3. 3
1.0
Preliminary
Need
The
preliminary
need
provided
was,
“For
a
conventional
rear-‐wheel
drive
sporty
vehicle,
there
is
a
need
for
a
4-‐speed
manual
transmission
that
can
operate
in
any
gear
over
an
infinite
life.”
While
designing
this
transmission
it
will
be
crucial
to
take
the
strength
and
torque
requirements
of
all
the
components
into
consideration.
2.0
Preliminary
Specifications
The
specifications
that
were
provided
are
as
follows:
1. Design
Point
(continuous):
40
HP
at
2000
RPM
(input
shaft)
2. Design
factor
(countershaft):
nd
=
1.75
3. Design
factor
(gears):
ng
=
1.25
4. Design
factor
(keys):
nk
=
1.1
5. Size
(Max):
a. Length:
2
ft
(target
1
ft
on
main
body,
1
ft
on
tail
piece)
b. Width:
14
in
c. Height:
14
in
6. Target
Ratios:
a. 1st
gear:
3.2
to
1
(+/-‐
5%)
b. 2nd
gear:
2.2
to
1
(+/-‐
5%)
c. 3rd
gear:
1.6
to
1
(+/-‐
2%)
d. 4th
gear:
direct
drive
2.1
Constraints
Imposed
by
Boss
The
constraints
given
by
the
boss
were:
1. Assume
constant
torque
at
the
design
point
(even
though
it
isn’t).
2. Input-‐output
shafts
collinear
(so
4th
can
be
direct
connect).
3. Helical,
parallel
axis
gearing,
all
gears
same
diametral
pitch.
4. Simplification:
no
reverse
gear
here.
5. Gears
Keyed
to
Countershaft
(idler
shaft).
6. Leave
3
in
spacing
between
1st
&
2nd
gears
and
also
between
3rd
&
4th
for
synchronizers.
7. Gear
teeth
(counts)
should
be
chosen
for
even
wear.
8. Passenger
side
view
of
countershaft,
maintain
ordering
of
gears:
starting
on
left:
1st,
2nd,
3rd
gear
with
countershaft
input
driver
gear
on
right.
9. Objective
is
to
fully
design
the
countershaft
(idler
shaft)
of
the
transmission.
The
case,
the
input,
output
shafts,
the
synchronizers
are
not
being
designed
in
this
project.
For
those
parts,
include
enough
mainshaft
detail
(input,
output
shafts)
and
case
detail
to
allow
for
simulation,
provide
space
for
synchronizers,
use
solid
works
mates
to
simulate
synchronizers.
4. 4
3.0
Design
The
above
figure
is
the
overall
design
of
the
4-‐speed
transmission.
The
view
is
oriented
from
the
passenger
side
of
the
vehicle
showing
the
input
shaft
on
the
top
left
and
the
output
on
the
top
right.
The
shaft
assembly
on
the
bottom
is
the
countershaft
with
a
total
of
four
gears.
The
gear
on
the
far
right
is
the
idler
gear,
which
is
driven
by
the
driving
gear
directly
above
it
at
2000
RPM.
To
the
left
of
the
idler
are
the
1st,
2nd,
and
3rd
gears
(left
to
right),
which
are
all
attached
to
their
corresponding
output
gears.
This
design
incorporates
multiple
snap
rings,
key
ways,
bearings,
and
spacers
to
keep
the
gears
in
place.
The
above
illustration
is
used
to
declare
important
values
of
the
gears
and
overall
transmission
design.
The
image
provides
the
face
width
values
for
the
output
and
driving
gears
along
with
the
tooth
numbers
for
all
gears.
Gear
teeth
numbers
were
chosen
in
order
to
achieve
the
desired
gear
ratios.
In
addition
the
pitch
for
all
gears
was
established
to
be
5
T/in
and
a
diametral
pitch
of
7.011.
Also,
a
centerline
distance
from
the
mainshaft
to
countershaft
was
chosen
to
be
5.9
in
which
allowed
gears
enough
clearance
to
work
properly
while
achieving
the
desired
torque
values.
5. 5
Since
location
A
only
experiences
radial
loading
and
spins
at
an
RPM
greater
than
100,
we
chose
to
use
a
1in
roller
bearing
at
this
location.
