Analysis of structures

Analysis of Structures
Prof. Dr. Ahmed Hasan Zubydan

Analysis of Structures
B
Prof. Dr. Ahmed Hasan Zubydan
Professor of Structural Engineering
Civil Engineering Department, Faculty of Engineering
PortSaid University.

Contents
Chapter 1 Statics of Structures
1.1 Supports and Reactions
1.2 Stability and Determinancy
1.3 Condition Equations
1.4 Computation of Reactions
Chapter 2 Trusses
2.1 Introduction
2.2 Trusses Configuration
2.3 Condition Equations
2.4 Stability and Determinancy
2.5 Methods of Analysis of Simple and Compound Trusses
2.6 Method of Joints
2.7 Method of Sections
2.8 Complex Trusses
2.9 Graphical Method of Joints (Maxwell Diagram)
Chapter 3 Statically Determinate Beams and Frames
3.1 Introduction
3.2 Axial Force, Shear Force, and Bending Moment Diagrams
3.3 Relationship between Load, Shear, and Bending Moment
Chapter 4 Influence Lines for Statically determinate Structures
4.1 Introduction
4.2 Applications of Influence Lines
4.3 Distributed Load
4.4 Concentrated Load
4.5 Maximum Absolute Bending Moment
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1
Statics of Structures
1.1 SUPPORTS AND REACTIONS
Reaction forces are developed at supports of a structure to equilibrate the effect of the
applied forces. The supports through which a structure transmits its load to the supporting
foundation can be classified as follows:
Roller support:
This support provides restraint against linear movement in the direction normal to supporting
surface as shown in Fig. 1.1a. The magnitude of the resultant reaction force at node a is given
directly by Ra. The direction of Ra is perpendicular to the roller surface and the point of
application is at point a.
Hinge support:
It constrains all translations at the point of attachment, but the structure is free to rotate about
this point. The unknown independent reaction components are Ya and Xa as shown in Fig 1.1b.
Fixed support:
It constrains the point of attachment from translation and rotation. With this type of supports,
there are three unknowns. The unknowns are two reaction forces Xa and Ya in addition to a
fixing moment Ma as shown in Fig. 1.1c.
Guide support:
It constrains the structure from rotation and translation normal to the guide slot as shown in
Fig. 1.1d. Thus, while there is no reactive forces parallel to the slot, there is an unknown force
normal to the slot (Ra) and an unknown bending moment at the support (Ma).
(a) Roller support
Ra
a
Ra
a

Structural analysis I, by Prof. Dr. Ahmed Zubydan.2
(b) Hinge support
(c) Fixed support (d) Guide support
Fig. 1.1 Symbols for various support types
1.2 STABILITY AND DETERMINANCY
A structure is considered geometrically stable if for any incipient movement, due to any
conceivable system of applied loads, an elastic resistance to this movement is immediately
developed. A structure is considered statically determinate if the external reactions and the
internal member forces can be determined only by considering the equations of statical
equilibrium without any references to the elastic properties of the structural members.
A rigid planar structure can be acted upon by a general system of forces that can be
combined into a single force and a single moment. For statical stability and determinancy,
the number and arrangement of reactions must be such that they keep the external forces in
equilibrium. Since there are three equations of static equilibrium (  Fx = 0,  Fy = 0,  M
= 0) for planar structures, there must be three components of reactions, which are neither
parallel nor concurrent. If the components of reactions are less than three, the structure is
unstable. If the number of reaction components is three, and if those components are neither
parallel nor concurrent, the structure is externally stable and determinate. If the number of
reaction components is greater than three, and if the components are arranged such that they
are neither parallel nor concurrent, the structure is externally stable and statically
indeterminate. The degree of external indeterminancy is defined as the number of unknown
reactions in excess of the number of the available equations of statics plus the number of
equations of construction.
The conditions are summarized in the following criteria:
r < 3; structure is statically unstable externally
r = 3; structure is statically determinate externally
r > 3; structure is statically indeterminate externally
where r is the total number of unknown reaction components.
The preceding criteria must be applied with some discretion. If r < 3, the structure is
definitely unstable; however, in cases where r  3, the structure is not necessarily stable. It is
Ya
Xa
aXa
Ya
a
Ya
Xa
Ma
a
Ma
Xa
a

Statics of structures 3
o
o
Link
possible that the reaction components are not properly arranged to ensure external stability.
These structures category are geometrically unstable.
Figures 1.2 to 1.4 gives the statical classification of several structures. The simple beam
shown in Fig. 1.2a is unstable because the number of reaction components is less than three (r
= 2). Since the beam shown of Fig 1.2b has no reaction to equilibrate the horizontal component
of the applied loading, the structure will move horizontally if any horizontal load acts on the
beam. On the other hand, the structures shown in Figs. 1.2c and 1.2d have the supports
arranged so that the lines of action of all three reactions intersect at one point. So, the structures
shown in Figs. 1.2a to 1.2d are geometrically unstable. The structures shown in Figs. 1.3a to
1.3f have three reaction components that are neither parallel nor concurrent. Therefore, they
are stable and statically determinate. Finally, the structures shown in Figs 1.4a to 1.4c have
more than three reaction components. These reaction components are neither parallel nor
concurrent, so these structures are stable and statically indeterminate.
(a) (b)
(c) (d)
Fig. 1.2 Unstable structures
(a) (b)
(c) (d)
Fig. 1.3 Stable and statically determinate structures
(a) (b)
(c)
Fig. 1.4 Stable and statically indeterminate structures

1.3 CONDITION EQUATIONS
Some structures incorporate unique internal features of construction that significantly
affect the behavior of the system and require special attention from the structural engineer.
For example, the beam shown in Fig. 1.5a has an internal hinge at b. This hinge results in no
bending moment transmission through the beam at point b. The free body diagrams for the
two beam segments ab and bc indicates that  Mb = 0 for the two individual free-body
diagrams. The one independent equation introduced by the condition of construction is
referred to as an equation of condition. For the intermediate roller of Fig. 1.5b, only the force
normal to the axes of the beams will be transmitted. This gives two equations of condition
since the force parallel to the longitudinal axis of the beam and the moment are both zero at
the roller.
Thus for a structure with c condition equations, the following criteria is applied:
r < 3 + c ; structure is statically unstable externally
r = 3 + c ; structure is statically determinate externally
r > 3 + c ; structure is statically indeterminate externally
(a) (b)
Fig. 1.4 Condition equations
The number of condition equations given by an intermediate hinge depends
on the number of sides that is connected by this hinge. Generally, the number
of condition equations c is given as
c = n – 1 ………………………………………………(1.1)
in which n is the number of sides which is connected with the hinge. For example, the hinge
given in Fig. 1.6a gives one condition equation because this hinge connects two sides. On the
other hand, the hinges shown in Figs 1.6b and 1.6c give two and three condition equations,
respectively.
(a) One condition equation
(b) Two condition equations (c) Three condition equations
Fig. 1.6 Number of condition equations given by intermediate hinge.
Hinge
a b c Roller
n = 2; c= 1 n = 2; c = 1n = 2; c = 1 n = 2; c = 1
n = 3; c= 2 n = 3; c= 2 n =4; c =3

The frames that contain closed panels require an internal classification in addition to the
external classification. Each closed panel requires three internal condition equations to be
solved. Let e be the number of internal condition equations for a closed panel. Then, the
following condition prevail for each closed panel.
e > 3; structure is statically unstable internally
e = 3; structure is statically determinate internally
e < 3; structure is statically indeterminate internally
The stability of the frames that contain closed forms can also be determined by cutting the
structure more than one part and by determining the stability of each portion, the whole stability
of the structure is evaluated.
EXAMPLE 1.1
Classify each of the following structures with respect to stability and statical determinancy
(a) (b)
(c) (d)
SOLUTION
Structure (a)
3 + c = 3 + 1 = 4
r = 3
3 + c > r unstable
Structure (b)
3 + c = 3 + 2 = 5
r = 5
3 + c = r stable and statically determinate
Structures (c) and (d)
3 + c = 3 + 1 = 4
r = 4
3 + c = r
a b c
b'
a
b
c
r1 r2

For structure (c), the node b can moves a short distance perpendicular to the line ac without
any resistance of members ab or bc. Hence, the structural system (c) is geometrically unstable
since the shape of the structure can be altered without any change in length of any member.
The structure (c) is said to be the critical form. This situation does not exist in the stable
structure (d) since the two arcs r1 and r2 have no common tangent and any incipient movement
of b will generate an immediate resistant from the various parts of the structure.
EXAMPLE 1. 2
(a) (b)
(c) (d)
SOLUTION
Structure (a)
Structure (a) has no resistance against horizontal loads, so this structure is unstable
Structure (b)
3 + c = 3 + 2 = 5
r = 4
3 + c > r unstable
Structure (c)
3 + c = 3 + 1 = 4
r = 4
Structure (d)
3 + c = 3 + 0 = 3
r = 5
3 + c < r stable and statically indeterminate to
the second degree

EXAMPLE 1.3
(a) (b)
(c) (d)
SOLUTION
Structures (a), (b), and (c)
3 + c = 3 + 2 = 5
r = 5
3 + c = r
It is noticed that the part abc of the structure (b) has a critical form, so this structure is
unstable. On the other hand, structures (a) and (c) are stable and statically determinate.
Structure (d)
3 + c = 3 + 3 = 6
r = 5
3 + c > r unstable
EXAMPLE 1.4
(a) (b)
a b c

a b
a
b
(c) (d)
SOLUTION
Structures (a), (b), and (c)
These structures are unstable because the reaction components are concurrent.
Structure (d)
3 + c = 3 + 2 = 5
r = 5
EXAMPLE 1.5
(a) (b)
(c) (d)
SOLUTION
Structure (a)
3 + c = 3 + 1 = 4
r = 4
Structure (b)
3 + c = 3 + 1 = 4
r = 4

a
b c
d
Structure (c)
3 + c = 3 + 1 = 4
r = 4
3 + c = r
The part ab couldn’t resist any load has a vertical component, so this part is unstable.
Structure (d)
3 + c = 3 + 2 = 5
r = 6
3 + c < r
The part ab couldn’t resist any load has a horizontal component, so this part is unstable.
EXAMPLE 1.6
(a) (b)
(c) (d)
SOLUTION
Structure (a)
3 + c = 3 + 0 = 3
r = 3
3 + c = r externally stable and statically determinate
e = 3 internally stable and statically determinate

a
b c
e = 3 e = 3
c
Structure (b)
The part abcd couldn’t resist any lateral loads. So the
structure is unstable. Another way for detecting the stability of
the structure by dividing it into two portions as shown in figure
and assuming that the upper portion resting on the lower one. It
is noticed that the upper portion is stable and statically
determinate while part ab of the lower portion is unstable. So
the structure is unstable
Structure (c)
When the structure is divided into two portions and
assuming that the upper portion is resting on the lower one as shown
in figure. It is noticed that, portion bdef is stable and statically
determinate. On the other portion, part ahg is a critical form, so the
structure is unstable.
Structure (d)
When the structure is divided into two portions and assuming
that the upper portion is resting on the lower one as shown in
figure. It is noticed that, each of the two portions is stable and
statically determinate. So, the structure is stbale and statically
determinate.
EXAMPLE 1.7
Classify each of the following structures with respect to stability and statical determinancy.
(a) (b)
(c) (d)
SOLUTION
a
b
c
d
3 + c = 4
r = 4
a
b
b
d
e
f
f
gh
3 + c = 4
r = 4
3 + c = 4
r = 4
3 + c = 4
r = 4
3 + c = 4
r = 4

e =3
e =3
Structures (a) and (b)
3 + c = 3 + 3 = 6
r = 6
3 + c = r
The part abc on the structure (b) has a critical form, so this structure is unstable. On the other
hand, structure (a) is stable and statically determinate.
Structure (c)
3 + c = 3 + 1 = 4
r = 4
external condition equation is given by hinge at c
It is also noticed that each closed form has three condition equations (e = 3 ), so the structure
is internally stable and statically determinate.
Structure (d)
When the structure is divided into two portions and
assuming that the upper portion carrying the lower one
as shown in figure. It is noticed that, each of the two
portions is stable and statically determinate. So, the
structure is stable and statically determinate.
EXAMPLE 1.8
(a) (b)
(c) (d)
SOLUTION
Structure (a)
3 + c = 3; r = 3
3 + c = 5
r = 5

3 + c = 3 + 0 = 3
r = 3
Since each closed form has three condition equations, then the structure is internally stable
and statically determinate
Structure (b)
Dividing the structure into three portions and assuming that each
portion resting on the lower one as shown in figure. It is noticed that,
each portion is stable and statically determinate. So, the structure is
stable and statically determinate.
Structure (c)
3 + c = 3 + 4 = 7
r = 7
Structure (d)
3 + c = 3 + 1 = 4
r = 4
EXAMPLE 1.9
(a) (b)
(c) (d)
SOLUTION
3 + c = 4
r = 4
3 + c = 4
r = 4
3 + c = 3
r = 3
3 + c = 3
r = 3

a
b c
e = 3
d
e
Structure (a)
Dividing the structure into three portions as shown in
figure. It is noticed that, each portion is stable and statically
determinate. So, the structure is stable and statically
determinate.
Structure (b)
3 + c = 3 + 3 = 6
r = 6
Structure (c)
Dividing the structure into two portions and assuming that the
upper portion carrying the lower one as shown in figure. It is
noticed that, each portion is stable and statically determinate. So,
the structure is stable and statically determinate.
Structure (d)
3 + c = 3 + 3 = 6
r = 7
3 + c < r stable and statically indeterminate to the first degree
EXAMPLE 1.10
(a) (b)
(c)
(d)
SOLUTION
3 + c = 4
r = 4
3 + c = 4
r = 4
3 + c = 3
r = 3
3 + c = 3
r = 3
3 + c = 5
r = 5

a
b
Structure (a)
Dividing the structure into three portions as shown in
determinate.
Structure (b)
Dividing the structure into two portions and assuming that
the upper portion is resting on the lower one as shown in figure.
It is noticed that, the lower portion is stable and statically
determinate while the upper portion is stable and statically
indeterminate to the second degree. By summing the two
portions once again, the structure is stable and statically
indeterminate to the second degree.
Structure (c)
Dividing the structure into two portions and
assuming that the upper portion resting on the lower one as
shown in figure. It is noticed that, the upper portion is
stable and statically determinate while lower portion is
unstable. Then, the structure is unstable.
Structure (d)
3 + c = 3 + 1 = 4
r = 4
external condition equation is given by the hinge at d
3 + c = r externally stable and statically determinate.
e = 3 internally stable and statically determinate.
EXAMPLE 1.11
3 + c = 4
r = 4
3 + c = 4
r = 4
3 + c = 3
r = 3
3 + c = 3
r = 3
3 + c = 5
r = 7
3 + c = 4
r = 3
3 + c = 4
r = 4

