(^250) „„„„„ 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
5kN
AC sin 30 0.5
P ===
°
(Compression)
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 4.33 kN
BC 3.75 tan 30 0.577
P
×

°
(Tension)
...(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
Example 13.2. Fig 13.8 shows a Warren girder consisting of seven members each of 3 m
length freely supported at its end points.
Fig. 13.8.
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.
Solution. Taking moments about A and equating the same,
RD × 6 = (2 × 1.5) + (4 × 4.5) = 21
∴
21
3.5 kN
D 6
R ==
and RA = (2 + 4) – 3.5 = 2.5 kN
† For details, please refer to the foot note on last page.