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Chapter 9
Static Equilibrium; Elasticity and Fracture
Torque and Two Conditions For Equilibrium
An object in mechanical equilibrium must
satisfy the following conditions:
1.
2.
The net external force must be zero:
ΣF = 0
The net external torque must be zero:
Στ = 0
Two Conditions of Equilibrium
 First
Condition of Equilibrium
○ The net external force must be zero
 F  0 or
 Fx  0 and  Fy  0
 Necessary, but not sufficient
 Translational equilibrium
Two Conditions of Equilibrium

Second Condition of Equilibrium
○ The net external torque must be zero
Στ = 0 or
Στx = 0 and Στy = 0
 Rotational equilibrium
 Both conditions satisfy mechanical
equilibrium
Two Conditions of Equilibrium
 Objects
in mechanical
equilibrium
 Rock
on ridge
 See-saw
Examples of Objects in Equilibrium
Draw a diagram of the system
1.

Include coordinates and choose a rotation
axis
Isolate the object being analyzed and
draw a free body diagram showing all
the external forces acting on the object
2.

For systems containing more than one
object, draw a separate free body diagram
for each object
Examples of Objects in Equilibrium
Apply the Second Condition of Equilibrium
Στ = 0

This will yield a single equation, often with
one unknown which can be solved
immediately
4.
Apply the First Condition of Equilibrium
ΣF = 0

This will give you two more equations
4.
Solve the resulting simultaneous equations for
all of the unknowns

Solving by substitution is generally easiest
3.
Examples of Objects in Equilibrium

Examples of Free Body Diagrams
(forearm)
• Isolate the object to be analyzed
• Draw the free body diagram for that object
• Include all the external forces acting on the object
Examples of Objects in Equilibrium

FBD - Beam
• The free body
diagram includes
the directions of
the forces
• The weights act
through the
centers of gravity
of their objects
Fig 8.12, p.228
Slide 17
Examples of Objects in Equilibrium

FBD - Ladder
• The free body diagram shows the normal
force and the force of static friction acting on
the ladder at the ground
• The last diagram shows the lever arms for the
forces
Examples of Objects in Equilibrium

Example:
Stress and Strain

So far, studying rigid bodies
 the rigid body does not ever stretch,
squeeze or twist
However, we know that in reality this
does occur, and we need to find a way
to describe it.
 This is done by the concepts of stress,
strain and elastic modulus.

Stress and Strain

All objects are deformable
 All objects are spring-like!
It is possible to change the shape or size
(or both) of an object through the
application of external forces
 When the forces are removed, the object
tends to its original shape

 This is a deformation that exhibits elastic
behavior (spring-like)
Elastic Properties

Stress is the force per unit area causing the
deformation

Strain is a measure of the amount of
deformation
Elastic Modulus

The elastic modulus is the constant of
proportionality between stress and strain
 For sufficiently small stresses, the stress is
directly proportional to the strain
 The constant of proportionality depends on the
material being deformed and the nature of the
deformation

Can be thought of as the stiffness of the
material
 A material with a large elastic modulus is very
stiff and difficult to deform
○ Analogous to the spring constant
Young’s Modulus:
Elasticity in Length

Tensile stress is the
ratio of the external
force to the crosssectional area
 Tensile is because
the bar is under
tension

The elastic
modulus is called
Young’s modulus
Young’s Modulus, cont.

SI units of stress are Pascals, Pa
 1 Pa = 1 N/m2

The tensile strain is the ratio of the
change in length to the original length
 Strain is dimensionless
F
L
Y
A
Lo
s t r e s s = E la s t ic m o d u lu s × s t r a in
Stress and Strain, Illustrated

A bar of material, with a force F
applied, will change its size by:
ΔL/L =  = /Y = F/AY


Strain is a very useful number,
being dimensionless
Example: Standing on an
aluminum rod:







Y = 70109 N·m2 (Pa)
say area is 1 cm2 = 0.0001 m2
say length is 1 m
weight is 700 N
 = 7106 N/m2
 = 104  ΔL = 100 m
compression is width of human
hair
L
F
F
A
L
 = F/A
 = ΔL/L
 = Y·
Young’s Modulus, final


Young’s modulus
applies to a stress of
either tension or
compression
It is possible to exceed
the elastic limit of the
material
 No longer directly
proportional
 Ordinarily does not return
to its original length
Breaking

If stress continues, it surpasses its ultimate
strength
 The ultimate strength is the greatest stress the
object can withstand without breaking

The breaking point
 For a brittle material, the breaking point is just
beyond its ultimate strength
 For a ductile material, after passing the ultimate
strength the material thins and stretches at a
lower stress level before breaking
Shear Modulus:
Elasticity of Shape




Forces may be
parallel to one of the
object’s faces
The stress is called a
shear stress
The shear strain is the
ratio of the horizontal
displacement and the
height of the object
The shear modulus is
S
Shear Modulus, final
F
shear stress 
A
x
shear strain 
h
F
x
S
A
h


S is the shear
modulus
A material having a
large shear modulus
is difficult to bend
Bulk Modulus:
Volume Elasticity

Bulk modulus characterizes the
response of an object to uniform
squeezing
 Suppose the forces are perpendicular to,
and act on, all the surfaces
○ Example: when an object is immersed in a
fluid

The object undergoes a change in
volume without a change in shape
Bulk Modulus, cont.

Volume stress, ΔP,
is the ratio of the
force to the surface
area
 This is also the
Pressure

The volume strain is
equal to the ratio of
the change in
volume to the
original volume
Bulk Modulus, final
V
P  B
V
A material with a large bulk modulus is
difficult to compress
 The negative sign is included since an
increase in pressure will produce a
decrease in volume

 B is always positive

The compressibility is the reciprocal of the
bulk modulus
Notes on Moduli
Solids have Young’s, Bulk, and Shear
moduli
 Liquids have only bulk moduli, they will
not undergo a shearing or tensile stress

 The liquid would flow instead
Ultimate Strength of
Materials
The ultimate strength of a material is the
maximum force per unit area the
material can withstand before it breaks
or factures
 Some materials are stronger in
compression than in tension

Stress and Strain

Example: