COMPUTATIONAL METHODS IN ENGINEERING AND SCIENCE
EPMESC X, Aug. 21-23, 2006, Sanya, Hainan, China
©2006 Tsinghua University Press & Springer
A 3-Dimensional Assumed Stress Hybrid Element with Drilling Degrees
of Freedom
A. P. Wang 1*, Z. S. Tian 1, X. Q. Zhang 2
1
2
Department of Mechanics, Graduate School, Chinese Academy of Sciences, PO Box 2706, Beijing, 100080 China
Institute of Engineering Thermophysics, Chinese Academy of Sciences, Beijing, 100080 China
Email: [email protected]
Abstract A new 3-dimensional 8-node assumed stress hybrid finite element with drilling degrees of freedom — which
contains a traction-free cylindrical surface — has been developed based on a modified complementary energy
principle. The assumed stress field satisfies the homogeneous equilibrium equations in the element and the
traction-free conditions over the cylindrical boundary. It also satisfies the compatibility equations when the element is
degenerated into 2-dimensional case. Numerical results have demonstrated that the special element can provide much
more accurate stress concentration factors and the distributions of circumferential stress along the rims of some cutouts
than those obtained by using other methods.
Key words: 3-D hybrid stress element, drilling degrees of freedom, stress concentration
INTRODUCTION
For stress analysis of solid with a traction-free cylindrical surface by the finite element method it is more advantageous
to use special solid elements which contain traction-free surfaces. For this purpose the assumed stress hybrid element
is most suitable. While it has been shown [1-4] that for the problems of stress concentrations in the vicinity of different
circular holes and notches, in general, the convergence rate for the finite element method is dominated by the nature of
the solution in the vicinity of steep stress gradient. The regular assumed stress elements obtained by using polynomials
of high order as interpolating functions cannot improve the rate of convergence. Unless using extremely fine meshes,
it will be difficult to obtain the reasonable stresses.
In the paper a new 3-dimensioanl 8-node hybrid stress element with drilling degrees of freedom which contains a
traction-free cylindrical surface is derived based on Hellinger-Reissner principle. The use of such special element in
the finite element solution has been shown to be highly accurate when only a very coarse element mesh is used near the
circular surfaces or holes.
ELEMENT STIFFNESS MATRIX
In formulating the element stiffness matrix by the Hellinger-Reissner principle [5], the energy functional is given by
equation:
{
Π HR (σ , u) = Σ
n
∫
Vn
⎡ 1 T
T
T ⎤
T
T
⎢− 2 σ S σ + σ (Du ) − F u⎥ dV − ∫Sun T (u − u) dS − ∫Sσn T u dS} = Stationary
⎣
⎦
where
Vn: volume of the nth individual element;
σ: the stresses;
S: the elastic compliance matrix;
u : displacements;
Du: strains;
F : body force;
⎯ 728 ⎯
(1)
u : prescribed displacement;
T: boundary traction ( T = vσ , v : the direction cosine of the normal to the boundary);
T : prescribed boundary traction;
S un : prescribed displacement boundary;
Sσ n : prescribed stress boundary;
n: element number.
In the finite element implementation of equation (1), we interpolate u in terms of nodal displacements q
u = Nq
(2)
and expressed σ by stress parameters β
σ = Pβ
(3)
Substituting equations (2) and (3) into equation (1), from the variation of Π HR , β can be expressed in terms of q, the
strain energy can be expressed, in turn, in terms of q, and the element stiffness matrix is
k = G T H −1G
(4)
where
H = ∫ P T S P dV
Vn
G = ∫ P T ( DN ) dV
(5)
Vn
Figure 1: Geometry of the 8-node element with a traction-free cylindrical surface
Figure 2: 8-node hexahedron element
with “drilling” DOF at each node
Figure 3: 20-node hexahedron element
COMPATIBLE DISPLACEMENT FIELD
The 8-node solid element with a traction-free cylindrical surface shown in Figure 1 has 6 degrees of freedom (DOF) at
each node, i.e. three translations ui , vi , wi and three “drilling” rotations ωxi, ωyi and ωzi. The method of forming the
shape functions of displacement is given in references [6-8]. The displacement field for a hexahedron element (Figure
2) with 48 DOF is derived from the 20-node (60 DOF) hexahedron element (Figure 3).
