Tl CO NA
`
`SNAP-FITS FOR
`ASSEMBLY AND DISASSEMBLY
`
`Presented by:
`
`Tim Spahr
`Technical Service Engineer
`
`November 1991
`(Revised 1100)
`
`~ Celanese
`l 1C1,..41C111.,...M
`
`AMT Exhibit 2002
`CORPAK v. AMT IPR2017-01990
`Page 1 of 18
`
`

`

`GENERAL DISCUSSION
`
`An economical and quick method of joining plastic parts is by a snap-fit joint A snap-fit joint can be designed
`so it is easily separated or so that it is inseparable, without breaking one of its components. The strength of the
`snap-fit joint depends on the material used, its geometry and the forces acting on the joint
`
`Most all snap-fit joint designs share the common design features of a protruding ledge and a snap foot
`Whether the snap joint is a cantilever or a cylindrical fit, they both function similarly.
`
`When snap-fit joints are being designed, it is important to know the mechanical stresses to be applied to the
`snap beams after assembly, the required mechanical stresses or strains on the snap beams during assembly,
`the number of times the snap joint will be engaged and disengaged, and the mechanical limits of the material(s)
`to be used in the design.
`
`Reasons to Use Snap-Fits:
`Reduces assembly costs.
`Are typically designed for ease of assembly and are often easily automated.
`Replaces screws, nuts, and washers.
`Are molded as an int.egral component of the plastic part.
`No welding or adhesives are required.
`They can be engaged and disengaged.
`
`Things To Be Aware of When Using Snao-Fits:
`Some designs require higher tooling cost
`They are susceptible to breakage due to mishandling and abuse prior to assembly.
`Snap-fits that are assembled Wlder stress will creep.
`It is difficult to design snap-fits with hermetic seals. If the beam and/or ledge relaxes, it could decrease
`the effectiveness of the seal.
`
`TYPES OF SNAP-FIT JOINTS
`
`There are a wide range of snap-fit joint designs. In their basic form, the most often used are the cantilever
`beam (snap leg), Figure 1, and the cylindrical snap-fit joint, Figure 2. For this reason, these two designs and
`designs derived from these basics are covered in this text
`
`SL/ (1
`
`I r
`L
`
`Figure 1
`Cantilever Snap
`
`~
`
`..
`
`-
`
`.
`
`-
`
`~
`
`Figure2
`Cvlindrial Snap
`
`l
`
`AMT Exhibit 2002
`CORPAK v. AMT IPR2017-01990
`Page 2 of 18
`
`

`

`Cantilever Snap Beams
`
`Using the standard beam equations, we can calculate the
`stress and strain during assembly of the snap beam. If
`we stay below the elastic limit of the material, we know
`the flexing beam will return to its original position.
`However, for such designs, there is usually not enough
`holding power with the low forces or small deflections
`involved.
`
`1 b r
`E3
`
`Figure3
`
`Therefore, much higher deformations are generally used.
`With most plastic materials, the bending stress calculated
`by using simple linear bending methods (Equations 1-3)
`can far exceed the recognized yield strength of the
`material. This is particularly true when large deflections
`are used and when the assembly occurs rapidly.
`
`Therefore, it often appears as if the beam momentarily
`passes through the maximwn deflection or strain, greatly
`exceeding its yield strength while showing no ill effects
`from the event.
`
`What actually happens is described later.
`
`For the present, simply note that the snap beams are
`usually designed to a stain rather than a stress.
`
`FL
`G=--
`Z
`
`FL 3
`Y=-
`3EI
`
`a=Es
`
`Equation 1
`
`Equation2
`
`Equation3
`
`, Section Modulus
`
`Where:
`a= Maxbn.um stress on beam
`F =Force on the beam
`L =Length of the beam
`I
`Z = -
`c
`d
`c = 2 = Half the beam height
`bd3
`I= -
`12
`h = Beam height
`b = Beam width
`E = Maximum strain on beam
`Y = Beam deflection
`
`, Moment of inertia
`
`The strain should not exceed the allowable dynamic
`strain for the particular material being used. By
`combining Equations 1-3, the design equation (Equation
`4) can be produced. Note that the strain is written in
`terms of the height, length, and deflection of the beam.
`
`3YH
`2 L 2
`
`Equation4
`
`AMT Exhibit 2002
`CORPAK v. AMT IPR2017-01990
`Page 3 of 18
`
`

