Mechanical Engineering Principles

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52 MECHANICAL ENGINEERING PRINCIPLES

4.5 The method of sections


(a mathematical method)


In this method, an imaginary cut is made through
the framework and the equilibrium of this part of
the structure is considered through a free body
diagram. No more than three unknown forces can
be determined through any cut section, as only three
equilibrium considerations can be made, namely

(a) resolve forces horizontally

(b) resolve forces vertically


(c) take moments about a convenient point.

Worked Problem 11 demonstrates the method of
sections.

Problem 11. Determine the unknown
member forcesF 2 ,F 5 andF 6 of the truss of
Figure 4.34, Problem 10, by the method of
sections (whereF 2 ,F 5 andF 6 are defined in
the solution on pages 49 to 51).

Firstly, all members will be assumed to be in tension
and an imaginary cut will be made through the
framework, as shown by Figure 4.40.

B

A 30 ° 30 ° 30 ° 30 °

60 ° 60 °

4 kN

3 kN

5 kN

R 1 = 5.75 kN R 2 = 6.25 kN

Imaginary
cut

Figure 4.40

Taking moments about B;see Figure 4.41.

Clockwise moments=anti-clockwise moments

Hence, 5.75 kN×2m=F 2 × 1 .155 m

where 2 tan 30°= 1 .155 m (from Figure 4.41)

i.e. F 2 =

5. 75 × 2
1. 155

= 9 .96 kN(tensile) ( 4. 28 )

1.155 m

4 kN
B

F 6

F 5

F 2

R 1 = 5.75 kN

A

2 m

30 °

30 °

30 °

Figure 4.41

Resolving forces vertically:

5 .75 kN+F 6 sin 30°=F 5 sin 30°+4kN

i.e. F 5 =F 6 +

5. 75
0. 5


4
0. 5
i.e. F 5 =F 6 + 3. 5 ( 4. 29 )

Resolving forces horizontally:

0 =F 2 +F 5 cos 30°+F 6 cos 30°

from which, F 5 cos 30°=−F 2 −F 6 cos 30°

and F 5 =−

F 2
cos 30°

−F 6 ( 4. 30 )

Substituting equation (4.28) into equation (4.30)
gives:

F 5 =−

9. 96
0. 866

−F 6

or F 5 =− 11. 5 −F 6 ( 4. 31 )

Equating equation (4.29) to equation (4.31) gives:

F 6 + 3. 5 =− 11. 5 −F 6

from which, 2F 6 =− 11. 5 − 3. 5 =− 15

and F 6 =−

15
2
=− 7 .5kN(compressive)
( 4. 32 )

Substituting equation (4.32) into equation (4.31)
gives:

F 5 =− 11. 5 −(− 7. 5 )

=−4kN(compressive) ( 4. 33 )
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