Title: Engineering Mechanics: STATICS
1Engineering MechanicsSTATICS
 Anthony Bedford and Wallace Fowler
 SI Edition
Teachings Chapter 5 Objects in Equilibrium
2Chapter Outline
 The Equilibrium Equations
 2Dimensional Applications
 Statically Indeterminate Objects
 3Dimensional Applications
 2Force 3Force Members
 Computational Mechanics
35.1 The Equilibrium Equations
 When an object acted upon by a system of forces
moments is in equilibrium, the following
conditions are satisfied  1. The sum of the forces is zero
 S F 0
(5.1)  2. The sum of the moments about any point is
 zero
 S Many point 0
(5.2)
45.1 The Equilibrium Equations
 Eqs. (5.1) (5.2) imply that the system of
forces moments acting on an object in
equilibrium is equivalent to a system consisting
no forces no couples  If the sum of the forces on an object is zero
the sum of the moments about 1 point is zero,
then the sum of the moments about every point is
zero
55.1 The Equilibrium Equations
 For an object subjected to concurrent forces F1,
F2, , FN no couples
 Moment about point P is zero
 The only condition imposed by equilibrium on a
set of concurrent forces is that their sum is
zero  F1 F2 FN 0 (5.3)
65.1 The Equilibrium Equations
 To determine the sum of the moments about a line
L due to a system of forces moments acting on
an object
 Choose any point P on the line determine the
sum of moments S MP about P  The sum of the moments about the line is the
component of S MP parallel to the line
75.1 The Equilibrium Equations
 If the object is in equilibrium, S MP 0
 The sum of the moments about any line due to the
forces couples acting on an object in
equilibrium is zero
85.2 2Dimensional Applications
 Supports
 Forces couples exerted on an object by its
supports are called reactions, expressing the
fact that the supports react to the other
forces couples or loads acting on the object  E.g. a bridge is held up by the reactions exerted
by its supports the loads are the forces
exerted by the weight of the bridge itself, the
traffic crossing it the wind  Some very common kinds of supports are
represented by stylized models called support
conventions if the actual supports exert the same
reactions as the models
95.2 2Dimensional Applications
 The Pin Support
 Figure a a pin support
 a bracket to which an object (such as a beam) is
attached by a smooth pin that passes through the
bracket the object  Figure b side view
105.2 2Dimensional Applications
 To understand the reactions that a pin support
can exert  Imagine holding the bar
 attached to the pin support
 If you try to move the bar without rotating it
(i.e. translate the bar), the support exerts a
reactive force that prevents this movement  However, you can rotate the bar about the axis of
the pin  The support cannot exert a couple about the pin
axis to prevent rotation
115.2 2Dimensional Applications
 Thus, a pin support cant exert a couple about
the pin axis but it can exert a force on the
object in any direction, which is usually
expressed by representing the force in terms of
components
 The arrows indicate the directions of the
reactions if Ax Ay are positive  If you determine Ax or Ay to be negative, the
reaction is in the direction opposite to that of
the arrow
125.2 2Dimensional Applications
 The pin support is used to represent any real
support capable of exerting a force in any
direction but not exerting a couple
 Used in many common devices, particularly those
designed to allow connected parts to rotate
relative to each other
135.2 2Dimensional Applications
 The Roller Support
 A pin support mounted on wheels
 Like a pin support, it cannot exert a couple
about the axis of the pin  Since it can move freely in the direction
parallel to the surface on which it rolls, it
cant exert a force parallel to the surface but
can exert a force normal (perpendicular) to this
surface
145.2 2Dimensional Applications
 Other commonly used conventions equivalent to the
roller support  The wheels of vehicles wheels supporting parts
of machines are roller supports if the friction
forces exerted on them are negligible in
comparison to the normal forces
155.2 2Dimensional Applications
 A plane smooth surface can also be modeled by a
roller support  Beams bridges are sometimes supported in this
way so that they will be free to undergo thermal
expansion contraction
165.2 2Dimensional Applications
 These supports are similar to the roller support
in that they cannot exert a couple can only
