chapter 2 tutorial .pdf
Short Description
Solid Mechanics - Strain...
Description
Problem 2-1 An air-filled rubber ball has a diameter of 150 mm. If the air pressure within it is increased until the ball's diameter becomes 175 mm, determine the average normal strain in the rubber. d0 150mm
Given:
d 175mm
Solution: H H
Sd Sd0 Sd0 0.1667
mm mm
Ans
Problem 2-2 A thin strip of rubber has an unstretched length of 375 mm. If it is stretched around a pipe having an outer diameter of 125 mm, determine the average normal strain in the strip. L 0 375mm
Given: Solution:
L S ( 125) mm H H
SL SL0 SL 0 0.0472
mm mm
Ans
Problem 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. Given:
a 3m
L CE 4m
b 4m
L BD 4m
'LCE 10mm Solution:
'LBD
§ a · 'L ¨ © a b ¹ CE H CE
H BD
'LCE L CE 'L BD LBD
'LBD
4.2857 mm
H CE
0.00250
mm mm
Ans
H BD
0.00107
mm mm
Ans
Problem 2-4 The center portion of the rubber balloon has a diameter of d = 100 mm. If the air pressure within it causes the balloon's diameter to become d = 125 mm, determine the average normal strain in the rubber. d0 100mm
Given:
d 125mm
Solution: H
H
Sd Sd0 Sd0 0.2500
mm mm
Ans
Problem 2-5 The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam is displace 10 mm downward, determine the normal strain developed in wires CE and BD. Given:
a 3m
b 2m
c 2m
L CE 4m
L BD 3m
' tip 10mm Solution: 'LBD a
=
'L CE
'LCE
ab
ab
=
' tip abc
'LCE
§ a b · ' ¨ © a b c ¹ tip
'LCE
7.1429 mm
'LBD
§ a · ' ¨ © a b c ¹ tip
'LBD
4.2857 mm
Average Normal Strain: H CE
H BD
'LCE L CE 'L BD LBD
H CE
0.00179
mm mm
Ans
H BD
0.00143
mm mm
Ans
Problem 2-6 The rigid beam is supported by a pin at A and wires BD and CE. If the maximum allowable normal strain in each wire is Hmax = 0.002 mm/mm, determine the maximum vertical displacement of the load P. Given:
a 3m
b 2m
L CE 4m
L BD 3m
H allow 0.002
mm mm
c 2m
Solution: 'LBD a
=
'L CE
'LCE
ab
ab
=
' tip abc
Average Elongation/Vertical Displacement: 'LBD L BD H allow
'LCE LCE H allow
'LBD
6.00 mm
' tip
§ a b c · 'L ¨ © a ¹ BD
' tip
14.00 mm
'LCE
8.00 mm
' tip
§ a b c · 'L ¨ © a b ¹ CE
' tip
11.20 mm
(Controls !)
Ans
Problem 2-7 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. Given: a 300mm
T 30deg
' A 2mm Solution: Consider the triangle CAA': I A 180deg T L CA' L CA'
' CA ' CA
IA 2
2
150 deg
a ' A 2 a ' A cos I A 301.734 mm L CA' a a 0.00578
mm mm
Ans
Problem 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 T = 0.3°, determine the normal strain in the cable. Originally the cable is unstretched. Given: a 400mm
b 300mm
c 300mm
T 0.3deg Solution: L AB
2
2
a b
L AB
500 mm
Consider the triangle ACB': I C 90deg T L AB'
2
IC
2
a b 2 a b cos I C
L AB'
501.255 mm
H AB
