Modul XIII
SAMPLE PROBLEM 13.1
Determine the forces in members DE and HJ of the truss shown.
Solution. Considering the entire truss as a free body, we determine the reactions at A
and L.
A = 160 kips
L = 160 kips
Force in Member HJ.
Section nn is passed through the truss that it intersects member HJ and only two
additional members. After the intersected members have been removed, we choose the
right-hand portion of the truss as a free body. Three unknown forces are involved, to
eliminate the two forces passing through point I, we write
+ ΣMI = 0:
(160kips)(60ft) - (60 kips)(30 ft) - (4 kips)(30 ft) + FHJ(20ft) =0
FHJ = -384 kips
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Dr. Ir. Abdul Hamid M.Eng.
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The sense of FHJ was chosen assuming member HJ to be in tension; the negative sign
obtained indicates that the member is in compression.
FHJ = 384 kips (C )
Force in Member DE. Section mm is passed through the truss so that it intersects
member DE and only two additional members. After the intersected members have been
removed, the left-hand portion of the truss is chosen as a free body. Three unknown
forces are again involved; since the equation ΣFy = 0 involves only FDE as an unknown,
we write:
ΣFy = 0:
+160 kips - 60 kips - 4 kips - 60 kips + FDE = 0
FDE =-36 kips
FDE = 36 kips ( C ).
SAMPLE PROBLEM 13.2
Determine the forces in members FH, GH, and CI of the roof truss shown.
Solution. Section nn is passed through the truss as shown. The right-hand portion of the
truss will be taken as a free body. Since the reaction at L acts on this free body, the
value of L must be calculated separately, using the entire truss as a free body; the equation ΣMA = 0 yields L = 7.50kN.
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Dr. Ir. Abdul Hamid M.Eng.
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Force in Member GI.
Using the portion HLI of the truss as a free body, the value of
FGI is obtained by writing
ΣMH = 0 :
(7.50kN)(10m) - (1kN)(5m) – FGI(5.33 m) = 0
FGI = + 13.13 kN
FGI = 13.13 kN (T).
Force in Member FH. The value of FFH is obtained from the equation ΣMG = 0. We
move FFH along its line of action until it acts at point F, where it is resolved into its x and
y components. The moment of FPH with respect to point G is now equal to (FFH cos α) (8
m).
ΣMG = 0:
(7.50kN)(15m) - (1kN)(10m) - (1kN)(5m) + (FFH cos α)(8 m) = 0
FFH = - 13.82 kN
FFH = 13.82 kN (C ).
Force in Member GH. The value of FGH is determined by first resolving the force FGff into
x and y components at point G and then solving the equation ΣML =0.
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Dr. Ir. Abdul Hamid M.Eng.
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ΣML =0:
(lkN)(10m) + (1kN)(5m) + (FGH cos β)(15 m) = 0
FGH = -1.372 kN
FGH = 1.372 kN ( C ).
PROBLEM 13.
13.1
Determine the force in members BD and DE of the truss shown.
Fig. P 13-1 ,P 13-2 and P 13-3
13.2 Determine the force in members FH and DH of the truss shown.
13.3 Determine the force in members DF and EG of the truss shown.
13.FRAMES AND MACHINES
13-5. Structures Containing Multiforce Members. Under Trusses, we have
considered structures consisting entirely of pins and of straight two-force members. The
forces acting on the two-force members were known to be directed along the members
themselves. We shall now consider structures in which at least one of the members is a
multiforce member, i.e., a member acted upon by three or more forces. These forces will
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Dr. Ir. Abdul Hamid M.Eng.
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generally not be directed along the members on which they act; their direction is
unknown, and they should be represented therefore by two unknown components.
Frames and machines are structures containing multiforce members. Frames are
designed to support loads and are usually stationary, fully constrained structures.
Machines are designed to transmit and modify forces; they may or may not be stationary
and will always contain moving parts.
