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# Statics with MATLAB Dan B. Marghitu Mihai Dupac Nels H. Madsen

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```a drawing of a part of a complete system, isolated in order
to determine the forces acting on that rigid body. The vector Fij represents the force
exerted by link i on link j and Fij = \u2212Fji. Figure 4.9 shows the joint reaction forces
for a pin joint, Fig. 4.9a, and a slider joint, Fig. 4.9b. Figure 4.10 shows various
free-body diagrams that are considered in the analysis of a slider-crank mechanism
Fig. 4.10a. In Fig. 4.10b, the free body consists of the three moving links isolated
from the frame 0. The forces acting on the system include an external driven force
F, and the forces transmitted from the frame at joint A, F01, and at joint C, F03.
Figure 4.10c is a free-body diagram of the two links 1 and 2 and Fig. 4.10d is a
free-body diagram of the two links 0 and 1. Figure 4.10e is a free-body diagram of
crank 1 and Fig. 4.10f is a free-body diagram of slider 3.
The force analysis can be accomplished by examining individual links or a sub-
system of links. In this way the joint forces between links as well as the required
input force or moment for a given output load are computed.
For three-dimensional systems, the forces and moments are related by six scalar
equilibrium equations
\u2211
Fx = 0,
\u2211
Fy = 0,
\u2211
Fz = 0,
\u2211
Mx = 0,
\u2211
My = 0,
\u2211
Mz = 0.
(4.6)
158 4 Equilibrium
1
A
B
C
2
3
0
0
F
1
A
B
C
2
3
F
F01
F03
(a)
(b)
1
A
B
C
2
F01
F32
(c)
1
A
B
F01
F21
(d)
C 3
F
F03
F23
(e)
1
A
B
0
F21
(f)
Fig. 4.10 Free-body diagrams of a slider-crank mechanism
One can evaluate the sums of the moments about any point. Although one can obtain
other equations by summing the moments about additional points, they will not
be independent of these equations. For a three-dimensional free-body diagram one
can obtain six independent equilibrium equations and one can solve for at most six
unknown forces or couples.
A body has redundant supports when the body has more supports than the mini-
mum number necessary to maintain it in equilibrium. Redundant supports are used
whenever possible for strength and safety. Each support added to a body results in
additional reactions. The difference between the number of reactions and the number
of independent equilibrium equations is called the degree of redundancy.
4.3 Free-Body Diagrams 159
A body has improper supports if it will not remain in equilibrium under the action
of the loads exerted on it. The body with improper supports will move when the loads
are applied.
4.4 Two-Force and Three-Force Members
A body is a two-force member if the system of forces and moments acting on the
body is equivalent to two forces acting at different points.
For example a body is subjected to two forces, FA and FB, at A and B. If the body
is in equilibrium, the sum of the forces equals zero only if FA = \u2212FB. Furthermore,
the forces FA and \u2212FB form a couple, so the sum of the moments is not zero unless
the lines of action of the forces lie along the line through the points A and B. Thus
for equilibrium the two forces are equal in magnitude, are opposite in direction, and
have the same line of action. However, the magnitude cannot be calculated without
A body is a three-force member if the system of forces and moments acting on
the body is equivalent to three forces acting at different points.
Theorem If a three-force member is in equilibrium, the three forces are coplanar
and the three forces are either parallel or concurrent.
Proof Let the forces F1, F2, and F3 acting on the body at A1, A2, and A3. Let \u3c0 be
the plane containing the three points of application A1, A2, and A3. Let \u394 = A1A2
be the line through the points of application of F1 and F2. Since the moments due to
F1 and F2 about \u394 are zero, the moment due to F3 about \u394 must equal zero,
[n · (r × F3)] n = [F3 · (n × r)] n = 0,
where n is the unit vector of \u394. This equation requires that F3 be perpendicular to
n × r, which means that F3 is contained in \u3c0. The same procedure can be used to
show that F1 and F2 are contained in \u3c0, so the forces F1, F2, and F3 are coplanar.
If the three coplanar forces are not parallel, there will be points where their lines of
action intersect. Suppose that the lines of action of two forces F1 and F2 intersect at
a point P. Then the moments of F1 and F2 about P are zero. The sum of the moments
about P is zero only if the line of action of the third force, F3, also passes through
P. Therefore either the forces are concurrent or they are parallel.
The analysis of a body in equilibrium can often be simplified by recognizing the
two-force or three-force member.
160 4 Equilibrium
Fig. 4.11 Basic element of a
plane truss, a triangle
Fig. 4.12 Link in tension (T )
and compression (C)
C
C
T
T
4.5 Plane Trusses
A structure composed of links joined at their ends to form a rigid structure is called
a truss. Roof supports and bridges are common examples of trusses. When the links
of the truss are in a single plane, the truss is called a plane truss. Three bars linked by
pins joints at their ends form a rigid frame or noncollapsible frame. The basic element
of a plane truss is the triangle, Fig. 4.11. Four, five or more bars pin-connected to
form a polygon of as many sides form a nonrigid frame. A nonrigid frame is made
rigid, or stable, by adding a diagonal bars and forming triangles. Frameworks built
from a basic triangle are known as simple trusses. The truss is statically indeterminate
supports which are not necessary for maintaining the equilibrium configuration are
called redundant.
Several assumptions are made in the force analysis of simple trusses. First, all the
links are considered to be two-force members. Each link of a truss is straight and has
two nodes as points of application of the forces. The two forces are applied at the
ends of the links and are necessarily equal, opposite, and collinear for equilibrium.
The link may be in tension (T ) or compression (C), as shown in Fig. 4.12.
The weight of the link is small compared with the force it supports. If the weight
of the link is not small, the weight W of the member is replaced by two forces, each
W/2 one force acting at each end of the member. These weight forces are considered
as external loads applied to the pin connections. The connection between the links
are assumed to be smooth pin joints. All the external forces are applied at the pin
connections of the trusses. For large trusses, a roller support is used at one of the
4.5 Plane Trusses 161
A
B
DE
C A
B
DE
C
F F
1
2 3
4
5
6
7
F0A
F0C
0 0
(a) (b)
Fig. 4.13 Simple truss
supports to provide for expansion and contraction due to temperature changes and
for deformation from applied loads. Trusses and frames in which no such provision
is made are statically indeterminate. Two methods for the force analysis of simple
trusses will be given. Each method will be explained for the simple truss shown in
Fig. 4.13a. The length of the links are AB = BE = ED = BC = CD = a. The
external force at E is given and has the magnitude of F. The free-body diagram of
the truss as a whole is shown in Fig. 4.13b. The external support reactions are usually
determined first, by applying the equilibrium equations to the truss as a whole. The
reaction force of the ground 0 on the truss at the pin support A is F0A = F/2 and the
reaction force of the ground 0 on the truss at the roller support C is F0C = F/2.
Method of Joints
This method for calculating the forces in the members consists of writing the con-
ditions of equilibrium for the forces acting on the connecting pin of each joint. The
method deals with the equilibrium of concurrent forces,```
Arthur fez um comentário
Muito bom! achei que nunca ia achar um pdf desse livro, fazia tempo que estava procurando! muito obrigado!!!
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