Page 1 of 13
`
`BOREALIS EXHIBIT 1061
`
`Page 1 of 13
`
`BOREALIS EXHIBIT 1061
`
`

`
`
`
`
`CONVERSIONS BETWEEN U.S. CUSTOMARY UNITS AND SI UNITS
`
`_
`Times conversion factor
`
`Eqilals 31 “mt
`U.S. Customary unit
`
`
`Accurate
`
`
`2
`meter per second squared m/s2
`meterper second squared m/S _l
`2
`square meter
`m 2
`square millimeter
`mm
`kilogram per cubic meter
`kg/m
`newton per cubic meter
`N/m3
`kilonewton per cubic
`meter
`kN/m3
`
`3
`
`'-
`
`0305
`0.0254
`0.0929
`645
`515
`157
`271
`
`1.36
`0.113
`3.6
`1055
`
`Acceleration (linear)
`foot per second squared
`inchper second squared
`Area
`square foot
`square inch
`Density (mass)
`slug per cubic foot
`r Density (weight)
`pound per cubic foot
`pound per cubic inch
`
`“
`4
`
`ft/s2
`in./s2
`rt?
`in.2
`slug/ft?’
`lb/ft3
`lb/in.3
`
`I‘
`
`-1
`
`L.
`
`I
`
`_
`
`0.3048*
`0.0254*
`0.0929o304*
`645.16*
`515.379
`157.087
`271.447
`
`1.35582
`0.112985
`3.6*
`1055.06
`4.44822
`4.44822
`14.5939
`175.127
`14.5939
`175.127
`0.3048*
`25.4*
`1.609344*
`14.5939
`1.35582
`0.112985
`1.35582
`0.112985
`
`I
`
`ft-lb
`in.-lb
`kWh
`Btu
`lb
`k
`lb/ft
`lb/in.
`k/ft
`k/in.
`ft
`in
`mi
`lb-s2/ft
`lb—ft
`lb—in.
`k—ft
`k—in.
`
`L.
`
`Page 2 of 13
`
`
`
`Energy; work
`foot-pound
`inch-pound
`kilowatt-hour
`British thermal unit
`Force
`pound
`kip (1000 pounds)
`Force per unit length
`pound per foot
`pound per inch
`kip per foot
`kip per inch
`Length
`foot
`inch
`mile
`Mass
`slug
`Moment of a force; torque
`pound-foot
`pound-inch
`kip—foot
`kip-inch
`
`joule (N-m)
`joule
`megajoulc
`joule
`lewlnn (kg~m/S2)
`iloncwiun
`_ newton per meter
`1 wlon per meter
`tiloncwmn per meter
`kilonewton per meter
`meter
`millimeter
`kilometer
`kilogram
`newton meter
`newton meter
`kilonewton meter
`kilonewton meter
`
`_
`
`J
`J
`MJ
`1
`N
`kN
`N/m
`N/m
`kN/m
`kN/m
`m
`mm
`km
`kg
`N~m
`N-m
`kN-m
`kN-m
`
`.
`
`_'
`
`‘
`
`-1 ‘"17"
`
`'
`
`__.
`
`_
`__
`.
`175
`13¥'fi_'3,:I7' " *-'1
`175
`0.305
`25.4
`1.61
`14.6
`1.36
`0.113
`1.36
`0.113
`
`
`
`Page 2 of 13
`
`

