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Mechanics of Materials 8th Edition Hibbeler Solutions Manual Full Download: http://alibabadownload.com/product/mechanics-of-materials-8th-edition-hibbeler-solutions-manual/ © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2–1. An air-filled rubber ball has a diameter of 6 in. If the air pressure within it is increased until the ball’s diameter becomes 7 in., determine the average normal strain in the rubber. d0 = 6 in. d = 7 in. e =
pd - pd0 7 - 6 = = 0.167 in./in. pd0 6
Ans.
2–2. A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5 in., determine the average normal strain in the strip. L0 = 15 in. L = p(5 in.) e =
L - L0 5p - 15 = = 0.0472 in.>in. L0 15
Ans.
2–3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
D
E
4m
P
¢LBD ¢LCE = 3 7
A
3 (10) = 4.286 mm 7 ¢LCE 10 = = = 0.00250 mm>mm L 4000
3m
¢LBD = eCE
eBD =
Ans.
¢LBD 4.286 = = 0.00107 mm>mm L 4000
Ans.
1
This sample only, Download all chapters at: alibabadownload.com
B
C
2m
2m
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*2–4. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire.
C 300
œ = 23002 + 22 - 2(300)(2) cos 150° = 301.734 mm LAC
eAC = eAB
œ - LAC LAC 301.734 - 300 = = = 0.00578 mm>mm LAC 300
mm
30⬚
Ans.
30⬚
300
A
P
mm
B
•2–5. The rigid beam is supported by a pin at A and wires BD and CE. If the distributed load causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
E D 2m 1.5 m 3m
2m A
B
C
w
Since the vertical displacement of end C is small compared to the length of member AC, the vertical displacement dB of point B, can be approximated by referring to the similar triangle shown in Fig. a dB 10 = ; dB = 4 mm 2 5 The unstretched lengths of wires BD and CE are LBD = 1500 mm and LCE = 2000 mm. dB 4 Ans. = = 0.00267 mm>mm A eavg B BD = LBD 1500
A eavg B CE =
dC 10 = = 0.005 mm>mm LCE 2000
Ans.
2
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2–6. Nylon strips are fused to glass plates. When moderately heated the nylon will become soft while the glass stays approximately rigid. Determine the average shear strain in the nylon due to the load P when the assembly deforms as indicated.
y 2 mm P
3 mm 5 mm 3 mm 5 mm 3 mm
g = tan - 1 a
2 b = 11.31° = 0.197 rad 10
x
Ans.
2–7. If the unstretched length of the bowstring is 35.5 in., determine the average normal strain in the string when it is stretched to the position shown. 18 in.
6 in. 18 in.
Geometry: Referring to Fig. a, the stretched length of the string is L = 2L¿ = 2 2182 + 62 = 37.947 in. Average Normal Strain: eavg =
L - L0 37.947 - 35.5 = = 0.0689 in.>in. L0 35.5
Ans.
3
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u
*2–8. Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes it to rotate by u = 0.3°, determine the normal strain in the cable. Originally the cable is unstretched.
D
P 300 mm
B
AB = 24002 + 3002 = 500 mm
300 mm
AB¿ = 2400 + 300 - 2(400)(300) cos 90.3° 2
2
A
C
= 501.255 mm eAB =
AB¿ - AB 501.255 - 500 = AB 500
400 mm
= 0.00251 mm>mm
Ans.
•2–9.
Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm>mm, determine the displacement of point D. Originally the cable is unstretched.
u D
300 mm
B
AB = 23002 + 4002 = 500 mm
300 mm
AB¿ = AB + eABAB A
= 500 + 0.0035(500) = 501.75 mm
C
501.752 = 3002 + 4002 - 2(300)(400) cos a a = 90.4185°
400 mm
p (0.4185) rad u = 90.4185° - 90° = 0.4185° = 180° ¢ D = 600(u) = 600(
P
p )(0.4185) = 4.38 mm 180°
Ans.
4
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2–10. The corners B and D of the square plate are given the displacements indicated. Determine the shear strains at A and B.
y A
16 mm D
B
3 mm 3 mm 16 mm
16 mm
Applying trigonometry to Fig. a f = tan - 1 a
13 p rad b = 39.09° a b = 0.6823 rad 16 180°
a = tan - 1 a
16 p rad b = 50.91° a b = 0.8885 rad 13 180°
By the definition of shear strain,
A gxy B A =
p p - 2f = - 2(0.6823) = 0.206 rad 2 2
Ans.
A gxy B B =
p p - 2a = - 2(0.8885) = -0.206 rad 2 2
Ans.
