W2L3
Stresses in Engineering Components
(problems 14, 15, 16)
(combining elastic moduli with geometry Æ elastic behaviour)
(Courseware pg 43-46)
→ Free Body Analysis: common technique to develop
stress equations for components – basically Free
Body Diagrams for static equilibrium cases
τdA
τdA =
Procedure:
1. Pass a cutting plane perpendicular to the line of
action of the load (force)
2. Net internal reaction will be equal to the external
force (otherwise not in static equilibrium), i.e.
(Internal force)
external force =
∫ τdA
cut cross
section
(essentially the
internal force across
the cut section)
1
We use this free body analysis technique mainly as
a tool to enable us to derive equations for stresses in
components that are subject to external forces.
We are going to consider two different cases:
1. Centroidally Loaded Case: This is when the
externally applied force must be symmetrical
about centroidal axis. An example is a
pressurized gas cylinder where the “external
applied force” is the internal pressure. It is
applied symmetrically about the centroid axis
(along the length of the cylinder). We will
consider use the free body analysis technique to
develop equations for:
- axial stresses in pressure vessels
- hoop stresses in pressure vessels
We will start this today (W2L3) and also will work
on it in the next lecture (W3L1)
2. Beam Bending: this is not a case of centroidal
loading. The free body analysis technique is a
handy way to determine what the stress equations
look like for beams of different shapes, sizes and
materials. This takes some time to do - we will
start this in W3L2 and will finish in W4L1.
2
Centroidally Loaded Case:
Axial stresses in Pressure /Vacuum Vessels
(Courseware pg 45)
→ P = internal pressure;
assume pipe is thin walled
so rinside ≈ routside = r
Make the imaginary cut
here – perp to the axial
direction (because we
want to develop an
expression for axial
stress)
wall thickness (t)
Flat,rigid
end cap
P
cut
Internal force
of the wall at the cut
F = (τaxial)(area) = (τaxial)(2πrt)
external force
(due to pressure
acting on end)
= πr2P
→ If nothing is moving (static equilibrium) then the
internal and external forces must be equal:
τ axial (2πrt ) = πr 2 P
τ axial
Pr
=
2t
→ WE HAVE JUST DEVELOPED AN EXPRESSION
FOR CALCULATING AXIAL WALL STRESS IN
A CYLINDER!
3
→ problem
4
Centroidally Loaded Case:
Hoop Stress in Cylindrical Pressure Vessels
(Courseware pg 45)
As with the Axial stress case in the last slide, the first step
is to make a cut perpendicular to the direction that you
want to get a stress expression for. Since you want a
HOOP stress you make a cut perpendicular to the hoop,
which is along the length. The drawing shows the cylinder
cut along its length.
L
external force = due to
the internal pressure
“P” which is (radial)
as shown
τhoop
internal reaction force=
hoop force acting on cut
wall
P
τhoop
wall
thickness t
Internal reaction force (hoop)
side view – can
think of it as:
F1
F2
(vertical components
of pressure will cancel)
F1
External force
5
What is ΣF1 ?
total internal reaction force (right)
What is ΣF2 ?
total force (left)
(external)
Wall cross section
= area (τH)
= (2Lt) τH
= sum of horizontal components of force
due to pressure
How do we get this? Obtain an expression
for the small amount of horizontal force on
each small part of the wall ds (see diagram
below). Then sum up the total force over
the entire semicircular section by integrating.
horizontal force on ds = (Phoriz at ds)(area of ds)
Remember this
= (Pcosθ)(Lds)
extends into the
page
= (Pcosθ)(Lrdθ)
So ΣF2 =
θ=
π
2
∫ Pr L cos θdθ
θ =−
ΣF2 =
Pr L sinθ
π
2
π
2
−
π
= 2PrL
2
Now for static equilibrium ΣF1 = ΣF2
2Lt τH = 2PrL
Pr
τH =
t
This is a formula for
6
calculating the hoop
stress in the wall
Comparing Hoop and Axial Stresses:
τ axial =
→
Pr
2t
Pr
τH =
t
Interesting note here: For a given pressure the
axial stress is half of the hoop stress – this is
why pressure cylinders fail by splitting
lengthwise rather than by the ends blowing off!
