Unit 4 stiffness-anujajape

SRES’s Sanjivani College of Engineering, Kopargaon
Structural Analysis-II, Prepared by Prof. Jape A. S. Page 1
Stiffness Matrix Method
a) Fundamental concepts of stiffness method of analysis, formulation of stiffness
matrix, application to trusses by member approach. Application to beams by
structure approach only, (Involving not more than three unknowns).
b) Application to rigid jointed rectangular portal frames by structure approach only
(Involving not more than three unknowns).
Introduction_______________________________________________________
There are several methods available in the literature for the analysis of continuous
beams such as slope deflection method, moment distribution method, flexibility
matrix method, stiffness matrix method, three moment theorem etc. However
among all these methods Stiffness Matrix Method is program oriented method. In
this method, the given indeterminate structure is first made kinematically
determinate by introducing constraints at the nodes. The required number of
constraints is equal to non-zero joint displacements is called as degrees of freedom
or kinematic indeterminacy. Stiffness matrix method is advanced tool for analyzing
determinate structures. This method is a matrix version of classical generalized
slope-deflection method. Conditions of joint equilibrium are used to form
equations in unknown displacements.
Steps for the solution of Indeterminate Beams by Stiffness Method_________
1. Determine the degree of kinematic indeterminacy. Let n= Dki=DOF
2. Assign the co-ordinate numbers to the unknown displacements.
3. Impose the restraints in the co-ordinate directions to get the fully restrained
structure.

4. Determine the forces developed due to loads on members in each of the co-
ordinate directions in the fully restrained structure. These forces are noted in
{ADL}.
5. Determine the stiffness matrix [S] by giving unit displacement to the
restrained structure in each of the co-ordinate direction and find forces
developed in all co-ordinate directions.
6. Observing the final forces in various co-ordinate directions, note down the
final forces {AD}.
7. Equilibrium equation       D DLA A S D 
where
 DA =Action corresponding to unknown displacements in the original
structure
 DLA =Action corresponding to unknown displacements in the restrained
structure (If sinking is given, add sinking moments also in this vector)
 S = Stiffness matrix
 D =Vector of unknown displacements
Example 4.1: Analyse the beam as shown in figure using stiffness matrix method.
Take EI = constant.
Solution:
I) Dki=02 ( B C,  )
Note: 1) Action corresponding to translation is reaction
2) Action corresponding to rotation is moment

II)  
0
0
B
D
C
A


 
  
 
(No moments acting at joint B and joint C)
III) Restrained Structure
 
75 50 25
50 50
B
DL
C
A


     
    
    
(Values of moments at B and C)
IV) Derivation of Stiffness matrix
1BLet  
Moment at B = 1.33 EI + EI = 2.333 EI
Moment at C = 0.5 EI
1CLet  
Moment at B = 0.5 EI
 
1 1
2 333 0
K = EI
5
0 5 1 0
B C
B
C
. .
. .
 


 
 
 
 
V) Equation of Equilibrium
      D DLA A K D 

0 25 2 333 0 5
0 50 0 5 1 0
. .
. .


       
             
B
C
EI
B C
50
0 0 and
EI
.  
VI) Moment Calculations
      M ML MDA A A D 
75 0 67 0 75
75 1 33 0 0 751
50 1 0 0 5 50 75
50 0 5 1 0 0
AB
BA
BC
CB
M .
M .
EI
M . . EI
M . .
       
                        
       
            
Example 4.2: Analyse the beam using stiffness matrix method if support B sink by
25mm. Take EI = 3800 kN.m2
Solution:
I) Dki=02 ( B C,  )
II)  
0
0
B
D
C
A


 
  
 
(No moments acting at joint B and joint C)
Sinking Moments: Sinking is given in mm. Put this in m while calculating sinking
moments. Since both the element are having same length. Sinking moment of both
the elements will be same.

2 2
6 6 3800 0.025
Sinking moments 15.833 .
6
EI
kN m
L
  
  
 
30 26 67 15 833 15 833 3 333
13 38 15 833 29 213
B
DL
C
. . . .
A
. . .


       
    
     
1BLet  
Moment at B = 0.67 EI + 0.67 EI = 1.333 EI
1CLet  
 
1 1
1 333 0 333
0 333 0
= EI
7
K
6
B C
B
C
. .
. .
 