`

`Strain Guidelines
`
`Fi re4
`
`0.02!!1
`
`JG to J Rat.io
`
`Fi re5
`
`Generally speaking, an unfilled material can withstand a strain level of around 6% and a filled material of
`around 1.5%. As a reference, a 6% strain level could be a beam with a thickness that is equal to 20% of its
`length (a 5: 1 I.lbo) and a deflection that is also equal to 200/o of its length (see Figure 5). A 1.5% strain level
`could be a beam with a thickness that is equal to 100/o of its length (10: 1 I.lbo) and a deflection that is equal to
`100/o ofits length (see Figure 4). lfusing a beam that is tapered so the thickness at the base of the snap foot is
`500/o that of the base of the beam, the length of the beam will approximately 78% (0.7819346 calculated) the
`length of the 6% and 1.5% beams with uniform thickness.
`
`A more accurate guideline for the allowable dynamic strain curve of the material may be obtained from the
`material's stress strain curve. The allowable dynamic strain, for most thennoplastics materials with a definite
`yield point, may be as high as 70% of the yield point strain (see Figure 7). For other materials, that break at
`low elongations without yielding, a strain limit as high as 500/o of the strain at break may be used (see Figure
`6). If the snap joint is required to be engaged and disengaged more than once, the beam should be designed
`to 60% of the above recorrnnended strain levels. However, the best source for allowable dynamic strain is the
`material supplier.
`
`Jfaterlal ll'J tnt Jlettni te rierd PeJ:nt
`
`Kate.rial Ii.th Dcfinif.~ Yi1d41oi.a&
`
`f T"rt'"f'1~~~rrrl"Tl~~~Trr~
`------------~
`----~--~-------a
`- - ~ - 3
`i-----~7,;?"'~~i
`. ---- --1--~ ---
`• .a----
`~ ~t--- 11,--1g-1.•---
`tt -t--
`I -:A---~
`x=- -1;--11-~---1
`!1: -
`-.-~~~rr~~j~Trr~
`
`l!i'traln (11)
`
`AMT Exhibit 2002
`CORPAK v. AMT IPR2017-01990
`Page 4 of 18
`
`

`

`Modulus of Elasticity vs. Secant Modulus
`
`Tensile Stress vs. Strain
`
`Fi re8
`
`Before going finther, we need to examine the actual s~ and forces developed in a snap finger. Figure 8
`shows a stress strain curve for a brittle thermoplastic material. The straight line portion of the curve is the
`region where stress is proportional to strain. Line A is drawn tangent to this region
`
`The slope of Line A is generally reported as the modulus of elasticity (Young's modulus or initial modulus) of
`the material. Many plastics do not possess this straight-line region. For these mat.erials, Line A is constructed
`tangent at the origin to obtain the modulus of elasticity. Ifwe designed a snap beam at 1.5% strain for this
`material in Figure 8 using Equations 1-4 and a modulus of elasticity of 1.6 x 106 psi (given by the material) as
`determined from Line A, the resulting stress would be 24,000 psi, Point A on Line A However, from the
`stress strain curve it can be seen that the true stress at 1.5% strain is about 18,000 psi, Point B on the curve.
`In addition, the deflection force predicted by Equations 1-4 will be high by the same proportions.
`
`Now, to make our math easier, we need some method to force Equations 1-4 to give us the proper stress and
`force results. If we construct a secant line from the origin to Point B, Line B, the slope of Line B is the
`material modulus just as the slope of Line A is the modulus of elasticity. The slope of Line B is the secant
`modulus for the material at Point Band is approximately 18,000 psi divided by 1.5% strain or 1.2 x 106 psi.
`Obviously, the secant modulus can be calculated for any point on the stress-strain curve. Plots of the secant
`modulus vs. strain (or stress) can then be produced, if desired. Obviously, at the lower strains, the Scant
`Modulus should approach the modulus of elasticity of the material.
`
`4
`
`AMT Exhibit 2002
`CORPAK v. AMT IPR2017-01990
`Page 5 of 18
`
`

