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Answers to Problems
CHAPTER 22.1 (b) SijSij ¼ 28, (c) SjiSji ¼ SijSij ¼ 28, (d) SjkSkj ¼ 23, (g) Snmaman ¼ Smnaman ¼ 59.
2.3 (a) b1 ¼ 2; b2 ¼ 2; b3 ¼ 2. (b) s ¼ 6.
2.4 (c) Eij ¼ BmiCmkFkj.
2.7 i ¼ 1 ! a1 ¼ @v1@t
þ v1@v1@x1
þ v2@v1@x2
þ v3@v1@x3
; etc:
2.10 d1 ¼ 6; d2 ¼ �3; d3 ¼ 2.
2.12 (2) For i ¼ k; LS ¼ RS ¼0 if j 6¼ l0 if j ¼ l ¼ i1 if j ¼ l 6¼ i
8<:
2.20 (b) T½ � ¼0 0 1
0 0 �1
�1 1 0
24
35.
2.21 (c) T aþ bð Þ ¼ 10e1.
2.22 T½ � ¼2 0 �1
0 1 3
1 3 0
24
35.
2.23 T½ � ¼�1=2 0 1=2�1=2 0 1=20 0 0
24
35.
2.24 (a) T½ � ¼1 0 0
0 �1 0
0 0 1
24
35, (b) T½ � ¼
1 0 0
0 1 0
0 0 �1
24
35.
2.25 (a) R½ � ¼1 0 0
0 cos y �sin y0 sin y cos y
24
35, (b) R½ � ¼
cos y 0 sin y0 1 0
�sin y 0 cos y
24
35.
2.26 (b) T½ � ¼ 1
3
1 �2 �2
�2 1 �2
�2 �2 1
24
35, (c) Ta ¼ � 3e1 þ 2e2 þ e3ð Þ.
2.27 ½T� ¼ 1
3
1 �2 �2
�2 1 �2
�2 �2 1
24
35.
Copyright © 2010, Elsevier Ltd. All rights reserved.
2.28 (b) n ¼ e1 þ e2 þ e3ð Þ=ffiffiffi3
p.
2.29 T½ � ¼ 1
3
1þ 2 cos y 1� cos yð Þ � ffiffiffi3
psin y 1� cos yð Þ þ ffiffiffi
3p
sin y1� cos yð Þ þ ffiffiffi
3p
sin y 1þ 2 cos yð Þ 1� cos yð Þ � ffiffiffi3
psin y
1� cos yð Þ � ffiffiffiffi3
psin y 1� cos yð Þ þ ffiffiffi
3p
sin y 1þ 2 cos yð Þ
24
35.
2.30 (b) RA ¼ sin yE.
2.31 (a) S½ � ¼1 0 0
0 �1=ffiffiffi2
p �1=ffiffiffi2
p0 �1=
ffiffiffi2
p1=
ffiffiffi2
p
24
35; (b) T½ � ¼
1 0 0
0 �1=ffiffiffi2
p1=
ffiffiffi2
p0 1=
ffiffiffi2
p1=
ffiffiffi2
p
24
35, (d) c½ � ¼
1
1=ffiffiffi2
p5=
ffiffiffi2
p
24
35.
2.37 a ¼ 2e 01.
2.38 (b) a ¼ e 01 þ
ffiffiffi3
pe 02.
2.39 T 011 ¼ 4=5; T 0
12 ¼ �15=ffiffiffi5
p; T 0
31 ¼ 2=5.
2.40 (a) T 0ij
h i¼ T½ � 0 ¼
0 �5 0
�5 1 5
0 5 1
24
35.
2.42 (b) TijTij ¼ 45, (c) T½ � 0 ¼2 5 1
2 3 1
0 0 1
24
35.
2.48 (a) TS� � ¼ 1 3 5
3 5 7
5 7 9
24
35, TA
� � ¼ 0 �1 �2
1 0 �1
2 1 0
24
35, (b) tA ¼ e1 � 2e2 þ e3.
2.50 (d) For l ¼ 1; n ¼ a1e1 þ a2e2 � a1 þ a2ð Þe3½ �=ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia21 þ a22 þ a23
q.
2.55 y ¼ 120o.
2.56 (c) For l ¼ 1; n ¼ �e3. (d) For l ¼ �1; n ¼ a1e1 þ a2e2; a21 þ a22 ¼ 1, (e) y ¼ p.
2.59 (a) For l1 ¼ 3; n1 ¼ �e3. For l2 ¼ �3; n2 ¼ � e1 � 2e2ð Þ=ffiffiffi5
p.
2.60 (a) For l1 ¼ 3; n1 ¼ �e1. For l2 ¼ 4; n2 ¼ � e2 þ e3ð Þ=ffiffiffi2
p.
2.61 For l1 ¼ 0; n1 ¼ � e1 � e2ð Þ=ffiffiffi2
p. For l2 ¼ l3 ¼ 2; n ¼ � ae1 þ ae2 þ a3e3ð Þ; 2a2 þ a23 ¼ 1.
2.65 (b) At ð0; 0; 0Þ; ðdf=drÞmax ¼ jrfj ¼ 2 in the direction of n ¼ e3.
At ð1; 0; 1Þ; ðdf=drÞmax ¼ jrfj ¼ 17 in the direction of n ¼ 2e1 þ 3e2 þ 2e3ð Þ=ffiffiffiffiffi17
p.
2.67 (a) q ¼ �3k e1 þ e2ð Þ; (b) q ¼ � 3ke1 þ 6ke2ð Þ.2.69 (a) ½rv�ð1;1;0Þ ¼ 2½I�, (b) rvð Þv ¼ 2e1, (c) div v ¼ 2; curl v ¼ 2e1, (d) dv ¼ 2ds e1 þ e3ð Þ.
