10.1
J =
dx
ds
u = a1 + a2s + a3s2
(1) – (2) gives
21
xx
(3) and (4) into (1)
a3 = x1 – x3 +
21
2
xx
=
1 2 3
2
2
x x x
2
21
xx
2
1 2 3
2
x x x s
2
10.2 Using Equation (10.1.1 b)
For Figure (a)
(a) s = [x –
12
( + )
2
xx
]
(10 + 20)
2
s = 0.5
1 0.5
2
1
4
1 0.5
2
3
4
(c) uA =
0.10
1/ 4 3 / 4 0.175 mm
0.20



(d)
x
0.10
11 0.0025
0.20
40 40






2
(5) – (6) gives
3
1
a
or a3 =
2
3
(x1 + x4 – x2 – x3) (7)
(7) into (5)
x1 + x4 = 2a1 +
2
23
(x1 + x4 – x2 – x3)
14
1 4 2 3
33
x x x x
(2) – (3) x2 – x3 = – a2 –
4
4
a
(10)
(9) – 2 (10) gives
4
3
a
427
1
2
3
4
x
x
x
x
N1 =
32
221
3 3 6 6
s
ss
N2 =
32
4 2 4 2
3 3 3 3
s s s
N3 =
32
4 2 4 2
3 3 3 3
s s s
N4 =
32
2 2 1
3 3 6 6
s
ss
(2)
du
ds
=
2 2 2
2
4 1 4 4 4 4
2 4 4
3 6 3 3 3 3
41
236
s s s s s s
ss
1
2
3
4
u
u
u
u
Differentiating (13)
dx
ds
241
244
+
244
433
ss
x3 +
241
236
ss
x4
Simplifying
= 2s2 (x4 – x1) +
4
3
s(x4 + x1) –
1
6
(x4 – x1)
1
2
4
4
ds
2
Now
du
dx
du
ds
dx
ds
du
dx
x =
2 2 2 2
33
22
12 8 1 12 4 4 12 4 4 12 8 1
33
LL
s s s s s s s s
LL
1
2
3
4
u
u
u
u
2 2 2 2
12 8 1 12 4 4 12 4 4 12 8 1
s s s s s s s s
33
22
33
LL
LL
10.6 For Figure (a)
(a) Using Equation (10.5.6)
2
2 0 0 2 2 (1)
(b) N1 =
– 1 – 1
22
s s
=
0.5 0.5 1
2
0.5 0.5 1
2
2
= 0.375
N3 = 1 – s2 = 1 – 0.52
= 0.75
N ’s = – 0.125 + 0.375 + 0.75
= 1.0
(c) By Eq. (10.5.11)
21
A3
22
12
2
3
2
2
uu
u u (0.5) (0.5)
22
uu
(0.5) (0.5)
22
2u (0.5)
2
0.002
2
0.002
0 (0.5)
2
2(0.001) (0.5)
2


x
2(0.5) 1
( )(0)
2
2(0.5) 1
( )(0.002)
14
33
14
33
4 16
33
Using Equations (4) and (1) in (A) and applying boundary condition u1 = 0,
4
3
16
3
=
2
4
3
2.393 – 2.732 10
– 2.732 5.468
u
u
(5)
x
s
1
4
x
t
=
1
4
[– x1 + s x1 – x2 – s x2 + x3 + s x3 + x4 – s x4]
y
s
=
1
4
[– y1 + t y1 + y2 – t y2 + y3 + t y3 – y4 – t y4]
y
t
4
1
1
4
J12 =
1
4
(– y1 + t y1 + y2 – t y2 + y3 + t y3 – y4 – t y4)
J21 =
1
4
(– x1 + s x1 – x2 – s x2 + x3 + s x3 + x4 – s x4)
1
4
16
+ 2x2y3 + 2s x2y3 – 2s x2y4 – 2t x2y4 + 2s x3y1 – 2t x3y1 – 2x3y2 – 2s x3y2
Factor out xi’s
1
8
+ x2 (– y1 + t y1 + y3 + s y3 – s y4 – t y4) + x3 (s y1 – t y1 – y2 – s y2 + y4 + t y4)
+ x4 (y1 – s y1 + s y2 + t y2 – y3 – t y3)]
1
2 2 3 3 4 4
1 1 3 3 4 4
y t y ty s y s y y
y t y y s y s y t y
1
8
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
(0) 1 ( 1)
1 (0) (0) 1
1 (0) 1
1 1 (0)
y y t y t s y s
y t y y y t
y s t y s y y t
y s y s t y t y
| J | =
1
8
[x1 x2 x3 x4]
1
2
3
4
0 1 1
1 0 1
1 0 1
1 1 0
y
t t s s
y
t s s t
y
s t s t
y
s s t t
10.10
[J] =
y
x
ss
y
x
tt
x and y from Problem 10.9
x
s
=
1
4
(– 1) (1 – t) x1 +
1
4
(1 – t) x2 +
1
4
(1 + t) x3 +
1
4
(– 1) (1 + t) x4
1, 2, 3, 4,
t t t t
N N N N
33
44
xy
xy
10.11 (a)
[ J ] =
y
x
ss
y
x
tt
12
34
2, 2
2, 2



xx
xx
x
1
1
1
1
y
1
1
1
1
8
x
1
1
1
1