ME 343: Mechanical Design-3
Design of Shaft (continue)
Dr. Aly Mousaad Aly
Department of Mechanical Engineering
Faculty of Engineering, Alexandria University
Objectives
At the end of this lesson, we should be able to
• Understand design method for variable load
• Define equivalent stresses on shaft
• Design shaft based on stiffness and
g y
torsional rigidity
• Understand critical speed of shafts
Lecture 2: Design of Shaft
2
Shaft design based on fatigue
• Any rotating shaft loaded by stationary
bendingg and torsional moments will be
stressed by completely reversed bending stress
while the torsional stress will remain steady
(i.e., Ma = M; Mm = 0; Ta = 0; Tm = T).
• A design method
h ffor variable
bl lload (f
(fatigue),
)
like Soderberg, Goodman or Gerber criteria
can be followed.
Lecture 2: Design of Shaft
3
Shaft design based on fatigue
Design under variable normal load (fatigue)
Lecture 2: Design of Shaft
4
Shaft design based on fatigue
A is the design point for which the stress
p
is σa and the mean stress is σm. In
amplitude
the Soderberg criterion the mean stress
material property is the yield point Sy,
whereas in the the Goodman and Gerber
criteria the material property is the ultimate
strength Sut. For the fatigue loading, material
property is the endurance limit Se in reverse
bending.
Lecture 2: Design of Shaft
5
Shaft design based on fatigue
SSoderberg
d b
criterion
it i
FS σ a FS σ m
Se
+
Sy
=1
mod-Goodman
dG d
criterion
it i
FS σ a FS σ m
Se
+
Sut
=1
Gerber criterion
2
FS σ a ⎛ FS σ m ⎞
Se
+⎜
⎝ Sut
⎟ =1
⎠
where
σa = stress amplitude (alternating stress); Se =
endurance limit (fatigue limit for completely reversed
loading); σm = mean stress; Sy = yield strength; σut =
g and FS= factor of safety.
y
ultimate tensile strength
Lecture 2: Design of Shaft
6
Shaft design based on fatigue
Design under variable shear load (fatigue)
Lecture 2: Design of Shaft
7
Shaft design based on fatigue
• It is most common to use the Soderberg
criterion.
FS K f σ a
Se
Sy K f σa
Se
Sy K f σ a
Se
FS σ m
+
=1
Sy
+σm =
Sy
FS
+ σ m = σ eq
Normal Stress Equation
Lecture 2: Design of Shaft
Kf = fatigue stress
concentration factor
S ys K fs τ a
Ses
+ τ m = τ eq
Shear Stress Equation
8
Shaft design based on fatigue
Sy K f σ a
Se
+ σ m = σ eq
S ys K fs τ a
Ses
+ τ m = τ eq
• σeq and τeq are equivalent to allowable stresses
(Sy/FS) and
d (Sys/FS),
/FS) respectively.
i l
• Effect of variable stress has been effectivelyy
defined as an equivalent static stress.
• Conventional
C
ti
l ffailure
il
th
theories
i can b
be used
d tto
complete the design.
Lecture 2: Design of Shaft
9
Shaft design based on fatigue
• Max. shear stress theory + Soderberg line
((Westinghouse
g
Code Formula))
⎛ σ eq ⎞
2
τ max = τ allowable = ⎜
2
τ
+
⎟
eq
2
⎝
⎠
2
⎛ Sy K f σ a
⎞ ⎛ S y K fs τ a
⎞
= ⎜
+σm ⎟ + ⎜
+τm ⎟
2 FS
Se
Se
⎝
⎠ ⎝
⎠
Sy
2
2
32 FS ⎛ K f M a M m ⎞ ⎛ K fs Ta Tm ⎞
+
+ ⎟
d =
⎜⎜
⎟⎟ + ⎜⎜
⎟
π
S
S
S
S
e
y
e
y
⎝
⎠ ⎝
⎠
2
3
Lecture 2: Design of Shaft
10
Shaft design based on rigidity
• Deflection is often the more demanding
constraint. Manyy shafts are well within
specification for stress but would exhibit too
much deflection to be appropriate
appropriate.
