremely useful as a general-purpose steel. To
justify our problem statement, we choose A36 steel because of its relatively low cost and as
mentioned the mechanical properties make it particularly suited for structural applications.
Many bridges is using this material because of its high strength and toughness.
As for the objectives of the design, we strongly believe that we had achieved it because
we had successfully design a bridge that requires 1.5m span, 0.5 height and overcome the trench
that is fore mentioned. Lastly, we did measure the tensile and compressive forces in each
members and the ultimate load-carrying capacity of the truss using appropriate software.
17
REFERENCES
Books
1. Gere, J.M. and Goodno, B.J., 2009. “Mechanics of Materials”, 7th Edition, Cengage
Learning.
2. Beer, F.P., Johnston, E. R. and Deworlf, J.T., 2009. “Mechanics of Materials”, 5th
Edition, Mc Graw Hill.
3. Hibbeler, R.C., 2008. “Mechanics of Materials”, 7th Edition, Pearson Prentice Hall.
4. Ugural, A.C., 2008. “Mechanics of Materials”, John Wiley & Sons Inc.
5. Riley, W.F., Sturges, L.D., and Morris, D.H., 2007. “Mechanics of Materials”, 6th
Edition, John Wiley & Sons Inc.
6. Meriam, J.L. And Kraige, L.G., 2007, “Engineering Mechanics: Statics”, 6th S.I.
Edition, John Wiley & Sons, Inc. Call number: TA350. M47 2007.
7. Hibbeler, RC., 2007, “Engineering Mechanics: Statics”, 12th Edition, Prentice-Hall
International. Call number: TA351. H52 2009.
Internet
1. http://www.federicobonfigli.com/EN/TrussSolver.aspx
2. http://www2.latech.edu/~kcorbett/LWTL/home/sophomore/engr220/design_project/re
quired_tables.htm
3. https://www.efunda.com/formulae/solid_mechanics/columns/calc_column_critical_lo
ad.cfm#calc
4. http://www.wikiengineer.com/Structural/AxialStress
5. http://www.ah-engr.com/som/3_stress/text_3-1.htm
6. https://www.engineersedge.com/strength_of_materials.htm
18
APPENDIX 1
1) Warren Truss design
The Warren Truss is a very common design for both real and model bridges. It’s exact
history and origination is a little muddled, however. James Warren patented a design in 1848
(in England), which many attribute the name “Warren Truss”. His patent was more about
the methodology of building rather than a “design”. Regardless, the Warren Truss has been
around a while and has been very popular. Examples of it can be found everywhere in the
world. This simple truss design consists of two parallel chords and equally sized triangles
placed in between. This effective design is popular not only in construction but also in
production countless other machines and systems. For example, early two-winged airplanes
used lightweight Warren truss mesh to reinforce the structure of the wings.
Warren Truss design
19
2) Howe Truss design
The Howe Truss was designed by William Howe in 1840. It used mostly wood in
construction and was suitable for longer spans than the Pratt truss. A very popular truss type
in which features triage diagonals that slope upward toward the center. Many smaller
bridges and architectural solutions for homes feature this simple design. Therefore, it
became very popular and was considered one of the best designs for railroad bridges back
in the day. Many Howe truss bridges exist in the North West United States, where wood is
plentiful.
Howe Truss design
20
3) Pratt Truss design
The Pratt Truss originated from Caleb and Thomas Pratt (father and son) when they
applied for a patent in 1844. It’s a very popular truss design where diagonal supports slope
down toward center (while in Howe trusses are pointing in the opposite direction). This
design enables the creation of structures that have spans of 76 meters between anchor points.
Bridges with this design were very commonly made between a middle of 19th and early
20th century.
Pratt Truss design
21
APPENDIX 2
DESIGN 1 (PRATT TRUSS DESIGN)
22
DESIGN 3 (HOWE TRUSS DESIGN)
23
APPENDIX 3
Reasons for choosing Warren Truss design
As a team, we decided that the overall bridge design we wanted to pursue would be
derived from a Warren Truss. We decided that the simplest way to adapt the Warren Truss for
our bridge was to build the bridge as two individual Warren trusses (See figure 2.4) with
bracing members connecting the trusses on the top and bottom. As such, our Warren Truss
bridge design is a very simple, common design that is used fairly regularly, originally patented
in the nineteenth century by James Warren and Willoughby Theobald Monzani.
Figure 4.1: Warren Truss design
There reasons why we pursued the Warren Bridge design are the truss of this model
are equilateral triangles which can minimize the forces to only compression and tension. These
equilateral triangles will make the bridge become stronger. Furthermore, this model is more
easy to fabricate and assemble due to its simple design compare to the other two sketch.
Moreover, this model is lighter than other two model (Howe Truss and Pratt Truss) as it used
lesser materials to construct out and save materials.
24
Advantages and disadvantages of Warren Truss
Advantages:

This bridge type uses ridged triangles in the design, which makes it very strong.

Warren truss bridges require lesser building material than most other bridge designs.

It can be constructed piece by piece, which makes cost lesser than conventional ones
which require the entire framework to be set up before building. This also increases the
ways in which the bridge can be built, giving access to many patterns.

The open nature of the bridge means that the view is not blocked.
Disadvantages:

The joints and fittings of a Warren truss bridge need to be checked regularly, and
maintenance can be expensive.

Bridges made over a long span may have many deflection flaws, which need to be
corrected during the building process.

Calculating the load-bearing capability can be complicated.

If the bridge is not designed properly, a lot of material can be wasted, because some of
the parts will not contribute to the bridge in any way.

Some studies have suggested that the design may be distracting to drivers.

Many consider these bridges to be visually unattractive.
25