At
location
B,
axial
and
radial
loading
occurs
and
thus
we
need
to
a
1.25in
tapered-‐roller
bearing
to
account
for
this
extra
force
component.
As
for
the
keys,
we
chose
to
use
the
same
key/keyway
for
all
the
gears.
We
chose
to
use
a
woodruff
key
with
a
length
of
0.75in
and
a
radius
of
0.303
in.
The
keyways
were
sized
accordingly
to
this
size
with
a
depth
of
0.06in
and
length
of
0.44
in.
The
keys
are
made
out
of
316
stainless
steel.
In
order
to
fix
bearing
A
to
the
countershaft,
two
1in
snap
rings
are
used
on
both
sides
of
the
bearing.
Bearing
B
is
held
in
place
by
the
1.5
in
shoulder
and
a
1.25in
snap
ring.
There
is
nothing
to
hold
the
gears
in
place
axially,
which
is
a
design
flaw
in
our
design.
3.1
Countershaft
Assembly
The
countershaft
can
easily
be
assembled
due
to
its
design
by
following
a
series
of
simple
steps.
The
first
step
in
the
countershaft
assembly
is
to
insert
all
the
keys
into
their
mating
key
ways,
since
all
the
keys
are
the
same
it
doesn’t
matter
which
key
goes
to
which
key
way.
Once
all
the
keys
are
installed,
gear
3
would
be
slid
over
the
countershaft
followed
by
a
0.125in
spacer
and
gear
2
would
be
slid
over
the
countershaft
and
against
the
spacer.
Following
this,
a
3in
spacer
and
gear
1
would
be
slid
over
the
counter
shaft.
Now
the
idler
gear
along
with
a
spacer
can
be
slid
onto
the
end
by
gear
3
followed
on
the
end
by
a
1.25in
snap
ring.
Finally
the
1in,
1.25in
roller
bearings
would
be
held
in
place
onto
the
ends
of
the
counter
shaft
with
the
appropriately
sized
snap
rings.
3.2
Countershaft
Layout
The
image
above
is
a
passenger
view
of
the
countershaft
alone.
On
both
the
far
left
and
far
right
of
the
countershaft
are
flat
ends
which
is
where
the
specific
rolling
bearing
will
be
implicated
in
order
to
allow
the
shaft
to
spin
freely
within
the
case.
Starting
from
the
left
6. 6
side
of
the
shaft
we
have
the
two
1in
snap
rings
(to
hold
the
roller-‐bearing
in
place)
followed
by
the
1st
gear,
2nd
gear,
3rd
gear
keys
ways
followed
by
the
1.25in
shoulder
and
idler
gear
key
way
followed
by
the
1.50
in
shoulder
and
finally
the
1.25
in
snap
ring
to
hold
the
tapered
roller
bearing
in
place.
4.0
Theoretical
Overview
4.1
Theoretical
Loads
The
following
figures
are
the
fbd’s
of
the
countershaft
gears.
All
dimensions
are
detailed
along
with
locations
of
all
the
forces.
All
images
are
a
passenger
view
of
the
countershaft
and
its
gears.
The
figure
above
is
the
fbd
of
the
1st
gear.
7. 7
The
figure
above
is
the
fbd
of
the
2nd
gear.
The
figure
above
is
the
fbd
of
the
3rd
gear.
8. 8
The
following
table
shows
the
calculations
used
to
find
the
various
tooth
and
reaction
forces
along
with
torques
from
the
three
countershaft
fbd’s
above.
Table
I
Diameter
of
drive
gear
𝑑! = 7𝑖𝑛
Pressure
angle
ϕ! = 14.5°
Helix
angle
ψ = 45°
Length
L=15.62in
B
coordinate
𝑥! = 14.4575𝑖𝑛
Input-‐shaft
torque
𝑇! =
40 𝐻𝑃
2000 𝑟𝑝𝑚
5252 𝑙𝑏𝑓 𝑓𝑡 12 𝑖𝑛 𝑟𝑝𝑚
𝑓𝑡 𝐻𝑃
= 1260 𝑙𝑏𝑓 𝑖𝑛
Counter-‐shaft
torque
𝑇! =
𝑁!