2.71 ru½ � ¼�A=r2 �B 0
B A=r2 0
0 0 0
24
35.
2 Answers to Problems
2.72 div u ¼ 3A.
2.73 ru½ � ¼A� 2B=r3 0 0
0 Aþ B=r3 0
0 0 Aþ B=r3
24
35.
2.77 div Tð Þr ¼ div Tð Þy ¼ div Tð Þz ¼ 0.
CHAPTER 3
3.1 (b) v1 ¼ kx11þ kt
; v2 ¼ 0; v3 ¼ 0.
3.2 (a) v1 ¼ a; v2 ¼ v3 ¼ 0; a1 ¼ a2 ¼ a3 ¼ 0;(b) y ¼ A atþ X1ð Þ. Dy=Dt ¼ Aa, (c) y ¼ BX2; Dy=Dt ¼ 0.
3.3 (b) v1 ¼ 0; v2 ¼ 2bX21t; v3 ¼ 0 and a1 ¼ 0; a2 ¼ 2bX2
1; v3 ¼ 0,
(c) v1 ¼ 0; v2 ¼ 2bx21t; v3 ¼ 0 and a1 ¼ 0; a2 ¼ 2bx21; a3 ¼ 0.
3.4 (b) v1 ¼ 2bX22t; v2 ¼ kX2; v3 ¼ 0 and a1 ¼ 2bX2
2; a2 ¼ 0; a3 ¼ 0,
(c) v1 ¼ 2bx22t=ð1þ ktÞ2; v2 ¼ kx2=ð1þ ktÞ; v3 ¼ 0; a1 ¼ 2bx22= 1þ ktð Þ2; a2 ¼ a3 ¼ 0.
3.5 (b) v1 ¼ k sþ X1ð Þ; v2 ¼ 0; v3 ¼ 0 and a1 ¼ 0; a2 ¼ 0; a3 ¼ 0,
(c) v1 ¼ k sþ x1ð Þ= 1þ ktð Þ; v2 ¼ 0; v3 ¼ 0 and a1 ¼ 0; a2 ¼ 0; a3 ¼ 0.
3.6 (b) For X1;X2;X3ð Þ ¼ 1; 3; 1ð Þ and t ¼ 2; v1 ¼ �4 3ð Þ2 2ð Þ ¼ �72; v2 ¼ �1; v3 ¼ 0:(c) For x1; x2; x3ð Þ ¼ 1; 3; 1ð Þ and t ¼ 2; v1 ¼ �200; v2 ¼ �1; v3 ¼ 0:
3.7 (a) For X1;X2;X3ð Þ ¼ 1; 1; 0ð Þ and t ¼ 2; v1 ¼ 2k; v2 ¼ 2k; v3 ¼ 0:(b) For x1; x2; x3ð Þ ¼ 1; 1; 0ð Þ and t ¼ 2; v1 ¼ 2k= 1þ 4kð Þ; v3 ¼ 0:
3.8 (a) t ¼ 2 ! x1 ¼ 5; x2 ¼ 3; x3 ¼ 0, (b) X1 ¼ �3; X2 ¼ 1; X3 ¼ 2,
(c) a1 ¼ 18; a2 ¼ 0; a3 ¼ 0, (d) a1 ¼ 2; a2 ¼ 0; a3 ¼ 0.
3.9 (b) ai ¼ 0.
3.10 (a) a ¼ �4xex � 4yey, (b) x2 þ y2 ¼ constant ¼ X2 þ Y2.
Or, x ¼ �Y sin 2tþ X cos 2t and y ¼ Y cos 2tþ X sin 2t.
3.11 (a) a ¼ k2 xex þ yey� �
, (b) x ¼ Xekt; y ¼ Ye�kt. Or xy ¼ XY.
3.12 Material description: a ¼ 2k2 x2 þ y2� �
xex þ yey� �
.
3.14 (b) a1 ¼ 0; a2 ¼ �p2 sin ptð Þ sin p X1ð Þ; a3 ¼ 0.
3.15 (b) a ¼ �ða2ffiffiffi2
p=4Þer; DY=Dt ¼ 2ak.
3.16 (b) a ¼ �ða2ffiffiffi2
p=4Þer; DY=Dt ¼ 0.
3.17 (b) ds1=dS1 ¼ ð1=ffiffiffi2
pÞ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ kð Þ2 þ 1
q¼ ds2=dS2,
cos p=2� gð Þ ¼ sin g ¼ f� 1þ kð Þ2 þ 1g=f 1þ kð Þ2 þ 1g.
Answers to Problems 3
(c) For k ¼ 1; ds1=dS1 ¼ ds2=dS2 ¼ffiffiffiffiffiffiffiffi5=2
p; sin g ¼ �3=5.
For k ¼ 10�2; ds1=dS1 ¼ ds2=dS2 � 1:005; g ¼ �0:0099 rad: (d) 2E 012 ¼ �0:01.
3.19 (a) E½ � ¼0 0 k=20 k 0k=2 0 0
" #, (b) 10�5=2.
3.20 (a) E11 ¼ 5k ¼ 5� 10�4; E22 ¼ 2k ¼ 2� 10�4; 2E12 ¼ k ¼ 10�4rad.
3.21 (a) E 011 ¼ 10�4=3.
3.22 (a) E 011 ¼ ð58=9Þ � 10�4, (b) 2E 0
12 ¼ ð32=ffiffiffiffiffi45
pÞ � 10�4rad:
3.23 (a) E 011 ¼ ð37=25Þ � 10�4, (b) 2E 00
12 ¼ ð72=25Þ � 10�4rad:
3.24 (a) I1 ¼ 11� 10�4; I2 ¼ 31� 10�8; I3 ¼ 17� 10�12.
