Transcript Static Equilibrium. Supports, Loads, Driven Oscillations

Strength of Materials
9/18/07
Topics to Cover
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Pressure
Stress
Strain
Scale and Square-Cube Effect
Tension
Compression
Shear
Effect of force on non-rigid objects
• Besides the force, what else is important?
• Examples to consider:
– Ice on a pond
– Hammering a nail into wood
• Pressure!
• Area
– The AREA over which it is applied; the larger
the area, the lower the pressure.
Utilizing pressure
• Many simple devices
– knives, picks, snowshoes, bulletproof vests, all simply
change the AREA over which a force is applied.
• Distribution of Force
(http://www.physics.unl.edu/outreach/football.html)
• Bed of Nails (http://video.google.com/videoplay?docid=1891333794864062153)
• Bed of Nails (wileyplus – ch. 6)
Pressure
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What are the units of pressure?
Hint – force per unit area
Pounds per square inch, abbreviated psi.
Inflating a tire; a car tire typically takes ~ 30 psi.
Pressure of the air around us is about 14 psi.
Pressure – con’t
• Pressure can take several forms in the material.
• Pulling the material apart, material is in
TENSION, the pressure is called the stress.
• Pushing together, material is in
COMPRESSION, and it’s again a stress.
• Examples
– strings can have tension, but not compression
– Springs or rigid objects can do both
Think – Pair - Share
• Consider a bed of nails with 600 nails
• A person weighs 150 lbs lies on it
• Will it puncture the skin? (It takes 2 lbs
per nail to puncture the skin)
• How much weight must be applied to
puncture the skin?
http://phun.physics.virginia.edu/demos/nail_bed.html
Think – Pair - Share
• High-heeled shoes can cause tremendous
pressure to be applied to a floor.
• Suppose the radius of a heel is 6.00 × 10-3 m.
• At times during a normal walking motion, nearly
the entire body weight acts perpendicular to the
surface of such a heel.
• Find the pressure that is applied to the floor
under the heel because of the weight of a
50.0 kg woman.
Answer

F m  g 50kg  9.8m / s
P 

2
2
A  r
  0.006m
2
  4,330kPa
Think – Pair - Share
• What exerts more pressure-per-square inch when
walking a 100 lb woman in high heels or a 6,000 lb
elephant in bare feet?
• [At the moment when only the heel rests on the ground.]
• Stiletto heels have an area of about 1/16 of a square
inch.
• Elephants, unlike humans, walk with two feet on the
ground at a time.
• Each foot is about 40 square inches.
Answer
• Pressure
F
P
A
• Woman’s heels
100lbs
lbs
P
 1600 2  1600psi
2
1 in
in
16
• Elephant’s feet
6000lbs
P
 75 psi
2
2  40in
• The heels exert more pressure!
Stress
• Internal distribution of forces within a body that
balances and reacts to the loads applied to it
• Tension, compression, or shear
Strain
• The amount of deformation in a material.
If we are stretching a bar, then the strain is
the percent change in its length.
• Simple case of strain proportional to stress
• Small forces
• can’t model objects breaking, but it does
work for the amount a bridge sags when a
car drives on it
Equations
Stress  E  Strain
Stress
E
Strain
• E is the modulus of elasticity
• Units of pressure (N/m2 or psi)
Stress and Strain – con’t
• What is the amount a bar or wire can be
stretched or compressed in one dimension?
(Hooke’s Law).
• Table 2.1 on page 21 of our book has a list
of constants for Moduli of Elasticity
• Conversion from N/m2 to Psi is 6830:1
L
Strain 
L
Stress
E
Strain
Stress
E
L
L
Behavior of Materials
• Elastic Behavior
– Recovers shape – no deformation
• Plastic Behavior
– Range between elastic limit and breaking
– Strain is not proportional to stress
• The Ultimate Tensile (or Compressive) Strength is where
a stretched material breaks.
– sample values in N/m2 are
• 4e7 for concrete or brick,
• 1e8 for Al,
• 2.3e8 copper,
• 5e8 for steel
• Bone 12e7 tension, 17e7 compression.
Stiffness
• Ability to resist strain
• Steels
– Same E
– different yield points
– different ultimate strengths
Stress and Strain – example 1
• How far can you stretch a 20 ft steel cable
before it snaps?
• E for steel is 2*1011
• Ultimate tensile strength for steel is 5.2*108
Stress Stress Stress * L
E



