Mechanical Engineering Principles

FORCES IN STRUCTURES 45

Problem 7. Determine the internal forces in

the pin-jointed truss of Figure 4.20.

2 m

4 kN

3 kN

5 kN

Joint (^1) Joint 2
R 1 R 2
2 m 2 m 2 m
30 ° 30 ° 30 ° 30 °
Figure 4.20
In this case, there are more than two unknowns
at every joint; hence it will first be necessary to
calculate the unknown reactionsR 1 andR 2.
To determineR 1 , take moments about joint (2):
Clockwise moments about joint( 2 )=counter-clock-
wise (or anti-clockwise) moments about joint (2)
i.e. R 1 ×8m=4kN×6m+3kN×4m
+5kN×2m
= 24 + 12 + 10 =46 kN m
Therefore, R 1 =
46 kN m
8m
= 5 .75 kN
Resolving forces vertically:
Upward forces=downward forces
i.e. R 1 +R 2 = 4 + 3 + 5 =12 kN
However,R 1 = 5 .75 kN, from above,
hence, 5 .75 kN+R 2 =12 kN
from which, R 2 = 12 − 5. 75
= 6 .25 kN
Placing these reactions on Figure 4.21, together with
the spaces between the lines of action of the forces,
we can now begin to analyse the structure.
R 1 = 5.75 kN R 2 = 6.25 kN
4 kN 5 kN
3 kN
B
A
C
D
E
F
GH
J
Figure 4.21
Starting at either jointAFEor jointDEJ, where
there are two or less unknowns, the drawing to
scale of the vector diagram can commence. It must
be remembered to work around each joint in turn,
in a clockwise manner, and only to tackle a joint
when it has two or less unknowns. The complete
vector diagram for the entire structure is shown in
Figure 4.22.
a
b
c
d
f e
g
h
j
R 1 = 5.75 kN
R 2 = 6.25 kN
Figure 4.22
The table below contains a summary of all the
measured forces.
Member Force (kN)
af −11.5
fe 10.0
jd −12.5
ej 10.9
bg −7.5
gf −4.0
ch −7.6
hg 4.6
jh −5.0
R 1 5.75
R 2 6.25