Finite Element Analysis - UNIT-2

TO
ByBy
Dr.G.PAULRAJDr.G.PAULRAJ
Professor & HeadProfessor & Head
Department of Mechanical EngineeringDepartment of Mechanical Engineering
Vel Tech (Owned by RS Trust)Vel Tech (Owned by RS Trust)
Chennai-600 062.Chennai-600 062.

04/03/19 Dr.G.PAULRAJ,
Professor&Head(Mech.),Avadi,
VTRS,Chennai.
2
UNIT-II-ONE DIMENSIONAL
PROBLEMS

UNIT II ONE-DIMENSIONAL
PROBLEMS
 One Dimensional Second Order Equations –
Discretization – Element types- Linear and Higher
order Elements – Derivation of Shape functions and
Stiffness matrices and force vectors- Assembly of
Matrices - Solution of problems from solid mechanics
and heat transfer. Longitudinal vibration frequencies
and mode shapes. Fourth Order Beam Equation –
Transverse deflections and Natural frequencies of
beams.
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ONE DIMENSIONAL FINITE
ELEMENT ANALYSIS
 The geometry and other parameters of bar and beam element can be
defined in terms of only one spatial co-ordinate, the element is called as
one dimensional element.
 A bar is a member which resist only axial loads.
 Where u1 & u2 are the nodal variables.
 u(x) is the nodal displacement.
 A beam can resist transverse and twisting loads.
 Where v1 & v2 are the transverse displacements
of nodal variables.
 θ1 & θ2 are the slopes (rotations) of the nodes.
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SHAPE FUNCTION FOR ONE-
DIMENSIONAL ELEMENT
 Consider a one-dimensional element (line segment) of length l
with two nodes, one at each end, as shown in Figure. Let the
nodes be denoted as i and j and the nodal values of the field
variable as Φi and Φj.ϕ
The variation of inside theϕ
element is assumed to be linear
as ϕ(x) = α1 + α2x …(1)
where α1 and α2 are the
unknown coefficients.
By using the nodal conditions
(x) = Φi at x = xiϕ
(x) = Φj at x = xjϕ
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and Eq. (1), we obtain
Φi = α1 + α2xi
Φj = α1 + α2xj
 The solution of these equations gives
α1 = Φixj −Φjxi
l …(2)
α2 = Φj −Φi
l
 where xi and xjdenote the global coordinates of nodes i and j,
respectively. By substituting Eq. (2) into Eq. (1), we obtain
 Φ(x) = Φixj −Φjxi + Φj −Φi x ……(3)
l l
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 This equation can be written, after rearrangement of terms, as
Φ(x) = Ni(x)Φi +Nj(x)Φj = [N(x)] Φ(e)
….(4)
[N(x)] = [Ni(x) Nj(x)] … (5)
Ni(x) = xj −x
l …(6)
Nj(x) = x −xi
l
and Φ(e) = Φi = vector of nodal unknowns of elements e …(7)
Φj
 Note that the superscript e is not used for Φi and Φj for simplicity.
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 The linear functions of x defined in Eq. (6) are called
interpolation or shape functions.
 Note that each interpolation function has a subscript to denote
the node to which it is associated.
 Furthermore, the value of Ni(x) can be seen to be 1 at node i (x =
xi) and 0 at node j (x = xj). Likewise, the value of Nj(x) will be 0
at node i and 1 at node j.
 These represent the common characteristics of interpolation
functions.
 They will be equal to 1 at one node and 0 at each of the other
nodes of the element.
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STIFFNESS MATRIX
 The primary characteristics of a finite element are
embodied in the element stiffness matrix.
 For a structural finite element, the stiffness matrix
contains the geometric and material behavior
information that indicates the resistance of the element
to deformation when subjected to loading.
 Such deformation may include axial, bending, shear,
and torsional effects.
 For finite elements used in nonstructural analyses, such
as fluid flow and heat transfer, the term stiffness matrix is
also used, since the matrix represents the resistance of the
element to change when subjected to external influences.
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STIFFNESS MATRIX …,
we know that,
 Strain, {e} = [B] {u*}
 {e}T
= [B]T
{u*}T
where,
 {e} is a strain matrix
 [B] is strain- displacement matrix
 {u*} is a degree of freedom
we know that,
 Stress, {δ} = [E] {e}
 {δ} = [D] {e}
 where [E] =[D] = Young’s modulus.
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 Strain energy expression is given by,

U = ∫½ {e}T
{δ} dv
v
Substitute {e}T
and {δ} values,

U = ∫½ [B]T
{u*}T
[D] {e} dv
v

= ½ {u*}T
∫[B]T
[D] {e} dv
v
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 Substitute {e} values,

U = ½ {u*}T
∫[B]T
[D] [B] {u*} dv
v

U = ½ {u*}T
∫[B]T
[D] [B] dv {u*}
v

So, Stiffness matrix, [K] = ∫[B]T
[D] [B] dv
v
where,
 [B] is strain- displacement relationship matrix
 [D] Elasticity matrix or Stress-strain relationship matrix
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 In 1D problem,
 Strain, e = du/dx
 Where, u = Displacement function
 [D] = [B] = E = Young’s modulus.
 In Beam problem, Strain, e = Curvature = d2
u/dx2
 [D] = [EI] = Flexural rigidity
 Properties of stiffness Matrix:
1. It is symmetric.
2. The sum of elements in any column is equal to zero.
3. It is an unstable element. So, the determinants is equal to zero.
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STIFFNESS MATRIX FOR
1-D LINEAR BAR ELEMENT
 Consider a 1-D bar element with nodes 1 and 2 as shown in figure. Let
u1 and u2 be the nodal displacement parameters or otherwise known as
degrees of freedom.
 We know that,
Stiffness matrix, [K] = [B]T
[D] [B] dv
v
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1 2
l
u1 u2
x

 In 1-D Bar element,
 Displacement function u = N1u1 + N2u2
Where , N1 = l-x/L and N2= x/L
We know that, Strain displacement matrix, [B] = dN1/dx dN2/dx
= -1/L 1/L
Substitute [B], [B]T
and [D] values in stiffness matrix equation[Limits is 0 to l]
L
[K] = -1/L x E x -1/L 1/L dv
1/L
0
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l
1/L2
-1/L2
[K] = E dv
-1/L2
1/L2
0
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l
1/L2
-1/L2
[K] = E A dx [ dv= A dx]
-1/L2
1/L2
0
l
1/L2
-1/L2
1/L2
-1/L2
L 1 -1
[K] = E A dx = AE [x] = AE/L
-1/L2
1/L2
-1/L2
1/L2
0 -1 1
0

 The properties of stiffness matrix are satisfied.
1. It is symmetric.
2. The sum of elements in any column is equal to zero.
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FINITE ELEMENT EQUATION
FOR 1-D BAR ELEMENT
 We know that, General force equation is,
[k] {u} = {F} ----------- (1)
 where [k] is stiffness matrix
 {u} is displacement vector and
 {F} is force vector in the coordinate directions
 The element kij of stiffness matrix maybe defined as the force at
coordinate i due to unit displacement in coordinate direction j.
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 For 1-D bar element, stiffness matrix [k] is given by,
 For two noded bar element,
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1 -1
[K] = AE/L
-1 1
F1
{F} = &
F2
u1
{u} = , Substitute [K] {F} {u}
values
u2 in equ. (1)
1 -1 u1
{F}=AE/L This is a finite element equation for 1D 2-noded bar element
-1 1 u2

