Method for beam 20 1-2020

A simple beam example with two degree of freedom
 Each node of the beam will have two degrees of freedom:
a vertical displacement (corresponding to shear) and
a rotation (corresponding to bending moment)
•In this method counterclockwise (anticlockwise) moments and counter-clockwise rotations
are taken as positive
• Hence, there are two nodes and 4 degrees of freedom for a beam element.
resulting stiffness matrix is of the order 4x4

Stiffness matrix
• The elements of the stiffness matrix indicate the forces exerted on
the member by the restraints at the ends of the member when
unit displacements are imposed at each DOF of the member.
Impart unit displacement at 1: see the reaction force at other DOF (figure 1)
Impart unit displacement at 3:see the reaction at other DOF (figure 2)

Member stiffness
6EI/L2: moment developed due to unit displacement at 1,
moment AT X-Y plane (positive z direction)

Provide unit displacement at translation DOF (1,3)
3 3
2 2
3 3
2 2
12 12
0 0
6 6
0 0
12 12
0 0
6 6
0 0
EI EI
L L
EI EI
L L
K
EI EI
L L
EI EI
L L
 
 
 
 
 
  
 
 
 
 
 
1DOF: Translation at node 1
3: translation at Node 2
2 DOF: rotation at node 1
4: Rotation at node 2
*Note that 1 and 3 column ( corresponding to 1 &3dof) is filled
• The elements of the stiffness matrix indicate the forces exerted on the
member by the restraints at the ends of the member when unit displacements
are imposed at each DOF of the member.
+ Force: Upward
+ Moment: anti clockwise

Provide unit displacement at translation DOF (1,3)
3 3
2 2
3 3
2 2
12 12
0 0
6 6
0 0
12 12
0 0
6 6
0 0
EI EI
L L
EI EI
L L
K
EI EI
L L
EI EI
L L
 
 
 
 
 
  
 
 
 
 
 
1DOF: Translation at node 1
3: translation at Node 2
2 DOF: rotation at node 1
4: Rotation at node 2
*Note that 1 and 3 column ( corresponding to 1 &3dof) is filled
+ Force: Upward
+ Moment: anti clockwise
12EI/L3: restrained action developed in the direction of translation
6EI/L2: moment developed due to unit displacement at 1, moment AT X-Y plane
(positive z direction)

Due to unit rotation at node j, resisting moment (4EI/L) will
developed, the direction of moment will be z, (moment will act in x-y
plane), 6EI/L2 :Reactive force will developed at j

• The elements of the stiffness matrix indicate the forces exerted on
the member by the restraints at the ends of the member when
unit displacements are imposed at each DOF of the member.
• KAA is the moment require for unit rotation at A
• KBA is the moment Developed at B due to unit rotation at A

2 2
2 2
6 6
0 0
4 2
0 0
6 6
0 0
2 4
0 0
EI EI
L L
EI EI
L L
K
EI EI
L L
EI EI
L L
 
 
 
 
 
  
  
 
 
 
 
Provide unit Rotation at Rotational DOF (2,4)
*Note that 2 and 4 column ( corresponding to 3 &4 dof) is filled

Combination
3 2 3 2
2 2
3 2 3 2
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
EI EI EI EI
L L L L
EI EI EI EI
L L L L
K
EI EI EI EI
L L L L
EI EI EI EI
L L L L
 
 
 
 
 
  
   
 
 
 
 
3 3 2 2
2 2
3 3 2 2
2 2
12 12 6 6
0 0 0 0
6 6 4 2
0 0 0 0
12 12 6 6
0 0 0 0
6 6 2 4
0 0 0 0
EI EI EI EI
L L L L
EI EI EI EI
L L L L
K
EI EI EI EI
L L L L
EI EI EI EI
L L L L
   
   
   
   
   
    
     
   
   
   
   

Stiffness matrix for a beam Element
For the beam element, One column has a stiffness values correspond to two different
degrees of freedom (u and θ) and the equilibrium of two different DoF is not defined and
therefore the sum of every single column of the element stiffness matrix will never be
zero.

