(^248) A Textbook of Engineering Mechanics
Solution. 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
2.5 kN
C 5
R ==
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.
Methods of Joints
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).
Fig. 13.6.
Resolving the forces vertically and equating the same,
PAB sin 60° = 7.5
or
7.5 7.5
AB sin 60 0.866
P ==
°^
=8.66 kN(Compression)
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 8.66
sin 60 0.866
===−−−
AB °
P kN
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).