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An air -filled rub ber ball mm. An air-filled air-fil air-filled led rubber rubber ball has hasaadiameter diameterofof150 diameter 6 in. If If the pressurewithin withinit itisisincreased increased until until the the ball’s ball’s the airair pressure mm,determine determinethe theaverage average normal diameter becomes becomes175 becomes 7 in., the normal strain in the rubber. rubber. 2–1.
d0 ==150 mm 6 in. d ==175 mm 7 in. e5 e =
pd
– pdp0d 175 –-150
d -
p
5 0
pd0
d0
p
=
7 6 = 0.167 mm/mm = 0.167 in./in. 150 6
Ans.
A thin strip of rubber has an unstretched unstretched length length of of 15 If it is aroundaround around a pipe having outer an outer diameter 375in.mm. Ifstretched it is stretched a pipe an having outer of diameter 5 in., determine of 125 mm, the average determine normal the average strain innormal the strip. strain in the strip. 2–2.
L0 = 15 in. L0 = 375 mm L =
(5 in.)
p
L = p(125 mm)
e e
5 =
L –-LL0 125 5pp – 15 - 375 L 0 5=
LL00
15 375
= 0.0472 mm/mm in.> in. = 0.0472
Ans.
The rigid beam is supported by a pin at A and wires and CE. If the the load P on the beam causes the end C to be displaced 10 mm downward, downward, determine the normal strain developed in wires CE and BD.
2–3. BD
D
E
4m
P
¢ LBD
3
=
¢ LBD =
eCE
=
eBD
=
¢ LCE
A
7 3 (10) 7
¢ LCE
L ¢ LBD
L
=
4.286 mm
=
10 4000
=
4.286 4000
=
=
B
3m
0.00250 mm> mm
Ans.
0.00107 mm> mm
Ans.
73
C
2m
2m
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The two wires are connected together at A. If th the e force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire. *2–4.
œ
LAC =
2 300 3002
+
22
-
2(300)(2) cos 150°
3 0 0 m m
301.734 mm
�
30
œ
eAC = eAB =
=
C
LAC - LAC LAC
=
301.734 - 300 300
=
0.00578 mm> mm
Ans.
�
30
A
P
m 0 m 0 3 B
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. 2–5.
•
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 LCE = 2000 mm. =
A eavg B
BD =
A eavg B
CE =
dB
LBD dC
LCE
LBD =
1500 mm and
=
4 1500
=
0.00267 mm> mm
Ans.
=
10 2000
=
0.005 mm> mm
Ans.
74
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2–6. Nylon strips are fused fused to glass glass plates plates.. Whe When n moderately heated the nylon will become soft while the glass stays approxima approximately tely rigid. rigid. Determ Determine ine 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 b 2 10
=
11.31°
=
0.197 rad
x
Ans.
2–7. If the theunstretched unstretched length the bowstring is 2–7. length of theofbowstring is 35.5 in., 887.5 mm, determine the average normal strain in the s tring determine the average normal strain in the string when it is when it is to stretched to theshown. position shown. stretched the position 18 mm in. 450
6 in. 150 mm 18 mm in. 450
Geometry: Referring to Fig. a, the stretched length of of the string is L =
2L ¿
=
22 418 58 0 2 2 1
62 2 =37.947 948.68in. mm
+ + 150 =
Average Normal Strain: eavg =
L - L0 L0
=
>
37.947 –-887.5 35.5 948.68 0.0689mm/mm in. in. == 0.0689 35.5 887.5
450 mm
450 mm
150 mm
75
Ans.
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Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a forc force e 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. *2–8.
u
D
P
300 mm
B
AB =
4002 2 400
AB ¿ =
400 2 400
2
+
3002
=
500 mm
+
2
-
2(400)(300) cos 90.3°
300
300 mm A
C
501.255 mm
=
AB ¿ - AB
eAB =
AB
=
501.255 - 500 500
=
400 mm
0.00251 mm> mm
Ans.
2–9. Part of a control linkage for an airplane consists of a rig rigid id membe memberr CBD and a flexible cable AB. If a force force is applied to the end D of the member and causes a normal strain strai n in the cabl cable e of 0.0035 mm> mm, dete determin rmine e the displacement of point D. Originally the cable is unstretched. unstretched.
u
•
300 mm
4002
+
=
500 mm 300 mm
AB ¿ = AB + eABAB
501.752 a = u =
=
500
=
3002
+
0.0035(500)
+
4002
-
=
A
501.75 mm
C
2(300)(400) cos a
90.4185°
400 mm
90.4185°
-
90°
600(u)
=
600(
¢D =
P
B
2 300 3002
AB =
D
=
0.4185° p
180°
=
p
180°
)(0.4185)
=
(0.4185) rad
4.38 mm
Ans.