a
(c) (d)
SOLUTION
Structure (a)
3 + c = 3 + 3 = 6
r = 5
3 + c > r unstable.
Structure (b)
3 + c = 3 + 3 = 6
r = 6
3 + c = r
In this structure, the column ab couldn’t resist any lateral loading, so the structure is unstable.
Structure (c)
3 + c = 3 + 1 = 4
r = 3
(the condition equation is given by the hinge at a)
3 + c > r unstable.
It is noticed that the condition equation given by the hinge at a couldn’t be used to find the
additional reaction component. Moreover, the part abc couldn’t resist any lateral loading. So,
the structure is unstable
Structure (d)
Dividing the structure into two portions as shown in
determinate.
1.4 COMPUTATION OF REACTIONS
The equations of static equilibrium play a vital role in determining the reactions for any
structure. If the structure is statically determinate externally, these equations provide all whate
e = 3
3 + c = 3
r = 3
3 + c = 4
r = 4

is needed for the solution, whereas for statically indeterminate cases, the equations of
equilibrium remain applicable, but they are not sufficient for the solution.
In determining the reactions, a free-body diagram (FBD)of the entire structure is
constructed. All given forces are shown to act in their prescribed direction, and each unknown
reaction component is assumed to act in a specified direction. The equations of static
equilibrium are applied in consistency with the assumed directions, and these equations are
solved for the unknown reactions. If the solution yields a positive reaction component, then
the assumed direction is correct, whereas a negative result indicates that the opposite direction
is correct. In the latter case, care must be exercised in subsequent calculations. If the negative
sign is retained, then the originally assumed direction must also be retained. However, the
direction can be reversed on the free-body diagram and a positive sign can then be used in
subsequent steps.
EXAMPLE 1.12
Calculate the reactions for the beam shown.
SOLUTION
Free-Body Diagram
The reaction at b has two unknown components, whereas the reaction at point e has only a
single unknown component as shown in the figure.
Determination of Reactions
 Px = 0: (→+) Xb = 0
 Me = 0: Yb  9 – 20  12 – 50  6 – 30  4 = 0
Yb = 73.3 kN
 Py = 0: (+) Ye + Yb – 20 – 50 –30 = 0
Ye + 73.3 – 20 – 50 –30 = 0
Ye = 26.7 kN
The positive sign indicates that the reaction components act in the assumed direction on the
free-body diagram.
20 kN 50 kN 30 kN
Yb Ye
Xb
b e
a b
dc
3 m 4 m
20 kN 50 kN 30 kN
3 m 2 m
e

EXAMPLE 1.13
SOLUTION
Free-Body Diagram
The distributed load is replaced by a resultant concentrated force applied at its centroidal
position. The unknown reaction components are also shown.
 Px = 0: (→+) Xd = 0
 Md = 0: Ya  6 – 60  5 + 60 = 0
Ya = 40 kN
 Py = 0: (+) Yd + Ya – 60 = 0
Yd + 40 – 60 = 0
Yd = 20 kN
EXAMPLE 1.14
SOLUTION
Free-Body Diagram
The inclined force is shown resolved into two components.
60 kN.m
60 kN
Ya
Yd
1m
a
Xdd
40 kN
Xa
Ya Ye
48 kN
36 kN
50 kN
a
e
2.5 m
a
b
d
c
3 m 2 m 1 m
60 kN.m
40 kN/m
4
3 fed
a
b
2 m 3 m
60 kN40 kN 10 kN/m
2 m3 m 1 m

 Px = 0: (→+) – Xd + 36 = 0
Xd = 36 kN
 Me = 0: Ya  8 – 40  5 – 48  2 + 50  0.5 = 0
Ya = 33.9 kN
 Py = 0: (+) Ye + Ya – 40 – 48 – 50 = 0
Ye + 33.9 – 40 – 48 – 50 = 0
Ye = 104.1 kN
EXAMPLE 1.15
SOLUTION
Free-Body Diagram
The structure is separated into three separate free bodies, and the unknown reactions and
internal forces at the hinges are shown in the figure.
Free body ce
The free body diagram ce is treated as a simply supported beam and thus inherently use the
equilibrium conditions. The reactions (internal forces transferred at hinges) are
Xc = Xe = 0 ; Yc = Ye = 7.5 kN
Free body abc
 Px = 0: (→+) Xa = 0
15 kN
10 kN 15 kN
Ya Y b
Y f Yi
Xa
Yc
Ye
Xc Xe
Xc Xe
40 kN
4m
a b c e
f i
c e
b
2
f i
hgedc
2 3 m
15 kN 10 kN 15 kN
5 kN/m
6 m
a
2 2 2 2

 Mb = 0: Ya  6 – 40  2 + Yc  2 = 0
Ya  6 – 40  2 + 7.5  2 = 0
Ya = 10.8 kN
 Py = 0: (+) Yb + Ya – Yc – 40 = 0
Yb + 10.8 – 7.5 – 40 = 0
Yb = 36.7 kN
Free body efi
 Mi = 0: Yf  7 – Ye  9 – 10  5 – 15  2 = 0
Yf  7 – 7.5  9 – 10  5 – 15  2 = 0
Yf = 21.1 kN
 Py = 0: (+) Yi + Yf – Ye – 10 –15 = 0
Yi + 21.1 – 7.5 – 10 –15 = 0
Yi = 11.4 kN
EXAMPLE 1.16
SOLUTION
Free-Body Diagram
The structure is broken into two separate free bodies, and the unknown reactions and internal
forces at the hinge are shown in figure.
Free body def
 Px = 0: (→+) Xd – 20 = 0
Xd = 20 kN
 Me = 0: Yd  5 + 40  2 – 37.5  1.67 = 0
40 kN
40 kN
20 kN
Y a
Xc
Yc
Y e
40 kN
Yd
Xd
Xd
a
c
d
d
e
37.5 kN
2m
1.67m
f
a
10 kN/m
d
f
e
b
c
4 m 4 m 5 m 2 m
40 kN 40 kN
20 kN
15 kN/m
1

Yd = – 3.5 kN
 Py = 0: (+) Ye + Yd – 37.5 – 40 = 0
Ye – 3.5 – 37.5 – 40 = 0
Ye = 81 kN
Free body acd
 Px = 0: (→+) Xc – Xd = 0
Xc – 20 = 0
Xc = 20 kN
 Mc = 0: Ya  8 + Yd  1 – 40  6 – 40  4 = 0
Ya  8 –3.5  1 – 40  6 – 40  4 = 0
Ya = 50.4 kN
 Py = 0: (+) Yc + Ya – Yd – 40 – 40 = 0
Yc + 50.4 + 3.5 – 40 – 40 = 0
Yc = 26.1 kN
EXAMPLE 1.17
Calculate the reactions for the structure shown.
SOLUTION
Free-Body Diagram
The reactions of the link members are shown in their components that are function of the force
in the link member and the inclination of the link.
2.83
2.83
2/2.83 F e
40 kN.m 20 kN
Ya
2/2.83 F e
a c
e
2/2.83 F d
2/2.83 F d
Fd
Fe
d
f
2
2
2
2
b
20 kN
3 m 2 m 2 m
2 m
2 m40 kN.m
a c
d
e
f

 Mc = 0: Ya  5 + 20  2 – 40 = 0
Ya = 0
 Me = 0: 2/2.83 Fd  4 – 40 = 0
Fd = 14.1 kN
 Px = 0: (→+) – 2/2.83 Fe + 2/2.83 Fd = 0
Fe = Fd
Fe = 14.1 kN
EXAMPLE 1.18
SOLUTION
Free body diagram
 Mf = 0: 4/4.47 Fa  10 + 40 – 60  8 = 0
Fa = 49.2 kN
 Px = 0: (→+) – Xf + 2/4.47 Fa = 0
– Xf + 2/4.47  49.2 = 0
Xf = 22 kN
 Py = 0: (+) Yf + 4/4.47 Fa – 60 – 120 = 0
Yf + 4/4.47  49.2 – 60 – 120 = 0
Yf = 136 kN
2 m 5 m 3 m 3 m
4 m
40 kN.m 60 kN 20 kN/m
a
bc d e
f
40 kN.m 60 kN
2/4.47 F a
4/4.47 F a
Xf
Yf
Fa
a
b
d
120 kN 3 m
f
2
4
4.47
4
3
5

EXAMPLE 1.19
Calculate the reactions for the frame shown.
SOLUTION
Free body diagram
 Px = 0: (→+) Xa – 30 – 10 = 0
Xa = 40 kN
 Mf = 0: Ya  10 – 20  12 – 72  7 – 30  5.5 – 10  4= 0
Ya = 94.9 kN
 Py = 0: (+) Yf + Ya – 20 – 72 = 0
Yf + 94.9 – 20 – 72 = 0
Yf = – 2.9 kN
EXAMPLE 1.20
SOLUTION
2 m 6 m 4 m
4 m
1.5 m
10 kN
20 kN 30 kN12 kN/m on Hal
a
bc
d
e
f
1.5 m
10 kN
20 kN 30 kN
Xa
Ya
Yf
fa
72 kN
3 m
1 m
4 m
15 kN 50 kN
10kN/m
a
b c
d
3 m 3 m

Free-Body Diagram
 Mc (left)= 0: Yb  6 – 15  7 – 50  3 = 0
Yb = 42.5 kN
 Mc (bottom)= 0: Xd  4 – 40  2= 0
Xd = 20 kN
 Py = 0: (+) Yd + Yb – 15 – 50 = 0
Yd + 42.5 – 15 – 50 = 0
Yd = 22.5 kN
 Px = 0: (→+) Xb + Xd – 40 = 0
Xb + 20 – 40 = 0
Xb = 20 kN
EXAMPLE 1.21
SOLUTION
Free body diagram
15 kN 50 kN
40 kN
Xb
Yb
Xd
Yd
2 m
b
c
d
 Mc (left) = 0
 Mc (bottom) = 0
2 m 4 m
1 m
4 m
20 kN
15 kN/m 18 kN/m
a
b c d
e
20 kN
60 kN
Xa
Ya
Xe
Ye
36 kN
8/3
2 m
30 kN
1 m
a
e
c
 Mc (left) = 0

 Mc (left)= 0: Ya  2 – Xa  5 – 30  1 = 0 ……………………….………(a)
 Me = 0: Ya  6 – Xa  1 + 20  4 – (15  6)  3 – 36  8/3 = 0 ……(b)
Solving (a) and (b),
Xa = 14 kN; Ya = 50 kN
 Py = 0: (+) Ye + Ya – (15  6) = 0
Ye = 40 kN
 Px = 0: (→+) – Xe + Xa + 20 – 36 = 0
Xe = – 2 kN
EXAMPLE 1.22
Calculate the reactions for the frame shown.
SOLUTION
Free body diagram
 Me (bottom)= 0: Xg  6 + 80  3 = 0
Xg = – 40 kN
 Md (right)= 0: Yg  6 – Xg  9 - 80  6 – 35  8 – 53.67  3 = 0
Yg  6 + 40  9 – 80  6 – 35  8– 53.67  3 = 0
Yg = 93.5 kN
2 m 6 m 2 m
3 m
20 kN
30 kN
35 kN
40 kN
80 kN
8 kN/m
3 m 3 m
3 m
3 m
a
b
c
d
e f
g
20 kN
30 kN
35 kN
40 kN
80 kN
Xa
Ya
Ma
Xg
Yg
a
e
g
d
53.67 kN
3 m
 Me (bottom)
= 0
 Md (right) = 0

 Ma = 0: Ma – Yg  12 – 80  3 – 35  8 +35  14 + 53.67  9+ 30  6 +
40  3 – 20  2 = 0
Mg = 129 kN.m
 Py = 0: (+) Ya + Yg – 20 – 40 –30 – 53.67 – 35 = 0
Ya = 85.2 kN
 Px = 0: (→+) Xa – Xg – 80 = 0
Xa = 40 kN
EXAMPLE 1.23
SOLUTION
Free-Body Diagram
Free Body defg
 Mo = 0: 2/3.61 Fcf  4.5 – 3/3.61 Fcf  7 + 30  3 + 40  7 – 15  4.5 = 0
Fcf = 90.9 kN
 Px = 0: (→+) – 2/3.61 Fbe + 2/3.61 Fcf – 15 = 0
– 2/3.61 Fbe + 2/3.61  90.9 – 15 = 0
Fbe = 63.9 kN
3 m 2 m 2 m 2 m
3 m
30 kN 40 kN
15 kN
8 kN/m
a b c
d e f
g
3/3.61Fbe 3/3.61Fcf
2/3.61Fcf2/3.61Fbe
e f
2/3.61Fbe
3/3.61Fbe
2/3.61Fcf
3/3.61Fcf
b c
2
3
3.61
30 kN 40 kN
15 kN
40 kN
Yd
Xa
Ya
Ma
3/3.61Fcf
2/3.61Fcf
3/3.61Fbe
2/3.61Fbe
a
d
e f
b c
2/3.61Fbe
3/3.61Fbe 3/3.61Fcf
2/3.61Fcf
g
o
4.5m
2.5 m

 Py = 0: (+) Yd + 3/3.61 Fcf – 3/3.61 Fbe – 30 – 40 = 0
Yd + 3/3.61  90.9 – 3/3.61  64 – 30 – 40 = 0
Yd = 47.5 kN
Free Body abc
 Ma = 0: Ma – 40  2.5 + 3/3.61 Fbe  5 – 3/3.61 Fcf  9 = 0
Ma – 40  2.5 + 3/3.61  64  5 – 3/3.61  90.9  9 = 0
Ma = 515 kN.m
 Px = 0: (→+) Xa + 2/3.61 Fbe – 2/3.61 Fcf = 0
Xa + 2/3.61  64 – 2/3.61  90.9 = 0
Xa = 15 kN
 Py = 0: (+) Ya – 40 + 3/3.61  Fbe – 3/3.61  Fbe = 0
Ya + 40 + 3/3.61  64 – 3/3.61  90.9 = 0
Ya = 62.5 kN
EXAMPLE 1.24
SOLUTION
Free body diagram
 Mi (right)= 0: Yk  4 – 48  2 = 0
Ya = 24 kN
 Mh = 0: Ya  9 – 15  11 – (8  9)  4.5 – 10  7 + 10  2 + 48  4
– Yk  6 = 0
Ya  9 – 15  11 – (8  9)  4.5 – 10  7 + 10  2 + 48  4
– 24  6 = 0
Ya = 54.6 kN
2 m 3 m 6 m 2 m 4 m
4 m
3 m
15 kN 10 kN
10 kN
8 kN/m
12 kN/m
a
bc
d e f
g
h
i
j
k
15 kN 10 kN
10 kN
48 kN
Xa
Ya Yk
Xh
Yh
a h
i
k
48 kN
2m
24 kN
e
1.5m
3m
 Mi (right) = 0
 Me (left) = 0