LAB' L AB L AB
H AB
90.3 deg
0.00251
mm mm
Ans
Problem 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. a 400mm
Given:
H AB 0.0035
b 300mm
c 300mm
mm mm
Solution: 2
L AB
2
a b
L AB
L AB' L AB 1 H AB
500 mm
L AB'
501.750 mm
Consider the triangle ACB': I C = 90deg T L AB' =
2
2
a b 2 a b cos I C
ª a2 b2 L 2º AB' » I C acos « 2 a b ¬ ¼ IC
90.419 deg
T I C 90deg
' D ( b c) T
T
0.41852 deg
T
0.00730 rad
'D
4.383 mm
Ans
Problem 2-10 The wire AB is unstretched when T = 45°. If a vertical load is applied to bar AC, which causes T = 47°, determine the normal strain in the wire. Given: T 45deg
'T 2deg
Solution: 2
2
LAB =
L L
LCB =
( 2L) L
2
2
LAB =
2L
LCB =
5L
From the triangle CAB: I A 180deg T
IA
135.00 deg
IB
18.435 deg
I' B
20.435 deg
sin I B sin I A = L LCB
§ L sin I A ·
I B asin ¨
©
5L
¹
From the triangle CA'B: I' B I B 'T
sin I' B sin 180deg I A' = L LCB
§ 5 L sin I' B ·
I A' 180deg asin ¨
©
I A'
L
¹
128.674 deg
I' C 180deg I' B I A' I' C
30.891 deg
sin I' B sin I' C = L L A'B
LA'B
H AB
L sin I' B
sin I' C
L A'B LAB LAB
LAB
H AB
2L
0.03977
Ans
Problem 2-11 If a load applied to bar AC causes point A to be displaced to the left by an amount 'L, determine the normal strain in wire AB. Originally, T = 45°. Given:
T 45deg 'L L
H AC =
Solution:
2
2
L AB =
L L
L CB =
( 2L ) L
2
2
L AB =
2L
L CB =
5L
From the triangle A'AB: I A 180deg T
IA
2
135.00 deg
L A'B =
'L LAB 2 'L L AB cos I A
L A'B =
'L 2 L 2 'L L
H AB =
2
2
2
L A'B LAB LAB 2
H AB =
'L 2 L 2 'L L 2
2L
2L 2
H AB =
ª1 § 'L ·º 1 § 'L · 1 « ¨ » ¨ ¬2 © L ¹¼ ©L ¹
Neglecting the higher-order terms,
§ 'L · H AB = ¨ 1 L ¹ © ª ¬
H AB = «1 H AB =
0.5
1
1 § 'L · º ¨ .....» 1 2 © L ¹ ¼
1 § 'L · ¨ 2 © L ¹
(Binomial expansion)
Ans
Alternatively, H AB =
L A'B LAB LAB
H AB = H AB =
'L sin T 2L 1 § 'L · ¨ 2 © L ¹
Ans
Problem 2-12 The piece of plastic is originally rectangular. Determine the shear strain Jxy at corners A and B if the plastic distorts as shown by the dashed lines. Given: a 400mm
b 300mm
'Ax 3mm
'Ay 2mm
'Cx 2mm
'Cy 2mm
'Bx 5mm
'By 4mm
Solution: Geometry : For small angles, D
E
\
T
'Cx
D
0.00662252 rad
E
0.00496278 rad
\
0.00662252 rad
b 'Cy 'By 'Cy
a 'Bx 'Cx 'Bx 'Ax
b 'By 'Ay 'Ay a 'Ax
T
0.00496278 rad
Shear Strain : 3
J xy_B E \
J xy_B
11.585 u 10
J xy_A T \
J xy_A
11.585 u 10
rad
3
rad
Ans Ans
Problem 2-13 The piece of plastic is originally rectangular. Determine the shear strain Jxy at corners D and C if the plastic distorts as shown by the dashed lines. Given: a 400mm
b 300mm
'Ax 3mm
'Ay 2mm
'Cx 2mm
'Cy 2mm
'Bx 5mm
'By 4mm
Solution: Geometry : For small angles, D
E
\
T
'Cx
D
0.00662252 rad
E
0.00496278 rad
\
0.00662252 rad
b 'Cy 'By 'Cy
a 'Bx 'Cx 'Bx 'Ax
b 'By 'Ay 'Ay a 'Ax