13-6. Analysis of a Frame. As a first example of analysis of a frame, we shall consider
again the crane described in preceding chapter, which carries a given load W(Fg. 135a). The free-body diagram of the entire frame is shown in Fig. 13-5b. This diagram may
be used to determine the external forces acting on the frame. Summing moments about
A, we first determine the force T exerted by the cable; summing x and y components,
We then determine the components Ax and Ay of the reaction at the pin A.
Fig. 13-5a
Fig. 13-5b
In order to determine the internal forces holding the various parts of a frame together, we
must dismember the frame and draw a free-body diagram for each of its component
parts (Fig. 13-5c). First, the two-force members should be considered. In this frame,
member BE is the only two-force member. The forces acting at each end of this member
must have the same magnitude, same line of action, and opposite sense. They are
therefore directed along BE and will be denoted, respectively, by FH, and FBE. Their
sense will be arbitrarily assumed as shown in Fig. 13-5c, and the correctness of this
assumption will be checked later by the sign obtained for the common magnitude FBE of
the two forces.
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Next, we consider the multiforce members, i.e., the members which are acted upon by
three or more forces. According to Newton's third law, the force exerted at B by member
BE on member AD must be equal and opposite to the force FBE exerted by AD on BE.
Similarly, the force exerted at E by member BE on member CF must be equal and
opposite to the force FBE
'.\erted by CF on BE. The forces that the two-force member BE verts on AD and CF are
therefore respectively equal to — FBE iiidFBE; they have the same magnitude FBE and
opposite sense,
•.lid should be directed as shown in Fig. 6,20c.
At C two multiforce members are connected. Since neither 'lie direction nor the
magnitude of the forces acting at C is known, these forces will be represented by their x
and y components. The components Cx and Cy of the force acting on member AD will
be arbitrarily directed to the right and upward. Since, according to Newton's third law, the
forces exerted by member AFon AD and by member AD on CF are equal and opposite,
the components of the force acting on member CF must be directed to the left and
downward; they will be denoted, respectively, by Cx and Cy . Whether the force Cx is
actually directed to the right and the force Cy is actually directed to the left will be
determined later from the sign of their common magnitude Cx, a plus sign indicating that
the assumption made was correct, and a minus sign that it was wrong. The free-body
diagrams of the multiforce' members are completed by showing the external forces
acting at A, D, and F.
The internal forces may now be determined by considering the free-body diagram of
either of the two multiforce members. Choosing the free-body diagram of CF, for
example, we write the equations ΣMC = 0, ΣME = 0, and ΣFx = 0, which yield the values
of the magnitudes FBE, Cy, and Cx respectively. These values may be checked by
verifying that member AD is also in equilibrium.
It should be noted that the free-body diagrams of the pins were not shown in Fig. 6.20c.
This was because the pins were assumed to form an integral part of one of the two
members they connected, This assumption can always be used to simplify the analysis
of frames and machines. When a pin connects three or more members, however, or
when a pin connects a support and two or more members, or when a load is applied to a
pin, a clear decision must be made in choosing the member to which the pin will be
assumed to belong. (If multiforce members are involved, the pin should be attached to
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Dr. Ir. Abdul Hamid M.Eng.
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one of these members.) The various forces exerted on the pin should then be clearly
identified.
13-7. Frames Which Cease to Be Rigid When De tached from Their Supports. The
crane analyzed in Sec 13-5 was so constructed that it could keep the same shape with- :
out the help of its supports; it was therefore considered as a rigid t body. Many frames,
however, will collapse if detached from their supports; such frames cannot be
considered as rigid bodies, Consider, for example, the frame shown in Fig. 6.21a, which
consists of two members AC and CB carrying loads P and Q at their midpoints; the
members are supported by pins at A and B and are connected by a pin at C. If detached
from its supports, this frame will not maintain its shape; it should therefore be considered
as made of two distinct rigid parts AC and CB.
