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Unformatted text preview: Chapter 9
Elasticity and Fracture Static Equilibrium
Static Equilibrium A special case of mechanics
− When the net force and net torque on an object,
or system of objects, are both zero.
− Both linear acceleration and the angular
acceleration of the object or system is zero.
− The object is either at rest or its center of mass is
moving at a constant velocity.
moving Static Equilibrium
Static Equilibrium Statics
− The calculation of the forces acting on and within
structures that are in equilibrium.
structures The Conditions of The Conditions of Equilibrium Equilibrium
− Objects in daily life have at least one force acting
− If they are at rest, then there are other forces
acting on them as well so that the net force is
− If all the net forces acting on an object are zero,
then the object is in equilibrium.
equilibrium. The Conditions of The Conditions of Equilibrium First Condition for Equilibrium
− For an object to be at rest, Newton’s second law
says that the forces acting on it must be zero.
− Since force is a vector, the components of the net
force must also be zero.
force The Conditions of The Conditions of Equilibrium Second Condition for Equilibrium
− Although the net force on it is zero, the ruler will
− A pair of equal forces acting in opposite
directions but at different points on an object is
referred to as a couple.
couple. The Conditions of The Conditions of Equilibrium Second Condition for Equilibrium
− The sum of the torques acting on an object,
calculated about any axis, must be zero.
calculated − This ensures that the angular acceleration is zero.
− If the object is not rotating, then it will not start
rotating. Stability and Balance
Stability and Balance Equilibrium
− An object in static equilibrium, if left
undisturbed, will undergo no translational or
rotational acceleration since the sum of all the
forces and the sum of all the torques acting on it
is Stability and Balance
Stability and Balance Equilibrium
− If an object is displaced slightly, it can:
− Return to its original position = stable equilibrium.
− Move even farther from its original position =
− Remain in its new position = neutral equilibrium.
neutral Stability and Balance
Stability and Balance Balance
− An object that is in stable equilibrium.
− An object whose center of gravity is above its base of
support will be stable if a vertical line projected
downward from the center of gravity falls within the base
of Elasticity; Stress and Elasticity; Stress and Strain Elasticity
− Any object changes shape under the action of
− If the forces are great enough, the objects will
break, or fracture.
fracture. Elasticity; Stress and Elasticity; Stress and Strain Elasticity and Hooke’s Law
− If the amount of change Δl is small relative to the
length of the object, then Δl is proportional to the
force exerted on the object defined by Hooke’s
Law: Elasticity; Stress and Elasticity; Stress and Strain Elasticity and Hooke’s Law
− Hooke’s Law is valid for almost any solid up to a
− If the force is too great, then the object will
break. Elasticity; Stress and Elasticity; Stress and Strain Elasticity and Hooke’s Law
− Applied force versus elongation will be linear up
to a point called the proportional limit.
− Past the proportional limit, the relationship
becomes more complicated.
− Slightly past the proportional limit, up to the
elastic limit, the object will return to its original
shape if the force is removed.
shape Elasticity; Stress and Elasticity; Stress and Strain Elasticity and Hooke’s Law
− The region between the origin and the elastic
limit is called the elastic region.
− Beyond the elastic region, the object enters the
plastic region where it will NOT return to its
original shape after the force is removed.
− The maximum change in length is reached at the
− The maximum force that can be applied without
breaking the object is called the ultimate
strength. Elasticity; Stress and Elasticity; Stress and Strain Elasticity and Hooke’s Law Elasticity; Stress and Elasticity; Stress and Strain Stress = force applied to an object per area
area A Strain = the change in length of the object as a
result of the stress
change in length ∆l
lo Elasticity; Stress and Elasticity; Stress and Strain Young’s Modulus
− The amount of elongation also depends on the
type of material from which an object is made.
− It is independent of the object’s original size and
shape. Elasticity; Stress and Elasticity; Stress and Strain Tension, Compression, and Shear Stress
− An object in tension has tensile stress.
− A force is pulling one down and an equal force is
pulling up to balance.
− Tensile stress extends objects. Elasticity; Stress and Elasticity; Stress and Strain Tension, Compression, and Shear Stress
− When forces act inwardly on an object, it is
− Compressive stresses shorted objects. Elasticity; Stress and Elasticity; Stress and Strain Tension, Compression, and Shear Stress
− An objects under shear stress has equal and
opposite forces applied across its opposite faces.
opposite Elasticity; Stress and Elasticity; Stress and Strain Tension, Compression, and Shear Stress
− Shear stress changes the shape of an object.