We need to consider the fixed end moment that is acting on the beam

Load in Beam
• Load directly acting at the joint: as load vector
• Load on the member: when the load acting on the member then
the load can be taken into account by calculating fixed end action
(Shearing force and bending moment)
• Fixed end action ((Shearing force and bending moment)) can be
transformed into joint loads along with actual load acting at the
joint as follows
            
      
      
1
:
:
EA EA
EM
EF
f K x f x K f f
for momnet
M K f
for force
F K U f


    
 
 

Solution Steps
 Step1 : find the global stiffness matrix
 SteP 2: Load: two types
(i) active force acting at the joint
(ii) if the load act on the member( in between the joint): load needs to
convert equivalent joint end (fixed end action) consider it as active force
      EAF K U f 
 Step3: Now solve for reaction: (two part):
Part1 : FD (force obtained from force-deformation relation)
Part 2: fixed end moment (FE)
So total reaction: FR= FD – Ftotal
            1
EA EAf K x f x K f f

    

Fixed end action (Equivalent joint load)
0
0
4
34 2
2
;
8
8
4
0 2
2
0
8
8
FA
P
PLP
PL P
P
P PLF PL
PL
P
P
PL
PL
 
   
    
    
    
    
       
       
     
     
     
     
           
 
(Fixedend action)re
Equivalent nodalload
4 40
3 30
2 20
8 8
0
2 2
8 8
EA T
P P
PL PL
P P
F F F PL PL
PL PL
P
P P
P
PL PL
   
   
   
    
    
    
        
         
      
     
          
   
   
      

3 2 3 2
2 2
3 2 3 2
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
EI EI EI EI
L L L L
EI EI EI EI
L L L L
K
EI EI EI EI
L L L L
EI EI EI EI
L L L L
 
 
 
 
 
  
   
 
 
 
 
3 2 3 2
2 2
3 2 3 3 2 2 3 2
2 2 2 2
3 2 3 2
2
12 6 12 6
0 0
6 4 6 2
0 0
12 6 12 12 6 6 12 6
6 2 6 6 4 4 6 2
12 6 12 6
0 0
6 2
0 0
G
EI EI EI EI
L L L L
EI EI EI EI
L L L L
EI EI EI EI EI EI EI EI
L L L L L L L L
K
L L L L L L L L
EI EI EI EI
L L L L
EI EI
L


   
        
   

   
      
   
  
2
6 4EI EI
L L L
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

4 4 2
2 4
GR
EI EI EI
L L L
K
EI EI
L L
  
  
  
 
  
4
3
2
8
2
8
Total
P
PL
P
F PL
PL
P
P
PL
 
 
 
 
 
 
  
  
 
 
 
 
 
 
  
8
8
R
PL
PL
F
PL
 
  
  
 
  
Reduced Global Matrix Reduced Global Force Vector
P=100, EI=1000, L=1

4 4 2
2 4
GR
EI EI EI
L L L
K
EI EI
L L
  
  
  
 
  
8
8
R
PL
PL
F
PL
 
  
  
 
  
P=100, EI=1000, L=1
4
6
900
8000 2000 112.58
2000 4000 100 12.5
8
U
U
 
      
       
      
  
4 0.0152
6 0.0045
U
U
   
   
   

Reaction
3 2 3 2
2 2
3 2 3 3 2 2 3 2
2 2 2 2
3 2 3 2
2
12 6 12 6
0 0
6 4 6 2
0 0
12 6 12 12 6 6 12 6
6 2 6 6 4 4 6 2
12 6 12 6
0 0
6 2
0 0
G
EI EI EI EI
L L L L
EI EI EI EI
L L L L
L L L L L L L L
K
L L L L L L L L
EI EI EI EI
L L L L
EI EI
L


   
        
   

   
      
   
  
2
6 4EI EI
L L L
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
0
0
0
0.0152
0
0.0045
U
 
 
 
 
  
 
 
 
 
Reaction = FD -Ftotal
FD= KU FD= force obtained from force deformation relation

Reaction at support
    
 
6
0.0152
2
0.0152
6
0 0.0152 0.0045
8 2
0.0152 0.0045
6 6
0.0152 0.0045
2 4
0.0152 0.0045
D
EI
L
EI
L
EI
L
f K x
EI EI
L L
EI EI
L L
EI EI
L L
 