76
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The corners B and D of the square plate are given the displacements indicated. Determine the shear strains at A and B. 2–10.
y A
16 mm
D
B
3 mm 3 mm 16 mm
16 mm
Applying trigonometry to Fig. a f
=
tan
a
=
tan
-
-
1
a 1316 b
1
a 1613 b
=
39.09°
rad a 180° b
=
50.91°
rad a 180° b
p
=
0.6823 rad
=
0.8885 rad
p
By the definition of shear strain,
A gxy B A A gxy B B
p =
2
-
2f
-
2a
p =
2
p =
2
-
p =
2
-
2(0.6823) 2(0.8885)
=
=
0.206 rad -
Ans.
0.206 rad
Ans.
77
C
16 mm
x
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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. 2–11.
y A
16 mm
D
B
3 mm 3 mm 16 mm
16 mm
Referring to Fig. a, LAB =
162 2 16
+
162
=
512 mm 2 512
LAB¿ =
162 2 16
+
132
=
425 mm 2 425
LBD =
16
LB¿D¿ =
13
+ +
16
=
13
32 mm
=
26 mm
Thus,
A
eavg
B
AB
=
A
eavg
B
BD
=
LAB¿ - LAB LAB LB¿D¿ - LBD LBD
=
=
425 2 425
-
512 2 512
512 2 512 26
-
32
32
= - 0.0889
= - 0.1875
mm> mm
Ans.
mm> mm
Ans.
78
C
16 mm
x
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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 l ines. *2–12.
u1
=
tan u1
=
u2
=
tan u2
=
gxy
+
2 300 3 400
=
u1
=
0.006667
=
0.006667 rad
=
0.0075 rad
y 3 mm
400 mm
u2
A
0.0075
+
0.0142 rad
=
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
=
2 (400) (400)2
=
tan
AB ¿
-1
a b 3 400
2 (300) (300)2
=
w
=
tan
-1
a
=
90°
-
(3)2
+
=
+
a b 2 300
0.42971°
-
2 mm
(400.01125)2 2 (400.01125)
D¿B¿
=
496.6014 mm
eDB
=
eAD
=
(300)2 2 (300)
+
3 mm
C D
400.01125 mm
=
400 mm
300.00667
0.381966°
=
=
x
y
A
0.381966°
D¿B¿
DB
B
0.42971°
(2)2
=
=
300 mm
Ans.
•
AD ¿
C D
+
(400)2
B 2 mm
=
89.18832°
(300.00667)2
=
300 mm
-
2(400.01125)(300.00667) cos (89.18832°)
500 mm
>
496.6014 - 500 = - 0.00680 mm mm 500 400.01125 - 400 -3 = 0.0281(10 ) mm mm 400
Ans.
>
Ans.
79
x
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Two bars are used to support a load. loa d. When unloaded, AB isis125 C is AB AC A has 5 in. in. mm long, lon long, g, A is 200 is 8 in.mm long, lon long, g, and and the thering ringatat A has A, the normal coordinates coordinates (0, (0,0). 0). If a load P acts acts on the ring at A normal PAB AB becomes strain in AB becomes e AB ==0.02 0.02mm/mm, in..> in in in.., and the normal normal AC A C becomes in AC becomese AC Determine the the PAC strain in becomes = =0.035 0.035 mm/mm. in..> in in in.. Determine coordinate position coordinate positionof ofthe thering ringdue dueto tothe theload. load. 2–14.
y
B
60
C C
�
1255minm.
