Mechanical Engineering Principles

FORCES IN STRUCTURES 51

Resolving forces vertically:

0 = 3 +F 6 sin 30°+F 7 +F 8 sin 30°

i.e. 0= 3 + 0. 5 F 6 +F 7 + 0. 5 F 8 ( 4. 25 )

Substituting equations (4.23) and (4.24) into equa-
tion (4.25) gives:

0 = 3 + 0. 5 ×− 7. 5 +F 7

+ 0. 5 ×− 7. 5

or F 7 =− 3 + 0. 5 × 7. 5 + 0. 5 × 7. 5

=− 3 + 7. 5

from which, F 7 = 4 .5kN(tensile) ( 4. 26 )

Consider joint (5); see Figure 4.39.

F 4 = −12.5 kN

F 8 = −7.5 kN

F 9

5 kN

30 ° 30 °

30 °

Figure 4.39

Resolving forces horizontally:

F 8 cos 30°+F 9 cos 30°=F 4 cos 30°

i.e. F 8 +F 9 =F 4 ( 4. 27 )

Substituting equations (4.24) and (4.16) into equa-
tion (4.27) gives:

− 7. 5 +F 9 =− 12. 5

i.e. F 9 =− 12. 5 + 7. 5

i.e. F 9 =−5kN(compressive)

The results compare favourably with those obtained
by the graphical method used in Problem 7; see the

table below.

Member Force (kN)

F 1 −11.5

F 2 9.96

F 3 10.83

F 4 −12.5

F 5 −4.0

F 6 −7.5

F 7 4.5

F 8 −7.5

F 9 −5.0

Now try the following exercise

Exercise 21 Further problems on the

method of joints

Using the method of joints, determine the

unknown forces for the following pin-jointed

trusses:

1. Figure 4.23 (page 46)
   ⎡

⎢

⎢

⎣

R 1 = 3 .0kN, R 2 = 1 .0kN,

1–2, 1.73 kN, 1–3,− 3 .46 kN,

2–3,− 2 .0kN

⎤

⎥

⎥

⎦

2. Figure 4.24 (page 46)
   ⎡

⎢

⎢

⎣

R 1 =− 2 .61 kN, R 2 = 2 .61 kN,

H 2 = 6 .0kN, 1–2,− 1 .5kN,

1–3, 3.0kN, 2–3,− 5 .2kN

⎤

⎥

⎥

⎦

3. Figure 4.25 (page 46)
   ⎡

⎢

⎢

⎣

R 1 = 5 .0kN, R 2 = 1 .0kN,

H 1 = 4 .0 kN, 1–2, 1.0 kN,

1–3,− 7 .07 kN, 2–3,− 1 .41 kN

⎤

⎥

⎥

⎦

4. Figure 4.26 (page 46)
   ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
   R 1 = 5 .0kN, R 2 = 7 .0kN,
   1–3,− 10 .0kN, 1–6,− 8 .7kN,
   3–4,− 8 .0kN, 3–6,− 2 .0kN,
   4–6, 4.0 kN 4–5, 8.0 kN,
   5–6,−6kN, 5–2,−14 kN,
   6–2, 12.1 kN

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