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8/3/2019 Fasteners Lecture 11-20-03
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ME 351
Power Screws, Fasteners,
and Connections
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Threads and Connections
Today we will start off discussing the
mechanics of screw threads. Next, powerscrews & threaded fasteners will be
examined. Since threaded fasteners are
often used to make connections, we will end
with that topic.
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The Inclined Plane
Truly one of the worlds great inventions!
By inspection, a steeper angle gains youelevation more quickly, but the appliedforce must increase.
QfN
N
W
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Helically-Inclined Planes
Differential element of one thread transferring force tothe mating thread. The helix or lead angle = the slope
of the ramp, and is a critical design parameter. is the
thread angle, and is another important parameter.
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, , andf On a screw thread, the helix angle
controls the distance traveled per revolutionand the force exerted.
, the thread angle, effects the friction forceresisting motion. Sometimes friction isdesirable (e.g., so that threads wontloosen), and sometimes it is not.
fis the coefficient of friction, and plays animportant role in all threads.
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and
, the helix angle, is given by
tan = L/(dm) (eq. 15.2)where,
L = the lead or pitch (threads per unit length)
dm = the mean dia. of the thread contactsurface.
, the thread angle, is determined by thedesign of the threads.
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Thread Friction Examples
Acme Threads
Bolt Threads
Pipe Threads
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Power
ScrewsForce F acts on
moment arm a to
produce a torque T.
Table 15.3 in the text
shows standard sizes
of power screw
threads.
In this drawing, only
the nut rotates.
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Power Screw Thread Types
Acme: in
wide use, but
less efficient.
Square: most
efficient, but
hard to make.
Modified Square:
compromise.
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Power Screws
Equations 15.6 through 15.13 in the text are
the governing equations for torque and
efficiency, given the geometry of thethreads.
However, as in the case of many previous
problems, often you are presented with thatinformation and must solve for other
variables.
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Power Screw Efficiency
Plot of equation
15.13; note thewide range as a
function of both
fand .
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Problem
15.62 square thread
power screw lifts W of
50 kips at 2 fpm.
Find rpm n, and the
HP required if the
efficiency is 85%, and
f= 0.15.
Neglect the collar
friction.
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Threaded Fasteners
Nomenclature
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Threaded Fasteners
Thread Forms
Note that the crests & roots may be either flat or rounded
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Threaded Fasteners: UNS & ISO
UNS = Unified National Standard. Threads arespecified by the bolt or screw diameter (alsocalled the major diameter)in inches, and the
number of threads per inch. ISO = International Standards Organization.
Threads are specified by the major diameter in
mm, and the pitch, or, number of mm perthread.
Generally UNS and ISO threads are NOTinterchangeable. (3mm is close to 1/8.)
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Threaded FastenersUNS
The specification is written in the format
Dia threads/inUNC or UNFclass andinternal or externalRH or LH.
UNC = Unified National Coarse
UNF = Unified National Fine
Class ranges from 1 (cheap & inaccurate) to 3
(expensive & precise). Class 2 is common.
A = external, B = internal
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Threaded FastenersUNS
RH = right hand threads, LH = left hand
Example thus would be:
13 UNC2ARH
Notes:
1. UNF and UNC are redundant information.
2. For diameters less than , a numeric size isspecified instead of the diameter. (00012?)
3. Summarized on page 602 in text, Table 15.1
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Threaded FastenersISO
Metric designations are a little simpler.Preceded by an M, then the diameter in
mm, then the pitch (mm per thread, notthreads per mm). There are also coarse andfine threads in the ISO system.
Examples: M10 x 1.5
M10 x 1.25
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Coarse Versus Fine Threads
Coarse threads are fine for normalapplications. They are easier to assemble, a
little more forgiving of dings, possibly cheaperto make, and for a given size of bolt, they exertless force than do fine threads; good for softermaterials bolted together.
Fine threads develop greater force per appliedtorque, and are more effective at resistingvibration-induced loosening.
