preservation, maintanence and improvement of industrial organism.pptx
Introduction
1. SOLID ELEMENT:
TETRAHEDRON
DEPARTMENT OF STRUCTURAL
ENGINEERING
BIRLA VISHVAKARMA
MAHAVIDHYALAYA
Presented by
Harshil Bhuva (140080720004)
Krutarth Patel (140080720009)
3. INTRODUCTION
For 3D solids, all the field variables are dependent of x, y
and z coordinates – most general element.
The element is often known as a 3D solid element or
simply a solid element.
A 3D solid element can have a tetrahedron and
hexahedron shape with flat or curved surfaces.
At any node there are three components in the x, y and z
directions for the displacement as well as forces.
5. TETRAHEDRON
It is the basic element for three dimensional
problems.
It is having four nodes, one at each corner.
6. SHAPE FUNCTION
As we know that tetrahedron is a four noded element and
so its having four values of shape function.
We define four lagrange type shape function N1, N2,N3
and N4 where shape function N1 has a value of 1 at node
1 and zero at other three nodes.
7. Using master element as shown in fig. We
can define the shape functions as
N1 = ξ , N2 = η , N3 = ƍ , N4 = 1- ξ- η- ƍ
8. The displacements u,v and w at x can be
written in terms of the unknown nodal values
as,
u=Nq
where,
1 2 3 4
1 2 3 4
1 2 3 4
node 1 node 2 node 3 node 4
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
N N N N
N N N N
N N N N
N
9. It is easy to see that shape function given by
above equation can be used to design the
coordinates x,y,z of the point at which
displacements are u,v and ware interpolated.
The isoparametric transformation is given by
x= N1x1 + N2x2 + N3x3
y= N1y1 + N2y2 + N3y3
z= N1z1 + N2z2 + N3z3
10. Using the chain rule for derivatives, say, of u, we have,
Thus the partial derivatives with respect to ξ , η and ƍ are
related to x,y, and z derivatives by the foregoing
relationship.
The jacobian of the transformation is given by,
du/d ξ
du/d η
du/d ƍ
= J
du/d x
du/d y
du/d z
11. The volume of element is given by ;
Ve = Idet JI ∫ ∫ ∫ d ξ d η d ƍ
Now using the polynomial integral formula we get
Ve = 1/6 Idet JI
du/d ξ
du/d η
du/d ƍ
du/d ξ
du/d η
du/d ƍ
du/d ξ
du/d η
du/d ƍ
=
X14 y14 z14
X24 y24 z24
X34 y34 z34
J =
0
1
0
1- ξ
0
1- ξ- η
12. • Now we can write that
Where J-1 = 1/det J
du/d ξ
du/d η
du/d ƍ
J -1 =
du/d x
du/d y
du/d z
Y24z34-y34z24 y34z14-y14z34 y14z24-y24z14
Z24x34-z34x24 z34x14-z14x34 z14x24-z24x14
X24y34-x34y24 x34y14-x14y34 x14y24-x24y14
14. Element Stiffness
The element stiffness matrix K is given by
In which Ve is the volume of the element
given by 1/6*|det J|.
e
T T
e eV
dV V k B cB B cB
15. The Force Term
The potential term associated with body force is
ʃe uTfdV= qT ʃʃʃ NT f det J dɳdζdx
Using the integration formula we have
Fe = Ve/4[fx,fy,fz,fx,fy,fz,…,fz]T
Let us now consider uniformly distributed traction on
the boundary surface. The boundary surface of a
tetrahedron is a triangle.
ʃ uTTdA = qT ʃ NTTdA = qTTe
The energy and Galerkin approaches yield the set of
equations
KQ = F
16. Stress Calculations
After solving previous equation, the nodal displacement
can be obtained.Since s cε and ε=Bq therefore stress is
s cBq
The three principal stresses can be calculated by using
the three invariants of the stress tensor:
I1= sx+ sy + sz
I2=sx sy +sy sz +sx sz -τ2
yz- τ2
xz- τ2
xy
I3=sx sysz + 2τyzτxz τxy -sx τ2
yz- sy τ2
xz- sz τ2
xy
17. a = I1
2/3 – I2
b = -2(I1 /3)3 + I1I2 /3 – I3
c = 2 (a/3)^0.5
Ѳ = 1/3 cos-1 ( -3b/ac)
The principal stresses are given by
sy = I1 /3 + c cos Ѳ
sy = I1 /3 + c cos (Ѳ+ 2ᴫ/3)
sy = I1 /3 + c cos (Ѳ+ 4 ᴫ/3)