The midside node displacements of the 20-node hexahedron element can be expressed as
⎯ 729 ⎯
uk =
y − yi
z − zi
1
(ui + u j ) + j
(ω zj − ω zi ) + j
(ω yi − ω yj )
2
8
8
vk =
z j − zi
x j − xi
1
(vi + v j ) +
(ω xj − ω xi ) +
(ω zi − ω zj )
2
8
8
wk =
x j − xi
y j − yi
1
( wi + w j ) +
(ω yj − ω yi ) +
(ω xi − ω xj )
2
8
8
(6)
If the displacements at the midside node of each side for the 20-node element are
um = [u9
u10 K u 20 ]
v m = [v9
v10 K v20 ]
w m = [w9
T
T
w10 K w20 ]
(7)
T
The midside nodal displacements are expressed by using nodal displacements qH8 of the 8-node solid element with
drilling DOF as
{um
vm
w m } = [Tu Tv Tw ] T qH8
T
(8)
[
q H8 = u1 K u8 v1 K v8 w1 K w8 ω x1 K ω x8 ω y1 K ω y 8 ω z1 K ω z 8
]
T
where Tu , Tv and Tw are each a 12×48 transformation matrix for um , v m and w m . Finally, the compatible
displacement field for the 8-node 48 DOF hexahedron element can be written as
⎧u ⎫
u = ⎪⎨ v ⎪⎬ = N H 20 T q = N H 20
(3×60)
(3×60) (60×48 ) (48×1)
⎪ w⎪
⎩ ⎭ H8
⎡I 01 01 0 2 ⎤
⎥
⎢
Tu
⎥
⎢
⎢01 I 01 0 2 ⎥
⎥ q = Nq
⎢
T
v
⎥ (48×1)
⎢
⎢01 01 I 0 2 ⎥
⎥
⎢
Tw
⎥⎦
⎢⎣
(9)
where N H20 is the usual shape function of the 20-node hexahedron element; I is a 8×8 identity matrix; 01 and 0 2 are
8×8 and 8×24 null matrices, respectively.
THE STRESS ASSUMPTIONS
The 3-dimensional homogeneous equilibrium equations are
∂σ r 1 ∂τ rθ ∂τ rz σ r − σ θ
+
+
+
=0
∂r r ∂θ
∂z
r
∂τ rθ 1 ∂σ θ ∂τ zθ 2τ rθ
+
+
+
=0
∂r
r ∂θ
∂z
r
(10)
∂τ zr 1 ∂τ zθ ∂σ z τ zr
+
+
+
=0
∂r
r ∂θ
∂z
r
By using four of stress functions [ φi (r ,θ ) , i = 1,…,4] for six stress components and an additional function [ V (r ,θ ) ]
for stress σ z , the stress field is expressed as
⎯ 730 ⎯
σr =
⎛ 1 ∂φ 2 1 ∂ 2φ 2 ⎞
1 ∂φ1 1 ∂ 2φ1
⎟
⎜⎜
+ 2
z
+
+ 2
2 ⎟
r ∂r r ∂θ 2
⎝ r ∂r r ∂θ ⎠
σθ =
∂ 2φ 2
∂ 2φ1
+
z
∂r 2
∂r 2
τ rθ =
⎛ 1 ∂φ 2 1 ∂ 2φ 2
1 ∂φ1 1 ∂ 2φ1
⎜⎜ 2
−
z
+
−
r 2 ∂θ r ∂r∂θ
⎝ r ∂θ r ∂r∂θ
τ rz =
1 ∂φ3 z ∂ 2φ4
+
r ∂ r r ∂r∂θ
τ θz =
⎛ ∂ 2φ
1 ∂ 2φ 3
2 ∂φ 4 ⎞
⎟
+ z ⎜⎜ 24 +
r ∂r∂θ
r ∂r ⎟⎠
⎝ ∂r
⎞ ∂ϕ 4
⎟⎟ −
⎠ ∂r
(11)
3
⎛ z ∂ 2φ 3 z ∂ 3φ 3 ⎞
1 ∂ 2φ 4
2 ⎛ 1 ∂ φ4
⎜
⎟
+
z
+
−
2
⎜ r ∂r 2 ∂θ r 2 ∂r∂θ
r 2 ∂r∂θ 2 ⎟⎠
⎝
⎝ r ∂r
σ z = V (r ,θ ) − ⎜⎜
⎞
⎟⎟
⎠
Such that the equilibrium equations (10) are satisfied automatically.