exert a force normal to a particular direction
(friction is neglected)
175.2 2Dimensional Applications
 In these supports, the supported object is
attached to a pin orr that can move freely
in 1 direction but is constrained in the
perpendicular direction  Unlike the roller support, these supports can
exert a normal force in either direction
185.2 2Dimensional Applications
 The Fixed Support
 The fixed support shows the supported object
literally built into a wall (builtin)  To understand the reactions
 Imagine holding a bar attached to the fixed
support
195.2 2Dimensional Applications
 If you try to translate the bar, the support
exerts a reactive force that prevents translation  If you try to rotate the bar, the support exerts
a reactive couple that prevents rotation  A fixed support can exert 2 components of force
a couple
205.2 2Dimensional Applications
 The term MA is the couple exerted by the support
the curved arrow indicates its direction  Fence posts lampposts have fixed supports
 The attachments of parts connected so that they
cannot move or rotate relative to each other,
such as the head of a hammer its handle, can be
modeled as fixed supports
215.2 2Dimensional Applications
 Table 5.1 summarizes the support conventions
commonly used in 2D applications
225.2 2Dimensional Applications
 FreeBody Diagrams
 By using the support conventions, we can model
more elaborate objects construct their
freebody diagrams in a systematic way  Example
 a beam with a pin support at the left end a
roller support on at the right end is loaded
with a force F  The roller support rests on a surface inclined at
30
235.2 2Dimensional Applications
 To obtain a freebody diagram of the beam,
isolate it from its supports  Complete the freebody diagram by showing the
reactions that may be exerted on the beam by the
supports  Notice that the reaction at B exerted by the
roller support is normal to the surface on which
the support rests
245.2 2Dimensional Applications
 Example
 The object in this figure has a fixed support at
the left end  A cable passing over a pulley is attached to the
object at 2 points  Isolate it from its supports complete the
freebody by showing the reactions at the fixed
support the forces exerted by the cable
255.2 2Dimensional Applications
 Dont forget the couple at the fixed support
 Since we assume the tension in the cable is the
same on both sides of the pulley, the 2 forces
exerted by the cable have the magnitude T  Once you have obtained the freebody diagram of
an object in equilibrium to identify the loads
reactions acting on it, you can apply the
equilibrium equations
265.2 2Dimensional Applications
 The Scalar Equilibrium Equations
 When the loads reactions on an object in
equilibrium form a 2D system of forces
moments, they are related by 3 scalar equilibrium
equations  S Fx 0
(5.4)  S Fy 0
(5.5)  S Many point 0
(5.6)
275.2 2Dimensional Applications
 More than 1 equation can be obtained from Eq.
(5.6) by evaluating the sum of the moments about
more than 1 point  But the additional equations will not be
independent of Eqs. (5.4)(5.6)  In other words, more than 3 independent equations
cannot be obtained from a 2D freebody diagram,
which means we can solve for at most 3 unknown
forces or couples
285.2 2Dimensional Applications
 The seesaw found on playgrounds, consisting of a
board with a pin support at the center that
allows it to rotate, is a simple familiar
example that illustrates the role of Eq. (5.6)  If 2 people of unequal weight sit at the seesaws
ends, the heavier person sinks to the ground
295.2 2Dimensional Applications
 To obtain equilibrium, that person must move
closer to the center  Draw the freebody diagram of the seesaw showing
the weights of the people W1 W2 the reactions
at the pin support
305.2 2Dimensional Applications
 Evaluating the sum of the moments about A, we
find that the equilibrium equations are  S Fx Ax 0
(5.7)  S Fy Ay W1 W2 0
(5.8)  S Mpoint A D1W1 D2W2 0
(5.9)  Thus, Ax 0, Ay W1 W2 D1W1 D2W2
 The last condition indicates that the relation
between the positions of the 2 persons necessary
for equilibrium
315.2 2Dimensional Applications
 To demonstrate that an additional independent
equation is not obtained by evaluating the sum of
the moments about a different point  Sum the moments about the right end of the seesaw
 S Mright end (D1 D2)W1 D2Ay 0
325.2 2Dimensional Applications
 This equation is a linear combination of Eqs.
