Mechanical Engineering Principles

FORCES IN STRUCTURES 49

Resolving forces vertically:

F 2 sin 60°+R 2 = 0

i.e. R 2 =−F 2 sin 60° ( 4. 11 )

Substituting equation (4.6) into equation (4.11) gives:

R 2 =−(− 2 )× 0. 866

i.e. R 2 = 1 .732 kN(acting upwards)

Resolving forces horizontally:

F 3 +F 2 cos 60°+H 2 = 0

i.e. H 2 =−F 3 −F 2 × 0. 5 ( 4. 12 )

Substituting equations (4.6) and (4.10) into equa-
tion (4.12) gives:

H 2 =−(− 3 )–(− 2 )× 0. 5

i.e. H 2 =4kN

These calculated forces are of similar value to those
obtained by the graphical solution for Problem 6, as
shown in the table below.

Member Force (kN)

F 1 3.47

F 2 −2.0

F 3 −3.0

R 1 −1.73

R 2 1.73

H 2 4.0

Problem 10. Solve Problem 7, Figure 4.20

on page 45, by the method of joints.

Firstly, assume all unknown member forces are in
tension, as shown in Figure 4.34.

30 ° 30 ° 30 ° 30 °

60 ° 60 °

3 kN

4 kN 5 kN

R 1 = 5.75 kN R 2 = 6.25 kN

Joint 1

Joint 3

Joint^4

Joint^5

Joint 2

Figure 4.34

Next, we will isolate the forces acting at each

joint by making imaginary cuts around each of the

five joints as shown in Figure 4.34.

As there are no joints with two or less unknown

forces, it will be necessary to calculate the unknown

reactionsR 1 andR 2 prior to using the method of

joints.

Using the same method as that described for the

solution of Problem 7, we have

R 1 = 5 .75 kNandR 2 = 6 .25 kN

Now either joint (1) or joint (2) can be considered,

as each of these joints has two or less unknown

forces.

Consider joint (1); see Figure 4.35.

R 1 = 5.75 kN

F 2

F 1

30 °

Figure 4.35

Resolving forces vertically:

5. 75 +F 1 sin 30°= 0

i.e. F 1 sin 30°=− 5. 75

or 0. 5 F 1 =− 5. 75

i.e. F 1 =−

5. 75

0. 5

i.e. F 1 =− 11 .5kN(compressive)

( 4. 13 )

Resolving forces horizontally:

0 =F 2 +F 1 cos 30°

i.e. F 2 =−F 1 cos 30° ( 4. 14 )

Substituting equation (4.13) into equation (4.14)

gives:

F 2 =−F 1 cos 30°=−(− 11. 5 )× 0. 866

i.e. F 2 = 9 .96 kN(tensile)

Consider joint (2); see Figure 4.36.