Transcript LEC 35 - KFUPM Open Courseware

ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 1
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 2
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 3
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 4
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 5
ME 307
Machine
Design I
8-1
8-2
8-3
8-4
8-5
8-6
8-7
8-8
8-9
8-10
8-11
8-12
8-13
8-14
8-15
Thread Standards and Definitions
The Mechanics of Power Screws
Strength Constraints
Joints-Fasteners Stiffness
Joints-Member Stiffness
Bolt Strength
Tension Joints-The External Load
Relating Bolt Torque to Bolt Tension
Statically Loaded Tension Joint with Preload
Gasketed Joints
Fatigue Loading of Tension Joints
Shear Joints
Setscrews
Keys and Pins
Stochastic Considerations
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 6
ME 307
Machine
Design I
8-3
8-4
8-5
8-6
Strength Constraints
Joints-Fasteners Stiffness
Joints-Member Stiffness
Bolt Strength
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 7
ME 307
Machine
Design I
Example-3
A single-threaded 20 mm power screw is 20 mm in diameter
with a pitch of 5 mm. A vertical load on the screw reaches a
maximum of 3 kN. The coefficients of friction are 0.06 for
the collar and 0.09 for the threads. The frictional diameter
of the collar is 45 mm. Find the overall efficiency and the
torque to "raise" and "lower" the load.
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 8
ME 307
Machine
Design I
Example-3
Given
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 9
ME 307
Machine
Design I
Example-3
(Cont.’d)
Solution
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 10
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Example-2
(Cont.’d)
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 11
ME 307
Machine
Design I
Power Screws Stress Analysis
The following stresses should be checked on both the nut and
the screw:
1.
Shearing stress in screw body.
2. Axial stress in screw body
16T

πd r3
(8-7)
F
4F
 

A
πd r2
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
(8-8)
CH-8
LEC 35 Slide 12
ME 307
Machine
Design I
3.
Power Screws Stress Analysis
Thread bearing stress
F
2F
B  

πd m nt p / 2
πd m nt p
(8-10)
where nt is the number of engaged
threads.
Figure 8-8
Geometry of square thread useful in
finding bending and transverse shear
stresses at the thread root
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 13
ME 307
Machine
Design I
Power Screws Stress Analysis
4.
Thread bending stress
The bending stress at the root of the thread
is given by
I 1 12  πd r nt  p 2 
π


d r nt p 2
C
24
 P / 2 2
3
Fp
M 
4
M
b 
I c
Dr. A. Aziz Bazoune
6F
 Fp  24



2
 4  πdr nt p πd r nt p
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
(8-11)
LEC 35 Slide 14
ME 307
Machine
Design I
Power Screws Stress Analysis
5. Transverse shear stress at the
center of the thread root
3V 3
F
3F



2 A 2 πd m nt p 2 πd m nt p
(8-12)
Notice that the transverse shear
stress at the top of the root is zero
 0
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 15
ME 307
Machine
Design I
Power Screws Stress Analysis
The state of stress at top of
root “plane” is
6F
x 
πd r nt p
y  0
4F
z   2
 dr
 Von-Mises Stress at top of
root plane is calculated
 xy  0
 yz 
16T
 d r3
using Eq. (6-14) of Sec. (6-5)
and failure criteria applied
(see example 8-1).
 zx  0


2
2
1
2
 '
 x   y    y   z    z   x   6  xy2   yz2   zx2 

2
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
12
LEC 35 Slide 16
ME 307
Machine
Design I
Power Screws Stress Analysis
 The engaged threads cannot share the load equally. Some
experiments show that
the first engaged thread carries 0.38 of the load
the second engaged thread carries 0.25 of the load
the third engaged thread carries 0.18 of the load
the seventh engaged thread is free of load
 In estimating thread stresses by the equations above, substituting
nt to 1 will give the largest level of stresses in the thread-nut
combination
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 17
ME 307
Machine
Design I
Power Screws Buckling
Assuming that the column (screw) is a Johnson column
F
 
