Chapter Three
Equilibrium
θ3 = 90◦ − 60.3◦ = 29.7◦
β = 180◦ − θ3 =180◦ − 29.7◦ = 150.3◦
γ = 180◦ − (α + β) = 180◦ − (12.1◦ + 150.3◦) = 17.6◦
Applying the law of sines, we obtain
R
P
20
= A = BC
sin α sin β sin γ
Substituting the values for α, β, and γ into this equation yields the same
values for PBC and RA as given in Part 1.
Example 19: The lever ABC is pin supported at A and
connected to a short link BD as shown in Figure (a). If
the weight of the members is negligible, determine
the force of the pin on the lever at A.
Solution: Free-Body Diagram: As shown in Figure
(b), the short link BD is a two-force member.
Although the magnitude of the force is unknown, the
line of action is known since it passes through B and
D.
Lever ABC is a three-force member, and
therefore, in order to satisfy moment equilibrium, the
three nonparallel forces acting on it must be
concurrent at O, Figure (c).
Equilibrium Equations: The angle θ which defines the
line of action of FA can be determined from
trigonometry
 0.7 
o
 = 60.3
 0.4 
θ = tan −1 
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Chapter Three
Equilibrium
Applying the force equilibrium equations,
ΣFx = 0 + ;
ΣFy = 0 + ;
Solving, we get
FA cos60.3o - F cos45o + 400 = 0
FA sin60.3o - F sin45o = 0
FA = 1.07 kN
F = 1.32 kN
Note: We can also solve this problem by representing the force at A by its two
components Ax and Ay and applying ΣMA = 0, ΣFx = 0, ΣFy = 0 to the lever. Once Ax
and Ay are determined, we can get FA and θ.
Example 20: A man raises a 10-kg joist, of length
4 m, by pulling on a rope. Find the tension T in
the rope and the reaction at A.
Solution: Free-body Diagram: The joist is a three-force body, since it is acted
upon by three forces:
W = mg = (10)(9.81) =98.1 N
Since the joist is a three-force body, the forces
acting on must be concurrent. The reaction R , therefore,
will pass through the point of intersection C of the lines
of action of the weight W and the tension force T.
Drawing the vertical BF through B and the
horizontal CD through C, we note that
AF = BF = (AB) cos45o = 4 cos45o =2.828 m
CD = EF = AE =
1
(AF) = 1.414 m
2
BD = (CD) cot(45o + 25o) = 1.414 tan20o = 0.515 m
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Chapter Three
Equilibrium
CE = DF = BF – BD = 2.828 – 0.515 = 2.313 m
tan α =
We write
CE 2.313
=
= 1.636
AE 1.414
α = 58.6o
We know the direction of all the forces acting
on the joist.
Force Triangle: Using the law of sines, we write
T
R
98.1
=
=
o
o
sin 31.4
sin 110
sin 38.6o
Solving, we get
T = 81.9 N,
R = 147.8 N α = 58.6o
General Problems
3.10
The beam consists of two bars connected by a pin at B. Neglecting the
weight of the beam compute the support reactions at A.
3.11
For the frame shown, determine the magnitude of the pin reaction at B.
Neglect the weight of the frame.
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Chapter Three
3.12
Equilibrium
The structure consists of two identical bars joined by a pin at B. Neglecting
the weights of the bars find the magnitude of the pin reaction at C.
3.13
The bars AB and AC are joined by a pin at A and a horizontal cable. The
vertical cable carrying the 200-kg mass is attached to the pin at A. Determine the
tension in the horizontal cable. Neglect the weights of the bars.
3.14 Neglecting
the weights of the members, determine the magnitude of the pin
reaction at D when the frame is loaded by the 200-N ·m couple.
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Chapter Three
3.15
Equilibrium
The bars AB and BC of the structure are each of length L and weigh W and
2W, respectively. Find the tension in cable DE in terms of W, L, and the angle θ.
3.16
Determine the magnitude of the pin reaction at A as a function of P. The
weights of the members are negligible.
3.17
Neglecting friction and the weights of the members, compute the
magnitudes of the pin reactions at A and C for the folding table shown.
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Chapter Three
Equilibrium
3.18 Neglecting
3.19 Calculate
the weight of the frame find the tension in cable CD.
the reactions at the built-in support at C, neglecting the weights of
the members.
3.20 Compute
the magnitudes of all forces acting on member CDE of the frame.
3.21 Calculate
all forces acting on member CDB.
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Chapter Three
Equilibrium
ANALYSIS OF PLANE TRUSSES
A truss is a structure that is made of straight, slender bars that are joined
together to form a pattern of triangles. Trusses are usually designed to transmit
forces over relatively long spans; common examples are bridge trusses and roof
trusses. A typical bridge truss is shown in Figures below.
The analysis of trusses is based on the following three assumptions:
1. The weights of the members are negligible. A truss can be classified as
a lightweight structure, meaning that the weights of its members are
generally much smaller than the loads that it is designed to carry.
2. All joints are pins. In practice, the members at each joint are usually
riveted or welded to a plate, called a gusset plate, as shown in Figure (b)
below.
3. The applied forces act at the joints. Because the members of a truss are
slender, they may fail in bending when subjected to loads applied at
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Chapter Three
Equilibrium
locations other than the joints. Therefore, trusses are designed so that the
major applied loads act at the joints.
Simple truss
If three members are pin connected at their ends they form a triangular truss will
be rigid, Figure (a). If four members are connected by smooth pins at their ends,
Figure (b), the resulting structure will not be rigid and will collapse when a load
P is applied. If an additional member from A to C is added to the structure, as
indicated in Figure (c), the five members form a rigid body and will support the
load P without collapse. The addition of the member AC in figure (b) below
reduces the structure to two triangles, and in general, a truss can be constructed
by starting with three members arranged in the form of a triangle and adding
members in noncollinear pairs for each additional joint. If a truss can be
constructed by expanding the basic triangular truss this way, it is called a simple
truss.
(a)
(b)
(c)
The free-body diagram for any member of a truss will contain only two
forces (the forces exerted on the member by the pin at each end). Therefore,
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Equilibrium
each member of a truss is a two-force body. When dealing with the internal
force in a two-force body, engineers commonly distinguish between tension and
compression. Tensile forces elongate (stretch) the member, whereas
compressive forces compress (shorten) it. Because the forces act along the
longitudinal axis of the member, they are often called axial forces. Note that
internal forces always occur as equal and opposite pairs on the two faces of an
internal cross section.
The two common techniques for computing the internal forces in a truss
are the method of joints and the method of sections, each of which is
discussed in the following articles.
1. Method of Joints
When using the method of joints to calculate the forces in the members of a
truss, the equilibrium equations are applied to individual joints (or pins) of the
truss. Because the members are two-force bodies, the forces in the FBD of a
joint are concurrent. Consequently, two independent equilibrium equations are
available for each joint.
ΣFx = 0,
ΣFy = 0
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