`

`Radii and Stress
`
`Jn designing the soap beam, it is very llnportant to
`avoid any sharp comers or sttuctural discontinuities,
`as s1ress will concentrate in such areas. To avoid
`such problems, inside comers should be designed
`with a minimum radius of 0.020 in. and where
`necessary to mainmin a unifonn wall, a radius equal
`to the inside radius plus the wall thic~ should be
`placed on the outside of the comer. As indicated in
`Figure 9, an inside radius of 500/o of the wall
`thickness is considered a good design standard.
`
`Therefore, it is recommended that when posSlble a
`:fillet radius equal to Yi the beam thickness be added
`to the base of a cantilever beam.
`
`3.[J
`
`2.5
`
`Tapered Beams
`(same length/stiffer snap)
`
`An improved method of designing cantilever beam
`snap-fits is to use a tapered beam. The beam is
`tapered from the root to the base of the snap foot
`Stresses in a straight beam concentrate at its base, as
`shown in Figure 10. Where as s1resses in the tapered
`beam are dis1ributed more unifonnly 1hrougb its leng1h,
`therefore reducing sttess, as shown in Figure 11.
`
`The taper effectively decreases the beam's strain and
`allows for a deflection great.er than 1hat of a s1nUgbt
`beam with the same base thickness. Another use for a
`tapered snap beam is that when the s1nUgbt beam is
`not stiff enough, the base of the beam can sometimes
`be increased to create a stiffer tapered beam.
`
`Figure9
`
`Figure 10
`
`Figure 11
`
`AMT Exhibit 2002
`CORPAK v. AMT IPR2017-01990
`Page 6 of 18
`
`

`

`Formulas for Cantilever Beams
`
`Straight Beams
`
`Where
`Y = Beam deflection
`h0 = Beam thickness at base
`L = Beam length
`b =Beam width
`
`rb1
`
`I
`i
`b
`
`I
`I
`I
`
`Figure 12
`
`--
`
`.:...
`
`I
`I
`I
`
`_,, ·-
`_.,,,1_,_,--
`
`.
`-1
`I
`I
`I
`I
`
`~
`
`== ""'="
`lb.n ~
`
`(lj_
`I li
`~
`
`Equations 5-8 can be used to calculate the following properties:
`
`Strain level in a straight beam
`
`3Yh 0
`e = ------
`2L2
`
`Deflection of a straight beam
`
`Length of a straight snap beam
`
`Thickness of a straight snap beam
`
`2L2e
`h =-----
`3Y
`
`0
`
`Equations
`
`Equation6
`
`Equation7
`
`Equation8
`
`AMT Exhibit 2002
`CORPAK v. AMT IPR2017-01990
`Page 7 of 18
`
`

`

`Tapered Beams
`
`To calculate the geometry of a tapered beam, the ratio of the thickness of the beam at the snap foot vs. the
`base (hLfho) must be known. On Table 1, find the hLlho (Column 1) value and the corresponding K factor
`(Column 2). K is a geometry factor and is required in all of the formulas related to the tapered beam.
`Example: A hL Ibo of 0.50, equates to a thickness at the snap foot of 500/o he base. The K factor for an hr,
`Ibo of 0.50 is 1.636.
`
`Table (K Values)
`
`hr/ho
`0.33
`0.34
`0.35
`0.36
`0.37
`0.38
`0.39
`0.40
`0.41
`0.42
`0.43
`0.44
`0.45
`0.46
`0.47
`0.48
`0.49
`
`K
`2.137
`2.098
`2.060
`2.024
`1. 989
`1. 956
`1.924
`1.893
`1.863
`1.834
`1.806
`1.780
`1.754
`1. 729
`1. 704
`1. 681
`1. 658
`
`hr/ho
`0.50
`0.51
`0.52
`0.53
`0.54
`0.55
`0.56
`0.57
`0.58
`0.59
`0.60
`0.61
`0.62
`0.63
`0.64
`0.65
`0.66
`
`K
`1.636
`1.614
`1.593
`1. 573
`1.553
`1.534
`1. 515
`1. 497
`1. 479
`1. 462
`1. 445
`1.429
`1. 413
`1.399
`1. 382
`1. 367
`1.352
`
`hr/ho
`0.67
`0.68
`0.69
`0.70
`0. 71
`0. 72
`0.73
`0.74
`0.75
`0.76
`0.77
`0.78
`0.79
`0.80
`0.81
`0.82
`0.83
`
`K
`1.338
`1.324
`1. 310
`1. 297
`1. 284
`1.272
`1.259
`1. 247
`1.235
`1.223
`1. 212
`1. 201
`1.190
`1.179
`1.168
`1.158
`1.148
`
`hr/ho
`0.84
`0.85
`0.86
`0.87
`0.88
`0.89
`0.90
`0.91
`0.92
`0.93
`0.94
`0.95
`0.96
`0.97
`0.98
`0.99
`1. 00
`
`K
`1.138
`1.128
`1.118
`1.109
`1.100
`1. 091
`1. 082
`1. 073
`1. 064
`1. 056
`1. 047
`1. 039
`1.031
`1.023
`1. 015
`1. 008
`1. 000
`
`Equations 9-12 can be used to calculate the following properties:
`
`Strain level in a straight beam
`
`Deflection of a straight beam
`
`Length of a straight snap beam
`
`F = bh: x E 8e
`6
`L
`
`d
`
`Thickness of a straight snap beam
`
`ho=
`
`2L2Ke
`3Y
`
`Equation 9
`
`Equation 10
`
`Equation 11
`
`Equation 12
`
`7J
`
`AMT Exhibit 2002
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`Page 8 of 18
`
`