3.25 I1 ¼ 0; I2 ¼ �t2; I3 ¼ 0.
3.26 At 1; 0; 0ð Þ; lmax ¼ 3k ¼ 3� 10�6.
3.27 (a) D dVð Þ=dV ¼ 0, (b) k1 ¼ 2k2.
3.28 (b) At 1; 2; 1ð Þ; E 011 ¼ k; (c) max elongation ¼ 4k; (d) DV ¼ k.
3.32 E11 ¼ a; E22 ¼ c; E12 ¼ b� aþ cð Þ=2.3.33 (a) E12 ¼ �100� 10�6: (b) For l1 ¼ 261:8� 10�6; y ¼ �31:7o, or
n ¼ 0:851e1 � 0:525e2. For l2 ¼ 38:2� 10�6; y ¼ 58:3o, or n ¼ 0:525e1 þ 0:851e2.
3.34 (a) E12 ¼ 0, (b) Prin. strains are 10�3 in any direction lying on the plane of e1 and e2.
3.35 E11 ¼ a; E22 ¼ 2bþ 2c� að Þ=3; E12 ¼ b� cð Þ=ffiffiffi3
p.
3.36 E11 ¼ 2� 10�6; E22 ¼ 1� 10�6; E12 ¼ ½1=ð2ffiffiffi3
p� � 10�6.
3.37 E11 ¼ 2� 10�3; E22 ¼ 2� 10�3; E12 ¼ 0.
3.38 (a) D½ � ¼0 kx2 0kx2 0 00 0 0
24
35; W½ � ¼
0 kx2 0�kx2 0 00 0 0
24
35, (b) D nð Þ nð Þ ¼ 3k.
3.39 D11 ¼ �a 1þ kð Þ; D 011 ¼ 1þ kð Þ=2.
3.40 (a) D12 ¼ p cos t cos px1ð Þ=2; W12 ¼ �W21 ¼ � p cos t cos px1ð Þ=2.(b) D11 ¼ 0, D22 ¼ 0, D 0
11 ¼ p=2.
3.42 (a) Drr ¼ �1=r2; Dyy ¼ 1=r2; other Dij ¼ 0; W½ � ¼ 0½ �. (b) Drr ¼ �1=r2.
3.43 At r ¼ 2; ar ¼ �18; ay ¼ 0; (b) D½ � ¼ 0 �1
�1 0
� �.
3.44 (a) ar ¼ �ðAr þ B=r2Þ2sin2y=r; ay ¼ �cos y sin yðAr þ B=r2Þ2=r; af ¼ 0.
(b) Drf ¼ �ð3B=2r3Þsiny; Dyf ¼ 0.
4 Answers to Problems
3.45 W½ � ¼ 0:
3.49 k ¼ 1.
3.50 f ¼ g yð Þ=r.3.51 vy ¼ � k=2ð Þsin y=
ffiffir
p.
3.53 v1 ¼ f x2ð Þ; v2 ¼ 0:
3.54 (a) r ¼ ro 1þ ktð Þ�a=k; (b) r ¼ r*xo=x1.
3.55 r ¼ roe�at2 .
3.60 (b) 2kX1X2 ¼ f X2;X3ð Þ þ g X1;X3ð Þ.
3.62@r@t
þ vr@r@r
þ vyr
@r@y
þ vz
@r@z
� �þ r
@vr@r
þ vrrþ 1
r
@vy@y
þ @vz@z
¼ 0.
3.63 (b) U½ � ¼1 0 0
0 2 0
0 0 3
24
35; (c) B½ � ¼
9 0 0
0 1 0
0 0 4
24
35, (d) R½ � ¼
0 0 1
�1 0 0
0 �1 0
24
35,
(e) E*� � ¼ 0 0 0
0 3=2 0
0 0 4
24
35, (f) e*
� � ¼ 4=9 0 0
0 0 0
0 0 3=8
24
35, (g)
DVDVo
¼ 6, (h) dA ¼ �3e3.
3.64 (b) U½ � ¼1 0 0
0 2 0
0 0 3
24
35, (c) B½ � ¼
4 0 0
0 9 0
0 0 1
24
35, (d) R½ � ¼
0 1 0
0 0 1
1 0 0
24
35,
(e) E*� � ¼ 0 0 0
0 3=2 0
0 0 4
24
35, (f) e*
� � ¼ 3=8 0 0
0 4=9 0
0 0 0
24
35, (g)
DVDVo
¼ 6, (h) dA ¼ 3e1.
3.65 (b) U½ � ¼1 0 0
0 2 0
0 0 3
24
35, (c) B½ � ¼
1 0 0
0 9 0
0 0 4
24
35, (d) R½ � ¼
1 0 0
0 0 1
0 �1 0
24
35,
(e) E*� � ¼ 0 0 0
0 3=2 0
0 0 4
24
35, (f) e*
� � ¼ 0 0 0
0 4=9 0
0 0 3=8
24
35, (g)
DVDVo
¼ 6,
(h) dA ¼ �3e3.
3.66 (b) U½ � ¼1 0 0
0 2 0
0 0 3
24
35, (c) B½ � ¼
4 0 0
0 1 0
0 0 9
24
35, (d) R½ � ¼
0 1 0
�1 0 0
0 0 1
24
35,
(e) E*� � ¼ 0 0 0
0 3=2 0
0 0 4
24
35; (f) e*
� � ¼ 3=8 0 0
0 0 0
0 0 4=9
24
35, (g)
DVDVo
¼ 6,
(h) dA ¼ 3e1.