 
 
 
 
      D DLA A K D 
0 3 333 1 333 0 333
0 29 213 0 333 0 67
. . .
. . .


       
             
B
C
EI

B C
9 45 48 189
EI EI
. .
 

 
      M ML MDA A A D 
30 15 833 0 33 0 42 714
75 15 833 0 67 0 9 45 20 4911
26 67 15 833 0 67 0 33 48 189 20 491
13 38 15 833 0 33 0 67 0
AB
BA
BC
CB
M . . .
M . . . .
EI
M . . . . . .EI
M . . . .
       
                          
        
             
Example 4.3: A continuous beam ABC is loaded as shown in fig. It has constant
flexural rigidity. Fixed support at A, roller support at B and guided support at C.
Analyze the beam using stiffness matrix method.
Solution:
I) Dki=02 ( B C,  )
II)  
0
0
B
D
C
A
 
  
 
(No moments acting at joint B and point load at C)
 
20
10
B
DL
C
Moment at B
A
Reaction at C
 
  
 

1BLet  
Reaction at C = -0.09375 EI
1CLet  
Moment at B = -0.0937 EI
Reaction at C = 0.0234 EI
 
1 1
1 0 0 0937
0 0937 0 02
K =
4
EI
3
B C
B
C
. .
. .


  
 
   
      D DLA A K D 
0 20 1 0 0 0937
0 10 0 0937 0 0234
. .
. .
        
              
B
C
EI
B C
32 078 555 8
and
EI EI
. .

 
  
      M ML MDA A A D 
40 0 25 0 31 98
40 0 5 0 32 078 56 041
20 0 5 0 0937 555 8 56 04
20 0 25 0 0937 24 05
AB
BA
BC
CB
M . .
M . . .
EI
M . . . .EI
M . . .
       
                         
        
             

Example 4.4: Analyze the continuous beam using stiffness matrix method. Take
EI constant
Solution:
I) Dki=03 ( B C C, ,   )
(Clockwise moments acting at joint B, no moment and point load at C)
II)  
30
0
0
B
D C
C
A


 
 
  
  
 
20
0
0
B
DL C
C
Moment at B
A Moment at C
Reaction at C


 
 
  
  
1BLet  
1CLet  

1CLet  
Moment at B = -1.5 EI
Moment at C = -1.5 EI
Reaction at C = 2.5 EI
 
3 0 1 0 1 5
K = EI 1 0 2 0 1 5
1 5 1 5 2
1
5
1 1B C C
B
C
C
. . .
. . .
. . .
 


 
 




  
 

      D DLA A K D 
3 0 1 0 1 5
1 0 2
30 20
0 0 0 1 5
1 5 1 5 2 50 0
. . .
. . .
. . .


      
     
 
 
 
    
 

        
 
   
B
C
C
EI
B C C
4 782 2 608 0 434
and
EI EI EI
. . .
, 
 
   
      M ML MDA A A D 
Note: Spring support is provided at support C, spring force developed due to
deformation of spring is added in the reaction at C

20 0 5 0 0 17 60
4 782
20 1 0 0 0 24 7821
0 434
0 2 0 1 0 1 5 5 218
2 608
0 1 0 2 0 1 5 0
AB
BA
BC
CB
M . .
.
M . .
EI .
M . . . .EI
.
M . . .
       
                          
                      

Example 4.5: Analyze the indeterminate beam as shown in figure using stiffness
matrix method. The beam is fixed at A, C and has internal hinge at B. Take EI
constant.
Solution:
I) Dki=03 ( B BA BC, ,  )
II)  
0
0
0
B
D BA
BC
A 

 
 
  
 
 
 
90
5
20
B
DL BA
BC
Reaction B
A Moment BA
Moment BC


 
 
  
 
 
1BLet  
Reaction at B = 12 EI + 1.5 EI = 13.5 EI
Moment BA = -6.0 EI
Moment BC = 1.5 EI

1BALet  
Reaction at B = -6.0 EI
Moment BA = 4.0 EI
Moment BC = 0
1BCLet  
Reaction at B = 1.5 EI
Moment BA = 0
Moment BC = 2 EI
 
13 5 6 0 1 5
K = EI 6 0 4 0 0
1 5 0
1
0
1
2
1B BA BC
B
BA
BC
. . .
. .
. .
 


