Civil Engineering Reference
In-Depth Information
2
8
5
3
7
1
6
F
IGURE
4.22
Force polygon for the truss of Ex. 4.5
4
insert arrows on the members AB and AE in Fig. 4.21 to indicate compression and
tension, respectively.
We next consider joint Bwhere there are now just twounknownmember forces sincewe
have previously determined the force in the member AB; note that, moving clockwise
round B, this force is represented by the vector 62, whichmeans that it is acting towards
B as it must since we have already established that AB is in compression. Rather than
construct a separate force polygon for the joint B we shall superimpose the force
polygon on that constructed for joint A since the vector 26 (or 62) is common to both;
we thereby avoid repetition. Thus, through the point 2, we draw a vector 23 vertically
downwards to represent the 2 kN load to the same scale as before. The force in the
member BC is represented by the vector 37 parallel to BC (or CB) while the force in
the member BE is represented by the vector 76 drawn in the direction of BE (or EB);
this locates the point 7 in the force polygon. Hence we see that the force in BC (vector
37) acts towards B indicating compression, while the force in BE (vector 76) acts away
from B indicating tension; again, arrows are inserted in Fig. 4.21 to show the action of
the forces.
Now we consider joint C where the unknown member forces are in CD and CE. The
force in the member CB at C is represented in magnitude and direction by the vector
73 in the force polygon. From the point 3 we draw a vector 34 vertically downwards to
represent the 3 kN load. The vectors 48 and 87 are then drawn parallel to the members
CD and CE and represent the forces in the members CD and CE, respectively. Thus
we see that the force in CD (vector 48) acts towards C, i.e. CD is in compression,
while the force in CE (vector 87) acts away from C indicating tension; again we insert
corresponding arrows on the members in Fig. 4.21.
Finally the vector 45 is drawn vertically upwards to represent the vertical reaction
(
=
3
.
25 kN) at D and the vector 58, which must be parallel to the member DE, inserted
