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Bend and crush: strengthlimited design

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So far, most of our discussion has concerned designing for a given stiffness. ... for instance a 2x4 on edge is a better floor support than a square 3x3 joist. ... – PowerPoint PPT presentation

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Title: Bend and crush: strengthlimited design


1
Bend and crush strength-limited design
2
The story so far So far, most of our discussion
has concerned designing for a given stiffness.
In many cases, parts can be made stiff enough
just by making them thicker, but we can select
the best material on the basis of minimizing
weight for a given thickness. The criterion of
the best choice will depend on whether the part
is to be used in tension, bending or
torsion. This approach is OK for simple shapes,
but even these can be modified by modifying the
shape, such as by changing from a rod to a tube,
by going from a rectangular beam to an I-section
or by adding porosity where the load is
least. More complexity can come from going to
composite parts with different materials in
different places. The price of this improved
performance, however, is usually more expense in
production. There are no well-defined ways of
finding the best answer except to very simple
problems. There are ways to hunt for the best
answer using various mapping methods or genetic
algorithms. Obviously, computer-based methods
that combine finite-element analysis with search
methods are important in letting us assess many
designs without having to build models and test
them. Next we discuss the design of parts which
are limited by strength rather than stiffness.
Tough materials are limited by their yield
strength. Brittle materials are limited by their
fracture strength.
3
Yield occurs at a critical stress. Read http//en
.wikipedia.org/wiki/Yield_(engineering)
4
Yield Strength The crystal structure of metals
can be viewed as layers of atoms which can be
induced to slide over one another if the shear
stress is sufficiently high. The actual process
occurs by the motion of dislocations See http//
en.wikipedia.org/wiki/Dislocation
5
Yield in tension A bar in tension will yield
when the stress exceeds the yield stress. Simple
tension along a bar can also be thought of as a
shear stress at 45 degrees to the length.
Dislocations move on these 45 degree planes so
that the bar forms the characteristic neck. In
compression, short fat bars can yield to get
shorter and fatter, if they are not constrained
by friction at the pushing surfaces. Longer bars
tend to buckle elastically first and then yield
in bending. In parts with complex shapes or
multiple loads, yield occurs when the combination
of stresses results in a shear stress exceeding
the yield stress as defined by the yield criteria
discussed in the Wiki article.
6
  • Yielding occurs when the stress exceeds the
    yield stress. For a beam loaded in bending. The
    stress state is shown on the right for purely
    elastic loading (a), the onset of plasticity (b),
    and full plasticity (c). Yield starts in the
    highest stress regions on the top and bottom and
    then moves in through the whole bar.

7
  • The plastic bending of beams. Red zone has
    yielded. The moment is FxL in the top case,
    FxL/4 (1./2 the force acts on the pivots each L/2
    away) in the middle and FxL/8 in the bottom case.

8
  • The area A, section modulus Ze and fully plastic
    modulus Zp for three simple sections. The
    elastic section modulus multiplied by the yield
    strength gives the (force times length) at which
    the beam starts to deform plastically on the top
    and bottom surfaces . This correspond to the
    moment at which the stress in the top surface
    exceeds the yield stress. The plastic section
    modulus multiplied by the yield strength gives
    the moment at which the beam finally gives
    plastically.

9
  • Elastic torsion of shafts. The stress in the
    shaft depends on the torque T and the polar
    moment of area K. Helical springs are a special
    case of torsional loading.
  • The torque on a circular shaft produces a shear
    stress which increases with radius from the
    center
  • t T.r/K G?r/K
  • ? is the angle of twist per unit length and G is
    the shear modulus
  • K ?r4/2 for a shaft and K (?/2) (ro4 - r14) for
    a hollow cylinder

10
  • Spinning disks, as in flywheels and gyroscopes,
    carry radial tensile stress caused by centrifugal
    force. Big flywheels are limited by the strength
    of the steel. Aluminum is better because it is
    lighter, so the stress is less..