`

`Snap Beam Assembly
`
`Y = Beam Deflection
`h,, = Beam thickness at its base
`µ = Coefficient of friction
`e = beam fiber strain
`F,, =Assembly force
`
`b = Width of beam
`E, =Secant Modulus
`a = Assembly angles
`Fd = Deflection force
`
`F;.
`
`v ~---.J? .. :..!L.._._I•
`
`f
`
`Fi re 13
`
`To calculate the deflection force of a straight or
`tapered cantilever beam, use Equation 13.
`
`bh~ ESB
`F =--x--
`6
`L
`d
`
`Equation 13
`
`The ~ly angle along with deflection force and coefficient of friction between the mating parts detennines
`the assembly force. The greater the angle and/or coefficient of friction, the higher the assembly force. It may
`not be possible to assemble parts with assembly angles 45° and a high coefficient of friction It is
`recommended that assembly angles between 15° and 300 be used.
`
`To calculate the assembly force of a straight or
`tapered cantilever beam, use Equation 14.
`
`F =F. µ+tana
`d 1 -µtan.a
`"
`
`Equation 14
`
`The retaining force is determined by the angle of the mating surfaces of the snap foot and ledge. To a point,
`the greater the angle, the greater the holding strength of the snap. This is 1rue only up to the shear strength of
`the snap and the effects of bending moments applied to the beam. It is a common belief that a retaining angle
`of 900 will prevent the snap joint beam from failing. However, the forces on the snap foot can create a
`bending moment that is high enough to rotate the snap foot back and disengage the snap joint without a shear
`failure (beam retention is discussed later). For detachable joints, it is recommended that a retaining angle
`between 30° and 45° be used.
`
`To calculate the retaining force, use Equations 13 and 14 and substitute the retaining angle for the assembly
`angle.
`
`When designing snap-fits that require the ability to be engaged and disengaged repeatedly, a safety
`factor is needed to predict the beam peiformance. Therefore, when designing such snap-fits, replace £
`in Equations 5 through 13 with 0.6£.
`
`AMT Exhibit 2002
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`Page 9 of 18
`
`

`

`Designing Cantilever Snae-Fit Joints From Beams of Circular Sections
`
`The following examples can be used in cases where snap-fits are used on a circular part, such as a boss.
`
`Half circle cross section 0 A
`One third circle cross section: Q ~
`One quarter circle cross section: Q ~ e
`
`or
`
`Figure 14
`
`C("
`
`Figure 15
`
`or
`
`Fivure 16
`
`L2
`Y...,. =0.578s-
`r
`
`'
`
`e
`Y.,,.. =0580£ -
`r
`
`Y.,.. =0.555£-
`r
`
`Equation 15
`
`Equation 16
`
`Equation 17
`
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`Page 10 of 18
`
`

`

`Snap-fit Retention
`
`Figure 17
`
`It is often thought that if the retaining angles (the interfacing surface between the snap foot and ledge) are 900
`:from their base and parallel, the snap foot or ledge must shear to fail. However, the forces that are applied to
`the snap foot after it is engaged will create a bending moment (Figure 17) on the beam. In some cases, this
`bending moment will roll the foot back off of the ledge.
`
`To reduce the risk of this occuning, the following are two recommendati.om.
`
`The Snap-Fit Loop
`
`r
`
`Figure 18
`
`The snap loop in Figure 18 is one alternative. Since the force is directly in-line with the snap loop, the failme
`modes, after assembly, are shearing of the support ledge or a tensile failure of the snap loop. The roll off
`failure, as with the standard cantilever beam, is eliminated.
`
`The length, deflection, thickness, and beam taper, (if required) are all calculated in the same manner as the
`cantilever beam snap-fit. To reduce stress concentrations, the inside comers should have fillet radii. To
`accommodate the radii on the snap loop, the ledge may have mating radii or clearance for the radii.
`
`l Q
`
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`
`