  
 
 






      D DLA A K D 
13 5 6 0 1 5
6 0 4 0 0
1 5 0 2 0
0 90
0 5
0 20
. . .
. .
. .


    
     
       
     
   
 
 
 
 
B
BA
BC
EI
BA BC
20 28 75 5
and
EI EI EIB
.
,  
 
   

Exercise
Example: For the following beam, find the vertical deflection and rotation at joint
B using Stiffness Matrix Method. Take EI = 12×103
kN.m2
Example: Determine the unknown joint displacements of the beam as shown in
figure using Stiffness Matrix Method. Take EI constant.
Example: Analyse the beam using Stiffness Matrix Method if support B is sink by
25mm. Take EI = 3800 kN.m2
Example: A continuous beam has fixed support at node 1 and roller supports at
nodes 2 and 3. Analyse the beam using Stiffness Matrix Method and draw SFD and
BMD. Take E = 200 GPa and I=4×106
mm4
.
Example: Analyze the continuous beam ABC as shown in Figure using Stiffness
Matrix Method. Take EI constant.

Example: Analyse the prismatic beam ABC loaded and supported as shown in
Figure using finite element approach. Support B is sink by 25 mm. Draw SFD and
BMD. Take EI constant.
Example: Determine support reactions of continuous beam ABC if support B sink
by 10 mm. Take EI = 6000 kN.m2
. Use Stiffness Matrix Method.
Example: Determine support reactions of continuous beam ABC as shown in
Figure 1 if support B sink by 10 mm. Take EI = 6000 kN.m2
. Use Stiffness
Matrix Method.

Stiffness Matrix Method for portal frame
 The plane frame is a combination of plane truss and beam. All members are
connected by rigid joints in case of frame.
 A frame is having three degrees of freedom at each node i.e. displacement in
x-direction, displacement in y-direction and rotation. Therefore the size of
stiffness matrix of frame element is 6 6 .
Example 4.6: Analyze the portal frame as shown in figure using Stiffness Matrix
Method. Take EI constant.
Solution:
I) Dki=03 ( B C, ,  )
II)  
0
0
0
B
D CA


 
 
  
   
(No moments and horizontal force acting at joint B and C)
10.4 Moment at B
{ } 1.6 Moment at C
4.72 Sum of Horizontal Reaction at B and C
B
DL CA


 
 
  
   

1BLet   1CLet  
Moment at B = 1.6EI + 4.0EI = 5.6 EI Moment at B = 2.0 EI
Moment at C = 2.0 EI Moment at C = 4.0EI+0.8EI=4.8EI
Reaction at B = 0.24 EI Reaction at C = 0.24 EI

1Let  
Reaction at B+C=0.048EI+0.096EI=0.144EI
 
5 6 2 0 0 24
K = EI 2 0 4 8 0 24
0 24 0 24 0 144
1 1 1B C
B
C
. . .
. . .
. . .
 


 
 
 
  


 
      D DLA A K D 
5 6 2 0 0 20 10 4
0 1 6
0 4
4
2 0 4 8 0 24
0 24 0 24 0 42 47 1
B
C
. . .
EI .
.
. . .
.. . .


     
     
       
          
 
 

   

1.0419 1.803 34.046
; ;B Crad rad m
EI EI EI
     
9 6 0 8 0 0 24 18 604
14 4 1 6 0 0 24 4 562
1 0419
4 4 2 0 4 5621
1 803
4 2 4 0 9 128
34 046
2 4 0 0 8 0 24 9 128
3 6 0 0 4 0 24 3 849
AB
BA
BC
CB
CD
DC
M . . . .
M . . . .
.
M .
EI .
M .EI
.
M . . . .
M . . . .
      
              
      
         
       
      
     
    



 

 
 
 
 

Example 4.7: Analyze the rigid frame by using Stiffness Matrix Method. Take EI
constant.
Solution:
I) Dki=03 ( B C, ,  )
II)  
0
0
50
B
D CA


 
 
  
   
(No moments acting at joint B and C, horizontal force at B)
75 Moment at B
{ } 75 Moment at C
0 Sum of Horizontal Reaction at B and C
B
DL CA


 
 
  
   

1BLet   1CLet  
Moment at B = 2.0 EI Moment at B = 0.5 EI
Moment at C = 0.5 EI Moment at C = 2.0EI
Reaction at B = 0.375 EI Reaction at C = 0.375 EI
1Let  

Reaction at B+C=0.375EI
 
2 0 0 5 0 375
1 1
K = EI 0 5 2 0 0 375
0 375 0 375 0 37
1
5
B C
B
C
. . .
. . .
. . .
 


