Transcript 1 PHYSICS 231 Lecture 20: material science and pressure

PHYSICS 231
Lecture 20: material science and pressure
gas
liquid
solid
Remco Zegers
Walk-in hour: Tue 4-5 pm
Helproom
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quiz
A child runs toward a merry-go-round and jumps on the
edge. As a results, the merry-go-round starts to spin.
Next, she moves towards the center of the
merry-go-round. If no net torque is applied to the child or
merry-go-round then:
a) they will start to spin faster while she moves to the center
b) the rotational velocity will not change while she moves to
the center
c) the will start to spin less fast while she moves to the
the center
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Solid
States of matter
Liquid
Gas
Plasma
difficult
to
deform
difficult
to
compress
difficult
to flow
easy to
deform
easy to
deform
easy to
deform
difficult
to
compress
easy to
flow
easy to easy to
compress compress
easy to
flow
easy to
flow
not
charged
not
charged
not
charged
charged
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Phase transformations
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solids
amorphous
ordered
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crystalline
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The deformation of solids
Stress: Tells something about the force causing the
deformation
Strain: Measure of the degree of deformation
For small stress, strain and stress are linearly correlated.
Strain = Constant*Stress
Constant: elastic modulus
The elastic modulus depends on:
• Material that is deformed
• Type of deformation (a different modulus is defined for
different types of deformations)
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The Young’s modulus
tensile stress
Y
tensile strain
2
tensile stress : F/A [N/m  Pascal (Pa)]
tensile strain : L/L 0
FL0
F/A
Y

L / L0 AL
Beyond the elastic limit an object
is permanently deformed (it does
not return to its original shape if
the stress is removed).
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example
An architect wants to design a 5m high circular pillar with
a radius of 0.5 m that holds a bronze statue that weighs
1.0E+04 kg. He chooses concrete for the material of the
pillar (Y=1.0E+10 Pa). How much does the pillar compress?
F/A
Y

L / L0
2
M statueg /(R pillar
)
L / L0
R=0.5 m L0=5m Y=1.0E+10 Pa M=1.0E+04 kg
5m
L=6.2E-05 m
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question
F/A
Given: Y 
L / L0
A rod with a cross section A of 1 cm2 is stretched by 1 mm
if a force of 1000 N is applied. If on a rod of the same
material but which is 2x loner and has a cross section of
2 cm2 a force of 2000 N is applied, it will stretch by:
a)
b)
c)
d)
1 mm
2 mm
4 mm
8 mm
Fx2, Ax2, L0x2 so L must be twice
larger to keep Y constant
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Ultimate strength
Ultimate strength: maximum force per unit area a material
can withstand before it breaks or fractures.
Different for
compression and tension.
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example
A builder is stacking 1 m3 cubic concrete blocks. Each blocks
weighs 5E+03 kg. The ultimate strength of concrete for
compression is 2E+07 Pa. How many blocks can he stack
before the lowest block is crushed?
The force on the low end of the lowest block
is: F=Nmblockg.
N: total number of blocks
mblock=mass of one block
g=9.81 m/s2
Ultimate strength: 2E+07=F/A
=Nmblockg/1
N=408 blocks.
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The Shear Modulus
x
shear stress
S
shear strain
2
shear stress : F/A [N/m  Pascal (Pa) ]
shear strain : x/h
F/A
Fh
S

x / h Ax
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Moving earth crust
100 m
30 m
A tectonic plate in the lower crust (100 m deep) of
the earth is shifted during an earthquake by 30m.
What is the shear stress involved, if the upper layer
of the earth does not move? (S=1.5E+10 Pa)
shear stress F / A
S

shear strain x / h
F/A=4.5E+09 Pa
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Bulk Modulus
volume stress
B
volume strain
volume stress : F/A [N/m 2  Pascal (Pa)]
volume strain : V/V0
F / A
P
B

V / V0
V / V0
P  pressure
Compressibility: 1/(Bulk modulus)
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example
What force per unit area needs to be applied to compress
1 m3 water by 1%? (B=0.21E+10 Pa)
F / A
B
V / V0
V/V0=0.01 so, F/A=2.1E+07 Pa !!!
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Some typical elastic moduli
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Density
M

V
3
(kg / m )
 specific   material /  water ( 4
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Pressure
Pressure=F/A (N/m2=Pa)
Same Force, different pressure
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example
A nail is driven into a piece of wood with a force of 700N.
What is the pressure on the wood if Anail=1 mm2?
A person (weighing 700 N) is lying on a bed of such
nails (his body covers 1000 nails). What is the pressure
exerted by each of the nails?
Pnail=F/Anail=700N/1E-06m2=7E+08 Pa
Pperson=F/(1000Anail)=700/1E-03=
7E+05 Pa
(about 7 times the atmospheric
pressure).
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Force and pressure
Air (P=1.0E+5 Pa) A
P=0
Vacuum
F
What is the force needed to move the lit?
Force due to pressure difference: Fpressure=PA
If A=0.01 m2 (about 10 by 10 cm) then
a force F=(1.0E+5)*0.01=1000N is needed to pull
the lit.
Fpressure-difference= PA
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Magdeburg’s hemispheres
Otto von Guericke (Mayor of Magdeburg, 17th century)
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