GARZOTA II MZ 156 SOLAR 18-ISLA SANTA
CALLE 10 DE AGOSTO S/N Y CALLE
We shall now consider special but important force systems and will establish
the resultants possible for each case. Examples will serve to illustrate
the method of procedure.
Case A. Coplanar Force Systems.
In Fig. 4.14, a system of forcesand couples is shown in plane A. By moving the forces a common point a
in plane A, we will form additional couples in the plane. The force portion of the equivalent system at such a point will he given as
= (4.6)
Figure 4.14. Coplanar force system.
The couple moment portion of the equivalent system can be given
(using the right hand rule proper signs):
=
+
+
..
+
+ . .
. (4.7)where etc., are perpendicular distances from point to the lines of
action of the noncouple forces, and C,, etc.. are the values of the given
couple moments. resultant at is shown in Fig. 4.15.
0 ) we can move the force
from to a new parallel position as to introduce a second couple
moment to cancel Fig. 4.15 i n the described earlier in Section
4.2. Since the and directions used are arhitrary, except for the condition
that they be in plane of the forces, we can make following conclusion.
in direction in plane add rhun zero,
may replace the entire by a specific
= Without a force at point
a, we can no longer eliminate a couple in plane A . Thus, our second conclu- sion is that if
be
In coplanar case, therefore, the simplest equivalent force system
must be a single force along specific line of action, a single couple moment,
or a null vector. The following example used to illustrate the method of such a resultant directly without the intermediate steps followed
in this discussion.
If 0, if 0
Figure at point
What happens if = and
P
SECTION 4.4 SIMPLEST RESULTANTS OF SPECIAL FORCE SYSTEMS
107
Example
Consider a coplanar force system shown in Fig. 4.16. The simplest resul-
tant is to be fnund. Since are not zero, we know that we
can replace the system by a single force, which is
I' P
We now need find the line of action in the plane that will make this sin-
gle force equivalent to the given system. To be equivalent for rigid-body mechanics, this force without a couple moment must have the same
ing about any point or axis in space as that of the given system.
Now the simplest resultant force must intercept the axis at some point
We can by equating the moment of the resultant force without
a couple moment about the origin with that of the original system of forces
and couples. Using the vector as a position vector from the origin to the
line of action of (see Fig. we accordingly have
X (6i =
+
2 j )(6i
+
3 j )+
(Si+
3 j ) - 30kCarrying out the cross products,
24k 12k
+
- 30k =Hence.
By specifying the intercept, we fully determine the line of action of the
simplest resultant force. We could have also used the intercept with t h e y axis, for this purpose. In that case, the position vector from the origin
to the line of action is and we have, on equating moments about 0
of the resultant without a couple moment with that of the original system:
X (6i
+
= 2 j ) X (6i+
3 j )+
(Si+
3 j ) X -Figure 4.16. Find simplest resultant
Figure 4.17. Simplest resultant.
CHAPTER 4 EQUIVALENT FORCE SYSTEMS
Example 4.7
Compute the resultant for the loads shown acting on the i n Fig. 4. I intercept with the axis.
.
75
Figure 4.18. Find resultant.
I t i s immediately apparent on inspection of the diagram that
-
75jhe the intercept with the axis of the line o f action of when this line o f action corresponds to zero couple moment Fig. 4. I n Fig. we have decomposed along this line of action into gular components so as to permit simple calculations moments about the origin (here we mean moments about the axis). Accordingly, equating moments about the axis of without a couple moment, with that the original system o f loads, we get,
= 50
= 2.37 m
Thus, the simplest resultant i s a force N intercepting the axis at a position = 2.37
As pointed out earlier, i n the instance wherein we
hly have as the simplest resultant a couple moment normal to the o f the coplanar force system. There i s the possibility that there i s zero couple moment, i n which the coplanar system
cancel each other’s effects on a rigid body. To find the couple moment for
case where = we take o f the coplanar force system ahout i n space. This moment, if i t i s not equal to zero, i s clearly the couple-moment vector sought.
SECTION 4.4 SIMPLEST RESULTANTS OF SPECIAL FORCE SYSTEMS
109
Example 4.8
What is the simplest resultant for the forces shown acting on beam AB in
Fig.
Figure 4.19. Cuplanar loading a simply beam
Our first step will he compute the resultant force adding up the
force vectors. Thus
=
-
-+
-Collecting terms, we have
(-666.2
+
1,373.3 -+
- = 0The simplest resultant clearly must he either a couple moment or be a null
vector. For this information, we shall take moments about point A.
= - -
+
-+
+
-=
+
- -+
=We have a couple moment in the minus direction having any arbitrary
line of action the simplest resultant.
It is important thet the nature of the equivalence just instituted be
clearly understood. Thus, for finding the supporting force system, we can use
the undeformed geometry and hence the single force replacement. However, for finding the deflection of the heam, it should be obvious that the replace- ment is invalid. Note, finally, that there is only one point on the beam will allow for a single force to he equivalent to the original system for pur- poses of rigid-body considerations.
4 SYSTEMS
4.20. Parallel system
Figure 4.21.
parallel system.