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DESCRIPTION
Teori membran
Citation preview
x,u
y,v
z,w
Fx Fx
Fy
Fy
PELAT MEMBRAN
M A T R IK H O O K E [H ] & [C ]
+
+
+
)2(100000
0)2(10000
00)2(1000
0001--
000-1-
000--1
E
1
υ
υ
υ
υυ
υυ
υυ
−
2
2-100000
02
2-10000
002
21000
000-1
000-1
000-1
υ
υ
υυυυ
υυυ
υυυ
)2-)(1(E
E
υυ+
[C] =
[H] =
ρ = massa volumik; E= modulus Young; υ = koefisien Poisson
pres. 7
P L A N E S T R A IN
HIPOTESA : u = u (x,y) ; v = v(x,y) ; w = 0εz = ε xz = ε yz = 0
matrik Hooke menjadi :
−
−
+
2
2-100
01
01
)2-)(1(1
E
υυυ
υυ
υυ
−
−+
200
01
01
E
1υυ
υυυ
[ C ] = [ H ]-1 =
[H] =
y
z x Penampang dianalisa sbg plane
strain
PRES. 8
PLANE STRAIN
y,v
x,u
z
Kondisi Plane Strain
Retainning Wall
{ }
+
+
+=
=
=
yz
xz
xy
z
y
x
yz
xz
xy
zz
yy
xx
w,v,
,wu,
v,u,
w,
v,
u,
2
2
2
yz
xz
xy
zz
yy
xx
γ
γ
γ
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
xy
yy
xx
2ε
ε
ε
xy
yy
xx
γ
ε
ε
+ xy
y
x
v,u,
u,
u,
{ε } = = =
DEFORMASI LINIERDEFORMASI LINIERDARI TEORI ELASTIS 3 DDARI TEORI ELASTIS 3 D
PROBLEM 2D :PROBLEM 2D :
PRES. 5
H U B U N G A N
T E G A N G A N - R E G A N G A N
{σσ } = {H} { εε}
{H} = Matrik Hooke
atau :
{ εε } = [ H ]-1{σσ } = [C] { σσ }
εx = )( zyE1 νσνσσ −−x
ε y = )( zyE1 νσσνσ −+− x
εz = )( zyE1 σνσνσ +−x
xyG1
xy σγ = ; xzG1
xz σγ = ; yzG1
yz σγ =
G = ) 2(1
E
υ+
MATERIAL ISOTROP :
PRES. 6
PLANE STRESS
Struktur Membrane
x
y
z
xy
σ
σ
σ
σ
σ
σ
σσ
xy
xy
xy
x
x
y
y
σ
z
y
σ
σ
y
y
z= 0x
dx
dy
dA
Komponen tegangan suatu pelat tipis dibebani sejajar pelat
h
L
i
b
a
ba
elemen ke i
nodal elemen
P
P
P
P
σ z = σ xz = σ yz = 0
P L A N E S T R E S S
HIPOTESA :
+
−
)2(100
01
0-1
E1
υ
υ
υ
2
-100
01
01
υυ
υ
)-(1E
2υ
[C] =
[H] =
PRES. 9
X
Y
P
P
HUBUNGAN TEGANGAN REGANGAN
{ σ } = [Hσ] {ε}
[ ]HE
συ
υυ
υ=
− −
1
1 01 0
0 0 12
2
{ }σσστ
=
xyxy
{ }εεε
γ=
xyxy
AXISYMMETRIC
r
r,u
z,v
θθ,w
MATRIKS HOOKEUNTUK MATERIAL ISOTROP-HOMOGEN & LINIER
ELASTIS
Plane Stress
[ ]HE
=− −
1
1 01 0
0 01
2
2υ
υ
υυ
Plane Strain
[ ]
( )( )( )
( )
( )( )
( )
HE
=−
+ −
−
−−
−
11 1 2
11
0
11 0
0 01 2
2 1
υυ υ
υυ
υυ
υ
υ
Axisymmetric
[ ]( )
( )( )
( ) ( )
( )
( )
HE
Simetric=
−+ −
− −
−
−−
11 1 2
11 1
0
1 1 01 0
1 22 1
υυ υ
υυ
υυ
υυ
υυ
E = Modulus Young ; υ = Poisson’s ratio
X
Y
σ x
σ yτyx
τxy
σ
τ
τxy
τxy0
σ y
σ σx y+
2
σ σx y−
2
2θ
σ x
σ 1
σ 2A
B
OC
R
D
F
E
( )Rx y
xy=−
+
σ στ2
22
( )
σσ σ
τσ σ
1 2
1 2
2
2
,
max
=+
±
=−
x y
xy
R
LINGKARAN MOHR
E L E M E N 2 DIM E N SI
TRIANGULAR
QUADRILATERAL
3 nodal 6 nodal9 nodal
4 nodal 8 nodal12 nodal
PRES. 10
η
λ=1-ξ-η
ξ
3
1
2
N1 = λ =1-ξ-η
N2 = ξN3 = η
1
ξ
2 3
46
5η
N1=λ(2λ-1)
N2=4λ ξ
N3=ξ(2ξ-1)
N4=4ξ η
N5=η (2η -1)
N6=4λ η
Triangular 6 nodal
η
ξ
9
8
7
6
5 10
1 2 3 4
λ = 1-ξ -η N1= ½ λ (3λ -1)(3λ -2)
N2 =½ 9λ ξ (3λ -1)
N3=½ 9λ ξ (3ξ -1)
N4=½ ξ (3ξ -1)(3ξ -2)
N5=½ 9ξ η (3ξ -1)
PRES. 11
Triangular 3 nodal
Triangular 10 nodal
F U N G S I B E N T U KE L E M E N S E I T I G A
N6=½ 9ξ η (3η -1)
N7=½ η (3η -1)(3η -2)
N8=½ 9 λ η (3η -1)
N9=½ 9λ η (3λ -1)
N10=½ 54λ ξ η