ITU FACULTY OF ARCHITECTURE, DEPARTMENT OF ARCHITECTURE

STRUCTURAL AND EARTHQUAKE ENGINEERING WG

MIM 208E STRUCTURAL ANALYSIS& DESIGN I

MIM 271E THEORY OF STRUCTURES

Analysis of Statically Determinate Trusses

2023-2024 Dr. Gülseren Erol Soyöz

 Trusses
2
Structural members with solid sections for crossing over big
spans may not be economic. With the increasing loading and
stress along the cross-section, the heights of the structural
elements will also increase. So more economic and safe
sections are introduced such as I sections and castellated
beams.
When the height of the section needs to be bigger than
producible/economical I-sections, trusses are preferred.
 Trusses
3
A truss is a structure composed of slender members joined
together at their end points.
The points where the truss members are joined at are named
joints.
The members commonly used in construction consist of timber
struts or metal bars.
The joint connections are usually formed by bolting or welding
the ends of the members to a common plate called gusset.
 Common Types of Trusses
4

 A truss is one of the major types of engineering structures which

provides a practical and economical solution for many engineering
constructions, especially in the design of bridges and buildings
that demand large spans.
 Trusses can be arranged as planar trusses or space trusses.
 Planar trusses lie in a single plane & is often used to support roof
or bridges.
Planar Trusses

Planar trusses are the systems where all the members, joints and
loads lie in a single plane.
Planar trusses used to support roofs and bridges. Roof load is
transmitted to the truss at joints by means of a series of purlins.
The analysis of the forces developed in the truss members is 2D.
Similar to roof truss, the bridge truss loading is also coplanar.
 Common Types of Trusses
6
 Roof Trusses
 Roof load is transmitted to the truss at the joints by means of
a series of purlins.
 To keep the frame rigid & thereby capable of resisting
horizontal wind forces, knee braces are sometimes used at
the supporting column.
 Common Types of Trusses
7
 Roof Trusses
 Common Types of Trusses
8
 Roof Trusses
 Common Types of Trusses
9
 Bridge Trusses
 The main structural elements of a
typical bridge truss are shown in figure.
Here it is seen that a load on the deck
is first transmitted to stringers, then to
floor beams, and finally to the joints of
the two supporting side trusses.
 The top and bottom cords of these side
trusses are connected by top and
bottom lateral bracing, which serves to
resist the lateral forces caused by wind
and the sideway caused by moving
vehicles on the bridge.
 Additional stability is provided by the
portal and sway bracing. As in the case
of many long-span trusses, a roller is
provided at one end of a bridge truss to
allow for thermal expansion.
 Common Types of Trusses
10
 Bridge Trusses
 Common Types of Trusses
11
 Bridge Trusses
 In particular, the Pratt, Howe, and
Warren trusses are normally used for
spans up to 61 m in length. The most
common form is the Warren truss with
verticals.
 For larger spans, a truss with a polygonal
upper cord, such as the Parker truss, is
used for some savings in material.
 The Warren truss with verticals can also
be fabricated in this manner for spans up
to 91 m.
 Common Types of Trusses
12
 Bridge Trusses
 The greatest economy of material is
obtained if the diagonals have a slope
between 45°and 60°with the
horizontal. If this rule is maintained,
then for spans greater than 91 m, the
depth of the truss must increase and
consequently the panels will get longer.
 This results in a heavy deck system and,
to keep the weight of the deck within
tolerable limits, subdivided trusses
have been developed. Typical
examples include the Baltimore and
subdivided Warren trusses.
 The K-truss shown can also be used in
place of a subdivided truss, since it
accomplishes the same purpose.
 Definition of a Truss
13

• Most structures are made of several trusses joined

together to form a space framework. Each truss
carries those loads which act in its plane and may
be treated as a two-dimensional structure.

• Bolted or welded connections are assumed to be

pinned together. Forces acting at the member ends
reduce to a single force and no couple. Only two-
force members are considered.