(5.8) (5.9)  (D1 D2)W1 D2Ay D2(Ay W1 W2)
 (D1W1
D2W2) 0
335.2 2Dimensional Applications
 Until now we have assumed in examples problems
that the tension in a rope or cable is the same
on both sides of a pulley
 Consider this pulley
 In its freebody diagram, we do not assume that
the tensions are equal  Summing the moments about the center of the
pulley  S Mpoint A RT1 RT2 0
345.2 2Dimensional Applications
 The tensions must be equal if the pulley is in
equilibrium  However, notice that we have assumed that the
pulleys support behaves like a pin support
cannot exert a couple on the pulley  When that is not true for example, due to
friction between the pulley the support the
tensions are not necessarily equal
35Example 5.1 Reactions at Pin Roller Supports
 The beam in Fig. 5.13 has a pin at A roller
supports at B is subjected to a 2kN force.
What are the reactions at the supports
36Example 5.1 Reactions at Pin Roller Supports
 Strategy
 To determine the reactions exerted on the beam
by its supports, draw a freebody diagram of the
beam isolated from the supports. The freebody
diagram must show all external forces couples
acting on the beam, including the reactions
exerted by the supports. Then determine the
unknown reactions by applying equilibrium
equations
37Example 5.1 Reactions at Pin Roller Supports
 Solution
 Draw the FreeBody Diagram
 Isolate the beam from its supports show the
 loads the reactions that may be exerted by the
 pin roller supports.
There are 3 unknown reactions 2 components of
force Ax Ay at the pin support a force B at
the roller support
38Example 5.1 Reactions at Pin Roller Supports
 Solution
 Apply the Equilibrium Equations
 Summing the moments about point A
 S Fx Ax Bsin 30 0
 S Fy Ay Bcos 30 2 kN 0
 S Mpoint A (5 m)(Bcos 30) (3 m)(2 kN)
0  Solving these equations, the reactions are
 Ax 0.69 kN, Ay 0.80 kN B 1.39 kN
39Example 5.1 Reactions at Pin Roller Supports
 Solution
 Confirm that the equilibrium equations are
satisfied  S Fx 0.69 kN (1.39 kN)sin 30 0
 S Fy 0.80 kN (1.39 kN)cos 30 2 kN
0  S Mpoint A (5 m)(1.39 kN)cos 30 (3 m)(2 kN)
0  Critical Thinking
 In drawing freebody diagrams, you should try to
choose the correct directions of the reactions
because it helps to develop your physical
intuition
40Example 5.1 Reactions at Pin Roller Supports
 Critical Thinking
 However, if you choose an incorrect direction for
a reaction in drawing the freebody diagram of a
single object, the value you obtain from the
equilibrium equations for that reaction will be
negative, which indicates that its actual
direction is opposite to the direction you chose  E.g. if we draw the freebody diagram of the beam
with the component Ay pointed downward
41Example 5.1 Reactions at Pin Roller Supports
 Critical Thinking
 Equilibrium equations
 S Fx Ax Bsin 30 0
 S Fy Ay Bcos 30 2 kN 0
 S Mpoint A (5 m)(Bcos 30) (3 m)(2 kN)
0  Solving, we obtain
 Ax 0.69 kN, Ay 0.80 kN B 1.39
kN  The negative value of Ay indicates that the
vertical force exerted on the beam by the pin
support at A is in the direction opposite to the
arrow, i.e. the force is 0.80 kN upward
42Example 5.2 Reactions at a Fixed Support
 The object in Fig. 5.14 has a fixed support at
A is subjected to 2 forces a couple. What are
the reactions at the support
43Example 5.2 Reactions at a Fixed Support
 Strategy
 Obtain a freebody diagram by isolating the
object from the fixed support at A showing the
reactions exerted at A, including the couple that
may be exerted by a fixed support. Then determine
the unknown reactions by applying the equilibrium
equations.