 
 
 
  
   

      D DLA A K D 
2 0 0 5 0 375
0 5 2 0 0 375
0
0 75
3
0 75
50 0 75 0 375 0 375
B
C
. . .
EI . . .
. . .


     
     
       
   
 
 
 
 
      
0 0 5 0 0 375 32 143
0 1 0 0 0 375 7 143
78 571
75 1 0 0 5 0 7 1431
21 428
75 0 5 1 0 0 92 857
190 476
0 0 0 5 0 375 92 857
0 0 1 0 0 375 8
AB
BA
BC
CB
CD
DC
M . . .
M . . .
.
M . . .
EI .
M . . .EI
.
M . . .
M . .
     
             
      
        
       
      
     
     2 143.
 
 
 
 
 
 
 
 
 

Example 7.8: Determine the rotation of joint B, and the horizontal displacements
of joints B and C. Take EI constant.
Solution:
I) Dki=03 ( B C, ,  )
II)  
0
0
0
B
D CA


 
 
  
   
13.33 Moment at B
{ } 40 Moment at C
40 Horizontal Reaction at B
B
DL CA


 
 
  
   

1BLet  
1CLet  
1Let  
 
2 0 0 5 0 375
K = EI 0 5 1 0 0
0 375 0 0 187
1 1 1B C
B
C
. . .
. .
. .
 


 
 
 
 


  


      D DLA A K D 
2 0 0 5 0 375
0
0 13 33
0 40
0 40
5 1 0 0
0 375 0 0 187
B
C
. . .
EI .
.
.
. .


     
     
       
      
 
 
 
       
113.77 96.88 442.05
; ;B Crad rad m
EI EI EI
 

   
26 67 0 5 0 0 375 135 55
113 77
26 67 1 0 0 0 375 25 331
96 88
40 1 0 0 5 0 25 33
442 05
40 0 5 1 0 0 0
AB
BA
BC
CB
M . . . .
.
M . . . .
EI .
M . . .EI
.
M . .
       
                         
         
             

Example 4.9: Analyze the rigid jointed portal frame shown in Figure using
Stiffness Matrix Method. Take EI constant.
Solution:
I) Dki=03 ( B C,  )
II)  
0
0
B
D
C
A


 
  
 
 
4 0 Moment at B
20 Moment at C
B
DL
C
.
A


 
  
 

1BLet  
1CLet  
 
1 1
1 8 0 5
0 5 1 0
B C
B
C
. .
K EI
. .
 


 
 
  
 

      D DLA A K D 
0 4 0 1 8 0 5
0 20 0 5 1 0
B
C
. . .
EI
. .


       
             
3.870 21.935
;B Crad rad
EI EI
 

 
36 0 4 0 34 45
24 0 8 0 3 870 27 091
20 1 0 0 5 21 935 27 09
20 0 5 1 0 0
AB
BA
BC
CB
M . .
M . . .
EI
M . . . .EI
M . .
       
                         
       
            
Example 4.10: Determine global stiffness matrix of the frame ABC shown in
figure using Stiffness Matrix Method. Take EI constant. Neglect axial deformation.
Example 4.11: Analyse the frame shown in Figure using Stiffness Matrix Method
and draw bending moment diagram. Neglect axial deformation.
(Ans. B AB BA BC CB1.2/ EI, M 0.8kN.m,M 1.6kN.m, M 18.4kN.m, M 14.8kN.m      )

Example 4.12: Determine the unknown joint displacements of the portal frame as
shown in Figure using Stiffness Matrix Method. Take EI constant. Neglect axial
deformation.
Example 4.13: Analyse the portal frame as shown in Figure using Stiffness Matrix
Method. Neglect axial deformation.
Example 4.14: Determine the unknown joint displacements of the portal frame as
shown in Figure using Stiffness Matrix Method. Take EI constant. Neglect axial
deformation.