P1: A two noded bar element is shown in figure. The nodal displacement are
u1=5mm and u2= 8mm. Calculate the displacement at x= l/4, l/3 and l/2.
Solution:
Displacement function for 2-noded bar element is given by,
u = N1u1 + N2u2
Where, N1 = l-x/L and N2 = x/L => u =[ l-x/L] u1 + [x/L]u2
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1 2
L
u1= 5mm u2=8mm
x

 Substitute x= l/4, u1 = 5 and u2 = 8 in the above equation,
 u = 5.75 mm at x=l/4
 u = 6 mm at x=l/3
 u = 6.5 mm at x=l/2
 Result: u = 5.75 mm at x=l/4, u = 6 mm at x=l/3 & u = 6.5 mm at x=l/2
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P2: A one dimensional bar is shown in figure. Calculate the following:
(i) Shape function N1 and N2 at point P.
(ii) If u1=3mm and u2=-5mm, calculate the displacement u at ponit P.
 We know that,
Actual length of the bar, L=x2-x1=36-20=16mm
The distance between point 1 and point P is, x= 24-20=4mm
Displacement function for two noded bar element is given by,
u = N1u1+ N2u2
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X1
2
L
X1= 20mm
u1= 5mm
X2=36mm
u2=-5mm
P
X=24mm

 Where, N1 = l- x/L and N2 = x/L
N1 = 16- 4/16 = 0.75mm
N2 = 4/6 = 0.25mm
 Substitute N1, N2, u1 and u2 values in the above equation,
u = N1u1+N2u2
= (0.75)(3)+0.25(-5)
u = 1mm
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THE QUADRATIC BAR
ELEMENT
Determination of Shape Function:
 The Three noded quadratic bar element is shown in figure.
 The displacement at any point within the element is now interpolated
from the three nodal displacement using the shape functions as follows:
 u(x) = N1(x)u1 + N2(x)u2+ N3(x)u3 ---------------- (1)
 The shape functions Ni vary quadratically with in the element.
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X=0 X=lX=l/2
1
→u1
23
X
→u3
→u2

 Let u(x) be given by the complete quadratic polynomial
 u(x) = a0 + a1x + a2x2
-------------- (2)
 We know that u(0) =u1, u(l) = u2 and u(l/2) = u3,. Hence, from the above
equation,
 u1 = a0, u2 = a0 + a1l + a2l2
and u3 = a0 + a1(l/2) + a2(l/2)2
 Solving for ai, we obtain
a0 = u1
a1 = (4u3 – u2 – 3u1)/l
a2 = (2u1 + 2u2 + 4u3)/l2
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 Thus we have
u(x) = u1 +[ (4u3 – u2 – 3u1 ) / l] x +[ (2u1 + 2u2 – 4u3 ) / l2
]x2
------------(3)
Rearranging the terms, we get
u(x) = u1( 1 – 3x/l + 2x2
/l2
) + u2(-x/l + 2x2
/l2
) + u3(4x/l – 4x2
/l2
) -----(4)
Comparing equations (1) and (4)
N1 = 1 – 3x/l + 2x2
/l2
,
N2 = -x/l + 2x2
/l2
,
N3 = 4x/l – 4x2
/l2
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STIFFNESS MATRIX FOR
QUADRATIC BAR ELEMENT
 Given the shape functions, we get the strain- displacement relation
matrix [B] as
 e = du/dx = d/dx [N] {u}
 Where [B] = [dN1/ dx dN2/dx dN3/dx]
 = [2/l2
(2x – 3 l/2) 2/l2
(2x – l/2) -4/l2
(2x - l)]
 We know that the stiffness matrix of the element
 Stiffness matrix, [K] = [B]T
[B] EA dx
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1
2/l2
(2x - 3l/2)
=
2/l2
(2x - l/2) [2/l2
(2x - 3l/2) 2/l2
(2x - l/2) -4/l2
(2x - l)] EA dx
0
-4/l2
(2x - l)
On evaluating all the integrals, we obtain the element stiffness matrix as
7 1 -8
[k] = AE/3l 1 7 -8
-8 -8 16
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 A steel bar of length 800mm is subjected to an axial load of 3KN as
shown in figure. Find the elongation of the bar, neglecting
self weight. Take E = 2x 105
N/mm2
, A = 300mm2
.
 Solution:
 We can divide the bar into two elements as shown in
Figure.
 Now the bar has 2elements with 3 nodes.
 Displacement at
node 1 = u1
node 2 = u2
node 3 = u3
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3 x 103
N
800mm
400mm
400mm1
3
1
2
1
2
3 x 103
N
800mm
400mm
400mm1
u3
u1
u2
1
2

 For 1D two noded bar element, the finite element equation is,
 For element 1: (node 1, 2)
 Finite element equation is,
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F1 1 -1 u1
=AE/L
F2
-1 1 u2
1 -1 u1 F1
A1E/L1 =
-1 1 u2 F2
400mm
u2
u1
1

------------ (1)
 For element 2 (Nodes 2,3)
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1 -1 u1 F1
300x2x105
/400 =
-1 1 u2 F2
1 -1 u1 F1
150x103
=
-1 1 u2 F2
3 x 103
KN
400mm
u3
u2
2
1 -1 u2 F2
A2E/L2 =
-1 1 u3 F3

---------------- (2)
Assemble the finite element equations (1) and (2).
-------- (3)
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1 -1 u2 F2
150x103
=
-1 1 u3 F3
1 -1 0 u1 F1
150x103
-1 2 -1 u2 = F2
0 -1 1 u3 F3

 Applying boundary conditions:
 u1 = 0
 F3 = 3x103
N
 Substitute u1, F1,F2 and F3 values in equation (3)
 Here u1=0. So, neglect first row and first column of [K] matrix. Hence
the final reduced equation is,
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1 -1 0 0 0
150x103
-1 2 -1 u2 = 0
0 -1 1 u3 3x103
2 -1 u2 0
150x103
=
-1 1 u3 3x103

 150 x 103
(2u2 – u3) = 0 ----------- (4)
 150 x 103
(-u2 + u3) 3 x 103
-----------(5)
 Solving, 150 x 103
(u2) = 3 x 103
u2 = 0.02mm
Substitute u2 value in equation (4),
u3 = 0.04mm
Result:
1. Displacement at node 1 , u1 = 0
2. Displacement at node 2 , u2 = 0.02 mm
3. Displacement at node 3 , u3 = 0.04 mm
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 Consider a taper steel plate of uniform thickness, t = 25 mm as shown in
figure. The Young’s modulus of the plate, E = 2 X 105
N/mm2
and weight
density, ρ = 0.82 x 10-4
N/mm3. In addition to its self-weight, the plate is
subjected to a point load p = 100N at its mid point. Calculate the
following by modeling the plate with
two finite elements:
(i) Global force vector {F}.
(ii) Global stiffness matrix [K].
(iii) Displacement in each element.
(iv) Stresses in each element.
(v) Reaction force at the support.
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300 mm
600 mm
75 mm
p
150 mm
x

 In this problem, the area of the element is varying at each c-s. If we
consider this area variation, the problem will be tedious. So the given
taper bar is considered as stepped bar as shown in figure.
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300 mm
600 mm
W3 = 75 mm
p
W1 = 150 mm
x
W2
3
2
1
W1 = 150 mm
300 mm
300 mm
p
75 mm
1
2