`

`Snap Through Support
`
`Square Snap Foot
`End View
`
`Conformed Snap Foot
`End View
`
`Figure 19
`
`,..
`
`The snap through support is a beam fixed on both ends (Figure 19) creating a hole for a snap foot to pass
`through. The inside comers of this hole should have full radii in its comers to reduce stress, just as with a
`cantilever beam. When using a snap through support, the cantilever beam does not have to deflect; the
`support does all the bending. In areas that are too short for the design of a cantilever beam, this approach is
`ideal. The deflection of the support beam is dependent on the width of the cantilever beam. The closer the
`snap foot and snap through hole are in width, the less deflection is allowable. That is, if the cantilever beam
`and the support beam are equal in width, the support beam has very little deflection before its strain limit is
`reached.
`
`Alternately, the snap foot of the cantilever beam can be designed to conform to the flexure of the support
`beam; i. e., use a radius profile instead of a square one (Figure 19). If a profiled snap foot is used, the
`supported beam will deflect the same as if the forces are a point cont.act in the center.
`
`The only caution in using this method of beam retention is that it usually has a weld line in the support beam.
`By increasing the wall section in the support beam, the strength of the weld line can be improved, but the
`allowable deflection would decrease. After assembly, the possiblefailure modes for a fixed support beam
`system are the weld line, shear at one end, or a failure of the cantilever beam due to tensile stress which
`disengages the snap-fit.
`
`11
`
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`
`

`

`Calculating the deflection of the support beam:
`
`To calculate the deflection of the fixed support beam with a square snap foot, use Equation 18.
`
`Where:
`b = Cantilver beam width
`L =Length of .fu:ed beam
`Tb.,,,,. = Thicknes of .fu:ed beam
`E = Strain in Fixed Beam
`Ymax =Maximum deflection of .fixed beam
`
`Equation 18
`
`---------------------------------------------------,
`b ---c:...i
`
`I
`
`__ , __ _L ____ _
`f-==r~
`
`L--------------------------------------------------
`
`Figure20
`
`To calculate the deflection of the fixed support beam with a snap foot that conforms to the deflection of the
`support beam, use Equation 19.
`
`Equation 19
`
`1
`L2
`=---X----·X8
`12
`TBeam
`
`Y1
`
`11111
`
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`
`

`

`CYLINDRICAL SNAP JOINTS
`
`t JI
`ln ........ J_LD-~
`
`Figure 21
`
`A cylindrical snap joint consists of a cylindrical part with an external lip (snap foot) which engages a cylindrical
`part with a corresponding internal lip (ledge) as shown in Figure 21. Generally, the shaft is considered rigid
`and the hub elastic. Variable Yin cylindrical snap joints is the total allowable diametral interference. Strain
`applies to cylindrical snap joints in the same manner as for cantilever beams.
`
`Dynamic Strain in a Cylindrical Snap Design
`
`Y
`6 = -------------= ------
`
`Dhub - D shaft
`
`Dshaft
`
`Dshaft
`
`= Inside diameter of the hub
`= Total diametral interference
`= Material strain
`= Outside diameter of the shaft
`
`E
`
`The difference in the largest diameter of the shaft and the smallest diameter of the hub is the deflection Y.
`
`Unlike the cantilever beam, the assembly force of the cylindrical snap fit can be only roughly estimated. This is
`because the length A of the hub, Figure 24, deformed during assembly, is difficult to predict Length A depends
`on both the wall thickness of the hub and the depth of the undercut (YzY). As it is difficult to predict a reference,
`a dimension of twice the width b of the hub should be used
`
`13
`
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`
`

