Structures and Stiffness

Structures and Stiffness
B. Furman
K. Youssefi
20SEP2007
K. Youssefi and B. Furman
Engineering 10, SJSU
1
Outline
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Newton’s 3rd Law
Hooke’s Law
Stiffness
Area moment of Inertia
Orientation of cross section and stiffness
Comparison of cross sections
Materials and stiffness
K. Youssefi and B. Furman
Engineering 10, SJSU
2
Newton’s 3rd Law
• Lex III: Actioni contrariam semper et
æqualem esse reactionem: sive corporum
duorum actiones in se mutuo semper esse
æquales et in partes contrarias dirigi.
• To every action there is always opposed an
equal reaction: or the mutual actions of two
bodies upon each other are always equal,
and directed to contrary parts.
K. Youssefi and B. Furman
Engineering 10, SJSU
3
Newton’s 3rd Law - example
T, tension
Free body diagram
T, tension
M
•Isolate the body of
interest
•Put back the forces
that are acting
M
M*g
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Engineering 10, SJSU
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Hooke’s Law
• Robert Hooke (1635-1702)
– Materials resist loads (push or pull back) in
response to applied loads
• This ‘resistance’ is accomplished by deformation of
the material (changing its shape)
– Tension (stretching)
– Compression (shortening)
– Stretching or shortening of chemical bonds in atoms
• The science of Elasticity concerns forces and
deformations in materials
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Engineering 10, SJSU
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Hooke’s Law, cont.
• Hooke found that deflection
was proportional to load
Load, N
Slope of Load-Deflection curve:
load
k=
deflection
slope, k
Deflection, mm
The “Stiffness”
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Engineering 10, SJSU
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Stiffness
• Stiffness in tension and compression
– Forces F applied, length L, cross-sectional area, A,
and material property, E (Young’s modulus)
A
F
F
L
k=
F
δ
K. Youssefi and B. Furman
F
=
FL
AE
FL
δ=
AE
AE
k=
L
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Stiffness, cont.
• Stiffness in bending
F
Ri
A
Ro
B
• How does the material resist the applied load?
– Think about what happens to the material as the beam
bends
• Inner “fibers” (A) are in compression (radius of curvature, Ri)
• Outer “fibers” (B) are in tension (radius of curvature, Ro)
K. Youssefi and B. Furman
Engineering 10, SJSU
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Review Question 1
•
Stiffness is defined as:
A.
B.
C.
D.
E.
Force/Area
Deflection/Force
Force/Deflection
Force x Deflection
Mass/area
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Engineering 10, SJSU
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Concept of Area Moment of Inertia
The Area Moment of Inertia is an important parameter in determine
the state of stress in a part (component, structure), the resistance to
buckling, and the amount of deflection in a beam.
The area moment of inertia allows you to tell how stiff
a structure is.
The Area Moment of Inertia, I, is a term used to describe the
capacity of a cross-section (profile) to resist bending. It is always
considered with respect to a reference axis, in the X or Y
direction. It is a mathematical property of a section concerned with
an area and how that area is distributed about the reference axis.
The reference axis is usually a centroidal axis.
Mathematically, the area moment of inertia appears in the denominator
of the deflection equation, therefore;
The higher the area moment of inertia, the less a
structure deflects (higher stiffness)
K. Youssefi and B. Furman
Engineering 10, SJSU
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Mathematical Equation for Area Moment of Inertia
Ixx = ∑ (Ai) (yi)2 = A1(y1)2 + A2(y2)2 + …..An(yn)2
A (total area) = A1 + A2 + ……..An
A2
A1
y1
y2
X
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Area, A
X
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Moment of Inertia – Comparison
1
Load
Maximum distance of
4 inch to the centroid
2
Load
2 x 8 beam
I2
I1
Maximum distance of 1 inch
to the centroid
2 x 8 beam
I2 > I1 , orientation 2 deflects less
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Engineering 10, SJSU
12
Moment of Inertia Equations for Selected Profiles
d
Round solid section
do
Round hollow section
di
π (d)4
I=
64
π
[(do)4 – (di)4]
I=
64
Rectangular solid section
Rectangular hollow section
b
I=
1
bh3
12
h
h
b
I=
B
I=
1
hb3
12
H
b
1
1
BH3 bh3
12
12
h
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Engineering 10, SJSU
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Example – Optimization for Weight & Stiffness
Consider a solid rectangular section 2.0 inch wide by 1.0 high.
1.0
I = (1/12)bh3 = (1/12)(2)(1)3 = .1667 , Area = 2
2.0
Now, consider a hollow rectangular section 2.25 inch wide by 1.25 high
by .125 thick.
b
B = 2.25, H = 1.25
h
b = 2.0, h = 1.0
H
B
I = (1/12)bh3 = (1/12)(2.25)(1.25)3 – (1/12)(2)(1)3= .3662 -.1667 = .1995
Area = 2.25x1.25 – 2x1 = .8125
(.1995 - .1667)/(.1167) = .20 = 20% less deflection
Compare the weight of the two parts (same material and length),
compare areas. Material and length is the same for both profiles.
(2 - .8125)/(2) = .6 = 60% lighter
So, for a slightly larger outside dimension section, 2.25x1.25 instead
of 2 x 1, you can design a beam that is 20% stiffer and 60 % lighter
K. Youssefi and B. Furman
Engineering 10, SJSU
14
Review Question 2
•
Which cross section has the larger I?
A.
Rectangular
Horizontal
B.
Rectangular
Vertical
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Stiffness Comparisons for Different sections
Square
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Box
Rectangular
Horizontal
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Rectangular
Vertical
16
Material and Stiffness
E = Elasticity Modulus, a measure of material deformation under a load.
Deflection of a Cantilever Beam
Support
L = length
F = force
Y = deflection = FL3 / 3EI
Fixed end
The higher the value of E, the less a structure
deflects (higher stiffness)
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Engineering 10, SJSU
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Modulus of Elasticity (E) of Materials
Steel is 3 times
stiffer than
Aluminum and
100 times stiffer
than Plastics.
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Engineering 10, SJSU
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Density of Materials
Plastic is 7 times
lighter than steel
and 3 times lighter
than aluminum.
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Wind Turbine Structure
The support structure should be optimized for weight and stiffness.
Support
Structure
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Engineering 10, SJSU
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Review Question 3
•
Which material has the higher stiffness?
A.
B.
C.
D.
E.
Steel
Aluminum
Alumina ceramic
Nylon
Unobtanium
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Examples of Achieving Structural
Stiffness
‘Boss’
‘Gusset’
‘Ribbing’
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Examples of Achieving Structural
Stiffness, cont.
http://en.wikipedia.org/wiki/Image:FT_Rail.jpg
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Engineering 10, SJSU
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Examples of Achieving Structural
Stiffness, cont.
Welded ‘box’
construction
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Engineering 10, SJSU
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Examples of Achieving Structural
Stiffness, cont.
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Engineering 10, SJSU
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Examples of Achieving Structural
Stiffness, cont.
‘Flange’
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Examples of Achieving Structural
Stiffness, cont.
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Engineering 10, SJSU
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