Chapter 6

Bar and Beam Elements

Hari R. Parajuli
Dr. of Eng., Post Doctorate
Lecturer of Civil Engineering
 Bar Element
Shape function for natural coordinate system in ξ

Relation between global

coordinate system x and natural
coordinate system ξ is given by
Displacement field is function of shape
function and nodal displacements
 Bar Element
Strain is function of displacement
 Bar Element
Strain energy is given by
 Bar Element
Strain energy is given by
 Bar Element
Nodal load vector-Body Force
 Bar Element
Nodal load vector-Surface Traction
 Bar Element in 2D space
Let u and are global coordinate
systems and u’ and v’ are local
coordinate system as shown in
the fig. θ be the rotation angle
between two coordinates
systems. I and j are two nodes
of the element of length L. Thus

Where, c=cosθ and s=sinθ

 Bar Element in 2D space
In matrix form  u i'   c s 0 0   ui 
 '   
 vi   − s c 0 0   v i 
 '=  
u j   0 0 c s  u j 
 v 'j   0 0 −s
 
c v j 
 
Global stiffness matrix is given by
c −s 0 0 1 0 −1 0  c s 0 0
s c 0 0  AE  0 0 0 0  − s c 0 0
{K } = 
0 0 c − s  L  −1 0 1 0  0 0 c s
    
0 0 s c 0 0 0 0  0 0 −s c 
 Bar Element in 2D space

c 0 −c 0  c s 0 0
s 0 −s 0 −s c 00
AE 
{K } = 
L −c 0 c 0  0 0 c s
  
 −s 0 s 0  0 0 −s c 

c2 cs −c2 − cs 
 2
AE  s2 − cs −s 
{K } =
L  c2 cs 
 2 
 s 
 Hermite Interpolation Function
H(ξ) = ∑A (ξ) f (ξi ) +∑B (ξ) f ' (ξi )
n n

i=1 i=1
i i

Ai (ξ ) = ∑ 1− 2(ξ −ξi ) L'i (ξi )  L2i (ξ )

n

i =1

Bi (ξ ) = ∑(ξ −ξi ) L2i (ξ )

n

i=1

H (ξ ) = ∑1− 2(ξ −ξi ) L (ξi )  L (ξ ) f (ξi ) + ∑(ξ −ξi ) L2i (ξ ) f ' (ξi )
n n
' 2
i i
i=1 i=1
Hermite Interpolation Function
Hermite Interpolation Function
Hermite Interpolation Function

N1 N2

N3
N4
Beam Element

v1 v2
θ1 θ2
ξ
1 2
ξ=-1 ξ=1
 Beam Element
Moment curvature relationship is given by v1 v2
θ1 θ2
d 2v
M = EI 2 ξ
dx 1 2
 dv   dv2  ξ=-1 ξ=1
v (ξ ) = H 1v1 + H 2  1  + H 3 v2 + H 4  
 dξ   dξ 
L
Where, v= θ
2
L L
v (ξ ) = H 1v1 + H 2θ1 + H 3 v2 + H 4θ 2
2 2

In matrix form ( v ) = [ H ]{δ }

[ H] = H1
L 
[δ ] = [v1 θ1 v2 θ2 ]
L T
Where, H2 H3 H4 
 2 2 
 Beam Element
,

d2v ∂2H1 L ∂2H2 ∂2H3 L ∂2H4 

2 { } [ ] { }
, = 2 δ =B δ v1 v2
dx  ∂x
2
2 ∂x 2
∂x2
2 ∂x  θ1 θ2
,
ξ
∂ H1 L ∂ H2 ∂ H3 L ∂ H4 
2 2 2 2 1 2
2 { }
M = EI  2 δ ξ=-1 ξ=1
 ∂x 2 ∂x ∂x 2 ∂x 
2 2

M = [ D][ B]{δ }

∂2H1 L ∂2H2 ∂2H3 L ∂2H4 

[ B] =  2 2 
 ∂x 2 ∂x2
∂x2
2 ∂x 
∂2H1 3 ∂2H2 1
= ξ = ( −1 + 3ξ )
∂ξ 2
2 ∂ξ 2
2
∂2 H3 3 ∂2 H4 1
= − ξ = (1 + 3ξ )
∂ξ 2
2 ∂ξ 2 2
 Beam Element
, v1 v2
θ1 θ2
, ξ
1 2
, Strain energy is given by
ξ=-1 ξ=1