44Example 5.2 Reactions at a Fixed Support
 Solution
 Draw the FreeBody Diagram
 Isolate the object from its support show the
reactions at the fixed support.  There are 3 unknown reactions 2 force
components Ax Ay a couple MA. (Remember that
we can choose the directions of these arrows
arbitrarily)  Also resolve the 100N force into its
components.
45Example 5.2 Reactions at a Fixed Support
 Solution
 Draw the FreeBody Diagram
46Example 5.2 Reactions at a Fixed Support
 Solution
 Apply the Equilibrium Equation
 Summing the moments about point A
 S Fx Ax 100cos 30 N 0
 S Fy Ay 200 N 100sin 30 N 0
 S Mpoint A MA 300 Nm (2 m)(200 N)
 (2 m)(100cos 30 N)
 (4 m)(100sin 30 N) 0
 Solving these equations, Ax 8.86 N, Ay 150.0
N MA 73.2 Nm.
47Example 5.2 Reactions at a Fixed Support
 Critical Thinking
 Why dont the 300 Nm couple the couple MA
exerted by the fixed support appear in the first
2 equilibrium equations  A couple exerts no net force
 Also, because the moment due to a couple is the
same about any point, the moment about A due to
the 300 Nm counterclockwise couple is 300 Nm
counterclockwise
48Example 5.3 Reactions on a Cars Tires
 The 14 000N car in Fig. 5.15 is stationary.
Determine the normal forces exerted on the front
rear tires by the road.
49Example 5.3 Reactions on a Cars Tires
 Strategy
 Draw the freebody diagram of the car, showing
the forces exerted on its tires by the road at A
B apply the equilibrium equations to
determine the forces on the front rear tires.
50Example 5.3 Reactions on a Cars Tires
 Solution
 Draw the FreeBody Diagram
 Isolate the car show its weight the
reactions exerted by the road.  There are 2 unknown reactions the forces A
B exerted on the front rear tires.
51Example 5.3 Reactions on a Cars Tires
 Solution
 Apply the Equilibrium Equations
 The forces have no x components.
 Summing the moments about point B
 S Fy A B 14000 N 0
 S Mpoint B (2 m)(14000 N) (3 m)A
0  Solving these equations, the reactions are A
9333 N  B 4667 N.
52Example 5.3 Reactions on a Cars Tires
 Critical Thinking
 This example doesnt fall within our definition
of a 2D system of forces moments because the
forces acting on the car are not coplanar
 From the oblique view of the freebody diagram of
the car, you can see the total forces acting on
the individual tires
53Example 5.3 Reactions on a Cars Tires
 Critical Thinking
 Total normal force on the front tires
 AL AR A
 Total normal force on the rear tires
 BL BR B
 The sum of the forces in the y direction
 S Fy AL AR BL BR 14000 N 0
 A B 14000 N 0
54Example 5.3 Reactions on a Cars Tires
 Critical Thinking
 Since the sum of the moments about any line due
to the forces couples acting on an object in
equilibrium is zero, the sum of the moments about
the z axis due to the forces acting on the car is
zero  S Mz axis (3 m)(AL AR) (2 m)(14000 N)
 (3 m)A (2 m)(14000 N) 0
 Thus we obtain the same equilibrium equations we
did when we solved the problem using a 2D
analysis
55Example 5.4 Choosing the Point About Which to
Evaluate Moments
 The structure AB in Fig. 5.16 supports a
suspended 2Mg (megagram) mass. The structure is
attached to ar in a vertical slot at A
has a pin support at B. What are the reactions at
A B
56Example 5.4 Choosing the Point About Which to
Evaluate Moments
 Strategy
 Draw the freebody diagram of the structure
the suspended mass by removing the supports at A
B. Notice that the support at A can exert only
a horizontal reaction. Then use the equilibrium
equations to determine the reactions at A B.
57Example 5.4 Choosing the Point About Which to
Evaluate Moments
 Solution
 Draw the FreeBody Diagram
Isolate the structure mass from the supports
show the reactions at the supports the force
exerted by the weight of the 2000kg mass. The
slot at A can exert only a horizontal force on
ther.