`

`Cylindrical Snap-Fit Assembly
`
`T b •
`-~a
`-
`
`-
`
`Fi ure 22
`
`To roughly estimate the assembly forces for a cylindrical
`snap, we must first calculate the geometry factor K, using
`Equation 30. It will be assumed that the shaft is rigid and
`that the total part interference is accommodated by the hub.
`Equation 30 shows the geometry factor K as a :function of
`the diameter ratio of Doh IDshaft
`
`2
`~ +1
`
`D ( J
`K= (~:)-!+I
`
`Dshaft
`
`Equation 30
`
`The joint pressure (p) can be calculated using Equation 31.
`
`p = _o_.5Y_ x E x ..!._
`s K
`Dh.,b
`
`Equation31
`
`Then using factor p, the assembly and pull-out forces can
`be calculated, using Equation 32.
`
`F= pxrc xD
`
`0
`
`,, x 2j µ+ tana )
`\J-µ xtana
`
`Equation 32
`
`Where
`= Inside diameter of the hub
`Dhub
`= Total diametral interference
`Y
`= Material strain
`e
`Dshaft = Outside diameter of theshaft
`0
`= Coefficient of friction
`!!P
`= Assembly an d retaining angle
`
`14
`
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`
`

`

`APPENDIX
`
`Example of a Cantilever Snap-Fit Calculation
`
`Straight Beam
`
`The material specified has an allowable fiber strain of 1.5% (0.015), and at a 1.5% strain the secant modulus
`is 1,200,000 psi. The length of the beam has to be 0.50 inches, the width of the beam is 0.200 inches, the
`required deflection is 0.030 inches, and the assembly angles are both 30. (Figme 23)
`
`Fi re23
`
`Using Equation 8, the beam's thickness can be calculated:
`
`ho
`
`2L2e
`= --
`3Y
`
`.085;;;;;;
`
`2( 0.502 )(0.015)
`3(0.030)
`
`The thickness of the straight beam
`is 0.085 inches.
`
`Using Equation 13, the deflection force of the beam can be calculated:
`
`F = bh: x Ese
`6
`L
`d
`
`8.7 = {020)(0.0852
`) • {1200000X,015)
`0.50
`6
`
`The calculated force deflection for
`the straight cantilever snap beam is
`8.7 lbs.
`
`Using Equation 14, the assembly forces can be calculated:
`
`F =F. µ+tana
`"
`"1
`-µtana
`
`7376 = 8.9 x
`
`(.17 + tan30)
`1-(0.17 • tan30)
`
`The calculated assembly force for
`the straight cantilever beam is 7.4
`lbs.
`
`15
`
`AMT Exhibit 2002
`CORPAK v. AMT IPR2017-01990
`Page 16 of 18
`
`

`

`Tapered Beam
`
`To increase the strength of this beam and maintain the same deflection and strain level, the beam can be
`tapered (Figure 24).
`
`0.030
`
`Fi re24
`
`There are now two unknowns:
`
`1.
`2.
`
`the thickness at the base
`the thickness at the snap foot
`
`These two unknowns are related by the K factors in Table 1. A decision on the base thickness and thickness
`ratio QiL/h0 ) is needed A good starting point is with a base twice as thick as the straight beam with a
`thickness ratio of 50%. For an hLlho ratio of0.5, the K factor is 1.636.
`
`Using Equation 12, the base thickness can be calculated:
`
`2L2 Ke
`h =
`3Y
`
`0
`
`222 = (0.2)(136') x. (1200000)(0.015)
`6
`0.5
`
`The calculated deflection force for
`the tapered beam is 22.20 lbs.
`
`Using Equation 13, the deflection force of the beam can be calculated:
`
`F = bh! x Ese
`6
`L
`
`d
`
`18.4 = 22.2 x
`
`0.17 +tan JO
`1-(0.17)(tan30)
`
`The calculated deflection force for
`the tapered beam is 22.20 lbs.
`
`Using Equation 14, the assembly forces can be calculated:
`
`F =F. µ+tana
`ti l
`-µtana
`
`a
`
`18.4 = 22.2 x
`
`0.17 + tan30
`1-(0.17)(tan30)
`
`The calculated assembly force for the
`tapered beam is 18.4 lbs.
`
`The strength of the beam's deflection may be increased by decreasing the h L /h0 value or decreased by
`increasing the hL lh0 value. An hL lh0 equal to 1 would be the same as calculating the straight cantilever
`beam.
`
`16
`
`AMT Exhibit 2002
`CORPAK v. AMT IPR2017-01990
`Page 17 of 18
`
`