 Me (left)= 0: Xa  7 – Ya  3 + 15  5 + 24  1.5 = 0
Xa  7 – 54.6  3 + 15  5 + 24  1.5 = 0
Xa = 7.5 kN
 Px = 0: (→+) – Xh + Xa – 10 = 0
– Xh + 7.5 – 10 = 0
Xh = – 2.5 kN
 Py = 0: (+) Yh + Ya + Yk – 15 – (8  9) – 10 – 48 = 0
Yh + 54.6 + 24 – 15 – (8  9) – 10 – 48 = 0
Yh = 66.4 kN

a
b
c d e f
2 m 2 m 2 m 4 m 2 m
20 kN
30 kN.m
15 kN/m1
1 a
b c d
1.5 m 2.0 m
30 kN 40 kN
15 kN/m
e
1.5 m 1.5 m
Problems
Calculate the reactions for the following structures:
(1) (2)
(3) (4)
(5) (6)
(7) (8)
(9) (10)
10 kN/m
a c
b
7 m 3 m
20 kN
1
2
6 kN/m
a
b
5 m 2 m
c
b
f
ged
c
3 m 3 m
20 kN 30 kN
8 kN/m
6 m 5 m 2 m 2 m
a
bc
d
1 m 4 m
2 m
10 kN
8 kN/m
2 m6 m 3 m
3 m
40 kN.m
40 kN
10 kN/m
a b
c
d e f
2 m
a b
c
d
3 m 2 m 4 m
3 m
40 kN
10 kN/m
20 kN
3 m
f
e
20 kN
a
b
c d
e
2 m 4 m
2.5 m
40 kN
60 kN
20 kN/m
2.5 m
3 m 3 m 4 m
a
b c d
e
f
g
1.5 m 5 m 3 m 2 m
6 m
60 kN
80 kN
30 kN/m

(11) (12)
(13) (14)
(15) (16)
(17) (18)
a
b
c
d
e
f
g
h
3 m 4 m 3 m 4 m 2 m
4 m
30 kN.m
20 kN
20 kN
10 kN/m
a
b
c
d
e
2 m 4 m
2 m
2 m
40 kN
20 kN/m
10kN/m
a
b
c
d
e
f
g
6 m 2 m 4 m 2 m
4 m
15 kN
60 kN 30 kN/m
20 kN
a
bc
d
e f
g
h
i
1 m 8 m 2 m
2 m
2 m
2 m
30 kN
20 kN/m
10kN/m
20 kN
1 m
4 m
20 kN
10kN/m
a
b
c
d
2 m 4 m
10 kN/m
e
a
b
c
d
e
f
g
3 m 2 m 4 m 2 m
4 m
60 kN 30 kN/m
20 kN
3 m
2 m 6 m 4 m
4 m
20 kN 30 kN20 kN/m on Hal
a
bc
d
e
f
1.5 m
1.5 m10 kN
a
b c
d
e
f
3 m 3 m 2 m
4 m
25 kN
20 kN/m
3 m

(19) (20)
2 m 4 m 3 m
2 m
3 m
3 m
20 kN 20 kN
90 kN 20 kN
/m
a
b
cd
e
f g
h
i
1.5 1.5 a b
c d
e
f g
h
i
j
3 m 3 m
1.5 m
2 m
2 m
20 kN
10 kN
40 kN
20 kN/m
1.5 m

2
Statically Determinate Trsses
2.1 Introduction
Trusses consist of straight members joined together at their ends by joints assumed
incapable of resisting any moment, that is, frictionless pins. However, the joints of actual
trusses used in construction do possess a certain amount of rigidity leading to a certain amount
of restraint rotation. The assumption of joints behaving as frictionless pins greatly simplifies
the analysis of a truss. Furthermore, extensive theoretical and experimental investigations
indicate that the error introduced by this assumption is insignificant when loaded at their joints.
To avoid eccentricity at joints, the centroidal axis of each member is straight and coincides
with the line connecting the joint centers at all ends of the members. Simple analysis of an
idealized truss results in purely axial forces in the members of the truss that is either tensile or
compressive.
2.2 TRUSSES CONFIGURATION
The simplest form of a stable planar truss is a triangle arrangement of three members as
shown in Fig. 2.1a. Each additional joint can be obtained by adding two bars to the joints
already formed. Such a truss is called a simple truss. These trusses are always composed of
triangular shapes as shown in Figs. 2.1a and 2.1b.
(a) (b) (c)
Fig. 2.1 Examples of simple trusses
A compound truss consists of two or more simple trusses joined so that relative motion of
the components is constrained. Three conditions of connectivity are required to prevent
relative translation and rotation. Two simple plane trusses are combined together by three
nonparallel nonconcurrent bars to form a compound truss as shown in Fig. 2.2a. Since two
a b
c
a b
c
d
e
a b
c
d
e
f

32 Structural analysis I, by Prof. Dr. Ahmed Zubydan.
bars can be replaced by a common hinge, a compound truss can also be formed by joining
together two plane trusses with a common hinge and bar, as shown in Figs. 2.2b and 2.2c.
Trusses that cannot be classified as either simple or compound are called complex trusses as
shown in Fig.2.2d.
(a) (b)
(c) (d)
Fig. 2.2 Compound and complex trusses
2.3 CONDITION EQUATIONS
As discussed in Chapter 1, three independent reaction components can be determined for
a planar system through the systematic application of the equations of static equilibrium.
Additional reaction components can be determined if there are internal conditions that permit
the expression of special condition equations. In the case of truss structures, the conditions
might take a number of forms.
The structure of Fig. 2.3a is a statically determinate stable truss, and the three reaction
components can be determined from statics. If an additional part dc is added to truss ab, then
an additional condition is required to evaluate the additional reaction component at c. This
condition, which is Md = 0 for the left or right portion of the truss, is provided from the
connected hinge at d.
If an additional reaction component is added to point d, as shown in Fig. 2.3c, the
intermediate hinge at e provides an additional condition to allow the determination of the
additional reaction component.
The absence of diagonal member in a truss panel, under proper support conditions,
introduces a condition as shown in Fig. 2.3d. In this case, a section through panel bc reveals
that no vertical force component can be transferred through this panel. This condition, coupled
with the overall equations of equilibrium, enables one to determine the reaction components.
a
b c
d
e f
Hinge
b c
a

Statically determinate trusses 33
(a) (b)
(c) (d)
Fig. 2.3. Condition equations for trusses
2.4 STABILITY AND DETERMINANCY
Based on the forgoing discussion, criteria can be developed concerning the statical
determinancy of truss structures. First, the truss can be checked for external determinancy and
stability according to the criteria developed in Chapter 1. In which, if r is taken as the actual
number of reaction components and c is the number of condition equations, then the following
criteria can be:
r < 3 + c; truss is unstable externally
r = 3 + c; truss is statically determinate externally
r > 3 + c; truss is statically indeterminate externally
These conditions are necessary but not sufficient conditions for statical classification. The
condition equations should be properly added to enable the determination of the additional
reaction components.
For internal classification, in addition to the above definition for r and c, let m be the total
number of bars and j the total number of joints. Then, the following conditions prevail:
m + c + 3 < 2j; truss is unstable internally
m + c + 3 = 2j; truss is statically determinate internally
m + c + 3 > 2j; truss is statically indeterminate internally
where c + 3 is the minimum required components of reactions. Care must be exercised in using
the above conditions. If m + c + 3 < 2j, the truss is definitely unstable, but if m + c + 3  2j, it
doesn’t necessarily follow that the truss is stable. It is possible that m bars are not properly
arranged to ensure internal stability. Such trusses are said to have critical form.
a b a b c
d
a
b c
d
e
a c
b e

EXAMPLE 2.1
Determine the conditions of stability and determinancy of the trusses shown.
Truss 1 Truss 2
Truss 3 Truss 4
SOLUTION
Truss 1
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 7 + 0 +3 = 10
2j = 10
m + c + 3 = 2j internally stable and statically determinate
Trusses 2, 3, and 4
3 + c = 3 + 1 = 4
r = 4
3 + c = r
m + c + 3 = 6 + 1 +3 = 10
2j = 10
m + c + 3 = 2j
Although the stability conditions are theoretically satisfied for Trusses 2, 3, and 4, Truss 2 is
geometrically unstable because it has a critical form. In this form, point b can move
perpendicular to line ac without any resistance. Also, the intermediate hinge at b couldn’t be
used to determine the additional horizontal reaction at c. On the other hand Trusses 3 and 4
are stable and statically determinate.
EXAMPLE 2.2
a
b
c
d e
a
b
c
d e
a
b
c
d e
a b c
d e
a
b
c
d
e
f
a
b
c
d
e
f

Truss 1 Truss 2
Truss 3 Truss 4
SOLUTION
Truss 1
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 9 + 0 +3 = 12
2j = 12
Trusses 2, 3, and 4
3 + c = 3 + 1 = 4
r = 4
3 + c = r
m + c + 3 = 8 + 1 +3 =
12
2j = 12
m + c + 3 = 2j
Studying the condition equations of trusses 2, 3, and 4 leads to the following conclusion:
Condition equation
of Truss 2
Condition equation
of Truss 3
Condition equation
of Truss 4
The additional horizontal reaction at support a can only be determined from the condition
equation of Truss 3, so, Truss 3 is stable and statically determinate while Trusses 2 and 4 are
unstable.
a
b
c
d
e
f
a
b
c
d
e
f
a
b
c
 Fy = 0
a
b
co
 Mo = 0
a
b
c
o
 Mo = 0

EXAMPLE 2.3
Truss 1
Truss 2
Truss 3 Truss 4
SOLUTION
Truss 1
3 + c = 3 + 1 = 4
r = 3
3 + c > r externally unstable
Truss 2
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 13 + 0 +3 = 16
2j = 16
Truss 3
3 + c = 3 + 1 = 4
r = 4
m + c + 3 = 12 + 1 +3 = 16
2j = 16
Truss 4
3 + c = 3 + 2 = 5
r = 5
m + c + 3 = 11 + 2 +3 = 16
2j = 16
a
b
c
d
e
f
g
h
i
 Me= 0
a
b
c
d
e
f
g
h
i
a
b
c
d
e
f
g
h
i
 Me= 0
a
b
c
d
f
g
h
i
 Mc= 0,  Mg= 0

EXAMPLE 2.4
Truss 1 Truss 2
Truss 3 Truss 4
SOLUTION
Truss 1
3 + c = 3 + 1 = 4
r = 3
3 + c > r geometrically unstable
Truss 2
3 + c = 3 + 1 = 4
r = 4
m + c + 3 = 9 + 1 + 3 = 13
2j = 12
m + c + 3 > 2j internally stable and statically indeterminate to the first
degree
Trusses 3 and 4
3 + c = 3 + 0 = 3
r = 3
3 + c = r externally stable and determinate
m + c + 3 = 9 + 0 + 3 = 12
2j = 12
m + c + 3 = 2j internally stable and determinate
EXAMPLE 2.5
Truss 1 Truss 2
a b
c
d
e f
 Fx = 0
a b
c
d
e
f
 Fx = 0
a b
c
d
e f
a b
c
d
e f
a
b
c
d
e
 Mc = 0
a
b
c
d
e
 Md
= 0

Truss 3 Truss 4
SOLUTION
Trusses 1 and 2
3 + c = 3 + 1 = 4
r = 4
3 + c = r
m + c + 3 = 7 + 1 + 3 = 11
2j = 10
m + c + 3 > 2j internally statically indeterminate to the first degree
Although m, c, r, and j are the same for the two trusses, Truss 1 is unstable because its
geometry produces a critical form. In this case, small movement of node c can occur
perpendicular to the line ae without any resistance and also the condition equation at c couldn’t
be used to determine the additional horizontal reaction at e. On the other hand, Truss 2 is stable
and internally indeterminate to the first degree.
Truss 3
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 7 + 0 + 3 = 10
2j = 10
Truss 4
3 + c = 3 + 1 = 4
r = 4
m + c + 3 = 6 + 1 + 3 = 10
2j = 10
EXAMPLE 2.6
Truss 1 Truss 2
a
b
c
d
e a
b
c
d
e Mb = 0

Truss 3
Truss 4
SOLUTION
Truss 1
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 5 + 0 + 3 = 8
2j = 8
Truss 2
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 9 + 0 + 3 = 12
2j = 12
Truss 3
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 12 + 0 + 3 = 15
2j = 16
m + c + 3 < 2j internally unstable
Truss 4
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 9 + 0 + 3 = 8
2j = 12

EXAMPLE 2.7
Truss 1 Truss 2
Truss 3 Truss 4
SOLUTION
Truss 1
3 + c = 3 + 1 = 4
r = 4
m + c + 3 = 14 + 1 + 3 = 18
2j = 18
Truss 2
3 + c = 3 + 1 = 4
r = 4
m + c + 3 = 8 + 1 + 3 = 12
2j = 12
Truss 3
3 + c = 3 + 0 = 3
r = 4
3 + c < r externally stable and statically indeterminate to the
first degree
m + c + 3 = 16 + 0 + 3 = 19
2j = 18
m + c + 3 > 2j internally stable and statically indeterminate to the
first degree
Truss 4
3 + c = 3 + 0 = 3
r = 4
a
 Ma = 0 a
b
c
d e
f
 Mb = 0 determinate

3 + c < r externally stable and statically indeterminate to
the first degree
m + c + 3 = 10 + 0 + 3 = 13
2j = 12
m + c + 3 > 2j internally stable and statically indeterminate to
the first degree
EXAMPLE 2.8
Truss 1 Truss 2
Truss 3 Truss 4
SOLUTION
Truss 1
3 + c = 3 + 1 = 4
r = 4
m + c + 3 = 12 + 1 + 3 = 16
2j = 16
Truss 2
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 12 + 0 + 3 = 15
2j = 16
Truss 3
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 26 + 0 + 3 = 29
2j = 28
first degree
a Ma = 0

Truss 4
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 17 + 0 + 3 = 20
2j = 20
EXAMPLE 2.9
Truss 1 Truss 2
Truss 3 Truss 4
SOLUTION
Truss 1
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 15 + 0 + 3 = 18
2j = 20
Truss 2
3 + c = 3 + 1 = 4
r = 4
3 + c = r
m + c + 3 = 14 + 1 + 3 = 18
2j = 18
m + c + 3 = 2j internally stable and determinate
It is noticed that the condition equation  Fy = 0 at the shown vertical section is useful for
determining the additional reaction component at c. On the other hand, the static equations (
 Fy = 0
a
b c
o
 Mo = 0