58Example 5.4 Choosing the Point About Which to
Evaluate Moments
 Solution
 Apply the Equilibrium Equations
 Summing the moments about point B
 S Fx Ax Bx 0
 S Fy By (2000)(9.81) N 0
 S Mpoint B (3 m)A (2 m)(200 N )(9.81)
N 0  The reactions are
 A 13.1 kN, Bx 13.1 kN By 19.6
kN.
59Example 5.4 Choosing the Point About Which to
Evaluate Moments
 Critical Thinking
 Although the point about which moments are
evaluated in writing equilibrium equations can be
chosen arbitrarily, a careful choice can often
simplify your solution  In this example, point B lies on the lines of
action of the 2 unknown reactions Bx By  By evaluating moments about B, we obtained an
equation containing only 1 unknown, the reaction
at A
60Design Example 5.5 Design for Human Factors
 Fig. 5.17 shows an airport luggage carrier
its freebody diagram when it is held in
equilibrium in the tilted position. If the
luggage carrier supports a weight W 250 N, the
angle 30, a 0.2 m, b 0.4 m d 1.2 m,
what force F must the user exert
61Design Example 5.5 Design for Human Factors
 Strategy
 The unknown reactions on the freebody diagram
are the force F the normal force N exerted by
the floor. If we sum moments about the center of
the wheel C, we obtain an equation in which F is
the only unknown reaction.
62Design Example 5.5 Design for Human Factors
 Solution
 Summing moments about C
 S M(point C) d(F cos ) a(W sin ) b(W cos
) 0  Solving for F

(1)  Substituting the values of W, , a, b d yields
the  solution F 59.3 N.
63Design Example 5.5 Design for Human Factors
 Design Issues
 Design that accounts for human physical
dimensions, capabilities characteristics is a
special challenge  This art is called design for human factors
 Here, we consider a simple device, the airport
luggage carrier in Fig. 5.17 show how
consideration of its potential users the
constraints imposed by equilibrium equations
affect its design
64Design Example 5.5 Design for Human Factors
 Design Issues
 The user moves the carrier by grasping the bar at
the top, tilting it walking while pulling the
carrier  The height of the handle (the dimension h) needs
to be comfortable  Since h R d sin , if we choose values of h
the wheel radius R, we obtain a relation between
the length of the carriers handle d the tilt
angle 
(2)
65Design Example 5.5 Design for Human Factors
 Design Issues
 Substituting the expression for d into Eq. (1)

(3)  Suppose that based on statistical data on human
dimensions, we decide to design the carrier for
convenient use by persons up to 1.9 m, which
corresponds to a dimension h of approximately 0.9
m  Let R 0.075 m, a 0.15 m b 0.3 m
66Design Example 5.5 Design for Human Factors
 Design Issues
 The resulting value of F/W as a function of is
shown in Fig. 5.18
67Design Example 5.5 Design for Human Factors
 Design Issues
 At 63, the force the user must exert is
zero, which means the weight of the luggage acts
at a point directly above the wheels  This would be the optimum solution if the user
could maintain exactly that value of  However, inevitably varies the resulting
changes in F make it difficult to control the
carrier  In addition, the relatively steep angle would
make the carrier awkward to pull
68Design Example 5.5 Design for Human Factors
 Design Issues
 From this point of view, it is desirable to
choose a design within the range of values of
in which F varies slowly, say 30 45
(even though the force the user must exert is
large in this range of in comparison with
larger values of , it is only about 13 of the
weight)  Over this range of , the dimension of d varies
from 1.65 m to 1.17 m  A smaller carrier is desirable for lightness
ease of storage, so we choose d 1.2 m for our
preliminary design
69Design Example 5.5 Design for Human Factors
 Design Issues
 We have chosen the dimension d based on
particular values of the dimensions R, a b  In an actual design study, we should carry out
the analysis for expected ranges of values of
these parameters  Our final design would also reflect decisions
based on safety (e.g. there must be adequate
means to secure luggage no sharp projections),
reliability (the frame must be sufficiently
strong the wheels must have adequate reliable
bearings) the cost of manufacture