LESSON 4:
ANALYSIS OF PIN JOINTED PLANE FRAMES(METHOD OF JOINTS)
ENGINEERING MECHANICS
A pin-jointed frame or truss is a triangulated structure made up
of slender members, which are pin connected at the ends and
carry axial forces only. Examples of such structures are roof
trusses, bridge trusses, transmission line towers etc. Both the
ends of the members are connected by nailing, bolting,
welding, riveting etc.
direction and can be analysed by making use of the equations
of equilibrium.
Plane Frame
All members lie in a single plane - roof trusses .
Space Frame
Members do not lie in a single plane -transmission line towers.
Perfect, Deficient and Redundant Frames
A pin jointed frame which has got just sufficient number of
members to resist the loads without undergoing appreciable
deformation in shape is called a perfect frame. Triangular frame
is the simplest perfect frame and it has three joints and three
members (Fig 4.1). Perfect frames with four and five joints are
shown in Fig. 4.2 and 4.3 respectively.
A frame is said to be deficient/ unstable, if the number of
members in it are less than that required for a perfect frame.
Such frames cannot retain their shape when loaded. A deficient/
unstable frame is shown in Fig. 4.5. It is called ‘Mechanism’.
It may be observed that to increase one joint in a perfect frame,
two more members are required. Hence the following
expression may be written down as the relationship between
number of joints j, and the number of members m. in a
perfect frame, which is stable and determinate.
m=2j–3
4.1
However, the above equation gives only a necessary, but not a
sufficient condition of a perfect frame. For example though the
two frames, as shown in Fig 4.4 (a) and (b) have the same
number of members and joints, the frame shown in fig 4.4(a)
is perfect whereas the frame shown in Fig. 4.4(b) is not capable
of retaining its shape if loaded at the joint no. 6, thus it is
unstable. Therefore, the necessary and sufficient condition of a
perfect frame, apart from satisfying above eqn. 4.1, is that it
should retain its shape when load is applied at any joint in any
A frame is said to be indeterminate if the number of members
in it are more than that required in a perfect frame and such
frames cannot be analysed by making use of the equations of
equilibrium alone. Thus, a indeterminate frame can be analysed
using the 3 equations of equilibrium.. Each extra member adds
one degree of indeterminacy. For the analysis of such frames
the consistency of deformation, which results in to more
number of required equations is to be considered. The truss
shown in the Fig 4.6 is a typical indeterminate truss. In this
truss one diagonal member in each panel is extra. Hence it is a
two-degree indeterminate frame.
m = 11, j = 6,
2j – 3 = 9,
m (=11) > (2j – 3) (=9)
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Assumptions
The Following Assumptions are Made
i. The ends of the members are pin-connected (hinged), i.e.
Moments are zero at the joints/ ends of the members. The
members carry only axial force-compression or tension.
ii. The loads act only at the joints.
iii. Self weights of the members are negligible;
iv. Cross-section of the members is uniform;
If at all the cross-section varies, the centre of gravity of the
section is assumed to be located along the same longitudinal
line.
In reality the members are connected by bolting, riveting or by
welding. No special care is taken to ensure perfect pin-
connections. However, experiments have shown that assuming
pin-connected ends in a triangulated frames, such as trusses is
quite satisfactory.
In most of the frames the loads act at the joints. Even if a load
is not acting at a joint, it can be replaced by its reaction at the
joint and a local bending effect on the member. The frame may
be analysed for the joint loads and the local bending effect on
the member superposed in the design of that member. In
most of the trusses, the self-weight is really small compared to
the external loads they carry. Hence self-weight of the members
may be neglected.
Because of the assumption of pin-connected ends, it is more
appropriate to call the theory that is going to be developed in
this chapter as analysis of pin-connected plane trusses. Analysis
of rigid frames is not covered in this course.
Types of Trusses
Different types of trusses used in the various industries are
shown in the following Fig.
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ENGINEERING MECHANICS
(m – 2j-3) = 2 is the degree of indeterminacy
ENGINEERING MECHANICS
Nature of Forces in Members
The members of a truss are subjected to either tensile or
compressive forces; a typical truss ABCDE loaded at joint E is
shown in Fig 4.7(a). The member BC is subjected to
compressive force C as shown in Fig. 4.7(b). Effect of this force
on the joint B (or C) is equal and opposite to the force C as
shown in Fig. 4.7(b).
If there are j number of joints, 2j number of equations can be
formed. There will be 3 reactions in general determinate truss.
The force in each member is unknown. Hence, if there are m
number of members, the total number of unknown will be m
+ 3. A problem can be analysed if there are as many equations
as there are unknowns. Hence, a frame analysis problem is
determined if eqn. 4.1 is satisfied (2j = m + 3). This is same as
that, for a perfect frame. Hence, a perfect frame is determinate.
If m > 2j-3, then the number of unknowns is more than
number of equations, then it is an indeterminate frame. In a
frame if, m < 2j-3, then the number of equations is more than
the number of unknowns, hence a set of solutions can satisfy
such equations i.e. there would be more than one solution,
which shows instability of the structure. Hence a deficient frame
is not stable.
Let us work out to determine the forces in all the members of
the truss shown 4.8 (a) and indicate the magnitude and nature
of forces on the background of the truss. Length of each
member is 2 m.
The member AE is subjected to tensile force T. Its effect on the
joints A and E are as shown in Fig 4.7 (b) In the analysis of
frame we mark the forces on the joints, instead of the forces in
the members as shown in fig. 4.7 (c). Note that compressive
force in a member is represented in a Fig by two arrows going
away from each other and a tensile force by two arrows coming
Fig. 4.8(a)
Now, we cannot find a joint with only two unknown forces
without finding reactions.
Consider the equilibrium of the entire frame.
R x4 40 x1 60x2 51x3 0
D
towards each other. This is because the markings on the
R 77.5
members represent the internal reactive forces developed which
D
are opposite in direction to the applied forces.
kN
H 0
H 0
A
Reaction at A is vertical only,
 V