L
Strain
L
L
L * Stress
L 
E
20 ft * 5.2 108
L 
 0.05 ft  0.6inches
11
2 10
Materials
• Wood and metal pieces respond about the
same to compression and tension.
• Stone pieces do NOT
– strong in compression (hard to crush)
– weak in tension (easy to pull apart)
Space Elevator
• Discussion of Space Elevator
http://science.nasa.gov/headlines/y2000/ast07sep_1.htm
• Basically a cable goes up to geosynchronous
orbit with ballast mass (asteroid) slightly above
it.
• Best calculations are that a Tensile Strength of
6.2 x 1010 N/m2 = 9.1 x 106 psi (about 100 times
stronger than steel) is required.
• Carbon nanotubes have a tensile strength 3
times this (Young’s Modulus is 1012 N/m2),
• Growing them long enough is a real concern.
How tall can a mountain
on Earth be?
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Rock
– Compressive Strength of 2 x 108 N/m2= 3 x 104 psi
– Specific gravity ~ 3
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Pressure at the bottom of a column is therefore weight/area
Weight = weight_water*specific gravity*height*area
areas cancel
Pressure at bottom = (0.036 pounds/in3)*(about 3)*height = Ultimate
Strength.
Height = (3 x 104 psi)/((.036 pounds/in3)*3) is about 3 x 105 inches, or
25,000 feet (over 4 miles)
Height of Mt. Everest is 29,000 feet – pretty good guess
If gravity was weaker, then the weight would be less, and mountains could
grow taller.
Mars has a gravity about 1/3 of Earth’s, and its tallest mountain (Mount
Olympus) is about three times as big as Everest.
Scale
• Imagine a simple building, as a cube 10’ on a
side.
– Volume is 10’ x 10‘ x 10’ = 1,000 ft3
– Area is 10’ x 10’ = 100 ft2
• If you double each dimension, what happens to
the VOLUME and AREA of the cube?
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Volume is 20’ x 20‘ x 20’ = 8,000 ft3
8 fold increase in volume
Area is 20’ x 20’ = 400 ft2
4 fold increase in area
Scale – con’t
• Weight of a structure scales with volume, and the strain
on the supports scale with area
• Building larger and larger with the same materials leads
to more and more strain and eventually collapse.
• Thus larger buildings need thicker supports:
• If you double the size, for example, then AFTER
doubling you need to make the supports twice the area,
or 40% larger in each direction.
• Of course eventually this becomes impractical, which is
why one needs stronger materials to build larger
structures: stone instead of wood, steel instead of stone.
Scale – con’t
• If you could somehow make a mouse as big as a
gorilla, it would probably just collapse
• Its bones (the support system) would be too thin
to support the increased weight at that size.
• Any sci-fi movie with a “shrinking ray” is
nonsense.
Shear Modulus
• Shear Modulus (ripping apart)
• Shear refers to cases where we are not directly
pushing in or pulling out on a material, but are
applying forces perpendicular to the surface.
• (Example: thick book)
• Effectively shear is a combination of tension on
one axis and compression on another, causing
the material to deform or twist.
Stress/Area Relationship Problem Compresion
• Given:
a 6400# weight is supported by an
8" x 8" post.
Determine:
a. the compression stress in the 8" x 8" post
b. the bearing stress between the post and the steel plate
c. the bearing stress between the steel plate and the concrete.
Solution:
a. The compressive force in the post : 6400 pounds/ 64 in2 = 100 psi
b. The stress at the point between the post and the steel bearing plate is dictated
by the smallest area being loaded. The stress between the two is again 6400
pounds divided by 64 square inches for a total stress of 100 psi.
c. The stress between the plate and the concrete floor is also determined by the
smallest area. In this case it is the 10" x 10" plate: 6400 pounds / 100 in2. The
resulting stress is 64 psi, a clear reduction in the stress.
Think – Pair - Share
Stress/Area Relationship Problem - Tension
• Given:
• A 1000 lb load suspended from
a ceiling by a 1"x 1" member.
Determine:
a. the stress in the member
b. the stress if the size of the
member is increased to 2" x 2".
Solution:
a. The internal force in that member is 1000 lb tension and the tension stress
(intensity of the force per unit of area) is 1000 pounds divided by the area of 1
square inch. This is a stress of 1000 psi.
b. The 1000 lb load is now distributed evenly across an area of 4 square inches;
thus, 1000 lb divided by 4 in2 is a stress of 250 psi. This clearly demonstrates the
