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FUNDAMENTAL CONCEPTS 1.1 We use the first three steps of Eq. 1.11
EEE
EEE
EEE
zyxz
zyxy
zyxx
σ+
σν−
σν−=ε
σν−
σ+
σν−=ε
σν−
σν−
σ=ε
Adding the above, we get
( )zyxzyx Eσ+σ+σ
ν−=ε+ε+ε
21
Adding and subtracting E
xσν from the first equation,
( )zyxxx EEσ+σ+σ
ν−σ
ν+=ε
1
Similar expressions can be obtained for εy, and ε z. From the relationship for γyz and Eq. 1.12,
( ) yzyzE
γν+
=τ12
etc.
Above relations can be written in the form σ = Dε where D is the material property matrix defined in Eq. 1.15. 1.2 Note that u2(x) satisfies the zero slope boundary condition at the support.
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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full file at http://testbankinstant.com 1.3 Plane strain condition implies that
EEE
zyxz
σ+
σν−
σν−==ε 0
which gives ( )yxz σ+σν=σ
We have, 0.3 psi 1030 psi 10000 psi 20000 6 =ν×=−=σ=σ Eyx . On substituting the values, psi 3000=σ z 1.4 Displacement field
( )( )( ) ( )
( )
−=ε
==
∂∂
+∂∂∂∂∂∂
=ε
+=∂∂
×=∂∂
+=∂∂
+−=∂∂
−+=
++−=
−
−−
−−
−
−
962
10
0,1at
2610 103
6410 6210
63106210
4
44
44
24
224
yxxv
yu
yvxu
yyv
xv
xyyuyx
xu
yyxvxyyxu
1.5 On inspection, we note that the displacements u and v are given by u = 0.1 y + 4 v = 0 It is then easy to see that
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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1.0
0
0
=∂∂
+∂∂
=γ
=∂∂
=ε
=∂∂
=ε
xv
yu
yvxu
xy
y
x
1.6 The displacement field is given as u = 1 + 3x + 4x3 + 6xy2
v = xy − 7x2
(a) The strains are then given by
xyxyxv
yu
xyv
yxxu
xy
y
x
1412
6123 22
−+=∂∂
+∂∂
=γ
=∂∂
=ε
++=∂∂
=ε
(b) In order to draw the contours of the strain field using MATLAB, we need to create a
script file, which may be edited as a text file and save with “.m” extension. The file for plotting εx is given below
file “prob1p5b.m”
[X,Y] = meshgrid(-1:.1:1,-1:.1:1); Z = 3.+12.*X.^2+6.*Y.^2; [C,h] = contour(X,Y,Z);
clabel(C,h); On running the program, the contour map is shown as follows:
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
4
4
4
6
6
6
6
6
6
8
88
8
8
8
810
10
1010
10
10
12
12
12
12
12
12
14
14
14
14
14
14
16
16
16
16
18
18
18
18
Contours of εx Contours of εy and γxy are obtained by changing Z in the script file. The numbers on the contours show the function values.
(c) The maximum value of εx is at any of the corners of the square region. The maximum value is 21.
1.7
a) 0.2 0.2 01
u y u y v= ⇒ = =
b) 0 0 0.2x y xyu v u vx y y x
ε ε γ∂ ∂ ∂ ∂= = = = = + =∂ ∂ ∂ ∂
(x, y) (u, v)
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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.
