x
61 0.3 0 0 0 10 0 10 0
30 10 0.3 1 0 0 20 0 0 0 20
(0.91)(200) 0 0 0.35 20 0 0 10 20 10
3
2
0
0
2.5 10
1.376 10
0
0












x
4123


p =
1
2
tan–1
2 7938
4123 1237




Using element (2)
61 0.3 0
30 10 0.3 1 0
1
2 2062


351
0.0000E+00 0.0000E+00
0.0000E+00 0.0000E+00
INPUT TABLE 4. ELEMENT DATA
GLOBAL INDICES OF ELEMENT NODES
ELEMENT 1 2 3 4 MATERIAL
1 1 5 4 4 1
2 1 2 5 5 1
OUTPUT TABLE 1. NODAL DISPLACEMENTS, m
1 0.00000000E+00 0.00000000E+00
2 0.5624948E–04 – 0.6597062E–04
OUTPUT TABLE 2. STRESSES AT ELEMENT CENTROIDS
ELEMENT X Y SIGMA (X) SIGMA (Y) TAU (X, Y)
1 0.083 0.125 0.0613E+07 3.1840E+06 3.3431E+06
2 0.250 0.042 1.6384E+07 1.5239E+07 – 6.9854E+06
2.2820E+07 8.8026E+06 – 4.2657E+01
3.6135E+07 6.5251E+06 – 6.5798E+01
2.2412E+07 2.4221E+06 – 6.5993E+01
(c)
352
INPUT TABLE 1. BASIC PARAMETERS
NUMBER OF NODAL POINTS. . . . . . . . . . . . . . . 5
NUMBER OF ELEMENTS. . . . . . . . . . . . . . . . . . . 4
BODY FORCES (1 = IN – Y DIREC., 0 = NONE) 0
INPUT TABLE 2. MATERAL PROPERTIES
MATERIAL MODULUS OF POISSON’S MATERIAL MATERIAL
INPUT TABLE 3. NODAL POINT DATA
POINT TYPE X Y
1 0 0.0000E+00 0.0000E+00
4 3 0.0000E+00 0.4000E+00
5 0 0.2000E+00 0.2000E+00
OR LOAD OR LOAD
0.0000E+00 0.0000E+00
0.0000E+00 0.0000E+00
INPUT TABLE 4. ELEMENT DATA
353
GLOBAL INDICES OF ELEMENT NODES
ELEMENT 1 2 3 4 MATERIAL
1 1 2 5 5 1
OUTPUT TABLE 1. NODAL DISPLACEMENTS (m)
NODE U = X–DISP. V = Y–DISP.
1 – 0.3303082E–05 – 0.25009076E–04
5 0.27411561E–12 – 0.32558982E–04
OUTPUT TABLE 2. STRESSES AT ELEMENT CENTROIDS
2
N
m
ELEMENT X Y SIGMA(X) SIGMA(Y) TAU(X, Y)
1 0.20 0.07 5.9891E+05 –3.7840E+06 4.0454E–01
SIGMA(1) SIGMA(2) ANGLE
5.9891E+05 –3.7840E+06 5.2883E–06
6.15
356
6.16
(a)
kN

f5y = –32.12 N
(c)
Equation (6.3.6)
(1)
1
(1)
1
(1)
2
Bx
By
Bx
f
f
f
0
154.2
0(0.4 m)(0.2 m)(0.005m)




0
10.28
0




nb = n0 (m + 1)
(a) nb = 2(3 + 1) = 8
8 12
1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8
00
0 0 0 0
00
0 0 0 0 0
00
Symmetry 00
Symmetry
bb
nn
6.22
By (6.2.10)
2
2 3 3
b b h
2
33
bh A
3
h
2
23
bh
2
33
bh A
2 3 3 2
b h h b
2
33
bh A
359
b
T
vb
X
12
23
A
A
1
3
1
3
or
0
b
i
X
N
1
3
0
vb
Y
{
i
b
f
} =
3
b
b
XV
Y
Similarly
j
b
3
b
b
XV
Y
6.25
4
b x h y
bh
4
b x h y
bh
N3 =
4
b x h y
bh
, N4 =
4
b x h y
bh
(1)
at center (x = 0, y = 0)
1
4
1
4
1
4
1
4
N1 + N2 + N3 + N4 = 1
at point
,
22
bh
xy
22
bh
bh
1
16
16
16
N1 + N2 + N3 + N4 = 1
1 0 1 0 1 0 1 0
6.26
{
} = [D] [B] {d}
1
4bh
( ) 0 0 0 0
h y h y h y h y
b x h y b x h y b x h y b x h y
At center (x = 0, y = 0)
1
1 0 1 0 1 0 1 0
x
y
xy
18.54
7.21