 Solution:
 Area at node 1, A1 = Width X Thickness = W1 X t1 = 150 X 25 = 3750 mm2
 Area at node 2, A2 = Width X Thickness = W2 X t2 = [(W1 + W2)/2] X t2
A2 = 2812.5 mm2
 Area at node 3, A3 = Width X Thickness = W3 X t3= 75 X 25 = 1875 mm2
 Average area of element (1):
 A1 = (Area of node 1 + Area of node 2) /2 = 3281.25 mm2
 Average area of element (2):
 A2 = (Area of node 2 + Area of node 3) /2 = 2343.75 mm2
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 The steel plate is subjected to self-weigth.So, we have to findout the
body force acting at nodal points 1,2 and 3.
1
 We know that, Body force vector, {F} = (ρ A L) /2
1
F1 1
 For element (1) : Force vector = (ρ1 A1 L1) /2
F2 1 1
= (0.82 X 10-4 X 3281.25 X 300)/2 X
F1 40.359 1
= --------- (1)
F2 40.359
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F2 1
 For element (2) : Force vector = (ρ2 A2 L2) /2
F3 1 1
= (0.82 X 10-4 X 2343.75 X 300)/2 X
F2 28.828 1
= --------- (2)
F3 28.828
Assembling the force vector i.e. assemble the equation (1) and (2)
F1 40.359
F2 = 69.187
F3 28.828
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 A point load of 100N is acting at node 2 as shown in figure. So, add
100N in F2 vector.
F1 40.359
Global Force Vector, F2 = 169.187
F3 28.828
Finite element equation for 1D- plate element is given by
F1 1 -1 u1
= A E /L
F2 -1 1 u2
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1 2
10.937 -10.937 1 u1 F1
2 X 105
= ------- (4)
-10.937 10.937 2 u2 F2
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1 -1 u1 F1
A1E/L1 =
-1 1 u2 F2
300mm
u2
u1
1

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2 3
7.8125 - 7.8125 2 u2 F2
2 X 105
= ------- (5)
- 7.8125 7.8125 3 u3 F3
1 -1 u2 F2
A2E/L2 =
-1 1 u3 F3
300mm
u3
u2
2
p

 Assemble the finite element equations (4) and (5)
 Apply the B.Cs i.e., at node 1, Displacement u1 = 0. Substitute u1, F1, F2
and F3 value in equ. (6)
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10.937 -10.937 0 u1 F1
2x105
-10.937 18.749 -7.8125 u2 = F2 ------- (6)
0 -7.8125 7.8125 u3 F3
10.937 -10.937 0 0 40.359
2x105
-10.937 18.749 -7.8125 u2 = 169.187
0 -7.8125 7.8125 u3 28.828

 In the above equation, u1 = 0. So, neglect first row and first column of
[K] matrix. The reduced equation is,
2x105
(18.749u2 - 7.8125 u3) = 169.187 ------------- (7)
2x105
(-7.8125u2+7.8125u3) = 28.828 ------------- (8)
Solving, 2x105
x (10.936)u2 = 198.015
u2 = 9.053 x 10-5
mm
 Substitute u2 value in equ. (7),
u3 = 10.898 x 10-5
mm
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2x105
18.749 -7.8125 u2 = 169.187
-7.8125 7.8125 u3 28.828

 Stresses :
 Stress, σ1 = E ε12 = E (u2-u1)/ L1 = 0.060 N/mm2
 Stress, σ2 = E ε23 = E (u3-u2)/ L1 = 0.0123 N/mm2
 Reaction force:
 We know that, {R} = [K]{u} – {F}
 2x105
(-10.937 x 9.053 x 10-5
) = 40.359
R1 = -238.379 N04/03/19 Dr.G.PAULRAJ,
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10.937 -10.937 0 0 40.359
{R} =2x105
-10.937 18.749 -7.8125 u2 - 169.187
0 -7.8125 7.8125 u3 28.828

 In figure, a load P = 60 X 103
N is applied as shown. Determine the
displacement field, stress and support reactions in the body.
Take E = 20 X 103
N/mm2
.
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Wall
P
1.2mm
150mm
150mm
250mm2

 Solution:
 In this problem, we should first determine whether contact occurs
between the bar and the wall. To do this, assume that the wall does not
exit. The deformation at node is given by,
 δ = PL/AE = (60 X 103
) x 150 /(250 x 20 X 103
) = 1.8 mm
 The gap between the wall and node 3 is 1.2 mm. So, that contact occurs
between the bar and the wall.
 Displacement at point 3 ,(u3) = 1.2 mm. From this result, we see that
contact does not occur.
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1.2mm
P
150mm
150mm
Wall
u3u2
u1

 The problem has to be resolved, since the B.Cs are now different. The
displacement at u3 is specified to be 1.2 mm. Consider the 2 element
finite element model in figure. The B.Cs are u1= 0 and u3 = 1.2mm.
 The structural stiffness matrix [K1
] and [K2
] are:
1 -1 1 -1
 [k12
] = AE/L =(250 x 20 x 103
) /150
-1 1 -1 1
 The global unreduced stiffness matrix is ,
1.667 -1.667 0
 [K] = 20 x 103
-1.667 3.333 -1.667
0 -1.667 1.667
The finite element equation can be written,
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3P
150mm
150mm
250mm2
1
2
1 2

1.667 -1.667 0
 [K] = 20 x 103
-1.667 3.333 -1.667
0 -1.667 1.667
Retaining 2 equation yields,
20x103
(3.33u2 -1.67 x 1.2) = 60 x 103
=> u2 =1.5 mm
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1.667 -1.667 0 0 0
20x103
-1.667 3.333 -1.667 u2 = 60 x 103
0 -1.667 1.667 1.2 0
P
150mm
u2
u1 1
150mm
u3
u2 2
1.2
mm

 Stresses:
For element 1:
 Stress, σ1 = E ε12 = E (u2-u1)/ L1 = 2000 N/mm2
For element 2:
 Stress, σ2 = E ε23 = E (u3-u2)/ L1 = -400 N/mm2
 Reaction force:
We know that, {R} = [K]{u} – {F}
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R1 1.667 -1.667 0 0 0
R2 =20x103
-1.667 3.333 -1.667 1.5 - 60 x 103
R3 0 -1.667 1.667 1.2 0

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R1 -50010 0 => R1 = -50010 N
R2 = 59982 - 60 X 103
=> R2 = 0
R3 -10002 0 => R3 = -10002 N
Verification : R1 + R2 + R3 = -50010 + 0 -10002 = -60012 N =60 X103
N
(Applied force)

SPRING ELEMENT
 Consider a system of springs connected in series as shown in Figure.
The analysis of the system (to find the nodal displacements under a
prescribed set of loads) can be conducted using the finite element
method.
 For this, the equilibrium relations of a typical spring element are to be
derived first.
 Let the stiffnesses of the springs be denoted k1, k2, …, kn. Let Fi and Fj be
the forces applied at nodes i and j, and ui and uj be the displacements of
nodes i and j, respectively.
 The relations between the nodal forces and nodal displacements of a
typical spring element e can be derived using the relation:
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52

 Force = Spring stiffness × Net deformation of the spring
 Thus, the forces at nodes i and j can be expressed as
Fi = ke ui −uj -------- (1)
 (where ui is assumed to be larger than uj so that the compressive force Fi
leads to a decrease in the length of the spring), and
Fj = ke uj −ui -------- (2)
 (where uj is assumed to be larger than ui so that the tensile force Fj leads
to an increase in the length of the spring). Equations (1) and (2) can be
expressed in matrix form as04/03/19 Dr.G.PAULRAJ,
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1 −1 ui Fi
 ke = = -------- (3)
−1 1 uj Fj
From Eq. (3), the characteristic (stiffness) matrix of the element can be
identified as
i j
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1 -1 i
[K] = k
-1 1 j l
K
Fi
i j
Fj

 where the degree of freedom numbers corresponding to the rows and
columns of the stiffness matrix [K] are also indicated. Equation (3) can
be expressed as
[K] u= F ------------ (4)
 where u and F denote the vectors of nodal displacements and nodal
forces, respectively.
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56