𝑁!
𝑇! =
35
24
1260 𝑙𝑏𝑓 𝑖𝑛 = 1838 𝑙𝑏𝑓 𝑖𝑛
Tangential
tooth
force
at
drive
gear
𝐹!" = 2
𝑇!
𝑑!
= 2
1838 𝑙𝑏𝑓 𝑖𝑛
7 𝑖𝑛
= 525 𝑙𝑏𝑓
Normal
tooth
force
at
drive
gear
𝐹! =
𝐹!"
cos ϕ! cos (ψ)
=
525 𝑙𝑏𝑓
cos 14.5 cos (45)
= 767 𝑙𝑏𝑓
Radial
tooth
force
at
drive
gear
𝐹!" = 𝐹! sin ϕ! = 767 𝑙𝑏𝑓 sin 14.5 = 192 𝑙𝑏𝑓
Axial
tooth
force
at
drive
gear
𝐹!" = 𝐹! cos ϕ! sin ψ = 767 𝑙𝑏𝑓 cos 14.5 sin 45 = 525
Drive
gear
coordinate
𝑥! = 𝑥! − 𝑥! = 11.975 𝑖𝑛
Drive
gear
pitch
radius
𝑦! =
7𝑖𝑛
2
= 3.5𝑖𝑛
Diameter
of
1st
gear
𝑑! = 3.6𝑖𝑛
Tangential
tooth
force
at
1st
gear
𝐹!! =
𝑑!
𝑑!
𝐹!" =
7 𝑖𝑛
3.6 𝑖𝑛
525 𝑙𝑏𝑓 = 1021 𝑙𝑏𝑓
Normal
tooth
force
at
1st
gear
𝐹! =
𝐹!!
cos ϕ! cos (ψ)
=
1021 𝑙𝑏𝑓
cos 14.5 cos (45)
= 1491 𝑙𝑏𝑓
Radial
tooth
force
at
1st
gear
𝐹!! = 𝐹! sin ϕ! = 1491 𝑙𝑏𝑓 sin 14.5 = 373 𝑙𝑏𝑓
Axial
tooth
force
at
1st
gear
𝐹!! = 𝐹! cos ϕ! sin ψ = 1491 𝑙𝑏𝑓 cos 14.5 sin 45
= 1021 𝑙𝑏𝑓
1st
gear
torque
𝑇! = 𝐹!!
𝑑!
2
= 525 𝑙𝑏𝑓
3.6𝑖𝑛
2
= 1838 𝑙𝑏𝑓𝑖𝑛
1st
gear
output
torque
𝑇!! = 𝐹!!
𝑑!!
2
= 525 𝑙𝑏𝑓
8.2𝑖𝑛
2
= 2153 𝑙𝑏𝑓𝑖𝑛
1st
gear
coordinate
𝑥! = 2.02125𝑖𝑛
1st
gear
pitch
radius
𝑦! =
𝑑!
2
=
3.6𝑖𝑛
2
= 1.8𝑖𝑛
Right-‐side
bearing
reaction
(axial)
𝑅!" = 𝐹!! − 𝐹!" = 496 𝑙𝑏𝑓
Right-‐side
bearing
reaction
(vertical)
𝑅!" =
1
𝑥!
𝑥! 𝐹!! + 𝑦! 𝐹!! + 𝑥! 𝐹!" − 𝑦! 𝐹!" = 211 𝑙𝑏𝑓
Right-‐side
bearing
reaction
(horizontal)
𝑅!" =
1
𝑥!
𝑥! 𝐹!" − 𝑥! 𝐹!! = 292 𝑙𝑏𝑓
Left-‐side
bearing
reaction
(vertical)
𝑅!" = 𝐹!! + 𝐹!! − 𝑅!" = 354 𝑙𝑏𝑓
Left-‐side
bearing
reaction
(horizontal)
𝑅!" = 𝐹!" − 𝐹!! − 𝑅!" = −788 𝑙𝑏𝑓
Diameter
of
2nd
gear
𝑑! = 4.6𝑖𝑛
Tangential
tooth
force
at
2nd
gear
𝐹!! =
𝑑!
𝑑!
