CHAPTER 5
Statically Determinate
Plane Trusses
TYPES OF ROOF TRUSS
TYPES OF ROOF TRUSS
ROOF TRUSS SETUP
ROOF TRUSS SETUP
OBJECTIVES
• To
determine
the
STABILITY
and
DETERMINACY of plane trusses
• To analyse and calculate the FORCES in truss
members
• To calculate the DEFORMATION at any joints
INTRODUCTION
• A truss is a structure of slender members
joined at their end points by bolting or
welding to gusset plate.
• Members commonly used: wooden struts,
metal bars, angles or channels
ROOF TRUSSES
• Transmitting loading from roof to columns – by
means of series of purlins.
• Trusses used to support roofs are selected on the
basis of the span, slope and roof materials.
Metal Gusset
Top Chord
Truss Web
Bottom Chord
ROOF TRUSSES
ROOF TRUSSES
BRIDGE TRUSSES
BRIDGE TRUSSES
ASSUMPTIONS IN DESIGN
• The members are joined together by smooth pins
(NO friction)
• All loadings and reactions are applied at the joints
• The centroid for each members are straight and
concurrent at a joint
 Each truss member acts as an axial force member.
 If the force tends to elongate → tensile (T)
 If the force tends to shorten → compressive (C)
SIGN CONVENTION
Tensile (positive)
Compressive (negative)
STABILITY & DETERMINACY
• In terms of stability, the most simple truss can be
constructed in triangle using three members.
• This shape will provide stability in both x and y
direction. Each additional element of two
members will increase one number of joint
STABILITY & DETERMINACY
There are 3 types of stable trusses:
1. Simple Truss
2. Compound Truss – combination of two or more
simple trusses together
3. Complex Truss – one that cannot be classified as
being either simple or compound
STABILITY & DETERMINACY
Simple Truss
STABILITY & DETERMINACY
Compound Truss
STABILITY & DETERMINACY
Complex Truss
DETERMINACY
• Let consider a simple truss.
• There will be 6 unknown values: 3 internal member
forces and 3 reactions

m+R → m+3
• And for every joint, 2 equilibriums can be written (Fx
= 0 and Fy =0) – no rotation or moment at joint  2
j
• By comparing the total unknowns with total number
of available equation, we can check the determinacy.
DETERMINACY
• The determinacy of truss should be checked internally
and externally
• The external determinacy is given by:
R = 3 (provided that the support reactions have no
lines of action that are either concurrent or
parallel)
• If R > 3  Statically indeterminate (external)
R = 3  Statically determinate (external)
R < 3  Unstable truss system
DETERMINACY
• The internal determinacy is given by
m = 2j – 3 (provided that the components of the
truss do not form a collapsible mechanism)
• If m > 2j – 3  Statically indeterminate (internal)
m = 2j – 3  Statically determinate (internal)
m < 2j – 3  Unstable truss system
EXAMPLE 1
Determine the stability and determinacy of the
truss shown in the figure below.
EXAMPLE 1 – Solution
Externally:
R = 3 ; R – 3 = 3 – 3 = 0  OK
Internally:
m = 9, j = 6,
9 = 2(6) – 3 = 9  OK
Therefore, the truss is determinate
(externally and internally)
EXAMPLE 2
Determine the stability and determinacy of the
truss shown in the figure below.
EXAMPLE 2 – Solution
Externally:
R = 3 ; R – 3 = 3 – 3 = 0  OK
Internally:
m = 9, j = 6,
9 = 2(6) – 3 = 9  OK
Therefore, the truss is determinate
(externally and internally)
EXAMPLE 3
Determine the stability and determinacy of the
truss shown in the figure below.
EXAMPLE 3 – Solution
Externally:
R = 4 ; R – 3 = 4 – 3 = 1  1 degree redundant
Internally:
m = 10, j = 6,
10 > 2(6) – 3 : 10 > 9  1 degree redundant
Therefore, the truss is internally and externally
indeterminate (1 degree redundant)
MEMBER FORCES
There are several methods of calculating the member
forces for the truss
i. Method of Joints
ii. Method of Sections
iii. Method of Force Resolution
METHOD OF JOINTS
• Suitable to be used to determine all the member
forces in the truss
• In this method, every joint will be analysed by drawing
the Free Body Diagram, limiting the unknown values
to TWO only.