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57

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58

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59

BEAM ELEMENT
Selection of Nodal D.O.F
 A beam is a straight bar element that is primarily subjected to transverse
loads.
 The deformed shape of a beam is described by the transverse
displacement(v) and slope (rotation-θ) of the beam is shown figure.
Typical Beam Section and
its deformation
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M
X
Z

 Hence, the transverse displacement and rotation at each end of the beam
element are treated as the unknown degrees of freedom.
 For a beam element of length L lying in the xz plane, as shown in
Figure, the four degrees of freedom in the local (xz) coordinate system
are indicated as vi, θi, vj, and θj.
 According to Euler- Bernoulli theory of beam bending, the entire c-s has
the same transverse deflection v as the neutral axis. The Euler- Bernoulli
beam element is shown in figure.
The Euler- Bernoulli beam element
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x=Lx=0
vj
vi
θi θj

 Hence the deflection are small and we assume that the rotation of each
section is the same as the slope of deflection curve at that point. i.e.,
(dv/dx)
 Therefore, the axial deformation of any point P away from the neutral
axis (due to bending alone) is given by uP = -(z)(dv/dx).
 Thus, if v(x) is determine, then the entire state of deformation of the
body is completely determined.
 Abeam element is a simple line element, representing the neutral axis of
the beam.
 To ensure the continuity of the deformation at any point ( i.e., on the
neutral axis), to ensure that v and dv/dx are continuous.
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 Taking two nodal D.O.F viz., v and θ (= dv/dx).
 These nodal D.O.F permits to readily assign prescribed essential B.Cs
i.e., v= 0= dv/dx at fixed end.
 A prescribed value of moment load can readily taken into account with
the rotational D.O.F θ.
 At all nodes, the following sign conversions are used:
(i) Moments are +ve in the CCW direction.
(ii) Rotations are positive in the CCW direction.
(iii) Forces are +ve in the +ve z direction.
(iv) Displacements are +ve in the +ve z direction
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SHAPE FUNCTION OF BEAM
ELEMENT
 The displacement field v(x) assumed for the beam element should be
such that it takes on the values of deflection and the slope at either end
as given by the nodal values vi, θi, vj, θj. Let v(x) be given by
 v(x) = c0 + c1x + c2x2
+ c3x3
------- (1)
 Differentiating w.r.t x, we have
 dv/dx = c1 + 2c2x + 3c3x2
---------- (2)
 At x= 0 and L , we have
 vi = c0, θi = c1, vj = c0 + c1L + c2L2
+ c3L3
, θj = c1 + 2c2L + 3c3L2
----- (3)
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 We can solve for the coefficients c0, c1, c2, c3 in terms of the nodal D.O.F.
vi, θi, vj, θj and substitute in equation (1) we get,
v(x) = N1vi + N2 θi + N3vj + N4 θj ---------- (4)
 Where the shape function N are given by,
 N1 = 1 – 3x2
/L2
+ 2X3
/L3
, N2 = x – 2x2
/L + x3
/L2
 N3 = 3x2
/L2
-2x3
/L3
, N4 = -x2
/L + x3
/L2
vi
θi
 Rewrite equation , v(x) = [N] {δ} = [N1 N2 N3 N4] vj ------------ (5)
θj
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STIFFNESS MATRIX FOR BEAM
ELEMENT
 The strain at any point in the beam is given by
 { ε} = εx = du/dx = d/dx [-z (dv/dx)] = -z (d2
v/dx2
)
 From equation (5) we have
 εx = -z [d2
N1/dx2
d2
N2/dx2
d2
N3/dx2
d2
N4/dx2
]{δ}
= -z [((-6/L2
) + (12x/L3
)) ((-4/L) + (6x/L2
)) ((6/L2
)- (12x/L3
)) ((-2/L) + (6x/L2
))]{δ}
Thus the strain displacement relation matrix [B] is given by
[B] = -z [((12x/L3
) - (6/L2
)) ((6x/L) – (4/L2
)) ((6/L2
) - (12x/L3
)) ((6x/L2
) – (2/L))]
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 We can obtain the element stiffness matrix as

[k] = ∫[B]T
[B] E dv
v
L (12x/L3
) - (6/L2
)
= ∫ ∫(-z) (6x/L) – (4/L2
) x (E)(-z) [((12x/L3
) - (6/L2
)) ((6x/L) – (4/L2
))
A 0 (6/L2
) - (12x/L3
) ((6/L2
) - (12x/L3
)) ((6x/L2
) – (2/L))] (dA) dx
(6x/L2
) – (2/L)
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
Noting that ∫z2
dA = I, the second moment of area of the C-S and
A
 carrying out the integration w.r.t x, we get
12 6L -12 6L
6L 4L2
-6L 2L2
[k] = EI/L3
-12 -6L 12 -6L
6L 2L2
-6L 4L2
 The element nodal force vectors are now obtained for specific cases of
loading
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P1: A beam, fixed at one end and supported by a roller at the other end, has
a 20KN concentrated load applied at the centre of the span, as shown in
figure. Calculate the deflection under the load and construct the shear
force and bending moment diagrams for the beam.
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69
20KN
500 cm
E= 20 X 106
N/cm2
I = 2500 cm
500 cm

 Solution:
 Let the beam is divided into 2 elements having nodes 1,2,3.
 Let v1, θ1, F1, M1- Vertical deflection, Slope, Shear force and Bending
moment at node 1.
 v2, θ2, F2, M2- Vertical deflection, Slope, Shear force and Bending
moment at node 2.
moment at node 3.
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E= 20 X 106
N/cm2
I = 2500 cm
θ1 M1
v1, F1 v2, F2
θ2 M2
v3,F3
θ3 M3
1 2
321

 (a) Determination of nodal deflections:
Element (1):
 It has nodal displacements v1, θ1 ,v2, θ2 at nodes 1 and 2.
 The element stiffness matrix for element (1) is given by
12 6L -12 6L
6L 4L2
-6L 2L2
[k1] =E1I1/L1
3
-12 -6L 12 -6L
6L 2L2
-6L 4L2
 Now E1I1/L1
3
= (20 x 106
x 2500) / 5003
= 400 N/cm
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v1 θ1 v2 θ2
12 3000 -12 3000 v1
3000 106
-3000 5x105
θ1
[k1] =400 -12 -3000 12 -3000 v2
3000 5x105
-3000 106
θ2
Element (2):
v2 θ2 v3 θ3
12 3000 -12 3000 v2
3000 106
-3000 5x105
θ2
[k2] =400 -12 -3000 12 -3000 v3
3000 5x105
-3000 106
θ3
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 The global stiffness matrix can be written as,
v1 θ1 v2 θ2 v3 θ3
12 3000 -12 3000 0 0 v1
3000 106
-3000 5x105
0 0 θ1
[k2] =400 -12 -3000 24 0 -12 3000 v2
3000 5x105
0 2 x106
-3000 5x105
θ2
0 0 -12 -3000 12 -3000 v3
0 0 -3000 5x105
-3000 106
θ3
The finite element equation is given by
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 [k] {u} = {F}
12 3000 -12 3000 0 0 v1 F1
3000 106
-3000 5x105
0 0 θ1 M1
i.e., 400 -12 -3000 24 0 -12 3000 v2 = F2 ----- (1)
3000 5x105
0 2 x106
-3000 5x105
θ2 M2
0 0 -12 -3000 12 -3000 v3 F3
0 0 -3000 5x105
-3000 106
θ3 M3
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 The nodal conditions are,
 v1 = 0, θ1 = 0, v3 = 0, F2 = -20000 N, M2 = 0, M3 = 0. Applying the nodal
conditions in the above F.E. equation (1), we get,
12 3000 -12 3000 0 0 0 F1
3000 106
-3000 5x105
0 0 0 M1
i.e., 400 -12 -3000 24 0 -12 3000 v2 = -20000
3000 5x105
0 2 x106
-3000 5x105
θ2 0
0 0 -12 -3000 12 -3000 0 F3
0 0 -3000 5x105
-3000 106
θ3 M3
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 Since v1 = 0, θ1 = 0, v3 = 0, neglect 1st
, 2nd
, 5th
rows and columns in the
global stiffness matrix. From the remaining terms, the F.E. equation can
be written as
24 0 3000 v2- 20000
400 0 2 x106
5x105
θ2 = 0
3000 5x105
106
θ3 0
 θ2 = -0.003125 rad, θ3 = 0.0125 rad and v2 = -3.646 cm
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 (b) Determination of shear force and bending moment:
 The shear force at node 1 is given by
 F1 = 400 (-12v2 + 3000 θ2 ) = 13750 N
 Bending moment at node 1 is
 M1 = 400 (-3000 v2 + 5 x 105
θ2) = 3750200 N-cm
 The shear force at node 3 is
 F3 = 400 (-12v2 - 3000 θ2 - 3000 θ3) = 6250 N
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78
13750
F1 F2
F3 Shear Force Diagram
3750200 N-
cm
0
0
-6250
+ + +
- - -
+ +
+