inverse relationship between stress and area.
Solution
• a. the stress in the member
– The internal force in that member is 1000 lb tension
and the tension stress (intensity of the force per unit
of area) is 1000 pounds divided by the area of 1
square inch. This is a stress of 1000 psi.
• b. the stress if the size of the member is
increased to 2" x 2".
– The 1000 lb load is now distributed evenly across an
area of 4 square inches; thus, 1000# divided by 4 in^2
is a stress of 250 psi. This clearly demonstrates the
inverse relationship between stress and area.
Think – Pair - Share
• Given:
• A 5000 lb load on 1” diameter rod.
Determine:
a. Is the rod in tension or compression?
• B. What is the stress in the rod?
b. What is the stress if the diameter of the rod is increased to 2".
Solution
Review
• How materials respond to applied forces
• When we apply a force over some area (a
pressure), we create stress in the material,
which then undergoes strain and changes
shape in response (tension, compression,
and shear).
• Scaling – pressure scales as area, weight
as volume, so a larger object needs
proportionally thicker supports
Lecture for 2/8/07
Deformations from Temperature
• Atomic chemistry reacts to changes in
energy
• Solid materials
– Can contract with decrease in temperature
– Can expand with increase in temperature
• Linear change can be measured per
degree
Thermal Deformation
•  - the rate of strain per degree
• UNITS:
°F,
• length change:
• thermal strain:
°C
 T   (T ) L
 T   (T )
• - no stress when movement allowed
Changes in Size
• Similar to the stress-strain relationship, assume:
– temperature changes are not too extreme
– change in the object is linear with temperature
• Simple case is a solid bar of material with a length,
width, and depth.
• When the object changes temperature by an
amount T, the length of the bar becomes
increases.
L  L0    T
T is thechangein temperature,  is hotter
 is rateof changeper degree
Expansion and Contraction
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Example
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a long bar of Aluminum 2” x 6” x 15 feet, with and  = 1.3 x 10-5 per °F
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heat from room temperature of 70 F to 130 F, what are its new dimensions?
Solution
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First step: T=130 – 70 = 60°F
Multiply  x T = 1.3 x 10-5 /°F x 60°F = 0.00078
Consider each dimension
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2” x 0.00078 = 2.002”
6” x 0.00078 = 6.005”
180” x 0.00078 = 180.14”
expands a bit more than an eight of an inch length-wise.
This is proportional to the length, so for a long bridge it could shift by an inch or even several
inches – enough to make a fixed connection untenable.
Hence a roller support can be an advantage for a material that expands and contracts – by
allowing one side to translate, there is no compression in the horizontal direction for the
bridge.
Demo
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http://www.physics.umd.edu/lecdem/services/demos/demosi1/i1-11.htm
Superposition Method
• Can remove a support
• Replace the support with a reaction
• Enforce the geometry constraint
Superposition Method – con’t
• Total length change
restrained to zero
• Constraint:
T   P  0
sub :
PL
P  
 T   (T ) L
AE
PL

  (T ) L  0
AE
P
f     (T ) E
A
Masonry
• Like all masonry, bricks are strong in compression but
weak in tension. You can see this with a single brick.
How?
Demonstration
Compression and Tension
• DEMO: compressible foam
– A force pushing down on a broad surface will clearly
cause compression beneath it.
– If the force is applied in the middle, then clearly
everywhere will be compressed, though the sides less
than the middle.
– However, if the force is applied at one edge, the top
will “see-saw” upwards, so that as we compress one
end we actually put the other side into tension.
– How far out can I go and not cause any tension?
Middle Third Rule
• Must stay within the “middle third” of the block to prevent
the opposite side from going into tension
– this is the important MIDDLE THIRD RULE.
• Why is it important?
– Because bricks/stone/etc are much stronger in compression than
in tension; need to keep them in compression or they can break!
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• To get around this problem
– reinforce concrete with a steel bar
– the steel can take the tension when the concrete can’t.
• We’ll go over this in more detail in Chapter 8.