5
C
16 mm
x
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2–11. The corners B and D of the square plate are given the displacements indicated. Determine the average normal strains along side AB and diagonal DB.
y A
16 mm D
B
3 mm 3 mm 16 mm
16 mm
Referring to Fig. a, LAB = 2162 + 162 = 2512 mm LAB¿ = 2162 + 132 = 2425 mm LBD = 16 + 16 = 32 mm LB¿D¿ = 13 + 13 = 26 mm Thus,
A eavg B AB =
LAB¿ - LAB 2425 - 2512 = = -0.0889 mm>mm LAB 2512
Ans.
A eavg B BD =
LB¿D¿ - LBD 26 - 32 = = -0.1875 mm>mm LBD 32
Ans.
6
C
16 mm
x
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*2–12. The piece of rubber is originally rectangular. Determine the average shear strain gxy at A if the corners B and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines.
y 3 mm
C D
2 = 0.006667 rad 300 3 u2 = tan u2 = = 0.0075 rad 400 u1 = tan u1 =
400 mm
gxy = u1 + u2
A
= 0.006667 + 0.0075 = 0.0142 rad
Ans.
•2–13.
The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD.
f = tan
B 2 mm
3 mm
C D
400 mm
3 b = 0.42971° a 400
AB¿ = 2(300)2 + (2)2 = 300.00667 w = tan - 1 a
x
y
AD¿ = 2(400)2 + (3)2 = 400.01125 mm -1
300 mm
A
2 b = 0.381966° 300
a = 90° - 0.42971° - 0.381966° = 89.18832° D¿B¿ = 2(400.01125)2 + (300.00667)2 - 2(400.01125)(300.00667) cos (89.18832°) D¿B¿ = 496.6014 mm DB = 2(300)2 + (400)2 = 500 mm 496.6014 - 500 = -0.00680 mm>mm 500 400.01125 - 400 = = 0.0281(10 - 3) mm>mm 400
eDB =
Ans.
eAD
Ans.
7
300 mm
B 2 mm
x
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2–14. Two bars are used to support a load. When unloaded, AB is 5 in. long, AC is 8 in. long, and the ring at A has coordinates (0, 0). If a load P acts on the ring at A, the normal strain in AB becomes PAB = 0.02 in.>in., and the normal strain in AC becomes PAC = 0.035 in.>in. Determine the coordinate position of the ring due to the load.
y
B
C
60⬚
5 in.
8 in.
A
x
P
Average Normal Strain: œ = LAB + eAB LAB = 5 + (0.02)(5) = 5.10 in. LAB œ = LAC + eACLAC = 8 + (0.035)(8) = 8.28 in. LAC
Geometry: a = 282 - 4.33012 = 6.7268 in. 5.102 = 9.22682 + 8.282 - 2(9.2268)(8.28) cos u u = 33.317° x¿ = 8.28 cos 33.317° = 6.9191 in. y¿ = 8.28 sin 33.317° = 4.5480 in. x = -(x¿ - a) = -(6.9191 - 6.7268) = -0.192 in.
Ans.
y = -(y¿ - 4.3301) = -(4.5480 - 4.3301) = -0.218 in.
Ans.
8
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2–15. Two bars are used to support a load P. When unloaded, AB is 5 in. long, AC is 8 in. long, and the ring at A has coordinates (0, 0). If a load is applied to the ring at A, so that it moves it to the coordinate position (0.25 in., -0.73 in.), determine the normal strain in each bar.
y
B
C
60⬚
5 in.
8 in.
A
x
P
Geometry: a = 282 - 4.33012 = 6.7268 in. LA¿B = 2(2.5 + 0.25)2 + (4.3301 + 0.73)2 = 5.7591 in. LA¿C = 2(6.7268 - 0.25)2 + (4.3301 + 0.73)2 = 8.2191 in. Average Normal Strain: eAB =
=
eAC =
=
LA¿B - LAB LAB 5.7591 - 5 = 0.152 in.>in. 5
Ans.
LA¿C - LAC LAC 8.2191 - 8 = 0.0274 in.>in. 8
Ans.
9
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*2–16. The square deforms into the position shown by the dashed lines. Determine the average normal strain along each diagonal, AB and CD. Side D¿B¿ remains horizontal.
y 3 mm D¿
B¿
B
D
Geometry: AB = CD = 2502 + 502 = 70.7107 mm 53 mm
C¿D¿ = 2532 + 582 - 2(53)(58) cos 91.5°
50 mm 91.5⬚
= 79.5860 mm C
B¿D¿ = 50 + 53 sin 1.5° - 3 = 48.3874 mm
A
x
C¿
AB¿ = 2532 + 48.38742 - 2(53)(48.3874) cos 88.5°
50 mm 8 mm
= 70.8243 mm Average Normal Strain:
eAB =
=
eCD =
=
AB¿ - AB AB 70.8243 - 70.7107 = 1.61 A 10 - 3 B mm>mm 70.7107
Ans.