280i0n.mm
A
x
P
Average Normal Strain: Strain: œ LAB
=
LAB
+ eAB LAB =
125 (0.02)(125) = 127.5 = 5.10 5 + +(0.02)(5) in. mm
œ LAC
=
LAC
+ eACLAC =
(0.035)(200) = 207.0 + + = 8.28 8200 (0.035)(8) in. mm
Geometry:
2 82 a =
=
a
2 2 2 -20 =.26.7268 4.3301 in. 0 − 108 5 = 168.17
mm
2 2 2 2 2 2 = 9.2268 + 8.28 - 2(9.2268)(8.28) 5.10127.52 cos u = 230.67 + 207.02 – 2(230.67)(207.0)
33.317° u = 33.317°
u =
x¿
=
= 6.9191 8.28 33.317° in. mm x9 = cos 207.0 cos 33.317° = 172.98
y¿
=
8.28 33.317° in. mm = 4.5480 y9 = sin 207.0 sin 33.317° = 113.70
x
x ¿ ( x a)a) = -x( = -9 – = - (6.9191 - 6.7268) = = –(172.98 – 168.17) = 0.192 –4.81 in. mm
y
Ans.
y ¿ ( y 4.3301) = -9 – y( = 108.25) = - (4.5480 - 4.3301) = = –(113.70 – 108.25) =0.218 –5.45in. mm
Ans.
230.67 mm 62.5 mm
108.25 mm 125 mm
127.5 mm
207.0 mm
200 mm
80
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T Two wo bars bars are used to support a load P. W When hen AC C is 8 AC A ABis unloaded, AB is5 125 in. long long, mm, A long, in. long long, is 200 , and mm thelong, the ring at and has the coordinates coordinat ring at A has es (0,coordinates 0). If a load (0, is applied 0). If a to load theisring the applied at A, to so that the ring it moves at A, soitthat to itthe moves coordinate it to the coordinate position (0.25 position in., in.), –18.25 determine normal strain normal in eachstrain bar. in each bar. - 0.73mm, (6.25 mm),the determine the normal bar. 2–15.
y y
B
C
60
�
1255min m.
inm . m 2800
A A
x x
P P
Geometry: a = a = = 168.17 2 8220-024.3301 in. mm − 1082. 2=52 6.7268 L L A = B = A¿9B
2 2 2 (6(2.5 .5 ++6.2 5) +2 (+ 10(4.3301 8. 25 + 18+. 20.73) 5) 2 (22.5 0.25)
= 143.98 5.7591mm in. 2 2 2 16(6.7268 .17 − 6.-250.25) ) + (210+8.(4.3301 25 + 18. 2+5) 0.73) L L A (86.7268 = 205. 48 mm = (2 A9¿C C =
==205.48 8.2191mm in. Average Normal Normal Strain: eAB
=
=
eAC
=
=
LA¿B - LAB LAB
5.7591 –-125 5 143.98 = in.> in. = 0.152 mm/mm 5 125
Ans.
LA¿C - LAC LAC
8.2191 –-200 8 205.48 = = 0.0274 in. mm/mm > in. 8 200
Ans.
230.67 mm 62.5 mm
108.25 mm
108.25 mm 125 mm
200 mm 18.25 mm 62.5 mm
6.25 mm
81
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*2–16. The square deforms into the position shown by the dashed lines. lines. Determine the average normal strain along each diagonal, AB and CD. Side D ¿ B ¿ remain remainss horiz horizontal ontal..
y 3 mm ¿
D
B
D
Geometry:
2 50 502
AB = CD = C¿ D¿ =
¿
B
2 53 532
+
582
+
502
70.7107 mm
=
53 mm
2(53)(58) cos 91.5°
-
50 mm �
91.5
= B¿ D¿ = AB¿¿ = AB =
79.5860 mm 50
+
53 sin 1.5°
532 2 53
48.38742
+
3
-
C
48.3874 mm
=
A 50 mm
2(53)(48.3874) cos 88.5°
-
x
¿
C 8 mm
70.8243 mm
Average Normal Strain: Strain:
eAB =
=
eCD =
=
AB¿¿ - AB AB AB
70.8243 - 70.7107 70.7107
=
1.61 A 10 - 3 B mm> mm
Ans.
=
126 A 10 - 3 B mm> mm
Ans.
C ¿ D ¿ - CD CD
79.5860 - 70.7107 70.7107
2–17. The three cords are attached to the ring at B.When a force 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 Provided ed these strain strainss are small, deter determine mine the CB is PCB. Provid normal strain in DB. No Note te th that at AB and CB remain horizontal and vertical, respectively respectively,, due to the roller guides at A and C .