MPa 393.24
MPa 713.13
MPa 213.6
MPa 607.35
get we1.8 Eq. From2
121
21
MPa 10 MPa 15 MPa 30
MPa 30 MPa 20 MPa 40
T
=
++=σ=
σ+τ+τ=−=
τ+σ+τ==
τ+τ+σ=
=
=τ=τ−=τ
=σ=σ=σ
zzyyxxn
zzyyzxxzz
zyzyyxxyy
zxzyxyxxx
xyxzyz
zyx
nTnTnT
nnnT
nnnT
nnnT
n
1.9 From the derivation made in P1.1, we have
( )( ) ( )[ ]
( )( ) ( )[ ]
( ) yzyz
vxx
zyxx
E
E
E
γν+
=τ
νε+εν−ν−ν+
=σ
νε+νε+εν−ν−ν+
=σ
12
and
21211
form in the written becan which
1211
Lame’s constants λ and µ are defined in the expressions
( )( )
( )ν+=µ
ν−ν+ν
=λ
µγ=τµε+λε=σ
12
211
,inspectionOn
2
E
E
yzyz
xvx
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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full file at http://testbankinstant.comµ is same as the shear modulus G. 1.10
( ) MPa 6.69106.3
/1012GPa 200
30102.1
0
40
06-
0
5
−=ε−ε=σ×=∆α=ε
×=α
==∆
×=ε
−
−
ET
CE
CT
1.11
+=
+==δ
+==ε
∫2
0
3
0
2
321
32
21
LL
xxdxdxdu
xdxdu
LL
x
1.12 Following the steps of Example 1.1, we have
( )
=
−
−++5060
808080504080
2
1
Above matrix form is same as the set of equations: 170 q1 − 80 q2 = 60 − 80 q1 + 80 q2 = 50 Solving for q1 and q2, we get q1 = 1.222 mm q2 = 1.847 mm
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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When the wall is smooth, 0xσ = . T∆ is the temperature rise.
a) When the block is thin in the z direction, it corresponds to plane stress condition. The rigid walls in the y direction require 0yε = . The generalized Hooke’s law yields the equations
yx
yy
TE
TE
σε ν α
σε α
= − + ∆
= + ∆
From the second equation, setting 0yε = , we get y E Tσ α= − ∆ . xε is then calculated
using the first equation as ( )1 Tν α− ∆ . b) When the block is very thick in the z direction, plain strain condition prevails. Now we
have 0zε = , in addition to 0yε = . zσ is not zero.
0
0
y zx
y zy
y zz
TE E
TE E
TE E
σ σε ν ν α
σ σε ν α
σ σε ν α
= − − + ∆
= − + ∆ =
= − + + ∆ =
From the last two equations, we get 1 2
1 1y zE T E Tα νσ σ α
ν ν− ∆ +
= = − ∆+ +
xε is now obtained from the first equation.
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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full file at http://testbankinstant.com1.14 For thin block, it is plane stress condition. Treating the nominal size as 1, we may set the
initial strain 00.11
Tε α= ∆ = in part (a) of problem 1.13. Thus 0.1y Eσ = − .
1.15
The potential energy Π is given by
∫∫ −
=Π
2
0
2
0
2
21 ugAdxdx
dxduEA
Consider the polynomial from Example 1.2,
( )( ) ( ) 33
23
1222
2
axaxdxdu
xxau
+−=+−=
+−=
On substituting the above expressions and integrating, the first term of becomes
322 2
3a
and the second term
3
2
0
2
0
32
3
2
0
34
3
a
xxaudxugAdx
−=
+−== ∫∫
Thus
( )
21 0
34
33
32
3
−=⇒=∂Π∂
+=Π
aa
aa
this gives ( ) 5.01221
1 =+−−==xu
1.16
x=0
x=2
g=1 E=1 A=1
f = x3
E = 1 A = 1
x=0 x=1 Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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We use the displacement field defined by u = a0 + a1x + a2x2. u = 0 at x = 0 ⇒ a0 = 0 u = 0 at x = 1 ⇒ a1 + a2 = 0 ⇒ a2 = − a1
We then have u = a1x(1 − x), and du/dx = a1(1 − x). The potential energy is now written as
( ) ( )
( ) ( )
306
61
51
34
241
21
44121
12121
21
12
1
12
1
1
0
1
0
541
221
1
0
1
01
3221
1
0
1
0
2
aa
aa
dxxxadxxxa
dxxxaxdxxa
fudxdxdxdu
−=
−−
+−=
−−+−=
−−−=
−
=Π
∫ ∫
∫ ∫
∫ ∫
0301
3 0 1
1
=−⇒=∂Π∂ aa
This yields, a1 = 0.1 Displacemen u = 0.1x(1 − x) Stress σ =E du/dx = 0.1(1 − x) 1.17 Let u1 be the displacement at x = 200 mm. Piecewise linear displacement that is
continuous in the interval 0 ≤ x ≤ 500 is represented as shown in the figure. 0 200 500
u1
u = a3 + a4x u = a1 + a2x