• Deflection analysis at even a single point of
interest requires complete
l
geometry
information for the entire shaft.
Lecture 2: Design of Shaft
11
Shaft design based on rigidity
• It is desirable to design the dimensions at
critical locations to handle the stresses, and fill
in reasonable estimates for all other
dimensions before performing a deflection
dimensions,
analysis.
• Deflection
fl
off the
h shaft,
h f b
both
h llinear and
angular, should be checked at gears and
bearings.
Lecture 2: Design of Shaft
12
Shaft design based on rigidity
• Sl
Slopes, llateral
t ld
deflection
fl ti off th
the shaft,
h ft and/or
d/ angle
l off ttwist
it
of the shaft should be within some prescribed limits.
Crowned
tooth
Diametral pitch, P = number of teeth/pitch diameter. 1 in = 25.4 mm.
Lecture 2: Design of Shaft
13
Shaft design based on rigidity
• In case off sleeve
l
b
bearings,
i
shaft
h f d
deflection
fl i across
the bearing length should be less than the oil-film
thi k
thickness.
Twist angle:
θ ( radd ) =
T ( N .mm ) L ( mm )
G ( N / mm 2 ) J ( mm 4 )
G: shear modulus; J: polar moment of inertia
• The limiting value of θ varies from 0.3
0 3 deg/m to 3
deg/m for machine tool shaft to line shaft
respectively.
respectively
Lecture 2: Design of Shaft
14
Shaft design based on rigidity
L t l deflection:
Lateral
d fl ti
• Double integration
• Moment-area
• Energy (Castigliano Theorem)
δ= f (applied load, material property, moment of
inertia and given dimension of the beam).
From the expression of moment of inertia, and
k
known
design
d
parameters, including
l d δ
δ, shaft
h f
dimension may be obtained.
Lecture 2: Design of Shaft
15
Double Integration Method
d2y
M
=
2
dx
EI
y(x) =
∫∫
M ( x)
dy
θ ( x) = = ∫
dx
dx
EI ( x)
M (x) 2
dx
EI (x)
Use boundary conditions to obtain integration
constants
Lecture 2: Design of Shaft
16
Conjugate beam method
•
•
•
•
•
Was developed by Otto Mohr in 1860
Slope (real beam) = Shear (conj. beam)
Deflection (real beam) = Moment (conj. beam)
Length of conj. beam = Length of real beam
The load on the conjugate beam is the M/EI
diagram of the loads on the actual beam
Lecture 2: Design of Shaft
17
Conjugate beam method
R lb
Real
beam
Conjugate beam
Real beam
Conjugate beam
Lecture 2: Design of Shaft
18
Example 1
Determine the slope
and deflection at the
tip of a cantilever using
the conj
conj. beam
method.
Sol:l
2
PL
θB =
2 EI
2EI
PL3
yB =
3EI
Lecture 2: Design of Shaft
19
Example 2
Compute the
h maximum
i
d
deflection
fl i and
d the
h
slopes at the bearings. EI is constant.
(a)
Lecture 2: Design of Shaft
(b)
20
Critical speed of rotating shaft
• Critical speed of a rotating shaft is the speed
y
y unstable.
where it becomes dynamically
• The frequency of free vibration of a nonrotating shaft is same as its critical speed
speed.
60
N critical ( RPM ) =
2π
Lecture 2: Design of Shaft
g
( w1δ1 + w2δ 2 + .... + wnδ n )
2
2
2
w
δ
+
w
δ
+
....
+
w
δ
( 11 22
n n )
21
Critical speed of rotating shaft
• W1, W2…. : weights of the rotating bodies (N)
• δ1, δ2 …. : deflections of the respective bodies
(m)
• For
F a simply
i l supported
d shaft,
h f h
half
lf off iits weight
i h
may be lumped at the center for better
accuracy.
• For a cantilever shaft,
shaft quarter of its weight
may be lumped at the free end.
Lecture 2: Design of Shaft
22
Shaft design: general considerations
• Axial
i l thrust
h
lloads
d should
h ld b
be taken
k to ground
d
through a single thrust bearing per load
direction.