 For the beam and loading shown in figure, determine (1) the slopes at 2
and 3 and (2) the vertical deflection at the midpoint of the distributed
load.
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79
12KN/m
1 m
E= 200 GPa
I = 4 x 106
mm4
1 m
1 2

 Solution:
 Element (1):
 Let v1, θ1, F1, M1- Vertical deflection, Slope, Shear force and Bending
moment at node 1.
moment at node 2.
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1 m
1
v1 F1
θ2 M2
v2 F2
θ1 M1

 Now E1I1/L1
3
= (200 x 109
x 4 x 10-6
) / 13
= 800 x 103
N/m
v1 θ1 v2 θ2
12 6L -12 6L v1
[k1] =800 x 103
6L 4L2
-6L 2L2
θ1 --------- (1)
-12 -6L 12 -6L v2
6L 2L2
-6L 4L2
θ2
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Element (2):
 Now E2I2/L2
3
= (200 x 109
x 4 x 10-6
) / 13
= 800 x 103
N/m
v2 θ2 v3 θ3
12 6L -12 6L v2
[k2] =800 x 103
6L 4L2
-6L 2L2
θ2 --- (2)
-12 -6L 12 -6L v3
6L 2L2
-6L 4L2
θ3
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12KN/m
1 m
2
θ2 M2 θ3 M3
v2, F2 v3,F3
32

 The global stiffness matrix can be written as,
v1 θ1 v2 θ2 v3 θ3
12 6 -12 6 0 0 v1
6 4 -6 2 0 0 θ1
[K] = 800 x103
-12 -6 24 0 -12 6 v2 ------- (3)
6 2 0 8 -6 2 θ2
0 0 -12 -6 12 -6 v3
0 0 -6 2 -6 4 θ3
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 The finite element equation is given by
[k] {u} = {F}
12 6 -12 6 0 0 v1 F1
6 4 -6 2 0 0 θ1 M1
800 x103
-12 -6 24 0 -12 6 v2 = F2 ---(4)
6 2 0 8 -6 2 θ2 M2
0 0 -12 -6 12 -6 v3 F3
0 0 -6 2 -6 4 θ3 M3
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 The nodal conditions are,
 For uniformly distributed load
 F2 = -wL/2 ; M2 = -wL2
/12 ;
 F3 = -wL/2 ; M3 = wL2
/12 [w=12 x 103
N/m]
 F2 = -6000 N ; M2 = -1000 N-m ;
 F3 = -6000 N ; M3 = 1000 N-m
 v1=0, θ1 =0, v2=0,v3 =0, F1=0,F2 =-6000 N, M2=-1000 N-m, F3=-6000 N
;M3=1000 N-m
 Applying the nodal conditions in the above F.E. equation (4), we get,
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12 6 -12 6 0 0 0 0
6 4 -6 2 0 0 0 0
800 x103
-12 -6 24 0 -12 6 0 = -6000 ---(4)
6 2 0 8 -6 2 θ2 -1000
0 0 -12 -6 12 -6 0 -6000
0 0 -6 2 -6 4 θ3 1000
 In the above equation v1, θ1 ,v2, v3 =0 ,so delete 1st
, 2nd
, 3rd
, and 5th
row and
column. Hence the equation reduces to
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8 2 θ2 -1000
800 x103
=
2 4 θ3 1000
Slope or rotation at 2, θ2 = -2.679 x 10-4
rad
Slope or rotation at 3, θ3 = 4.464x 10-4
rad
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 The vertical deflection at the midpoint of the distributed load.
 Consider the element (2) and its loading arrangements as shown in
figure.
 For any beam element the vertical deflection can be obtained by using
the relation,
 v(x) = N1v1 + N2 θ1 + N3v2 + N4 θ2
 Where the shape function N are given by,
 N1 = 1 – 3x2
/L2
+ 2X3
/L3
, N2 = x – 2x2
/L + x3
/L2
 N3 = 3x2
/L2
-2x3
/L3
, N4 = -x2
/L + x3
/L2
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12KN/m
1 m
2
θ2 M2 θ3 M3
v2, F2 v3,F3
21

 Also in order to simplify the analysis, consider the element (2) as the
separate element and its nodes 1 and 2.
 v1 = 0, θ1 = -2.679 x 10-4
rad, v2 = 0, θ2 = 4.464x 10-4
rad
 Now at x= 0.5m (i.e., at mid point of element (2))
 N1v1 = 0, N2 θ1 = [0.5 – (2(0.5)2
/1) + ((0.5)3
/12
)] (-2.67 x 10-4
)
 = 3.338 x 10-5
 N3v2 =0 , N4 θ2 = [(– (0.5)2
/1) + ((0.5)3
/12
)] (4.46x 10-4
)
 = 5.575 x 10-5
 v = 0-3.338 x 10-5
+ 0 – 5.575 x 10-5
= -8.913 x 10-5
m = -0.08913 mm
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 i.e., Displacement at the midpoint is directing downwards ass shown in
figure.
v = -0.08913 mm
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2

FRAME ELEMENT
 The bar element can permit only axial deformation and the beam
element can permit only transverse deflection.
 By combining the bar and the beam elements, we obtain the “frame
element” which enables us to model typical problems of framed
structures which in general involve both types of deformation.
 A typical planar frame element is shown in figure, and the
corresponding element equations are
[k] {δ} = {f}
Plane frame element
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ujui
x=Lx=0
vj
vi
θi θj