`
`
`
`Jf
`}‘
`
`1-
`J[
`
`—+-
`
`I
`If
`‘:2
`
`
`
`"
`
`[-
`“
`
`7
`
`
`
`mm4
`
`“I
`
`NITS AND SI UNITS (Continued)
`
`CONVERSIONS BETWEEN U.S. CUSTOMARY U
`Times conversion factor
`Equals SI unit
`~
`'
`U.S. Customary unit
`1
`Accurate
`
`Moment of inertia (area)
`inch to fourth power
`in.4
`416,231
`416,000
`millimeter to fourth
`power
`m4 2|
`inchto fourth power
`in.'4
`0.416231 x 10“5
`0.416x 10‘6 meterto fourthpower
`Moment of inertia (mass)
`kg~m2
`slugfootsquared
`slug-ft2
`1.35582
`1.36
`kilogrammetersquared
`Power
`W
`foot-pound per second
`ft-lb/s
`1.35582
`1.36
`watt (J/s or N~m/s)
`W
`foot—pound per minute
`ft-lb/rnin
`0.0225970
`0.0226
`watt
`W
`horsepower (550 ft-lb/s)
`hp
`745.701
`746
`Lwatt
`r—Pressure‘ stress
`Pa
`pound per square foot
`psf
`47.8803
`47.9
`pascal (N/m2)
`Pa
`pound per square inch
`psi
`6894.76
`6890
`pascal
`kPa
`kip per square foot
`ksf
`47.8803
`47.9
`kilopascal
`MPa
`kip per square inch
`ksi
`6.89476
`6.89
`megapascal
`—|
`-1
`Section modulus
`inch to third power
`in.-3
`16,387.1
`16,400
`millimeter to third power mm3
`inchtothirdpower
`in.3
`16.3871 X 10-6
`16.4X 10-6 _‘_metertothirdpower
`m3
`Velocity (linear)
`foot per second
`ft/s
`0.3048*
`0.305
`meter per second
`rn/s
`inch per second
`in./s
`0.0254*
`0.0254
`meter per second
`In/s
`mile per hour
`mph
`O.44704*
`0.447
`meter per second
`m/s
`mile per hour
`mph
`l.609344*
`1.61
`J kilometer per hour
`km/h
`Volume
`cubic foot
`ft3
`0.0283168
`0.0283
`cubic meter
`m3
`cubic inch
`111.3
`16.3871 x 10-6
`16.4 x 10-6
`cubic meter
`m3
`cubic inch
`in.3
`16.3871
`16.4
`cubic centimeter (cc)
`cm3
`gallon (231 111.3)
`gal.
`3.78541
`3.79
`liter
`L
`E31110“ (231 in-3)
`gal.
`0.00378541
`0.00379
`-1 cubic meter
`m3
`
`*An asterisk denotes an exact conversion factor
`Note: To convert from SI units to USCS units, divide by the conversion factor
`
`Temperature Conversion Formulas
`
`T(°C) = %[T(°F) — 32] = T(K) - 273.15
`T(K) = -3-[T(°F) — 32] + 273.15 = T(°C) + 273.15
`T(°F) = L:-T(°C) + 32 = -E-T(K) ~ 459.67
`
` 4
`
`Page 3 of 13
`
`Page 3 of 13
`
`

`
`Mechanics of
`
`‘- FIFTH EDITION
`
`James M. §ere
`I
`Professor Emeritus, Stanford University
`
`Brookslcole
`
`Thomson Learning“.
`Australia - Canada 0 Mexico - Singapore - Spain - United Kingdom - United States
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`Page 4 of 13
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`Page 4 of 13
`
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`Gere. James M.
`Mcchniiics of materials / James M. Gere.—— 5th ed.
`p. cm.
`Includes bibliographical references and index.
`ISBN 0 534—37133—7
`1. Strength of materials.
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`