A¿¿ A
•
A
B
L
Coordinates of B (L cos u, L sin u) Coordinates of B ¿ (L cos u LDB DB¿¿ =
2 (L cos u
u
+ eAB L cos u, L sin u + eCB L sin u)
2 + eAB L cos u) +
cos2 u(1 + LDB DB¿¿ = L 2 cos
2eAB
(L sin u
2 + eAB) +
sin2 u(1
C ¿ C ¿ D
2 + eCB L sin u)
+
2eCB
2 + eCB)
Since eAB and eCB are small, LDB DB¿¿ = L 2 1 +
(2 eAB cos2 u
+
2eCB sin2 u)
Use the binomial theorem, 1 (2 eAB cos2 u 2
LDB DB¿¿ = L (
1
+
= L (
1
2 + eAB cos u + eCB
Thus, eDB
=
L(
1
+
2eCB sin2 u)) sin2 u)
2 2 + eAB cos u + eCB sin u) - L
L 2
eDB = eAB cos u + eCB
B¿
2
sin
Ans.
u
82
C
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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 l ines. 2–18.
y
5 mm
2 mm 2 mm
B
4 mm
C
300 mm
Geometry: For small angles, a
b
=
=
c
u
=
=
2 mm D
2 302
=
2 403
=
400 mm
0.00662252 rad
A
x
3 mm
0.00496278 rad
Shear Strain:
(gB)xy
(gA)xy
=
a
=
0.0116 rad
+
b =
= -
(u
= -
0.0116 rad
+
11.6 A 10 - 3 B rad
Ans.
c) = -
11.6 A 10 - 3 B rad
Ans.
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 l ines. 2–19.
y
5 mm
2 mm 2 mm
B
4 mm
C
300 mm 2 mm D
400 mm
A
3 mm
Geometry: For small angles,
2 403 2 b = u = 302 Shear Strain: a
=
c
(gC)xy
(gD)xy
=
=
0.00496278 rad
=
0.00662252 rad
= -
(a
= -
0.0116 rad
+
b)
=
u
=
0.0116 rad
+
= -
11.6 A 10 - 3 B rad
Ans.
c =
11.6 A 10 - 3 B rad
Ans.
83
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 =
2 400 4002
DB ¿ =
2 405 4052
A¿ C¿ =
2 401 401
2
+
3042
+
3002
+
300
500 mm
300 mm
506.4 mm
=
2
=
=
2 mm D
500.8 mm
400 mm
x
A
3 mm
Average Normal Strain: Strain: eAC
A ¿ C ¿ - AC
=
AC
0.00160 mm> mm
=
eDB
DB ¿ - DB
= =
500.8 - 500 500
=
DB
=
=
1.60 A 10 - 3 B mm> mm
Ans.
506.4 - 500 500
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° ab abou outt pin pin A. Deter Determine mine the the average average normal normal strain developed develo ped in the wire. Origin Originally ally,, the wire is unstretched. unstretched. •
D
600 mm
Geometry: Referring to Fig. a, the stretched length of LB¿D can be determined using the consine law, LB¿D =
(0.6 cos 45°)2 2 (0.6
=
+
(0.6 sin 45°)2
-
=
LB¿D - LBD LBD
C
�
45
B
0.6155 m
Average Normal Strain: The unstretched length of wire BD is obtain eavg
A
2(0.6 cos 45°)(0.6 sin 45°) cos 93°
=
0.6155 - 0.6 0.6
=
LBD =
0.0258 m> m
0.6 m . We
Ans.
84
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A square piece of material is deformed into the dashed position. Determine the shear strain gxy at A. 2–22.
y
15.18 mm B
Shear Strain:
(gA)xy
=
p
2 =
-
¢ ≤ 89.7° 180°
p
C
15.24 mm
15 mm
5.24 A 10 - 3 B rad
Ans.
89.7
�
A
A square piece of material is deformed into the dashed parallelogram. Determine the average normal strain AC C and BD. that occurs along the diagonals A 2–23.
15 mm 15.18 mm
x
D
y
15.18 mm B
C
15.24 mm
15 mm 89.7
�
A
Geometry: AC
BD
=
AC ¿
B¿D¿
=
152 2 15
=
15.182 2 15.18
=
21.5665 mm
+
+
152
15.242
=
2 15.18 15.182
=
21.4538 mm
= -
15.242
+
-
21.2132 mm 2(15.18)(15.24) cos 90.3°
2(15.18)(15.24) cos 89.7°
Average Normal Normal Strain: eAC =
= eBD =
=
AC ¿ - AC AC
=
21.5665 - 21.2132 21.2132
0.01665 mm> mm B ¿ D ¿ - BD BD
=
=
16.7 A 10 - 3 B mm> mm
Ans.