 A crit
2
 Sy l  1
 Sy  

2

k

 CE
where
l  L  L'
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 18
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Example-3
(Example 8-1 in Textbook)
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 19
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 20
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Example-3
(Example 8-1 in Textbook)
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 21
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Example-3
(Example 8-1 in Textbook)
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 22
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Example-3
(Example 8-1 in Textbook)
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 23
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 24
ME 307
Machine
Design I
Types of fasteners
Three types of threaded fastener: (a) Screw (b)Bolt and nut; (c) Stud and
nut, (d) Threaded rod and nuts
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 25
ME 307
Machine
Design I
Threaded Fasteners
A- BOLTS:
Purpose:
to clamp two or more
members together.
Parts:
(1) Head (Square or
Hexagonal)
(2) Washer (dw=1.5d)
(3) Threaded part
(4) Unthreaded part
Dr. A. Aziz Bazoune
Dimensions of square and hexagonal bolts are
given in TABLE A-29
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 26
Threaded Fasteners
ME 307
Machine
Design I
The diameter of the washer face
is the same as the width across
the flats of the hexagon. The
thread length (LT) is :
English
1

2
D

in

4
LT  
2 D  1 in

2
L  6 in
L  6 in
Metric (in mm)
2 D  6

L T  2 D  12
2D+25

Dr. A. Aziz Bazoune
L  125 D  48
125  L  200
L  200
D is the nominal diameter
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 27
Threaded Fasteners
ME 307
Machine
Design I
B- NUTS:
Same material as that of a screw
Table A-31 gives dimensions of Hexagonal nuts
Good Practice:
1. Never re-use nuts;
2. Tightening should be done such that 1 or 2 threads come out of the nut;
3. Washers should always be used under bolt head to prevent burr stress
concentration.
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 28
ME 307
Machine
Design I
Threaded Fasteners
Common Cap screws
Used for clamping members
same as bolt except that 1
member should be threaded.
The head of a hexagon-head
cup screw H cap is slightly
thinner than that of a
hexagonal head bot H bolt .
Figure 8-10
Typical cap-screw heads: (a)
fillister head; (b) flat head; (c)
hexagonal socket head
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 29
ME 307
Machine
Design I
Threaded
Fasteners
Figure 8.11:
Types of heads used in
machine screws
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 30
ME 307
Machine
Design I
Joints: Fastener Stiffness
In joint under tension the members are under compression
and the bolt under tension: kb = equivalent spring constant
of bolt composed of threaded (kt) and unthreaded (kd) parts
acting as springs in series
1
1
1


kb k d kt
kb 
kd 
From Ch 5
k d kt
k d  kt
km
Ad E
AE
kt  t
ld
lt
Ad At E
kb 
Ad lt  AChapter
Dr. A. Aziz Bazoune
t l d 8: Screws, Fasteners and the Design of Nonpermanent Joints
For short bolts kb= kt
CH-8
LEC 35 Slide 31
ME 307
Machine
Design I
Joints:
Fastener
Stiffness
To find different
parameters use
table 8-7
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 32
Joints: Member Stiffness
ME 307
Machine
Design I
Members act as springs under compression
Frustum
m
Compression stress distribution from experimental data
Pdx
d 
Equivalent spring constant km
EA
1
1
1
1
1
Integrating from 0 to l
 
  ....
P
(2t tan   D  d )( D  d )
k m k1 k 2 k3
ki
 
ln
Ed tan  (2t tan   D  d )( D  d )
P
Ed tan 
k  
 ln (2t tan   D  d )( D  d )
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints CH-8 LEC 35 Slide 33
(2t tan   D  d )( D  d )
ME 307
Machine
Design I
Joints: Member Stiffness
For Members made of Aluminum, hardened steel and cast iron 25 <a< 33°
For a= 30°
0.5774 Ed
k 
(1.155 t  D  d )( D  d )
ln
(1.155 t  D  d )( D  d )
(8  20 )
If Members have same E with symmetrical frusta (l= 2t) they act as 2
identical springs km = k/2
For a= 30° and D = dw = 1.5d
km 
0.5774 Ed
0.5774 l  0.5d
2 ln 5
0.5774 l  2.5d
Dr. A. Aziz Bazoune
(
)
(8  22 )
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 34
ME 307
Machine
Design I
Joints: Member Stiffness
Other equations
From Finite element analysis results, A and B from
km
table 8.8
 A exp( Bd / l )
for standard washer
Ed
Faces and members
Of same material
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
(8  23)
CH-8
LEC 35 Slide 35
ME 307
Machine
Design I
Bolt Strength
Bolt strength is specified by minimum proof strength Sp or minimum proof
load, Fp and minimum tensile strength, Sut
The SAE specifications are given in
Table 8-9 bolt grades are numbered
according to minimum tensile
strength
The ASTM Specs for steel bolts
proof strength
(structural) are in Table 8-10.
Metric Specs are in table 8-11.
If Sp not available use Sp =0.85 Sy
Fp = At Sp
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 36
Tension Joints
ME 307
Machine
Design I
Static Analysis
a) External Load
External Load P is shared by bolt and
members
Equilibrium P  P  P
b
m
1
Compatibility    b   m
(2)
P
P
Pk
m
b
b m