Example 4.15: Analyze the rigid jointed portal frame shown in Figure 6 using
Stiffness Matrix Method. Take EI constant. Draw BMD. Neglect axial
deformation.
Stiffness matrix method for analysis of trusses
The truss may be statically determinate or indeterminate. All members are
subjected to only direct stresses (tensile or compressive). Joint displacements are
selected as unknown variables. Here we select two noded bar element for the
formulation of stiffness matrix of truss element. Since the members are subjected
to only axial forces, the displacements are only in the axial directions of the
members. Therefore, the nodal displacement vector for the bar element is
  1
2
e
u'
x'
u'
 
  
 

where, 1 2and' '
u u are the displacements in axial direction of the element. The
stiffness matrix of a bar element is
1 2
1
2
1 1
1 1
'
u' u'
u'AE
K
u'L
 
       
Transformation matrix for the truss:
' 'x y = Local coordinate
systems
x, y = global coordinate
system
1 2u' ,u' = Displacements in
local coordinate system
1 1 2 2u , v ,u , v = Displacements
in global coordinate system
 =Angle measured in
anticlockwise sense w.r.t.
positive x-axis.
Since axial directions of all members of truss are not same, hence in global
coordinate system (x-y) there are two displacement components at every node.
Hence the nodal displacement vector for typical truss element is
 
1
1
2
2
e
u
v
x
u
v
 
 
 
  
 
  
Refereeing above figure,
At Node 1, At Node 2,
1 1 1cos sin'
u u v   2 2 2cos sin'
u u v  

Therefore, in matrix form above relation are
1
11
22
2
cos sin 0 0
0 0 cos sin
'
'
u
vu
uu
v
 
 
 
     
    
    
  
    '
e
x L x
where,
 '
e
x = vector of local unknowns
 x = vector of global unknowns
 L = Transformation matrix =  
0 0
0 0
l m
L
l m
 
  
 
where, 2 1 2 1
cos or sin or
x x y y
l l m m
L L
 
 
   
Stiffness matrix of truss element in global coordinate system
      '
T
K L K L
 
0
0 1 1 0 0
0 1 1 0 0
0
l
m l mAE
K
l l mL
m
 
            
 
 
 
0
0
0
0
l
m l m l mAE
K
l l m l mL
m
 
           
 
 
 
1 1 2 2
2 2
1
2 2
1
2 2
2
2 2
2
u v u v
ul lm l lm
vlm m lm mAE
K
uL l lm l lm
vlm m lm m
  
 
  
  
 
  

Example 1: Analyze the truss as shown in figure. Cross-sectional area of members
are AB=1000 mm2
, BC=800 mm2
, CA= 800 mm2
. Take E = 2 × 105
MPa
Solution: Step 1: Degrees of freedom: 06 ( A A B B c cu ,v ,u ,v ,u ,v )
Assume x-axis horizontal through point c and vertical through point A. The
coordinate of node A(0, 1.5), B(4, 1.5) and C (2, 0). Take E in GPa
Member x2-x1 y2-y1 L l m AE/L (kN/mm) B.C.
AB 4 0 4 1 0 50 uA = 0Av 
BC -2 -1.5 2.5 -0.8 -0.6 64 0Bv 
CA -2 1.5 2.5 -0.8 0.6 64 ---
Step 2: Element stiffness matrices
Stiffness matrix of element AB: Stiffness matrix of element BC:
 
1 0 1 0
0 0 0 0
50
1 0 1 0
0 0 0 0
A A B B
A
A
AB
B
B
u v u v
u
v
K
u
v
 
 
 
 
 
 
 
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
64
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
B B c c
B
B
BC
c
c
u v u v
u. . . .
v. . . .
K
u. . . .
v. . . .
  
  
 
  
 
  
Stiffness matrix of element CA:
 
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
64
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
c c A A
c
c
CA
A
A
u v u v
u. . . .
v. . . .
K
u. . . .
v. . . .
  
  
 
  
 
  
Step 3: Global stiffness matrix (Total DOF are 06, size of stiffness matrix 6×6)

 
90 96 30 72 50 0 40 96 30 72
30 72 23 04 0 0 30 72 23 04
50 0 90 96 30 72 40 96 30 72
0 0 30 72 23 04 30 72 23 04
40 96 30 72 40 96 30 72 81 92 0
30 72 23 04 30 72 23 04 0 46 08
A A B B c c
A
A
B
B
c
u v u v u v
u. . . .
v. . . .
u. . . .
K
v. . . .
u. . . . .
. . . . .
   