• The selected joints must only consisted concurrent
and coplanar forces
• Using the equilibrium of Fx = 0 and Fy =0, we can
start and solve the problems.
EXAMPLE 4
Determine all the member forces for the given truss
below.
B
100 kN
C
E
A
G
F
D
50 kN
150 kN
8m
6m
3m
3m
EXAMPLE 4 – Solution
1. Check the Stability and Determinacy
Externally:
R = 3 ; R – 3 = 3 – 3 = 0  OK
Internally:
m = 11, j = 7,
11 = 2(7) – 3 = 11  OK
Therefore, truss is determinate (externally and
internally)
EXAMPLE 4 – Solution
2. Calculate the Reactions at the Support
+ MA = 0
100 (6) + 150 (8) + 50 (11) – RG (14) = 0
RG = 167.9 kN ( )
+ Fy = 0 ; RA – 150 – 50 – 167.9 kN ;
RA = 32.1 kN ( )
+ Fx = 0 ; HA = 100 kN ( )
EXAMPLE 4 – Solution
3. Analyse Every Joints
At Joint A
FAB
+Fx = 0 ;
100
A
32.1
FAD
+ Fy = 0 ;
100 + FAD = 0
FAD = 100 kN (T)
FAB + 32.1 = 0
FAB = 32.1 kN (C)
EXAMPLE 4 – Solution
At Joint B
100
B
32.1
+ Fx = 0 ;
FBC
+Fy = 0 ; 32.1 – FBD (6/10) = 0
FBD = 53.5 kN (T)
FBD
FBC + 100 + 53.5 (8/10) = 0
FBC = 142.8 kN (C)
EXAMPLE 4 – Solution
At Joint C
C
142.8
FCD
+ Fy = 0 ;
+Fx = 0 ; 142.8 + FCE (3/√18) = 0
FCE = 201.9 kN (C)
FCE
FCD – (201.9) (3/√18) = 0
FCD = 142.8 kN (T)
EXAMPLE 4 – Solution
At Joint D
142.8
53.5
FDE
D
100
FDF
+Fy = 0 ; 142.8 + 53.5 (6/10) – 150
+ FDE (3/√18) = 0
FDE = 35.2 kN (C)
150
+ Fx = 0;
100 – 53.5 (8/10) + (-35.2)(3/√18) + FDF = 0
FDF = + 166.7 kN (T)
EXAMPLE 4 – Solution
At Joint E
142.8
E
35.2
FEF
+ Fy = 0;
FEG
+Fx = 0 ; FEG (3/√18) + 201.9 (3/√18)
+ 35.2 (3/√18) = 0
FEG = 237.1 kN (C)
FEF  201.9 (3/√18) + 35.2 (3/√18) –
(237.1)(3/√18) = 0
FEF = 49.8 kN (T)
EXAMPLE 4 – Solution
At Joint F
+Fx = 0 ; 167.7 + FFG = 0
FFG = 167.7 kN (T)
49.8
166.7
F
50
FFG
EXAMPLE 4 – Solution
At Joint G (Checking)
237.1
166.7
G
+Fy = 0 ;
167.9 – 237.1 (3/√18) = 0
167.9
+ Fx = 0; 167.7 + 237.1 (3/√18) = 0 OK !
OK !
EXAMPLE 4 – Solution
Summary: Internal Member Forces
Member
AB
AD
BC
BD
CD
CE
Force
32.1
+100
C
142.8
+53.5
+142.8
C
T
T
C
201.9
T
Member
DE
DF
EF
EG
FG
Force
35.2
+167.7
C
T
+49.8
T
C
T
237.1
+167.7
METHOD OF SECTIONS
• When only some of the member forces need to be
calculated, it is suitable to use this method. However, it
can also used to determine all the member forces in
truss.
• The method of sections consists of cutting through the
truss into two parts, provided that the unknown values
are not more than three
• The unknown forces will be assumed to be either in
tension or compression
• Three equilibriums (Fx = 0, Fy = 0, M = 0) will be
used to solve the problems.
EXAMPLE 5
Determine the member forces for BD, DE and CE.
