Strength of Materials II
(SOM2601)
Chapter 1: Simple Trusses
Chapter 1: Simple trusses
• Trusses:
Chapter 1: Simple trusses
• Definition:
 A truss can be defined as a two dimensional assemblage of members, each member
being joined at its ends to the foundation or to other members by frictionless pin
joints.
 There are different types of trusses, such as the Pratt, Parker and Warren
 Trusses are statically determinant
 Applications:
There are roof trusses, bridge trusses and stadium trusses
Chapter 1: Simple trusses
• Definition (continued):
Chapter 1: Simple trusses
• Definition (Continued):
• In a simple truss, m = 2j – 3 where
• m is the total number of members and
• j is the number of joints
Chapter 1: simple trusses
• Analysis Methods:
1. Method of Joints
2. Method of Sections
CHAPTER 1: Simple trusses
• Direct Force: Two types of direct forces: tensile and
compressive force.
P
P
Tensile
Compressive
P
P
Chapter 1: Simple trusses
• Moment of a Force: The moment 𝑀𝐴 of a force P about a point A is equal to the
force multiplied by the shortest distance between the point A and the line of action of the
force, illustrated in the figure below.
•
•
x
A
P
Chapter 1: Simple trusses
• Conditions for Static Equilibrium:
• The sum of all the forces in any direction must be zero 𝑃 = 0
• The sum of all the moments of the forces about any point must be zero 𝑀 = 0
• The two conditions are normally written in this manner:
𝑃𝑥 = 0 (sum of the forces in the x-direction=0 )
𝑃𝑦 = 0 (sum of the forces in the y-direction=0)
𝑀 = 0 (sum of the moments=0)
Chapter 1: Simple trusses
• Method of Joints:
• The method of joints is one of the simplest methods for determining the force
acting on the individual members of a truss because it only involves two force
equilibrium equations.
Chapter 1: Simple trusses
Chapter 1: Simple trusses
• Analysis of Trusses by the Method of Joints with free body diagrams :
Chapter 1: Simple trusses
Chapter 1: Simple trusses
• Analysis of Trusses by the Method of Joints (continued):
• Use conditions for equilibrium for the entire truss to solve for the
reactions RA and RB.
Chapter 1: Simple trusses
• Analysis of Trusses by the Method of Joints (continued):
• Dismember the truss and create a free body diagram for each member
and pin.
• The two forces exerted on each member are equal, have the same line of
action, and opposite sense.
• Forces exerted by a member on the pins or joints at its ends are directed
along the member and equal and opposite.
Chapter 1: Simple trusses
• Conditions of equilibrium on the pins provide 2j
equations for 2j unknowns. For a simple truss, 2j =
m + 3. May solve for m member forces and 3
reaction forces at the supports.
• Conditions for equilibrium for the entire truss
provide 3 additional equations which are not
independent of the pin equations
Chapter 1: Simple trusses
• Assumptions:
• The following assumptions are usually made in the analysis of trusses:
1. Buckling of members will not occur.
2. The force in the member due to the mass of the structure are small compared to the loads
and may be ignored.
3. All members are connected by frictionless pin joints. In practice the members are bolted,
riveted or welded together, thus making the structure more rigid. This assumption simplifies
the computations involved and gives conservative results.
4. All external forces, including the reactions at the supports, are applied at the joints and the
structure is statically determinate.
Chapter 1: Simple trusses
• Methods of Sections:
• When the force in only one member
or the forces in a very few members
are desired, the method of sections
works well.
• To determine the force in member BD,
pass a section through the truss as
shown and create a free body diagram
for the left side (or right side).
Chapter 1: Simple trusses
• With only three members cut by the section, the
equations for static equilibrium may be applied to
determine the unknown member forces, including
FBD
Chapter 1: Simple trusses
• Procedure:
• Finding the reaction forces;
• Make a cut through the members where the
unknown forces are applied;
• Establish equilibriums for one of the sections: Σ Px
= 0; Σ Py = 0; ΣM = 0;
• If your result comes out to be negative, then it
means you assumed a wrong direction of the force.
Chapter 1: Simple trusses
• Method of Sections (Continued):
• Applications:
• The method of sections is commonly used when the
forces in only a few particular members of a truss are
to be determined;
• The method of sections is always used together with
the method of joints to analyse trusses.
Chapter 1: Simple Trusses
• IMPORTANT NOTES:
• For a truss to be properly constrained:
– It should be able to stay in equilibrium for any
combination of loading.
– Equilibrium implies both global equilibrium and
internal equilibrium.
• Note that if 2j > m + r, the truss is most
definitely partially constrained (and is unstable to certain loadings). But 2j ≥ m + r, is no
guarantee that the truss is stable.
• If 2j < m + r, the truss can never be statically
determinate.
Chapter 1: Simple trusses
• Simple Space Trusses:
• Space trusses are three dimensional structures the simplest
stable space truss consists of six members joined to form a
tetrahedron.
Chapter 1: Simple trusses
Chapter 1: Simple trusses
• Example 1:
• Which of the frames shown would not be classified as a truss? Why?
Chapter 1: Simple trusses
• Solution:
• Frame (b) cannot be treated as a truss because the
load is not applied at a joint.
• Frame (c) cannot be treated as a truss since to resist
the load at E, CDE must be a single member and is
therefore not joined at its end alone.
Chapter 1: Simple trusses
• Example 3:
• Determine the forces in the members in the following truss.
Chapter 1: Simple trusses
• Example 3(continued):
• Solution:
• As we are solving the problem using the method of joints, we take equilibrium at each point. As we have assumed
the forces in all the members are tensile, the direction of the reaction force they exert on the hinges are as shown
1
• tan 𝛽 = 1.5; sin 𝛽 =
2
;
13
cos 𝛽 =
3
13
• Equilibrium at B:
•
•
𝐹𝑦 =0
𝐹𝐵𝐶 sin 𝛽 + 𝐵𝑦 =0
𝐹𝐵𝐶 =-10.62kN
𝐹𝑥 =0
𝐹𝐴𝐵 + 𝐹𝐵𝐶 cos 𝛽=0
𝐹𝐴𝐵 = 8.84kN
• Equilibrium at A:
•
𝐹𝑦 =0
𝐹𝐴𝐶 sin 𝛽 + 𝐴𝑦 =0
𝐹𝐴𝐶 =-2.13kN
Chapter 1: Simple trusses
•
•
Example 3 (Continued):
𝐹𝐵𝐶
𝛽
B
𝐵𝑦 = 5.89 kN
𝐹𝐵𝐶
𝐴𝑥 = 7.07kN
A
𝐴𝑦 = 1.18 kN
𝛽
𝐹𝐴𝐵
Chapter 1: Simple trusses
• Example 3 (continued):
• We assumed that all the forces in the members were tensile. But
we got some of them negative. So, the negative sign indicates
that the forces in the members are compressive
• 𝐹𝐴𝐵 = 8.84 kN (T)
• 𝐹𝐵𝐶 = 10.62 kN (C)
• 𝐹𝐴𝐶 = 2.13 kN (C)
Chapter 1: Simple trusses
•
•
Example 4:
Find the force in member BC,BH and GH
10k
A
20k
B
40k
C
D
E
10m
F
G
H
I
10m
J
20k
10k
Chapter 1: Simple trusses
• Example 4 (Continued):
• Finding the reaction forces first:
• 𝐹𝑥 =0;
• 𝐹𝑦 = R(𝐹𝑦 )+ R(𝐽𝑦 )-40k-20k-20k-10k-10k=0;
• 𝑀 𝐹 = 0;
• R(𝐽𝑦 )×40m-20k×10m-40k×20m-20k×30m-10k×40m=0
• Find R(𝐹𝑦 )= R(𝐽𝑦 )=50kN
Chapter 1: Simple trusses
• Example 4 (Continued):
• Since the forces of particular members laying in the middle of
the truss are asked, we use the method of section. Cut the truss
through the members with the unknown forces:
• Assume the direction of the forces as shown in the diagram
Chapter 1: Simple trusses
• Example 4 (Continued):
•
•
•
•
•
•
•
•
Write the equation of equilibrium for the section:
Σ 𝐹𝑥 = 𝐹𝐵𝐶 + 𝐹𝐵𝐺 x sin45 + 𝐹𝐺𝐻 =0
Σ 𝐹𝑦 = 50k – 10k – 20k –𝐹𝐵𝐺 cos45 = 0
Σ 𝑀𝐵 = 𝐹𝐺𝐻 10m + 10k x 10m – 50k x 10m = 0
Solve the equilibrium, we find:
𝐹𝐺𝐻 =40k (in tension)
𝐹𝐵𝐺 = 28.3k (in tension)
𝐹𝐵𝐶 = -60k (the negative sign means we assumed the wrong direction. BC
is in compression)
Chapter 1: Simple trusses
• Problem 1:
• Classify each of the structures as completely, partially, or improperly
constrained, further classify it as statically determinate or
indeterminate.
Chapter 1: Simple trusses
• Problem 2:
• Classify each of the structures as completely, partially, or
improperly constrained, further classify it as statically
determinate or indeterminate.
Chapter 1: Simple trusses
• Problem 3:
• Answer the following:
• State two main conditions to be satisfied for static equilibrium
• Distinguish briefly between statically determinate and statically indeterminate
structures
• When is a plane truss considered to be statically determinate?
• Describe the procedure for solving the internal forces in a plane truss using the
method of sections. Elaborate on each step. You can do this by explaining each step
in a solved example either in the study guide or textbook.
Chapter 1: Simple trusses
• Problem 4:
• Using methods of joints, determine the forces in the members of the trusses
shown:
Chapter 1: Simple trusses
• Problem 5:
• A Fink roof truss is loaded as shown in the figure below. Use
the method of sections to determine the force in members (a)
BD, CD, and CE (b) FH, FG, EG
Chapter 1: Simple trusses
 Instructions:
 Work through Chapter 1 on Simple Trusses from the prescribed textbook by J G Drotsky.
 Make use of the prescribed textbook and study guide
 Work through examples from this slide
 Work out problems 1 through to problem 6.
 (Note: Kindly participate by working through these problems, as they will assist in your understanding of the
module, as you actively participate, it will also make your assignment preparation easier)
Chapter 1: Simple trusses
Any questions?
• Questions can be directed to my e-mail address on:
mlangph@unisa.ac.za
Chapter 1: Simple trusses
• End of Lesson Chapter on Simple Trusses
Thank You