Fx = 0,  Fy = 0,  M = 0) couldn’t be used to determine the vertical reaction at a or b because
the line of action of vertical reaction at b passes through support a. For this reason, the truss is
unstable.
Truss 3
3 + c = 3 + 0 = 3
r = 4
first degree
m + c + 3 = 13 + 0 + 3 = 16
2j = 16
Truss 4
3 + c = 3 + 1 = 4
r = 4
m + c + 3 = 11 + 1 + 3 = 15
2j = 16
EXAMPLE 2.10
Truss 1
Truss 2
Truss 3 Truss 4
SOLUTION
Truss 1
3 + c = 3 + 2 = 5
r = 5
m + c + 3 = 23 + 2 + 3 = 28
2j = 28
a
b
c
d
e
 Md = 0,  Me = 0
a
 Ma = 0

It should be noticed that the reaction components at b could be analyzed as internal forces in
members ba and bc by the method of joint. This is necessary to activate the first condition
equation ( Md (left) = 0).
Truss 2
3 + c = 3 + 0 = 3
r = 4
first degree
m + c + 3 = 21 + 0 + 3 = 24
2j = 24
Truss 3
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 11 + 0 + 3 = 14
2j = 14
Truss 4
3 + c = 3 + 1 = 4
r = 4
m + c + 3 = 10 + 1 + 3 = 14
2j = 14
EXAMPLE 2.11
Truss 1
Truss 2
Truss 3
Truss 4
SOLUTION
a

Truss 1
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 19 + 0 + 3 = 22
2j = 20
second degree.
Truss 2
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 9 + 0 + 3 = 12
2j = 14
m + c + 3 < 2j unstable
Truss 3
3 + c = 3 + 0 = 3
r = 3
m + c + 3 = 19 + 0 + 3 = 22
2j = 22
m + c + 3 = 2j internally stable and statically determinate.
Truss 4
In this truss, it is noticed that the lines of action of all reactions intersect at point a. So, the
truss is geometrically unstable.
2.5 METHODS OF ANALYSIS OF SIMPLE
AND COMPOUND TRUSSES
Three methods of analysis that are most common for simple and compound trusses.
These are the method of joints, method of sections, and graphical method. The method of
joints and the graphical method are useful when the forces in all or in the majority of the truss
members are required, whereas the method of sections is more convenient if the forces in few
members of the truss are required. Sometimes a combination of the method of section with
the other two methods is useful, especially in the analysis of compound trusses. In some
trusses, the use of the method of section is required for determining some member forces
before using the other two methods.
2.6 METHOD OF JOINTS
The method of joints is frequently a simple way of analyzing a statically determinate truss.
In common cases, the unknown reactions can be first determined by taking the entire truss as a

free body. Next, the joints containing only two unknown member forces are analyzed using
their free body diagrams. The two equations of static equilibrium involving summation of
forces in two orthogonal directions are applied to determine the two unknown member forces
at the joint. The procedure continues until all internal axial forces are computed. Since, the
total number of joint equations of equilibrium is more than the number of unknown member
forces, there will be surplus equations for checking purposes after member axial forces have
been obtained.
A number of frequently occurring special conditions can be used to shorten the calculation.
These conditions are summarized in the following rules:
Rule 1
With reference to Joint a, it should be noticed that members ab and ac are in
the same direction. Since the joint is not loaded, it can be concluded that the
force in the third member ad is zero (Fad = 0) and Fab = Fac. This fact can be
verified by constructing the free-body diagram for joint a and summing the
forces first in the vertical direction and then in the horizontal direction.
Rule 2
With reference to the unloaded Joint e which has only two members, it can be
concluded that the forces in the two members are equal to zero (Fef = 0, Feg =
0 ). This fact can be verified by summing the forces in the vertical and
horizontal directions, respectively.
Rule 3
For Joint k, the forces in opposite members must be equal; that is, Fkl = Fkm
and Fkn = Fko. This can be demonstrated by summing the forces first in the
vertical direction and then in the horizontal direction
Example 2.12
Calculate the forces in all members of the shown truss using the method of
joints.
SOLUTION
From symetry: Ya = Yb = 40/2 = 20 kN
 Fx = 0: Xa = 0
a
Fab
Fad
Fac
e
Fef
Feg
k
Fkl
Fkn
Fkm
Fko
1.5 m 1.5 m
4 m
2 m
40 kN
a b
dc
e
1.5
2
2.5
3
4
5
40 kN
Ya=20 kN Yb=20 kN
a
b
c d
e
Xa =0

Joint e
 Fx = 0 (→ +):
 Fy = 0 (+):
1.5/2.5 Fde – 1.5/2.5 Fce = 0 ………….(a)
2/2.5 Fde + 2/2.5 Fce + 40 = 0 ………..(b)
Fce = – 25kN , Fde = – 25 kN
Joint c
 Fx = 0 (→ +):
 Fy = 0 (+):
From symmetry:
3/5 Fbc – 1.5/2.5  25 = 0
Fbc = 25 kN
Fac + 4/5 Fbc + 2/2.5  25 = 0
Fac + 4/5 25 + 2/2.5 25 = 0
Fac = – 40 kN
Fad = Fbc = 25 kN
Fbd = Fac = – 40 kN
Joint a
 Fx = 0 (→ +):
 Fy (+):
Fab + 3/5 25 = 0
Fab = – 15 kN
40 – 4/5 25 – 20 = 0 ……………….(Check)
Member forces
40 kN
e
Fce
Fde
c
Fbc
Fac
Fce = 25
a
Fac
= 40
20 kN
Fab
Fad = 25
40 kN
-15
-40
2525
-40
-25
-25
20 kN 20 kN

Example 2.13
Calculate the forces in all members of the shown truss using
the method of joints.
SOLUTION
 Mb =0:
 Fx (→+):
 Fy (+):
Xa  9 – 40 6 –20 12 = 0
Xa = 53.3 kN
– Xb + Xa = 0
– Xb + 53.3 = 0
Xb = 53.3 kN
Yb – 40 – 20 = 0
Yb = 60 kN
Joint e
 Fy = 0 (+):
 Fx = 0 (→ +):
4.5/7.5 Fde – 20 = 0
Fde = 33.3 kN
– Fce - 6/7.5 Fde = 0
– Fce – 6/7.5  33.3 = 0
Fce = – 26.7 kN
Joint c
 Fy = 0 (+):
 Fx = 0 (→ +):
Fcd – 40 = 0
Fcd = 40 kN
– Fac – 26.7 = 0
Fac = – 26.7 kN
Joint d
 Fy = 0 (+):
 Fx = 0 (→ +):
4.5/7.5 Fbd – 4.5/7.5 Fad – 40
– 4.5/7.5  33.3 = 0 ……………………(a)
– 6/7.5 Fbd – 6/7.5 Fad
+ 6/7.5  33.3 = 0 ………………..….…(b)
Fbd = 66.7 kN ; Fad = – 33.3 kN
6 m 6 m
4.5 m
4.5 m
40 kN 20 kN
a
b
d
c
e
6
4.57.5
6
7.54.5
40 kN 20 kNXa=53.3 kN
Xb =
53.3 kN
Yb=60 kN
a
c
e
d
b
20 kN
Fce
Fde
e
40 kN
c
26.7Fac
Fcd
d
33.3
40
Fad
Fbd

Joint b
 Fy = 0 (+):
 Fx = 0 (→ +):
– Fab – 4.5/7.5  66.7 + 60 = 0
Fab = 20 kN
6/7.5  66.7 – 53.3 = 0 …………(Check)
Joint a
 Fy = 0 (+):
 Fx = 0 (→ +):
20 – 4.5/7.5  33.3 = 0 ……………(Check)
53.3 – 26.7 - 6/7.5  33.3 = 0 ……..(Check)
Member forces
EXAMPLE 2.14
Calculate the forces in all members of the shown truss using the
method of joints.
SOLUTION
 Mb =0:
 Fy (+):
 Fx (→+):
Ya  6 – 50 3 = 0
Ya = 25 kN
Yb – Ya – 50 = 0
Yb – 25 – 50 = 0
Yb = 75 kN
Xa = 0
53.3 kN
60 kN
66.7
Fab
b
53.3 kN
33.3
a
20
26.7
40 kN 20 kN
-26.7
-33.3
20
-26.7
40
33.3
66.7
53.3 kN
53.3 kN
60 kN
1.5 1.5 3 m 3 m
3 m
3 m
3 m
f g
h
b
d
c
e
50 kN
a
3
3
4.24
1.5
3
3.35
3
4.5
5.41
50 kN
Ya
=25 kN Yb= 75 kN
Xa
=0
a b
c d
e
f
g
h

Joint g
 Fy = 0 (+):
 Fx = 0 (→ +):
3/4.24 Fgh – 50 = 0
Fgh = 70.7 kN
– Ffg – 3/4.24 Fgh = 0
– Ffg – 3/4.24  70.7 = 0
Ffg = – 50 kN
Joint h
 Fx = 0 (→ +):
 Fy = 0 (+):
– 3/4.24 Feh + 3/4.24  70.7 = 0
Feh = 70.7 kN
– Ffh – 3/4.24 Feh – 3/4.24  70.7 = 0
– Ffh – 3/4.24  70.7 – 3/4.24  70.7 = 0
Ffh = – 100 kN
Joint f
 Fx = 0 (→ +):
 Fy = 0 (+):
– Fef – 50 = 0
Fef = – 50 kN
– Fdf – 100 = 0
Fdf = – 100 kN
Joint e
 Fx = 0 (→ +):
 Fy = 0 (+):
– 1.5/3.35 Fce + 3/4.24 Fde + 3/4.2470.7
– 50 = 0 ……………………………………..(a)
– 3/3.35 Fce – 3/4.24 Fde
+ 3/4.24  70.7 = 0 ……………………..….(b)
Solving (a), (b),
Fce = 37.3 kN ; Fde = 23.6 kN
Joint d
 Fx = 0 (→ +):
 Fy = 0 (+):
– Fcd – 3/4.243  23.6 = 0
Fcd = – 16.7 kN
– Fbd – 100 + 3/4.24  23.6 = 0
Fbd = – 83.3 kN
Joint c
50 kN
gFfg
Fhg
h
70.7
Ffh
Feh
f
100
50
Fef
Fdf
e
70.7
50
Fde
Fce
d
10023.6
Fbd
Fcd

 Fx = 0 (→ +):
 Fy = 0 (+):
4.5/5.41 Fbc – 1.5/3.35 Fac + 1.5/3.35  37.3
– 16.7 = 0 …………………………………..(a)
– 3/5.41 Fbc – 3/3.35 Fac + 3/3.35  37.3 = 0
……………………………..…………..……(b)
Fbc = 15 kN ; Fac = 28 kN
Joint b
 Fx = 0 (→ +):
 Fy = 0 (+):
– Fab – 4.5/5.41  15 = 0
Fab = –12.5 kN
75 +3/5.41  15 – 83.3 = 0 ……….(Check)
Joint a
 Fx = 0 (→ +):
 Fy = 0 (+):
1.5/3.35  28 – 12.5 = 0 ………….(Check)
3/3.35  28 – 25 = 0 ………………(Check)
Member forces
Example 2.15
SOLUTION
c
37.3
16.7
FbcFac
b
83.3
75 kN
15
Fab
a 12.5
28
25 kN
50 kN
-12.5
28
15
-83.3
-16.7
37.3
23.6
-100
-50
70.7
-50
-100
70.7
25 kN 75 kN
3 m 4 m 3 m
3 m
3 m
10 kN
10 kN
10 kN
10 kN 10 kN
f
a b
d
c
e
6
4
7.21
3
3
4.24

 Mb :
 Fy (+):
 Fx (→+):
Ya  4 + 10  (3 – 7 + 6 – 4)= 0
Ya = 5 kN
Yb + Ya – 10 – 10 – 10 = 0
Yb + 5 – 10 – 10 –10 = 0
Yb = 25 kN
– Xa + 10 +10 = 0
Xa = 20 kN
Joint c
 Fx = 0 (→ +):
 Fy = 0 (+):
3/4.24 Fcd + 3/4.24 Fac + 10 = 0 ……………(a)
3/4.24 Fcd – 3/4.24 Fac – 10 = 0 …………….(b)
Fcd = 0 ; Fac = – 14.1 kN
Joint f
According to rule 2; Fef = Fbf = 0
Joint e
 Fx = 0 (→ +):
 Fy = 0 (+):
Fde = 0
– Fbe – 10 = 0
Fbe = – 10 kN
Joint d
 Fx = 0 (→ +):
 Fy = 0 (+):
4/7.21 Fbd + 10 = 0
Fbd = – 18 kN
– Fad – 6/7.21 Fbd – 10 = 0
Fad = 5 kN
Joint b
 Fx = 0 (→ +):
 Fy = 0 (+):
– Fab + 4/7.21  18 = 0
Fab = – 10 kN
25 – 10 – 6/7.21  18 = 0 ……….(Check)
10 kN
10 kN
10 kN
10 kN 10 kN
Xa =20 kN
Ya=5 kN Yb=25 kN
c
a b
f
e
d
10 kN c
10 kN
Fcd
Fac
Fef
f
Fbf
e
10 kN
0
Fbe
Fde
d
10 kN
10 kN 0
0 Fbd
Fad
b
10
018
Fab
25 kN

Joint a
 Fx = 0 (→ +):
 Fy = 0 (+):
10 + 3/4.24  14.1 – 20 = 0 ………(Check)
5 + 5 – 3/4.24  14.1 = 0 …………(Check)
Member forces
EXAMPLE 2.16
Calculate the forces in all members of the shown
truss using the method of joints.
SOLUTION
From symmetry:
 Me (left) = 0:
 Fx = 0:
Ya = Yi = 100 kN
Xa  14 + 100  10 – Ya  20 = 0
Xa = 71.4 kN
Xi = Xa = 71.4 kN
Joint a
 Fx = 0 (→ +):
 Fy = 0 (+):
10/11.18 Fab + 71.4 = 0
Fab = – 79.9 kN
Fac + 5/11.18 Fab + 100 = 0
Fac – 5/11.18  79.9 + 100 = 0
Fac = – 64.3 kN
a
20 kN
514.1
5 kN
10
10 kN
10 kN
10 kN
10 kN 10 kN
10
-14.1
5
-18
-10
20 kN
5 kN 25 kN
0
0
0
0
10.2
10
13.45
11.1 8
4 at 10 m = 40m
5 m
5 m
2 m
2 m
100 kN 100 kN
f
g
h
ia
b
d
c
e
5
10
9
10
2
5
10
11.1 8
100 kN 100 kN
Xa =71.4
Ya=100 kN
Xi=71.4
Yi=100 kN
a
b
c
d
e
g
h
f
i
a
100 kN
71.4 kN
Fac
Fab