C¿D¿ - CD CD 79.5860 - 70.7107 = 126 A 10 - 3 B mm>mm 70.7107
Ans.
•2–17.
The three cords are attached to the ring at B. When a force is applied to the ring it moves it to point B¿ , such that the normal strain in AB is PAB and the normal strain in CB is PCB. Provided these strains are small, determine the normal strain in DB. Note that AB and CB remain horizontal and vertical, respectively, due to the roller guides at A and C.
A¿
B¿
A
B
L
Coordinates of B (L cos u, L sin u)
u
Coordinates of B¿ (L cos u + eAB L cos u, L sin u + eCB L sin u)
C¿ D
LDB¿ = 2(L cos u + eAB L cos u) + (L sin u + eCB L sin u) 2
2
LDB¿ = L 2cos2 u(1 + 2eAB + e2AB) + sin2 u(1 + 2eCB + e2CB) Since eAB and eCB are small, LDB¿ = L 21 + (2 eAB cos2 u + 2eCB sin2 u) Use the binomial theorem, LDB¿ = L ( 1 +
1 (2 eAB cos2 u + 2eCB sin2 u)) 2
= L ( 1 + eAB cos2 u + eCB sin2 u) Thus, eDB =
L( 1 + eAB cos2 u + eCB sin2 u) - L L
eDB = eAB cos2 u + eCB sin2 u
Ans.
10
C
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2–18. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners A and B if the plastic distorts as shown by the dashed lines.
y 5 mm 2 mm 2 mm
B
4 mm
C 300 mm
Geometry: For small angles, 2 mm D
2 a = c = = 0.00662252 rad 302 b = u =
A
x
400 mm 3 mm
2 = 0.00496278 rad 403
Shear Strain: (gB)xy = a + b = 0.0116 rad = 11.6 A 10 - 3 B rad
Ans.
(gA)xy = -(u + c) = -0.0116 rad = -11.6 A 10 - 3 B rad
Ans.
2–19. The piece of plastic is originally rectangular. Determine the shear strain gxy at corners D and C if the plastic distorts as shown by the dashed lines.
y 5 mm 2 mm 2 mm
B
4 mm
C 300 mm 2 mm D
A 400 mm 3 mm
Geometry: For small angles, 2 = 0.00496278 rad 403 2 = 0.00662252 rad b = u = 302 Shear Strain: a = c =
(gC)xy = -(a + b) = -0.0116 rad = -11.6 A 10 - 3 B rad
Ans.
(gD)xy = u + c = 0.0116 rad = 11.6 A 10 - 3 B rad
Ans.
11
x
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*2–20. The piece of plastic is originally rectangular. Determine the average normal strain that occurs along the diagonals AC and DB.
y 5 mm 2 mm 2 mm
Geometry:
B
4 mm
C
AC = DB = 24002 + 3002 = 500 mm
300 mm
DB¿ = 24052 + 3042 = 506.4 mm
2 mm D
A¿C¿ = 2401 + 300 = 500.8 mm 2
2
x
A 400 mm 3 mm
Average Normal Strain: eAC =
A¿C¿ - AC 500.8 - 500 = AC 500
= 0.00160 mm>mm = 1.60 A 10 - 3 B mm>mm eDB =
Ans.
DB¿ - DB 506.4 - 500 = DB 500
= 0.0128 mm>mm = 12.8 A 10 - 3 B mm>mm
Ans.
•2–21.
The force applied to the handle of the rigid lever arm causes the arm to rotate clockwise through an angle of 3° about pin A. Determine the average normal strain developed in the wire. Originally, the wire is unstretched.
D
600 mm
Geometry: Referring to Fig. a, the stretched length of LB¿D can be determined using the consine law, A
LB¿D = 2(0.6 cos 45°)2 + (0.6 sin 45°)2 - 2(0.6 cos 45°)(0.6 sin 45°) cos 93°
B
= 0.6155 m Average Normal Strain: The unstretched length of wire BD is LBD = 0.6 m. We obtain eavg =
C
45⬚
LB¿D - LBD 0.6155 - 0.6 = 0.0258 m>m = LBD 0.6
Ans.