Answers to Problems 5
3.67 (a) C½ � ¼1 3 0
3 10 0
0 0 1
24
35: (b) For l1 ¼ 10:908326; n1 ¼ 0:289785e1 þ 0:957093e2.
For l2 ¼ 0:0916735; n2 ¼ 0:957093e1 � 0:289784e2: For l3 ¼ 1; n3 ¼ e3,
(c) U½ �ni ¼3:30277 0 0
0 0:302774 0
0 0 1
24
35, (d) U½ �ei ¼
0:554704 0:832057 0
0:832057 3:05087 0
0 0 1
24
35,
½U�1�ei ¼3:050852 �0:832052 0
�0:832052 0:554701 0
0 0 1
24
35, (d) R½ �ei ¼
0:55470 0:83205 0
�0:83205 0:55470 0
0 0 1
24
35.
3.70 (a) 3, 2 and 0.6, (b) ds=dSð Þ ¼ffiffiffiffiffiffiffiffiffiffi13=2
p, (c) cos y ¼ 0. No change in angle.
3.71 (a) U½ � ¼ 1ffiffiffi5
p2 1 0
1 3 0
0 0ffiffiffi5
p
24
35, (b)
ffiffiffiffiffiffiffiC22
p¼
ffiffiffi2
p, (c)
ds
dS¼
ffiffiffiffiffiffiffiffi5=2
p, (d) cos y ¼ 1ffiffiffi
2p :
3.72 (a) U½ � ¼ 1ffiffiffi2
p1 1 0
1 3 0
0 0ffiffiffi2
p
24
35, (b)
ffiffiffiffiffiffiffiC22
p¼
ffiffiffi5
p, (c)
ds
dS¼
ffiffiffi5
p, (d) cos y ¼ 2ffiffiffi
5p :
3.77 B�1rr ¼ @ro
@r
2
þ ro@yo@r
2
þ @zo@r
2
; B�1yy ¼ @ro
r@y
2
þ ro@yor@y
2
þ @zor@y
2
.
3.80 C�1royo ¼
ro@yo@r
@ro@r
þ ro@yo
r@y
@ror@y
þ ro@yo
@z
@ro@z
.
3.81 Bry ¼ @r
@X
r@y@X
þ @r
@Y
r@y@Y
þ @r
@Z
r@y@Z
.
3.82 B�1ry ¼ @X
@r
@X
r@y
þ @Y
@r
@Y
r@y
þ @Z
@r
@Z
r@y
.
3.84 (a) B½ � ¼1 0 0
0 1þ rkð Þ2 rk0 rk 1
24
35; (b) C½ � ¼
1 0 0
0 1 rk0 rk 1þ rkð Þ2
24
35.
3.85 (a) B½ � ¼a=rð Þ2 0 0
0 r=að Þ2 0
0 0 1
24
35, (b) det B ¼ 1, no change of volume.
3.86 C½ � ¼f ðXÞð Þ2 0 0
0 gðYÞð Þ2 0
0 0 hðZÞð Þ2
24
35.
6 Answers to Problems
CHAPTER 44.1 (a) 1 MPa; 4 MPa; 0 MPa: (b) 3:61MPa; 5:39 MPa; 5:83 MPa.
4.2 (a) t ¼ 1=3ð Þ 5e1 þ 6e2 þ 5e3ð Þ: (b) Tn ¼ 3 MPa; Ts ¼ 0:745 MPa:
4.3 (a) t ¼ 3:47e1 � 2:41e2: (b) Tn ¼ 2:21 MPa; Ts ¼ 3:60 MPa:
4.4 t ¼ 25ffiffiffi3
pe1 þ 25e2 � 25
ffiffiffi3
pe3.
4.5 (a) t ¼ e3. (b) n21 � n22 ¼ 0, including n ¼ e3; n ¼ e1 þ e2ð Þ=ffiffiffi2
p; n ¼ e1 � e2ð Þ=
ffiffiffi2
p.
4.6 T 011 ¼ �6:43 MPa; T 0
13 ¼ 18:6 MPa.
4.7 (a) te1 ¼ ax2e1 þ be2. (b) FR ¼ 0e1 þ 4 be2; Mo ¼ � 4a=3ð Þe3.4.8 (a) te1 ¼ ax22e1. (b) FR ¼ 4a=3ð Þe1; Mo ¼ 0.
4.9 (a) te1 ¼ ae1 þ ax3e2. (b) FR ¼ 4ae1; Mo ¼ � 4a=3ð Þe1.4.10 (a) tn1 ¼ 0, tn2 ¼ ax3e2 � ax2e3; tn3 ¼ �ax3e2 þ ax2e3: (b) FR ¼ 0; Mo ¼ 8pae1.
4.11 (b) FR ¼ 0; Mo ¼ �p= 2ffiffiffi2
p �.
4.12 (a) tr S ¼ 0. (b) S½ � ¼0 500 �200
500 �300 400
�200 400 300
24
35kPa:
4.13 (a) 4.
4.14 (b) T12 ¼ T21.
4.17 fmax ¼ 2.
4.21 (a) Ts ¼ 0: (b) For Tmax ¼ 100 MPa; n1 ¼ e1 þ e2ð Þ=ffiffiffi2
p. For Tmin ¼ �100 MPa;
n2 ¼ e1 � e2ð Þ=ffiffiffi2
p: (c) Tsð Þmax ¼ 100 MPa, on the planes e1 and e2.