• Do not split
p axial loads between thrust
bearings as thermal expansion of the shaft can
g
overload the bearings.
• Shaft length should be kept as short as
possible to minimize both deflections and
possible,
stresses.
Lecture 2: Design of Shaft
23
Shaft design: general considerations
• A cantilever beam will have a larger deflection
p y supported
pp
((straddle mounted))
than a simply
one for the same length, load, and cross
section.
section
• Hollow shafts have better stiffness/mass ratio
and higher
h h naturall ffrequencies than
h solid
l
shafts, but will be more expensive and larger
in diameter.
Lecture 2: Design of Shaft
24
Shaft design: general considerations
• Slopes, lateral deflection of the shaft, and/or
g of twist of the shaft should be within
angle
some prescribed limits.
• First natural frequency of the shaft should be
at least three times the highest forcing
f
frequency.
((A ffactor off ten times or more is
preferred, but this is often difficult to achieve).
Lecture 2: Design of Shaft
25
Example 3
Determine the diameter of a shaft of length L
y g a load of 5 kN at the center if the
=1m, carrying
maximum allowable shaft deflection is 1mm.
What is the value of the slope at the bearings
bearings.
Calculate the critical speed of this shaft if a disc
weighting 45 kg is placed at the center
center. E=209
E 209
GPa. ρst = 8740 kg/m^3.
Lecture 2: Design of Shaft
26
Example 3: solution
Maximum deflection:
PL3
y=
48 EI
1=
5000 × (1000 )
48 × ( 209 × 10 ) ×
3
3
π
64
d4
∴ d = 56.45 mm
From a standard shaft size, d = 58 mm
Lecture 2: Design of Shaft
27
Example 3: solution
Slope at bearings:
5000 × (1000 )
PL
θ1 = θ 2 =
=
16 EI 16 × 209 × 103 × π × 58 4
(
) 64 ( )
= 0.0027 rad
2
2
which is much for tapered and cylindrical
roller bearings. However, this value may be
acceptable for deep-groove ball bearing.
Lecture 2: Design of Shaft
28
Example 3: solution
Critical
C
iti l speed:
d
Shaft weight = rho x A x L = 7840 x A x 1 = 20.7 kg
0.5 × 20.7 + 45 ) × 9.807 × (1000 )
(
PL3
=
= 0.097 mm
δ=
π
4
48 EI
48 × ( 209 × 103 ) × × ( 58 )
64
3
N critical
60
=
2π
9.807
= 3030
−3
0.097 × 10
RPM
The operating speed of the shaft should be well
p )
below this value ((sayy less than 1000 rpm).
Lecture 2: Design of Shaft
29
Example 4
FFor th
the same problem
bl
solved
l d iin th
the previous
i
llesson,
determine the following:
1 Lateral deflection at D (using Conj
1.
Conj. Beam method)
2. Slope at bearings A and B (using Conj. Beam method)
3. Critical speed
A
C
B
D
.
m
Lecture 2: Design of Shaft
m
m
d sh = 66 mm
30
Shaft design: summary
• Shaft design means material selection,
ggeometric layout,
y
stress and strength
g (static
(
and fatigue), deflection and slope at bearings.
• Conjugate beam method for slope and
deflection calculations.
• Some design considerations: (axial load, shaft
length, support layout, hollow shafts, slopes
and deflection, operating speed)
Lecture 2: Design of Shaft
31
Report
¾ Write
W it a short
h t reportt on material
t i l selection
l ti for
f
shaft manufacturing. Talk briefly about heat
treatment processes that may be considered.
considered
¾ Deadline: Thursday, April 7, 2011
¾ How to submit:
• Send attachment to mosaadaly2000@yahoo.com
Pl
Please
give
i a title
titl to
t th
the emailil as your student
t d t
number followed by your name, e.g.
999 Name Surname
• Print out: not recommended
Lecture 2: Design of Shaft
32
Slides and sheets
Available for download at:
http://www engr uconn edu/~aly/Courses/ME343/
http://www.engr.uconn.edu/
aly/Courses/ME343/
Lecture 2: Design of Shaft
33