`
`622
`
`CHAPTER 9 Deflections of Beams
`
`Example 9-3
`
`orts a concentrated load P acting at distances a and b
`A simple beam AB supp
`ht—hand supports, respectively (Fig. 9—l2a).
`from the left—hand and rig
`Determine the equations of the deflcction curve,
`the angles of rotation
`6A and 0,; at the supports, the maximum deflection Smx, and the deflection 5
`at the midpoint C of the beam (Fig. 9-l2b). (Note: The beam has length L Eng
`constant flexural rigidity E1.)
`
`
`
`FIG.9-12 Example 9-3. Deflections of
`a simple beam with a concentrated load
`
`(a)
`
`(b)
`
`solution
`Bending marrtenis in. the beam. In this example the bending moments are‘
`expressed by two equations. one for each part of the beam. Using the free-body
`diagrams ot'Fig. 9-13,, we arrive -at the following equations:
`M=5%» msxéw
`
`wmw
`
`(aSx_<_L)
`M=£%—P(x—a)
`Differential equatiarts of the deflectian. curve. The differential eqU'dli°,"
`for the two parts of the beam are obtained by substituting the bending-m0““?”
`expressions (Eqs. 9-273 and b) into Eq. (9~1Za). The results are
`Pbx
`EIv":T (0_<_x_<.a)
`
`
`
`,,
`Pb.
`EIv =71-P(x—a.)
`
`(aSxSL)
`
`fthe two ‘N
`
`Slopes and deflecrians of the beam. The first integrations 0
`ential equations yield the following expressions for the slope‘-41
`2
`EIv’=P:Z +c1
`
`(0SxSu)
`__
`2
`E[V/=:§I’f:_ +C2
`
`(gs;cSL)
`
`W
`
`(a)
`
` km“
`
`V
`
`BE
`L
`
`x
`,\:>a
`
`(13)
`
`FIG.9-13 Free-body diagrams used in
`determining the bending moments
`(Example 93)
`
`Page6of13
`
`Page 6 of 13
`
`

`
`SECTION 9.3 Deflections by Integration of the Bending-Moment Equation
`
`623
`
`in which C1 and C2 are constants of in
`gives the deflectionsz
`
`3
`EIv=P6bZ +C1x+C3
`__
`3
`
`EIv_Pbx3_P(x 41
`
`tegration. A second pair of integrations
`
`(osxsg)
`
`(3')
`
`. of four constants to be evaluate .
`Constants of integration. The four constants of integration can be found
`from the following four conditions:
`
`(1) At X = a, the slopes v’ for the two parts of the beam are the same.
`(2) At x = a, the deflections v for the two parts of the beam are the same.
`(3) Atx = 0, the deflection v is zero.
`(4) At x = L, the deflection v is zero.
`
`Condition (1) means that the slopes determined from Eqs. (h) and (i) must
`be equal when x = a; therefore,
`
`Pb 2
`2g +C1=
`
`Pb 2
`
`+C2 QT C1:C2
`
`Condition (2) means that the deflections found fro
`equal when x = a; therefore,
`
`II1 Eqs. (j) and (k) must be
`
`Pba3
`Pba3
`6L +C,a+c3= 6L +C2a+C4
`
`Inasmuch as C1 = C2, this equation gives C3 = C4.
`Next, we apply condition (3) to Eq. (3') and obtain C3 = 0; therefore,
`C3 = C4 : 0
`
`Finally, we apply condition (4) to Eq (k) and obtain
`PbL2
`P123
`6
`— T + CZL : 0
`
`‘_
`
`Therefore,
`
`C1: C2 =
`
`
`/ Pb(L2 — 122)
`6L
`
`(1)
`
`(In)
`
`Egmm'm'zs of the deflecrtfon curve. We now subs'u'tu.te the constants of inte-
`gration (Eqs. I and In) into the equations for the deflections (Eqs. j and K) and
`obtain the deflection equations for the two parts of the beam. The resulting
`equations, after at slight rearrangement, are
`
`continued
`
`Page 7 of 13
`
`Page 7 of 13
`
`

