Contents

Chapter 13 : Analysis of Perfect Frames (Analytical Method)

Though there are many types of frames, yet from the analysis point of view, the frames may be
classified into the following two groups:
1. Perfect frame. 2. Imperfect frame.

A perfect frame is that, which is made up of members just suf-

ficient to keep it in equilibrium, when loaded, without any change in
its shape.
The simplest perfect frame is a triangle, which contains three
members and three joints as shown in Fig. 13.1. It will be intersting to
know that if such a structure is loaded, its shape will not be distorted.
Thus, for three jointed frame, there should be three members to prevent
any distortion. It will be further noticed that if we want to increase a
joint, to a triangular frame, we require two members as shown by dot-
ted lines in Fig. 13.1. Thus we see that for every additional joint, to a
triangular frame, two members are required.
The no. of members, in a perfect frame, may also be expressed by the relation :
n = (2j – 3)
n = No. of members, and
j = No. of joints.

An imperfect frame is that which does not satisfy the equation :

n = (2j – 3)
Or in other words, it is a frame in which the no. of members are more or less than (2j – 3). The
imperfect frames may be further classified into the following two types :
1. Deficient frame. 2. Redundant frame.

A deficient frame is an imperfect frame, in which the no. of members are less than (2j – 3).

A redundant frame is an imperfect frame, in which the no. of members are more than (2j – 3).
In this chapter, we shall discuss only perfect frames.

When a body is acted upon by a force, the internal force which is transmitted through the body is
known as stress. Following two types of stress are important from the subject point of view :
1. Tensile stress. 2. Compressive stress.
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A Textbook of Engineering Mechanics

Sometimes, a body is pulled outwards by two equal and opposite forces and the body tends to
extend, as shown in Fig 13.2. (a). The stress induced is called tensile stress and corresponding force
is called tensile force.

Sometimes, a body is pushed inwards by two equal and opposite forces and the body tends to
shorten its length as shown in Fig. 13.2 (b). The stress induced is called compressive stress and the
corresponding force is called compressive force.

Following assumptions are made, while finding out the forces in the members of a perfect
frame:
1. All the members are pin-jointed.
2. The frame is loaded only at the joints.
3. The frame is a perfect one.
4. The weight of the members, unless stated otherwise, is regarded as negligible in
comparison with the other external forces or loads acting on the truss.
The forces in the members of a perfect frame may be found out either by analytical method
or graphical method. But in this chapter, we shall discuss the analytical method only.

The following two analytical methods for finding out the forces, in the members of a perfect
frame, are important from the subject point of view :
1. Method of joints. 2. Method of sections.

In this method, each and every joint is treated as a free body in equilibrium as shown in Fig.
13.3 (a), (b), (c) and (d). The unknown forces are then determined by equilibrium equations viz.,
V = 0 and H = 0. i.e., Sum of all the vertical forces and horizontal forces is equated to zero.
1. The members of the frame may be named either by Bow’s methods or by the joints at their
ends.
2. While selecting the joint, for calculation work, care should be taken that at any instant, the
joint should not contain more than two members, in which the forces are unknown.
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Chapter 13 : Analysis of Perfect Frames (Analytical Method)

This method is particularly convenient, when the forces in a few members of a frame are
required to be found out. In this method, a section line is passed through the member or members, in
which the forces are required to be found out as shown in Fig. 13.4 (a). A part of the structure, on any
one side of the section line, is then treated as a free body in equilibrium under the action of external
forces as shown in Fig. 13.4 (b) and (c).

The unknown forces are then found out by the application of equilibrium or the principles
of statics i.e., 0.
1. To start with, we have shown section line 1-1 cutting the members AB and BC. Now in
order to find out the forces in the member AC, section line 2-2 may be drawn.
2. While drawing a section line, care should always be taken not to cut more than three
members, in which the forces are unknown.

Finally, the results are tabulated showing the members, magnitudes of forces and their
nature. Sometimes, tensile force is represented with a + ve sign and compressive force with a
– ve sign.
The force table is generally prepared, when force in all the members of a truss are
required to be found out.
The truss ABC shown in Fig. 13.5 has a span of 5 metres. It is carrying a
load of 10 kN at its apex.

Find the forces in the members AB, AC and BC.

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A Textbook of Engineering Mechanics

From the geometry of the truss, we find that the load of 10 kN is acting at a
distance 1.25 m from the left hand support i.e., B and 3.75 m from C. Taking moments about B and
equating the same,
RC × 5 = 10 × 1.25 = 12.5
12.5
RC 2.5 kN
5
and RB = 10 – 2.5 = 7.5 kN
The example may be solved by the method of joints or by the method of sections. But we shall
solve it by both the methods.