77.5 40 60
RA72.5
kN
 0
RA
Analysis of Forces in Members of Frames
Joint A
Method of Joints

At each joint the forces in the members meeting and the loads
acting, if any, constitute a system of concurrent forces. Hence,
two equations of equilibrium can be formed at each joint. We
select a joint where there are only two unknown forces. Many
 0
F sin
F 
AB
AB

F 
AE
times, before starting with the joints we find out the reaction at
the supports by considering the equilibrium of the entire frame.
By using these two equations of equilibrium at that joint the
two unknown forces are found then, the next joint is selected
for analysis where there are only two un known forces. Thus,
the analysis proceeds from joint to joint to find the forces in all
the members.
V
F AB
50
H
 72 .5
60   R
83 .7158 kN ( comp )
A
 0

83.7158 cos 60
0
41.8579 kN (Tension )
Fig. 4.8(b)
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
Joint D 
 V
(Ans: F = - 17.32 kN; F = + 5 kN; F = - 20 kN; F = 17.32 kN; F = + 20 kN; F = - 15 kN; F = - 30 kN)
AB
 0
DC
H
0
F DE 

87 .4893 cos 60 0
kN Tension )
F  44
.7446
(
Joint B
Fig. 4.8(b)
DE
Fig. 4.10

0
F BE sin 60   F AB sin 60 

o

 F BE  72 . 5

40
0
40  37 . 5278 kN ( Tension )
o
(Ans: F = 60 kN; F = 51.96 kN; F = - 20 kN; F = - 40
AB
AC
BC
BD
kN; F = = + 40 kN; symmetry)
CD
sin 60
 0
 F
F
BD
DE
D
DC
H
BC
CE
60   R  77 .5
89 . 4893 kN ( Comp )
F sin
 F 
V
AC
CD
BC
cos 60oF
AB
o
BE
cos 60
FBC(83 .7158
37. 5274 ) 0.5
FBC 60. 6218 kN (Comp )
0
Fig. 4.8(b)
Fig. 4.11
Joint C
V
(Ans: F = - 4.5 13 kN; F = + 13.5 kN; F = + 6 kN;
F = +13.5 kN; F = - 0.5 kN; F = - 4kN; F = +8 kN)
AC
0
BD
o
F sin 60
CE
F sin 60
0
DC
77.5
F

CE

sin60 o
Notes
50
kN
31.7543(
)
Fig. 4.8(b)
Now the forces in all the members are known. A check can be
performed by analysing joint E, which will give the same result
as calculated earlier. The results are shown on the diagram of
the truss in Fig. 4.8(f).
Problems
Determine the forces in all the members of the frames shown
in Fig. 4.9 to 4.11. Indicate the nature of forces also. (Tension as
+ve and compression as –ve).
AB
CD
BC
CE
DE
ENGINEERING MECHANICS
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