LESSON 5:
ANALYSIS OF PIN JOINTED PLANE FRAMES (METHOD OF SECTION)
ENGINEERING MECHANICS
Having analysed the pin jointed frames using the method of
joints in the previous lesson, we shall apply the method of
section for analyzing the pin jointed frames.
Method of Section
In the method of section, after determining the reactions, a
section line is drawn passing through not more than 3
members in which forces are not known or required to be
found such that, the frame is cut into two separate parts. Each
part should be in equilibrium under the action of loads,
reactions & forces in the members that are cut by the section
line. Equilibrium of any one of these two parts is considered
and the unknown forces in the members cut by the section line
are determined.
Features of Method of Joints vis-à-vis Method of
Section
 The system of forces acting on either part of truss
constitutes a non-concurrent force system. Since in a nonconcurrent system of forces, there are only three independent
equations of equilibrium, we can have only three unknown
forces. Hence, in this method it is an essential condition that
the section line should pass through not more than three
members in which forces are not known and it should
separate the frame in to two parts.
 Thus, the method of section is the application of nonconcurrent force system analysis where as the method of
joints is the analysis of concurrent force system.
 Under the following two conditions the method of section
is preferred over the method of joints:
i. A large truss in which forces are to be calculated in a few
members only.
ii. In the situation where the method of joints fails to start
with the analysis.
Let us find out the magnitude & nature of forces in a few
selected members of a loaded truss shown in Fig. 5.1(a). We will
find the forces in members U3U4, L3L4 and U4L3 of the truss
shown in the figure.
R x 36 - 200 x 6 – 200 x 12 - 150 x 18 - 100 x 24 - 100 x 30 = 0,
R = 325 kN
2
2
V 0
gives, R = 200 + 200 + 150 + 100 + 100 – 325
= 425 kN
Take the section 1-1 and consider the right hand side part.
1
Fig. 5.1(b)
1 
2
6
2
= 6.0828 ,
Now, U3U4 =
FL3L4 x 8 – 325 x 12 + 100 x 6 = 0,
FL3L4 = 412.5 kN (Tension)
M = 0,FU4U3 x cos x 9 + 100 x 6 + 100 x 12 – 325 x 18 = 0
L3
1
FU4U3 = 456.2072 kN (Comp)
H = 0, FU4U3sin - FU4U3 cos  + PL4L3= 0
2
F 4L3
U
1
456.2072x0.9864 412.5
0.6
62 kN Tension)
.5
(
Problems
1. Find the force in the member FG of the triangular truss
shown in Fig. 5.2.
(Hint: Take section A – A and find force in FD. Then analyse
joint F)
[Ans. + 28 kN]
Fig. 5.1(a)
M
Now,
0;and V 0
LO
M
0 gives:
Fig. 5.2
LO
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ENGINEERING MECHANICS
2. Determine the forces in the members AB, AC, DF and CE
of the scissors truss shown in Fig. 5.3.
(Hint: Find reaction R and analyse joint A. Take
A
section A – A and find forces in DF and CE)
[Ans. F = - 6.25 W; F = 4.51 W; F = - 3.75 W; F = +
2.75 W]
AB
AC
DF
CE
Fig. 5.3
3. Find the force in member KL of the French truss shown in
Fig. 5.4.
(Hint: Take section A – A and find FLE and FDE. From
joint D find FDL. Then analyse joint L to get FKL)
[Ans. + 41.96 kN]
Fig. 5.4
Notes
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