&
P

m
Relation P-
k m kb
kb
3
Pb 
kb
P  CP
kb  k m
Pm 
km
P  (1  C ) P (4)
kb  k m
C
kb
kb  k m
C is the stiffness constant of the joint,
For typical values of C see table 8-12
Most of external Load P is taken by members
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 37
Tension Joints
ME 307
Machine
Design I
Static Analysis
b) Resultant Bolt & member loads: Fb& Fm
Fb  Pb  Fi  CP  Fi
Fm<0
Fm  Pm  Fi  (1  C ) P  Fi
Fi is preload; high preload is desirable in tension connections
c) Torque required to give preload Fi F = 0.75 F For re-use
Fd
T  i m
2
 l  πfdm sec 

 πdm  fl sec 
i
 Fi f c d c
 
2

Fi = 0.90 Fp For permanent joint
 t an   f sec 

 l  f t an sec 
d c  ( d  d w ) / 2  ( d  1.5d ) / 2  1.25d
t an  
[
l
πdm
T 
p
Fi d m
2
]
 Fi f c d c
 
2

K is torque coefficient
T 
K values are given in
table 8-15 (Average
d m  t an  f sec  

  0.625 f
K A. Aziz Bazoune
T  KFi d
Dr.
Chapter 8: Screws,
Joints= CH-8
Value
0.2) LEC 35 Slide 38
2d  l  f t an
 sec   Fastenerscand the Design of Nonpermanent
d m  t an  f sec 

2d  l  f t an  sec 

  0.625 f c Fi d

Tension Joints
ME 307
Machine
Design I
Static Analysis
d) Joint Safety
Failure of Joint occurs when:
1) Bolt yields
b 
Fb CP Fi


At
At At
Failure starts  b  S P
C (nP) Fi

 SP
At
At
or 2) Joint separates
Let P0 be external load causing separation Fm=0
Fm  (1  C ) P  Fi
Fm  0
n
Fi
P1  C 
0 = (1-C) P0-Fi
S P At  Fi
CP
Fi
S P At  Fi
For n bolts n 
For
N Bazoune
Bolts nChapter
 8: Screws, Fasteners and the Design of NonpermanentPJoints
Dr. A. Aziz
/ N 1
 CLEC 35 Slide 39
CH-8
CP / N
Load factor n 
Tension Joints
ME 307
Machine
Design I
Static Analysis
e) Gasketed Joints
If a full gasket is present in joint
The gasket pressure p is:
p
Fm
Ag / N
With load factor n
Fm  (1  C ) nP  Fi
p  [ Fi  nP(1  C )]
N
Ag
To maintain uniformity of pressure adjacent bolts
should not be placed more than 6 nominal
= maintain
(1-C) P0-F
i
diameters apart on bolt circle.0To
wrench
clearance bolts should be placed at least 3 d apart
3
Dr. A. Aziz Bazoune
Db
Nd
6
D is the diameter of the bolt circle
CH-8
Chapter 8: Screws,
b Fasteners and the Design of Nonpermanent Joints
LEC 35 Slide 40
Tension Joints
ME 307
Machine
Design I
Fatigue Analysis
Fa
Fmax=
Fb
Fa
Fi
Pb
Fm
In general, bolted joints are subject
to 0-Pmax,e.g pressure vessels,
flanges, pipes, …
Fb  Fi
Fba 
2
Fb  Fi
Fbm 
2
Dr. A. Aziz Bazoune
 ba 
Fb  Fi (CP  Fi )  Fi CP


