Strength of Materials - Technical symposium.

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Transcript Strength of Materials - Technical symposium.

Notes on Lesson
Faculty Name
:
V. KIRAN KUMAR
Code
:
AU28
Subject Name
:
DESIGN OF MACHINE
Code
:
ME2303
ELEMENTS
Year
:
III
Semester
:
V
Degree & Branch
:
B.E. – AUTOMOBILE
Section
:
-
Failure Theories
 Stress in machine components should be
accurately computed.
 Designer must understand material limits to
ensure a safe design.
Design Factor
 Factor of Safety (N)
ExpectedStress
N
Stress at ComponentFailure
 Suitable values depend on inherent danger,
certainty of calculations, certainty of
material properties, etc.
Static Stresses - Brittle Materials
 Percent elongation < 5%
•
•
N
sut
N
suc


1 1  2


•
N sut suc
for parts in tension
for parts in compression
for parts with general stress
Example
The Gray Cast Iron (Grade 40) cylinder carries
an axial compressive load of 75,000 lbs and a
torque of 20,000 in lbs. Compute the resulting
design factor.
R0.25”
R0.25”
Ø4.00”
Ø5.00”
Static Stresses - Ductile Materials
 Percent elongation > 5%
 Distortion Energy Theory
Define von Mises Stress
 '   12   22   1 2
sy
• For nominal stress
N
• For localized stress
su
N
'
'
Static Stresses - Ductile Materials
 Percent elongation > 5%
 Maximum Shear Stress Theory
• For nominal stress
• For localized stress
N
s ys
 max
N
sus
 max

sy
2 max
Example
Specify a diameter for the middle portion of
the rod, if it is to be made from AISI 1040-hot
rolled steel.
450
5000 lbs
Example
For the seat support shown, specify a standard
structural tube to resist static loads shown.
The tube has properties similar to AISI 1020
hot-rolled steel. Use a design factor of 3.
400 lb
14”
200 lb
20”
Stress
Repeated Loads
alt
mean
Time
Example
The notched bar is machined from AISI 1020
steel. This bar is subjected to a load that
varies from 2000 lb to 3000 lb. Determine the
mean and alternating nominal stresses.
1.25”
0.1” R
1”
.75”
Fatigue Strength
 R.R. Moore Test
Motor
Alternating Stress, a
Endurance Strength, sn
103
104
105
106
107
108
Cycles to Failure, N (log)
Endurance Strength
 sn = Endurance strength
 Listed in tables
 If no information is available, use
• sn  0.5 su
(Steel)
• sn  0.4 su
(Aluminum)
Adjusted Endurance Strength
The data from the standard R.R. Moore
test is adjusted for a particular
application.
sn’ = Adjusted endurance strength
= (Cs) (Cm) (Cst) (CR) (sn)
Size and Stress Type Factors
– Cs = Size Factor
• D< 0.4 in
Cs = 1.0
• 0.4 < D 2.0 in
Cs = (D/0.3)-0.068
• 2.0 < D 10.0 in
Cs = D-0.19
For rectangular sections, D=.808(h b)1/2
– Cst = Stress Type Factor
• = 1.0 for bending
• = 0.80 for axial tension
• = 0.50 for torsion
Material and Reliability Factor
– Cm = Material Factor
• = 1.0 for wrought steel
• = 0.80 for cast steel
• = 0.70 for cast iron
– CR = Reliability Factor
• 50%
CR = 1.0
• 90%
CR = 0.90
• 99%
CR = 0.81
• 99.9%
CR = 0.75
Example
The notched bar is machined from AISI 1020
steel. This bar is subjected to a load that
varies from 2000 lb to 3000 lb. Determine the
endurance limit of the material.
1.25”
0.1” R
1”
.75”
Repeated Stresses - Ductile Materials
 Distortion Energy Theory
Define repeated von Mises Stress
 'm   12m   22m   1m 2 m
 'a   12a   22a   1a 2 a
• Solderberg criterion
1  'm K t  'a


N
sy
s 'n
Repeated Stresses - Ductile Materials
 Maximum Shear Stress Theory
1 ( m ) max Kt ( a ) max


