Modul IX:
SAMPLE PROBLEM 9.1
Three cables are attached to a bracket as shown.
cables with an equivalent force-couple system at A.
Replace the forces exerted by the
Solution. We first determine the vectors Δr joining point A with the points of application of
the forces and resolve the forces into rectangular components. Observing that FB = (350
lb)λBE where
BE 
BE 1
 (3i  6 j  2k )
BE 7
we have
rB  AB  = 3i + 2k
FB = 150i – 300j + 100k
rC  AC  3i - 2k
Fc = 354i
rD  AD  4i - 4j
FD = 300i + 520j
- 354k
The force-couple system at A equivalent to the given forces consists of a force R = ΣF
and a couple M AR = Σ(Δr x F). The force R is readily obtained by adding respectively
the x, y, and z components of the forces:
R =ΣF== (804 lb)i + (220 lb)j - (254 lb)k
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Dr. Ir. Abdul Hamid M.Eng.
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The computation of M AR will be facilitated if we express the moments Δr x F in the form
of determinants :
Adding the expressions obtained,we have:
M AR = Σ(Δr x F) = (600 Ib • in.)i + (354 Ib • in.)j + (2380 Ib • in.)k
The rectangular components of the force R and the couple M AR
shown in the adjoining sketch.
SAMPLE PROBLEM 9.2
A rectangular slab, 5 by 7.5 m, supports five columns which exerton the slab the forces
indicated. Determine the magnitude and poin of application of the single force equivalent
to the given forces.
Solution. We shall first reduce the given system of forces to force-couple system at
the origin O of the coordinates. This couple system consists of a force R and a couple
M OR defined as follows:
R
M O = Σ(r x F)
R = ΣF
The position vectors of the points of application of the various are determined and the
computations are arranged in tabular form:
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Dr. Ir. Abdul Hamid M.Eng.
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R
Since the force R and the couple vector M O are mutually perpendicular, the forcecouple system obtained may be reduced further single force R. The new point of
application of R will be selecldl the plane of the slab and in such a way that the moment
R
of R about O will be equal to M O . Denoting by r the position vector of the desired
point of application, and by x and z its coordinates, we write
R
r x R= M O
(xi + zk) x (-24j) = 76i - 82k
-24xk + 24zi = 76i - 82k
from which it follows that
-24x = -82
x = 3.42 m
24z = 76
z = 3.17m
We conclude that the resultant of the given system of forces is R = 24 kN , at x = 3.42 rn
, z = 3.17m.
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Dr. Ir. Abdul Hamid M.Eng.
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PROBLEMS
9.1 Two parallel forces P and Q are applied at the ends of a beam AB of length L. Find
the distance x from A to the line of action their resultant. Check the formula obtained by
assuming L =8 in. and (a) P = 10 Ib down, Q = 30 Ib down; (b) P - 10 Ib down, Q=30lb
up.
9.2 Determine the distance from point A to the line of action of the resultant of the three
forces shown when (a) a =1 m,(b) a=1.5m, (c) a = 2.5 m.
4.4. Equilibrium of a Rigid Body In Two Dimensions. The conditions stated in Sec.
4.1 for the equilibrium of rigid body become considerably simpler in the case of a two
dimensional structure. Choosing the x and y axes in the plane of the structure, we have
Fz = 0
Mx = My = 0
Mz = M0
for each of the forces applied to the structure. Thus, the six equations of equilibrium
derived in Sec. 4.1 reduce to
ΣFx = 0
ΣFy = 0
ΣM0 = 0 ………(9.1)
and to three trivial identities 0=0. Since the third of the equations (9.1) must be satisfied
regardless of the choice of Q origin O, we may write the equations of equilibrium for a
two dimensional structure in the more general form
ΣFx = 0
ΣFy = 0
ΣMA = 0………(9.2)
where A is any point in the plane of the structure. The three equations obtained may be
solved for no more than three unknowns.
We saw in the preceding section that unknown forces usual consist of reactions, and
that the number of unknowns correspending to a given reaction depends upon the type
of support or connection causing that reaction. Referring to Sec. 4.3 check that the
equilibrium equations (9.2) may be used to determine the reactions of two rollers and
one cable, or of one fixed support, or of one roller and one pin in a fitted hole, etc.