2 At
2 At
2 At
 bm 
Fb  Fi (CP  Fi )  Fi CP Fi



  ba   i
2 At
2 At
2 At At
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 41
Tension Joints
ME 307
Machine
Design I
Fatigue Analysis
High Preload is especially important in fatigue. i is a
constant the load line at Fi/At has a unit slope, r=1.0
To find sa use Goodman
for conservative
Sa Sm
 1 assessment of n:
Se Su
Sa 
S e ( Sut   i )
Sut  S e
or Gerber Eq. 8.42
or ASME-elliptic. Eq. 8.43
Safety factor (using Goodman)
Dr. A. Aziz Bazoune
nf 
Sa
i

2S e ( Sut At  Fi )
F
with  i  i
CP( Sut  Se )
At
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
Check for yielding also using proof strength: Eq. 8.48
LEC 35 Slide 42
Tension Joints
ME 307
Machine
Design I
Fatigue Analysis
In case of cut threads use the method described in chapter 7
with Kf values of table 8-16.
The fully corrected endurance limit for rolled threads is
given in table 8-17
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 43
ME 307
Machine
Design I
Fig.
8.19
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 44
Fig.
8.21
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 45
ME 307
Machine
Design I
Failure Modes of Riveted Fasteners
Under Shear
Failure modes due to shear loading of riveted fasteners. (a) Bending of
member; (b) shear of rivet; (c) tensile failure of member; (e) bearing of rivet
on member or bearing of member on rivet.
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 46
ME 307
Machine
Design I
Shear Joints
Centroid of pins, rivets or bolts
x' 
y' 
A1 x1  A2 x2  A3 x3  A4 x4  A5 x5 Ai xi

A1  A2  A3  A4  A5
Ai
A1 y1  A2 y 2  A3 y3  A4 y 4  A5 y5
Ai yi

A1  A2  A3  A4  A5
Ai
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 47
ME 307
Machine
Design I
Shear Joints
Shear of pins, rivets bolts due to eccentric loading
V & M statically indeterminate problem
4 steps (assuming same diameter bolts, load
shared equally)
1)
Direct load F’
Primary Shear
V
F'
N
2)
Centroid
Dr. A. Aziz Bazoune
Ai xi
x' 
Ai
Ai yi
y' 
Ai Fasteners and the Design of Nonpermanent Joints
Chapter 8: Screws,
CH-8
LEC 35 Slide 48
ME 307
Machine
Design I
Shear Joints
Shear of pins, rivets bolts due to eccentric loading
3)
Moment load M
Secondary Load
M1  FA"rA  FB "rB  FC "rC  FD "rD
The force taken by each bolt is proportional to its radial
distance from the centroid
FC "
FA "
FB "
FD "



rA
rB
rC
rD
Fn " 
4)
rA
2
M 1rn
2
2
 rB  rC  ...
Add Vectorially the direct
and moment loads If bolts are not same size only bolts with max. R should
Dr. A. Aziz Bazoune
be considered
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 49
ME 307
Machine
Design I
See Examples 8.6 and 8.7
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 50
Fig.
8.26
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 51
ME 307
Machine
Design I
Problem 8-50
½ in-13 UNC SAE 5
Dr. A. Aziz Bazoune
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 52
ME 307
Machine
Design I
1)
Problem 8-50
Direct load F’
F'
2)
3)
300
V

 150 lb
2
N
4)Add Vectorially the direct and moment loads
FA  1650  150  1500 lb
FB  1650  150  1800 lb
Centroid is at O
Secondary Load
FA "  FB " 
M 1rA

2
2
rA  rB
4950 x1.5
 1650 lb
1.5 2  1.5 2
Dr. A. Aziz Bazoune
Is As=Ad? LT=2d+1/4=1.25”
Hnut+2p+3/8=7/16+2(1/13)+3/8=0.97”<1.25”
So As=At = 0.1419in2 and tB=12.7 Kpsi;
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints CH-8 LEC 35 Slide 53
n=4.2
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Problem 8-50
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 54
ME 307
Machine
Design I
Dr. A. Aziz Bazoune
Problem 8-50
Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints
CH-8
LEC 35 Slide 55