11
For the material with the best combination of
strength and light weight in tension we want to
maximize the ratio of yield strength to density
sy/? For a sheet in bending sy1/2/? and for a
beam in bending sy2/3/? but we also have the
chance to optimize shape in this case, for
instance a 2x4 on edge is a better floor support
than a square 3x3 joist. Also the advantage
gained by changing shape will depend on the
material. The improvement on going from a square
beam to an I-beam is more in a denser material,
so steel I-beams make more sense than wooden
I-beams. (And anyway if you cut a wooden I-beam
you cant do much with the wood removed.)
12
  • The strengthdensity chart with the indices sy/r,
    s2/3y/r and sy1/2/r plotted on it.

13
Weight is not always important For a hinge, like
the lid of a plastic box, the best material will
bend through the smallest radius without
yielding. This corresponds to the highest strain
at yield, or M sy / E For a spring we want
to maximize the energy stored at yield, which
gives us a maximum M sy2/ E
14
  • Materials for elastic hinges and springs.
    Polymers are the best choice for the former.
  • High-strength steel, CFRP, and certain polymers
    and elastomers are the best choice for the latter.

15
  • Springs leaf, helical, spiral and torsion bar.
    Springs store energy. The best material for a
    spring, regardless of its shape or the way it is
    loaded, is that of a material with a large value
    of s2el/E.

16
Stress concentrations may lead to local
yielding Sometimes, this may blunt the corner or
crack and the part becomes stable. Under
repeated loading, the movement of dislocations
during yielding leads to local strain-hardening,
then formation of a fatigue crack which slowly
grows until failure occurs.
17
  • Stress concentrations. The change of section
    concentrates stress most strongly where the
    curvature of the surface is greatest. Reducing
    stress concentrations by rounding corners is a
    design issue.
  • The stress concentration factor depends on the
    length and radius of the defect.

In tension ? is 2, but it is ½ in torsion and
bending. For a circular defect, c? and the
stress concentration factor is 3 in tension.
18
Effect of holes shapes from FEA modeling Bar 3m
long by 1 m wide. 20 cm hole in center. Stress
in tension 5MPa Square hole max von Mises
stress10.2 MPa about 1/3 way along top and bottom
sides of hole Square hole with rounded corners
13.9MPa Chamfered corners 16.5MPa Shallow
chamfers along top and bottom 14.5MPa fill in
between chamfers 14.6MPa Circular hole 15.8 MPa
(3x original stress) Diamond hole
15.1MPa Series of elliptical holes across bar,
long axis 20 cm different short axes short axis
0.1 cm 23.7MPa, 0.08 cm 27MPa, 0.06cm 34.7MPa
0.04 cm 49.4 MPa c/? is 4, 6.25, 11, 25 giving
stress intensity factors of 5, 6, 7.6 and 11
which correspond pretty well to the equation on
the slide before.
19
  • Contact stresses are another form of stress
    concentration. This is important for bearings.
    If the material yields under load, the bearing
    surface will start to crack up over time. When
    elastic, the stresses and displacement of the
    surfaces towards each other can be calculated.
    Yield occurs at the red point. You see the same
    when a rock hits a window and a chip comes out
    the other side.

20
Bar 2m long, x0.1x0.1 with 10 notch Load 5 MPa
along -x
21
Shear stress plot Stresses are negative and
positive but equal
22
Von mises stress plot, max stress is near the
corners.
23
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24
In 3D
25
Bar 2m long, 0.1m high with a triangular
notch Note that the shear stresses (von Mises)
are highest at 45 degrees and so tend to open up
the crack in a ductile, shearing material The
tensile stresses Sxx are maximum below the crack
and so extend the crack in a brittle material
26
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27
Note the stress shadow at the top. Max stress is
about 2x-3x the average stress away from the
crack. Note that the defect is not a circle but
the stress concentration acts as if it were. The
point is that it is about as long as it is wide.
28
Smaller notch, lower stresses so there is a
critical crack size for local yield or fracture.
Here the width to depth ratio is increased
compared to the previous case.
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