Joint c
 Fx = 0 (→ +):
 Fy = 0 (+):
10/11.18 Fbc + 10/10.2 Fcd = 0 ……….(a)
– 5/11.18 Fbc + 2/10.2 Fcd + 64.3 = 0 .(b)
Fbc = 102.7 kN , Fcd = – 93.7 kN
Joint d
According to Rule 3: Fbd = – 100 kN , Fde = – 93.7 kN
Joint b
 Fx = 0 (→ +):
 Fy (+):
10/13.45 Fbe + 10/11.18  79.9
– 10/11.18  102.7 = 0
Fbe = 27.5 kN
9/13.45 Fbe + 5/11.18  79.9 + 5/11.18  102.7
– 100
= 9/13.45  27.5 + 5/11.18  79.9
+ 5/11.18  102.7 – 100 = 0 ………….(Check)
Member forces
EXAMPLE 2.17
Calculate the forces in all members of the shown truss using the
method of joints.
SOLUTION
c
Fcd
Fbc
64.3
d
Fde
Fbd
100 kN
93.7
b
100
79.9
102.7
Fbe
100 kN 100 kN
-79. 9
-64.3
-7 9.9
-64.3
1 02.7
-100
27.5
102.7
-100
27.5
-93.7 -93.7
-93.7 -93.7
71.4 kN
100 kN
71.4 kN
100 kN
4 11.31
8.94
4 m 4 m 4 m 4 m
8 m
8 m
100 kN
f
g
h
a
b
c
ed
8 8
8

From Symmetry:
 Fx (→+):
Ya = Yc = 100/2 = 50 kN
Xa = 0
Joint f
According to Rule 2: Ffg = Fdf = 0
Joint g
 Fx = 0 (→ +):
 Fy = 0 (+):
– 8/11.31 Fdg + 8/11.31 Feg = 0 …………..(a)
– 8/11.31 Fdg – 8/11.31 Feg – 100 = 0 …….(b)
Fdg = Feg = – 70.7 kN
Joint a
 Fy = 0 (+):
 Fx = 0 (→ +):
8/8.94 Fad + 50 = 0
Fad = – 55.9 kN
Fab – 4/8.94 Fad = 0
Fab + 4/8.94  55.9 = 0
Fab = – 25 kN
Joint b
According to rule 1: Fbd = Fbe = 0

Joint d
 Fx = 0 (→ +):
 Fy (+):
Fde – 4/8.94  55.9 – 8/11.31  70.7 = 0
Fde = 75 kN
8/8.94  55.9 – 8/11.31  70.7 = 0 …(Check)
100 kN
Ya=50 kN Yc
=50 kN
a c
b
d e
f
g
h
Xa
=0
Fdf
Ffg
f
100 kN
g
FegFdg
0 0
Fad
50 kN
a Fab
Fbd Fbe
b
25 25
Fbd
Fbe
b
d
0 70.7
Fde
0
55.9

Member forces
EXAMPLE 2.18
SOLUTION
 Mg =0:
 Md (left) = 0:
 Fy =0 (+):
 Fx =0 (→+):
Ya8 – Xa4 + 406 – 604 + 208
+ 403 + 104 = 0 ……………...(a)
Ya8 + Xa4 – 402 –604 = 0 ..(b)
Ya = 0 ; Xa = 80 kN
Yg + Ya – 60 – 40 = 0
Yg = 100 kN
Xg – Xa + 40 +20 + 10 = 0
Xg = 10 kN
100 kN
-25
-55.9
-25
-55.9
75
-70.7
-70.7
50 kN 50 kN
0 0
00
0 0
3
4
c 4.47
5.66
3
4 m 4 m 3 m
4 m
2 m
2 m
60 kN
40 kN
20 kN
10 kN
40 kN
f
g
a
b
d
e
4
5
4
5
4
2
4
60 kN
40 kN
20 kN
10 kN
40 kN
Xa =80
Xg
=10 kN
Ya=0
a
b
c
d
e f
g
Yg
=100 kN

Joint a
According to Rule 1: Fab = 80 kN , Fac = 0
Joint b
 Fx =0 (→+):
 Fy = 0 (+):
4/5.66 Fbd – 80 = 0
Fbd = 113.1 kN
Fbc + 4/5.66 Fbd – 60 = 0
Fbc = – 20 kN
Joint c
 Fx =0 (→+):
 Fy (+):
4/4.47 Fcd + 40 = 0
Fcd = – 44.7 kN
2/4.47 Fcd + 20
= – 2/4.47  44.7 + 20 = 0 ………. (Check)
Joint g
 Fx =0 (→+):
 Fy =0 (+):
3/5 Ffg + 10 = 0
Ffg = – 16.7 kN
Feg + 4/5 Ffg + 100 = 0
Feg – 4/5  16.7 + 100 = 0
Feg = – 86.7 kN
Joint e
According to rule 3: Fde = – 86.7 kN , Fef = – 10 kN
Joint f
 Fx =0 (→+):
 Fy (+):
– 3/5 Fdf + 10 + 3/5  16.7 = 0
Fdf = 33.3 kN
– 4/5 Fdf – 40 + 4/5  16.7
= – 4/5  33.3 – 40 + 4/5  16.7 = 0 …..
(Check)
Joint d
40 kN
c
0
20
Fcd
80 kN Fab
Fac
a
60 kN
b
80
Fbd
Fbc
10 kN
100 kN
g
Ffg
Feg
10 kN
e
Fde
Fef
86.7
40 kN
f
Fdf
10
16.7

 Fx (→+):
 Fy (+):
20 + 3/5  33.3 + 4/4.72  44.7
– 4/5.66  113.1 = 0 …………… (Check)
86.7 – 4/5  33.3 + 2/4.72  44.7
– 4/5.66  113.1 = 0 …………… (Check)
Member forces
2.7 MEHOD OF SECTIONS
When the method of joints leads to a situation where no subsequent joint contains only
two unknown member forces, the next approach would be to use a group of joints (called a
section of the truss) as a free body. The method of sections may be used in lieu of the method
of joints, even when the method of joints is possible. The three equations of statics ( Fx =0,
 Fy =0, and  M =0) can be then used to calculate the unknown member forces.
Example 2.19
Determine the forces in the marked members of the shown
truss using the method of sections.
SOLUTION
From symmetry:
 Fx = 0 (→+):
Ya = Yh = 20  3 /2 = 30 kN
Xa = 0
Section s1-s1
 Me = 0
 Mb = 0
 Fy (+):
Fbd  3 + 20  4 – 30  8 = 0; Fbd = 53.3 kN
Fce  3 + 30  4 = 0; Fce = -40 kN
3/5 Fbe + 30 – 20 = 0; Fbe = – 16.7 kN
20 kN
44.7
113.1
86.7
33.3
d
60 kN
40 kN
20 kN
10 kN
40 kN
80
-20
113.1
-44.7
-86.7
33.3
-10
-86.7
-16.7
80 kN
10 kN
100 kN
0
4 at 4 m=16m
3 m
20 kN 20 kN20 kN
f
g
ha b d
c e
4
35
20 kN 20 kN 20 kNYa=30 Yh=30
Xa=0
a b
c
d
e
f
g
h
s 1
s1
s2
s 2
20 kN30 kN
a
b
c e
Fbd
Fbe
Fce

Section s2-s2
 Mb = 0
 Ma = 0
3/5 Fac  4 + 30  4 = 0
Fac = – 50 kN
Fbc  4 – 20  4 –3/5  16.7  4 = 0
Fbc = 30 kN
Member forces
Example 2.20
Determine the forces in the marked members of the shown truss
using the method of sections.
SOLUTION
 Mc =0:
 Fy (+):
 Fx =0 (→+):
Ya  4 – 10  1.5 – 10  3
– 10  4.5 = 0
Ya = 22.5 kN
Yc – Ya = 0
Yc = 22.5 kN
– Xa + 10 + 10 + 10 = 0
Xa = 30 kN
Section s1-s1
 Mf = 0
 Md = 0
Fdg  4 + 30  1.5 – 22.5  4 = 0
Fdg = 11.3 kN
Ffi  4 + 22.5  4 – 30  1.5 = 0
Ffi = – 11.3 kN
20 kN30 kN
a
b 53.3
16.7Fac
Fbc
20 kN 20 kN 20 kN
40
-50
53.3
30
-16.7
53.3
20
40
-16.7
30
-50
-40 -40
30 kN 30 kN
2 m 2 m
1.5 m
1.5 m
1.5 m
10 kN
10 kN
10 kN
f
g h
i
a
b
d
c
e
j k
2
1.5
2.5
10 kN
10 kN
10 kN
Xa =30 kN
Ya =22.5 kN Yc =22.5 kN
a b
c
d e
f
g h i
j
k
s1 s 1
s2s2
10 kN
30 kN
22.5 kN 22.5 kN
a b
c
d
f
Fdg
Fde Fef
Ffi

Section s2-s2
 Mi = 0
 Fx (→+):
1.5/2.5 Feg  2 + 2/2.5 Feg  1.5+ 11.3  4
+ 30  3 – 10  1.5 – 22.5  4 = 0
Feg = – 12.5 kN
2/2.5 Fei – 2/2.5 Feg + 10 – 30 = 0
Fei = 12.5 kN
Member forces
Example 2.21
SOLUTION
 Mg =0:
 Fy (+):
 Fx =0 (→+):
Ya18 – 4015 – 5012 – 806 = 0
Ya = 93.3 kN
Yg + Ya – 40 – 50 - 80 = 0
Yg = 76.7 kN
Xa = 0
Section s1-s1
10 kN
30 kN
22.5 kN 22.5 kN
a b
c
d
f
11.3
Feg Fei
11.3
e
i
10 kN
10 kN
10 kN
30
22.5
-18.8
18.8
-22.5
5
11.3
-15
-12.5 12.5
-11.3
3.8
-10
-6.2 6.2
-3.8
-5
30 kN
22.5 kN 22.5 kN
0
-5
3 m 3 m 3 m 3 m 3 m 3 m
1.5 m
3 m
50 kN 80 kN
40 kN
f
ga
b
d
c e
3
3
4.24
3
4.5 5.41
1.5
6
6.18
50 kN
40 kN
93.3 kN
a c
b
Fce
Fcd
Fbd
o
9 m
d
50 kN 80 kN
40 kN
Ya =93.3 kN Yg=76.7
Xa =0
a
b
c
d
e
f
g
s1
s1
s2
s2

 Md =0:
 Mc =0:
 Mo =0:
Fce  4.5 + 93.3  9 – 40  6 – 50  3 = 0
Fce = – 100 kN
6/6.18 Fbd  3 + 1.5/6.18 Fbd  3 + 40  3 – 93.3  6 = 0
Fbd = 120.9 kN
4.5/5.41 Fcd  15 + 50  15 + 40  12 – 93.3  9= 0
Fcd = – 31.2 kN
Section s2-s2
 Mb =0:
 Fy (+):
Fac  3 + 93.3  3 = 0
Fac = – 93.3 kN
3/4.24 Fbc – 1.5/6.18  120.9 + 93.3 – 40 = 0
Fbc = – 33.9 kN
Member forces
Example 2.22
SOLUTION
 Mk =0:
 Fy =0 (+):
 Mf (right)=0:
 Fx =0 (→+):
Ya  40 – 20  10 – 10  20 = 0
Ya = 10 kN
Yk – Ya = 0
Yk = 10 kN
Xk  20 – Yk  20 = 0
Xk = 10 kN
– Xa – Xk + 10+ 20 = 0
Xa = 20 kN
40 kN
93.3 kN
a
b
120.9
Fac
Fbc
50 kN 80 kN
40 kN
-93.3
132
-100
-33.9
-31.2
-76.7
-65.1
-40.9
108.4
120.9 126.4
93.3 kN 76.7 kN
14
.14
14.14
10 m 10 m 10 m 10 m
10 m
10 m
20 kN
10 kN
j
k
f g
h
i
a
b
d
c
e
10
10
10
10
20 kN
10 kN
Xa =20
Ya =10 kN
Xk=10
Yk=10 kN
a
b
c
d e
f
g
h
i j
k
s1
s1
s2
s2

Section s1-s1
 Mc =0:
 Mf =0:
 Fy =0 (+):
Fde  10 + 10  10 + 20  10 – 10  10= 0
Fde = – 20 kN
Fce  10 – 20  10 + 20  20 – 10  20= 0
Fce = 0
10/14.14 Fcf + Fce - 10 = 0
Fcf = 14.1 kN
Section s2-s2
 Mc =0:
 Fx =0 (→+):
Fbd  10 + 20  10 – 10  10= 0
Fbd = – 10 kN
– 10/14.14 Fcd +10/14.14  14.1 + 20
– 20 = 0
Fcd = 14.1 kN
Member forces
Example 2.23
Determine the forces in the marked members of the shown truss
using the method of sections.
SOLUTION
 Mb =0:
 Fy =0 (+):
 Fx =0 (→+):
Ya  8 – 120  18 = 0
Ya = 270 kN
Yb – Ya – 120 = 0
Yb = 390 kN
Xa = 0
120 kN
Ya=270 Yb=390 kN
a
b
d
c e
f
g
h
Xa
=0
s1
s1
s2
s2
s3
s3
s4
s4
20 kN
10 kN
20 kN
10 kN
Fde
Fce
Fcf
e
f
d
c
b
a
20 kN
20 kN
10 kN
0 14.1
c
b
a
Fcd
Fbd
20 kN
10 kN
-10
28.3
-14.1
-20
-10
14.1
14.1
-14.1
-20 -20
20 kN
10 kN
10 kN
10 kN
0 0
0 0
0
0
0
a
cb
d e f g
i
h
j
k
0
8 m 6 m 6 m 6 m
8 m
5 m
3 m
120 kNf
g
h
a b
d
c
e
8
16
17.89
8
6
10
6
8
10
6
57.8 1
3
6
6.71