T
0.00496278 rad
Shear Strain : 3
J xy_D D T
J xy_D
11.585 u 10
J xy_C D E
J xy_C
11.585 u 10
rad
3
rad
Ans Ans
Problem 2-14 The piece of plastic is originally rectangular. Determine the average normal strain that occurs along th diagonals AC and DB. Given: a 400mm
b 300mm
'Ax 3mm
'Ay 2mm
'Cx 2mm
'Cy 2mm
'Bx 5mm
'By 4mm
Solution: Geometry : 2
2
2
2
L AC
a b
L DB
a b
L A'C' L A'C' L DB' L DB'
L AC
500 mm
L DB
500 mm
a 'Ax 'Cx 2 b 'Cy 'Ay 2 500.8 mm
a 'Bx 2 b 'By 2 506.4 mm
Average Normal Strain : H AC
LA'C' L AC LAC
H AC
1.601 u 10
H BD
LDB' L DB L DB
H BD
12.800 u 10
3 mm
mm 3 mm
mm
Ans
Ans
Problem 2-15 The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns of the frame tilt T = 2°. Determine the approximate normal strain in the wire when the frame is in thi position. Assume the columns are rigid and rotate about their lower supports. Given: a 4m
b 3m
c 1m
T 2deg
Solution: T=
§ T · S ¨ © 180 ¹
T
0.03490659 rad
Geometry : The vertical dosplacement is negligible. 'Ax c T
'Ax
34.907 mm
'Bx ( b c) T
'Bx
139.626 mm
L AB L A'B' L A'B'
2
2
a b
L AB
5000 mm
a 'Bx 'Ax 2 b2 5084.16 mm
Average Normal Strain : H AB H AB
LA'B' L AB LAB 16.833 u 10
3 mm
mm
Ans
Problem 2-16 The corners of the square plate are given the displacements indicated. Determine the shear strain along the edges of the plate at A and B. Given: ax 250mm 'v 5mm
ay 250mm 'h 7.5mm
Solution: At A :
§ T' A ·
tan ¨
© 2 ¹
=
ax 'h ay 'v
§ ax 'h ·
T' A 2 atan ¨
© ay 'v ¹
J nt_A
§ S · T' ¨ A © 2¹
T' A
1.52056 rad
J nt_A
0.05024 rad
I' B
1.62104 rad
Ans
At B :
§ I' B ·
tan ¨
© 2 ¹
=
ay 'v ax 'h
§ ay 'v ·
I' B 2 atan ¨
© ax 'h ¹
S J nt_B I' B 2
J nt_B
0.05024 rad
Ans
Problem 2-17 The corners of the square plate are given the displacements indicated. Determine the average normal strains along side AB and diagonals AC and DB. Given:
ax 250mm
ay 250mm
'v 5mm
'h 7.5mm
Solution: For AB : L AB L A'B'
2
ax ay
2
ax 'h 2 ay 'v 2
H AB
LA'B' L AB LAB
L AB
353.55339 mm
L A'B' H AB
351.89665 mm 4.686 u 10
3 mm
mm
Ans
For AC :
2 ay 'v
L AC 2 ay
L AC
L A'C'
L A'C'
H AC
LA'C' L AC LAC
500 mm 510 mm
H AC
20.000 u 10
L DB 2 ax
L DB
500 mm
L D'B'
L D'B'
3 mm
mm
Ans
For DB :
H DB
2 ax 'h
LD'B' L DB LDB
H DB
485 mm 3 mm
30.000 u 10
mm
Ans
Problem 2-18 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. Given: a 50mm
b 50mm
'Bx 3mm 'Cx 8mm
'Cy 0mm L AD' 53mm
T' A 91.5deg Solution: For AB :
'By L AD' cos T' A 90deg b 2
2
'By
2.9818 mm
L AB
a b
L AB
70.7107 mm
L AB'
a 'Bx 2 b 'By 2
L AB'
70.8243 mm
H AB H AB
LAB' L AB L AB
3 mm
1.606 u 10
mm
Ans
For CD : 'Dy 'By L CD L C'D'
2
2
a b
2.9818 mm
L CD
70.7107 mm
a 'Cx 2 b 'Dy 2 2 a 'Cx b 'Dy cos T'A
L C'D'
79.5736 mm
H CD
LC'D' L CD LCD
H CD
'Dy