12
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2–22. A square piece of material is deformed into the dashed position. Determine the shear strain gxy at A.
y 15.18 mm B
Shear Strain: (gA)xy =
89.7° p - ¢ ≤p 2 180°
C
15.24 mm
15 mm
= 5.24 A 10 - 3 B rad
Ans.
89.7⬚ A
2–23. A square piece of material is deformed into the dashed parallelogram. Determine the average normal strain that occurs along the diagonals AC and BD.
15 mm 15.18 mm
x
D
y 15.18 mm B
C
15.24 mm
15 mm 89.7⬚ A
Geometry: AC = BD = 2152 + 152 = 21.2132 mm AC¿ = 215.182 + 15.242 - 2(15.18)(15.24) cos 90.3° = 21.5665 mm B¿D¿ = 215.182 + 15.242 - 2(15.18)(15.24) cos 89.7° = 21.4538 mm Average Normal Strain: eAC =
eBD
AC¿ - AC 21.5665 - 21.2132 = AC 21.2132
= 0.01665 mm>mm = 16.7 A 10 - 3 B mm>mm
Ans.
= 0.01134 mm>mm = 11.3 A 10 - 3 B mm>mm
Ans.
B¿D¿ - BD 21.4538 - 21.2132 = = BD 21.2132
13
15 mm 15.18 mm
D
x
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*2–24. A square piece of material is deformed into the dashed position. Determine the shear strain gxy at C.
y 15.18 mm B
C
15.24 mm
15 mm 89.7⬚ A
(gC)xy =
15 mm 15.18 mm
x
D
p 89.7° - ¢ ≤p 2 180° = 5.24 A 10 - 3 B rad
Ans.
u ⫽ 2⬚
•2–25.
The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns of the frame tilt u = 2°. Determine the approximate normal strain in the wire when the frame is in this position. Assume the columns are rigid and rotate about their lower supports.
u ⫽ 2⬚
B
Geometry: The vertical displacement is negligible 3m
xA
2° = (1) ¢ ≤ p = 0.03491 m 180° A
2° xB = (4) ¢ ≤ p = 0.13963 m 180°
1m
x = 4 + xB - xA = 4.10472 m A¿B¿ = 232 + 4.104722 = 5.08416 m AB = 232 + 42 = 5.00 m Average Normal Strain: eAB =
=
A¿B¿ - AB AB 5.08416 - 5 = 16.8 A 10 - 3 B m>m 5
Ans.
14
4m
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2–26. The material distorts into the dashed position shown. Determine (a) the average normal strains along sides AC and CD and the shear strain gxy at F, and (b) the average normal strain along line BE.
y 15 mm C
25 mm D
10 mm B E
75 mm
90 mm
A
Referring to Fig. a, LBE = 2(90 - 75)2 + 802 = 26625 mm LAC¿ = 21002 + 152 = 210225 mm LC¿D¿ = 80 - 15 + 25 = 90 mm f = tan-1 ¢
25 p rad ≤ = 14.04° ¢ ≤ = 0.2450 rad. 100 180°
When the plate deforms, the vertical position of point B and E do not change. LBB¿ 15 = ; LBB¿ = 13.5 mm 90 100 LEE¿ 25 = ; 75 100
LEE¿ = 18.75 mm
LB¿E¿ = 2(90 - 75)2 + (80 - 13.5 + 18.75)2 = 27492.5625 mm Thus,
A eavg B AC =
LAC¿ - LAC 210225 - 100 = = 0.0112 mm>mm LAC 100
Ans.
A eavg B CD =
LC¿D¿ - LCD 90 -80 = = 0.125 mm>mm LCD 80
Ans.
A eavg B BE =
LB¿E¿ - LBE 27492.5625 - 26625 = = 0.0635 mm>mm LBE 26625
Ans.
Referring to Fig. a, the angle at corner F becomes larger than 90° after the plate deforms. Thus, the shear strain is negative. 0.245 rad
Ans.
15
80 mm
F
x
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2–27. The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF.
y 15 mm
25 mm D
C 10 mm
The undeformed length of diagonals AD and CF are
B E
LAD = LCF = 280 + 100 = 216400 mm 2
2
The deformed length of diagonals AD and CF are
75 mm
90 mm
LAD¿ = 2(80 + 25) + 100 = 221025 mm 2
2
LC¿F = 2(80 - 15)2 + 1002 = 214225 mm A
Thus,
A eavg B AD =
LAD¿ - LAD 221025 - 216400 = = 0.132 mm>mm LAD 216400
Ans.