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Bolts, Screws, and Studs
The same fastener could be a bolt or a screw, depending on
if a nut is used. Studs are threaded at both ends.
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Bolt Grades
Bolts (and nuts) are made from a variety of
materials. The SAE Grade is an indication
of the strength of the material, based on theproof stress, Sp (slightly less than the yield
stress). Sp ranges from 33 ksi for a grade1
bolt, up to 120 ksi for a grade 8 bolt. Theproof loadof a bolt is the load at which
permanent deformation commences.
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SAE Bolt Head Markings
SAE 2 SAE 5 SAE 7 SAE 8
http://raskcycle.com/techtip/webdoc14.html
Hexagonal bolt heads are stamped with radial lines to
indicate the grade. The grade = the number of lines + 2.
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Thread Manufacture
Threads are generally produced by either
rolling (forming with a specialized die) or
by cutting, as on a lathe. Rolled threads arestronger and have better fatigue properties
due to the cold work put into the material.
Power screw threads may be ground toachieve a very smooth surface to reducef.
Threads may also be cast into a part.
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Stresses in Threaded Fasteners
Due to imperfect thread spacing, most of the
load between a bolt and a nut is taken by the
first pair of threads. This is partiallyrelieved by bending and localized yielding,
however most thread failures occur in that
region. The stress concentration rangesfrom 2 to 4.
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Major-Diameter Stresses
Axial stress is given by the familiar
= P/A
For A, use either the root diameter for powerscrews, or tabulated values for fasteners.
Torsional stress is given by the familiar = T/J = 16T/d3
See p. 615 for interpretation of T and d. T isthe applied torque for power screws, or thewrench torque, for fasteners.
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Bearing Stress
Bearing stress, the compressive stressbetween the surfaces of the threads, is givenby b = P/(dmhne) (eq. 15.17)
P = load,
dm = pitch or mean screw thread diameter,
h = depth of thread, and
ne = number of threads in engagement.
b is usually not a limiting design factor.
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Nomenclature for Thread
Stress Analysis
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Direct Shear Stress on Threads
Then we have,
= 3P/(2 dbne), where,
d = root dia. for the screw or major dia. for thenut,
b = the thread thickness at the root, and
ne = the number of threads in engagement.
Note that can be a limiting factor.
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Bolt Tightening & Preload
Bolted joints commonly hold parts together
in opposition to both normal and shear
forces.In certain applications it is desirable to
tighten a bolted joint to a specified preload
Fi, which is some fraction of the bolts proofload, Fp.
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Bolt Tightening & Preload
An engineer would specify a preload in the
case of fatigue applications, in order to
minimize the relative magnitude of thealternating load Pa compared to the average
load Pmean. (Recall definitions from ch. 8.)
Preloading is also important in sealingapplications, as in a gasketed joint. Both
reasons are important for auto cylinder heads.
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Preload Values
The optimum preload is often given by eq. 15.20:
Fi = 0.75 Fp for connections to be reused, or
Fi = 0.90 Fp for permanent connections.
The proof load Fp is found from eq. 15.14 as,
Fp
= SpA
t, where the proof stress S
pis an SAE
specification (see Table 15.4 or 15.5), and tension
area At is found in Table 15.1 or 15.2.
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Tightening Torque
To develop the specified preload, the
tightening torque is given by eq. 15.21:
T = KdFi, whereT = the tightening torque,
d = the nominal bolt diameter (e.g., ),
Fi = the desired preload, andK = a torque coefficient
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Tightening Torque
Equation 15.21 is approximate, and applies
for standard threads.
For dry, unlubricated, or average threads,
K = 0.2. For lubricated threads, K = 0.15.
Rewrite eq. 15.21 as,
Fi = T/(Kd) to see that, for a given torque, Fiincreases with lubricated threads.
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Relaxation and Exactness
Most joints lose on the order of 5% of the
original preload over time, due to relaxation
effects (usually over the course of 100s or1000s of hours).
By now it should be clear that threaded
fasteners are extremely complex. Oftenextensive testing is done for critical
applications.