First of all consider joint B. Let the *directions of the forces PAB and PBC (or PBA and PCB) be
assumed as shown in Fig 13.6 (a).

Resolving the forces vertically and equating the same,

PAB sin 60° = 7.5
7.5 7.5
or PAB 8.66 kN (Compression)
sin 60 0.866
and now resolving the forces horizontally and equating the same,
PBC = PAB cos 60° = 8.66 × 0.5 = 4.33 kN (Tension)

The idea, of assuming the direction of the force PAB to be downwards, is that the vertical component of
the force PBC is zero. Therefore in order to bring the joint B in equilibrium, the direction of the force PAB must
be downwards, or in other words, the direction of the force PAB should be opposite to that of the reaction RB. If,
however the direction of the force PAB is assumed to be upwards, then resolving the forces vertically and equating
the same,
PAB sin 60° = –7.5 (Minus sign due to same direction of RB and PAB.)
7.5 7.5
PAB 8.66 kN
sin 60 0.866
Minus sign means that the direction assumed is wrong. It should have been downwards instead of
upwards. Similarly, the idea of assuming the direction of the force PBC to be towards right is that the horizontal
component of the reaction RB is zero. Therefore in order to bring the joint B in equilibrium, the direction of the
force PAB must be towards right (because the direction of the horizontal component of the force PAB is towards
left).
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Chapter 13 : Analysis of Perfect Frames (Analytical Method)

Now consider the joint C. Let the *directions of the forces PAC and PBC (or PCA and PCB) be
assumed as shown in Fig. 13.6 (b). Resolving the forces vertically and equating the same,
PAC sin 30° = 2.5
2.5 2.5
PAC 5.0 kN (Compression)
sin 30 0.5
and now resolving the forces horizontally and equating the same,
PBC = PAC cos 30° = 5.0 × 0.866 = 4.33 kN (Tension).
...(As already obtained)

First of all, pass section (1-1) cutting the truss into two parts (one part shown by firm lines and
the other by dotted lines) through the members AB and BC of the truss as shown in Fig 13.7 (a). Now
consider equilibrium of the left part of the truss (because it is smaller than the right part). Let the
directions of the forces PAB and PAC be assumed as shown in Fig 13.7 (a).
Taking** moments of the forces acting in the left part of the truss only about the joint C and
equating the same,
PAB × 5 sin 60° = 7.5 × 5
7.5 5 7.5
PAB 8.66 kN (Compression)
5 sin 60 0.866
and now taking moments of the forces acting in the left part of the truss only about the joint A and
equating the same,
PBC × 1.25 tan 60° = 7.5 × 1.25

7.5 1.25 7.5

PBC 4.33 kN (Tension)
1.25 tan 60 1.732

* For details, please refer to the foot note on last page.

** The moment of the force PAB about the joint C may be obtained in any one of the following two ways :
1. The vertical distance between the member AB and the joint C (i.e., AC in this case) is equal to
5 sin 60° m. Therefore moment about C is equal to PAB × 5 sin 60° kN-m.
2. Resolve the force PAB vertically and horizontally at B. The moment of horizontal component about
C will be zero. The moment of vertical component (which is equal to PAB × sin 60°) is equal to
PAB × sin 60° × 5 = PAB × 5 sin 60° kN-m.
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A Textbook of Engineering Mechanics

Now pass section (2-2) cutting the truss into two parts through the members AC and BC. Now
consider the equilibrium of the right part of the truss (because it is smaller than the left part). Let the
†direction of the forces PAC and PBC be assumed as shown in Fig 13.7 (b).
Taking moments of the force acting in the right part of the truss only about the joint B and
equating the same,
PAC × 5 sin 30° = 2.5 × 5
2.5 2.5
PAC 5 kN (Compression)
sin 30 0.5
and now taking moments of the forces acting in the right part of the truss only about the joint A and
equating the same,
PBC × 3.75 tan 30° = 2.5 × 3.75

2.5 3.75 2.5

PBC 4.33 kN (Tension)
3.75 tan 30 0.577
...(As already obtained)
Now tabulate the results as given below :
S.No. Member Magnitude of force in kN Nature of force
1 AB 8.66 Compression
2 BC 4.33 Tension
3 AC 5.0 Compression

Fig 13.8 shows a Warren girder consisting of seven members each of 3 m

length freely supported at its end points.

The girder is loaded at B and C as shown. Find the forces in all the members of the girder,
indicating whether the force is compressive or tensile.
Taking moments about A and equating the same,
RD × 6 = (2 × 1.5) + (4 × 4.5) = 21
21
RD 3.5 kN
6
and RA = (2 + 4) – 3.5 = 2.5 kN

† For details, please refer to the foot note on last page.