𝐹!" =
7 𝑖𝑛
4.6 𝑖𝑛
525 𝑙𝑏𝑓 = 799 𝑙𝑏𝑓
Normal
tooth
force
at
2nd
gear
𝐹! =
𝐹!!
cos ϕ! cos (ψ)
=
799 𝑙𝑏𝑓
cos 14.5 cos (45)
= 1167 𝑙𝑏𝑓
Radial
tooth
force
at
2nd
gear
𝐹!! = 𝐹! sin ϕ! = 1167 𝑙𝑏𝑓 sin 14.5 = 292 𝑙𝑏𝑓
20. 20
Drive
gear
results.
Input
gear
results.
When
comparing
the
torque
calculations
in
4.1
to
the
rush
gear
results
above
for
each
gear,
it
is
clear
that
the
gear
will
be
strong
enough
to
handle
the
transmitted
torque.
4.3
Countershaft
Bearings
Since our countershaft at the design point spins at 1371 RPMs, we must select bearings that are
rated for speeds greater than 1371RPMs. Since the radial loads at location A equals:
𝑅!" = 𝑅!"! + 𝑅!"! + 𝑅!"! = 788 lbf+359 lbf +247 lbf= 1394 lbf
21. 21
𝑅!" = 𝑅!"! + 𝑅!"! + 𝑅!"! = 354 lbf+197 lbf +156 lbf= 707 lbf
Since 1394 lbf >707 lbf, we need to select a bearing that can handle that is rated for more than
1394 lbf of dynamic loading. Since we have only radial loads at point A and 𝜔 > 100 RPMS, we
chose to use a double sealed ball bearing. The one chosen is rated up to 2500 RPMs, has a
maximum dynamic load capacity of 1480 lbf and its designed for a 1in diameter shaft.
Now we analyze the bearing at point B. Since point B has axial and radial loading, we chose to
use a tapered-roller bearing. First we will analyze the radial loading.
𝑅!" = 𝑅!"! + 𝑅!"! + 𝑅!"! = 292 lbf+85 lbf +116 lbf= 493 lbf
𝑅!" = 𝑅!"! + 𝑅!"! + 𝑅!"! =211 lbf+287 lbf +276 lbf= 774 lbf
Since 774 lbf >493 lbf, we need to choose a bearing that can handle more than 774 lbf radially.
Now we will analyze the axial loading
𝑅!" = 𝑅!"! + 𝑅!"! + 𝑅!"! =496 lbf+274 lbf +131 lbf= 901 lbf
So we need a bearing that can handle more than 901 lbf axially and 774 lbf radially. Because of
this we chose a tapered steel roller bearing with a maximum radial dynamic load rating of 2,130
lbf, a maximum axially axial dynamic load rating of 1,500 lbf and is designed for a shaft with a
diameter of 1.25 in.
4.4
Countershaft
Critical
Locations
Critical
locations
occur
at
locations,
which
undergo
high
bending
moments,
high
stress
concentrations,
small
cross
sectional
area,
and/or
heavy
torque.
Throughout
the
countershaft
design
there
are
several
critical
locations
that
can
be
recognized.
These
locations
include
key
ways,
snap
ring
grooves,
and
shoulders.
The
keyways
and
keys
themselves
have
a
very
small
cross
sectional
area
and
the
gears
are
torqueing
the
keys.
The
snap
rings
are
critical
locations
because
they
involve
keeping
the
gears
from
walking
around
on
the
countershaft.
4.5
Theoretical
Critical
Locations
The
Mcomp
graph
image
in
section
4.1
will
be
used
to
find
the
correct
moment
values
used
for
the
analysis
of
the
critical
locations.
This
section
is
used
primarily
to
determine
what
material
and
dimensions
of
the
countershaft
should
be
used
in
order
for
the
shaft
to
be
strong
enough
for
the
applied
forces.
Tables
and
equations
were
all
found
in
the
McGraw-‐
Hill
textbook.
First
we
will
analyze
the
critical
point
at
the
shoulder
to
the
right
of
gear
3.
The
material
used
for
the
countershaft
was
AISI
1045
steel
CD
with
a
diameter,
d=
1in.