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Tension Joints
Bolted joints are frequently used to clamp
together parts that themselves carry
additional loads: these additional loadsincrease the bolt tension. The engineer
often must determine acceptable loads for
such joints.We consider both the joints and the parts as
springs, with spring constants kb and kp.
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Tension Joints
After assembly with preload Fi, applied load P
will change the force in the bolt and the parts.
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Tension Joints
P = Fb + Fp, where
Fb = the increased tension in the bolt, and
Fp = the decreased compression force in the
parts. The deformations are given by
b = Fb/kb, and p = Fp/kp
Then compatibility requires that
Fb/kb = Fp/kp
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Joint Constant C
The joint stiffness factor, or joint constant, is
defined in eq. 15.22 as C = kb/(kb + kp).
Then the preceding equations yieldFb = CP and Fp = (1C)P
kb is usually small compared to kp, and so C is
a small fraction.
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Forces in Bolted Joints
When a load P is applied to a bolted joint, the
tensile force Fb in the bolt increases, and the
compressive force Fp in the parts decreases.As long as Fp > 0, the forces are:
Fb = CP + Fi (eq. 15.23)
and,
Fp = (1C)PFi (eq. 15.24)
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Determination of C
Review from Chapter 4 about 100 years ago:
Deflection is given by = PL/AE, and the
spring rate k is given by k = P/.Combining these we obtain
kb = AbEb/L (eq. 15.31),
and,
kp = ApEp/L (eq. 15.32)
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Determination of kb
In determining kb, the
threaded and the unthreaded
parts of the bolt are
considered as separate
springs in series. Equation
15.33 gives:
1/kb = Lt/AtEb + Ls/AbEb
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Determination of kp
kp is more complex:
the stress distribution
in the parts is clearly
non-uniform, anddepends on factors
like washers, etc.
It is approximated bythe double-cone
illustrated.
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Determination of kp
Estimate of kp for standard hex-head bolts and
washers is given by eq. 15.34:
kp = (.5 Epd)/{2 ln [5(.58L+.5d)/(.58L+2.5d)]}
d = bolt diameter and
L = grip (thickness of bolted assembly).
Alternatively, just use kp = 3kb !
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Example: Problem 15.11
Bolt diameter is 15 mm.
Grip length L = 50 mm.
Tightening torque foraverage threads is
T = 72 N-m by eq. 15.21
Find maximum P thatwill not loosen the initial
compression in the part
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Example: Problem 15.16
Bolt is M20 x 2.5 coarse thread.
Sy is 630 MPa.
Ep = Eb
L = 60 mm, and P = 40 kN
Determine:
Total force on bolt if joint is
reusable, and, the tightening torque
if the threads are lubricated.
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Typo!
On page 625, the equation for Pa is
incorrect. It should read,
Pa = (PmaxPmin)
(This is in section 15.12, Tension Joints
Under Dynamic Loading.)
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Some Rules of Thumb for
Threaded Fasteners Threaded depth: for a bolt diameter d, the
length of full thread engagement should be
1.0din steel, 1.5din cast iron, and 2.0dinaluminum.
In gasketed joints, bolts are arrayed in a bolt
circle or other pattern. The bolt-to-boltspacing should not exceed about 6dto
maintain uniform pressure.
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Rivets
Rivets often find application in larger structures such
as bridges and towers. They are also used
extensively in aircraft construction. A rivet starts offas a cylinder with one head (usually rounded). The
protruding cylinder is deformed to create a second
head, which locks the joint in compression.
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Joints Primarily in Shear
Both bolts and rivets are used in
connections that primarily experience shear
loading (separate from the case of axial ornormal loading which we just examined).
Such connections may experience any of
several failure modes, and the engineermust analyze for each mode.
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Shear Joint Failure Modes
Shearing Failure of Fastener: = 4P/d2
d = diameter of fastener (from Table 15.7)
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Shear Joint Failure Modes
Tensile Failure of Plate: t= P/(wd
e)t, where
de = effective hole dia., w = width, and t = thickness
of thinnest plate (from Table 15.7).
