(^318) „„„„„ A Textbook of Engineering Mechanics
S.No. Member Magnitude of force in kN Nature of force
1 BF 2.9 Compression
2 FE 3.3 Tension
3 CG 1.9 Compression
4 FG 2.3 Compression
5 GH 1.15 Tension
6 HD 2.3 Compression
7 HI 0—
8 ID 2.3 Compression
9 IE 1.33 Tension
14.11. METHOD OF SUBSTITUTION
Sometimes work of drawing the vector diagram is held up, at a joint which contains more than
two unknown force members and it is no longer possible to proceed any further for the construction
of vector diagram. In such a situation, the forces are determined by some other method. Here we shall
discuss such cases and shall solve such problem by the method of substitution.
Example 14.22. A french roof truss is loaded as shown in Fig. 14.53.
Fig. 14.53.
Find the forces in all the members of the truss, indicating whether the member is in tension or
compression.
Solution. Since the truss and loading is symmetrical, therefore both the reactions will be
equal.
∴ 12
100 200 200 200 200 200 200 200 100
N
2
RR
++++++++

= 800 N
First of all, draw the space diagram and name all the members according to Bow’s notations
and also name the joints as shown in Fig. 14.54 (a).
While drawing the vector diagram, it will be seen that the vector diagram can be drawn for
joint Nos. 1, 2 and 3 as usual. Now when we come to joint No. 4, we find that at this joint there are
three members (namely DP, PO and ON) in which the forces are unknown. So we cannot draw the
vector diagram for this joint.
Now, as an alternative attempt, we look to joint No. 5. We again find that there are also three
members (namely NO, OR and RK) in which the forces are unknown. So we can not draw the vector
diagram for this joint also. Thus we find that the work of drawing vector diagram is held up beyond
joint No. 3. In such cases, we can proceed by the substitution of an imaginary member.