(^272) „„„„„ A Textbook of Engineering Mechanics
Example 13.13. Figure 13.45 shows a pin-jointed frame carrying a vertical load at B and
a horizontal load at D
Fig. 13.45.
Find the forces in the members DF, HE and DH of the frame.
Solution. Since the frame is supported on rollers at the right hand support (E), therefore the
reaction at this support will be vertical (because of horizontal support). The reaction at the left hand
support (A) will be the resultant of vertical and horizontal forces and inclined with the vertical.
Taking moments about the joint* A and equating the same,
RE × 8 = (3 × 2) + (4 × 4.5) = 24
∴
24
3kN
E 8
R ==
From the geometry of the figure, we find that
3
tan 0.75
4
θ= = or θ = 36.9°
4.5
tan 2.25
2
α= = or α = 66°
The example may be solved either by the method of joints or method
of sections. But we shall solve it by the method of joints, as we can resolve
the force in the members at joint E in which the force are required to be
found out. Now consider the point E. Let the directions of the forces PDE and
PHE be assumed as shown in Fig. 13.46.
Resolving the forces horizontally and equating the same,
PDE cos 66° = PHE cos 36.9° = PHE × 0.8
∴
0.8 0.8
1.97
cos 66 0.4062
HE HE
DE HE
PP
PP
××

°
and now resolving the forces vertically and equating the same,
PDE sin 66° =PHE sin 36.9° +3
1.97 PHE × 0.9137 = (PHE × 0.6) + 3
1.2 PHE=3
or
3
2.5 kN (Tension)
HE 1.2
P ==
and PDE = 1.97 PHE = 1.97 × 2.5 = 4.93 (Compression)
Fig. 13.46.

• There are no need of finding out the vertical and horizontal reaction at A, as we are not considering this
  part of the truss.