FEM

1. SOLID ELEMENT:
TETRAHEDRON
DEPARTMENT OF STRUCTURAL
ENGINEERING
BIRLA VISHVAKARMA
MAHAVIDHYALAYA
Presented by
Harshil Bhuva (140080720004)
Krutarth Patel (140080720009)

2. Content
 INTRODUCTION
 SHAPE FUNCTION
 ELEMENT STIFFNESS
 STRESS CALCULATIONS

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.

4. TETRAHEDRON
ELEMENT

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

13.  ε= Bq where B is 6x12 matrx given by
B=
A11 0 0 A12 0 0 A13 0 0 -Ã1 0 0
0 A21 0 0 A22 0 0 A23 0 0 –Ã2 0
0 0 A31 0 0 A32 0 0 A33 0 0 –Ã3
0 A31 A21 0 A32 A22 0 A33 A2 3 0 –Ã3 –Ã2
A31 0 A11 A32 0 A12 A33 0 A13 –Ã3 0 -Ã1
A21 A11 0 A22 A12 0 A23 A13 0 –Ã2 -Ã1 0

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)

18. REFERENCE
 Ashok D.Belegundu , “Finite Elements in
Engineering” PHI Pvt. Ltd.