N
ssy
s'sn
• ssy = 0.5 sy
• s’sn = 0.5 sn
Example
The notched bar is machined from AISI 1020
steel. This bar is subjected to a load that
varies from 2000 lb to 3000 lb. Comment on
the robustness of the design.
1.25”
0.1” R
1”
.75”
Example
Comment on the robustness of a 1-1/4” round
bar made from AISI 1213 C-D steel. It carries
a constant tensile load of 1500 lbs, a bending
load that varies from 0 to 800 lbs at the senter
of the 48” length and a constant torque of
1200 in lbs.
48”
Shafts
 Connect power transmission components.
 Inherently subjected to transverse loads and
torsion.
Shaft Forces
 Gears
As before
Wt
Wr
T
Shaft Forces
 Chains
Ftight
Ftight
2T

D
D
T
Fslack = 0
Shaft Forces
 V-belts
Ftight
Ftight
2.5T

D
D
Fslack
T

2D
Fslack
T
Shaft Forces
 Flat belts
Ftight
3T
Ftight 
D
D
Fslack
T

D
Fslack
T
Material Properties
 For steady load (torsion)
sys=.5sy
 For fatique load ( bending)
sn’=cs cR sn
cT = 1 (bending)
cm = 1 (wrought steel)
Stress Concentrations
 Keyseats
– Sled Runner Kt = 1.6
– Profile
Kt = 2.0
– Woodruff
Kt = 1.5
Stress Concentrations
 Shoulders
– Sharp, Bearing (r/d .03) Kt = 2.5
– Round, Gear Bore (r/d .17) Kt = 1.5
 Grooves
– Retaining Rings Kt = 1.5
Try not to let Kt’s overlap.
Leave .10 - .15” between
Strength Analysis
Kt M c Kt M
 Bending stress  

I
S
For round sections
 Torsion stress
S
 D3
32
Tr T


J
2S
For round sections
J
I
2
r
c
Strength Analysis
 Mohr’s circle and Solderberg
3
2

K t M / sn '  T / s y 
1
4

N
S
2
Suggested Design Factors:
• N=2 smooth operation
• N=3 typical industrial operation
• N=4 shock or impact loading
Minimum Acceptable Diameter
 The designer must size the shaft.
– Solve for appropriate diameters
 32 N
D
 
3
2
K t M / sn '  T / s y  
4

2
Example
Determine a suitable diameter for a shaft
made from AISI 1144 OQT 1000. It is
subjected to a reversing bending moment of
3000 ft lbs and a steady torque of 1800 ft lbs.
The shaft has a profile keyway.
Example
The shaft shown is part of a grain drying system
 At A, a 34 lb. propeller-type fan requires 12 hp
when rotating at 475 rpm.
 A flat belt pulley at D delivers 3.5 hp to a screw
conveyor handling the grain.
 All power comes to the shaft through the v-belt at
C.
Using AISI 1144 cold drawn steel, determine the
minimum acceptable diameter at C.
Example
B
A
12”
Sheave C
D
C
10”
10”
E
4”
150
Sheave D
Shafts Accessories
 Components used to securely mount power
transmitting elements on a shaft.
Axial
Rotational
Keys
 Allow torque to be transferred from a shaft
to a power transmitting element (gear,
sprocket, sheave, etc.)
Key Design
 Use a soft, low strength material
H
L
(ie, low carbon steel)
W
 Standard size H=W=1/4 D
 Design length
based on strength
Standard Key Sizes
Shaft Dia. (in)
W (in)
W
T
H
S
D  H  D2  W 2
S
2
D  H  D2  W 2
T
 .005in.
2
Key Design
T
2T
F