Consider, for instance, the truss shown in Fig. 9.1, is subjected to the given forces P, Q,
and S. The truss is held in place by a pin at A and a roller at B. The pin prevents point A
from moving by exerting on the truss a force which may resolved into the components Ax
and Ay; the roller keeps truss from rotating about A by exerting the vertical force. The
free-body diagram of the truss is shown in Fig. 9.2(a) includes the reactions Ax, AY, and B
as well as the applied fo P, Q, S, and the weight W of the truss. Expressing that sum of
the moments about A of all the forces shown in 9.2(b) is zero, we write the equation
ΣMA=0, which may solved for the magnitude B since it does not contain Ax or Ay .
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Dr. Ir. Abdul Hamid M.Eng.
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Expressing, then, that the sum of the x components and y of the components of the
forces are zero, we write the expressionsΣFx=0 and
ΣFy =0, which
may be solved for the components Ax and AY, respectively.
Fig.9.1
Additional equations could be obtained by expressing that sum of the moments of the
external forces about points O.
We could write, for instance, ΣMB = 0. Such statement, however, does not contain any
new information, since it has already been established that the system of the forces
shown in Fig. 4.2b is equivalent to zero. The additional equation is not independent and
cannot be used to determine a fourth unknown. It will be useful, however, for checking
the solution obtained from the original three equations of equilibrium. While the three
equations of equilibrium cannot be augmented as additional equations, any of them may
be replaced by another equation. Therefore, an alternate system of equations of equilibrium is
ΣFx = 0 Σ,MA = 0
ΣMB = 0……. (9.3)
where the line AB is chosen in a direction different from the Fj direction (Fig.9.2b). These
equations are sufficient conditions or the equilibrium of the truss. The first two equations
indicate that the external forces must reduce to a single vertical force at A. Since the
third equation requires that the moment of this force be zero about a point B which is not
on its line of action, the force must be zero and the rigid body is in equilibrium. A third
possible set of equations of equilibrium is
ΣMA = 0
ΣMB = 0
ΣMC = 0 …….(9.4)
where the points A, B, and C are not in "a straight line (Fig. (4.2b). The first equation
requires that the external forces reduce to a single force at A; the second equation
requires that this force pass through B; the third, that it pass through C. Since the points
A, B, C are not in a straight line, the force must be zero, and the rigid body is in
equilibrium.
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Dr. Ir. Abdul Hamid M.Eng.
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The equation ΣMA = 0, which expresses that the sum of the moments of the forces about
pin A is zero, possesses a more definite physical meaning than either of the other two
equations (9.4). These two equations express a similar idea of balance, but with respect
to points about which the rigid body is not actually hinged. They are, however, as useful
as the first equation, and our choice of equilibrium equations should not be unduly influenced by the physical meaning of these equations. Indeed, it will be desirable in
practice to choose equations of equilibrium containing only one unknown, since this
eliminates the necessity of solving simultaneous equations. Equations containing only
one unknown may be obtained by summing moments about the point of intersection of
the lines of action of two unknown forces or, if these forces are parallel, by summing
components in a direction perpendicular to their common direction. In the case
Fig.9.3
of the truss of Fig. 9.3. For example, which held by roll at A and B and a short link at D,
the reactions al A and B Lc eliminated by summing x components. The reaction at, A
and D will be eliminated by summing moment about C the reactions al B and D by
summing moments about D. The equations obtained are:
ΣFx = 0 Σ,MC = 0
ΣMD = 0……. (9.5)
Each of these equations contains only one unknown.
4.5. Statically Indeterminate Reactions. Partial Constraints.
In each of the two examples considered in preceding section shown in Fig,.9.3 and 9.3b.
the type of supports were such that the rigid body could not possibly move under given
loads or under any other trading conditions. In such cases,the rigid body is laid to be
completely constraint . We also recall that the reactions corresponding to these support
involved three unknowned and couldnot determined by solving -the -equations of
equilibriumr When such a situation exists, the reactions are said to be statically
determinate.
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Dr. Ir. Abdul Hamid M.Eng.
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Fig. 9.4
Consider now the truss shown in Fig. 9.4 wich is held at pins at A and B. These supports
provide more constraints than are necessary to keep the truss from moving under the
GIVEN loads or under any other loading conditions. We also note that free-body diagram
of Fig. 9.4b that the corresponding reactions involve four unknowns. Since only three
independent equilibrium equations are available;there are more unknowns than
equations, and all the the unknowns cannot be determined. While the equations ΣMA =
0 and ΣMB yield the vertical components By and Ay.