• When forces tend to pull the member apart, it is

in tension. When the forces tend to compress the
member, it is in compression.
 Analysis of Planar Trusses
14
To design both the members and the connections of a truss, it is
necessary first to determine the force developed in each member
under given loading. Forces in the members are internal forces. For
external force members, equations of equilibrium will be applied.

Assumptions for Design

1. All loadings are applied at the joints.
- Weight of the members neglected.
2. There is no change at the length of the members.
3. The members are joined together by smooth pins.
- Assume connections provided the center lines of the
joining members are concurrent.
 Classification of Coplanar Trusses
15
 Simple , Compound or Complex Truss.
 Simple Truss:
 To prevent collapse, the framework of a truss must be rigid. The simplest
framework that is rigid or stable is a triangle.
 A simple truss is constructed by successively adding two members and one
connection to the basic triangular truss.
 Classification of Coplanar Trusses
16
 Simple Truss
 The basic “stable” triangle element is ABC.
 The remainder of the joints D, E & F are established in
alphabetical sequence.
 Simple trusses do not have to consist entirely of triangles.
 Classification of Coplanar Trusses
17
 Compound Truss
 Itis formed by connecting 2 or more simple truss together.
 Often, this type of truss is used to support loads acting over a
larger span as it is cheaper to construct a lighter compound
truss than a heavier simple truss.

 Type 1 : The trusses may be connected by a common joint &

bar.
 Type 2 : The trusses may be joined by 3 bars.
 Type 3 :The trusses may be joined where bars of a large
simple truss, called the main truss, have been
substituted by simple truss, called secondary trusses.
 Classification of Coplanar Trusses
18
 Compound Truss
 Classification of Coplanar Trusses
19
• Compound trusses are statically
determinant, rigid, and completely
constrained.
m = 2n − 3

• Truss contains a redundant member

m: no of truss members
n: no of joints
and is statically indeterminate.
r: no of support reactions m  2n − 3

• Additional reaction forces may be

necessary for a rigid truss.
• Necessary but insufficient
condition for a compound truss to
rigid be statically determinant, rigid, and
non-rigid completely constrained,
m  2n − 3 m = 2n − 4 m + r = 2n
 Classification of Coplanar Trusses
20
 Complex Truss
A complex truss is one that cannot be classified as being
either simple or compound
Arrangement of Statically Determinate Trusses
21
 Determinacy
 The total number of unknowns : the forces in m number of
bars of the truss and the total number of external support
reactions r.
 Since the truss members are all straight axial force members
lying in the same plane, the force system acting at each
joint is coplanar and concurrent.
 Consequently, rotational or moment equilibrium is
automatically satisfied at the joint (or pin).
Arrangement of Statically Determinate Trusses
22
 Determinacy
 Therefore only
 Fx = 0 and  Fy = 0

 Bycomparing the total unknowns with the total number of

available equilibrium equations, we have:

m + r = 2n statically determinate
m + r > 2n statically indeterminate

n: number of joints
m: number of bars of the truss
r: number of reaction forces at supports
Arrangement of Statically Determinate Trusses
23
 Stability
 If m + r < 2n => unstable

A truss can be unstable, if it is statically determinate or

statically indeterminate. It will collapse since there will be an
unsufficient number of bars or reactions to constrain all the
joints.