D/2 D
 Key Shear
F
2T
 
A DLW
 Failure Theory N 
 Length
4TN
L
DWs y
sy
2

s y LW
4TD
Example
Specify a key for a gear (grade 40, gray cast
iron) to be mounted on a shaft (AISI 1144, hot
rolled) with a 2.00 in. diameter. The gear
transmits 21000 lb-in of torque and has a hub
length of 4 in.
Retaining Rings
 Also known as snap rings
 Provides a removable shoulder to lock
components on shafts or in bores.
 Made of spring steel, with a high shear
strength.
 Stamped, bent-wire, and spiral-wound.
Retaining Ring Selection
 Based on shaft diameter & thrust force
Set Screws
 Setscrews are fasteners that hold collars,
pulleys, or gears on shafts.
 They are categorized by drive type and
point style.
Standard Set Screw Sizes
Set Screw Holding
Pins
 A pin is placed in double shear
 Holds torsion and axial loads
8T N
d
 D sy
D
d
 Hole is made slightly smaller than the pin
(FN1 fit)
Example
Specify a pin for a gear (grade 40, gray cast
iron) to be mounted on a shaft (AISI 1144, hot
rolled) with a 2.00 in. diameter. The gear
transmits 21000 lb-in of torque and has a hub
length of 4 in.
Roll Pins
 Easier disassembly
Collars
 Creates a shoulder on shaft without
increasing stock size.
 Held with either set screw or friction
(clamped)
Mechanical Couplings
 Couplings are used to join two shafts
 Rigid couplings are simple and low cost.
But they demand almost perfect alignment
of the mating shafts.
 Misalignment causes undue forces and
accelerated wear on the shafts, coupling,
shaft bearings, or machine housing.
Mechanical Couplings
 In connecting two shafts, misalignment is
the rule rather than the exception. It comes
from such sources as bearing wear,
structural deflection, thermal expansion, or
settling machine foundations.
 When misalignment is expected, a flexible
coupling must be used.
Mechanical Couplings
 Selection factors include:
 Amount of torque (or power & speed)
 Shaft Size
 Misalignment tolerance
Fasteners, Powers Screws,
Connections
Helical thread screw was an important invention.
Power Screw, transmit angular motion to liner motion
Transmit large or produce large axial force
It is always desired to reduce number of screws
Definition of important
Terminologies
Major diameter d, Minor diameter dr Mean dia or pitch diameter dp
Lead l, distance the nut moves for one turn rotation
Single and Double threaded screws
Double threaded screws are stronger and moves faster
Screw Designations
 United National
Standard UNS
 International Standard
Organization
Roots and crest can be either flat or round
Pitch diameter produce same width in the thread and space,
Coarse thread Designated by UNC
 Fine Thread UNF, is more resistance to
loosening, because of its small helix angle.
 They are used when Vibration is present
 Class of screw, defines its fit, Class 1 fits have
widest tolerances, Class 2 is the most commonly
used
 Class three for very precision application
 Example:1in-12 UNRF-2A-LH, A for Ext. Thread
and B for Internal, R root radius
 Metric M10x1.5 10 diameter mm major
diameter,1.5 pitch
Some important Data for UNC, UNF
and M threads
 Lets Look at the Table 8-1 on Page 398
Square and Acme Threads are used
for the power screw
Preferred pitch for Acme Thread
d, in
1/4
5/16
3/8
1/2
5/8
3/4
7/8
1
1 1/4
p,in
1/16
1/14
1/12
1/10
1/8
1/6
1/6
1/5
1/5
Mechanics of Power Screws
Used in design to change the angular motion to linear motion, Could
you recall recent failure of power screw leading to significant
causalities
What is the relationship between the
applied torque on power screw and
lifting force F
Torque for single flat thread
Fd m l  fd m
TR 
(
)
2 d m  fl
Fd m fd m  l
TL 
(
)
2 d m  fl
If the thread as an angle α, the torque will be
Fd m l  fd m sec
TR 
(
)
2 d m  fl sec
Wedging action, it
increases friction
Stresses in the power Screw
Shear stress in the base
of the screw
Bearing stress
Bending stress at the root
of the screw
Shear stress in the thread
nt number of engaged
thread

16T
d 3
B  
F
d m nt p / 2
6F
b 
d r nt p

3V
3F

2 A d r nt p
Loading to the fasteners and their
Failure considerations
Bolts are used to clamp two or more parts
It causes pre tension in the bolt
Grip length is the total thickness of parts
and washers
l
d
t
l
l
ld
h
t2
lt=L’- ld
L’ effective grip= h+t2 if t2<d
=h=d/2 for t2  d
Failure of bolted or riveted joints
Type of Joints
 Lap Joint (single Joint)
But Joint
Example 1
Example 2
Example 2
Example 3
Weld
Weld under Bending
Springs
Flexible machine elements
Used to:
 Exert force
 Store energy
Spring Rate
 Effective springs have a linear deflection
curve.
 Slope of the spring deflection curve is the
rate
Force
k
1
Deflection
F
k
L
Example
A compression spring with a
rate of 20 lb/in is loaded with
6 lbs and has a length of 1.5
in. Determine the unloaded
spring length (free length)
Geometry
Di
Dw
Dw
L
Do
 Wire diameter, Dw (Standard gages)
 Mean Diameter, Dm
Dm = Do - Dw
Spring Parameters
 Spring index
Dm
C
Dw
C > 5 (manufacturing limits)
 Active coils, Na
= N for plain ends
= N-1 for ground ends
= N-2 for closed ends
Deflection
 Deflection for helical springs
8FDm3 N a 8FC3 N a


