Transcript DESIGNING WITH HIGH-STRENGTH LOW

Designing for Stiffness
For beam which is supported at both ends, suffers max deflection, y
in its middle, when subjected to central load, L.
Can be represented as,
y =
L x l3
48 x E x I
l – length of beam, E – Young’s Modulus, I – moment of inertia of
the cross section with respect to the neutral axis.
• The deflection of the beam under load can be take as a measure of
stiffness.
• Stiffness component can be define as its resistance to elastic
deflection is a function of both shape and modulus of elasticity of
the material
Designing for Stiffness
Stiffness of material increase by increasing its second moment of
area, which computed from cross-sectional dimension and selecting a
high-modulus material for its manufacture.
An important characteristic of metallic materials is that their elastics
modulus are very difficult to change by changing the composition or
heat treatment.
On the other hand, the elastic modulus of composite can be changed
over wide range by changing the volume fraction and orientation of the
constituents.
When a metallic component is loaded in tension, compression, or
bending the Young’s Modulus E, is used in computing its stiffness.
Designing for Stiffness
When the loading is in shear or torsion the modulus of rigidity, G, is
used in computing its stiffness. The relation between these two elastics
constant is given by :
E
G
21   
μ is Poisson’s ratio
When initially straight beam is loaded, it becomes curved as a result of
its deflection.
As the deflection at a given point increase, the radius of curvature
at this point decreases.
The radius of curvature, r, is given by the relationship :
EI
r 
M
M = Weight/load applied along the beam length
The importance of stiffness arises in complex assemblies where
differences in stiffness could lead to incompatibilities and
misalignment between various components, thus delay their efficiency
or even causing failure.
Using high strength materials in attempts to reduces weight usually
comes at the expense of reduces cross-sectional area and reduced
second moment of area.
This could be adversely affect stiffness of the component if the elastic
constant of the new strong material does not compensate for the
reduced second moment of area.
Another solution to the problem of reduced stiffness is to change the
shape of the component cross section to achieve higher second
moment of area, I.
Figure 4.3 gives the formula for calculating I for some commonly
used shape and the value of I for constant cross-sectional area.
In application where both stiffness and weight of structure are
important, it becomes necessary to consider the stiffness /weight or
specific stiffness of the structure.
In the simple case of structural member under tensile or compressive
load, the specific stiffness is given by E/P, where E is the Young’s
Modulus of material and P is density.
In such case, the weight of a beam is given stiffness can easily shown
to be proportional to P/E.
This shown that the weight of the component cab be reduced equally
by selecting a material with lower density or higher elastic modulus.
The weight of the beam for square cross sectional area, w, can be
shown to be :
1/ 2
L
w  lxb xp 
x 
2  y
2
l
5/ 2
p
x 1/ 2
E
This shown that for given deflection y under load L, the weight of the
beam is proportional to (P/E1/2).
For solid round bar diameter D the second moment of area, I is given
as :
xD4
I 
64
The weight of the round bar is given by :
w  lx
xD2
4
2 xl 2 xLxb1/ 2  p 
xp 
 1/ 2 
1/ 2

