ME 343: Mechanical Design-3
Design of Shaft
Aly Mousaad Aly, PhD
Assistant Professor, Dept. of Mech. Eng.
Faculty of Engineering, Alexandria University
Outline
• Practical information
• Shaft design
Lecture1: Design of Shaft
2
Instructor
• Lecturer: Dr. Aly Mousaad Aly
• Office: Last floor, Dept. of Mech. Eng.,
Faculty of Eng., Alexandria University
• Office hours:
Thursday 8:30 to 9:35
Thursday,
Thursday, 11:00 to 12:05
Lecture 1: Introduction
3
Course Materials
Slides: Available online.
Available at the department copy center.
Course website:
www.engr.uconn.edu/~aly/ME343
References:
• Shigley’s Mechanical Engineering Design,
Eighth Edition, The McGraw−Hill Companies,
Inc., 2006.
Lecture 1: Introduction
4
Grading
•
•
•
•
•
Class participation
Assignments
Reports
Midterm exam
Final examination
Lecture 1: Introduction
5
Policy
• Attendance
d
to llectures and
d exercises
i
iis
compulsory.
• We may check the attendance at the
g
g of lessons.
beginning
• Everybody should attend his scheduled classes
according to his name and student number.
number
We will be VERY STRICT about this rule.
• Come
C
t lessons
to
l
about
b t 5 min
i b
before
f
th
the
starting time.
Lecture1: Design of Shaft
6
Outline
• Practical information
• Shaft design
Lecture1: Design of Shaft
7
Definition of shaft?
• It is
i a rotating
i member,
b iin general,l h
has a
circular cross-section and is used to transmit
power.
• The shaft mayy be solid or hollow. It is
supported on bearings and it rotates a set of
ggears or pulleys
p y for the purpose
p p
of p
power
transmission.
• The shaft is generally acted upon by bending
moments, torsion and axial forces.
Lecture1: Design of Shaft
8
Shaft versus axle and spindle
Axle is a non-rotating member used for
pp
g rotatingg wheels, etc., and do not
supporting
transmit any torque. Spindle is simply defined as
a short shaft
shaft. However
However, design method remains
the same for axle and spindle as that for a shaft.
Lecture1: Design of Shaft
9
What does it mean “shaft
shaft design”?
design ?
•
•
•
•
Material selection
Geometric layout
Stress and strength: static and fatigue
Deflection and rigidity: bending defl., torsional
g, slope
p at bearings
g and shafttwisting,
supported elements, and shear deflection
due to transverse loading on short shafts.
shafts
• Vibration: critical speed
Lecture1: Design of Shaft
10
Material selection
• M
Many shafts
h ft are made
d ffrom low
l carbon,
b
coldld
drawn or hot-rolled steel.
• Alloy steel:
steel Nickel,
Nickel chromium and vanadium are
some of the common alloying materials.
However alloy steel is expensive.
However,
expensive
• Shafts usually don’t need to be surface hardened
unless they serve as the actual journal of a
bearing surface.
• Hardening
g off surface
f
((wear resistant):
) case
hardening and carburizing ; cyaniding and
nitriding.
Lecture1: Design of Shaft
11
Geometric layout
Lecture1: Design of Shaft
12
Geometric layout
• The geometry of shaft is generally that of stepped
cylinder.
li d
• There is no magic formula to give the shaft geometry
f any given
for
i
d
design
i situation.
it ti
Lecture1: Design of Shaft
13
Geometric layout
• The
h b
best approach
h iis to llearn ffrom similar
i il
problems that have been solved and combining
th b
the
bestt tto solve
l your own problem.
bl
• A general layout to accommodate shaft elements,
e.g. gears, bearings, and pulleys, must be
specified early in the design process.
• Shoulders are used for axially locating shaft
elements and to carry any thrust loads.
• Common Torque Transfer Elements: keys, set
pins, p
press or shrink fits, tapered
p
fits.
screws, p
Lecture1: Design of Shaft
14
Geometric layout
• Small pinions are often machined onto shafts.
• Sequence of assembly should be thought.
• Use chamfers to ease assembly and avoid
i
interferences.
f
• Consider stress risers due to ggrooves and
sharp steps in shafts.