Fig.13-6
The equations ΣFx, = 0, ΣFy = 0, ΣM = 0 (about any given point) express the conditions
for the equilibrium of a rigid body ; we should use them, therefore, in connection with the
free-body diagrams of rigid bodies, namely, the free-body diagrams of members AC and
CB (Fig. 13-6b). Since these members are multiforce members, and since pins are used
at the supports and at the connection, the reactions at A and B and the forces at C will
each be represented by two components. In accordance with Newton's third law, the
components of the force exerted by CB on AC and the components of the force exerted
by AC on CB will be represented by vectors of the same magnitude and opposite sense;
thus, if the first pair of components consists of Cx and C , the second pair will be
represented by Cx and Cy. We note that four unknown force components act on free
body AC, while only three independent equations may be used to express that the body
is in equilibrium; similarly, four unknowns, but only three equations, are associated with
CB. However, only six different unknowns are involved in the analysis of the two members, and altogether six equations are available to express that the members are in
equilibrium. Writing ΣMA = 0 for free body AC and 2Mg = 0 for CB, we obtain two
simultaneous equations which may be solved for the common magnitude Cx of the
components Cx and — Cx, and for the common magnitude Cy of the components Cy and
Cy. Writing, then, 2^=0 and 2F = 0 for each of the two free bodies, we obtain
successively the magnitudes Ax, Ay, Bx, and By.
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Fig.13-6b
We shall observe now that, since the equations of equilibrium ΣFx = 0, ΣFy = 0, ΣM = 0
(about any given point) are satisfied by the forces acting on free body AC, and since
they are also satisfied by the forces acting on free body CB, they must be satisfied when
the forces acting on the two free bodies are considered simultaneously.
Fig.13-6c
Since the internal forces at C cancel each other, we find that the equations of equilibrium
must be satisfied by the external forces shown on the free-body diagram of the frame
ACB itself (Fig. 13-6c), although the frame is not a rigid body.
SAMPLE PROBLEM 13.4
In the small frame shown, members EBF and ABCD are connected by a pin at B and by
the cable EC. A 75-lb load is supported by second cable which passes over a pulley at F
and is attached to the vertical member at G. Determine the tension in cable EC at the
components of the pin reaction at B.
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Entire Frame. The external reactions on the frame involve three unknowns; these
reactions are determined by taking the entire frame (including the cables) as a free body.
ΣFy = 0 :
Dy - 75 Ib = 0
Dy= +75 Ib
ΣMD = 0 :
-(751b)(12ft) + A(15ft) = 0
A = +60 Ib
ΣFX = 0:
-60 Ib +DX = 0
Dx = +60 Ib
Since the values''obtained are positive, the forces are directed as assumed in the
diagram, that is, Dx to the right, Dy up, and A to the left.
Members. The frame is dismembered; since only two members are connected at B, the
components of the unknown forces acting on EBF and ABCD at B are, respectively,
equal and opposite. The forth exerted at E and C by the cable EC are equal and
opposite, and thdr direction is known. From the free-body diagram of the pulley, it is
seen that the force exerted at F by the pulley on member EBF may be resolved into two
75-lb components as shown. The cable also exert; a 75-lb force on ABCD at point G.
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Member EBF.
ΣME = 0 :
Using the free body EBF, we write:
By(4ft) - (75lb)(14ft) = 0
By = 263lb
ΣMB = 0 :
(Tcosα)(4ft) - (75lb)(10ft) = 0
T = +225Ib
ΣFx = 0:
Tsin α - Bx - 75 Ib = 0
Bx = +50.0 Ib
Since the values obtained are positive, the forces are directed as shown on the diagram:
the forces Bx and By acting on member ABCD are directed, respectively, to the right and
down, while the forces B, anc B acting on member EBF are directed, respectively, to the
left and up.
Member ABCD (Check). The computations are checked by considering the free body
ABCD. For example,
ΣFx = -60 Ib + 75lb + Bx - Tsinα + 60 Ib
= -60 Ib + 75 Ib + 50 Ib - (225 Ib) sin 33.7° + 60lb = 0 (checks).
PROBLEM 13.
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Dr. Ir. Abdul Hamid M.Eng.
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13-1Determine the components of the forces acting on each member of the frame
shown.
13-2.A 600-lb horizontal force is applied to pin A of the frame shown. Determine the
forces acting on the two vertical members of the frame.
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