Mechanical Engineering Principles

48 MECHANICAL ENGINEERING PRINCIPLES

4 kN

30 ° 60 °

Joint 1

Joint 3

Joint 2

R 1

R 2

H 2

Figure 4.30

Next, we will isolate each joint by making imagi-
nary cuts around each joint, as shown by the circles
in Figure 4.30; this will result in three free body
diagrams. The first free body diagram will be for
joint (1), as this joint has two or less unknown
forces; see Figure 4.31.

60 °

4 kN

30 °

F 2

F 1

Figure 4.31

Resolving forces horizontally:

F 1 cos 30°=4kN+F 2 cos 60°

i.e. 0. 866 F 1 = 4 + 0. 5 F 2

from which,F 1 =

4 + 0. 5 F 2

0. 866

or F 1 = 4 .619 kN+ 0. 577 F 2 ( 4. 4 )

Resolving forces vertically:

0 =F 1 sin 30°+F 2 sin 60°

i.e. 0 = 0. 5 F 1 + 0. 866 F 2

or −F 1 =

0. 866 F 2

0. 5

or F 1 =− 1. 732 F 2 ( 4. 5 )

Equating equations (4.4) and (4.5) gives:

4 .619 kN+ 0. 577 F 2 =− 1. 732 F 2

i.e. 4. 619 =− 1. 732 F 2 − 0. 577 F 2

=− 2. 309 F 2

Hence F 2 =−

4. 619

2. 309

i.e. F 2 =−2kN(compressive) ( 4. 6 )

Substituting equation (4.6) into equation (4.5) gives:

F 1 =− 1. 732 ×(− 2 )

i.e. F 1 = 3 .465 kN ( 4. 7 )

Consider next joint (2),as this joint now has two

or less unknown forces; see Figure 4.32.

F 3

R 1

30 °

F 1 = 3.465 kN

Figure 4.32

Resolving forces vertically:

R 1 +F 1 sin 30°= 0

or R 1 =−F 1 sin 30° ( 4. 8 )

Substituting equation (4.7) into equation (4.8) gives:

R 1 =− 3. 465 × 0. 5

i.e. R 1 =− 1 .733 kN (acting downwards)

Resolving forces horizontally:

0 =F 1 cos 30°+F 3

or F 3 =−F 1 cos 30° ( 4. 9 )

Substituting equation (4.7) into equation (4.9) gives:

F 3 =− 3. 465 × 0. 866

i.e. F 3 =−3kN(compressive) ( 4. 10 )

Consider next joint 3; see Figure 4.33.

F 2 = −2 kN

R 2

H 2

F 3 = −3 kN

60 ̊

Figure 4.33