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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full file at http://testbankinstant.com 0 ≤ x ≤ 200 u = 0 at x = 0 ⇒ a1 = 0 u = u1 at x = 200 ⇒ a2 = u1/200 ⇒ u = (u1/200)x du/dx = u1/200 200 ≤ x ≤ 500 u = 0 at x = 500 ⇒ a3 + 500 a4 = 0 u = u1 at x = 200 ⇒ a3 + 200 a4 = u1
⇒ a4 = −u1/300 a3 = (5/3)u1 ⇒ u = (5/3)u1 − (u1/300)x du/dx = − u1/200
12
121
1
21
2
21
1
1
500
200
2
2
200
0
2
1
100003002002
1
100003003002
12002002
1
1000021
21
uuAEAE
uuAEuAE
udxdxduAEdx
dxduAE
stal
stal
stal
−
+=
−
−+
=Π
−
+
=Π ∫∫
010000300200
0 121
1
=−
+⇒=
∂Π∂ u
AEAEu
stal
Note that using the units MPa (N/mm2) for modulus of elasticity and mm2 for area and
mm for length will result in displacement in mm, and stress in MPa. Thus, Eal = 70000 MPa, Est = 200000, and A1 = 900 mm2, A2 = 1200 mm2. On
substituting these values into the above equation, we get u1 = 0.009 mm This is precisely the solution obtained from strength of materials approach 1.18 In the Galerkin method, we start from the equilibrium equation
0=+ gdxduEA
dxd
Following the steps of Example 1.3, we get
∫∫ φ+φ
−2
0
2
0
dxgdxdxd
dxduEA
Introducing
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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( )( ) 1
21
2
2
and ,2
φ−=φ
−=
xxuxxu
where u1 and φ1 are the values of u and φ at x = 1 respectively,
( ) ( ) 02212
0
2
0
2211 =
−+−−φ ∫ ∫ dxxxdxxu
On integrating, we get
034
38
11 =
+−φ u
This is to be satisfied for every φ1, which gives the solution u1 = 0.5 1.19 We use
2at 00at 0
34
2321
====
+++=
xuxu
xaxaxaau
This implies that
4321
1
84200
aaaa
+++==
and
( ) ( )
( ) ( )4312
42
243
34
23
−+−=
−+−=
xaxadxdu
xxaxxau
a3 and a4 are considered as independent variables in
( ) ( )[ ] ( )∫ −−−−+−=Π2
043
2243 324312
21 aadxxaxa
on expanding and integrating the terms, we get
43432
42
3 6288.12333.1 aaaaaa ++++=Π
We differentiate with respect to the variables and equate to zero.
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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066.258
028667.2
434
433
=++=∂Π∂
=++=∂Π∂
aaa
aaa
On solving, we get a3 = −0.74856 and a4 = −0.00045.
On substituting in the expression for u, at x = 1, u1= 0.749 This approximation is close to the value obtained in the example problem. 1.20
(a) ( )udxxTAdxLL
∫∫ −εσ=Π00
T
21
dxduE =εε=σ and
On substitution,
( ) udxudxxdxdxdu
udxTudxTdxdxduEA
∫∫∫
∫∫∫
−−
×=Π
−−
=Π
60
30
30
0
60
0
26
60
30
30
0
60
0
2
30010106021
21
(b) Since u = 0 at x = 0 and x = 60, and u = a0 + a1x + a2x2, we have
( )
( )602
60
2
2
−=
−=
xadxdu
xxau
On substituting and integrating, 2
22
10 877500010216 aa +×=Π Setting dΠ/da2 = 0 gives
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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( )602935.60
1003125.2 62
−−==σ
×−= −
xdxduE
a
Plots of displacement and stress are given below:
0 10 20 30 40 50 60
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2x 10
-3
Displacement u
0 10 20 30 40 50 60
-4000
-3000
-2000
-1000
0
1000
2000
3000
4000
Stress
. 1.21 y = 20 at x = 60 implies that
( )210
210
1803120yields which ,36006020
aaaaaa
−−=++=
Substituting for k, h, L, and a0 in I, we get
( ) ( ) ( )[ ]
( ) ( )
76050001070210117104561200045600
3918350004410
80018031202521210
24
142
25
212
1
60
0
221
22
221
21
60
0
221
221
+×+×+×++=
+++++=
−−−++=
∫
∫
aaaaaaI
aadxaxaxaaI
aadxxaaI
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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0107021090612000
010117612000912000
42
51
2
421
1
=×+×+=
=×++=
aadadI
aadadI
On solving, a2 = 0.1699 a1 = −13.969 Substituting into the expression for a0, we get a0 = 246.538
. 1.22 Since u = 0 at x = 0, the displacement satisfying the boundary condition is u = a1x. Also
the coordinates are x2 = 1, and x3 = 3.