 
 
 
 
 
      
   
             
    
 
            
    

          
     



P=100, EI=1000, L=1
   
   
   
   
6000 0.0152 91.2
2000 0.0152 30.4
0 0.0152 6000 0.0045 27
8000 0.0152 2000 0.0045 112.6
6000 0.0152 6000 0.0045 64.2
2000 0.0152 4000 0.0045 12.4
   
   
   
       
   
      
         
   
       
Reaction at the support can be obtained by subtracting the (Ftotal) from the reaction
coming from the force deformation relation (FD)
          EA D EAf K x f f f   
Considering First Term

41
32
23
4
8
5
6 2
8
P
PL
f
Pf
f
Kx PL
f PL
f P
P
f
PL
 
 
 
   
   
   
   
    
    
   
   
   
 
 
 
          EA D EAf K x f f f   
 
91.2
30.4
27
112.6
64.2
12.4
Df Kx
 
 
 
 
   
 
 
 
 
1 91.2 100
2 30.4 25
3 27 150
4 100 112.6 12.5
5 100 64.2 50
6 12.4 12.5
1 191.2
2 55.4
3 123
4 0
5 114.2
6 0
f
f
f
f
f
f
f
f
f
f
f
f
     
     
     
     
      
      
      
     
     
   
   
   
   
   
   
   
   
   

, weave and Gere
K = zeros(2,2);
U = zeros(2,1);
F = zeros(2,1);
K(1,1)=8000;
K(2,1)=2000;
K(1,2)=2000;
K(2,2)=4000;
F(1,1)=112.5;
F(2,1)=12.5;
U = KF
p=100;
EI=1000;
L=1;
u1=-5*p*L*L/(EI*112);

Find the displacement and forces using Matrix
method and Draw SFD and BMD
          EA D EAf K x f f f   

For the beam shown, use the stiffness method to:
(a) Determine the deflection and rotation at B.
(b) Determine all the reactions at supports.
(c) Draw the quantitative shear and bending moment diagrams.

(a) Determine the deflection and rotation at B.
(b) Determine all the reactions at supports.
(c) Draw the quantitative shear and bending moment diagrams.

Fixed End Moment
 
4.5
6.75
4.5 5
6.75 3.75
5
3.75
EAf
 
 
 
 
 
  
 
 
 

3 2 3 2
1 1 1 1
2 2
1 1 1 1
3 2 3 3 2 2 3 2
1 1 1 2 1 2 2 2
2 2 2 2
1 1 1 2 1 2 2 2
3
2
12 6 12 6
0 0
6 4 6 2
0 0
12 6 12 12 6 6 12 6
6 2 6 6 4 4 6 2
12 6
0 0
G
EI EI EI EI
L L L L
EI EI EI EI
L L L L
L L L L L L L L
K
L L L L L L L L
EI E
L


   
        
   

   
      
   
  2 3 2
2 2 2
2 2
2 2 2 2
12 6
6 2 6 4
0 0
I EI EI
L L L
EI EI EI EI
L L L L
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
4.5
6.75
4.5 5
6.75 3.75
5
3.75
EAf
 
 
 
 
 
  
 
 
 

 
4.5
6.75
4.5 5
6.75 3.75
5
3.75
EAf
 
 
 
 
  
  
 
 
 

41 2 2
6
2 2
1
4 1 2 2
6
2 2
4 4 2
0 3
0 3.752 4
4 4 2
3
3.752 4
EI EI EI
xL L L
xEI EI
L L
EI EI EI
x L L L
x EI EI
L L

  
  
                  
 
  
  
  
            
 
  
 
4.5
6.75
4.5 5
6.75 3.75
5
3.75
EAf
 
 
 
 
  
  
 
 
 
4
6
0.779 /
2.423 /
bx EI
x c EI


     
      
     

3 2 3 2
1 1 1 1
2 2
1 1 1 1
3 2 3 3 2 2 3 2
1 1 1 2 1 2 2 2
2 2 2
1 1 1 2
12 6 12 6
0 0
6 4 6 2
0 0
12 6 12 12 6 6 12 6
6 2 6 6
1
2
3
4
5
6
EI EI EI EI
L L L L
EI EI EI EI
L L L L
L L L L L L L L
EI EI EI EI
L L L L
f
f
f
f
f
f
  