− The shear stress can be calculated by: − G is the Shear modulus. Elasticity; Stress and Elasticity; Stress and Strain Volume Change – Bulk Modulus
− If an object is subject to inward forces from all
sides, its volume decreases.
− The force per area on an object is called pressure
and is equivalent to stress.
and Elasticity; Stress and Elasticity; Stress and Strain Volume Change – Bulk Modulus
− The change is volume as a result of pressure is
proportional to the original volume and to the
change in pressure.
change − B is the proportionally constant called the bulk
modulus. Elasticity; Stress and Elasticity; Stress and Strain Elasticity; Stress and Elasticity; Stress and Strain Fracture
− If the stress on an object is too great, then is breaks, or
− A horizontal beam will be under both tensile and
compressive stress due to its own weight. Therefore, it
must be made of a material that is strong under both
compression and tension.
compression Elasticity; Stress and Elasticity; Stress and Strain Problems Example 1: A tower crane must always be carefully balanced so that
there is no net torque tending to tip it. A particular crane at
a building site is about to lift a 2800-kg air-conditioning
unit. (a) Where must the crane’s 9500-kg counterweight be
placed when the load is lifted from the ground? (Note that
the counterweight is usually moved automatically via
sensors and motors to precisely compensate for the load.)
(b) Determine the maximum load that can be lifted with this
counterweight when it is placed at its full extent. Ignore the
mass of the beam.
mass Problems Example 1: Problems
Problems Example 1:
• (a) See the free-body diagram. Calculate
torques about the pivot point P labeled in the
diagram. The upward force at the pivot will
not have any torque.
P x r
Mg FP d r
mg Problems Example 1:
• (a) The total torque is zero since the crane is
in ∑τ = Mgx − mgd = 0
M → ( 2800 kg ) ( 7.7 m )
( 9500 kg ) = 2.3m Problems Example 1:
• (b) Again we sum torques about the pivot
point. Mass m is the unknown in this case,
and the counterweight is at its maximum
distance from the pivot.
∑τ = Mgxmax − mmax gd = 0 → mmax = Mxmax
d = ( 9500 kg ) ( 3.4 m )
( 7.7 kg ) = 4200 kg Problems
Problems Example 2: Find the tension in the two wires supporting
the traffic light. Problems Example 2:
• Draw a free-body diagram of the junction of
the three wires.
FT1 53o r
mg Problems Example 2:
• The tensions can be found from the
conditions for force equilibrium.
conditions ∑F = FT1 cos 37° − FT2 cos 53° = 0 → FT2 = ∑F = FT1 sin 37° + FT2 sin 53° − mg = 0 x y cos 37°
cos 53° FT1 Problems Example 2:
FT1 sin 37° + FT1 = FT2 = cos 37°
cos 53° FT1 sin 53° − mg = 0 → ( 33 kg ) ( 9.80 m
sin 37° + cos 37°
cos 53° FT1 = cos 37°
cos 53° cos 37° s2 ) = 194.6 N ≈ 190 N sin 53° (
cos 53° ) 1.946 × 102 N = 258.3 N ≈ 260 N Problems
Problems Example 3: A refrigerator is approximately a uniform
rectangular solid 1.9 m tall, 1.0 m wide, and
0.75 m deep. If it sits upright on a truck with
its 1.0-m dimension in the direction of travel,
and if the refrigerator cannot slide on the
truck, how rapidly can the truck accelerate
without tipping the refrigerator over? Problems
Problems Example 3:
• Assume the truck is accelerating to the right.
We want the refrigerator to not tip in the
non-inertial reference frame of the truck.
Accordingly, to analyze the refrigerator in
the non-inertial reference frame, add a
pseudoforce in the opposite direction of the
actual acceleration. Problems
Problems Example 3:
• The free-body diagram is for a side view of the
refrigerator, just ready to tip so that the normal
force and frictional force are at the lower back
corner of the refrigerator. The center of mass is in
the geometric center of the refrigerator.
mg h Problems Example 3:
• Write the conditions for equilibrium, taking
torques about an axis through the center of
mass, perpendicular to the plane of the
paper. The normal force and frictional force
cause no torque about that axis.
cause Problems Example 3: ∑F
∑F = Ffr − ma truck = 0 → Ffr = matruck horiz
vert = FN − mg = 0 → FN = mg ∑τ = F (
Ffr = h
w = 1
2 w ) − Ffr ( h ) = 0 →
ma truck → atruck = g w
Ffr ( = h
w = 9.80 m s 2 1.0 m ) 1.9 m = 5.2 m s 2 Problems Example 4: How much pressure is needed to compress
the volume of an iron block by 0.10%?