 Stabilitywill have to be determined either through inspection

or by force analysis.
Arrangement of Statically Determinate Trusses
24
 Stability
 External Stability
◼A structure is externally unstable, if all of its reactions are
concurrent or parallel.
◼ The trusses are externally unstable since the support reactions
have lines of action that are either concurrent or parallel.
Arrangement of Statically Determinate Trusses
25
 Internal Stability
◼ The internal stability can be checked by careful inspection of the
arrangement of its members.
◼ If it can be determined that each joint is held fixed so that it cannot
move in a “rigid body” sense with respect to the other joints, then
the truss will be stable.
◼ A simple truss will always be internally stable.
◼ If a truss is constructed so that it does not hold its joints in a fixed
position, it will be unstable or have a “critical form”.
Arrangement of Statically Determinate Trusses
26
 Internal Stability
◼ To determine the internal stability of a compound truss, it is
necessary to identify the way in which the simple truss are
connected together. Joints should be held in a fixed position.
◼ The truss shown is unstable since the inner simple truss ABC is
connected to DEF using 3 bars which are concurrent at point O.
Arrangement of Statically Determinate Trusses
27
 Internal Stability
◼ Thus, an external load applied at A, B or C & cause the truss to
rotate slightly.
◼ For complex truss, it may not be possible to tell by inspection if it is
stable.
◼ The instability of any form of truss may also be noticed by using a
computer to solve the 2n simultaneous equations for the joints of
the truss.
◼ If inconsistent results are obtained, the truss is unstable or have a
critical form.
 Example 3.1
28
Classify each of the trusses as stable, unstable, statically determinate or
statically indeterminate. The trusses are subjected to arbitrary external
loadings that are assumed to be known & can act anywhere on the
trusses.
 Solution
29

For (a),
 Externally stable.
 Reactions are not concurrent or parallel.
 m = 19, r = 3, n = 11
 m + r =2n = 22
 Truss is statically determinate.
 By inspection, the truss is internally stable.
 Solution
30

For (b),
 Externally stable.
 m = 15, r = 4, n = 9

 m + r = 19 >2n

 Truss is statically indeterminate.

 By inspection, the truss is internally stable.

 Solution
31

For (c),
 Externally stable.
 m = 9, r = 3, n = 6

 m + r = 12 = 2n

 Truss is statically determinate.

 By inspection, the truss is internally stable.

 Solution
32

For (d),
 Externally stable.
 m = 12, r = 3, n = 8

 m + r = 15 < 2n

 The truss is internally unstable.

 Analysis of Planar Trusses
33

 The Method of Joints.

 The Method of Sections (Ritter Method).
 The Graphical Method (Cremona Method).
 The Graphical Method
(Cremona Method)
34

 This method deals mainly with the graphical representation of

equilibrium for each joint.
 The basic advantage that makes the method attractive, is its ability to
unify all the force polygons, resulting from graphical equilibrium of
each joint, into one only force polygon, known as Cremona’s
diagram. The method was created by the Italian mathematician Luigi
Cremona.
 The Graphical Method
(Cremona Method)
35
 At a joint, where not more than 2 unknown forces are present, the
unknowns can be determined using the equilibrium of joint that is
represented by the force polygons.
 The Graphical Method
(Cremona Method)
36
 The procedure:
 Draw the vector of the completely known force, in the proper
direction, scale, magnitude and sense.
 From one end of the vector, draw a line parallel to the direction of
one of the 2 forces, while from the other end draw a second line
parallel to the other direction.
 The vector and the point of section of the two lines define a
triangle.
 Now, following the path of the vector by laying out the 2 unknown
forces tip to tail, thus closing the force triangle, we find both the
magnitudes and the senses of the other 2 forces.
 The Graphical Method
(Cremona Method)
37
 Of course the completely known force can be considered as the
resultant of other known forces, through a force polygon. From this
procedure we realize that the basic characteristic which appears to be
common in the method of joints and Cremona’s diagram lies in the
main strategic.
 Compared to the analytical method of joints, the graphical method of
Cremona’s diagram is less precise. However, the ‘loss of precision’ is
unimportant and theoretical. Nevertheless, the speed and the
elegance of the method are the main characteristics that make it
popular and attractive by many designers.
 The Graphical Method
(Cremona Method)
38
 Analysis Order:
 Determine the reactions at supports by using equations of
equilibrium.
 Draw the truss to a scale. (ex: 1cm=1kN)
 The Graphical Method
(Cremona Method)
39

