As h 22 Materials

8/12/2019 As h 22 Materials

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Density ()

density = massvolume

= m / V

unit = kg m-3

Note: 1 g cm-3is the same

as 1000 kg m-3


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Density examplesdensity

/ kg m-3

density

/ kg m-3Interstellar medium iron

hydrogen lead

helium mercuryair uranium

wood (average) gold

lithiumwater Suns core

plastics neutron star

aluminium black hole

0.0989

0.1791.29

0.534

19 100

850to1400

10-25to 10-15

13 500

150 000

700

1000

2 700

7 900

11 300

22 610

19 300

1017

> 4 x 1017

osmium


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Question

Calculate the weight of a gold ingot ofdimensions (20 x 10 x 4) cm


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Answers

density mass volume

240 g 40 cm3

3000 kg m-3 4500 kg

0.80 g cm-3 80 cm3

9 kg 0.003 m3

Complete:


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Hookes law

The force(F ) needed to stretch a spring is directlyproportional to the extension(L ) of a spring from itsnatural length.

F L

Adding a constant of proportionality:

F = k Lkis called the spring constant

The spring constant is the force required to produce anextension of one metre.

unit = Nm-1


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Elastic limit

Up to a certain extension if the force isremoved the spring will return to its originallength. The spring is said to be behavingelastically.

If this critical extension is exceeded, known asthe elastic limit, the spring will be permanentlystretched.

Plasticbehaviour then occurs and Hookes lawis no longer obeyed by the spring.


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Tensile stress ()

A stretching force is also called a tensileforce.

Tensile stress = tensile force

cross-section area

= F/ A

unitPa (pascal) or Nm-2

Note: 1 Pa = 1 Nm-2


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Breaking stress

This is the stress required to cause a

material to break.


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Tensile strain ()

Tensile strain = extension

original length

=L / L

unitnone (its a ratio like pi)


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Question

A wire of natural length 2.5 m and diameter0.5 mm is extended by 5 cm by a force of

40 N. Calculate:

(a) the tensile strain(b) the tensile stress

(c) the force required to break the wire if its

breaking stress is 1.5 x 109Pa.


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The Young Modulus (E )

This is a measure of the stiffness of a material.

Young modu lus = tensi le stress

tensi le strain

E = /

unitpascal (same as stress)


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Also: tensile stress = F / A

and tensile strain =L / L

Therefore: E = (F / A )

(L / L)

which is the same as:

E = F L

A L


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Examples of Young Modulus

Material E/ x 109Pa

diamond 1200

titanium carbide 345

steel 210copper 130

brass 100

glass 80oak 12

rubber band 0.02


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Question 1

Calculate the tensile strain caused to a steelwire when put under 4.0 x 10 7Pa of stress.


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Question 2

A metal wire of original length 1.6m, crosssectional area 0.8 mm2extends by 4mm when

stretched by a tensile force of 200N.

Calculate the wires

(a) strain,

(b) stress

(c) Young Modulus.


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Measurement of E

With equal control and test weightsof 10N adjust the micrometer

attached to the test wire so that the

spirit level between the two wires is

horizontal.

Note the reading on the micrometer

and also the length of the test, L

wire using a metre ruler.

Use another micrometer to measurethe diameter of the test wire at

various places along the wire and

calculate an average value, D.

hinge

test

weights

rigid support

spiritlevel

long

wires

control

weight

micrometer


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Measurement of E

Calculate the average cross-section area of the wire, A fromA = D2/4

Add an additional load, Fof 5N

to the test wire.

Readjust the micrometer tobring the spirit level again andnote the new reading

hinge

test

weights

rigid support

spiritlevel

long

wires

control

weight

micrometer


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Measurement of ECalculate:(a) the extension,Lcaused by theaddition of 5N to the test wire.

(b) the tensile strain, produced in thewire using: = L/ L

(c) the tensile stress, applied to thewire using: = F / A

Repeat with additional 5Nloads.

hinge

test

weights

rigid support

spiritlevel

long

wires

control

weight

micrometer


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Measurement of E

Stop before the strain reaches0.01 in order to prevent the wireexceeding its limit ofproportionality (just before theelastic limit).

Draw a graph of stress againststrain. This should be a straightline through the origin.

Measure the gradient of thisgraph which will be equal to theYoung Modulus, Eof the testwire.

0

Stress, / Pa

Strain,

Gradient= / = E


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Stressstrain curves(a) Metal wire (e.g. steel)

P = Limit of

proportionality

Up to this point

the stress is

proportional to

the strain.

stress

strain

P


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E = Elastic limit

This is close to P

Beyond this point the

wire will become

permanently

stretched and sufferplastic deformation.

stress

P

E

strain


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Y1= Yield point

This is where the

wire weakens

temporarily.

BeyondY2, a small

increase in stress

causes a large

increase in strain asthe wire undergoes

plastic flow.

Y1stress

P

E

strain

Y2


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UTS = Ultimatetensile stress

Beyond the

maximum stress,(UTS), the wire losesits strength, extendsand becomesnarrower at itsweakest point whereit fractures at B

Y1stress

P

E

UTSbreaking

point B

strain

Y2


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Stressstrain curves(b) Brittle material (e.g. glass)

A brittle material does

not undergo plastic

deformation and willfracture at its

elastic limit.

stress

P

E

breaking

point B

strain


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Stressstrain curves

(c) Ductile material (e.g. copper)

A ductile material can bedrawn into a wire.

Both steel and copper areboth ductile but copper ismore ductile because itcan withstand a greaterstrain than steel beforebreaking although it is notas strong or as stiff assteel.

stress

copper

steel

strain


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Elastic strain energy

When a spring or wire is stretched potentialenergy is stored.

This form of potential energy is called elasticstrain energy.

Consider a spring of original length L

undergoing an extensionLdue to a tensileforce F.


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The graph opposite showshow the force varies as thespring extends.

The work done in extendingthe spring is given by:

work = force x distance0

force

extension


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= average tensile force x extension

= F L

= area under the curve

= energy sto red in the spr ing

and so:

elastic strain energy = F L

0

F

area = F L

L

force

extension


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Stretching rubber

The work done in stretching rubberup to extensionLis equal to thearea under the loading curve.

The unloading curve for rubber isdifferent from its loading curve.

When the rubber is unloaded onlythe energy equal to the area underthe unloading curve is returned.

The area between the two curves is

the energy transferred to internalenergy, due to which the rubberband becomes warmer.

0

unload ing

L

force

extension

loading

energy lost

to heating

the rubber


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Answers

tensile force extension strain energy

120 N 2 m

40 N 15 cm

3 kN 50mm 150 J

2MN 6 m 12 J

Complete:


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Question

A spring of original length 20cm extends to25cm when a weight of 4N is hung from it.

Calculate:

(a) the elastic strain energy stored in thespring,

(b) the spring constant

(c) the length of the spring when it isstoring 0.5 J of energy.

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As h 22 Materials