Quick review St St i dth i l ti hi Stress: (Sections 1.3~1.6) Stress, Strain and their relationship A F A F : stress Shear , : Stress Normal Strain: (Sections 2.1~2.2) Angle L xy : strain Shear , : Strain Normal Relating stress and strain—Hooke’s Law: (Sections 3.1, 3.2, 3.4, 3.7, 4.2) E FL G EA G
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Wh thi t h tt it d Th it d f thWhen something gets hotter, it expands. The magnitude of the expansion is proportional to the original size of the object and to the increase in temperature
TL THERMAL
to the increase in temperature.
α is the constant of proportionality, called the coefficient of thermal expansion. It is property of the material. L is the original length.∆T is the increase in temperature.
Th l St i 1)(ssdimensinleofunits:
C
ofUnits
o
Thermal Strain:
T
THERMAL
1)(Tof units
CC
oo
TL
THERMALTHERMAL
Typical values of α are measured in 10-6(oC)-1Typical values of α are measured in 10 6(oC) 1. For structural steel, 1))(612( CE o
Example 3:A brass bar and an aluminium bar are held between two rigid supports with a gap of
B
Aluminium
A brass bar and an aluminium bar are held between two rigid supports with a gap of 0.5 mm between their ends. The temperature is raised by the 60oC. Determine the stress in each bar and the elongation of the brass bar.
Brass
For the brass bar:L=300mm
For the aluminium bar:L=250mm
0.5mm
B
Aluminium
E=105GPaDiameter=50mmα=(18E-6)(oC)-1
E=70GPaDiameter=75mmα=(23E-6)(oC)-1
Brass
Touch? ?50 mm
0.5mm
Touch? ?5.0min mmGapiumTAluTBrass
Brass
Aluminium Touch! Load ExternalThermalTotal
Brass
PPiumPAluiumTAluiumAlu
PBrassTBrassBrass
minminmin
GapiumAluBrass min P
Example 3:Step 1: Determine if the bars will touch
Brass
Aluminium mmETLTBrass 324.060300)618(
Step 1: Determine if the bars will touch.
BrassmmETLiumTAlu 345.060250)623(min
iumTAluTBrass 345.0324.0min
0.5mmmmmm Gap
iumTAluTBrass
5.0669.0min
Touch!Touch!
Step 2: After the bars touch, a compressive force P will act between the bars. Calculate the combined elongation due to the temperature rise and to P.
Elongation due to temperature rise,
mmiumTAluTBrasseTemperatur 669.0min
Elongation due to the compressive force P,
AlAl
Al
BrassBrass
BrassiumPAluPBrassP AE
PLAE
PL min
Example 3: Brass
Aluminium
5.0 GapPeTemperatur
Step 3: To find the value of P by δTotal=δGap.
0.5mmGapPeTemperatur
5.0)6444418()970(
250)6751963()9105(
300669.0
EE
PEE
P)644.4418()970()675.1963()9105( EEEE
kNP 67.74
Step 4: Use the value of P to determine the stress in each bar
MPaEA
PBrass 03.38
751963)367.74(
Stress in the brass bar,
ABrassBrass 75.1963
Stress in the aluminium bar,
MPaEA
P 9.1644.4418
)367.74(
AluminiumAluminium
Example 3: Brass
Aluminium
S 5 U h li i d i hStep 5: Use the earlier equation to determine the elongation in each bar
El ti i th b b 0.5mmElongation in the brass bar,
EPBTBB
300)367.74(324.0
mmEEPBrassTBrassBrass
215.0109.0324.0)675.1963()9105(
324.0
Elongation in the aluminium bar,
E 250)36774(
mmEE
EiumPAluiumTAluiumAlu
285.0060.0345.0)644.4418()970(
250)367.74(345.0minminmin
Chapter 6: Chapter 6: Poisson’s Ratio (Section 3.6)Poisson’s Ratio (Section 3.6)P Poisson’s Ratio
lateral
P Poisson’s Ratio
B l t
v is a material property.
allongitudinP Bulge out
essdimensionlssdimensinlessdimensinle
of units of units ofUnits
Becomes thin
A block has a stress in the X direction only and the strain
Typical values of v are 0.25 to 0.4. For structural steel, v=0.32.Becomes thin
x
A block has a stress in the X-direction only, and the strain in the X direction will be:
σx
Ex z
yThe strain in the Y and Z directions:
x σx E
xxy
Ex
xz
MultiMulti--axial loadingaxial loading
I l σ
zyx
In general,z
x y
σy σx σz
EEEzyx
x σx
σy
σz EEEzyx
y
EEEzyx
z
EEE
Relationships between Young’s Modulus (E), Shear Relationships between Young’s Modulus (E), Shear M d l (G) d P i ’ R ti ( )M d l (G) d P i ’ R ti ( )
EG
Modulus (G) and Poisson’s Ratio (v)Modulus (G) and Poisson’s Ratio (v)