From
the
material
properties
in
Solid
Works,
Sut=
91kpsi,
and
Sy=
77kpsi.
In
order
to
find
Se
we
must
find
ka
&
22. 22
kb
first.
From
table
6-‐2
for
cold
drawn
finish
we
get,
a=
2.7
&
b=
-‐0.265.
From
Eq.
6-‐19,20
we
get:
𝑘! = 𝑎𝑆!"
!
𝑘! = 2.7(91𝑘𝑝𝑠𝑖)!!.!"#
𝑘! = 0.817
𝑘! = 0.879𝑑!!.!"#
𝑘! = 0.879(1𝑖𝑛)!!.!"#
𝑘! = 0.879
𝑆! = 𝑘! 𝑘!
𝑆!"
2
𝑆! = (0.817)(0.879)
91𝑘𝑝𝑠𝑖
2
𝑆! = 32.68 𝑘𝑝𝑠𝑖
For
the
shoulder
at
gear
3,
its
estimated
that
Ma=
2000lbf
in.
Using
!
!
= 1.5𝑖𝑛 &
!
!
=
0.1𝑖𝑛, gives
kt=
1.65,
and
kts=
1.425
when
using
table
A15-‐7,8.
Next
from
figure
6-‐20,21
q=
0.825,
and
qs=
0.84.
All
these
values
will
now
be
used
in
Eq.
6-‐32:
𝑘! = 1 + 𝑞 𝑘! − 1
𝑘! = 1 + 0.825 1.65 − 1
𝑘! = 1.53625
𝑘!" = 1 + 𝑞 𝑘!" − 1
𝑘!" = 1 + 0.84 1.425 − 1
𝑘!" = 1.357
Next
the
Alternating
and
Midrange
Von
Mises
Stresses
will
be
solved
for
with
all
the
given
values
from
above.
σ!
!
=
32𝑘! 𝑀!
π𝑑!
σ!
!
=
32(1.53625)(2000 𝑙𝑏𝑓 𝑖𝑛)
π(1𝑖𝑛)!
σ!
!
= 31.296 𝑘𝑝𝑠𝑖
σ!
!
= 3
16𝑘!" 𝑇!
π𝑑!
σ!
!
= 3
16(1.357)(1838 𝑙𝑏𝑓 𝑖𝑛)
π(1𝑖𝑛)!
σ!
!
= 22.002 𝑘𝑝𝑠𝑖
Finally
the
fatigue
factor
of
safety
and
Langer
will
be
calculated:
1
𝑛!
=
σ!
!
𝑆!
+
σ!
!
𝑆!"
23. 23
1
𝑛!
=
31.296 𝑘𝑝𝑠𝑖
32.68 𝑘𝑝𝑠𝑖
+
22.002 𝑘𝑝𝑠𝑖
91 𝑘𝑝𝑠𝑖
𝑛! = 0.834
1
𝑛!
=
σ!
!
𝑆!
+
σ!
!
𝑆!
1
𝑛!
=
31.296 𝑘𝑝𝑠𝑖
77 𝑘𝑝𝑠𝑖
+
22.002 𝑘𝑝𝑠𝑖
77 𝑘𝑝𝑠𝑖
𝑛! = 1.44
These
are
not
acceptable
factors
of
safety
for
the
shoulder
to
the
right
of
gear
3.
The
target
factor
of
safety
was
at
least
1.75.
In
order
to
obtain
this
factor
of
safety
the
material
of
the
countershaft
could
be
changed
to
a
stronger
material
but
more
importantly
the
diameter
of
1in
should
be
increased.
Next
we
will
find
the
Alternating/Midrange
Von
Mises
Stresses
and
fatigue/Langer
factors
of
safety
of
the
shoulder
to
the
right
of
the
drive
gear.
At
this
point
it
is
estimated
that
Ma=
690
lbf
in
and
the
midrange
stress
is
equal
to
the
previous
shoulders.
σ!
!
=
32𝑘! 𝑀!
π𝑑!
σ!
!
=
32(1.53625)(690 𝑙𝑏𝑓 𝑖𝑛)
π(1𝑖𝑛)!
σ!
!
= 10.8 𝑘𝑝𝑠𝑖
1
𝑛!