3 mm
125.340 u 10
mm
Ans
Problem 2-19 The square deforms into the position shown by the dashed lines. Determine the shear strain at each of its corners, A, B, C, and D. Side D'B' remains horizontal. a 50mm
Given:
b 50mm
'Bx 3mm 'Cx 8mm T' A 91.5deg Solution:
T' A
'Cy 0mm LAD' 53mm
1.597 rad
Geometry :
'By LAD' cos T' A 90deg b
'By
2.9818 mm
'Dy 'By
'Dy
2.9818 mm
'Dx
1.3874 mm
'Dx L AD' sin T' A 90deg In triangle C'B'D' :
'Cx 'Bx 2 b 'By 2
LC'B' LC'B'
54.1117 mm
LD'B' a 'Bx 'Dx
48.3874 mm
a 'Cx 'Dx 2 b 'Dy 2
LC'D' LC'D'
LD'B'
79.586 mm 2
2
2
LC'B' LD'B' LC'D' cos T B' = 2 L C'B' L D'B'
ª L 2 L 2 L 2º « C'B' D'B' C'D' » T B' acos « » ¬ 2 LC'B' LD'B' ¼
T B'
101.729 deg
T B'
1.7755 rad
T D' 180deg T' A
T D'
88.500 deg
T D'
1.5446 rad
T C' 180deg T B'
T C'
78.271 deg
T C'
1.3661 rad
Shear Strain :
0.5S T B' 0.5S T C' 0.5S T D'
3
J xy_A 0.5S T' A
J xy_A
26.180 u 10
J xy_B
J xy_B
204.710 u 10
J xy_C
204.710 u 10
J xy_D
26.180 u 10
J xy_C J xy_D
rad
3
3
3
rad
rad
rad
Ans Ans Ans Ans
Problem 2-20 The block is deformed into the position shown by the dashed lines. Determine the average normal strain along line AB. Given:
'xBA ( 70 30)mm 'xB'A ( 55 30)mm 'yB'A
2
2
'yBA 100mm
110 15 mm
Solution: For AB : 2
2
L AB
'xBA 'yBA
L AB'
'xB'A 'yB'A
H AB H AB
2
2
L AB
107.7033 mm
L AB'
111.8034 mm
LAB' L AB L AB 38.068 u 10
3 mm
mm
Ans
Problem 2-21 A thin wire, lying along the x axis, is strained such that each point on the wire is displaced 'x = k x2 along the x axis. If k is constant, what is the normal strain at any point P along the wire? 'x = k x
Given:
2
Solution: H=
d 'x dx
H = 2 k x
Ans
Problem 2-22 The rectangular plate is subjected to the deformation shown by the dashed line. Determine the averag shear strain Jxy of the plate. a 150mm
Given:
b 200mm
'a 0mm
'b 3mm
Solution:
§ 'b · © a¹
'T atan ¨ 'T
1.146 deg
'T
19.9973 u 10
3
rad
Shear Strain : J xy 'T J xy
3
19.997 u 10
rad
Ans
Problem 2-23 The rectangular plate is subjected to the deformation shown by the dashed lines. Determine the averag shear strain Jxy of the plate. Given:
a 200mm
'a 3mm
b 150mm
'b 0mm
Solution:
§ 'a · © b ¹
'T atan ¨ 'T
1.146 deg
'T
19.9973 u 10
3
rad
Shear Strain : J xy 'T J xy
19.997 u 10
3
rad
Ans
Problem 2-24 The rectangular plate is subjected to the deformation shown by the dashed lines. Determine the average normal strains along the diagonal AC and side AB. Given:
a 200mm
b 150mm
'Ax 3mm 'Bx 0mm
'Ay 0mm 'By 0mm
'Cx 0mm 'Dx 3mm
'Cy 0mm 'Dy 0mm
Solution: Geometry : 2
L AC
2
a b
L AB b L A'C L A'C L A'B L A'B
L AC
250 mm
L AB
150 mm
a 'Cx 'Ax 2 b 'Cy 'Ay 2 252.41 mm 2
'Ax b
2
150.03 mm
Average Normal Strain : H AC
LA'C L AC L AC
H AC
9.626 u 10
H AB
LA'B L AB L AB