Fig.9.5
The supports used to hold the truss shown in Fig.9.5a to consist of roller at A and B.
Clearly, the constraint provided by these this supports are not sufficient to keep the truss
from moving. While any vertical motion is prevented, the truss is free to move
horizontally. The truss is said to be partilally constrained. Turning our attention to Fig.
9.5b, we note that the reactions at A and B involve only two unknownfi. Since three
equations of equilibrium must still be satisfied, there are fewer unknowns than
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
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equations, and ONE of the equilibrium equations will not be satisfied. While the
equations ΣMA =0 and ΣMB = 0 can be satisfied by a proper choice of reactions at A and
B, the eiquation ΣFx = 0 will not be satisfied unless the sum of the horizontal the applied
forces happens to be zero. We thus check that the equilibrium of the truss of Fig. 9.5
cannot be maintained under general loading conditions.
Fig.9.6
Consider, for example, the truss shown in Fig.9.6a, which is held by rollers at A,B and E.
While there are three unknown reactions. A, B, and E (Fig.9.6a) , we find that the
equation ΣFx = 0 will not be satisfied unless the sum of the horizontal components of
the applied forces happens to be zero. There is a sufficient number of constraints, hut
these constraints are not properly arranged, and the truss is free to move horizontally.
Another example of improper constraints — and of static indeterminacy is provided by
the truss shown m Fig. 4.7.
Fig.9.7
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Dr. Ir. Abdul Hamid M.Eng.
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is held by a pin at A and by rollers at B and C, which altogelher involve four unknowns.
Since only three independent equilibrium equations are available, the reactions at the
supports are statically indeterminate. On the other hand, we note that equation ΣMA = 0
could not be satisfied under general loading conditions, since the lines of action of the
reactions B and C required to pass through A. We conclude that the truss rotate about A
and that it is improperly constrained.
SAMPLE PROBLEM 9.1
A fixed crane has a mass of 1000 kg and is used to lift a 2400-kg crate.It is held in
plaxce by a pin at A and a rocker at B.The centre of gravity of the crane is located at G.
Determine the components of the reactions at A and B.
Solution. A free-body diagram of the crane is drawn. Multiplying the masses ol the crane
and of the crate by g =9.81 m/ s2, we obtain the corresponding weights that is 9810 N or
9.81 kN, and 23500 N or 23.5 kN. The reaction at pin A is a force of unknown direction,
represented by its components Ax and Ay. The reaction at the rocker B is perpendicular
to the rocker surface; thus it is horizontal. We assume that Ax, ,Ay and B act in the
directions shown.
Determination of B.
We express that the sum of the moments of all external forces about point A is zero. The
equation obtained will contain neither Ax nor Ay since the moments ol Ax and Ay about A
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Dr. Ir. Abdul Hamid M.Eng.
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are zero. Multiplying the magnitude of each force by its perpendicular distance from B,
we write
ΣMA =0: +B(1.5 m)-(9.81 kN)(2m)-(23.5kN)(6m)=0
B=+107.1kN
Since the result is positive, the reaction is directed as assumed.
Determination of Ax. The magnitude Ax is determined by expressing that the sum of the
horizontal components of all external forces is zero.
ΣFx =0:
Ax + B = 0
Ax +107.1kN = 0
Ax= -107.1 kN
Since the result is negative, the sense ol Ax is opposite to that assumed originally.
Determination of Ay. The sum ol the vertical components must also equal zero.
+ ΣFy =0;
Ay – 9.81 kN - 23.5 kN = 0
Ay = +33.3 kN
Adding vectorially the components Ax end Ay, we find that the reaction at A is 112,2 kN
the angle of 17.30.
Check. The value obtained for the reactions may be checked by recalling that reaction
of the moments of all external forces about any point must be zero. For example,
considering point B, we write
ΣMB = -(9.81kN)(2m) – (23.5 kN)(6 m)+ (107.1 kN)(1.5 m) =0
PROBLEMS
9.1 The 40-ft boom AB weighs 2 kips; the distance from the axle A to the center of
gravity C of the boom is 20 ft. For the position shown, determine he tension T in the
cable and the reaction at A.
9.2 The 600-kg forklift truck is used to hold the 150-kg crate C in position shown.
Determines the reactions (a) at each of the two wheels A (one wheel on each side of
the truck), (b)at the single steerable wheel B.
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Dr. Ir. Abdul Hamid M.Eng.
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PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
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