100 kN
B
A
D
2m
2
1
80 kN
E
C
F
40 kN
4m
4m
4m
EXAMPLE 5 – Solution
Stability and Determinacy
Externally:
R = 3 ; R – 3 = 3 – 3 = 0  OK
Internally:
m = 9, j = 6,
9 = 2(6) – 3 = 9  OK
Therefore, the truss is determinate (externally and
internally)
EXAMPLE 5 – Solution
Calculate the Reactions at the Support
+ MF = 0
(2/√5)RA (12) + (2/√5)RA (2) – 100 (8) – 40 (4) = 0
RA = 82.6 kN ( )
+ Fy = 0 ; RF + (2/√5)(82.6) – 100 – 40 = 0 ;
RF = 66.2 kN ( )
+ Fx = 0 ; HF + (1/√5)(82.6) – 80 = 0 ;
HF = 43.1 kN ( )
EXAMPLE 5 – Solution
Section 1:
FDB
D
FDC
FEC
E
F
43.1
40 kN
66.2
MD = 0
FEC (2) – 43.1 (2) – 66.2 (4) = 0 ; FEC = 175.5 kN (T)
Fy = 0 ; 40 + 66.2 – FDC (2/√20) = 0 ; FDC = 58.6 kN (T)
Fx = 0 ; FDB – 58.6 (4/√20) – 175.5 + 43.1 = 0 ;
FDB = 184.8 kN (C)
EXAMPLE 5 – Solution
FED FFD
Section 2:
FEC
E
40 kN
MF = 0 ; FED (4) – 40 (4) = 0 ; FED = 40 kN (T)
F
66.2
43.1
METHOD OF FORCE RESOLUTIONS
• This is an extended version from the method of joints
• Every single joint is carefully analysed by considering
not more than two unknowns at each joints.
• In this method, we do not have to write all the
equations and calculations. All member forces are
solved directly on the diagram.
EXAMPLE 6
Determine the member forces of the truss
50 kN
100 kN
B
20 kN
50 kN
D
F
1.5 m
H
A
C
2m
G
E
2m
2m
2m
ZERO FORCE MEMBERS
• Truss analysis using method of joints can greatly be
simplified if one can first determine those member
that support no loading (zero force member)
• The zero-force members can be determine by
inspection of the joints. Normally, there are two cases
where zero-force member can be identified
ZERO FORCE MEMBERS
Case 1:
If only two members form a truss joint, and no external
load or support is applied, the members must be zeroforce members
Case 2:
If three members form a truss joint for which two of the
members are collinear, the third member will be a zeroforce member (provided no external load or support
reaction acting at the joint).
DEFORMATION OF STATICALLY
DETERMINATE PLANE TRUSS
• The deformation of statically determinate plane truss
can be determined using Virtual Work Method
• Consider to determine vertical deformation at joint C
• Due to external loads, point C will deform – producing 
DEFORMATION OF STATICALLY
DETERMINATE PLANE TRUSS
D
B
A
F
C
E
‘Actual Structure’
G
H
DEFORMATION OF STATICALLY
DETERMINATE PLANE TRUSS
• Now, eliminate all external loads and assign 1 unit load
(vertical) at joint C. This structure is known as ‘Virtual
Structure’
C
‘Virtual Structure’
DEFORMATION OF STATICALLY
DETERMINATE PLANE TRUSS
•
•
•
Both structures will then combined, thus producing the
concept of ‘work’. Therefore, the external work:
W=1.
Say P is the member force due to external loads and u is
the member force due to unit load
If we consider one of the truss members, having force
of P, this member will produce certain deformation
which can be calculated, given as:
 = PL/AE
DEFORMATION OF STATICALLY
DETERMINATE PLANE TRUSS
• Through combination, the amount of internal work is
given by =
𝑢
𝑃𝐿
𝐴𝐸
• According to Energy Work Method:
External load = Internal Load
1 ∙ ∆=
𝑃𝐿
𝑢
𝐴𝐸
SOLUTION PROCEDURE
1. Calculate the member forces for “Actual Structure”.
2. Eliminate all external loads and assign ONE (1) unit load
in the same direction of deformation. Then, calculate
the member forces of “Virtual Structures”.