H AB
199.980 u 10
3 mm
mm 6 mm
mm
Ans
Ans
Problem 2-25 The piece of rubber is originally rectangular. Determine the average shear strain Jxy if the corners B and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines Given:
a 300mm
b 400mm
'Ax 0mm 'Bx 0mm
'Ay 0mm 'By 2mm
'Dx 3mm
'Dy 0mm
Solution:
§ 'By ·
'T AB atan ¨
© a ¹
'T AB
0.38197 deg
'T AB
6.6666 u 10
3
rad
§ 'Dx ·
'T AD atan ¨
© b ¹
'T AD
0.42971 deg
'T AD
7.4999 u 10
3
rad
Shear Strain : J xy_A 'T AB 'T AD J xy_A
14.166 u 10
3
rad
Ans
Problem 2-26 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. Given:
a 300mm
b 400mm
'Ax 0mm 'Bx 0mm
'Ay 0mm 'By 2mm
'Dx 3mm
'Dy 0mm
Solution: Geometry : 2
L DB
2
a b
L AD b
L DB
500 mm
L AD
400 mm
L D'B'
a 'Bx 'Dx 2 b 'Dy 'By 2
L AD'
'Dx b 'Dy
2
2
L AD'
L D'B'
400.01 mm
Average Normal Strain : H BD
LD'B' L DB LDB
H BD
6.797 u 10
H AD
L AD' L AD L AD
H AD
28.125 u 10
3 mm
mm 6 mm
mm
Ans
Ans
496.6 mm
Problem 2-27 The material distorts into the dashed position shown. Determine (a) the average normal strains Hx and Hy , the shear strain Jxy at A, and (b) the average normal strain along line BE. Given:
a 80mm
E x 80mm
b 125mm
Bx 0mm By 100mm
'Ax 0mm 'Ay 0mm
'Cx 10mm 'Cy 0mm
'Dx 15mm 'Dy 0mm
E y 50mm
Solution: 'xAC 'Cx 'Ax 'yAC 'Cy 'Ay § 'Cx · 'T AC atan ¨ b
©
¹
'xAC 'yAC
10.00 mm
'T AC
79.8300 u 10
0.00 mm 3
rad
Since there is no deformation occuring along the y- and x-axis, H x_A 'yAC
H x_A 2
0
Ans
2
'xAC b b
H y_A
b
H y_A
Ans
0.00319
J xy_A 'T AC J xy_A
79.830 u 10
3
rad
Ans
Geometry : 'Bx 'Cx 'Ex 'Dx
=
'Bx
§ By · ¨ 'C ©b¹ x
'Bx
8 mm
'By 0mm =
L BE L B'E' H BE
By b Ey b
'Ex
§ Ey · ¨ 'D ©b¹ x
'Ex
6 mm
'Ey 0mm
Ex Bx 2 Ey By 2
L BE
a 'Ex 'Bx 2 Ey 'Ey 'By 2 L B'E' L BE L BE
H BE
L B'E'
3 mm
17.913 u 10
94.34 mm
mm
92.65 mm
Ans
Note: Negative sign indicates shortening of BE.
Problem 2-28 The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF. Given: a 80mm
E x 80mm
b 125mm
Bx 0mm By 100mm
'Ax 0mm 'Ay 0mm
'Cx 10mm 'Cy 0mm
'Dx 15mm 'Dy 0mm
E y 50mm
Solution: 'xAC 'Cx 'Ax 'yAC 'Cy 'Ay § 'Cx · 'T AC atan ¨ b
©
¹
'xAC 'yAC
10.00 mm
'T AC
79.8300 u 10
0.00 mm 3
rad
Geometry : L AD L CF L A'D' L A'D' L C'F L C'F
2
a b 2
a b
2
2
L AD
148.41 mm
L CF
148.41 mm
a 'Dx 'Ax 2 b 'Dy 'Ay 2 157.00 mm
a 'Cx 2 b 'Cy 2 143.27 mm
Average Normal Strain : H AD
L A'D' L AD LAD
H AD
57.914 u 10
H CF
LC'F L CF LCF
H CF
34.653 u 10
3 mm
mm
3 mm
mm
Ans
Ans
Problem 2-29 The block is deformed into the position shown by the dashed lines. Determine the shear strain at corners C and D. Given:
a 100mm
'Ax 15mm 'Bx 15mm
b 100mm
L CA' 110mm Solution: Geometry :
§ 'Ax ·
'T C asin ¨
© LCA' ¹ § 'Bx ·
'T D asin ¨
©
LCA'
¹
'T C
7.84 deg
'T C
0.1368 rad
'T D