 
 
  
 
 
              
     
 
    
 
 
 
 
 
  
2
1 2 2 2
3 2 3 2
2 2 2 2
2 2
2 2 2 2
0
0
0
0.779
4 4 6 2
0
12 6 12 6
0 0
6 2 6 4
0 0
2.423
4.5
6.75
9.5
EI EI EI EI EI
L L L L
EI EI EI EI
L L L L
EI EI EI EI
L L L L EI
  
  
  
  
  
  
  
  
    
  
                   
       
  
  
   
    
3
5
3.75
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
  



reaction
6
0.779
2
9
2
0.779
9
1 4.6
0 0.779 2.423
2 2
3
3
4 4 24
0.779 2.423
5 3
9 3
6
6
6
0.779 2.423
2 23
3
2 4
0.779 2.423
3 3
f
f
f
f
f
f


  
 
   
    
  
   
   
  
 
 
 
 
 
 
 
 
 
   
   
   
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
5
6.75
9.5
3
5
3.75


 
 
 
 
 
 
 
1 4.557 4.557
2 6.923 6.923
3 11.115 11.115
4 3 3 0
5 2.134 2.134
6 3.75 3.75 0
f
f
f
f
f
f
     
     
     
     
      
     
     
     
     

1 4.557 4.557
2 6.923 6.923
3 11.115 11.115
4 3 3 0
5 2.134 2.134
6 3.75 3.75 0
f
f
f
f
f
f
     
     
     
     
      
     
     
     
     

To get the reaction take moment
about node 1:
6.92-6.4-1*9*4.5=R2x9
R2=4.44
R1=9-4.44=4.56
1 4.557 4.557
2 6.923 6.923
3 11.115 11.115
4 3 3 0
5 2.134 2.134
6 3.75 3.75 0
f
f
f
f
f
f
     
     
     
     
      
     
     
     
     

To get the reaction take moment
about node 1:
6.92-6.4-1*9*4.5=R2x9
R2=4.44
R1=9-4.44=4.56

Q3
P=1000=P2y
M1=1000 N-m=M2
L1=1m=L2
(EI)=1000
(EI)2=2000
W=100 N per meter length
Find the deflection and support reaction at both the end

Conjugate beam Method and Carry over factor

Conjugate beam
Now consider a determinate beam, subjected to
moment M’ at end B and assume that rotation
(θA2, θB2)occur due to applied moment M’
First consider SS beam and apply
moment ‘M’ at A and draw
conjugate beam diagram

Conclusion
if a positive moment ‘M’ is applied to the hinged end of a beam then a positive moment of
(M/2) will be transferred to the fixed end

Alternate:
• Such support settlements induce fixed end moments in the beams
so as to hold the end slopes of the members as zero
• Applied load is zero, so fixed end
moment is Mf=0
• No rotation pure translation so θA,
θB =0
• MAB=-(6EI/L)Δ
• MBA= =(6EI/L)Δ

3 3 2 2
2 2
3 3 2 2
2 2
3 2 3 2
2 2
3
12 12 6 6
0 0 0 0
6 6 4 2
0 0 0 0
12 12 6 6
0 0 0 0
6 6 2 4
0 0 0 0
12 6 12 6
6 4 6 2
12 6
EI EI EI EI
L L L L
EI EI EI EI
L L L L
K
EI EI EI EI
L L L L
EI EI EI EI
L L L L
EI EI EI EI
L L L L
EI EI EI EI
L L L LK
EI E
L
   
   
   
   
   
    
     
   
   
   
   


  2 3 2
2 2
12 6
6 2 6 4
I EI EI
L L L
EI EI EI EI
L L L L
 
 
 
 
 
 
 
 
 
 
 
FINITE ELEMENT ANALYSIS UNIT - II
https://www.slideshare.net/ASHOKKUMAR27088700/me6603-finite-element-
analysis-unit-ii-notes-and-question-bank

Method for beam 20 1-2020

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