The potential energy for the problem is
2
3
2 2 3 30
12
duEA dx P u Pudx
π = − − ∫
We have u2 = a1, u3 = 3a1, E = 1, A = 1, and 1du adx
= . Thus
( )3 2 2
1 1 1 1 10
1 33 42 2
a dx a a a aπ = − − = −∫ .
For stationary value, setting 1
0ddaπ= , we get
3a1 – 4 = 0, which gives a1 = 0.75.
The approximate solution is u = 0.75x.
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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full file at http://testbankinstant.com 1.23 Use Galerkin approach with approximation 2u a bx cx= + + to solve
( )
3 0 1
0 1
du u x xdx
u
+ = ≤ ≤
=
The week form is obtained by multiplying by φ satisfying ( )0 0φ = . 1
03 0du u x dx
dxφ + − = ∫
We now set 21u bx cx= + + satisfying ( )0 1u = and 21 2a x a xφ = + . On introducing these
into the above integral,
( )( )( ) ( )
1 2 21 20
1 12 2 2 3 2 2 3 3 3 41 20 0
2 3 3 3 0
3 3 2 3 3 3 2 3 0
a x a x b cx bx cx x dx
a bx x x bx cx cx dx a bx x x bx cx cx dx
+ + + + + − =
+ − + + + + + − + + + + =
∫∫ ∫
On integrating, we get
1 2
1 2
3 1 2 3 1 3 31 02 2 3 3 4 3 4 4 2 5
3 17 7 13 11 3 02 12 6 12 10 4
b c c b b c ca b a
a b c a b c
+ − + + + + + − + + + =
+ + + + + =
This must be satisfied for every a1 and a2. Thus the equations to be solved are 3 17 7 02 12 613 11 3 012 10 4
b c
b c
+ + =
+ + =
The solution is b = –1.9157, c = 1.2048. Thus 21 1.9157 1.2048u x x= − + . 1.24
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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full file at http://testbankinstant.com The deflection and slope at a due to P1 are
31
3PaEI
− and 2
1
2Pa
EI− . Using this the deflection
and slope at L due to load P1 are
( )2311
1
21
1
3 2
2
Pa L aPavEI EI
PavEI
−= − −
′ = −
The deflection and slope due to load P2 are
32
2
22
2
3
2
P LvEI
P LvEI
= −
′ = −
We then get
1 2
1 2
v v vv v v= +′ ′ ′= +
1.25
(a) The displacement of B is given by (–0.1, 0.1) and A, C, and D remain in their original position. Consider a displacement field of the type
1 2 3 4
1 2 3 4
u a a x a y a xyv b b x b y b xy= + + += + + +
The four constants can be evaluated using the known displacements
(0,0) (1,0)
(1,1) (0,1)
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
download full file at http://testbankinstant.com
full file at http://testbankinstant.comAt A (0, 0) 1
1
00
ab==
At B (1, 0) 1 2
1 2
0.10.1
a ab b+ = −+ =
At C (1, 1) 1 2 3 4
1 2 3 4
00
a a a ab b b b+ + + =+ + + =
At D (0, 1) 1 3
1 3
00
a ab b+ =+ =
The solution is
a1 = 0, a2 = –0.1, a3 = 0, a4 = 0.1 b1 = 0, b2 = 0.1, b3 = 0, b4
This gives 0.1 0.1
0.1 0.1u x xyv x xy= − += −
(b) The shear strain at B is
( ) ( )
0.1 0.1 0.1
0.1 1 0.1 0.1 0 0.2B
u v x yy x
γ
γ
∂ ∂= + = + −∂ ∂
= + − =
Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.