Mechanical Engineering Principles

50 MECHANICAL ENGINEERING PRINCIPLES

F 3

F 4

R 2 = 6.25 kN

30 °

Figure 4.36

Resolving forces vertically:

R 2 +F 4 sin 30°= 0

i.e. R 2 + 0. 5 F 4 = 0

or F 4 =−

R 2

0. 5

( 4. 15 )

SinceR 2 = 6 .25,F 4 =−

6. 25

0. 5

i.e. F 4 =− 12 .5kN(compressive)
( 4. 16 )

Resolving forces horizontally:

F 3 +F 4 cos 30°= 0

i.e. F 3 =−F 4 cos 30° ( 4. 17 )

Substituting equation (4.16) into equation (4.17)
gives:

F 3 =−(− 12. 5 )× 0. 866

i.e. F 3 = 10 .83 kN(tensile)

Consider joint (3); see Figure 4.37.

F 5

F 6

4 kN

30 °

30 °

30 °

F 1 = −11.5 kN

Figure 4.37

Resolving forces vertically:

F 6 sin 30°=F 1 sin 30°+F 5 sin 30°+ 4

i.e. F 6 =F 1 +F 5 +

4

sin 30°

( 4. 18 )

Substituting equation (4.13) into equation (4.18)

gives:

F 6 =− 11. 5 +F 5 + 8 ( 4. 19 )

Resolving forces horizontally:

F 1 cos 30°=F 5 cos 30°+F 6 cos 30°

i.e. F 1 =F 5 +F 6 ( 4. 20 )

Substituting equation (4.13) into equation (4.20)

gives:

− 11. 5 =F 5 +F 6

or F 6 =− 11. 5 −F 5 ( 4. 21 )

Equating equations (4.19) and (4.21) gives:

− 11. 5 +F 5 + 8 =− 11. 5 −F 5

or F 5 +F 5 =− 11. 5 + 11. 5 − 8

i.e. 2 F 5 =− 8

from which, F 5 =−4kN(compressive)

( 4. 22 )

Substituting equation (4.22) into equation (4.21)

gives:

F 6 =− 11. 5 −(− 4 )

i.e. F 6 =− 7 .5kN(compressive) ( 4. 23 )

Consider joint (4);see Figure 4.38.

3 kN

30 ° 30 °

F 6 =−7.5 kN F^7 F

8

Figure 4.38

Resolving forces horizontally:

F 6 cos 30°=F 8 cos 30°

i.e. F 6 =F 8

but from equation (4.23),

F 6 =− 7 .5kN

Hence, F 8 =− 7 .5kN(compressive) ( 4. 24 )