3. The deformation at point C can be then calculated
using:
𝒖
𝑷𝑳
𝑨𝑬
VIRTUAL WORK METHOD
• When a structure is loaded, its stressed elements
deform. As these deformations occur, the structure
changes shape and points on the structure displace.
• Work – the product of a force times a displacement
in the direction of the force
VIRTUAL WORK METHOD
• External Work – when a force F undergoes a
displacement dx in the same direction as the force.
• Internal Work – when internal displacements δ occur
at each point of internal load u.
𝑷∙∆=
Work of
External
Load
𝒖∙𝜹
Work of
Internal
Load
VIRTUAL WORK METHOD
• When a bar is loaded axially, it will deform and store
strain energy u.
• A bar (as shown in the figure) subjected to the
externally applied load P induces an axial force F of
equal magnitude (F = P). If the bar behaves
elastically (Hooke’s Law), the magnitude of the strain
energy u stored in a bar by a force that increases
linearly from zero to a final value F as the bar
undergoes a change in length dL.
VIRTUAL WORK METHOD
𝑃𝐿
From Hooke’s Law: 𝑑𝐿 =
𝐴𝐸
F
F
L
P
F
dL
P
P
x

x
Δ
DISPLACEMENT OF TRUSSES
• We can use the method of virtual work to determine
the displacement of a truss joint when the truss is
subjected to an external loading, temperature
change, or fabrication errors.
• When a unit force acting on a truss joint, and
resulted a displacement of Δ, the external work =
1Δ.
DISPLACEMENT OF TRUSSES
• Due to the unit force, each truss member will carry
an internal forces of u, which cause the deformation
of the member in length dL. Therefore, the
displacement of a truss joint can be calculated by
using the equation of:
1.   u.dL
Virtual
loadings
Real
displacements
50 kN
B
A
C
STEPS FOR ANALYSIS
D
20 kN
1
1. Place the unit load on the truss at the joint where
the desired displacement is to be determined. The
load should be in the same direction as the specified
displacement; e.g. horizontal or vertical.
STEPS FOR ANALYSIS
2. With the unit load so placed, and all the real loads
removed from the truss, use the method of joints or
the method of sections and calculate the internal
force in each truss member. Assume that tensile
forces are positive and compressive forces are
negative.
3. Use the method of joints or the method of sections
to determine the internal forces in each member.
These forces are caused only by the real loads
acting on the truss. Again, assume tensile forces are
positive and compressive forces are negative.
STEPS FOR ANALYSIS
Member
Virtual
Force, u
Real Force, P
(kN)
AB
CB
DB
0.601
-0.507
0.515
50
-30
20
L (m)
u.PL
(kN.m)
1.8
1.8
1.8
Total
2
-3
5
4
• Apply the equation of virtual work, to determine the
desired displacement. It is important to retain the
algebraic sign for ach of the corresponding internal
forces when substituting these terms into the
equation.
STEPS FOR ANALYSIS
• If the resultant sum of displacement is positive, the
direction is same as the unit load or vice-versa.
• When applying any formula, attention should be paid
to the units of each numerical quantity. In particular,
the virtual unit load can be assigned any arbitrary
unit (N, kN, etc.).
EXAMPLE 6
Determine the vertical displacement of joint C of the
steel truss shown in Figure. The cross-sectional area of
each member is A = 300 mm2 and E = 200 GPa.