Section s1-s1
 Mf =0:
 Me =0:
 Mo =0:
Feg  3 + 390  12 – 270  20= 0
Feg = 240 kN
6/7.81 Fdf  8+ 270  14 + 390  6 = 0
Fdf = – 234.3 kN
3/6.71 Fef  9.6 + 270  23.6
– 390  15.6 = 0
Fef = – 67.1 kN
Section s2-s2
 Md =0:
 Mo =0:
Fce  8 +390  6 – 270  14= 0
Fce = 180 kN
Fde  9.6 + 390  15.6 – 270  23.6 = 0
Fde = 30 kN
Section s3-s3
 Mc =0:
 Mp =0:
6/10 Fbd  16 + 270  8 = 0
Fbd = – 225 kN
8/10 Fcd  12 + 270  20 – 390  12= 0
Fcd = – 75 kN
Section s4-s4
 Mb =0:
 Ma =0:
16/17.89  Fac  8 – 270  8 = 0
Fac = 301.9 kN
Fbc  8 + 390  8 – 8/10  225  8 = 0
Fbc = – 210 kN
270 kN 390 kN
Fce
Fde
234.3
9.6 m
o
a b
c
d
e
270 kN 390 kN
Feg
Fef
Fdf
9.6 m
o
a b
c
d
e
f
270 kN 390 kN
180
12 m
p
a b
c
Fcd
Fbd
270 kN 390 kN
a b
225Fac
Fbc

Member forces
Example 2.24
Determine the forces in the marked members of the
shown truss using the method of sections.
SOLUTION
 Md =0:
 Fy =0 (+):
 Fx =0 (→+):
Ya 12 + 54 – 2012 – 206 = 0
Ya = 28.3 kN
Yd + Ya – 20 – 20 = 0
Yd = 11.7 kN
– Xa + 5 = 0
Xa = 5 kN
Section s1- s1
 Mi =0:
 Ma =0:
 Fy =0 (+):
Fbc 6 + 52 + 206 – 28.36 – 56 = 0
Fbc = 11.7 kN
6/6.32 Fei  4 + 5  4 = 0
Fei = – 5.3 kN
2/2.83 Ffi + 2/6.32 Fei + 28.3 – 20 = 0
Ffi = – 9.4 kN
Section s2- s2
120 kN
-135
301.9
-225
-210
-75
30
-234.3
180
-67.1
240
-268.3
240
270 kN 390 kN
4 m 2 m 2 m 4 m
4 m
2 m
5 kN
20 kN
20 kN
f g h
i
a
b
d
c
e
6
26.32
2
2
2.83
4
4
5.66
5 kN
20 kN
20 kN
Xa=5 kN
Ya =28.3 kN Yd=11.7 kN
a b c
d
e
f g
h
is1
s1
s2
s2
s3
s3
5 kN
20 kN
5 kN
28.3 kN
b
f
e
i
Fbc
Ffi
Fei

 Me =0:
 Fx =0 (→+):
Fbf  4 + 11.7  4 – 5  4 = 0
Fbf = – 6.7 kN
Fef + 11.7 + 5 – 5 – 6/6.32  5.3 = 0
Fef = – 6.7 kN
Section s3- s3
 Mb =0:
 Fx =0 (→+):
Fae  4 + 28.3  4 = 0
Fae = – 28.3 kN
– 4/5.66 Fbe + 11.7 – 5 = 0
Fbe = 9.4 kN
Member forces
Example 2.25
shown truss using the method of sections.
SOLUTION
 Mf =0:
 Fy =0 (+):
 Fx =0 (→+):
Ya  10.5 – 40  3.5 = 0
Ya = 13.3 kN
Yf + Ya – 40 = 0
Yf = 26.7 kN
Xa = 0
5 kN
20 kN
5 kN
28.3 kN
a b
e
11.7
5.3
Fef
Fbf
5 kN
28.3 kN
a b
11.7
6.7
Fbe
Fae
5 kN
20 kN
20 kN
5
-28.3
11.7
9.4
-6.7
-11.7
16.5
-11.7
-6.7
-5.3
-9.4
-11.7
-16.5
5 kN
28.3 kN 11.7 kN
0
0
1.5 m 2 m 3.5 m 2 m 1.5 m
2 m
1.5 m
40 kN
fa
b
dc
e1.5
2
2.5
40 kN
Ya =13.3 kN Yf=26.7 kN
a
b
c d
e
fs 1
s 1
Xa = 0

Section s1-s1
 Mo =0:
 Mg =0:
1.5/2.5 Fab  1.5 – 2/2.5 Fab  6.375
– 26.7  2.625 = 0
Fab = – 16.7 kN
1.5/2.5 Fef  3.5 – 2/2.5 Fef  7.875
– 26.7  7.875 = 0
Fef = – 50 kN
Member forces
Example 2.26
SOLUTION
 Mb =0:
 Fy =0 (+):
 Fx =0 (→+):
Ya  16 + 40  14 – 100  8 = 0
Ya = 15 kN
Yb + Ya – 100 = 0
Yb = 85 kN
– Xa + 40 = 0
Xa = 40 kN
26.7 kN
b
c
f
o
Fcd
Fef
Fab
g
2.625 m1.125 m
40 kN-16.7
18.4
-8.9
6.1
-50
26.8
-20 -60
-40
13.3 kN 26.7 kN
6 m 2 2 6 m
2 m
4 m
8 m
100 kN
40 kN
f
a b
d
c
e
8
28.25
6
6
8.49
8
2
8.25
100 kN
40 kN
Xa=40
Ya=15 kN Yb=85
a
b
c
d
e
f
s1
s 1

Section s1-s1
 Md =0:
 Fy =0 (+):
 Fx =0 (→+):
8/8.25 Fef  4 + 8/8.25 Fbc  4 – 2/8.25 Fbc  2
– 100  2 = 0 ……………………..……. (a)
8/8.25 Fef – 2/8.25 Fbc – 6/8.49 Fad – 100 = 0…(b)
– 2/8.25 Fef + 8/8.25 Fbc – 6/8.49 Fad = 0 ……..(c)
Solving (a), (b), and (c),
Fef = 66 kN , Fbc = – 16.5 kN , Fad = – 45.3 kN
Member forces
EXAMPLE 2.27
shown truss using a combination of method of sections
and method of joints.
SOLUTION
 Mf =0:
 Fy =0 (+):
 Fx =0 (→+):
Ya  9 + 40  4 –50  6
– 50  3 = 0 ; Ya = 32.2 kN
Yf + Ya – 50 –50 = 0
Yf = 67.8 kN
– Xf + 40 = 0
Xf = 40 kN
100 kN
c
d
e
Fef
Fad Fbc
100 kN
40 kN
62.3
-45.3
19.6
-16.5
-93.3
35.8
71.6
-48
66
40 kN
15 kN 85 kN
a
c
b
d e
f
3 m 3 m 3 m
4 m
50 kN
40 kN
50 kN
fa
b
d
c e
9
49.85
4
3
5
50 kN
40 kN 50 kN
Ya
=32.2 kN
Xf=40
Yf=67.8 kN
a d
b
c
e
f
s1
s1

Section s1-s1
 Mf =0:
 Fy (+):
Fce  4 + 50  3 = 0
Fce = -37.5 kN
4/9.85 Fbf + 67.8 – 50 = 0
Fbf = – 43.8 kN
Joint f
 Fy (+):
 Fx (→+):
Fef + 67.8 – 4/9.85  43.8 = 0
Fef = – 50 kN
– Fdf – 40 + 9/9.85  43.8 = 0
Fdf = 0
Joint e
 Fy =0 (+):
 Fx (→+):
– 4/5 Fde + 50 = 0
Fde = 62.5 kN
– 3/5 Fde + 37.5
= – 3/5  62.5 + 37.5 = 0 ………(Check)
Member forces
EXAMPLE 2.28
Determine the forces in the marked members of
the shown truss using a combination of method of
sections and method of joints.
SOLUTION
50 kN
40 kN
67.8 kN
Fad
e
f
d
Fbf
Fce
40 kN
67.8 kN
f
43.8
Fef
Fdf
e37.5
Fde
50
50 kN
40 kN
50 kN
37.5
17.8
-62.5
62.5
-43.8
-50
-37.5
32.2 kN
40 kN
67.8 kN
0
0
7.21
6 at 4 m =24 m
3 m
3 m
3 m
80 kN
30 kN
j
k
l
f
g
h
i
a
b d
c
e
4
6
4
3
5

 Ml =0:
 Fy =0 (+):
 Fx =0 (→+):
Ya  24 – 80  16
– 30  6 = 0 ; Ya = 60.8 kN
Yl + Ya – 80 = 0
Yl = 19.2 kN
Xa – 30 = 0
Xa = 30 kN
Section s1-s1
 Mf = 0:
 Me =0:
 Ma =0:
4/5 Feg  6 + 3/5 Feg  4 – 80  4 + 60.8  12 = 0
Feg = – 56.9 kN
Fdf  6 + 30  6 – 60.8  8 = 0
Fdf = 51.1 kN
4/7.21 Fef  6 + 6/7.21 Fef  8 + 80  8 = 0
Fef = – 64.1 kN
Joint g
 Fx (→+):
 Fy (+):
4/5 Fgi + 4/5  56.9 = 0
Fgi = – 56.9 kN
– Ffg + 3/5  56.9 – 3/5 Fgi = 0
Ffg = 68.3 kN
Joint e
 Fx (→+):
 Fy (+):
– 4/5 Fce – 4/5  56.9 – 4/7.21  64.1 = 0
Fce = – 101.4 kN
– Fde + 6/7.21  64.1 – 3/5 Fce – 3/5  56.9 =
0
Fde = 80 kN
Member forces
80 kN
30 kN
j
k
l
f
g
h
i
a b d
c
e
Ya
=60.8 kN
Xa =30
Yl=19.2 kN
s 1
s1
80 kN
30 kN
60.8 kN
a b
c
e
d
f
Fef
Feg
Fdf
Ffg
56.9
Fgi
g
64.1
56.9
Fce
Fde
e
80 kN
30 kN
51.1
-101.4
51.1 51.1
80
25.6
-64.1
-18
68.3
25.6 25.6
-31.9
-101.4 -31.9
-56.9 -56.9
30 kN
60.8 kN 19.2 kN
0
00
0
0

EXAMPLE 2.29
SOLUTION
From symmetry:
 Fx =0 (→+):
Ya = Yn = 100 kN
Xa = 0
Section s1-s1
 Mf = 0:
 Md = 0:
Fdg  6 + 40  4 – 100  8 = 0
Fdg = 106.7 kN
Ffh  6 – 40  4 + 100  8 = 0
Ffh = – 106.7 kN
Joint e
 Fx (→+): 4/5 Feh + 4/5 Feg = 0
Feh = – Feg ……………………………….(a)
Section s2-s2
 Fy (+):
From symmetry:
3/5 Feh – 3/5 Feg + 100 – 40 – 40 = 0 ..(b)
Feh = – 16.7 kN
Feg = 16.7 kN
Fhj = Feh = – 16.7 kN (compression)
Joint h
 Fy (+): – Fhg + 3/5  16.7 + 3/5  16.7 = 0
Fhg = 20 kN
Joint f
6 at 4 m = 24m
3 m
3 m
40 kN 40 kN 40 kN 40 kN 40 kN
f
g
h
i
a
b d
c
e j
k
l
m
n
3
4
5
40 kN 40 kN100 kN
Ffh
Fef
Fde
Fdg
a
c
f
db
40 kN 40 kN 40 kN 40 kN 40 kN Yn=100Ya =100
a b d
c
e
f
g
h k
j
i l
m
nXa = 0
s 1
s 1 s 2
s 2
e
Fef
Fde
Feh
Feg
40 kN 40 kN100 kN
Ffh
Feh
Feg
Fdg
a
c
e
db
Fhg
16.7 16.7
h

 Fx (→+):
 Fy (+):
– 4/5 Fcf – 106.7 = 0
Fcf = – 133.3 kN
– Fef – 3/5 Fcf = 0
Fef = 80 kN
Member forces
EXAMPLE 2.30
truss using a combination of method of sections and
method of joints.
SOLUTION
From symmetry:
 Fx =0 (→+):
Ya = Ye = 150 kN
Xa = 0
Section s1-s1
 Mc = 0:
 Fy (+):
 Fx (→+):
Fjk  12.5 + 150  10 – 100  5 = 0
Fjk = – 80 kN
5/5.59 Fch + 150 – 100 –100 = 0
Fch = 55.9 kN
Fcd + Fjk + 2.5/5.59 Fch = 0
Fcd – 80 + 2.5/5.59  55.9 = 0
Fcd = 55 kN
Fcf
f
Fef
106.7
-166.7
40 kN 40 kN 40 kN 40 kN 40 kN
106.7
-33.3
133.3
60
-33.3
133.3
40
-16.7
80
100 kN 100 kN
-133.3
-106.7
-133.3
-16.7
133.3 106.7
16.7
-106.7
20
16.7
6080
133.3
40
-1 66.7
5.59
7.91
4 at 5 m = 20 m
5 m
7.5 m
100 kN 100 kN 100 kN
f g h i
a
b dc
e
j k
2.5
7.5
5
2.5
100 kN 100 kNYa=150 kN Ye =150100 kN
Xa=0
a b c d e
f g h i
j k
s1
s1
s2
s2
100 kN 100 kN150 kN
a
f
j
g
b c
Fjk
Fch
Fcd

Joint j
 Fy (+):
 Fx (→+):
– 7.5/7.91 Ffj – 7.5/7.91 Fgj = 0 ………….(a)
– 2.5/7.91 Ffj + 2.5/7.91 Fgj – 80 = 0 …….(b)
Ffj = – 126.5 kN ; Fgj = 126.5 kN
Section s2-s2
 Mb =0:
 Fy (+):
Ffg  5 +150  5 – 2.5/7.91  126.5  5
– 7.5/7.91  126.5  2.5= 0
Ffg = – 50 kN
5/5.59 Fbg + 150 – 100 – 7.5/7.91  126.5 = 0
Fbg = 78.3 kN
Member forces
EXAMPLE 2.31
shown truss using a combination of method of
SOLUTION
j 80
Ffj Fgj
100 kN150 kN
a
f
b
126.5
Ffg
Fbg
Fbc
100 kN 100 kN 100 kN
75
-167.7
55
33.5
78.3
55
55.9
55.9
75
78.3
33.5
-167.7
-50
-126.5
126.5
-50
126.5
-126.5
-80
150 kN 150 kN
3.61
6.71
3 m 1.5 m 1.5 m 3 m
4 m
2 m
20 kN
40 kN 40 kN
f g h i
a b
dc
e
2
3
2
1.5
2.5
6
3