The assumed stresses can then be obtained by expanding the stress functions as trigonometric functions along θ and
polynomials along r. The stress terms are also chosen to satisfy the traction-free condition along the cylindrical
surface.
σ r = (1 −
a2
a4
r
)
β
+
(
1
−
) [ β 2 cosθ + β 3sinθ ]
1
r4
r2
+ (1 − 4
a2
a4
a4
+
3
)[
β
cos
2
θ
+
β
sin
2
θ
]
+
(
1
−
4
) [ β 6 cos 2 θ + β 7 sin 2 θ ]
4
5
r2
r4
r4
+ r (1 − 5
a4
a6
r 2 a6
r
+
4
)
[
β
cos
3
θ
+
β
sin
3
θ
]
+
(
2
−
− ) [ β10 cos3θ + β11sin3θ ]
8
9
a2 r6
r4
r6
+ z{(1 −
a4
a2
r
+
(
1
−
) [ β13cosθ + β14sinθ ]
)
β
12
r4
r2
+ (1 − 4
a4
a2
a4
+
(
1
−
4
) [ β17 cos 2 θ + β18sin 2 θ ]
+
3
)[
β
cos
2
θ
+
β
sin
2
θ
]
15
16
r2
r4
r4
+ r (1 − 5
σ θ = (1 +
a4
a6
r 2 a6
+
4
)
[
β
cos
3
θ
+
β
sin
3
θ
]
+
(
2
−
− ) [ β 21cos3θ + β 22sin3θ ] }
r
19
20
r4
r6
a2 r6
a2
a4
r
)
β
+
(
3
+
)[ β 2 cosθ + β 3sinθ ]
1
r2
r4
− (1 + 3
− r (1 −
r 2 a4
a4
−
(
1
−
4
− )[ β 6cos2θ + β 7sin 2θ ]
)[
β
cos
2
θ
+
β
sin
2
θ
]
4
5
a2 r 4
r4
a4
a6
r 2 a6
r
+
4
)[
β
cos
3
θ
+
β
sin
3
θ
]
−
(
2
−
5
− )[β10cos3θ + β11sin3θ ]
8
9
a2 r 6
r4
r6
+ z{(1 +
a4
a2
r
)
β
+
(
3
+
)[β13cosθ + β14sinθ ]
12
r2
r4
− (1 + 3
a4
r 2 a4
)[
β
cos
2
θ
+
β
sin
2
θ
]
−
(
1
−
4
− )[ β17 cos2θ + β18sin 2θ ]
15
16
r4
a2 r 4
− r (1 −
a4
a6
r 2 a6
+
4
)[
β
cos
3
θ
+
β
sin
3
θ
]
r
−
(
2
−
5
− )[β 21cos3θ + β 22sin3θ ] }
19
20
r4
r6
a2 r 6
⎯ 731 ⎯
τ rθ = r (1 −
a4
)[ β 2 sinθ − β 3 cosθ ]
r4
(12)
a2
a4
r 2 a4
−
3
)[
β
sin
2
θ
−
β
cos
2
θ
]
−
(
1
−
2
+ )[ β 6sin 2θ − β 7 cos2θ ]
4
5
r2
r4
a2 r 4
− (1 + 2
r 2 a6
a4
a6
r
−
(
2
3
+ )[ β10 sin 3θ − β11cos3θ ]
−
4
)[
β
sin
3
θ
−
β
cos
3
θ
]
8
9
a2 r 6
r4
r6
− r (1 + 3
+ z{ r (1 −
− (1 + 2
a4
)[ β13sinθ − β14 cosθ ]
r4
a2
a4
r 2 a4
−
3
)[
β
sin
2
θ
−
β
cos
2
θ
]
−
(
1
−
2
+ )[β17sin 2θ − β18cos2θ ]
15
16
a2 r 4
r2
r4
r 2 a6
a4
a6
−
(
2
3
+ )[ β 21sin 3θ − β 22 cos3θ ] }
r
−
4
)[
β
sin
3
θ
−
β
cos
3
θ
]
19
20
a2 r6
r4
r6
− r (1 + 3
− r (1 −
a4
a2
2
r
)
β
−
(
1
−
)[β 40cosθ + β 41sinθ ]
39
r2
r4
− r (1 −
a4
a6
3
r
)[
β
cos
2
θ
+
β
sin
2
θ
]
−
(
1
−
)[ β 44 cos2θ + β 45sin 2θ ]
42
43
r4
r6
τ rz = (1 −
a2
a4
a4
)
β
+
(
1
−
)[
β
cos
θ
+
β
sin
θ
]
+
(
1
−
)[β 29 cos2θ + β 30sin 2θ ]
r
26
27
28
r2
r4
r4
+ r 2 (1 −
a6
a6
r
)[
β
cos
2
θ
+
β
sin
2
θ
]
+
(
1
−
)[ β 33cos3θ + β 34sin3θ ]
31
32
r6
r6
a8
1
a2
+ r (1 − 8 )[ β 35 cos3θ + β 36 sin3θ ] − 3 (1 − 2 )[ β 37 cos3θ + β 38 sin3θ ]
r
r
r
3
+ z{ r (1 −
a4
a4
+
2(
1
−
)[ − β 42 sin 2θ + β 43 cos 2θ ]
)[
−
β
sin
θ
+
β
cos
θ
]
40
41
r4
r4
+ 2r 2 (1 −
a6
)[− β 44 sin 2θ + β 45 cos 2θ ] }
r6