• What
Wh t can ffailil and
dh
how will
ill it h
happen??
Lecture1: Design of Shaft
15
Shaft design based on strength
Design
i iis carried
i d out so that
h stress at any llocation
i off
the shaft should not exceed material yielding.
Stress due to torsion:
τ
xy
T × r
1 6T
=
=
J
π d o3 (1 − c
τxy : Shear stress due to torsion
T : Torque on the shaft
Note:
4
)
7024 × hp
9549 × kW
T ≈
≈
N (RPM )
N (RPM )
Lecture1: Design of Shaft
16
Shaft design based on strength
Bending stress:
σ
b
M × y
32M
=
=
I
π d o3 (1 − c )
M : Bending moment at the point of interest
do : Outer diameter of the shaft
c: di/do
Lecture1: Design of Shaft
17
Shaft design based on strength
Axial
i l stress:
σ
a
Fa
4α Fa
=
=
A
π d o2 (1 − c 2 )
Fa
Fa
Fa: Axial force (tensile or compressive)
α: Column-action
C l
ti factor(=
f t ( 1
1.0
0 for
f tensile
t il lload)
d)
α arises due to the phenomenon of buckling of
l
long
slender
l d members
b which
hi h are acted
d upon b
by
axial compressive loads.
Lecture1: Design of Shaft
18
Shaft design based on strength
A i l stress
Axial
t
((continue):
ti
)
1
α =
,
1 − 0 .0 0 4 4 λ
λ 2 s yc
α =
,
2
π nE
(λ = L / r ) ≤ 1 1 5
λ > 115
n = 1.0 for hinged end; n = 2.25 for fixed end
n = 1.6 for ends p
partlyy restrained,, as in bearingg ,
L = shaft length
p
syc = yyield stress in compression
Lecture1: Design of Shaft
19
Shaft design based on strength
Maximum shear stress theory (ductile mat.):
Failure occurs when the maximum shear stress
at a point exceeds the maximum allowable
shear stress for the material.
material Therefore,
Therefore
2
τ max = τ allowable
τ allowable
⎛σx ⎞
2
= ⎜ ⎟ + τ xy
⎝ 2 ⎠
16
=
3
4
π d o (1 − c )
Lecture1: Design of Shaft
⎛
α Fa d o (1 + c
⎜M +
⎜
8
⎝
2
) ⎞⎟
⎟
⎠
2
+T2
20
Shaft design based on strength
Maximum normal stress theory (brittle mat.):
σ max = σ allowable
σ allowable
σx
2
⎛σx ⎞
=
+ ⎜ ⎟ + τ xy2
2
⎝ 2 ⎠
2
⎡
⎛
⎞
+
α
F
d
1
c
)
16
a o (
⎢
⎜M +
⎟
=
3
4
⎟
8
π d o (1 − c ) ⎢⎜⎝
⎠
⎣
⎛
α Fa d o (1 + c
+ ⎜M +
⎜
8
⎝
Lecture1: Design of Shaft
2
) ⎞⎟
⎟
⎠
2
⎤
⎥
+T2 ⎥
⎥⎦
21
Shaft design based on strength
Von Mises/ Distortion-Energy theory:
σ max = σ allowable = σ x2 + 3τ xy2
σ allowable
ll
bl
16
=
π d o3 (1 − c 4 )
Lecture1: Design of Shaft
⎛
α Fa d o (1 + c
⎜ 2M +
⎜
4
⎝
2
) ⎞⎟
⎟
⎠
2
+ 3×T 2
22
Shaft design based on strength
ASME design code (ductile material):
τ allowable
16
=
π d o3 (1 − c 4 )
⎛
α Fa d o (1 + c
⎜ km M +
⎜
8
⎝
2
) ⎞⎟
⎟
⎠
2
+ ( kt T )
2
where, km and kt are bending and torsion factors
where
accounts for shock and fatigue. The values of
th
these
ffactors
t are given
i
iin ASME d
design
i code
d ffor
shaft.