 Mb =0:
 Fy (+):
 Fx =0 (→+):
Ya  3 + 40  1.5 – 40  3
– 20  6 = 0 ; Ya = 60 kN
Yb – Ya – 40 – 40 = 0
Yb = 140 kN
– Xa + 20 = 0
Xa = 20 kN
Section s1-s1
 Mg =0:
 Mo =0:
 Fy (+):
Fab  6 + 60  1.5 – 20  6 = 0
Fab = 5 kN
Ffg  6 + 20  6 + 60  3 = 0
Ffg = – 50 kN
2/2.5 Fcg – 60 = 0
Fcg = 75 kN
Joint a
 Fx (→+):
 Fy (+):
– 3/6.71 Fae + 5 – 20 = 0
Fae = – 33.5 kN
Fac + 6/6.71 Fae – 60 = 0
Fac = 90 kN
Joint e
 Fy (+):
 Fx (→+):
– 2/3.61 Fce + 6/6.71  33.5 = 0
Fce = 54.1 kN
Fef + 20 + 3/3.61 Fce – 3/6.71  33.5 = 0
Fef = – 50 kN
Member forces
20 kN
40 kN 40 kN
Xa=20 kN
Ya=60 kN Yb=140 kN
a b
e
c d
f
g
h is1
s 1
20 kN
20 kN
60 kN
Fab
Ffg
gf
Fcg
c
e
a
3 m
o
20 kN
60 kN
5
a
FacFae
20 kN
Fef
e
Fce
33.5
20 kN
40 kN
40 kN
5
-33.5
90
11.2
-150
-50
54.1
-50 70
75
-125
70
-90.1
20 kN
60 kN 140 kN
0 0

EXAMPLE 2.32
shown truss using a combination of method of sections
and method of joints.
SOLUTION
 Mg =0:
 Fy (+):
 Fx =0 (→+):
Yf  8 + 10  5 + 20  4
– 40  4 = 0 ; Yf = 3.7 kN
Yg + Yf – 40 – 20 = 0
Yg = 56.3 kN
– Xg + 10 = 0
Xg = 10 kN
Section s1-s1
 Mc =0:
 Fy (+):
 Fx (→+):
4/4.12 Fhi  3 + 1/4.12 Fhi 4 + 10  3 + 3.7  4 = 0
Fhi = – 11.6 kN
2/4.47 Fcf + 1/4.12 Fhi + 3.7 = 0
Fcf = – 2.1 kN
Fbc + 4/4.47 Fcf + 4/4.12 Fhi + 10 = 0
Fbc = 3.1 kN
Joint h
 Fx (→+):
 Fy (+):
– 4/5 Fah – 4/4.12  11.6 + 10 = 0
Fah = – 1.6 kN
– Fbh – 3/5 Fah – 1/4.12  11.6 = 0
Fbh = – 1.9 kN
Joint a
 Fy (+):
 Fx (→+):
– 2/4.47 Faf – 3/5  1.6 = 0
Faf = – 2.1 kN
Fab + 4/4.47 Faf – 4/5  1.6 = 0
Fab = 3.1 kN
4 m 4 m 4 m 4 m
2 m
3 m
1 m
40 kN 20 kN
10 kN j
f g
h
i
a
b dc e
4
1
4.12 4
35
2
4
4.47
40 kN
20 kN
10 kN
Yf=3.7 kN
Xg=10 kN
Yg=56.3
a
b
f
c
h i j
d
g
e
s 1
s 1
10 kN
3.7 kN
Fcf
Fbc
Fhi
c
ba
f
h
h
11.6
Fbh
Fah
10 kN
a
Fab
Faf
1.6

Member forces
EXAMPLE 2.33
shown truss using a combination of method of
SOLUTION
From symmetry:
 Fx =0 (→+):
Ya = Yp = 300 kN
Xa = 0
Joint f
 Fx (→+): 10/11.18 Ffk + 10/11.18 Ffi = 0
Ffk + Ffi = 0 ………………………..….….(a)
Joint i
 Fx (→+): – 10/11.18 Ffi – 10/11.18 Fdi = 0
Ffi + Fdi = 0 ……………………….………(b)
Section s1-s1
40 kN
20 kN
10 kN
3.1
-2.1
-1.6
3.1
-1.9
-1.9
68.1
-2.1 -74.8
5.6
68.1
5.65.6
-63.6
-14.1
-11.6 -11.6
3.7 kN
10 kN
56.3 kN
11 .18
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
10 m 10 m 10 m 10 m
5 m
5 m
5 m
200 kN 200 kN 200 kN
5
10
Xa=0
Yp=300
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
200 kN 200 kN 200 kN
Ya=300
s 1
s 2
s1
s 2
f
Ffi
Ffk
Ffg
Fef
Fij
Ffi
Fdi
Fhi
i
a
b
c
d
e
g
300 kN
Fad
Fde
Fef
Ffg
Fgk

 Fy (+): 5/11.18 Ffk – 5/11.18 Ffi – 5/11.18 Fdi
+ 300 – 200 = 0
Ffk – Ffi + Fdi + 223.6 = 0 ……………..…(c)
Solving (a), (b), and (c),
Ffk = – 74.5 kN
Ffi = 74.5 kN
Fdi = – 74.5 kN
Section s2-s2
 Md = 0:
 Mg = 0:
Fgk  15 + 30010 = 0
Fgk = – 200 kN
Fad  15 – 300  10 = 0
Fad = 200 kN
Joint d
 Fy (+):
 Fx (→+):
Fde – 5/11.18  74.5 – 200 = 0
Fde= 233.3 kN
Fdh – 10/11.18  74.5 – 200 = 0
Fdh = 266.7 kN
Member forces
EXAMPLE 2.34
SOLUTION
a
b
c
d
e
f
g
200 kN300 kN
Fdh
Fdi
Ffi
Ffk
Fgk
Fde
200 d
200 kN
Fdh
74.5
100
166.7
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
200 kN 200 kN 200 kN
-200
200
-223.6
0
223.6
-223.6
0
233.3
266.7
-74.5
33.3
100
74.5
-74.5
-200
200
74.5
100
33.3
-74.5
74.5
-266.7
0
223.6
166.7
-33.3
-223.6
223.6
-200
-300-100
300 kN 300 kN
5
3 m 3 m 3 m 3 m
3 m
1 m
1 m20 kN
60 kN 40 kN
f
g
h
i
a b
d
c
e
3
1
3.16
3
4

 Mb =0:
 Fy (+):
 Fx =0 (→+):
Ya  6 + 603 – 403 – 204 = 0
Ya = 3.3 kN
Yb – Ya – 60 – 40 = 0
Yb = 103.3 kN
– Xb + 20 = 0
Xb = 20 kN
Group cdfe
 Fx (at d) =0:
 Fx (at e) =0:
 Fy (at d) =0:
 Fy (at e) =0:
Fcd = Fdf …………………………….(a)
Fce = Fef ……………………..….…..(b)
Fde = 1/3.16 Fcd + 1/3.16 Fde ……...(c)
Fde = 1/3.16 Fce + 1/3.16 Fef ……..(d)
 Fx (at c) =0:
 Fy (at c) :
From (a), (b), (c), and (d)
Fcd = Fdf = Fce = Fef
Fcd = 0.527 F
From (c), Fcd = 0.333 F
1/3.16 Fcd – 1/3.16 Fce =0;
Fcd = Fce
Then, the group cdfe is carrying only a horizontal force. So, the group can be replaced by a link
member as shown in figure.

Replacing member groups cdfe and fgih by link
members, the truss becomes as shown in figure
20 kN
60 kN 40 kN
Ya=3.3 kN
Xb=20
Yb=103.3
a b
e
f
h
c
d g
i
s1
s1
d
e
Fde
FdeFcd
FefFce
FF
c f
F FFF F
0.527 F
0.333F
0.527 F0.527 F
0.527 F
20 kN
60 kN 40 kN
3.3 kN
20 kN
103.3 kN
a
c f
i
b

Section s1-s1
 Ma =0:
 Mf =0:
 Fy (+):
Fcf  4 + 20  4 = 0 ; Fcf = – 20 kN
Then, from group cdfe, we get
Fcd = Fef = – 0.527  20 = – 10.5 kN, and
Fde = 0.333  20 = 6.7 kN
Fab  4 +3.3  3 = 0 , Fab = – 2.5 kN
4/5 Faf – 3.3 = 0 ; Faf = 4.2 kN
Member forces
EXAMPLE 2.35
Determine the forces in all members of the shown
truss using a combination of method of sections and
method of joints.
SOLUTION
 Mi =0:
 Fy (+):
 Fx =0 (→+):
Ya  24 – 80  18 – 60  15
– 100  12 – 60  9 – 60  3 = 0
Ya = 177.5 kN
Yi + Ya – 80 – 60 – 100 – 60
– 60 = 0 ; Yi = 182.5 kN
Xa = 0
The given compound truss is subdivided into a primary truss and four secondary trusses as
follow:
20 kN
3.3 kN
a
c fFcf
Faf
Fab
20 kN
60 kN 40 kN
-2.5
4.2
-79.2
-50
-10.5
-10.5
6.7
-10.5
-10.5
15.8
15.8
-10
15.8
15.8
3.3 kN
20 kN
103.3 kN
0
a
b c d e f g h
i
j
k
l m
n o p
3 m 3 m 3 m 3 m 3 m 3 m 3 m 3 m
3 m
3 m
80 kN
100 kN 60 kN60 kN
60 kN
Ya=177.5
Yi=182.5
a
b c d e f g h
i
j
k
l m
n o p
80 kN 100 kN 60 kN
60 kN
60 kN
Xa
=0

Primary truss Secondary trusses
The secondary trusses are first analyzed by the method of joints, and the forces in their
members are given in the figure. The secondary trusses are supported by the primary truss.
Loads that equal to the reactions of the secondary trusses must be applied at the corresponding
panel points of the primary truss in the opposite direction. For example,
At panel point c of the primary truss
Load acting on the compound truss at c = 80 kN ()
Reaction at c on the secondary truss cked = 30 kN ()
Therefore, load transferred from the secondary truss cked to panel point c of primary truss =
30 kN ()
Total load acting at panel point c of the primary truss = 80 + 30 = 110 kN ()
At panel point e of the primary truss
Load acting on the compound truss at e = 100 kN ()
Reaction at e on the secondary truss cked = 30 kN ()
Reaction at e on the secondary truss elgf = 30 kN ()]
Therefore, load transferred from the secondary trusses cked and elgf to panel point
e of primary truss = 30 +30 = 60 kN ()
Total load acting at panel point c of the primary truss = 80 + 60 = 110 kN ()
After all the panel-point loads are determined, the primary truss is analyzed by the method of
joints or sections.
The forces in the members of the given compound truss are obtained by algebraic
summation of the forces in the corresponding members of the primary and secondary trusses.
The final member forces are given as follows:
Final member forces
-258.1
-215.7
-42.4
-42.4
-251
-251
177.5 177.5
0
207.5
0
110 207.5
60
182.5
53
88.4
0
182.5
60
182.5
-42.4
60
182.5
60
95.5
130.8
-245 -245
177.5 kN 182.5 kN
80 kN 100 kN 60 kN60 kN 60 kN
000
a
j
c
b0 0
0
0
0
-42.4
-42.4
30 30
60
30 kN60 kN30 kN
c
k
e
d
-42.4
30 30
60
-42.4
30 kN60 kN30 kN
e
l
g
f
-42.4
30 30
60
-42.4
30 kN60 kN30 kN
g
m
i
h
-258.1
-215.7
-42.4
-42.4
-251
-251
177.5 177.5
0
207.5
0
110
207.5
60
182.5
53
88.4
0
182.5
60
182.5
-42.4
60
182.5
60
95.5
130.8
-245 -245
177.5 kN 182.5 kN
80 kN 100 kN 60 kN60 kN 60 kN

2.8 COMPLEX TRUSSES
The configuration of some trusses can be categorized as neither simple nor compound. For
example, the truss in Fig. 2.4 is a simple, as can be demonstrated by starting with triangle abc
and subsequently expanding the truss by adding two members and one joint at a time to locate
joints e, d, f, and g in that order. If member eb is removed and replaced by member ag, as
shown in Fig. 2.4, a different rigid truss is created that is neither simple nor compound but
complex. Since this truss is statically determinate, the method of joints and the method of
sections can be used for the analysis. In this case, several linear equations should be solved to
get the forces in the members. The solution of this set of equations is somewhat tedious.
An alternative approach is proposed by considering the force in one member of the truss
as P. The forces in other members can now be obtained in terms of P. The value of P can be
determined when resolving the last joint in the truss at which two different expressions of one
member force are found. Once P is calculated, then the forces in all members become known.
This method is illustrated numerically in Example 2.37.
The virtual work method may also be used for solving complex trusses. This method is
proceeded by removing one member and replacing it by another one to change the structure to
a simple truss. The new simple truss can be solved by one or more of the previous method.
Let the forces in the new simple truss, due to the applied forces, be F0. The applied forces are
then removed and two opposite unit loads (X = 1) are applied in the direction of the removed
member. The truss is now resolved for the new case of loading. The system of forces is termed
F1. Let the forces in the new member be F0 and F1 from systems F0 and F1, respectively. To
eliminate the effect of the new member, the final force should be zero. Therefore,
F0 + X F1 = 0 ……………………………………………(2.1)
(a) (b)
Fig. 2.4 Simple and complex trusses
From this condition, the value of X can be determined and consequently the forces in all
members can be easily determined by adding the two sets of forces F0 and X F1 as follow:
0
1
F
X
F
= − ………………………………………………(2.2)
F (final) = F0 + X F1 …………………………………….(2.3)
This approach is illustrated numerically in the following example.
a
b
d
c
e g
f
a
b
d
c
e g
f

EXAMPLE 2.36
complex truss.
SOLUTION
From symmetry:
 Fx =0 (→+):
Ya = Yc = 80/2 = 40 kN
Xa = 0
If member ef is removed and member dg is added, the following simple truss is obtained
Member forces (F0)
This simple truss can be analyzed using the method of joints and the members forces is called
F0
If the actual applied forces are removed and two equal and opposite unit forces are applied at
joints e and f (the ends of the removed member) and in the direction ef. The following internal
forces (F1) can be determined using the method of joints.
Member forces (F1)
The final forces (F) in the original truss members are calculated using the following
relationship:
F = F0 + X F1
4 m 4 m 4 m 4 m
4 m
2 m
80 kN
f
g
a
b
d
c
e
80 kNYa
=40 kN Yc=40 kN
Xa=0
a
b
c
d
e f
g
a
b
c
d
e f
g
80 kN
0
-40
0
0
89.4 89.4
0
-40
0
-80
0
40 kN 40 kN
a
b
c
d
e f
g
1.0 kN
0.50
0.25
-0.559
0.50
0 0
-0.559
0.25
0.559
-0.50
0.559
0 0

where X is a dimensionless correction factor which is the negative ratio between F0 and F1 of
the added member that is given as follow:
0
1
80
160
0.5dg
F
X kN
F
  − 
= − = − = −   −  
The final internal forces is in the original complex truss is given as follow:
Final member forces
EXAMPLE 2.37
complex truss.
SOLUTION
 Mb =0:
 Fy =0 (+):
 Fx =0 (→+):
Ya  20 + 20  5 + 10  15
– 60  10 = 0 ; Ya = 17.5 kN
Yb + Ya – 60 = 0
Yb = 42.5 kN
– Xa + 20 +10 = 0
Xa = 30 kN
The force in member ad is assumed to be P kN (Fad = P kN). Using the method of joints, the
forces in the other members are determine in terms of P as follows:
80 kN
-80
-80
89.4
-80
89.4 89.4
89.4
-80
-89.4
-160 -89.4
40 kN
40 kN
20.62
14.14
a b
c
d
e
f
10 m 10 m
5 m
5 m
5 m
20 kN
60 kN
10 kN
10
10
20
5
a
b
c
d
e
f
20 kN
60 kN
10 kN
Xa
=30 kN
Ya = 17.5 kN Yb
= 42.5 kN

