axial deformation 2
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pertemuan 3TRANSCRIPT
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Fall 2002 Axially Loaded Bars 1
Axial Deformation
Chapter 5Lecture 2
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Strain Distribution
Axial deformation is characterized by extensional strain that is not a function of
position in the cross section.
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Strain-Displacement Relationship
Length Initial
Length Initial Length Final -ε
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Strain-Displacement Relationship
x*
x x 0
0
x
x
x xlim
x
u x x u xlim
du
dx
x
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Strain-Displacement Equation
dx
duε
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Strain-Displacement Equation
L
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Material BehaviorAssume linear elastic behavior:
x xE
EE is a material property called the modulus of elasticity or Young’s modulus.
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Total Elongation
u L u
x dxL
0
0
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Summary
Axial Stress Formula
Axial Strain Formula
Axial Force-Deformation
Equation
x
x
L
0
x
0
P xx
A x
P xdu
dx EA x
P xe dx
EA x
Pu u 0 d
EA
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Uniform Axial Deformation
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Uniform Axial Deformation
constant
L L
0 0
F x 0 P x P
P x P PLdx dx
EA x EA EA
Force-Deformation Behavior.
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Uniform Axial Deformation
PL
EA
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Flexibility
f flexibility coefficie t
f
n
P
L
LP
EA
EA
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Stiffness
k stiffness coefficien
k
t
PLP
EA
EA
L
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Definitions1. The flexibility coefficient, f, is the
elongation produced when a unit force is applied to the member. It has dimensions of Length/Force (L/F )
2. The stiffness coefficient, k, is the force required to produce a unit elongation of the member. It has dimensions of Force/Length (F/L)
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Lateral Strain
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Lateral Strain
lat
dL dL
dL
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Lateral Strain
lat
lat lat
L 1 L
L L L
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Lateral Strain
lat
lat
0 0
0 0
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latD 1 D
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Poisson’s Ratio
Greek letter nu is
the Poisson's Ratio
lat
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Poisson’s Ratio
Poisson's Ratio is a material property
just as the modulus of elasticity is.
0 0.5
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Lateral Strain
lat
P
EA
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Example
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A prismatic bar with length L = 200 mm and a circular cross section with a diameter D = 10 mm is subjected to a tensile load P = 16 kN. The length and diameter of the deformed bar are measured and determined to be L’ = 200.60 mm and D’ = 9.99 mm. What are the modulus of elasticity and the Poisson’s ratio for the bar?
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Example
Stress:
Strain:
Lateral Strain
2
lat
P 16000N203.7 MPa
A .01m
4
L L 200.60mm 200mm0.003
L 200mm
D D 9.99mm 10mm0.001
D 10mm
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Example
lat
lat
203.7 MPa
0.003
0.001
203.7 MPaE 67.9 GPa
0.003
0.0010.333
0.003
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Example
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Example
The bar has cross sectional area of A = 0.4 in2
and a modulus of elasticity of E = 12 x 106 psi. If a 10 kip force is applied downward at B how far downward does point B move?
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Example
10 kip
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yF 10kip Psin 0
P 12.5kip
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6 22
P 12.5kip
L 20 in
12500lb 20inPL0.0521in
lbEA12 10 0.4in
in
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20 in
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small deformation approximation
2 2 2
2 2
2 2
12 16 v 20
32v v 40
v v
32v 40
v 0.0651 in
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Example
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Bars AB and AC each have a cross sectional area A = 60 mm2 and a modulus of elasticity E = 200 GPa The dimensions h = 200 mm If a downward force F = 40 kN is applied at A what is the resulting horizontal and vertical displacements of point A?
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40 kN
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o ox AB AC
o oy AB AC
AB
AC
F 0 P cos60 P cos45 0
F 0 P sin60 P sin45 40kN 0
P 29.28kN
P 20.70kN
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AB
AC
oAB AB
AB9 5 2
2
oAC AC
AC9 5 2
2
P 29.28kN
P 20.70kN
0.2m29280N
P L sin60.000563m
NEA200 10 6 10 m
m
0.2m20700N
P L sin45.000487m
NEA200 10 6 10 m
m
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2 2 2o oAB AB AB AB
o 2 o 2 2AB AB AB AB AB
o oAB
L sin60 v L cos60 u L
2vL sin60 v 2L ucos60 u 2 L
vsin60 ucos60
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2 2 2o oAC AC AC AC
o 2 o 2 2AC AC AC AC
o oAC
L sin45 v L cos45 u L
2vL sin45 v 2L ucos45 u 2 L
vsin45 ucos45
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mm
mm
mm
mm
AB
AC
o oAB
o oAC
0.563
0.487
vsin60 ucos60
vsin45 ucos45
u 0.0250
v 0.6652