7.84 deg
'T D
0.1368 rad
Shear Strain : J xy_C 'T C J xy_C
3
136.790 u 10
rad
Ans
J xy_D 'T D J xy_D
136.790 u 10
3
rad
Ans
Problem 2-30 The bar is originally 30 mm long when it is flat. If it is subjected to a shear strain defined by Jxy = 0.0 x, where x is in millimeters, determine the displacement 'y at the end of its bottom edge. It is distorte into the shape shown, where no elongation of the bar occurs in the x direction. Given:
L 300mm
J xy = 0.02 x unit 1mm
Solution: dy = tan J xy dx dy = tan ( 0.02 x) dx
'y
´ µ ¶0
L
´ 1 dy = µ tan ( 0.02 x) ( unit) dx ¶0 30
´ 'y µ ¶0 'y
tan ( 0.02x) ( unit) dx
9.60 mm
Ans
Problem 2-31 The curved pipe has an original radius of 0.6 m. If it is heated nonuniformly, so that the normal strain along its length is H = 0.05 cos T, determine the increase in length of the pipe. Given:
r 0.6m
H = 0.05 cos T
Solution: ´ 'L = µ H dL ¶ 90deg
´ 'L = µ ¶0
90deg
´ 'L µ ¶0 'L
0.05 cos T d rT
0.05 r cos T dT
30.00 mm
Ans
Problem 2-32 Solve Prob. 2-31 if H = 0.08 sin T . Given:
r 0.6m
H = 0.08 sin T
Solution: ´ 'L = µ H dL ¶ 90deg
´ 'L = µ ¶0
90deg
´ 'L µ ¶0 'L
0.08 sin T d rT 0.08 r sin T dT
0.0480 m
Ans
Problem 2-33 A thin wire is wrapped along a surface having the form y = 0.02 x2, where x and y are in mm. Origina the end B is at x = 250 mm. If the wire undergoes a normal strain along its length of H = 0.0002x, 2
§ dy · dx . determine the change in length of the wire. Hint: For the curve, y = f (x), ds = 1 ¨ © dx ¹ Given:
Bx 250mm
H = 0.0002 x
unit 1mm
Solution: y = 0.02x
2
dy = 0.04x dx 2
´ 'L = µ H ds ¶
§ dy · dx ¨ © dx ¹
ds =
1
ds =
1 ( 0.04x) dx
2
B ´ x 2 'L = µ ( 0.0002x) 1 ( 0.04x) dx ¶0 250
´ 'L ( unit) µ ¶0 'L
42.252 mm
2
( 0.0002x) 1 ( 0.04x) dx Ans
Problem 2-34 The fiber AB has a length L and orientation 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'. Solution: Geometry: L A'B' =
L cos T uA 2 L sin T vB 2
L A'B' =
L uA vB 2L vB sin T uA cos T 2
2
2
Average Normal Strain: H AB =
L A'B' L L 2
H AB =
1
2
uA vB L
2
2 vB sin T uA cos T 1 L
Neglecting higher-order terms uA2 and vB2 , H AB =
1
2 vB sin T uA cos T 1 L
Using the binomial theorem: H AB = 1 H AB =
1 ª 2 vB sin T uA cos T « 2¬ L
vB sin T L
uA cos T L
º» .. 1
Ans
¼
Problem 2-35 If the normal strain is defined in reference to the final length, that is,
H'
n
=
lim
po
§ 's' 's · ¨ p' © ' s' ¹
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, Hn - H'n = Hn H'n. Solution:
Hn =
' S' ' S 'S
§ 'S' 'S · § 'S' 'S · ¨ © 'S ¹ © 'S' ¹
H n H'n = ¨
2
'S' 2 'S 'S' ' S H n H'n = 'S 'S' H n H'n =
2
'S' 'S 2 'S 'S' § 'S' 'S · § 'S' 'S · ¨ © 'S ¹ © 'S' ¹
H n H'n = ¨
H n H'n = H n H'n
(Q.E.D.)
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