  
 
   
  
  
   
 
    cv



  
Step 4: Reduced stiffness matrix (Since uA= 0Av  , 0Bv  eliminate corresponding
rows and columns from global stiffness matrix)
 
90 96 40 96 30 72
40 96 81 9 0
30 72 0 46 08
B c c
B
c
c
u u v
. . . u
K . . u
. . v
  
  
 
  
Step 5: Equation of equilibrium
    K f 
90 96 40 96 30 72 0
40 96 81 9 0 0
30 72 0 46 08 120
B
c
c
. . . u
. . u
. . v
      
         
         
1 6 0 8 3 67B c cu . mm, u . mm, v . mm     
Example 2: Figure shows a plane truss with three members. Cross-sectional area
of all members 800 mm2
Young modulus is 200 KN/mm2
. Determine
deflection at loaded joint.
Solution:
Step 1: Degrees of freedom: 08 ( A A B B c c D Du ,v ,u ,v ,u ,v ,u ,v )

Assume origin support A (0, 0). The coordinates of other nodes B (1000, 0),
C(2000, 0) and D(1500, 1000)
AD 1500 1000 1802.8 0.832 0.555 88.75 0A Au v 
BD 500 1000 1118 0.447 0.894 143.112 0B Bu v 
CD -500 1000 1118 -0.447 0.894 143.112 0c cu v 
Step 2: Element stiffness matrices
Stiffness matrix of element AD:
 
61 43 40 98 61 43 40 98
40 98 27 34 40 98 27 34
61 43 40 98 61 43 40 98
40 98 27 34 40 98 27 34
A A D D
A
A
AD
D
D
u v u v
u. . . .
v. . . .
K
u. . . .
v. . . .
   
   
 
  
 
  
 
( 0A Au v  )
Stiffness matrix of element BD:
 
28 59 57 19 28 59 57 19
57 19 114 38 57 19 114 38
28 59 57 19 28 59 57 19
57 19 114 38 57 19 114 38
B B D D
B
B
BD
D
D
u v u v
u. . . .
v. . . .
K
u. . . .
v. . . .
   
   
 
  
 
  
 
( 0B Bu v  )
Stiffness matrix of element CD:
 
28 59 57 19 28 59 57 19
57 19 114 38 57 19 114 38
28 59 57 19 28 59 57 19
57 19 114 38 57 19 114 38
C C D D
C
C
CD
D
D
u v u v
u. . . .
v. . . .
K
u. . . .
v. . . .
   
   
 
  
 
  
 
( 0c cu v  )
Step 3: Reduced stiffness matrix
 
118 61 40 98
40 98 256 10
D D
D
D
u v
u. .
K
v. .
 
  
 
    K f 

118 61 40 98 200
40 98 256 10 0
D
D
u. .
v. .
    
    
    
1 785 0 286D Du . mm, v . mm  
Example 3: for the truss as shown in figure using stiffness matrix method,
determines deflections at loaded joints. The joint B is subjected to 50 kN
horizontal force towards left and 80 kN force vertically downward. Take cross-
sectional area of all members 1000 mm2
Young modulus is 200 GPa.
Solution: Step 1: Degrees of freedom: 06 ( A A B B c c D Du ,v ,u ,v ,u ,v ,u ,v ). Assume
origin point B. The coordinates of points areA (-4, 3), B (0,0), C (4,-3), D (-4, -3)
AB 4000 -3000 5000 0.8 -0.6 40 0A Au v 
DB 4000 -3000 5000 0.8 -0.6 40 0D Du v 
CB -4000 -3000 5000 -0.8 -0.6 40 0c cu v 
Step 2: Stiffness matrix of element AB:
 
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
40
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
A A B B
A
A
AB
B
B
u v u v
u. . . .
v. . . .
K
u. . . .
v. . . .
   
   
 
  
 
  
 
( 0A Au v  )
Stiffness matrix of element DB:

 
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
40
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
D D B B
D
D
DB
B
B
u v u v
u. . . .
v. . . .
K
u. . . .
v. . . .
   
   
 
  
 
  
 
( 0D Du v  )
Stiffness matrix of element CB:
 
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
40
0 64 0 48 0 64 0 48
0 48 0 36 0 48 0 36
C C B B
C
C
CB
B
B
u v u v
u. . . .
v. . . .
K
u. . . .
v. . . .
   