Express your answer in N/m2 and compare it
to atmospheric pressure (1 X 105 N/m2)
(1 Problems Example 4:
• The relationship between pressure change
and volume change is given by:
and ∆V = −V0
Patm = ∆P
B → ∆P = − 9.0 × 107 N m 2
1.0 × 105 N m 2 ∆V
V0 ( B = − 0.10 × 10−2 ) ( 90 × 10 = 9.0 × 102 , or 900 atmospheres 9 ) N m 2 = 9.0 × 107 N m 2 Problems Example 5: A steel cable is to support an elevator whose
total (loaded) mass is not to exceed 3100 kg.
If the maximum acceleration of the elevator
is 1.2m/s2 calculate the diameter of cable
required. Assume a safety factor of 8.0.
Problems Example 5:
• From the free-body diagram, write Newton’s
second law for the vertical direction. Solve
for the maximum tension required in the
cable, which will occur for an upwards
acceleration. ∑F y = FT − mg = ma → FT = m ( g + a ) r
mg Problems Example 5:
• The maximum stress is to be 1/8th of the
tensile strength for steel. The maximum
stress will occur for the minimum area, and
thus the minimum diameter.
stress max =
Amin = tensile strength 4 ( 8.0 ) m ( g + a ) 8.0 π ( tensile strength ) = → A1 = π ( 1 d ) =
2 ( 32 ( 3100 kg ) 11.0 m s 2 π ( 500 × 106 N m 2 ) ) 8.0 FT
tensile strength → = 2.6 × 10−2 m = 2.6 cm Problems
Problems Example 6: A 50-story building is being planned. It is to be
180.0 m high with a base 46.0 m by 76.0 m. Its total
mass will be about 1.8 X 107 kg and its weight
therefore about 1.8 X 107 N. Suppose a 200 km/h
wind exerts a force of 950 N/m2 over the 76.0 m
wide face. Calculate the torque about the potential
pivot point, the rear edge of the building, and
determine whether the building will topple. Assume
the total force of the wind acts at the midpoint of
the building’s face, and that the building is not
anchored in bedrock. Problems Example 6: Problems Example 6:
• Assume that the building has just begun to
tip, so that it is essentially vertical, but that
all of the force on the building due to contact
with the Earth is at the lower left corner.
FA 23.0 m
90.0 m r
FE y r
FE x Problems Example 6:
• Take torques about that corner, with counterclockwise
torques as positive.
torques ∑τ = F ( 90.0 m ) − mg ( 23.0 m )
= ( 950 N m ) ( 180.0 m ) ( 76.0 m ) ( 90.0 m ) − ( 1.8 × 10 kg ) ( 9.80 m s ) ( 23.0 m ) A 2 7 2 = −2.9 × 109 mg
N • Since this is a negative torque, the building will tend to
rotate clockwise, which means it will rotate back down to
the ground. Thus the building will not topple. Problems
Problems Example 7: The roof over a 9.0 m X 10.0 m room in a
school has a total mass of 13,600 kg. The roof
is to be supported by vertical “2 X 4’s”
(actually about 4 cm X 9 cm ) equally spaced
along the 10.0 m sides. How many supports
are required on each side, and how far apart
must they be? Consider only compression,
and assume a safety factor of 12. Problems
Problems Example 7:
• The number of supports can be found from
the compressive strength of the wood. Since
the wood will be oriented longitudinally, the
stress will be parallel to the grain. Problems Example 7:
Safety Factor ( # supports ) = Load force on supports = Area of supports = Weight of roof ( # supports ) ( area per support ) Weight of roof Safety Factor ( area per support ) Compressive Strength ( 1.36 × 10 kg ) ( 9.80 m s )
4 ( 0.040 m ) ( 0.090 m ) 2 12 ( 35 × 10 6 Nm 2 ) = 12.69 supports Problems
Problems Example 7: Since there are to be more than 12 supports,
and to have the same number of supports on
each side, there will be 14 supports, or 7
supports on each side. That means there will
be 6 support-to-support spans, each of which
would be given by
Spacing = 10.0 m
6 gaps = 1.66 m gap End of Chapter 9
End of Chapter 9 ...
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This note was uploaded on 08/10/2011 for the course GOV 102 taught by Professor Smithy during the Spring '11 term at CSU San Marcos.
- Spring '11