X = 0 HA = 0
MB = 0 VA = ( 4kN · 4m + 4kN · 2m + 2kN · 6m ) / 4m = 9kN VA = 9kN
Y = 0 VB = 2kN + 4kN + 4kN + 2kN – 9kN = 3kN VB = 3kN
 The Graphical Method
(Cremona Method)
40
Define the notation to designate each member and all the forces.
Each member or external load separates two areas.
Label the spaces external to the truss between points of application
of external forces with a number/letter in a clockwise order.
Label the triangular spaces within the truss in any order.
 The Graphical Method
(Cremona Method)
41
Using the scaled drawing of the truss, draw the equilibrium polygon
of external forces. (1 cm=1kN)
Each region is represented by a point at the force polygon. The
intersection of the parallel lines drawn from one region (point) to the
adjacent region gives the point corresponding to the adjacent region.
 The Graphical Method
(Cremona Method)
42

2cm

4cm
9cm

4cm

2cm
 The Graphical Method
(Cremona Method)
43
The force polygon of a joint with only 2
unknown forces is drawn first.
Construct a line through 2 with the known
direction of 2-7 and a line through 3 with the
known direction of 3-7. The intersection of
these lines locates point 7.
 The Graphical Method
(Cremona Method)
44

Once the diagram is constructed, the

magnitude of the force in any member is
obtained by measuring the distance between
the numbers that denote the member.
 The Graphical Method
(Cremona Method)
45

Determine whether the member

force is tension or compression.
 Joint with max 2 unknowns:
Turn at clockwise direction at
the joint. According to the no
of areas you are passing, place
the arrows to the Cremona
diagram.
 If the direction of the arrow is
away from the joint… tension

Tabulate all the member forces.

 Solution
46
 The Graphical Method
(Cremona Method)
47

A set square with integrated protractor

 Analysis of Trusses by the Method of Joints
48
• Dismember the truss and create a freebody diagram
for each member and pin. Force system acting at
each joint is coplanar and concurrent. ∑Fx = 0 and
∑Fy = 0 must be satisfied for equilibrium.
• 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.
• Conditions of equilibrium on the pins provide 2n
equations for 2n unknowns. For a simple truss,
2n = 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.
 Joints Under Special Loading Conditions
49
• Forces in opposite members intersecting in two
straight lines at a joint are equal.
• The forces in two opposite members are
equal when a load is aligned with a third
member. The third member force is equal to
the load (including zero load).

• The forces in two members connected at a

joint are equal if the members are aligned and
zero otherwise.
• Recognition of joints under special loading
conditions simplifies a truss analysis.
Procedure for Method of Joints

• Start with the joint where you have one known and two unknowns

50
Procedure for Method of Joints

• Determine the support reactions considering the whole truss as a rigid body

51
Procedure for Method of Joints

52
Zero-Force Members
• Zero-force members are used to increase the stability of the truss during
construction and to provide added support if the loading is changed.
• If only two members form a truss joint, these two members must be zero-force
members. (No external load or support reaction is applied)
• If only three members form a truss joint and two of the members are collinear,
the third member is a zero-force member. (No external load or support
reaction is applied)

AB, AF, DC, DE

53
 The Method of Sections
(Ritter Method)
54
 If the forces in only a few members of a truss are to be
found, the method of sections generally provide the most
direct means of obtaining these forces
 The method is created the German scientist August Ritter
(1826 - 1908).

 This method consists of passing an imaginary section through

the truss, thus cutting it into 2 parts
 Provided the entire truss is in equilibrium, each of the 2 parts
must also be in equilibrium
 Analysis of Trusses by the Method of Sections
55
• 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.
• With only three members cut by the section,
the equations for static equilibrium may be
applied to determine the unknown member
forces, including FBD.
 Analysis of Trusses by the Method of Sections
56
 Example 2
57
SOLUTION:
• Take the entire truss as a free body.
Apply the conditions for static equilib-
rium to solve for the reactions at A and L.
• Pass a section through members FH,
GH, and GI and take the right-hand
section as a free body.
• Apply the conditions for static
equilibrium to determine the desired
Determine the force in members FH, member forces.
GH, and GI.
 Example 2
58

SOLUTION:
• Take the entire truss as a free body.
Apply the conditions for static equilib-
rium to solve for the reactions at A and L.

 M A = 0 = −(5 m )(6 kN ) − (10 m )(6 kN ) − (15 m )(6 kN )

− (20 m )(1 kN ) − (25 m )(1 kN ) + (25 m )L
L = 7.5 kN 
 Fy = 0 = −20 kN + L + A
A = 12.5 kN 
 Example 2
59
• Pass a section through
members FH, GH, and GI and
take the right-hand section as
a free body.

• Apply the conditions for static

equilibrium to determine the
desired member forces.

MH = 0
(7.50 kN )(10 m ) − (1 kN )(5 m ) − FGI (5.33 m ) = 0
FGI = +13.13 kN

FGI = 13.13 kN T
 Example 2
60
FG 8 m
tan  = = = 0.5333  = 28.07
GL 15 m
 MG = 0
(7.5 kN )(15 m ) − (1 kN )(10 m ) − (1 kN )(5 m )
+ ( FFH cos  )(8 m ) = 0
FFH = −13.82 kN
FFH = 13.82 kN C
GI 5m
tan  = =2 = 0.9375  = 43.15
HI
3
(8 m )
ML = 0
(1 kN )(10 m ) + (1 kN )(5 m ) + (FGH cos  )(10 m ) = 0
FGH = −1.371 kN
FGH = 1.371 kN C
 Example 3.5
61

Determine the force in members CF and GC of the roof truss. State

whether the members are in tension or compression. The reactions at the
supports have been calculated.
 Solution
62

The free-body diagram of member CF can be obtained by considering

the section a-a,
A direct solution for FCF can be obtained by applying  M E = 0
Applying Principal of transmiss ibility,
FCF is slide to point C for simplicity .
With anti - clockwise moments as + ve,  M E = 0
− FCF sin 30 o (4) + 1.50(2.31) = 0
FCF = 1.73kN (C )
 Solution
63

The free-body diagram of member GC can be obtained by considering

the section b-b,
Moments will be summed about point A in order to eliminate the
unknowns FHG and FCD . Sliding to FCF point C, we have :
FCF is slide to point C for simplicity .
With anti - clockwise moments as + ve,  M A = 0
− 1.50(2.31) + FGC (4) − 1.73 sin 30 o (4) = 0
FGC = 1.73kN (T )
 Example 3.6
64

Determine the force in member GF and GD of the truss. State whether the
members are in tension or compression. The reactions at the supports have
been calculated.
 Solution
65

The distance EO can be determined by proportional triangles or realizing

that member GF drops vertically 4.5 – 3 = 1.5m in 3m.
Hence, to drop 4.5m from G the distance from C to O must be 9m
 Solution
66

The angles FGD and FGF make with the horizontal are
tan-1(4.5/3) = 56.3o
tan-1(4.5/9) = 26.6o

The force in GF can be determined directly by applying

MD = 0
FGF is slide to point O.
With anti - clockwise moments as + ve,  M D = 0
− FGF sin 26.6o (6) + 7(3) = 0
FGF = 7.83kN (C )
 Solution
67

The force in GD can be determined directly by applying

 MO = 0
FGD is slide to point D.
With anti - clockwise moments as + ve,  M O = 0
− 7(3) + 2(6) + FGD sin 56.3o (6) = 0
FGD = 1.80kN (C )
 References
68

 Hibbeler, R.C., Structural Analysis, Prentice Hall, 2011.

 Lecture Notes by Assistant Prof.Dr. Cenk Üstündağ.
 Lecture Notes by Dr. Gülseren Erol Soyöz.
 Lecture Notes by Prof.Dr. Serdar Soyöz.