So here is an interesting question – normally in a
system we have three main stresses – in the simple
case these stresses are orthogonal, so here these
would be axial, hoop and thickness stresses. Below
is shown a small piece of the pipe wall with these
stresses labelled.
τT
τH
τ ax
So why haven’t we discussed thickness stress yet?
Well, in a thin-wall cylinder these are essentially zero.
Why? Because in order to have a stress you need two
opposing forces. However, there is nothing in the
thickness direction to oppose the internal gas pressure
(only air on the other side) – the cylinder will just
expand until it is stopped by the hoop and axial stress
state. (for the record, a balloon has no stresses in the
thickness direction either)
--- thick wall cylinders DO have a thickness stress
7
because the cross section at the outside is greater than
that on the inside – creates a complex stress gradient.
FINALLY...
Do we have to consider Poisson’s ratio in a
cylinder?
Certainly! Usually (if the wall is thin) there are no
stresses in the thickness (radial) direction (the 3rd
orthogonal direction). But we still have strains in all
3 directions, because of Poisson. So lets write
down the strain expressions for all 3 components,
assuming the stress in the thickness direction is
zero.
Note – the format for the strains will be identical to
those for the 3 orthogonal xyz direction strains we
developed in W2L1.
There is no
stress in the
radial direction
so τr =0.
Given this, the
3 strains look
like:
εh =
εa =
τh
Y
τa
Y
ε r = −σ
−σ
−σ
τa
Y
τa
Y
τh
Y
−σ
τh
Y
Note here: these equations will NOT be on the
exam equation sheets. They are derived
easily from the xyz equations which ARE on
the formula sheet
8
problem
9
Ex. Gas and Oil Pipelines
Gas transmission lines like the TransCanada pipeline are
linked by compressor stations (about 100 km apart) where
gas is pressurized and pumped along. Gas pressures are
typically quite high, about 6.5 MPa. (vs 2.5 MPa for BBQ test)
What we try to avoid
For a typical 36” diameter pipe of 9mm wall thickness:
1. What is the hoop stress in the pipe wall due to its internal
pressure?
2. What is the axial stress in the pipe wall due to the internal
pressure?
3. What is the axial contraction due to Poisson’s ratio,
assuming it is free to contract?
10
1. Hoop stress in pipe wall?
τ hoop =
Pr 6.5 x (0.5 x 36" x 0.0254)
=
= 330 MPa
−3
t
9 x10
(note that the yield stress in this material is about 380MPa)
2. Axial stress in the pipe wall? ----no end caps
so there is effectively no axial stress due to the
end cap effects.
0.29 for steel
3. Axial contraction?
τa
⎛τh ⎞
εa = − σ ⎜ ⎟
Y
⎝Y ⎠
220GPa for steel
εa =
− 330 MPa
∆l
0
.
29
=
−
0
.
000435
=
220 x103 MPa
l
∆l = -(.000435)(100,000) = -43.5m
So if this pipe were free to contract it would do
so by 43.5m!! It is constrained at the
compressor station but stories are told of pipes11
disappearing when cut!
Wall Stresses in a Sphere (under pressure)
Cut again:
internal
reaction
= (2πrt)(τwall)
τwall force in the
P
wall
(to the right)
total force (to the left) due to pressure = πr2P
(not shown, but can be obtained by integrating horizontal terms over entire
hemispherical area)
equating:
2πrtτwall= πr2P
τ wall
Pr
=
2t
So this is the equation for
calculating wall stresses
in a sphere
Note – this is the same as a cylinder axial stress which
is half the cylinder hoop stress
Hmmm… this means
For a given pressure, the maximum cylinder wall stress
is twice that of a sphere!
So if I were designing a pressure vessel...
Spheres are better pressure vessels but fabrication
difficulties mean that they are only used in the 12
most
critical situations!!
problem
13
Further notes:
→ Interesting side note: bathyspheres suffered from
one main problem - cable breakage meant bye bye
crew! So in 1946 Swiss Physicist August Piccard
designed a bathyscaphe --which could surface and
sink as required. It is a short Ti-alloy cylinder that is
reinforced on the inner surface.
→ The current depth record of 10,915m (deepest ocean
point - Marianas Trench) was obtained in one of
these by August’s son Jacques Piccard (boldly
going….)
→ The Titanic sits at 3400 m and was explored by a
bathyscaphe
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