`
`624
`
`CHAPTER 9 Deflections of Beams
`
`Pbx
`2
`v = ~ GLEI (L — b2 — x2)
`
`(0 s x s a)
`
`(9-29a)
`
`,_
`
`v —
`
`Pb
`6L; (L2 — 192
`
`x2)
`
`Po — at
`6E1
`
`(a s x s L)
`
`(9-29b) 4.
`
`The first of these equations gives the deflection curve for the part of the beam to
`the left of the load P. and the second gives the deflection curve for the part of
`the beam. to the right of the load.
`The slopes for the two parts of the beam can be found either by substituting
`the values of C. and C2 into Eqs. (I1) and (1) or by taking the first derivatives of
`the deflection equations (Eqs. 9-2921 and b). The resulting equations are
`Pb
`v’ = —-fi(L2 * b2 — 3x2)
`
`(0 S x S a)
`P
`__
`2
`(9-30b) «-
`(a Ex 2 L)
`3x2) — ("MI")
`122
`v’ e — 6'Z’E[(L2
`The deflection and slope at any point along the axis of the beam can be calcu..'
`lated from Eqs. (9-29) and (9-30).
`Angles of rotation at the supports. To obtain the angles of rotation 0,1 and
`03 at the ends of the beam (Fig. 9-12b), we substitute x = 0 into Eq. (9-30a) and
`x = L into Eq. (9-30b):
`
`(9—30a) 4.
`
`9A — —v'(0) -
`
`Pb(L2 — bi) _ Pab(L + b)
`6LEI
`6LE1
`
`(9-3121) '
`
`(etiirrmx =
`
`(9~31b) "'
`
`-,
`
`
`Pb(2L2 — 3bL + bl) _ Pa.b(L + a)
`6LEI
`'
`6LEI
`63 = V (L) =
`Note that the angle 19,, is clockwise and the angle 03 is counterclockwise.”
`shown in Fig. 9—1?.b.
`‘
`The angles of rotation are functions of the position of the load and rent;
`their -largest values when the load is located near the midpoint of the beam‘ in
`theease of the angle of rotation 6,4. the rnaxirnum value of the ‘angle IF
`Prfx/37
`273}
`and occurs when E: =.m/*3 : 0.5771. (or a = 0.4231,}. This Vail‘? ‘ff
`obtained by taking the derivative of 9,1 with respect to F) (H5393 the f
`two expressions for 9,. in'Eq. 9'-31a) and then setting it equal to zero.
`Maximum deflecrton of-the beam. The maximum deflection
`an
`point D (Fig. 9-12b) where the-deflection curve has a liorizofltfii ‘3
`load is to the right of the midpoint. that is, if a > b, paint Dis I1} ‘hint sl0P°-"
`beam to the left of the toad. -We can locate this point by equating‘
`new ‘.
`from Eq. (9-30a) to zero and solving for the distance x. which We
`.17‘. In this manner we obtain the following formula for xii
`. L2 _ b2
`3
`
`x] =
`
`(a2b)
`
`
`
`Page 8 of 13
`
`Page 8 of 13
`
`

`
`625
`From this equation we see that as the load)" moves frotn the middle of the beam
`(/1 = L1?) to the right-hand end (I: = 0).
`the distance it. varies from M2 to
`L/V5 = 0.5771,. Thus, the maximum deflection occurs at a point very close to
`the midpoint of the beam, and thispoirtt is always between the midpoint of the
`beam and the load".
`The mexitnum deflection Sm“ is found by substituting x1 (from Eq. 9-33)
`into the deflection equation (Eq. 9—29a) and then inserting a minus sign:
`
`
`
`point
`ill of the ntidpoint (Fig. 9-l2b) is obtained by
`
`'
`
`._—.jf;____
`6 : _V(£) : Pb(3L2 — 4122)
`48EI
`‘
`
`mum deflection and the deflection at the midpoint is less than 3% of the
`maximum deflection, as demonstrated in Problem 9.3-ID.
`-Special case (load at the mfdpoint of the beam). An im
`portant special ease
`occurs when the load P acts at the midpoint of the beam (
`a = b 2 L/2). Then
`we obtain the following results from Eqs.
`(9-30a). (9~29a), (9-3]), and (9-34),
`respectively:
`
`1 4 _ P
`v —
`16EI(L
`
`2 _
`
`2
`4x)
`
`<0_<_xS 2
`
`_
`(9 36)
`
` (51217)
`
`(0 s x s 5)
`
`2
`
`V 2 — PX (312 — 42:2)
`
`48E]
`
`PL2
`9*‘ : 6'3 : 16EI
`PL3
`
`(9-37)
`
`(938)
`
`5C —
`amax
`,
`the mi
`Since the deflection curve is symmetric about
`the equations for 1:’ and 1: 3113 given only for the le-ft~|1and half of the beam
`(Eqs. 9-36 and 93-7]. If needed.
`the e'qttat'ions for the right-hand half can be
`-obtained front Eqs. (9-30b) and (9—29hJ by substituting a = 1: = M2.
`
`k—?——?
`Page 9 of 13
`
`Page 9 of 13
`
`