   
 
  
 
  
 
( 0c cu v  )
 
1 92 0 48
40
0 48 1 08
B B
B
B
u v
u. .
K
v. .
 
   
    K f 
1 92 0 48 50
40
0 48 1 08 80
B
B
u. .
v. .
     
         
0 00125 0 0024B Bu . mm, v . mm   
Example 3: Determine the deflections at loaded joint in two bar truss supported by
spring as shown in figure. Bar one has length of 5m and bar two a length of 10m.
The stiffness of spring is 2000 kN/m. Take A = 5×10-4
m2
and E = 200 GPa.

Solution:
Step 1: Degrees of freedom: 06 ( 1 1 2 2 3 3u ,v ,u ,v ,u ,v )
Take origin node 1. The coordinates of nodes are
1(0, 0), 2(-3.535, 3.535), 3(-10, 0)
1-2 -3.535 3.535 5 0.707 -0.707 200×102
2 2 0u v 
1-3 -10 0 10 -1 0 100×102
3 3 0u v 
1-4 --- --- --- --- --- --- ---
Step 2: Stiffness matrix of element 1-2:
 
1 1 2 2
1
12
1 2
2
2
0 5 0 5 0 5 0 5
0 5 0 5 0 5 0 5
200 10
0 5 0 5 0 5 0 5
0 5 0 5 0 5 0 5
u v u v
u. . . .
v. . . .
K
u. . . .
v. . . .

  
  
  
   
 
   
 
( 2 2 0u v  )
Stiffness matrix of element 1-3:
 
1 1 3 3
1
12
1 3
3
3
1 0 1 0
0 0 0 0
100 10
1 0 1 0
0 0 0 0
u v u v
u
v
K
u
v

 
 
  
  
 
 
 
( 3 3 0u v  )
Stiffness matrix of Spring element
 
1 4
1
3
4
1 1
2000
1 1
v v
v
K
v
 
    

 
1 1
1
1
20000 10000
10000 12000
u v
u
K
v
 
   
Step 4: Equation of equilibrium:     K f 

1
1
20000 10000 0
10000 12000 40
u
v
     
         
1 12 857 5 714u . mm, v . mm   
Example: For the plane truss shown in figure, determine the x and y components
of displacements at node 1. Take E = 70 GPa and A = 500 mm2
for all elements.
Length of member 1-3 is 2500mm.
Example: For the plane truss composed of three elements shown in figure
subjected to a downward force of 50 kN applied at node 1, determine the x and y
components of displacements at node 1. Take E = 200 GPa and A = 1000 mm2
for
all elements.
Example: Figure shows a plane truss with two members. Both the members are of
cross-sectional area 70.71 mm2
. Young’s modulus is 200 kN/mm2
. Determine
deflections of loaded joint and hence the member forces.

Example: A steel truss as shown in figure. The modulus of elasticity is 210 GPa.
The cross sectional area of member AB is 300 mm2
, BC is 400 mm2
and AC is 500
mm2
. Calculate the horizontal and vertical displacements at point ‘A’ using
stiffness matrix method.
Example: Figure shows a plane truss with three members. All members are of
length 1000 mm and cross-sectional area 600 mm2
. Young’s modulus is 150
kN/mm2
. Determine unknown joint displacements of the truss.
Example: For the two bar truss shown in figure determine the displacements at the
loaded joint using stiffness matrix method. Take A = 200 mm2
and E = 70
GPa.

Example: Find the vertical and horizontal deflection at point C for the two
member truss as shown in figure. Area of inclined member is 2000 mm2
whereas horizontal member is 1600 mm2
. Take E = 200 GPa
Example: Figure shows plane truss with three members. All members are of
length 1000mm and c/s area 600mm2. E=150 KN/mm2. Determine forces in
members of truss using stiffness matrix method.
Example: Analyze the two member truss shown in figure using stiffness matrix
method. Take c/s area of each member 1000 mm2
and E = 200 GPa. The length
of each member is 5m.

Example: For the plane truss structure shown in figure, determine the
displacements at the loaded joint using stiffness matrix method. Assume A =
2000 mm2
and E = 200 GPa.

Unit 4 stiffness-anujajape

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