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r
y (4)
Compression
Tension
Neutral Axis
strain
stressE then
The force acting on a differential area dA is dA
y
r
E (5)
from equation (5) ydAr
EdA
0
ydAr
EdA
The internal bending moment must be
equal to the external bending moment,M.
dAyr
EdAyM
2
dAyI2
However, Neglecting (-)
r
EIM (6)or
EI
M
r
1
dA dA The sum () of horizontal
forces is zero.F
Fy
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EI
M
dx
dy
dx
yd
r
2
32
2
2
1
1
Beam Deflection Equation:
112
3
2
dx
dysince is small,
dx
dy
EIM
dxyd 2
2
(7)
1cMdxdx
dyEI (8)
21 cxcMdxEIy (9)
General Slope Equation
General Curve Equation
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y
F
L
Solving Moment,Mat anyx location:
Moment Diagram
-FL0
-FL0
L
x
M
usingEI
M
dx
yd
2
2
(7)
xL
M
L
FL
)( xLFM or
)(2
2
xLFMdx
ydEI (10)
1
2
2
Cx
FFLx
dx
dyEI (11)
21
32
62CxC
xF
xFLEIy (12)
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y
F
L
(0,0)
2
2x
FFLxEIdx
dyEI General Slope Equation
Also, whenx=0,y=0.
62
32x
Fx
FLEIy Elastic Curve Equation
It can be seen that whenx=0, dy/dx==0.
x
1
2
2
Cx
FFLx
dx
dyEI (11) Thus, C1=0.
21
32
62CxC
xF
xFLEIy (12)Thus, C2=0.
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)3(662
232
xLFxx
Fx
FLEIy
100mm
F
700mm
Case 1
100mm
F
700mm
or
;Lx when
3
4109
700.0
1.010909.4101053
m
mmxPaxF
Material:
Diameter= 10mm;Length = 700mmy=100mm
Implant Grade (ASTM F136 Titanium)Modulus of Elasticity,E=
105GPa
Moment of Inertia, I
41044
10909.464
)01.0(
64
mx
mDI
3
3FLEIy
kgfNF 60.408.45
x
x
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100mm
45.08N
700mm
x
y
)3(6
2
xLFx
EIy
-120
-100
-80
-60
-40
-20
0
0 100 200 300 400 500 600 700 800
x- Location of Fn (mm)
y-deflection(mm)
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y
F
L
aCase 2
C
y
F
L
(0,0)
x
2
2x
FFLxEIdx
dyEI
2
2x
LxFEI
62
32x
Fx
FLEIy
xLFx
EIy 36
2
A
B
Case 1
x
xLFx
EIy BA 36
2
aL
xaFxEIy BA 36
2
when
?CBEIy since B-C is still a straight line
BBC
2
2FaEI B aLx CB
)(23
23
aLFaFaxyEIy CBBBCB
)3(6
2
aLFa
EIy CB
ax 0
LxorLa
aLx
when ax 3
3Fa
EIyB
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=
+
Method of Superposition
Deflection or Slope(A)
F1
F2
L
a
(A)
F2
a
(B)
F1
L
(C)
Deflection or Slope(B)
Deflection or Slope(C)
=
+
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0
5
10
15
20
25
0 100 200 300 400 500 600 700 800
x- Location of Fn (mm)
y-defle
ction(mm)
+F1
L
xn
+F2 +F3 +F4 +F5 +F6 +F7
-F1 -F2 -F3 -F4 -F5 -F6 -F7
Case 3
Force,N -100 -100 50 -300 10 -6 120
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-0.14
-0.12
-0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
0 100 200 300 400 500 600 700 800
x- Location of Fn (mm)
y-deflection(mm)
+F1
L
xn
+F2 +F3 +F4 +F5 +F6 +F7
-F1 -F2 -F3 -F4 -F5 -F6 -F7
Force,N 100 -50 10 0 0.02 0 0.258
Case 4
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+F1
L
xn
+F2 +F3 +F4 +F5 +F6 +F7
-F1 -F2 -F3 -F4 -F5 -F6 -F7
Force F1=? F2=? F3=? F4=? F5=? F6=? F7=?
yn mm1 mm2 mm3 mm4 mm5 mm5 mm6
Numerical Method for Solving Fn(xn, yn)
xn mm1 mm2 mm3 mm4 mm5 mm5 mm6
Solve