`
` _ —‘
`
`822
`
`CHAPTER 12 Review of Centroids and Moments of Inertia
`
`1 2.4 MOMENTS OF INERTIA OF PLANE AREAS
`The moments of inertia of a plane area (Fig. l2-9) with respect to the X
`and y axes, respectively, are defined by the integrals
`
`[x = Jy2dA
`
`I), = Jryfld/4
`
`(12-931))
`
`
`
`%
`L27
`B
`
`> B
`
`Lb
`E
`' 2
`.
`H
`FIG.12-10 Moments of inertia ofa
`rectangle
`
`in which at and y are the coordinates of the differential element of area
`dA. Because the element dA is multiplied by the square of the distance
`from the reference axis, moments of inertia are also called second
`moments of area. Also, we see that moments of inertia of areas (unlike
`first moments) are always positive quantities.
`To illustrate how moments of inertia are obt.ained by integration, we
`will consider a rectangle having width b and height h (Fig. 12—10)_ The X
`and y axes have their origin at the centroid C. For convenience, we use a
`differential element‘ of area dA in the form of a thin horizontal strip of
`width b and height dy (therefore, dA 2 bdy). Since all parts of the ele-
`mental strip are the same distance from the x axis, we can express the
`moment of inertia 1,, with respect to the x axis as follows:
`V
`M2
`[9113
`(U
`J‘
`-my
`y
`12
`1,: MA:
`Zbd =——
`In a similar manner, we can use an element of area in the form of a verti-
`cal strip with area dA = hdx and obtain the moment of inertia with
`respect to the y axis:
`2
`b/2
`2
`hba
`b.
`I). = x (1/1 =
`x hdx = ——
`(
`)
`ab/2
`12
`If a different set of axes is selected. the moments of inertia will lmw
`different values. For instance, consider axis BB at the base of the reciilfl‘
`gle (Fig. 12-10). If this axis is selected as the reference. we must deg‘
`y as the coordinate distance from that axis to the element Ufi1”’“
`'
`Then the calculations for the moment of inertia become
`it
`1
`by
`
`t
`
`(c)
`
`133 = Jy2dA : J‘ ylindy = —%'
`“
`Note that the moment of inertia with respect to
`moment of inertia with respect to the centrnidiil -"_
`moment of inertia increases as the reference axis IS 1“
`itself farther from the centroid.
`‘
`h refill“! [
`The moment of inertia of a composite area W1}
`mg parts W
`-eetmn .
`ticular axis is the sum of the moments of inertia "f
`to that same axis. An example is the hollow_b"" I; rtltruufih
`Fig. 12-lla, where the x and y axes are axes 0| 33”“
`
`,
`
`.
`
`.
`
`'
`
`an‘
`
`.-
`U th
`
`
`
`Page 10 of 13
`
`Page 10 of 13
`
`