4
GDw
GDw
G = Shear modulus
 Spring rate for helical springs
GDw
k 3
8C N a
Example
A helical compression spring is formed from
35 gage music wire with 10-1/4 turns and an
O.D. if 0.850 in. It’s ends are squared. The
free length is 2 inches. Determine the force to
press the spring solid.
Stress Analysis
 Spring wire is in torsion
V
T r 8K F C


2
J
 Dw
 Wahl factor, K
Accounts for the
curvature of the wire
4C  1 .615
K

4C  4
C
T
F
Example
A helical compression spring is formed from
35 gage music wire with 10-1/4 turns and an
O.D. if 0.850 in. It’s made from A228 and
the ends are squared. The free length is 2
inches.
If the spring is repeatedly compressed to 1.3
in, do you expect problems?
Design Procedure
 Select a material
 Compute required spring rate
 Estimate Dm based on size constraints
 Determine required Dw (use K=1.2)
 Select standard wire
 Verify actual stress is satisfactory.
 Compute number of coils required.
Example
Design a helical compression spring to exert a
force of 22 lbs when compressed to a length
of 1.75 in. When its length is 3.0 in, it must
exert 5 lb. The spring will be cycled rapidly.
Use ASTM A401 steel wire.
Rolling Element Bearings
 Provides support for machine elements,
while allowing smooth motion.
m=0.001 - 0.005
Types
Single-row
Radial Ball
Angular
Contact Ball
Radial Roller
Angular
Roller
Types
Spherical
Roller
Tapered Roller
Needle
Thrust
Ball Bearings
Stress Analysis
 Contact Stress
c=300,000 is not unusual
 Balls, rollers and races are made from
extremely high strength steel
ex. AISI 52100
sy = 260,000 psi
su=322,000 psi
Bearing Load/Life
 Test (fatigue) data
Radial Load (lbs)
Empirical relationship:
L2  P1 
  
L1  P2 
k
•k=3.0 (ball)
•k=3.33 (roller)
L10 Life (cycles)
Example
A bearing is mounted on a shaft rotating at
1200 rpm. The bearing has been tested to have
a L10 life of 300 hrs, when loaded with 500
lbs. Determine the expected L10 life, if the
load is increased to 700 lbs.
Manufacturer’s Data
 Vendors publish the
Basic Dynamic Load
rating (C) of a bearing at
an L10 life of 1 million
cycles.
Bearing Selection
 Determine the design life (in cycles)
 Determine the design load
Pd = V R
• V=1 for inner race rotation
• V=1.2 for outer race rotation
 Calculate the required
basic
dynamic
load
1
 Ld  k
Creq 'd  Pd  6 
 10 
 Select a bearing with (C > Creq’d) and a bore
that closely matches the shaft diameter.
Example
Specify suitable bearings for a shaft used in an
grain dryer. The shaft rotates at 1700 rpm.
The required supporting loads at the bearing
are
R =589 lb
Bx
RBy=164 lb
and the minimum acceptable diameter is
2.16”.
Mounting of Bearings
 Shaft/bearing bore has a light interference
fit.
 Housing/outer race has a slight clearance fit.
Check manufacturers catalog
 Match maximum permissible fillet radius.
 Shaft or housing shoulders not to exceed
20% of diameter.
Mounted Bearings
 Pillow block
 Bearing is inserted into a cast housing, with
base or flange slots, which can be readily
attached to a machine base.
Bearings with Varying Loads
 Compute a weighted average load based on
duty cycle.

  Fi  p N i
Fm  
  Ni




1
p
Fm=equivalent load
Fi= load level for
condition i
Ni= cycles for
condition i
p = exponent for
load/life
Example
Bearing 6211 is carrying the following load
cycle, while rotating at 1700 rpm.
Stage
Load (lbs)
Time (min)
1
600
480
2
200
115
3
100
45
Compute the bearing L10 life in minutes.
Radial & Thrust Loads
 Calculate an equivalent load
P=VXR +YT
T=thrust load
X
factors depending
=
Y
on bearing
Thrust factors, Y
 Deep -groove, ball bearings
X = 0.56 for all values of Y
Example
A bearing is to carry a radial load of 650 lb
and a thrust load of 270 lb. Specify a suitable
single-row, deep-groove ball bearing if the
shaft rotates at 1150 rpm and the design life is
20,000 hrs.