4.23 (a) Tsð Þmax ¼ 150 MPa; n ¼ ðe1 � e3Þ=ffiffiffi2
p; (b) Tn ¼ 250 MPa:
4.24 T33 ¼ 1 and T11 ¼ 1.
4.25 (a) Tn ¼ 800=9 ¼ 88:89 kPa; Ts ¼ 260 kPa;(b) Tsð Þmax ¼ 300 kPa:
4.26 (a) tn ¼ 5=ffiffiffi2
p �e1 þ e2ð Þ, (b) Tn ¼ 5 MPa;
(d) Tnð Þmax ¼ 5 MPa; n1 ¼ 1=ffiffiffi2
p �e1 þ e2ð Þ;
Tnð Þmin ¼ �3 MPa; n2 ¼ 1=ffiffiffi2
p �e1 � e2ð Þ. Tsð Þmax ¼ 4 MPa, on n ¼ e1 and n ¼ e2.
4.27 (a) For l1 ¼ t; n1 ¼ 1=ffiffiffi2
p �e1 þ e2ð Þ.
For l2 ¼ �t; n2 ¼ 1=ffiffiffi2
p �e1 � e2ð Þ.
For l3 ¼ 0; n3 ¼ e3. (b) Tsð Þmax ¼ t; n ¼ e1 and n ¼ e2.
Answers to Problems 7
4.29 T12 � T21 ¼ M*3; T13 � T31 ¼ M*
2 and T23 � T32 ¼ M*1.
4.30 (b) T12 ¼ 2x1 � x2 þ 3.
4.31 T33 ¼ 1þ rg=að Þx3 þ f x1; x2ð Þ.4.32 (a) C ¼ �1, (b) A ¼ 1;B ¼ 2.
4.36Tyrr
þ @Tyr@r
þ Tryr
þ 1
r
@Tyy@y
þ @Tyz@z
þ rBy ¼ ray.
4.39 B ¼ ðpo � piÞr2i r2o=ðr2o � r2i Þ; A ¼ ðpir2i � por2oÞ=ðr2o � r2i Þ.
4.41 A ¼ �ðpor3o � pir3i Þ=ðr3o � r3i Þ; B ¼ �ðpo � piÞr3or3i =½2ðr3o � r3i Þ�.
4.42 (a) To½ � ¼1000=16 0 0
0 0 0
0 0 0
24
35MPa; to ¼ 1000=16ð Þe1.
(b) ~T� � ¼ 1000=256 0 0
0 0 0
0 0 0
24
35MPa; ~t ¼ 1000=256ð Þe1.
4.44 (a) dV ¼ 1=4; dA ¼ 1=16ð Þe1,
(b) To½ � ¼100=16 0 0
0 0 0
0 0 0
24
35MPa; to ¼ 100=16ð Þe1MPa:
(c) ~T� � ¼ 100=64 0 0
0 0 0
0 0 0
24
35MPa; ~t ¼ 100=64ð Þe1MPa; d~f ¼ 100
64
e1.
4.45 (a) dV ¼ dVo ¼ 1; dA ¼ e1 � ke2,
(b) To½ � ¼�100k 100 0
100 0 0
0 0 0
24
35MPa; to ¼ 100 �ke1 þ e2ð Þ MPa; t ¼ 100ffiffiffiffiffiffiffiffiffiffiffiffiffi
1þ k2p �ke1 þ e2ð Þ.
(c) ~T� � ¼ �200k 100 0
100 0 0
0 0 0
24
35MPa; ~t ¼ 100 �2ke1 þ e2ð ÞMPa; d~f ¼ 100 �2ke1 þ e2ð Þ.
4.46 (a) dV ¼ 8dVo ¼ 8. dA ¼ 4e1
(b) To½ � ¼400 0 0
0 400 0
0 0 400
24
35MPa; to ¼ 400e1 MPa; t ¼ 100e1 MPa:
(c) ~T� � ¼ 200 0 0
0 200 0
0 0 200
24
35MPa; ~t ¼ 200e1 MPa; d~f ¼ 200e1.
8 Answers to Problems
CHAPTER 55.3 EY=l ! 0; m ! EY=3.
5.4 m ¼ EY
2 1þ nð Þ ; k ¼ 2m 1þ nð Þ3 1� 2nð Þ.
5.9 l ¼ 81:7 GPa 11:8� 106psi� �
; m ¼ 38:4 GPa 5:56� 106psi� �
; k ¼ 107:3 GPa 15:6� 106psi� �
.
5.10 n ¼ 0:27; l ¼ 89:1 GPa 12:9� 106psi� �
; k ¼ 140 GPa 20:3� 106psi� �
.
5.11 T½ � ¼17:7 1:9 4:751:9 18:4 0
4:75 0 16:0
24
35MPa:
5.13 (a) E½ � ¼0:483 0:253 0:3800:253 �1:41 0
0:380 0 1:12
24
35� 10�3,
(b) e ¼ 0:193� 10�3; DV ¼ 24:1� 10�3cm3.
5.14 DV ¼ 2:96� 10�3:
5.17 (a) T11 ¼ T22 ¼ T33 ¼ 0; T12 ¼ T21 ¼ 2mkx3; T13 ¼ T31 ¼ mk 2x1 þ x2ð Þ; T23 ¼ T32 ¼ mk x1 � 2x2ð Þ.
5.19 (a) T½ � ¼ 2klx3 mx3 mx2mx3 lx3 mx1mx2 mx1 lþ 2mð Þx3
24
35.
5.21 For n ¼ 1=3; cL=cT ¼ 2; n ¼ 0:49; cL=cT ¼ 7:14; n ¼ 0:499; cL=cT ¼ 22:4.
5.24 (c) a ¼ 1;(d) b ¼ np= 2ℓð Þ; n ¼ 1; 3; 5 . . .