21.4538 - 21.2132 21.2132
0.01134 mm> mm
=
11.3 A 10 - 3 B mm> mm
Ans.
85
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
=
p
-
2 =
15 mm 15.18 mm
x
D
¢ ≤
89.7° p 180°
5.24 A 10 - 3 B rad
Ans.
2–25. The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns columns of the frame tilt u = 2°. Determine the approximate normal strain in the wire when the frame is in this position. position. Assume the columns are rigid and rotate about their l ower supports.
u
•
2
�
u
B
Geometry: The vertical displacement is negligible
x
¢ ≤ ¢ ≤
=
=
2° (4) p 180°
xA
xB
3m
2° (1) p 180°
=
4
A¿B¿ AB
+
=
=
-
xB
2 32
2 32
+
0.03491 m A
xA +
=
=
0.13963 m
=
4.10472 m
4.104722
42
=
=
1m
5.08416 m
5.00 m
Average Normal Strain: Strain: eAB
=
=
A ¿B¿
-
AB
AB
5.08416 5
-
5
=
16.8 A 10 - 3 B m> m
Ans.
86
4m
2
�
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The material distorts into the dashed position shown. Determ shown. Determine ine (a) the average normal normal strains along along sides A (b) the the AC C and CD and the shear strain gxy at F , and (b) average normal strain along line BE. 2–26.
y 15 mm
C
25 mm
D
10 mm
B E
75 mm
90 mm
A
Referring to Fig. a, LBE
=
(90 - 75)2 2 (90
LAC¿
=
2 100 1002
LC¿D¿
=
80
f
=
tan-1
-
+
15
¢ ≤ 25 100
152
=
25
+
=
802
+
=
=
6625 mm 2 6625
2 10225 10225 mm 90 mm
14.04°
¢ ≤ p rad
=
180°
0.2450 rad.
When the plate deforms, deforms, the vertical position of point B and E do not change. LBB¿
90 LEE¿
75 LB¿E¿
=
15 ; 100
LBB¿
=
13.5 mm
=
25 ; 100
LEE¿
=
18.75 mm
=
2 (90 (90 - 75)2
+
(80
- 13.5 + 18.75)
2
=
2 7492.5625 7492.5625 mm
Thus,
A eavg B AC =
LAC¿
A eavg B CD =
LC¿D¿
A eavg B BE =
LB¿E¿
-
LAC
LAC LCD
-
LCD -
LBE
LBE
=
10225 2 10225
-
100
100
=
0.0112 mm> mm
=
90 - 80 80
=
7492.5625 - 2 6625 6625 2 7492.5625 2 6625 6625
=
0.125 mm> mm
=
Ans.
Ans.
0.0635 mm> mm
Ans.
Referring to Fig Referring Fig.. a, the angle angle at corner corner F becomes becomes larger larger than than 90° after the the plate plate deforms.Thus, the shear strain is negative. negative. 0.245 rad
Ans.
87
80 mm
F
x
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The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF . 2–27.
y 15 mm
25 mm
C 10 mm
The undeformed length of diagonals AD and CF are LAD = LCF =
2 80 802 + 1002
=
D
B E
2 16400 16400 mm
The deformed length of diagonals AD and CF are
75 mm
90 mm
LAD AD¿¿ =
2
2 (80 (80 + 25)
LC¿F =
(80 - 15)2 2 (80
+
2
100
=
2 21025 21025 mm
+
1002
=
14225 mm 2 14225 A
Thus,
A eavg B
AD
=
A eavg B
CF
=
LAD AD¿¿ - LAD LAD LC¿F - LCF LCF
2 21025 21025 - 2 16400 16400
=
=
2 16400 16400 2 14225 14225
-
2 16400 16400
2 16400 16400
=
>
0.132 mm mm
= - 0.0687
>
mm mm
80 mm
Ans.
P
xe
dx
L
¢L =
L c 12 d � c 12 xe
-x
2
dx
0
L
= -
e
2 -x
= -
0
=
1 [1 2
- e
e
2 -L
-
1 2
x2
�
x x L
2 -x
x
Ans.
*2–28. The wire is subjected to a normal strain that is 2 defined by P = xe - x , where x is in millimeters. millimeters. If the wire has an initial length L, determine the increase in its length.
dL = e dx = x e
F
d
2 -L
]
Ans.
