Mechanical Engineering Principles

FORCES IN STRUCTURES 47

Problem 8. Solve Problem 5, Figure 4.8 on

page 42, by the method of joints.

Firstly, assume all unknowns are in tension, as
shown in Figure 4.27.

Joint 1

F 3

F 2

F 1

3 kN

Joint 2^30 °^60 ° Joint 3

Figure 4.27

Next, make imaginary cuts around the joints, as
shown by the circles in Figure 4.27. This action
will give us three free body diagrams. The first we
consider is around joint (1), because this joint has
only two unknown forces; see Figure 4.28.

F 1 F 2

30 °

3 kN

60 °

Figure 4.28

Resolving forces horizontally at joint (1):

Forces to the left = forces to the right

i.e. F 1 cos 30°=F 2 cos 60°

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

from which, F 1 =

0. 5 F 2

0. 866

i.e. F 1 = 0. 577 F 2 ( 4. 1 )

Resolving forces vertically at joint (1):

Upward forces=downward forces

i.e. 0=3kN+F 1 sin 30°+F 2 sin 60°

i.e. 0= 3 + 0. 5 F 1 + 0. 866 F 2 ( 4. 2 )

Substituting equation (4.1) into equation (4.2) gives:

0 = 3 + 0. 5 × 0. 577 F 2 + 0. 866 F 2

i.e. − 3 = 1. 1545 F 2

from which, F 2 =−

3

1. 1545

i.e. F 2 =− 2 .6kN(compressive) ( 4. 3 )

Substituting equation (4.3) into equation (4.1) gives:

F 1 = 0. 577 ×(− 2. 6 )

i.e. F 1 =− 1 .5kN(compressive)

Consider next joint (2), as it now has two or less

unknown forces; see Figure 4.29.

R 1

F 3

F 1 = −1.5 kN

30 °

Figure 4.29

Resolving horizontally:

Forces to the left=forces to the right

i.e. 0 =F 1 cos 30°+F 3

However,F 1 =− 1 .5kN,

hence, 0 =− 1. 5 × 0. 866 +F 3

from which, F 3 = 1 .30 kN(tensile)

These results are similar to those obtained by the

graphical method used in Problem 5; see Figure 4.12

on page 43, and the table below.

Member Force (kN)

F 1 −1.5

F 2 −2.6

F 3 1.3

Problem 9. Solve Problem 6, Figure 4.15

on page 44, by the method of joints.

Firstly, we will assume that all unknown internal

forces are in tension, as shown by Figure 4.30.