5.25 (c) a ¼ 1;(d) b ¼ np=ℓ; n ¼ 1; 2; 3 . . .
5.28 (d) b ¼ np=ℓ; n ¼ 1; 2; 3 . . .
5.30 (a) u1 ¼ 3e5sin
2pℓf
� �; u2 ¼ 4e
5sin
2pℓf
� �; f x1; x2tð Þ ¼ 3x1
5þ 4x2
5� cLt� �
.
5.32 (a) a2 ¼ a3 ¼ 0, e2 ¼ e1, and (b) a3 ¼ 31:17o; e2 ¼ 0:742e1; e3 ¼ 0:503e1.
5.35 (b) e3=e1 ¼ �sin 2a1=cos a1 � a3ð Þ; e2=e1 ¼ cos a1 þ a3ð Þ=cos a1 � a3ð Þ.
5.38 (a) u1 ¼ a cosox1cL
þ tanoℓcL
sinox1cL
cos ot;
(b) oℓ=cL ¼ np=2; n ¼ 1; 3; 5 . . .
5.40 (a) u3 ¼ a cos ox1=cTð Þ � cot oℓ=cTð Þsin ox1=cTð Þ½ �cos ot,(b) o ¼ npcT=ℓ; n ¼ 1; 2; 3 . . .
5.42 (a) Tnð Þmax ¼ 71:4� 106N; Tsð Þmax ¼ 23:7� 106N; bð Þdℓ ¼ 1:39� 10�3m.
Answers to Problems 9
5.44 (a) Tn ¼ s cos2 a; Ts ¼ s sin 2a=2,(b) (i) a ¼ p=2; Ts ¼ Tn ¼ 0; and iið Þ a ¼ p=4; Ts ¼ Tn ¼ s=2;(c) s � 2to=sin 2a.
5.46 (a) To11 ¼ 2s=3; To
22 ¼ To33 ¼ �s=3; To
12 ¼ To13 ¼ To
23 ¼ 0;
(b) I1 ¼ 0; I2 ¼ �s2=3; I3 ¼ 2s3=27.
5.49 M1 ¼ Mtℓ2= ℓ1 þ ℓ2ð Þ; M2 ¼ Mtℓ1= ℓ1 þ ℓ2ð Þ.
5.51 Tnð Þmax ¼ ½sþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 þ 4b2r2
q�=2; Ts ¼ ½
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffis2 þ 4b2r2
q�=2; s ¼ P=A; b ¼ Mt=Ip.
5.53 (a) Mtð Þell= Mtð Þcir ¼ 2, (b) a 0ell=a
0cir ¼ 5=16.
5.54 (b) C ¼ a0=6a; (c) T12 ¼ 0; T13 ¼ 0 at all three corners; along x3 ¼ 0,
T12 ¼ 0; T13 ¼ ma0=2að Þ 2ax2 þ x22� �
, (d) Ts ¼ 3a=2ð Þma0 at x2; x3ð Þ ¼ a; 0ð Þ.
5.57 Mt ¼ ma0
3
2að Þ3 2bð Þ 1� 192
p5a
b
� X1n¼1;3;5
1
n5tan h
npb2a
" #.
5.60 Neutral axis in the direction of M2e2 þ I22=I33ð ÞM3e3.
5.63 (b) T11 ¼ 2a3; T12 ¼ �a2; T22 ¼ 2a1;(c) x1 ¼ 0; t ¼ �2a3e1 þ a2e2; x1 ¼ b; t ¼ 2a3e1 � a2e2;(d) T33 ¼ 2n a3 þ a1ð Þ; T13 ¼ T23 ¼ 0; Ei3 ¼ 0; E11 ¼ 2 1=EYð Þ 1� n2
� �a3 � n 1þ nð Þa1
� �,
(e) Ti3 ¼ 0; E13 ¼ E23 ¼ 0; E33 ¼ �2 n=EYð Þ a3 þ a1ð Þ; E11 ¼ 2 1=EYð Þ a3 � na1ð Þ.5.66 (b) T11 ¼ 2ax1 þ 6x1x2; T22 ¼ 0; T12 ¼ �2ax2 � 3x22;
ðcÞ a ¼ �3c=2; ðdÞ tx1¼0 ¼ 3x2 x2 � cð Þe2; tx1¼b ¼ 3b 2x2 � cð Þe1 � 3x2 x2 � cð Þe2; tx2¼0 ¼ 0.
5.67 u1 ¼ Px21x22EYI
þ nPx326EYI
� Px326mI
þ P
2mI
h
2
2
x2 � c1x2 þ c3.
5.69 (a) T12 ¼ 2Am� lmcð Þcos h lmcf gsin h lmx2 þ sin h lmc lmx2 cos h lmx2ð Þ
sin h 2lmcþ 2lmc
� �sin lmx1.
5.72 ur ¼ 1þ nð ÞEY
�A
rþ 2B 1� 2nð Þr ln r � Br þ 2C 1� 2nð Þr
� �þ H sin yþ G cos y,
uy ¼ 1=EYð Þ 4Bry 1� nð Þ 1þ nð Þ½ � þ H cos y� G sin yþ Fr:
5.74 r; r�1; r3 and r ln r.
5.84 Trr ¼ � 1� 2nð ÞzR3
þ 3r2z
R5; Tyy ¼ � 1� 2nð Þz
R3; Trz ¼ 1� 2nð Þr
R3þ 3rz2
R5
,
Tzz ¼ 3z3
R5þ 1� 2nð Þ z
R3.
5.87 Txx ¼ � Fz
2p3x2z
R5� 1� 2nð Þz
R3þ 1� 2nð ÞR Rþ zð Þ �
1� 2nð ÞR Rþ zð Þ
1
R
x2
Rþ zð Þ þx2
R2
� �� �.