`
`SECTION 12.4 Moments of Inertia of Plane Areas
`
`823
`
`1 =
`X
`
`up
`12
`
`— ——
`ban‘
`12
`
`(CD
`
`using the listed formulas in conjunction with the parallel-axis theorem.
`If an area is of such irregular shape that its moments of inertia cannot be
`obtained in this manner, men we can use numerical methods. The proce-
`dure is to divide the area into small elements of area AA,-, multiply each
`such area by the square of its tZliSltU1CC from the reference axis. and then
`sum the products.
`
`Radius of Gyration
`A distance known as the radius of gyration is occasionally encountered
`in mechanics. Radius of gyration of a plane area is defined as the square
`root of the moment of inertia of the area divided by the area itself‘: thus,
`
`(12—10a,b)
`
`in which r_, and r_,. denote the radii ofgyration with respect to the J.‘ and y
`axes. respectively. Since moment of inertia has units of length to the
`fourth power and area has units of length to the second power. radius ol'
`gyration has units of length.
`Although the radius of gyration of an area does not have an obvious
`physical meaning. we may consider it to be the distance (fmm the refer-
`
`(b)
`
`FIG.12-11 Composite areas
`
`Page 11 of 13
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`Page 11 of 13
`
`

`
`824
`
`CHAPTER 12 Review of Centroids and Moments of Inertia
`
`Example 12-3
`
`Determine the moments of inertia Ix and I), for the parabolic semisegment OAB
`shown in Fig. 12-12. The equation of the parabolic boundary is
`2
`
`y =f(x) = h(1—
`
`(E)
`
`(This same area was considered previously in Example 12-1.)
`
`Solution
`
`To determine the moments of inertia by integration. we will use Eqs._ (12_9a)
`and (12-9b). The differential element of area tie! is selected as a vertical strip of
`width dx and height y, as shown in Fig. 12-12. The area of this element is
`X2
`
`dA=ydx=h(1—F>dx
`
`(0
`
`
`
`
`
`ts.
`
`X
`
`F|G.12-12 Example 12-3. Moments of
`inertia of a parabolic semisegment
`
`the
`Since every point in this element is at the same distance from the y axis,
`moment of inertia of the element with respect to the y axis is x2dA. Therefore,
`the moment of inertia of the entire area with respect to the y axis is obtained as
`follows:
`
`1’
`
`2
`
`3
`
`[y = J‘x2d/1 = I x2h(l — "—2)ax = M’
`
`b
`
`15
`
`0
`
`(g) 4-
`
`To obtain the moment of inertia with respect to the x axis, we note that the
`differential element of area dA has a moment of inertia dlx with respect to the .t'
`axis equal to
`
`.3
`(11,. = %(dx)y3 = %dx
`as obtained from Eq. (c). Hence, the moment of inertia of the entire area Wilh-
`respect to the x axis is
`
`”
`
`3
`
`2
`
`3
`
`16b/3
`
`1, =
`
`”
`
`3
`
`0
`
`y_dx=.J‘ h_<1_£_)dx=;-1-
`(M ‘-
`I
`3
`0
`3
`[,2
`105
`element "1 ‘M
`[emcm -0
`These same results for Ix and I, can be obtained by using 3“
`form of a horizontal strip of area dA = x dy or by using a rectangulap t’«[ the pref
`area dA = dx dy and performing a double integration. Also, notitm ndix D‘.
`ceding formulas for 1,, and [y agree with those given in Case 17 Of P5“
`
`
`
`Page 12 of 13
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`Page 12 of 13
`
`

`
`axis.
`
`aiming to
`pa:-a!fe{~zm‘.r Jfrmrem,
`cm of inertia with respect to :1
`V with respect to any parallel
`
`
`
`shape with
`sets of coordinate axes:
`, and (3') :1 set of parallel .r_t~‘
`es between the two sets of
`
`FIG. 12-13 Derivation
`theorems
`
`of parallel-axis
`
`
`
`Equations {'i2~IIaJ and (DJ lb) re
`for moments of inertia: '
`m trrry a.-.'£.-; in its pfcttate '
`pmwllel cc:-t!:'r2id::I cm'.s' plus the. pmcmct
`the cl:'.s'rar:r.'e bett-wen the two £Ll'e’.t‘.
`
`third integral
`eretmc, the preceding equation reduces; to
`
`.
`L rt. E1Xl'i
`tn the area A melt
`
`[A = It + Ad?
`
`, we can determine the
`
`This result agrees
`integration (Eq. c
`
`
`
`Page 13 Of 13
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`Page 13 of 13