Mechanical Engineering Principles

42 MECHANICAL ENGINEERING PRINCIPLES

4.3 Graphical method

In this case, the method described in Chapter 3 will
be used to analyse statically determinate plane
pin-jointed trusses. The method will be described
with the aid of worked examples.

Problem 5. Determine the internal forces

that occur in the plane pin-jointed truss of

Figure 4.8, due to the externally applied

vertical load of 3 kN.

A B

C

R 1 R 2

D

3 kN

30 °^60 °

Figure 4.8

Firstly, we will fill the spaces between the forces
with upper case letters of the alphabet, as shown in
Figure 4.8. It should be noted that the only reactions
are the vertical reactionsR 1 andR 2 ;thisisbecause
the only externally applied load is the vertical load
of 3 kN, and there is no external horizontal load.
The capital lettersA,B,CandDcan be used to
represent the forces between them, providing they
are taken in a clockwise direction about each joint.
Thus the lettersAB represent the vertical load of
3 kN. Now as this load acts vertically downwards,
it can be represented by a vectorab,wherethe
magnitude of ab is 3 kN and it points in the direction
froma to b.Asabis a vector, it will have a
direction as well as a magnitude. Thus ab will
point downwards fromatobas the 3 kN load acts
downwards.
This method of representing forces is known as
Bow’s notation.
To analyse the truss, we must first consider the
jointABD; this is because this joint has only two
unknown forces, namely the internal forces in the
two members that meet at the jointABD. Neither
joints BCD and CAD can be considered first,
because each of these joints has more than two
unknown forces.

Now the 3 kN load is between the spacesAand

B, so that it can be represented by the lower case

lettersab, point fromatoband of magnitude 3 kN,

as shown in Figure 4.9.

a

b

d

30 °

3 kN

60 °

Figure 4.9

Similarly, the force in the truss between the spaces

BandD, namely the vectorbd, lies at 60°to the

horizontal and the force in the truss between the

spacesDandA, namely the vectorda, lies at 30°

to the horizontal. Thus, in Figure 4.9, if the vectors

bdandadare drawn, they will cross at the pointd,

where the pointdwill obviously lie to the left of

the vectorab, as shown. Hence, if the vectorabis

drawn to scale, the magnitudes of the vectorsbdand

dacan be measured from the scaled drawing. The

direction of the force in the member between the

spacesBandDat the jointABDpoint upwards

because the vector fromb tod points upwards.

Similarly, the direction of the force in the member

between the spacesDandAat the jointABDis

also upwards because the vector fromdtoapoints

upwards. These directions at the jointABD are

shown in Figure 4.10. Now as the framework is in

equilibrium, the internal forces in the membersBD

andDAat the joints (2) and (1) respectively, will

be equal and opposite to the internal forces at the

jointABD; these are shown in Figure 4.10.

A B

C

R 1 R 2

Joint

D

Joint 1 2

3 kN

−1.55 kN −2.6 kN

Figure 4.10

Comparing the directions of the arrows in

Figure 4.10 with those of Figure 4.3, it can be seen