10 Answers to Problems
5.88 Tzz ¼ qoz3
r2o þ z2� �3=2 � qo.
5.101 C11 ¼ 1
DE2E3
1� n32n23ð Þ; C12 ¼ 1
DE2E3
n21 þ n31n23ð Þ; C13 ¼ 1
DE2E3
n31 þ n21n32ð Þ,
C23 ¼ 1
DE1E3
n32 þ n31n12ð Þ; D ¼ 1� 2n13n21n32 � n13n31 � n23n32 � n21n12½ �E1E2E3
.
5.112 Brr ¼ a=rð Þ2; Byy ¼ rcð Þ2; Bzz ¼ 1; Bry ¼ 0; Brz ¼ 0, Byz ¼ 0.
5.113 Brr ¼ l21; Bry ¼ Brz ¼ 0; Byy ¼ l21 þ rKð Þ2; Bzz ¼ l23; Bzy ¼ Byz ¼ l3rK.
CHAPTER 6
6.1 RB ¼ 5:1� 104N:
6.2 h ¼ 2:48m.
6.3 h2 ¼ r1h1 � r3h3ð Þ=r2.6.5 (b) Fx ¼ g 2r2L
� �. Fx is 2r/3 above the ground. Fy is 4r/3p left of the diameter.
6.6 p� pa ¼ rðgþ aÞh:6.8 h ¼ aℓ=g.
6.10 h1 � h2 ¼ o2ðr21 � r22Þ=ð2gÞ.6.12 (A) for n 6¼ 1; p n�1ð Þ=n ¼ p�1= n�1ð Þ
o po � fðn� 1Þ=ngrog z� zoð Þ½ �n= n�1ð Þ.
(B) for n ¼ 1, p ¼ po exp �rop�1=no g z� zoð Þ
h i.
6.14 (b) Tn ¼ mk � p; T2s ¼ 0, (c) any plane n1; 0; n3ð Þ and n1; n2; 0ð Þ.
6.16 (a) �Tnð Þ � p ¼ 44m=5; (b) Ts ¼ 8m=5.
6.20 (a) x2 ¼ a2; (b) x1 ¼ 1þ kX2t
1þ kX2toX1 and x2 ¼ X2.
6.22 (a) x21 þ x22 ¼ a21 þ a22, (b) x21 þ x22 ¼ X21 þ X2
2, time history:
x1 ¼ X2 sin otþ X1 cos ot; x2 ¼ X2 cos ot� X1 sin ot.
6.23 (a) y ¼ yo, (b) y ¼ Y, time history: r2 ¼ R2 þ Qt=ðpÞ.6.26 v ¼ ða=2mÞðx2d � x22Þ þ vox2=d; Q ¼ ad3=ð12mÞ þ vod
2=ð2dÞ.6.27 mv ¼ rg sin y d � x2=2ð Þx2.
6.29 m1vtð Þ ¼ �a
x222� b
2
m2 � m1m1 þ m2
x2 � b2
m1m1 þ m2
� �.
6.32 (b) v ¼ 1
4mdp
dzr2 þ a2 � b2ð Þ
ln b=að Þ ln r þ b2ln a� a2ln bð Þln b=að Þ
� �.
Answers to Problems 11
6.34 A ¼ ba2b2=f2ða2 þ b2Þg; B ¼ �A.
6.36 wave length ¼ffiffiffiffiffiffi2p
p=103 ¼ 2:51� 10�3m.
6.38 (b) A ¼ Qm=ð2pÞ.
6.40 Y ¼ � m3
12ka2vodþ a
2m
d � 2x2ð Þ
� �4þ Cx2 þ D.
D ¼ Yo þ m3
12ka2vodþ ad2m
4
.
6.42 Y ¼ � mB2
kr2þ C ln r þ D; C ¼ r2o � r2i
r2i r2o
mB2
k
�ln
riro
:
6.44 (b) T11 ¼ �pþ 2mk; T22 ¼ �p� 2mk; T33 ¼ �p; T12 ¼ T13 ¼ T23 ¼ 0;
(c) a1 ¼ k2x1; a2 ¼ k2x2; a3 ¼ 0, (d) p ¼ �ðr=2Þðv21 þ v22Þ þ po, (f) F ¼ 4mk2;(h) the nonslip boundary condition at x2 ¼ 0 is not satisfied for a viscous fluid.
6.46 Ans. B ¼ � 1=mð Þ @p=@x1ð Þx2e3.
6.48 (c)dy
dx
’¼constant
dy
dx
c¼constant
¼ �1: (d) B ¼ � @2c@y2
þ @2c@x2
ez.
6.49@2c@y2
þ @2c@x2
¼ 0.
6.51 Q ¼ A1A2
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 p1 � p2ð Þ � rgh½ �= r A2
1 � A22
� �� �q:
CHAPTER 7
7.1
ðv � ndS ¼
ðdiv vdV ¼ 16.
7.3
ðdiv vdV ¼
ðv � ndS ¼ 64p.
7.9 (b) m ¼ 3roe�a t�toð ÞA; dm=dt ¼ �3aroe
�a t�toð ÞA.
7.11 (b) m ¼ kAro ln 3; dm=dt ¼ 0.
7.14 (a)dP
dt¼ � 9
2a2Aroe
�a t�toð Þe1; (b) 9Aroa2e�a t�toð Þ, (c) F ¼ 9
2Aroa
2e�a t�toð Þe1.
7.15 (a) P ¼ 2rokaAe1; (b) 2kroAa2; (c) F ¼ 2kroAa
2e1.