F
E
3m
B
A
3m
C
3m
20 kN
D
3m
20 kN
EXAMPLE 6 – Solution
Calculate the Member Forces due to “Virtual Force”
-0.333
-0.471
-0.471
0.333
0.333
0.333
-0.943
1
0.667
0.667
1.0
0.667
Virtual Structure: Virtual force, u
EXAMPLE 6 – Solution
Calculate the Member Forces due to “Actual Forces”
-20 kN
-28.3 kN
0
20 kN
20 kN
20 kN
-28.3 kN
20 kN
20 kN
20 kN
20 kN
20 kN
20 kN
Actual Structure: Real force, N
EXAMPLE 6 – Solution
Calculate the total deformation
Member
Virtual force, u
Real force, P (kN)
L (m)
u.PL (kN.m)
AB
0.333
20
3
20
BC
0.667
20
3
40
CD
0.667
20
3
40
DE
-0.943
-28.3
4.24
113
FE
-0.333
-20
3
20
EB
-0.471
0
4.24
0
BF
0.333
20
3
20
AF
-0.471
-28.3
4.24
56.6
CE
1
20
3
60
Σ
369.6
EXAMPLE 6 – Solution
Calculate the final deformation
1 ∙ ∆=
1 𝑘𝑁 ∙ ∆𝑐𝑣 =
𝑢 ∙ 𝑑𝐿
𝑢 ∙ 𝑃𝐿 369.6
=
𝐴𝐸
𝐴𝐸
369.6
∆𝑐𝑣 =
300 × 10−6 200 × 103
∆𝑐𝑣 = 6.16 𝑚𝑚
DISPLACEMENT OF TRUSSES
(Due to Temperature Changes & Fabrication Error)
In some cases, truss members may change their length due to temperature. If
α is the coefficient of thermal expansion for a member and ΔT is the change
in its temperature, the change in length of a member is:
𝟏 ∙ ∆=
𝒖 ∙ 𝜶 ∙ ∆𝑻 ∙ 𝑳
1 = External virtual unit load acting on the truss joint in the stated direction
of Δ
u = Internal virtual normal force in a truss member caused by the external
virtual unit load
Δ = External joint displacement caused by the temperature change
α = Coefficient of thermal expansion of member
ΔT = Change in temperature of member
L = Length of member
DISPLACEMENT OF TRUSSES
(Due to Temperature Changes & Fabrication Error)
Occasionally, errors in fabricating the lengths of the members of a truss may
occur. Also, in some cases truss member must be made slightly longer or
shorter in order to give the truss a camber. If a truss member is shorter or
longer than intended, the displacement of a truss joint from its expected
position can be determined from direct application:
𝟏 ∙ ∆=
𝒖 ∙ ∆𝑳
1 = External virtual unit load acting on the truss joint in the stated direction
of Δ
u = Internal virtual normal force in a truss member caused by the external
virtual unit load
Δ = External joint displacement caused by the fabrication errors
ΔL = Difference in length of the member from its intended size as caused by
a fabrication error
EXAMPLE 7
Determine
the
vertical
displacement of joint C of
the steel truss as shown in
the Figure. Due to radiant
heating from the wall,
member AD is subjected to
an increase in temperature
of ΔT = +60°C. Take α = 1.08
× 10-5/°C and E = 200 GPa.
The cross-sectional area of
each member is indicated in
the figure.
1.8 m
Wall
D
C
1200
2.4 m
mm2
1200 mm2
1200 mm2
900 mm2
A
300 kN
1200 mm2
B
400 kN
EXAMPLE 7 – Solution
Calculate the Member Forces due to “Virtual Force”
1.0
1
0.75
0.75
1
-1.25
0.75
0
Virtual force, u
0
EXAMPLE 7 – Solution
Calculate the Member Forces due to “Actual Force”
400 kN
600 kN
300 kN
600 kN
400 kN
400 kN
-500 kN
300 kN
Real force, N
0
400 kN
EXAMPLE 7 – Solution
Both loads and temperature affect the deformation,
therefore:
1 ∙ ∆=
𝑢 ∙ 𝑑𝐿 +
𝑢 ∙ 𝛼 ∙ ∆𝑇 ∙ 𝐿
0.75 600 1.8
1 400 2.4
1 𝑘𝑁 ∙ ∆𝑐𝑣 =
+
1200 × 10−6 200 × 106
1200 × 10−6 200 × 106
−1.25 −500 3
+
+ 1 1.08 × 10−5 60 2.4
−6
6
900 × 10 200 × 10
∆𝑐𝑣 = 0.0193 𝑚 = 19.3 𝑚𝑚
SUMMARY
Forces in Member
1. Method of Joints (Fx, Fy)
2. Method of Sections (Fx, Fy, M)
3. Force Resolution (Fx, Fy)
Stability and Determinacy
(External & Internal)
If OK, calculate reactions (3
equilibriums)
To determine
vertical/horizontal
displacement
A+
Virtual Work Method
• Actual structure (P)
• Virtual structure (u)