`

`N.QTICE TQ USERS~ TQ t_h~ ~1 Qf Wt' knQ.W~ tJic inf~iQn QQDl~M int~ p\lbli~iQn ii!;~~. hqwc:vc;nv~ dQ not IJUUIIJI;,
`eny liabili~~Y whats.~~ forth.e ~end C<t>mpl~,en~a Qfaudl. in(Qr.ma.tion.. Furth~, th~ ~s.i& t~1l~ in~~ iil. t_b.is,
`pubbion ace Qften simpJi&ationa and,, therefore,, approximate in nature, M ~ ~tous, analysis; toebniques, andtbr p:rotQiypc testing
`~ s.t'tQllgJy ~dc;d tQ vi:;rify s.US.f-.t~ pl!« perfQtmllll~ AnyQn~ intend.ihgtQ :t:1;W on sucoh ~W!.tiQn <1f tQ us~ any
`~~l.J'ltooissingt~~ Qt mattldaJ.. QD.t.iQn~ in. thil!; publieati.Qn sJi~uld s.Qtisfy t~~~ tllilt th~ ~.m.e« allapp,lbh~
`safc;ty and hc;alth stan4ants,
`
`11 il!; th~ sol~~Qnslbility oJ[th.e; qs,cn tQ, in:i.;estlftltc wh~.h«: any eids.tingpa1en1s l!le; in.ftioP by the llS.c Qfth.e mat~ qntiPn.~ in.
`thil!; pttblication.
`
`Any d.d.ermi.Da.t.iO.n. Qfthc suit.ability Qf a. p,art~·matetial! for any ~c QQD1empla1e:d by th.e; llS.et iS t.h.c S-Qlc l!es.po.ns.ibility ofth.c llS.er.
`Thcv,ser·lillJllt verify that thcm.itt~. • !ll.l.bs~uent,W p~s~ ~s thcn;q~t11; Qfthcpart:iCIJlarp~ WlJ&e• Thc lJ&Q':i:&:
`~ tQ te&t ptQtQtypes Ql': S-l!Dlplc;& of the produd: under· the barshes.t QQDd,itiQDS liki:W tQ, he en«OUDtered tQ, ddennine the
`s.uitahility Qf'th.e materials ..
`
`Material data and values in«luded in. thil!; pub~.ion are· either· bJtSod on test.ing oflaOOtatmy test sp~ and tepresent data that fall
`within. th~nQtmal.t~ Qf11rqpi;rt~ f~IDltwalIJUlt.Q. Qt w~ ~.ed. ftQID. vori'Qus. pub~he:d S.QlltG.Q, AU Hl';; Qi:W;v~ tQ he
`rqn:~entative. Co.JQJ1nt11; Qf Qther 1¥Wi~iv~ 1113}' e1tJS.c sJg!ille<ant variat.Kms. ID. <hlta. v~, Thc;S;c valQ.es. ~ nQt intend.c:d f<lf q£CJ in
`establiShing~minitnnm, or·~ ofvalue.s forsp~wnpurp,oses ..
`
`We s.tlongiy ttJWlllliletld. that, uain s.edt 11nd. ed.h«c to th~ manJU:~er··s <:tl': sup,pli.!ll:'ft ~t iils.ttlldims for handlin.g•. ma.terial
`thqf use~ Please <lall 1-800-833-48821 for additional tedmi<lal. information. Call Customer SeNkcs at: the number listed below for-the
`QPp~ril!t~ M.at:~Saf~ ll!J.t~ Sh=.11 (M.SDS) Qe.fQteaU1<1Dpt:i'ngto p~..£ thoi~ prvdud.&. M<ro<QV~, thet:Q ~I!~ tQ ~
`hlUllall ~Ql~ tQ man~ mat.~ tot.he; }Qw~t p.l'.aid.bl llini.t.s. in.~ Qf P.,Qlsib.k l!d.\linct eff~.s. .. T Qt~ ~ent that, .i~ haz.l!l!dft may
`ha:i.;c;; been mcnti'Qne<l in thil!; p,ttblicat.ion. we· nc;ithc;r- su~t IlQI"' &Jl1111Dt~ that suc;h hazards arc;; tbQ Qilly Qru!li, that: ~t ..
`
`11'
`
`AMT Exhibit 2002
`CORPAK v. AMT IPR2017-01990
`Page 18 of 18
`
`