=
σ!
!
𝑆!
+
σ!
!
𝑆!"
1
𝑛!
=
10.8 𝑘𝑝𝑠𝑖
32.68 𝑘𝑝𝑠𝑖
+
22.002 𝑘𝑝𝑠𝑖
91 𝑘𝑝𝑠𝑖
𝑛! = 1.75
1
𝑛!
=
σ!
!
𝑆!
+
σ!
!
𝑆!
1
𝑛!
=
10.8 𝑘𝑝𝑠𝑖
77 𝑘𝑝𝑠𝑖
+
22.002 𝑘𝑝𝑠𝑖
77 𝑘𝑝𝑠𝑖
𝑛! = 2.35
These
are
acceptable
factors
of
safety
for
the
shoulder
to
the
right
of
the
drive
gear.
The
target
factor
of
safety
was
at
least
1.75,
which
was
just
obtained.
Next
we
will
find
the
Alternating
Von
Mises
Stresses
and
fatigue/Langer
factor
of
safety
of
the
snap
ring
groove
closest
to
gear
1,
directly
to
the
right
of
the
bearing.
For
this
snap
ring
groove,
its
estimated
that
Ma=
200lbf
in.
Using
!
!
= 0.2𝑖𝑛 &
!
!
=
!.!
!.!
= 1.2, gives
kt=
4.35
when
using
table
A15-‐16.
Next
from
figure
6-‐20,21
q=
0.58,
when
using
r=
(0.2)(0.5)=
0.1.
All
these
values
will
now
be
used
in
Eq.
6-‐32:
𝑘! = 1 + 𝑞 𝑘! − 1
𝑘! = 1 + 0.58 4.35 − 1
𝑘! = 2.943
24. 24
Next
the
Alternating
Von
Mises
Stresses
will
be
solved
for
with
all
the
given
values
from
above.
σ!
!
=
32𝑘! 𝑀!
π𝑑!
σ!
!
=
32(2.943)(200 𝑙𝑏𝑓 𝑖𝑛)
π(1𝑖𝑛)!
σ!
!
= 6 𝑘𝑝𝑠𝑖
Finally
the
fatigue/Langer
factor
of
safety
will
be
calculated,
since
there
is
no
torque
acting
at
this
location
the
midrange
stress
is
equal
to
zero:
1
𝑛!
=
σ!
!
𝑆!
+
σ!
!
𝑆!"
1
𝑛!
=
σ!
!
𝑆!
1
𝑛!
=
6 𝑘𝑝𝑠𝑖
32.68 𝑘𝑝𝑠𝑖
𝑛! = 5.45
1
𝑛!
=
σ!
!
𝑆!
+
σ!
!
𝑆!
1
𝑛!
=
σ!
!
𝑆!
1
𝑛!
=
6 𝑘𝑝𝑠𝑖
77 𝑘𝑝𝑠𝑖
𝑛! = 12.8
These
exceed
the
acceptable
factor
of
safety
for
the
snap
ring
groove
to
the
right
of
the
bearing
closest
to
gear
1.
The
target
factor
of
safety
was
at
least
1.75.
Next
we
will
find
the
Alternating
Von
Mises
Stresses
and
fatigue/Langer
factor
of
safety
of
the
snap
ring
groove
to
the
right
of
the
bearing
closest
to
the
drive
gear.
At
this
point
it
is
estimated
that
Ma=
600
lbf
in.
σ!
!
=
32𝑘! 𝑀!
π𝑑!
σ!
!
=
32(2.943)(600 𝑙𝑏𝑓 𝑖𝑛)
π(1𝑖𝑛)!
σ!
!
= 18 𝑘𝑝𝑠𝑖
Finally
the
fatigue/Langer
factor
of
safety
will
be
calculated,
since
there
is
no
torque
acting
at
this
location
the
midrange
stress
is
equal
to
zero:
1
𝑛!
=
σ!
!
𝑆!
+
σ!
!
𝑆!"
25. 25
1
𝑛!
=
σ!
!
𝑆!
1
𝑛!
=
18 𝑘𝑝𝑠𝑖
32.68 𝑘𝑝𝑠𝑖
𝑛! = 1.82
1
𝑛!