88
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The pip m.If If it is Thecurved curvedpipe pipeehas pipe hasan anoriginal originalradius radiusofof0.6 2 ft. heated nonuniformly, nonuniformly, so that the normal strain along its length is P = 0.05 cos u, determine the increase in length of the pipe.
•2–29. •2–29.
0.05 cos u
e =
L L
¢L =
e
=
2 ft 0.6 m
dL
A
u
90°
(0.6 dud)u) (0.05 cos u)(2
0
L
90°90°
90° 90°
cos 0.100mft ud u] 0u� ]] = = =[0.1[sin = 0.03 cos 0.03[–sin uudu
0.1 0.03
=
Solve Prob. Prob. 2–29 if P
2–30.
Ans.
0
00
=
0.08 sin u.
dL == 0.6 2 dduue===== = 0.08 sin u u = 0.08esin
L L
¢L =
e
=
dL
90°
2 ft 0.6 m
dud)u) (0.08 sin u)(2 (0.6
0
90° 90°
90° 90°
L
sin sinu udd = 0.16[ 0.048[–cos ] ]0� = =0.16 0.048ft m u u= - cos uu
0.048 0.16
=
Ans.
0
00
The rubber band AB has an unstretched length of If it it is is fixed fixed surface at point point A A ¿9, 1 m. ft. If fix ed at B and and attached attached to the surface determine the average normal strain in the band.The surface y = is defined defined by the funct function function ion y (x22))m, ft, where x is is in in meters. feet.. feet = ( x 2–31.
y y
¿
1m ft
=
L A a b dy dx
1
+
=
x2 then
0
However y
L 2 1 1 C 2 2 1 4
2
dx
dy
=
dx
ft 1m
2x
B
1m ft
L
=
+
4 x2 dx
+
4 x2
0
=
=
x
+
ln A 2x
+
2 1
+
1m ft
4x2 B D �0
1.47894 ft m
Average Normal Normal Strain: eavg =
L
-
L0
2 x
�
A¿
Geometry: L
A
u
L0
=
1.47894 1
-
1
=
>
ft ft 0.479 m/m
Ans.
89
ft 1m
A
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 shape shown, where no elongation of the bar occurs in the x direction.
y
� y
x 300 mm
Shear Strain: dy dx
dy
tan gxy ;
=
tan (0.02 x)
=
dx
L
¢y
dy
=
0
L
300 mm
tan (0.02 x)dx
0
¢ y = - 50[ln =
mm cos (0.02x)]|300 0
2.03 mm
Ans.
2–33. 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 respectively the fiber when it is in position A ¿ B ¿ .
y
•
¿
B
vB
B L
Geometry: LA¿B¿
=
2 (L cos 2 L 3
=
u - uA)2 + (L sin u + yB)2 +
u2A
A
y2B + 2L(yB sin u - uA cos u)
+
Average Normal Strain: Strain: LA¿B¿ - L eAB =
L =
A
1
+
u2A
+ 2
y2B
+
L
2(yB sin u
-
uA cos u)
L
-
1
Neglecting higher terms u2A and y2B eAB =
B
1
+
2(yB sin u
-
uA cos u)
L
R
1 2
-
1
Using the binomial theorem: eAB = 1 +
=
¢
1 2yB sin u 2 L
yB sin u
L
-
-
2uA cos u L
≤
+
...
-
1
uA cos u
Ans.
L
90
¿
u A A
u
x
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If the normal strain is defined in reference to the final length, that is, is, 2–34.
œ
lim
Pn =
p : p¿
a
¢ s ¿ - ¢s ¢s¿
b
instead of in reference instead reference to the original original length, length, Eq. 2–2, show that the difference in these strains is represented as a second-order term, term, namely namely,, Pn - Pnœ = PnPnœ .
eB
eB
=
-
¢S ¿ - ¢ S ¢S œ
eA
=
=
=
=
=
¢ S ¿ - ¢S ¢S ¢S ¿
2
¢S ¿
2
¢S¿ - ¢S
-
¢ S¿
2 - ¢ S ¢ S ¿ - ¢S ¿ ¢ S + ¢S
¢S ¢S¿ 2
+ ¢S
-
2 ¢S ¿¢ S
¢S ¢S¿
( ¢S ¿
2 - ¢ S)
¢S ¢S ¿ œ
eA eB
¢
=
¢ S¿ - ¢S ¢S
≤¢
¢ S ¿ - ¢S ¢S ¿
≤
(Q.E.D)
91