τ zθ = − r (1 −
a4
a4
)[
β
sin
θ
−
β
cos
θ
]
−
2
(
1
−
)[ β 29 sin 2θ − β 30 cos 2θ ]
27
28
r4
r4
− 2r 2 (1 −
a6
a6
)[
β
sin
2
θ
−
β
cos
2
θ
]
r
−
3
(
1
−
)[ β 33sin3θ − β 34 cos3θ ]
31
32
r6
r6
− 3r 3 (1 −
a8
3
a2
)[
β
sin3
θ
−
β
cos3
θ
]
+
(
1
−
)[ β 37 sin3θ − β 38 cos3θ ]
35
36
r8
r3
r2
+ z{ (3 −
a2
) β 39 + 4r[ β 40 cosθ + β 41sinθ ]
r2
+ (3 +
a4
a6
2
)[
β
cos
2
θ
+
β
sin
2
θ
]
r
+
(5
+
)[ β 44 cos 2θ + β 45 sin 2θ ] }
42
43
r4
r6
⎯ 732 ⎯
σz = β 23 + r β 24 + θ β 25
−
2
4
z
{ (1 + a 2 ) β 26 + r (1 + 3 a 4 )[ β 27 cosθ + β 28sinθ ]
r
r
r
− (3 − 7
a4
a6
2
r
)[
β
cos
2
θ
+
β
sin
2
θ
]
−
(
1
−
7
)[ β 31cos 2θ + β 32 sin 2θ ]
29
30
r4
r6
− r (7 − 13
a8
a6
3
r
−
(
5
−
13
)[ β 35 cos3θ + β 36 sin3θ ]
)[
β
cos3
θ
+
β
sin3
θ
]
33
34
r8
r6
1
a2
+ 3 (11 − 13 2 )[ β 37 cos3θ + β 38 sin3θ ] }
r
r
− z 2{ (3 +
a4
)[ − β 40 sinθ + β 41cosθ ]
r4
a4
a6
4
+ (1 + 4 )[− β 42 sin 2θ + β 43 cos 2θ ] + 4r (2 + 6 )[− β 44 sin 2θ + β 45 cos 2θ ] }
r
r
r
Table 1 Stress parameters deleted for each element
Stress assumption
Stress parameters deleted*
No. of β
3-dimensional element
A
37-45
36
B
33-38, 44, 45
37
C
31, 32, 33-38
37
2-dimensional element
D
10-45
9
* Extracted from original 45 β-parameters given by Eq. (12)
The designation letters (A, B, C and D) and the total number of β terms deleted for each of the respective stress
assumptions are given in Table 1. When the element is degenerated into two-dimensional case, the stress field (case D
in the Table) just has 9-β which also satisfies the compatibility equations.
It is well known that the assumed stress terms need for suppressing any possible zero energy deformation modes, and
the suppression of zero energy deformation modes is the practical method for satisfying the mathematical form of a
stability criterion, named the LBB condition [9, 10]. One necessary criterion to satisfy this condition for an individual
element is
nβ ≥ nq – nr
(13)
where nβ is the number of stress parameters; nq, the number of displacement degree of freedom and nr the number of
rigid body modes.
For the present special hexahedron element, nq = 48 and nr = 6. It follows that nβ ≥ 42. It shows that the displacement
field [i. e. eq. (9)] induces 6 zero energy deformation modes in which all translational components of displacement
vanish and all corner rotations are taken up same non-zero value. These zero energy deformation mode can be easily
suppressed if any of the rotations is constrained in the finite element mesh concerned. So, the modes can be deleted.