=
σ!
!
𝑆!
+
σ!
!
𝑆!
1
𝑛!
=
σ!
!
𝑆!
1
𝑛!
=
18 𝑘𝑝𝑠𝑖
77 𝑘𝑝𝑠𝑖
𝑛! = 4.28
These
are
acceptable
factors
of
safety
for
the
snap
ring
groove
to
the
right
of
the
bearing
closest
to
the
drive
gear.
The
target
factor
of
safety
was
at
least
1.75.
4.6
Theoretical
Keys
Next
we
will
analyze
the
keyway
critical
location
at
gear
2,
all
keyways
are
critical
locations
but
the
keyway
here
has
the
highest
bending
moment.
At
this
point
its
estimated
that
Ma=
2500lbf
in.
Using
!
!
= 0.02𝑖𝑛, gives
kt=
2.14,
and
kts=
3.0
when
using
table
7-‐1.
Next
from
figure
6-‐20,21
q=
0.65,
and
qs=
0.7.
All
these
values
will
now
be
used
in
Eq.
6-‐32:
𝑘! = 1 + 𝑞 𝑘! − 1
𝑘! = 1 + 0.65 2.14 − 1
𝑘! = 1.741
𝑘!" = 1 + 𝑞 𝑘!" − 1
𝑘!" = 1 + 0.7 3.0 − 1
𝑘!" = 2.4
Next
the
Alternating
and
Midrange
Von
Mises
Stresses
will
be
solved
for
with
all
the
given
values
from
above.
σ!
!
=
32𝑘! 𝑀!
π𝑑!
σ!
!
=
32(1.741)(2500 𝑙𝑏𝑓 𝑖𝑛)
π(1𝑖𝑛)!
σ!
!
= 44.33 𝑘𝑝𝑠𝑖
σ!
!
= 3
16𝑘!" 𝑇!
π𝑑!
σ!
!
= 3
16(2.4)(1838 𝑙𝑏𝑓 𝑖𝑛)
π(1𝑖𝑛)!
26. 26
σ!
!
= 38.91 𝑘𝑝𝑠𝑖
Finally
the
fatigue
factor
of
safety
and
Langer
will
be
calculated:
1
𝑛!
=
σ!
!
𝑆!
+
σ!
!
𝑆!"
1
𝑛!
=
44.33 𝑘𝑝𝑠𝑖
32.68 𝑘𝑝𝑠𝑖
+
38.91 𝑘𝑝𝑠𝑖
91 𝑘𝑝𝑠𝑖
𝑛! = 0.56
1
𝑛!
=
σ!
!
𝑆!
+
σ!
!
𝑆!
1
𝑛!
=
44.33 𝑘𝑝𝑠𝑖
77 𝑘𝑝𝑠𝑖
+
38.91 𝑘𝑝𝑠𝑖
77 𝑘𝑝𝑠𝑖
𝑛! = 0.93
These
are
not
acceptable
factors
of
safety
for
the
keyway
at
gear
2.
The
target
factor
of
safety
was
at
least
1.1.
In
order
to
obtain
this
factor
of
safety
the
material
of
the
countershaft
could
be
changed
to
a
stronger
material
but
more
importantly
the
diameter
of
1in
should
be
increased.
Finally
we
will
calculate
the
static
failure
factor
of
safety
for
the
keys.
The
keys
used
had
a
length
and
width
of
0.065
and
0.25
inches.
The
key
is
made
out
of
316
stainless
steel
which
has
a
σy=
30
kpsi.
𝑉 =
2𝑇!
𝑑
=
2(1838 𝑙𝑏𝑓 𝑖𝑛)
1𝑖𝑛
= 3.7 𝑘𝑝𝑠𝑖
τ =
𝑉
ℎ𝑙
=
3.7 𝑘𝑝𝑠𝑖
(0.065𝑖𝑛)(0.25𝑖𝑛)
= 226 𝑘𝑝𝑠𝑖
𝑛!"" =
σ!
2τ
=
30 𝑘𝑝𝑠𝑖
2(226 𝑘𝑝𝑠𝑖)
= 15.1
This
factor
of
safety
exceeds
the
target
factor
of
safety
at
1.1.
