Engineering Mechanics

(^284) „„„„„ A Textbook of Engineering Mechanics
Solution. The example may be solved either by the method of joints or method of sections.
But we shall solve it by the method of sections, as we have to determine forces in three members of
the frame only.
First of all pass section (1-1) cutting the truss
through the members AB, EF and CD as shown in Fig
13.62. Now consider equilibrium of the upper por-
tion of the frame. Let the directions of the forces PAB
and PCD be assumed as shown in Fig 13.62. Now con-
sider the joint F. We know that horizontal compo-
nent of 15 kN load is zero. Therefore force in mem-
ber EF is also zero. Ans.
Now taking moments of the forces acting on
the upper portion of the frame about the joint A and
equating the same,
PCD × 3 = 15 × 2 = 30
or^30 10 kN
3
PCD==^ Ans.
and now taking moments of the forces about the joint D and equating the same,
PAB × 3 = 15 × 1 = 15
or
15
5kN
AB 3
P == Ans.
Example 13.21. A framed structure of 6 m span is carrying point loads as shown in
Fig 13.63.
Fig. 13.63
Find by any method the forces in the members AC, BD and FG of the structure.
Solution. First of all, from D draw DK perpendicular to AB as shown in Fig 13.63. From the
geometry of the figure, we find that
AD = AB cos 30° = 6 × 0.866 = 5.196 m
and DK= AD sin 30° = 5.196 × 0.5 = 2.598 m
Similarly AK = AD cos 30° = 5.196 × 0.866 = 4.5 m
∴
2.598
tan 0.7423
3.5
DK
EK
α= = = or α = 36.6°
Fig. 13.62.