A eavg B CF =
LC¿F - LCF 214225 - 216400 = = -0.0687 mm>mm LCF 216400
Ans.
*2–28. The wire is subjected to a normal strain that is 2 defined by P = xe - x , where x is in millimeters. If the wire has an initial length L, determine the increase in its length.
80 mm
P ⫽ xe⫺x
L
2
dL = e dx = x e-x dx L 2
L0
x e-x dx
L 1 1 1 2 2 = - c e-x d 冷 = - c e-L - d 2 2 2 0
=
x
2
x x
¢L =
F
1 2 [1 - e-L ] 2
Ans.
16
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•2–29. The curved pipe has an original radius of 2 ft. If it is heated nonuniformly, so that the normal strain along its length is P = 0.05 cos u, determine the increase in length of the pipe.
e = 0.05 cos u ¢L =
L
2 ft
e dL
=
A
u
90°
(0.05 cos u)(2 du)
L0
90°
= 0.1
90°
cos u du = [0.1[sin u] 0冷 ] = 0.100 ft
L0
Ans.
Solve Prob. 2–29 if P = 0.08 sin u.
2–30.
dL = 2 due = 0.08 sin u ¢L =
e dL
L
90°
=
2 ft
(0.08 sin u)(2 du)
L0
= 0.16
L0
90°
sin u du = 0.16[-cos u] 0冷 = 0.16 ft
Ans.
2–31. The rubber band AB has an unstretched length of 1 ft. If it is fixed at B and attached to the surface at point A¿, determine the average normal strain in the band. The surface is defined by the function y = (x2) ft, where x is in feet.
y y ⫽ x2
A¿
Geometry: 1 ft
L =
L0
A
1 + a
dy 2 b dx dx
However y = x2 then
1 ft
dy = 2x dx
B
1 ft
L =
=
L0
A
u 90°
21 + 4 x2 dx
1 1 ft C 2x21 + 4 x2 + ln A 2x + 21 + 4x2 B D 冷0 4
= 1.47894 ft Average Normal Strain: L - L0 1.47894 - 1 eavg = = = 0.479 ft>ft L0 1
Ans.
17
A 1 ft
x
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*2–32. The bar is originally 300 mm long when it is flat. If it is subjected to a shear strain defined by gxy = 0.02x, where x is in meters, determine the displacement ¢y at the end of its bottom edge. It is distorted into the shape shown, where no elongation of the bar occurs in the x direction.
y
⌬y x 300 mm
Shear Strain: dy = tan gxy ; dx
dy = tan (0.02 x) dx 300 mm
¢y
dy =
L0
L0
tan (0.02 x)dx
¢y = -50[ln cos (0.02x)]|0300 mm = 2.03 mm
Ans.
The fiber AB has a length L and orientation u. If its ends A and B undergo very small displacements uA and vB , respectively, determine the normal strain in the fiber when it is in position A¿B¿.
•2–33.
y B¿ vB B L
Geometry: LA¿B¿ = 2(L cos u - uA) + (L sin u + yB) 2
2
A
= 2L3 + u2A + y2B + 2L(yB sin u - uA cos u) Average Normal Strain: LA¿B¿ - L eAB = L =
A
1 +
2(yB sin u - uA cos u) u2A + y2B + - 1 L L2
Neglecting higher terms u2A and y2B 1
eAB
2(yB sin u - uA cos u) 2 = B1 + R - 1 L
Using the binomial theorem: eAB = 1 +
=
2uA cos u 1 2yB sin u ¢ ≤ + ... - 1 2 L L
yB sin u uA cos u L L
Ans.
18
uA A¿
u x
02 Solutions 46060
5/6/10
1:45 PM
Page 19
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2–34. If the normal strain is defined in reference to the final length, that is, Pnœ = lim a p : p¿
¢s¿ - ¢s b ¢s¿
instead of in reference to the original length, Eq. 2–2, show that the difference in these strains is represented as a second-order term, namely, Pn - Pnœ = PnPnœ .
eB =
¢S¿ - ¢S ¢S
œ = eB - eA
¢S¿ - ¢S ¢S¿ - ¢S ¢S ¢S¿
¢S¿ 2 - ¢S¢S¿ - ¢S¿¢S + ¢S2 ¢S¢S¿ 2 2 ¢S¿ + ¢S - 2¢S¿¢S = ¢S¢S¿ =
=
(¢S¿ - ¢S)2 ¢S¿ - ¢S ¢S¿ - ¢S = ¢ ≤¢ ≤ ¢S¢S¿ ¢S ¢S¿
= eA eBœ (Q.E.D)
19
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