 Where
 {δ} = {ui, vi, θi, uj, vj, θj}
ui uj
----------- (1)
vi θi vj θj
12EI/L3 6EI/L2 -12EI/L3 6EI/L2 vi
6EI/L2 4EI/L -6EI/L2 2EI/L θi --------(2)
[kbeam] = -12EI/L3 6EI/L2 12EI/L3 -6EI/L2 vj
6EI/L2 2EI/L -6EI/L2 4EI/L θj
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AE/L - AE/L ui
[kbar] =
- AE/L AE/L uj

 and the element stiffness matrix for frame element (using equs. 1 &2) is
ui vi θi uj vj θj
AE/L 0 0 -AE/L 0 0 ui
0 12EI/L3
6EI/L2
0 -12EI/L3
6EI/L2
vi
[k] = 0 6EI/L2
4EI/L 0 -6EI/L2
2EI/L θi
-AE/L 0 0 AE/L 0 0 uj
0 -12EI/L3
-6EI/L2
0 12EI/L3
-6EI/L2
vj
0 6EI/L2
-2EI/L 0 -6EI/L2
4EI/L θj
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 The beam elements also have different orientations. Figure shows the
inclined frame element.
 It have two displacement and a rotational deformation for each node.
 The nodal displacement vector is given by
{δ} = {ui, vi, θi, uj, vj, θj}
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ui
y
x i
j
uj
vj
θi
θj
ujy
ujx
uiy
uix
Fi sin θθi
Fi cos θ
Fi
Fj sin θθj
Fjcos θ
Fj
vi
vj

 For beam element in its own local coordinate frame, we have the force
deflection relations, viz
 The tranformation matrix would be
cos θ sin θ
[T] =
-sin θ cos θ
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1 -1 u1 Fi
AE/L =
-1 1 u2 Fj
ui = uix cos θ + uiy sin θ
uj = ujx cos θ + ujy sin θ

 P1: The framed structure made of steel shown in figure. To calculate the
deflection using the frame element.
Structure details
FE model
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20 mm dia.
Steel rod
4 m
3 m
Steel I-beam
A=2000mm2
I=106
mm4
10KN
Steel I-beam
A=2000mm2
I=106
mm4
20 mm dia.
Steel rod
10KN
Y
y2
X
x2
θ
y1
x1
3
21 4
1
2

 Element 1:
 L = 3m =3000mm, A = 2000mm2, E = 200,000 N/mm2, I = 106
mm4. The
element stiffness matrix in its coordinate frame is given in the form
ui vi θi uj vj θj
AE/L 0 0 -AE/L 0 0 ui
0 12EI/L3
6EI/L2
0 -12EI/L3
6EI/L2
vi
[k] = 0 6EI/L2
4EI/L 0 -6EI/L2
2EI/L θi
-AE/L 0 0 AE/L 0 0 uj
0 -12EI/L3
-6EI/L2
0 12EI/L3
-6EI/L2
vj
0 6EI/L2
-2EI/L 0 -6EI/L2
4EI/L θj
-------- (1)
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ui vi θi uj vj θj
0.133x106
0 0 -0.133x106
0 0 ui
0 0.889x102
0.133x106
0 -0.889x102
0.133x106
vi
[k1
] = 0 0.133x106
0.267x109
0 - 0.133x106
0.133x106
θi
- 0.133x106
0 0 0.133x106
0 0 uj
0 - 0.889x102
-0.133x106
0 0.889x102
-0.133x106
vj
0 0.133x106
0.133x106
0 - 0.133x106
0.267x109
θj
-------- (2)
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 The local coordinate frame of element 1 is identical to the global frame
and hence no transformation is required
 Element 2: L=5000 mm, A=314.16 mm2
, E = 200,000 N/mm2, I =
7,854mm4
. The element stiffness matrix in its coordinate frame is given
in the form
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ui vi θi uj vj θj
0.126x105
0 0 - 0.126x105
0 0 ui
0 0.151 0.377x10 3
0 -0.151 0.337x103
vi
[k2
] = 0 0.377x10 3
0.126x105
0 -0.377x103
0.628x106
θi
- 0.126x105
0 0 0.126x105
0 0 uj
0 - 0.151 -0.377x103
0 0.151 - 0.377x103
vj
0 0.337x103
0.628x106
0 - 0.377x103
0.126x107
θj
-------- (3)
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100

 The local coordinate frame is inclined w.r.t global reference frame and
hence we need to transform the element stiffness matrix into global
frame. For any orientation θ, the transformation matrix related local and
global d.o.f. can be readily shown to be given by
cos θ sin θ 0 0 0 0
-sin θ cos θ 0 0 0 0
[T] = 0 0 1 0 0 0
0 0 0 cos θ sin θ 0
0 0 0 -sin θ cos θ 0
0 0 0 0 0 1
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 Thus the element stiffness matrix in the global frame is written in the form
[K2
] = [T]T
[k2
] [T]
0.6 -0.8 0 0 0 0
0.8 0.6 0 0 0 0
[K2
] = 0 0 1 0 0 0
0 0 0 0.6 -0.8 0
0 0 0 0.8 0.6 0
0 0 0 0 0 1
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0.6 0.8 0 0 0 0
-0.8 0.6 0 0 0 0
0 0 1 0 0 0
0 0 0 0.6 0.8 0
0 0 0 -0.8 0.6 0
0 0 0 0 0 1
0.126x105
0 0 - 0.126x105
0 0
0 0.151 0.377x10 3
0 -0.151 0.377x103
0 0.377x10 3
0.126x105
0 -0.377x103
0.628x106
- 0.126x105
0 0 0.126x105
0 0
0 - 0.151 -0.377x103
0 0.151 - 0.377x103
0 0.377x103
0.628x106
0 - 0.377x103
0.126x107

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0.126x105
0 0 - 0.126x105
0 0
0 0.151 0.377x10 3
0 -0.151 0.377x103
0 0.377x10 3
0.126x105
0 -0.377x103
0.628x106
- 0.126x105
0 0 0.126x105
0 0
0 - 0.151 -0.377x103
0 0.151 - 0.377x103
0 0.377x103
0.628x106
0 - 0.377x103
0.126x107
0.6 0.8 0 0 0 0
-0.8 0.6 0 0 0 0
0 0 1 0 0 0
0 0 0 0.6 0.8 0
0 0 0 -0.8 0.6 0
0 0 0 0 0 1
0.452x104
0.603x104
-0.302x103
-0.452x104
-0.603x104
-0.302x103
0.603x104
0.804x104
0.226x10 3
-0.603x104
-0.804x104
0.226x10 3
-0.302x103
0.226x103
0.126x105
0.302x103
-0.226x103
0.628x106
-0.452x104
-0.603x104
0.302x103
0.452x104
0.603x104
0.302x103
-0.603x104
-0.804x104
-0.226x103
0.603x104
0.804x104
-0.226x103
0.302x103
0.226x10 3
0.628x106
0.302x103
-0.226x103
0.126x107

u1 v1 θ1 u2 v2 θ2
0.133x106
0 0 -0.133x106
0 0 u1
0 0.889x102
0.133x106
0 -0.889x102
0.133x106
v1
[k1
] =0 0.133x106
0.267x109
0 - 0.133x106
0.133x106
θ1
-0.133x106
0 0 0.133x106
0 0 u2
0 - 0.889x102
-0.133x106
0 0.889x102
-0.133x106
v2
0 0.133x106
0.133x106
0 - 0.133x106
0.267x109
θ2
u2 v2 θ2 u3 v3 θ3
0.452x104
0.603x104
-0.302x103
-0.452x104
-0.603x104
-0.302x103
u2
0.603x104
0.804x104
0.226x10 3
-0.603x104
-0.804x104
0.226x10 3
v2
[K2
] = -0.302x103
0.226x103
0.126x105
0.302x103
-0.226x103
0.628x106
θ2
-0.452x104
-0.603x104
0.302x103
0.452x104
0.603x104
0.302x103
u3
-0.603x104
-0.804x104
-0.226x103
0.603x104
0.804x104
-0.226x103
v3
0.302x103
0.226x10 3
0.628x106
0.302x103
-0.226x103
0.126x107
θ3
VTRS,Chennai.
104