Thus arises nβ ≥ 36.
It is noted that just 36 or 37 stress parameters are chosen in present elements A, B and C. The reason is that the authors
[11-13] have found that successful results can be obtained when the number of stress parameters is much smaller than
that is required based on the LBB condition for an individual element in the elastic crack and stress concentration
analyses in which the combination of special elements and ordinary elements are used. It can be argued that stability of
finite element solution is a global problem, and it is not necessary for the individual element to satisfy the LBB
condition, particularly when an element is connected at a large number of nodal points with neighboring elements.
⎯ 733 ⎯
NUMERICAL EXAMPLES
The stress distributions are analyzed to evaluate the following four arrangements:
(1) Combining the special assumed stress hybrid elements with the 8-node Allman-type displacement elements [7];
(2) Using the 8-node Allman-type displacement elements [7] everywhere;
(3) Using the 8-node ordinary assumed stress hybrid elements everywhere, which are derived by expanding in stresses
in natural coordinates with 18-β parameters;
(4) Using the 8-node conventional assumed displacement isoparametric elements everywhere.
1. A thin square plate with a circular hole A square plate (8R×8R×0.4R) with a center hole of radius equal to R
(Figure 4) is acted by uniform tension σ 0 along two opposite edges. The problem is analyzed by only one layer of
elements using the mesh (Figure 5) with 16 elements for 1/4 of the plate. The computed circumferential stresses σ θ
around the rim of the hole are shown in Table 2. The total DOF used for each arrangement are also given in the Table.
It can be seen that the present special element D yields the best stress concentration factors and the best circumferential
stresses σ θ along the rim of the hole.
Figure 4: A square plate with a circular hole under tension
Figure 5: Mesh pattern for 1/4 of plate
Table 2 Computed circumferential stresses around the rim of the hole (A thin square plate)
σθ /σ 0
No.
Type of element
DOF
1 Present special element D and
4-node Allman-type element
75
θ
σθ /σ 0
2 4-node Allman-type
displacement element
75
Error%
σθ /σ 0
3 Ordinary 8-node assumed
hybrid stress element
150
Error%
σθ /σ 0
4 8-node assumed displacement
isoparametric element
150
Error%
σθ /σ 0
Reference [14]
Error%
σθ /σ 0
90º*
3.447
60º
2.325
-3.7
3.288
-1.259
0.0
2.032
-8.2
3.091
-1.040
1.958
3.287
14.4
-12.6
-13.7
29.3
-1.005
-15.7
2.020
-8.2
3.580
0º
31.7
-1.109
-13.1
2.324
24.7
-1.472
* Location of maximum stress
2. A thin rectangular plate with symmetric fillets A rectangular plate (26R×W×0.4R) with symmetric fillets of
rounded corners is subjected to bending (Figure 6). The problem is analyzed just using the mesh for one-fourth of the
plate shown in Figure 7. Since there are no reference solutions for the bending problem, the result of 2h/W = 0.4
obtained by the use of ordinary 4-node assumed displacement element of a very fine mesh with 768 elements and 2199
degree-of-freedom for one-fourth of the plate is considered as the references. The stress concentration factors for
different ratios of 2h/W are calculated by the equation (14) and the results are given in Table 3. As shown, the results
of stress concentration factors of bending obtained by the present special element D are also very close to the
references even in the coarse mesh.
⎯ 734 ⎯
Figure 6: Thin rectangular plate with symmetric
fillets under bending (h/R=2)
SCF =
Figure 7: Mesh pattern for 1/4 of the plate (2h/W = 0.4)
2
max
σ θmax d t σ θ
=
6M
σ0
(14)
where σ θmax : the largest circumferential stress σθ ; t: the thickness of the plate.
Table 3 Computed SCFs for a thin plate with symmetric fillets under bending (h/R=2)
No
.
