Mechanical Engineering Principles

FORCES IN STRUCTURES 43

that the membersBDandDAare in compression
and are defined as struts. It should also be noted
from Figure 4.10, that when a member of the
framework, say,BD, is so defined, we are referring
to the top joint, because we mustalways work
around a joint in a clockwise manner; thus the
arrow at the top ofBDpoints upwards, because in
Figure 4.9, the vectorbdpoints upwards frombto
d. Similarly, if the same member is referred to as
DB, then we are referring to the bottom of this
member at the joint (2), because we must always
work clockwise around a joint. Hence, at joint (2),
the arrow points downwards, because the vectordb
points downwards fromdtobin Figure 4.9.
To determine the unknown forces in the horizontal
member between joints (1) and (2), either of these
joints can be considered, as both joints now only
have two unknown forces. Let us consider joint (1),
i.e. jointADC. Now the vectoradcan be measured
from Figure 4.9 and drawn to scale in Figure 4.11.

30 °

a

d c

Figure 4.11

Now the unknown force between the spacesD
andC, namely the vectordcis horizontal and the
unknown force between the spacesCandA, namely
the vectorcais vertical, hence, by drawing to scale
and direction, the pointccan be found. This is
because the pointcin Figure 4.11 lies below the
pointaand to the right ofd.
In Figure 4.11, the vectorcarepresents the mag-
nitude and direction of the unknown reactionR 1
and the vector dc represents the magnitude and
direction of the force in the horizontal member at
joint (1); these forces are shown in Figure 4.12,
whereR 1 = 0 .82 kN anddc= 1 .25 kN.

A B

C

R 1 =ca= 0.82 kN R 2 =?

D

3 kN

−1.55 kN −2.6 kN

+1.25 kN

Figure 4.12

Comparing the directions of the internal forces in
the bottom of the horizontal member with Figure 4.3,

it can be seen that this member is in tension and

therefore, it is a tie.

The reactionR 2 can be determined by considering

joint (2), i.e. jointBCD, as shown in Figure 4.13,

where the vectorbcrepresents the unknown reaction

R 2 which is measured as 2.18 kN.

b

d c

Figure 4.13

The complete vector diagram for the whole frame-

work is shown in Figure 4.14, where it can be seen

thatR 1 +R 2 =3 kN, as required by the laws of

equilibrium. It can also be seen that Figure 4.14 is a

combination of the vector diagrams of Figures 4.9,

4.11 and 4.13. Experience will enable this problem

to be solved more quickly by producing the vector

diagram of Figure 4.14 directly.

a

b

d c

Figure 4.14

The table below contains a summary of all the

measured forces.

Member Force (kN)

bd −2.6

da −1.55

cd 1.25

R 1 0.82

R 2 2.18

Problem 6. Determine the internal forces in

the members of the truss of Figure 4.15. due

to the externally applied horizontal force of

4 kN at the jointABE.