Result:
 u2 = -0.12mm; v2= 2.564 mm
 θ2 = 0.0013 rad θ3 = θ4 = 3.26 x 10-4
rad
VTRS,Chennai.
105

ONE-DIMENSIONAL HEAT
TRANSFER
 The GDE for the steady state one-dimensional conduction heat transfer
with convective heat loss from lateral surfaces is given by
k d2
T /dx2
+ q = [P/Ac] h (T - T∞) --------- (1)
Where
 k = Co-efficient of thermal conductivity of the material,
 T = temperature,
 q = internal heat source /unit volume,
 Ac= the C-S area,
 h = convective heat transfer co-efficient, and
 T∞ = ambient temperature.
VTRS,Chennai.
106

 The weighted residual statement can be written as
L
∫ W [k (d2
T /dx2
) + q – (P/Ac) h (T - T∞) ] dx = 0 --------- (2)
0
 By performing integration by parts, the weak form of the DE can be obtained
as
L L L
∫ W [k (d2
T /dx2
)+ ∫Wq dx –∫W(P/Ac) h (T - T∞) ] dx = 0
0 0 0
VTRS,Chennai.
107

L

∫ W d[k (dT /dx)]
0 u d v
L L L L
W k(dT/dx) -∫[k (dW/dx dT/dx)dx + ∫ Wq dx - ∫W(P/Ac) h (T - T∞)] dx= 0
0 0 0
0
VTRS,Chennai.
108
∫ u dv = uv - ∫ v du

 The weak form, for a typical mesh of n finite elements, can be written as,
VTRS,Chennai.
109
L L
∫k (dW/dx) (dT/dx)dx + ∫W(P/Ac) h (T) dx
0 L L 0 L
= ∫Wq dx +∫W(P/Ac) h (T∞) dx + W k (dT/dx)
0 0 0
n L L
Σ ∫[k (dW/dx) ( dT/dx) dx + ∫W(P/Ac)hT dx
k=1 0 n L 0 L
= Σ ∫W [q + (P/Ac ) hT∞ ] dx + Wk(dT/dx) --------- (3)
k=1 0 0

PROCEDURE FOR DEVELOPING THE
1-D HEAT TRANSFER BAR ELEMENT
 Figure shows the typical heat transfer element, with its nodes and nodal
d.o.f.
 Equation for 1D bar element heat transfer nodal temperature:
 T(x) = 1- (x/L) T1 + x/L T2 -------- (4)
 dT/ dx = (T2 – T1)/L -------- (5)
VTRS,Chennai.
110
Node 2Node 1
x=Lx=0
T2T1

 W1 = 1- (x/L), dW1/dx = -1/L
-------- (6)
 W2 = (x/L), dW2/dx = 1/L
 Now compute the elemental level contribution to Equ. (3)
 LHS first term (Equ. (3)). Similar to 1D bar finite element, k replaces AE
and T replaces u. Thus we have
VTRS,Chennai.
111
1 -1 T1
k/L
-1 1 T2

 LHS Second term (Equ. (3)). With W1,
L
∫ 1- (x/L) (P/Ac) h (1- (x/L) ) T1 + (x/L)T2 dx
0
L L
= (Ph/Ac) ∫[ 1- (x/L) ]2
dx T1 + ∫(x/L) (1- (x/L) ) T2 dx
0 0
= ((PhL)/(6Ac)) [2T1 + T2]
VTRS,Chennai.
112

 With W2,
L
∫ (x/L) (P/Ac) h (1- (x/L) ) T1 + (x/L)T2 dx = ((PhL)/(6Ac)) [T1 + 2T2]
0
We can now write the second term on the LHS in matrix notation as
2 1 T1
(PhL)/(6Ac)
1 2 T2
VTRS,Chennai.
113

 RHS First term. Similar to the RHS first term of the bar element equation
replacing q0 by q0 + (PhT∞)/(Ac) , we can write
L/2
q0 + (PhT∞)/(Ac)
L/2
RHS Second term. Similar to RHS second term of the bar element
-Q0
QL
 Where Q0, QL represent the heat flux at the ends of the element (nodes).
Thus the element level equations can be written as
VTRS,Chennai.
114

 LHS = Element conduction matrix
 RHS First term = Nodal heat flux
 RHS Second vector = Net heat flux at nodes
VTRS,Chennai.
115
1 -1 2 1 T1 L/2 -Q0
k/L +(PhL)/(6Ac) = q0 + (PhT∞)/(Ac) +
-1 1 1 2 T2 L/2 QL

 P1 : A 1 mm dia., 50mm long aluminium pin-fin as shown in figure,
used to enhance the heat transfer from a surface wall maintained at
3000
C. Use k = 200 W/m/0
C for ,aluminium, h = 20 W/m2
C , T∞ = 300
C.
Calculate the temperature distribution between in the fin.
A Pin- Fin.
VTRS,Chennai.
116
h T∞
Tw
Wall
L=50 mm

 With two equal elements, L1 = L2 =0.025m. Assembling the two
elemental level equations , we get the
1 -1 0 2 1 0 T1
200/0.025 -1 2 -1 + ((π)(0.001)(20) (0.025))/((6)(π) (0.0005)2
) 1 4 1 T2
0 -1 1 0 1 2 T3
0.0125 QWall
= ((π)(0.001)(20))/((π) (0.0005)2
) (30) 0.025 + 0
0.0125 Qtip
From the B.Cs , T1 = 3000
C, Qtip = 0.
Substituting and solving the equations, we get
T1 = 226.1850
C, T3 = 203.5480
C04/03/19 Dr.G.PAULRAJ,
VTRS,Chennai.
117
1
2
3
21
L1 L2

SOLUTION OF EIGENVALUE
PROBLEMS
 The general form of G.E for un-damped free vibration of the structure is
given by
 [k]nxn {ui}nx1 = ω2
i [m]nxn{ui}nx1, i= 1, 2, 3.., n ---(1)
 Where, [k] = stiffness matrices
[m] = mass matrices
ωi = Natural frequencies
{ui} = Mode shape
 Then equ (1) can be rewrite as
 [m]-1
[k] {ui} = ω2
i {ui}
 [A] {ui} = λi {ui}
 Where [A] = [m] λ-1
[k] and λi = ω2
i
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 118

 [A] {u} = λ {u} is known as Standard form of eigen value problem.
 [k] {u} = λ [m]{u} is known as non-standard form of eigen value
problem.
 The determinant | k – λm | = 0 is called as the characteristic equation.
 {k – λm } {u} = 0 is called Eigen vector.
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 119

 P1: Find the natural frequencies of longitudinal vibration of the
unconstrained stepped bar shown in figure.
 Solution:
 The dynamic equation of motion
for the un-damped free vibration
of whose system is given by
 [ [K] – [M] ω2
] (u) = 0 ------(1)
Where
[K] = Global stiffness matrix
[M] = Global mass matrix
ω = Natural frequency
{u} = Displacement vector
Dr.G.PAULRAJ,
VTRS,Chennai.
u3
A(1) = 2A
A(1) = A
Element 1 Element 2
u1 u2
x
1
2
3
l(1) = L/2 l(2) = L/2
Stepped bar with axial degree of freedom
04/03/19 120

 For two element stepped bar, the equation of motion(1) can be derived
as follows:
 For element (1): (Between nodes 1&2)
1 -1
 Element stiffness matrix [k1] = (A1E1)/l1
-1 1
1 -1 1 -1
= (2A E)/(L/2) = (4A E)/(L) ------(2)
-1 1 -1 1
2 1
Element stiffness matrix [m1] = (ρ1 A1 l1)/6
1 2
(Assumingconsistent mass matrix)Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 121