Joint d
 Fx =0 (→+):
 Fy =0 (+):
– 10/14.14 Fdf – 20/20.62 P = 0
Fdf = – 1.373 P kN ……………………………(a)
– Fbd + 10/14.14 Fdf – 5/20.62 P = 0
Fbd = – 1.212 P kN ……………………………(b)
Joint f
 Fx =0 (→+):
 Fy =0 (+):
– 10/14.14 Fcf – 10/14.14  1.373 P + 10 = 0
Fcf = (14.14 – 1.373 P) kN …………………….(c)
– Fef – 10/14.14 Fcf + 10/14.14  1.373 P – 60 = 0
Fef = (1.94 P – 70) kN …………………………(d)
Joint e
 Fx =0 (→+):
 Fy (+):
10/14.14 Fbe – 10/14.14 Fae = 0
Fbe = Fae
– 10/14.14 Fbe – 10/14.14 Fae + (1.94 P – 70) = 0
Fbe = Fae = (1.372 P – 49.49) kN ………….(e)
Joint a
 Fx =0 (→+):
 Fy = 0 (+):
20/20.62 Fad + 10/14.14  (1.372 P – 49.49) – 30 = 0
Fad = (67 – P) kN ……………………………….(f)
Fac + 5/20.62 Fad + 10/14.14  (1.372 P – 49.49)
+ 17.5 = 0
Fac = (1.254 – 0.728 P) kN ……………………..(g)
It is noticed that member ad has two values. These values are the original estimated value (P)
and the value obtained from Eq. (f) (67 – P). Since the two values must be equal, then
P = 67 – P
Or, P = 33.5 kN
The forces in the members can be obtained by substituting P = 33.5 in Eqs. (a) to (g) as follows:
From Eq.(a):
From Eq.(b):
From Eq.(c):
From Eq.(d):
From Eq.(e):
Fdf = – 1.373  33.5 = – 46 kN
Fbd = – 1.212  33.5 = – 40.6 kN
Fcf = 14.14 – 1.373  33.5 = – 31.8 kN
Fef = 1.94  33.5 – 70 = – 5 kN
d
Fbd
P
Fdf
e
Fae Fbe
(1.94P-70)
a
30 kN
17.5 kN
(1.372P-49.49)
Fad
Fac
Fef
f10 kN
60 kN
1.373P
Fcf

From Eq.(g):
Fbe = Fae = 1.372  33.5 – 49.49 = – 3.5 kN
Fac = 1.254 – 0.728  33.5 = – 23.1 kN
The force in the remaining member can be obtained by analyzing the forces at joint c as follow:
 Fx =0 (→+):
 Fy (+):
20/20.62 Fbc + 20 - 10/14.14  31.8 = 0
Fbc =2.6 kN
23.1 – 5/20.62 Fbc – 10/14.14  31.8 = 0 .(Check)
Final member forces
2.9 GRAPHICAL METHOD OF JOINTS (MAXWELL DIAGRAM)
Any group of forces in equilibrium, when plotted to scale successively by arrowhead lines,
should form a closed polygon. The external force polygon represents the reactions and external
forces. Each internal force polygon represents the equilibrium of the concurrent forces at each
joint. If the external forces and reactions is reading in a clockwise manner around the outside
of the truss, the internal forces must also be reading in a clockwise manner around each joint.
The procedure for the graphical method of joints is proposed as follows:
1. Assign a number to the space outside the truss that is separated by either a reaction or an
external force, proceeding clockwise around the truss.
2. Assign a number to each triangle within the truss.
3. Begin the construction of the Maxwell diagram by starting with the external forces system,
drawing a line parallel to and in the direction of a known external force and of a length to
represent its magnitude to scale. Proceed around the structure clockwise; the resulting
force polygon should close, as the external force system is in equilibrium.
4. Then applying superposition on the external force line polygon, such as 1-2-3-4-5 in the
next example, and beginning with joint containing only two unknowns, construct line
polygons for each joint in succession until all joints have been treated. At each joint
proceed from force to force clockwise around the joint.
5. To interpret the results, at a given joint a member is designated by two numbers, always
using the sequence obtained by moving clockwise around a joint. The “number-number”
sequence is the same sequence used on the Maxwell diagram to establish the directions of
the forces. The length of the line on the Maxwell diagram is proportional to the magnitude
of the force.
c
20 kN Fbc
23.1
31.8
a
b
c
d
e
f
20 kN
60 kN
10 kN-23.1
33.5
-3.5
2.6
-40.6
-3.5
-31.8
-46
-5
30 kN
17.5 kN
42.5 kN

EXAMPLE 2.38
Determine the forces in the members of the
shown truss by the graphical method of joints.
SOLUTION
The spaces between external forces and reactions are designated 1-2-3-4-5-1 clockwise
around the truss, so the force vector 1-2-3-4-5-1 forms a closed polygon. The four internal
triangles are numbered 6-7-8-9. These numbers are determined in the Maxwell diagram in
succession; for instance, point 6 is the intersection of 1-6 parallel to member ac and 2-6 parallel
to member ce.
The magnitude of each of axial force may be measured from the force diagram and its
direction acting on the joint (tension-compression) is determined by the indicated direction
when proceeding from point to point on the Maxwell diagram according to the name of the
member obtained by reading in a clockwise pattern around a joint of the truss diagram. For
example,
Fab = 7-4 on joint a = 4-7 on joint b = 20 kN (tension)
Fac = 1-6 on joint a = 6-1 on joint c = 14.1 kN (compression)
External forces
Maxwell force diagram
a b
c d
e f
3 m 4 m 3 m
3 m
3 m20 kN
30 kN
a b
c d
e f
20 kN
30 kN
30 kN
12.5 kN 7.5 kN
1
2
3
45
6
7
8
9
1
2 3
4
5
6
7
8
9

Member forces
EXAMPLE 2.39
Determine the forces in the members of the truss given in Example 2.19 by the graphical
method of joints.
SOLUTION
External forces
Member forces
EXAMPLE 2.40
method of joints.
SOLUTION
1
2
3
4
5
6
7,10
8
9
11
a b
c d
e f
20 kN
30 kN
20
-14.1
-2.5
21.2
-9
-15
14.1
21.2
15
30 kN
12.5 kN 7.5 kN
20 kN 20 kN 20 kN30 kN 30 kN
a b
c
d
e
f
g
h
1234
5
6
7
8 9
10
11
20 kN 20 kN 20 kN
40
-50
53.3
30
-16.7
53.3
20
40
-16.7
30
-50
-40 -40
30 kN 30 kN

External forces
Member forces
EXAMPLE 2.41
Determine the forces in the members of the shown truss
by the graphical method of joints
SOLUTION
External forces
50 kN 80 kN
40 kN
93.3 kN 76.7kN
a
b
c
d
e
f
g
12
3 4 5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
50 kN 80 kN
40 kN
-93.3
132
-100
-33.9
-31.2
-76.7
-65.1
-40.9
108.4
120.9 126.4
93.3 kN 76.7 kN
7 m 7 m 7 m 7 m
7 m
2 m
80 kN
100 kN 60 kN
f
g
ha
b d
c
e
80 kN
100 kN 60 kN125 kN 115 kN
123
4
5
6
7
8 9
10
11
a
b d f
h
g
e
c
1
2
3
4
5
6
7
8
9
10
11

Member forces
EXAMPLE 2.42
Determine the forces in the members of the truss in Example 2.28 by the graphical method of
joints
SOLUTION
External forces
Member forces
EXAMPLE 2.43
joints
SOLUTION
-176.8
80 kN
100 kN 60 kN
125 132.2
9.3
-11.8
132.2
100
115
82.1
-28.1
-162.6
-130 -119.6
125 kN 115 kN
80 kN
30 kN
j
k
l
f
g
h
i
a b d
c
e
60.8 kN
30 kN
19.2 kN
12
3
4
5
6 7
8
9
10 11
12
13
14 15
1
2
3,5
4
6,7,8
9
10
11
12,13
14,15
-56.9
-64.1
-101.4
80 kN
30 kN
51.1
-101.4
51.1 51.1
80
25.6
-18
68.3
25.6 25.6
-31.9
-31.9
-56.9
30 kN
60.8 kN 19.2 kN
0
0
0
0
0

External forces
Member forces
EXAMPLE 2.44
joints.
SOLUTION
External forces
1 2
3
4,9 5 6
7 8
10
1112
60 kN
40 kN
20 kN
10 kN
40 kN
80 kN
10 kN
a b
c
d
e f
g
100 kN
1
2
3
4
5 6
7
8
9
10
11
12
60 kN
40 kN
20 kN
10 kN
40 kN
80
-20
113.1
-44.7
-86.7
33.3
-10
-86.7
-16.7
80 kN
10 kN
100 kN
0
50 kN
25 kN 75 kN
a b
c d
e
f
g
h
1
2
3
4
5
6
7
8 9
1
2
3
4
5 6
7 8
9

Member forces
EXAMPLE 2.45
method of joints.
SOLUTION
External forces Maxwell force diagram
Member forces
10 kN
10 kN
10 kN
Xa=30 kN
Ya=22.5 kN Yc=22.5 kN
a b
c
d e
f
g h i
j
k
1
2
3
4
5
6
7 8
9
10
11
12
13
14
15
50 kN
-12.5
28
15
-83.3
-16.7
37.3
23.6
-100
-50
70.7
-50
-100
70.7
25 kN 75 kN
1,9
2,6
3 4 5
7
8
10 12
11,13 15
14
10 kN
10 kN
10 kN
30
22.5
-18.8
18.8
-22.5
5
11.3
-15
-12.5 12.5
-11.3
3.8
-10
-6.2 6.2
-3.8
-5
30 kN
22.5 kN 22.5 kN
0
-5

EXAMPLE 2.46
Determine the forces in the members of the shown
truss by the graphical method of joints.
SOLUTION
External forces
Member forces
EXAMPLE 2.47
Determine the forces in the members of the shown
truss by the graphical method of joints.
SOLUTION
External forces
5.2 m 5.2 m 5.2 m 5.2 m
1.5 m
1.5 m
3 m
80 kN
100 kN
50 kN
f
g
a b
d
c
e
80 kN
100 kN
50 kN
50 kN
80.6 kN 99.4 kN
123
4
5
6
7
8 9 10
11
a
b
e
c d
g
f
1
2
4
5 6
7
8
9
10,11
3
-397.9-380.2
-397.9-380.2
80 kN
100 kN
50 kN
394.8 358.7
137.3
-100
320
274.2
50 kN
80.6 kN 99.4 kN
0
a b
c
d e f g
h i j
3 m 8 m 8 m 3 m
4 m
3 m
30 kN 10 kN
80 kN
a b
c d e f
g
h i j30 kN 10 kN
80 kN
73.8 kN 46.2 kN
12
3
4
5
6
7 8 9 10 11 12
13
1
2
3,6
4
7
8
9,10
11
125,13

Member forces
EXAMPLE 2.48
method of joints.
SOLUTION
External forces
Member forces
EXAMPLE 2.49
Determine the forces in the members of the shown truss by the
graphical method of joints.
SOLUTION
1
2 3
4
56
7
8
9
10
11
1212
-73.8
42.4
a b
c d e f
g
h i j30 kN 10 kN
80 kN
0
-73.8
-46.2
0
-30 -30 -10
124.6
0
103.2
-10
-46.3
14.1
-86.7 -86.7
73.8 kN 46.2 kN
100 kN 100 kN
71.4 kN
100 kN
71.4 kN
100 kN
a
b
c
d
e
g
h
f
i
1
2
3
4
5
6
7
8 9 10 11
12
100 kN 100 kN
-79.9
-64.3
-79.9
-64.3
102.7
-100
27.5
102.7
-100
27.5
-93.7 -93.7
-93.7 -93.7
71.4 kN
100 kN
71.4 kN
100 kN
a b
c d
e
f g h
i
9 m
4 m
4 m
4 m20 kN
20 kN10 kN
9 m
4 m 4 m

External forces
Member forces
EXAMPLE 2.50
method of joints.
SOLUTION
External forces
a
b
c d
e
f g
h
i
20 kN 20 kN
10 kN
10 kN
13.3 kN
20 kN
26.7 kN
7
8
10
9
13
11 12
12
3
4 5
6 1,8
2 3
4
5
67
9
12
13 10,11
a b
c d
e
f g
h
i
20 kN 20 kN
10 kN
-16.7
0
-33.3
-20
-20.6
8.3
0
-8.3
-20.6
5 5
8.3
5
-8.3
10 kN
13.3 kN
20 kN
26.7 kN
120 kN
270 kN 390 kN
a b
d
c
e
f
g
h
1
2
3
4
5
6
7
8 9
1
2
3
4
5
6
7
8,9

Member forces
EXAMPLE 2.51
method of joints.
SOLUTION
External forces
Member forces
EXAMPLE 2.52
Determine the forces in the members of the shown truss
by the graphical method of joints.
SOLUTION
120 kN
-135
301.9
-225
-210
-75
30
-234.3
180
-67.1
240
-268.3
240
270 kN 390 kN
40 kN 40 kN 40 kN 40 kN 40 kN 100 kN100 kN
a b d
c
e
f
g
h k
j
i l
m
n
123456
7
8 9
10
11
13
12
14
15
16
17
18 19
1
2
3
4
5
6
7
8
9
10
11
19
18
17
15
12,16
14
13
-166.7
40 kN 40 kN 40 kN 40 kN 40 kN
106.7
-33.3
133.3
60
-33.3
133.3
40
-16.7
80
100 kN 100 kN
-133.3
-106.7
-133.3
-16.7
133.3 106.7
16.7
-106.7
20
16.7
60
80
133.3
40
-166.7
8 at 4 m 32 m
3 m
3 m
50 kN 50 kN50 kN50 kN 50 kN50 kN 50 kN
o
p
f
g
h
i
a
b d
c
e
j
k
l
m
n

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