E 
From the both weight equation shows that the weight of an
axisymmetric bar can be reduced by reduction p or increasing E of the
material.
However, reducing p is more effective as E is present as square root.
Cost per unit property method
In the simplest cases of optimizing the selection of materials, one
property stands out as the most critical service requirement.
In such simple case, the cost per unit property can be used as a criteria
for selecting the optimum material.
Considering the case of a bar of a given length, L, to support a tensile
force, F. the cross sectional area, A, of the bar is given by :
A
F
S
Where S=working stress of the material, which is related to its yield strength by an
appropriates factor of safety.
The cost of the bar (C’) is given by :
CxpxFxL
C '  CxpxAxL 
S
Where C=cost of the material per unit mass
p=density of the material
In comparing the different candidate materials, only the quality
[(C*p)/S], which is the cost of unit strength, need to be compared as F
and L are constant
The material with the lowest cost per unit strength is the optimum
material.
Example 1
It is required to select a structural material for the manufacture of the
tie rod of a suspension bridge. A representative rod is 10 m long and
should carry a tensile load 500 kN without yielding. The maximum
extension should not exceed 18 mm. Which one of the steels listed in
table 1 will give the lightest tie rod.
Table 1 : Candidate materials for suspension bridge tie rods
Material
YS(MPa)
E (GPa) Specific density (kg/m3)
ASTM A675 Grade 60
205
ASTM A572 Grade 50
345
ASTM A717 Grade 70
485
Maraging steel Grade 200 1400
Al 5052-H38
259
Catridge brass 70 % hard
441
temper
212
211
211
211
70.8
100.6
7.8
7.8
7.8
7.8
2.7
8.0
Solution
Table 1 : Candidate materials for suspension bridge tie rods
Material
YS(MPa) E (GPa)
ASTM A675 Grade 60
ASTM A572 Grade 50
ASTM A717 Grade 70
Maraging steel Grade 200
Al 5052-H38
Catridge brass 70 % hard
temper
205
345
485
1400
259
441
Specific
Area based on yield
density (kg/m3)
strength (x 10-3 m2)
212
211
211
211
70.8
100.6
7.8
7.8
7.8
7.8
2.7
8.0
2.44
1.45
1.03
3.6
1.93
1.13
For the present case, calculate the area based on YS :
F
Force
A

YS
YS
A can also be calculate based on deflection= (load x length)/(E x deflection)
Multiply A with length of the bar and specific density, get mass (kg)
Mass
(g)
190
113
102
102
106
221
Solution
Table 1 : Candidate materials for suspension bridge tie rods
Material
YS(MPa) E (GPa) Specific
Area based on
Mass
density (kg/m3) deflection (x 10-3 m2) (g)
ASTM A675 Grade 60
205
ASTM A572 Grade 50
345
ASTM A717 Grade 70
485
Maraging steel Grade 200 1400
Al 5052-H38
259
Catridge brass 70 % hard
441
temper
212
211
211
211
70.8
100.6
7.8
7.8
7.8
7.8
2.7
8.0
1.31
1.32
1.32
1.32
3.92
2.76
102
103
103
103
306
215
The larger of the area will be taken as the design area and
will be used to calculate the mass.
The result of the calculation shown in table 2 and show that
steel A717 Grade 70 and maraging steel Grade 200 give the
least mass.
As the former steel is more ductile and less expensive will
be selected
Example 2
What design change are required when
substituting HDPE for stainless steel in
making a fork for picnic set while maintaining
similar stiffness. The narrowest cross section
of the original stainless steel fork was
rectangular of 0.6 x 5 mm.
E for stainless steel = 210 GPa
E for HDPE = 1.1 GPa
Solution
E for stainless steel = 210 GPa
E for HDPE = 1.1 GPa
I for stainless steel section = 5x(0.6)3 = 0.09 mm4
12
EI should be kept constant for equal deflection under load :
EI for stainless steel = 210 x 0.09 = 18.9
EI for HDPE design = 1.1 x I
I for HDPE design = 17.2 mm4
Taking a channel section of thickness 0.5 mm, web height 4
mm and width 8 mm,
I = [8 x (4)3 – 7 x (3.5)3]/12 = 17.7 mm4
which means the required value.
Area of stainless steel section = 3 mm2
Area of the HDPE section = 7.5 mm2
The specific density of stainless steel is 7.8 and that of
HDPE is 0.96. Thus :
Relative weight of HDPE/stainless steel
= (7.5 x 0.96)/(3 x 7.8) = 0.3
Example 3
Given a 2 m long aluminum alloy tube of outer diameter 3
cm and inner diameter 2.6 cm, determine its second moment
of inertia.
a) What would the diameter of a solid rod have to be to have
the same flexural stiffness ?
b) How do the weight of the two structures compare ?
(p=2700 kg/m3)
c) What would the dimensions be if a square section was
used to give the same second moment of inertia.