Lecture1: Design of Shaft
23
Shaft design based on strength
ASME design code (brittle material):
σ allowable
2
⎡
⎛
⎞
α
F
d
1
c
+
)
16
a o (
⎢⎜ k m M +
⎟
=
3
4
⎟
8
π d o (1 − c ) ⎢⎜⎝
⎠
⎣
⎛
α Fa d o (1 + c
+ ⎜ km M +
⎜
8
⎝
Lecture1: Design of Shaft
2
) ⎞⎟
⎟
⎠
2
⎤
2 ⎥
+ ( kt T ) ⎥
⎥⎦
24
Shaft design based on strength
ASME design code:
Combined shock and fatigue factors
Stationary shaft Rotating shaft
Type of load
Gradualyy applied
pp
load
km
kt
km
kt
1
1
1.5
1
Suddenly applied load, minor shock 1.5-2 1.5-2
1.5-2 1-1.5
Suddenly applied load
load, heavy shock ---
23
2-3
Lecture1: Design of Shaft
---
153
1.5-3
25
Shaft design based on strength
ASME design
d i code:
d
Commercial steel shafting
τallowable = 55 MPa for shaft without keyway
τallowable = 40 MPa for shaft with keyway
y y
Steel under definite specifications
τallowable = 30% of the yield strength but not over
18% of the ultimate strength in tension for shafts
without keyways. These values are to be reduced by
25% for the presence of keyways.
Lecture1: Design of Shaft
26
Standard sizes of shafts
Typical sizes of solid shaft that are available in
the market are:
diameter
up to 25 mm
25 to
t 50 mm
50 to 100 mm
100 to 200 mm
Lecture1: Design of Shaft
increments
0.5 mm
1 0 mm
1.0
2.0 mm
5.0 mm
27
Example: problem
A pulley
ll d
drive
i iis ttransmitting
itti power tto a pinion,
i i
which in turn is transmitting power to some other
machine element.
element Pulley and pinion diameters are
400 mm and 200 mm respectively. Shaft has to be
designed
g
for minor to heavyy shock.
A
C
B
D
.
m
Lecture1: Design of Shaft
m
m
28
Example: solution
Torsion:
TD = 6000 x (Dpinion/2)
= 6000 x (200/2)
= 6x10e5 N.mm
OR
TC = (4000 -1000) x (Dpulley/2)
= 3000 x (400/2) = 6x10e5 N.mm
Lecture1: Design of Shaft
29
Example: solution
B di ((vertical
Bending
ti l plane):
l
)
A
200 mm C
RAV
400 mm
D
200 mm
B
RBV
1000 N
6000 N
RBV= (1000x200- 6000x(400+200))/(200+400+200) = -4250 N
MDV= -4250x200
4250x200 = -8
8.5e5
5e5 N.mm
N mm
MCV= 6000x400 - 4250x600 = -1.5e5 N.mm
Lecture1: Design of Shaft
30
Example: solution
B di (h
Bending
(horizontal
i t l plane):
l
)
A
200 mm C
400 mm
D
200 mm
RAH
5000 N
2200 N
B
RBH
RBH = (5000x200 + 2200x(400+200))/(200+400+200)
= 2900 N
MDH = 2900x200
2900 200 = 5.8e5
5 8 5 N.mm
N
MCH = 2900x600 - 2200x400 = 8.6e5 N.mm
Lecture1: Design of Shaft
31
Example: solution
Bending
di ((resultant):
l
)
MD =
Similarly
Similarly,
( M DV ) + ( M DH )
2
2
= 10.29 × 105 N .mm
MC =
(1.5 × 10 ) + (8.6 × 10 )
5 2
5 2
= 8.73 × 105 N .mm
Since TC = TD and MD > MC, section-D is critical.
Lecture1: Design of Shaft
32
Example: solution
ASME
S
code:
d
Under minor to heavy shock, let us consider
km= 2 and kt = 1.5. Also let us assume the shaft will
be fabricated from commercial steel, i.e. τallowable =
40 Mpa.
16
5 2
5 2
d =
2 × 10.29 × 10 ) + (1.5 × 6 × 10 )
(
40 × π
d o = 65.88
65 88 mm
3
o
The value of standard shaft diameter is 66 mm.
Lecture1: Design of Shaft
33