T
U = ∫ ([ B ]{δ }) EI [ B ]{δ } dx
1L
20
 Beam Element
, v1 v2
Relation between natural coordinate ξ and θ1 θ2
, global coordinate x is given by ξ
2
( x − x1 )
1 2
, ξ = −1 + ξ=-1
x 2 − x1 ξ=1

2
d ξ = dx
L
Changing integrand from dx to dξ
1 L
U = {δ } ∫ [ B ] EI [ B ]{δ } d ξ
T 1 T

2 −1 2
Using chain rule dv dv d ξ dv 2
= =
dx d ξ dx dξ L

d 2v 4 d 2v
= 2
dx 2
L dξ 2
 Beam Element
v1 v2
dv d ξ
,
dv dv 2 θ1 θ2
Using chain rule = = ξ
d ξ dx dξ L
,
dx 1 2
,
d 2v 4 d 2v ξ=-1 ξ=1
= 2
dx 2
L dξ 2
Since, {v}=[H]{δ}
2 T
d v2
T 16  d H   d H 
2
2

 2 
= {δ } 4  2   2 { }
δ
 dx  L  dξ   dξ 

 d 2 [ H ]   3ξ L  −1+ 3ξ  3ξ L  1+ 3ξ 
 =   −  
2 
 dξ   2 2 2  2 2  2 

 d2 [ H]  dξ 2 3ξ L  −1+3ξ  3ξ L 1+3ξ  16

[ B] =  2   =    −   4
 d ξ   
dx 2 2  2  2 2  2  L
 Beam Element
,  3ξ 
 2 
,  
 L  − 1 + 3ξ  
 
, 16  2  2    3ξ L  − 1 + 3ξ  3ξ L  1 + 3ξ 
[B] [B] = 4  −
T
    
L  3ξ  2 2 2  2 2 2 
−
 2 
 
 L  1 + 3ξ  

 2  2   

9 2 3Lξ  −1 + 3ξ  9 3Lξ  1 + 3ξ  
4ξ 
4  2 
 − ξ2
4

4  2  


 L2  −1 + 3ξ 
2
3Lξ  −1 + 3ξ  L2  −1 + 9ξ 2  
   −    
16  4 2  4  2  4 4 
[ ] [ ] 4
=
T
B B 
L
 9 2 3Lξ  1 + 3ξ  
ξ −  
 4 4  2 
 2 
 L2  1 + 3ξ  
  
4  2  
 Beam Element
,

1 T 1 L
U = EI {δ } ∫ [ B ] [ B ] dξ {δ }
T
,

,
2 −1 2

 9 2 3Lξ  −1+ 3ξ  9 2 3Lξ  1+ 3ξ  

4ξ   − ξ  
 4  2  4 4  2 
 L2  −1+ 3ξ 
2
3Lξ  −1+ 3ξ  L2  −1+ 9ξ 2  
   −    
1 T 8EI  4 2  4  2  4  4 
U = {δ } 3   dξ {δ }
2 L
 9 2 3Lξ  1+ 3ξ 
ξ −  
 4 4  2 
 2 
 L  1+ 3ξ  
2

  
4  2  
 Beam Element
,
1 2 1 1
∫ ξ dξ = ∫ ξ dξ = 0 ∫ dξ = 2
, 2

,
−1 3 −1 −1

1
U = {δ } [ k ]{δ }
T

2
Integration of above matrix yields stiffness matrix

12 6L − 12 6L 
 4 L2 −6 L 2 L2 
EI 
[k ] = 3  12 −6 L 
L
 2 
 4L 
 Beam Element
,
p
, Load vector

, 1 2
 L

∫ pvdx =  ∫ [ ]  {δ }
p H dx
L 0  pL/2 pL/2

 pL 1 
∫ pvdx =  ∫ [ H ] dξ  {δ }
L  2 −1  pL2/12 pL2/12

 PL 1  L L  
∫ pvdx =  ∫ 1 H H2 H3 H4 dξ {δ}
L  2 −1  2 2  

∫ pvdx = { f } {δ }
T

L
T
 pL 2
pL pL pL 
2
{f} = −
T

2 12 2 12 
 Beam Element
,

, Example

,
 Beam Element
,

, Example

,
12 6L − 12 6 L  v1
 4L2
−6 L 2 
2 L  θ1
EI 
[ k ]1 = 3
− 6 L  v2
L  12
 2 
 4L θ2

12 6L − 12 6 L  v2
 4 L2 −6 L 2 L2  θ 2
EI 
[ k ]2 = 3
− 6 L  v3
L  12
 2 
 4L  θ3
 Beam Element
,

, Example

,
 Beam Element
,

, Applying boundary condition

,
 Beam Element
,

, Applying boundary condition