7.17 ðrC2pr6o=3Þe1; ðprC2r6o=4Þe1.
7.19 gx ¼ xd2x
dt2þ dx
dt
2
.
12 Answers to Problems
7.21 Force from water to the bend is Fw ¼ 1100e1 � 282e2N:
7.22 rAv2o e1.
7.24 Force on the vane is Fvane ¼ rA vo � vð Þ2 1� cos yð Þe1 � sin ye2½ �.
CHAPTER 8
8.3 S ¼ Geþ �dedt
; S=eo ¼ GHðtÞ þ �dðtÞ.
8.4 (b)S
eo¼ �2oG
ð� þ �oÞ2e
�Gt�þ�o þ ��o
ð� þ �oÞdðtÞ.
8.5 S12 ¼ mvoð1� e�t=lÞ.
8.8 G 0 ¼ð1
l¼o
l2o2H lð Þl 1þ l2o2� � dl; G00 ¼
ð1l¼o
loH lð Þl 1þ l2o2� � dl .
8.11 (b) Ct½ � ¼1 0 0
0 1 0
0 0 1
24
35þ
0 k 0
k 0 0
0 0 0
24
35 t� tð Þ þ
2k2 0 0
0 0 0
0 0 0
24
35 t� tð Þ2
2; k dv=dx1.
8.12 (b) Ct½ � ¼e�2k t�tð Þ 0 0
0 e2k t�tð Þ 0
0 0 1
24
35 ¼
I½ � þ t� tð Þ�2k 0 0
0 2k 0
0 0 0
24
35þ
4k2 0 0
0 4k2 0
0 0 0
24
35 t� tð Þ2
2þ
�8k3 0 0
0 8k3 0
0 0 0
24
35 t� tð Þ3
3!þ . . . :
8.13 (b) Ct½ � ¼e2k t�tð Þ 0 0
0 e2k t�tð Þ 0
0 0 e�4k t�tð Þ
24
35 ¼ I½ � þ t� tð Þ
2k 0 0
0 2k 0
0 0 �4k
24
35
þ4k2 0 0
0 4k2 0
0 0 16k2
24
35 t� tð Þ2
2þ
8k3 0 0
0 8k3 0
0 0 �64k3
24
35 t� tð Þ3
3!þ . . . :
8.14 (a) x 0 ¼ x1 cos h k t� tð Þ þ x2 sin h k t� tð Þ; x 0 ¼ x1 sin h k t� tð Þ þ x2 cos h k t� tð Þ,
(b) Ct½ � ¼cos h2fk t� tð Þg þ sin h2fk t� tð Þg sin hf2k t� tð Þg 0
sin hf2k t� tð Þg sin h2fk t� tð Þg þ cos h2fk t� tð Þg 0
0 0 1
264
375
¼ I½ � þ0 2k 0
2k 0 0
0 0 0
24
35 t� tð Þ þ
4k2 0 0
0 4k2 0
0 0 0
24
35 t� tð Þ2
2þ
0 8k3 0
8k3 0 0
0 0 0
24
35 t� tð Þ3
6þ . . . :
Answers to Problems 13
8.20 (a) A1½ � ¼0 k rð Þ 0
k rð Þ 0 0
0 0 0
24
35; k ¼ dv
dr� v rð Þ
r
.
(b) rA1ð Þrry ¼ � 2k
r; rA1ð Þryy ¼ 0; rA1ð Þrzy ¼ 0; rA1ð Þyry ¼ 0;
rA1ð Þyyy ¼2k
r; rA1ð Þyzy ¼ 0; rA1ð Þzry ¼ 0; rA1ð Þzyy ¼ 0; rA1ð Þzzy ¼ 0.
(c) A2 ¼2k2 0 0
0 0 0
0 0 0
24
35.
8.23 m S12k
¼ �ð10
sf2 sð Þd; s1 ¼ S11 � S22 ¼ �k2ð10
s2f2 sð Þds; s2 ¼ S22 � S33 ¼ 0.
8.27 (a) Corotational stress rate is: T∘h i
¼ mk2 0
0 �mk2
� �;
(c) T∘ ¼ mk2
cos 2otð Þ sin 2otð Þsin 2otð Þ �cos 2otð Þ
� �; (d) T
∘*
h i¼ Q½ � T∘
h iQ½ �T .
8.28 (b) The corotational derivative of T: rk v22 � v21� �
I.
8.30 T½ � ¼ �p I½ � þ f1 k2=4; 0� � 0 k=2 0
k=2 0 0
0 0 0
24
35þ f2 k2=4; 0
� � k2=4 0 0
0 k2=4 0
0 0 0
24
35.
8.36 S12 ¼ mkAðkÞ ; s1 S11� S22 ¼ 2lmk2
AðkÞ ; s2 ¼ S22� S33 ¼�lmk2 1þ að ÞAðkÞ ; AðkÞ ¼ 1þ 1� a2
� �lkð Þ2
h i.
8.37 � kð Þ ¼ S12=k ¼ m; s1 ¼ T11 � T22 ¼ 2mk2 l1 � l2ð Þ; s1 ¼ T22 � T33 ¼ 0.
8.38 � kð Þ ¼ S12k
¼ m 1þ l2mok2ð Þ
1þ l1mok2ð Þ ; s1 ¼ T11 � T22 ¼ 2mk2 l1 � l2ð Þ1þ l1mok2ð Þ ; s2 ¼ T22 � T33 ¼ 0.
8.40 (a) Szr ¼ t kð Þ; Szz � Srr ¼ s1 kð Þ; Srr � Syy ¼ s2 kð Þ; Szy ¼ Sry ¼ 0.
14 Answers to Problems