2 1 2 1
= (ρ.2A. L/2)/6 = (ρAL)/6 ---(3)
1 2 1 2
For element (2): (Between nodes 2&3)
1 -1
 Element stiffness matrix [k2] = (A2E2)/l2
-1 1
1 -1 1 -1
= (A E)/(L/2) = (2A E)/(L) ------(4)
-1 1 -1 1
2 1
Element stiffness matrix [m2] = (ρ2A2 l2)/6
1 2
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 122

2 1 2 1
= (ρ.A. L/2)/6 =(ρAL)/12 ------(5)
1 2 1 2
By combining the equas.(2) & (4), we get global stiffness matrix as
2 -2 0
[K] = (2AE)/L -2 3 -1
0 -1 1
Similarly by combining the equas.(3) & (5), we get global mass matrix as
4 2 0 u1
[M]= (ρAL)/12 2 6 1 and {u} = u2
0 1 2 u3
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 123

 Now , by substituting the values of above matrices in equation (1) we get
2 -2 0 4 2 0 u1
(2AE)/L -2 3 -1 - (ρAL)/12 2 6 1 ω2
u2 = 0 --(6)
0 -1 1 0 1 2 u3
 We know that, for getting the non-zero solution of circular frequency ω
for {u}, the determinant of the coefficient matrix [ [K] – [M] ω2
] should be
zero.
 i.e., | [K] – [M] ω2
| = 0
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 124

2 -2 0 4 2 0
i.e., (2AE)/L -2 3 -1 - (ρAL)/12 2 6 1 ω2
= 0
0 -1 1 0 1 2
(or) 2 -2 0 4 2 0
-2 3 -1 -(ρL2
ω2
)/24E 2 6 1 = 0 ------(7)
0 -1 1 0 1 2
Let λ = (ρL2
ω2
)/24E
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 125

 Then, the equation (7) becomes,
(2-4λ) (-2-2λ) 0
(-2-2λ) (3 -6λ) (-1- λ ) = 0 ----(8)
0 (-1-λ) (1-2 λ)
i.e., (2-4λ) [(3 -6λ) (1-2 λ) - (-1-λ)(-1-λ)] - (-2-2λ) [(-2-2λ) (1-2 λ)-0] +0 = 0
i.e., -36 λ3
+ 90λ2
+ -36 λ = 0
i.e., 18λ(1-2λ) (λ-2) = 0
(or) λ(1-2λ) (λ-2) = 0
In the above equation either λ=0 or (1-2λ)=0 or (λ-2) = 0
When λ = 0 we get (ρL2
ω2
)/24E= 0 => ω2
= 0 (or) ω=0
i.e., the first natural frequency ω1 = 0
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 126

 When (1-2λ)=0, we get 1- ( 2 (ρL2
ω2
)/24E)= 0
 i.e., ω2
=12E/ ρL2
=> ω = 3.46( E/(ρL2
))1/2
 The second natural frequency ω2 = 3.46( E/(ρL2
))1/2
 When (λ-2)=0, we get (ρL2
ω2
)/24E) - 2= 0
 i.e., ω2
=48E/ ρL2
=> ω = 6.93( E/(ρL2
))1/2
 The third natural frequency ω3 = 6.93( E/(ρL2
))1/2
 That is the natural frequencies of unconstrained stepped bar are,
 ω1 = 0
 ω2 = 3.46( E/(ρL2
))1/2
rad/sec
 ω3= 6.93( E/(ρL2
))1/2
rad/sec
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 127

 P2: Find the mode shapes (i.e., eigenvectors) for the natural
frequencies of longitudinal vibration of the unconstrained stepped
bar shown in figure.
 Solution:
 The natural frequencies of the
unconstrained stepped bar as
described in the above problem are :
 ω1 = 0
 ω2 = 3.46( E/(ρL2
))1/2
rad/sec
 ω3= 6.93( E/(ρL2
))1/2
rad/sec
 The natural frequencies follow some well defined deformation patterns
called mode shapes. It is observed that the first frequency ω1 = 0
corresponds to rigid body mode, whereas the second and third
frequencies (ω2 , ω3) correspond to elastic deformation modes.
Dr.G.PAULRAJ,
VTRS,Chennai.
u3
A(1) = 2A
A(1) = A
Element 1 Element 2
u1 u2
x
1
2
3
l(1) = L/2 l(2) = L/2
Stepped bar with axial degree of freedom
04/03/19 128

 To find the mode shape corresponding to natural frequencies ωi, we
must solve the equation[ [K] – [M] ω2
] (u) = 0.
 The equation (6) of the above problem is given by
2 -2 0 4 2 0 u1
(2AE)/L -2 3 -1 -(ρAL)/12 2 6 1 ω2
u2 = 0 ------ I
0 -1 1 0 1 2 u3
 By selecting λ = (ρL2
ω2
)/24E the above equation can be simplified as
(2-4λ) (-2-2λ) 0 u1
(-2-2λ) (3 -6λ) (-1- λ ) u2 = 0 ---- II
0 (-1-λ) (1-2 λ) u3
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 129

 By equating the determinant of coefficient matrix for {u} to zero, we get
 λ(1-2λ) (λ-2) = 0 which implies λ = 0 (or) λ = 0.5 (or) λ = 2.
 i.e., the eigen values are
 λ1 0
λ2 = 0.5 writing in the increasing order.
λ3 2
 To find 1st
mode shape:
 Let λ = 0
 Now the equation II implies
2u1 – 2u2 = 0 ---(i)
-2u1 + 3u2 – u3 = 0 ---(ii)
-u2 + u3 = 0 ---(iii)
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 130

 Equa. (i) => u1 = u2 and Equa (iii) u2 = u3
 Hence u1 = u2 = u3
 For ω1 = 0 (or) λ = 0, corresponding to rigid body
 u1 = u2 = u3 = 1
 That is , 1st
mode shape is
u1 1
Um1 = u2 = 1 u1
u3 1
 To find 2nd
mode shape:
 Let λ = 0.5
 Now the equation II implies
– 3u2 = 0 ---(iv)
-3u1–1.3u3 = 0 ---(v)
-1.5u2 = 0 ---(vi)Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 131

 Equa(iv) => u2 = 0 and Equa(v) => 3u1 = -1.5u3 (or) 2u1 = -u3
 That is , 2nd
mode shape is
u1 1 1
Um2 = u2 = 0 = 0 u1
u3 -2u1 -2
 To find 3rd
mode shape:
 Let λ = 2
 The equation II implies
-6u1– 6u2 = 0 ---(vii)
-6u1-9u2–3u3 = 0 ---(viii)
- 3u2 -3u3= 0 ---(ix)
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 132

 Equa(vii) => u1= -u2 and Equa(ix) => u2 = -u3
 That is , 3rd
mode shape is
u1 u1 1
Um3 = u2 = -u1 = -1 u1
u3 u1 1
The three mode shapes corresponding to λ = 0, λ = 0.5 and λ = 2
(i.e.,corresponding to ω1 , ω2 and ω3) are shown in figure.
Dr.G.PAULRAJ,
VTRS,Chennai.
04/03/19 133

Types of mode-shapes
(Eigenvectors)
1st
mode shape
2nd
mode shape
Dr.G.PAULRAJ,
VTRS,Chennai.
u3u1
u21
2
3
L/2 L/2
x
1 1 1
u1 u2 u3
0 L/2 L
x
u1
u2
u3
0
L0
1
-2
x
u1
u2
u3
0 0
1 1
-1
3rd
mode shape
04/03/19 134

Finite Element Analysis - UNIT-2

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