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Effective Hole Diameter
In analyzing potential tensile failure of the
plate, the effective hole diameter is used
rather than the diameter of the fastener.de= the fastener diameter + 1/16 for drilled
holes, or,
de= the fastener diameter + 1/8 forpunched holes (this is usually used).
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Shear Joint Failure Modes
Bearing Failure of Plate or Fastener: b = P/dt, where d =diameter of fastener and t = thickness of the thinnest
plate. (from Table 15.7)
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Shear Joint Failure Modes
Shearing Failure of Plate: t = P/2at, where t =
thickness of thinnest plate and a = closest distance
from fastener to edge. (from Table 15.7)
a >= 1.5d
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Joint Efficiency
The efficiency of a joint is defined as:
e = Pall/Pt, (eq. 15.41)
where
Pall is the smallest of the allowable loads in
the preceding failure mode examples, and
Pt is the static tensile strength of the plate with
no holes. e is always less than 100%.
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Shear Example, Problem 15.22
Plate thickness is
3/8. Rivets are
diameter, holes
drilled 2 apart.Sall,tension = 22ksi;
Sall,bearing = 48 ksi, and
all = 15 ksi.
Find the joint
efficiency.
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Welded Joints
Welded joints are produced by localized
melting of the parts to be joined, in the
region of the joint. Often a filler metal (orplastic, in the case of plastics) is added,
creating a chemical bond in the parts that
may be stronger than the base material.There are many, many welding processesan
entire engineering major.
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Strength of Butt Welds
The height h does not include the crowned region;
generally it is just the plate thickness.
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Strength of Fillet Welds
Specified size is based on h, but stress is calculated
with t, the region of minimum cross sectional area.
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Factor of Safety for
Welds in ShearJust as many riveted or bolted joints are in
shear, so too are many welded joints. The
factor of safety for a welded joint is givenby:
n = Sys/ = 0.5Sy/ (eq. 15.44)
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Eccentric Loading of Welded
Joints (P.E. Question!)Determination of the exact stress
distribution is very complicated.
With some simplifying
assumptions, the following
procedure gives reasonably
accurate results.
Direct shear stress is given by
d = P/A, where A = the throat
area of all the welds.
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Eccentric Loading
of Welded Jointsd is taken to be uniformly distributed over thelength of all the welds.
Due to the eccentricity e, a torque T is developedabout the centroid C of the weld group: T = Pe.The torque causes an additional shear stress inthe welds:
t = Tr/J (eq. 15.46)
J = polar moment of inertia of the weld groupabout C, based on the throat area. (Continued:)
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Eccentric Loading
of Welded Jointst = Tr/J (eq. 15.46)
In this equation, r is the distance from C to
the point in the weld of interest. t is notuniform across the weld group, and one
point will experience the greatest stress
resultant: = (t
2 + d2) (eq. 15.47)
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Location of the
Centroid (Review)C is located at coordinates x-bar
and y-bar, where
x-bar = (Aixi)/Ai, and
y-bar = (Aiyi)/Ai, where i
denotes a given weld segment,
and the coordinate origin is
conveniently chosen. A key is
that the weld throat t is assumed
to be very small, sometimes 0.
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Moment of Inertia of a Weld and
the Parallel Axis Theorem.Use the familiar bh3/12,
substituting t and L for b and h as
appropriate. However, assume t3
= 0 to simplify.
Remember the parallel axis
theorem, Ix = Ix + Ay12, to find the
moment of inertia about thecentroid of the weld group. (So
even if Ix = 0, you still have A.)
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Polar Moment of Inertia
The polar moment of inertia is the sum of Ixand Iy for each weld about the centroid ofthe weld group. Knowing J, apply
t = Tr/J (eq. 15.46)to find t at a given point, and then use
= (t2 + d
2) (eq. 15.47)
to find the max , which is used to find therequired weld size.