Type of element
DOF
2h/W
0.3
0.4
0.5
0.6
1 Present special element D and
4-node Allman-type element
90
SCF
1.745
1.571
1.497
1.289
2 4-node Allman-type
displacement element
90
Error%
SCF
Error%
3 Ordinary 8-node assumed
hybrid stress element
180
4 8-node assumed displacement
isoparametric element
180
SCF
Error%
SCF
Error%
References
-2.7
1.535
-14.4
1.082
-39.7
1.381
-23.0
1.793
-3.5
1.373
-15.7
1.061
-34.8
1.205
-26.0
1.628
2.0
1.261
-14.0
1.004
-32.0
1.179
-20.1
1.476
-0.3
1.081
-16.4
1.006
-22.2
1.060
-18.0
1.293
3. A Square block with a center circular hole A square block (8R×8R×2R, Figure 8) with a center hole of radius
equal to R is acted by uniform tension σ 0 over two opposite faces. The Poisson’s ratio is taken as 0.25. The problem is
analyzed by using 64 elements over one-eighth of the block as shown in Figure 8. The resulting solutions for σ θmax and
σ zmax of θ = 90º at the face and the middle plane along the rim of the circular hole are given in Table 4. It is seen that the
use of present special elements around the rim of the hole yields the solutions also much closer to the reference
solutions in comparison to those by using ordinary hybrid stress elements and ordinary assumed displacement
elements.
Figure 8: Thick block with circular hole under uniform tension load,
mesh pattern for 1/8 of block
⎯ 735 ⎯
Table 4 Computed σ θmax / σ 0 and σ zmax / σ 0 at the rim of the circular hole
in thick square block under tension (ν = 0.25)
No.
1
Type of element
Middle
Point A
σ θmax / σ 0
σ θmax / σ 0
Face
Error% Point B
Middle
Point A
Error%
Present special element A and
8-node Allman-type element
3.545
-4.2
3.284
-2.1
0.27
Present special element B and
8-node Allman-type element
3.564
-3.6
3.190
-4.9
0.34
Present special element C and
8-node Allman-type element
3.566
-3.6
3.182
-5.1
0.35
2
8-node Allman-type element
3.314
-10.4
3.091
-7.8
0.40
3
Ordinary 8-node assumed
hybrid stress element
3.094
-16.4
2.773
-17.3
0.30
4
8-node assumed displacement
isoparametric element
3.393
-8.3
3.042
-9.3
0.43
Reference [1]
3.699
3.353
0.27
4. A rectangular block with symmetric fillets A rectangular block (26R × 10R × 6R) with symmetric fillets is acted
by uniform tension σ 0 over two opposite faces. The mesh pattern for one-eighth of the block is shown in Figure 9. The
computed stresses for σ θmax and σ zmax of the face and the middle plane along the rim of the fillets are given in Table 5.
The results obtained by using ordinary 8-node assumed displacement isoparametric elements of a very fine mesh with
10256 elements and 33864 degree-of-freedom for one-eighth of the block are considered as the references. As is
shown the results obtained by using present special element A are the closest to the reference solutions, and those
obtained by the use of present elements B and C are also in good agreement with the reference solutions.
Figure 9: Thick block with symmetric fillets under uniform tension load,
mesh pattern for 1/8 of block (h/R = 2)
CONCLUSIONS
Three kinds of 8-node special solid elements with drilling degrees of freedom based on the assumed stress hybrid
model for analyzing solids with traction-free circular surfaces have been developed and compared. In terms of
predicting stress concentration factors and stress distributions for solids with circular holes, numerical results
demonstrate that the special elements are far superior to the conventional assumed displacement and ordinary assumed
stress elements. The special elements are also found to be better than the 8-node displacement element with drilling
degrees of freedom.
Acknowledgement
This work was supported by the Knowledge Innovation Program of Chinese Academy of Sciences.
⎯ 736 ⎯
Table 5: Computed σ θmax / σ 0 and σ zmax / σ 0 at the rim of the fillets
in a thick block under tension (ν = 0.25)
σ θmax / σ 0
Middle
Point A
Error%
Present special element A and
8-node Allman-type element
3.449
-0.1
3.171
3.7
0.40
Present special element B and
8-node Allman-type element
3.434
-0.5
3.137
2.6
0.52
Present special element C and
8-node Allman-type element
3.435
-0.5
3.142
2.8
0.46
2
8-node Allman-type element
3.046
-11.7
2.640
-13.6
0.62
3
Ordinary 8-node assumed
hybrid stress element
2.417
-30.0
2.226
-27.2
0.27
4
8-node assumed displacement
isoparametric element
2.557
-25.9
2.325
-23.9
0.40
References
3.451
No.
Type of element
1
Face
Point B
σ zmax / σ 0
3.057
Error%
Middle
Point A
0.46
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