Unit 3 flexibility-anujajape

SRES’s Sanjivani College of Engineering, Kopargaon
Structural Analysis-II, prepared by Prof. Jape A. S. Page 1
Flexibility Matrix Method
a) Fundamental concepts of flexibility method of analysis, formulation of
flexibility matrix, application to pin jointed plane trusses (Involving not more than
three unknowns).
b) Application of flexibility method to beams and rigid jointed rectangular portal
frames (Involving not more than three unknowns).
Introduction_______________________________________________________
Most of the structures in the world are statically indeterminate structures.
Determinate structures can be analyzed by using equations of statics alone. But in
indeterminate structures where number of reactions is greater than number of
equilibrium equations, either reactions or internal force cannot be determined by
statics equations. Hence, in addition to static equilibrium equations, compatibility
equations are essential to determine the reactions or internal forces in
indeterminate structures. For any structure as a whole, it needs to satisfy
i. Equilibrium equations(Structure is in equilibrium)
ii. Compatibility equations(Continuity of structure without any break)
iii. Force- displacement relation(how displacements are related to force)
Basically there are two methods of analysis of indeterminate structures considering
high degree of indeterminate structures and development of computers known as
Flexibility Matrix Method and Stiffness Matrix Method.
Flexibility Matrix Method____________________________________________
A systematic development of consistent deformation method is also known as
flexibility matrix method or force matrix method. In this method, the basic
unknowns to be determined are redundant forces. Hence, the degree of static
indeterminacy of the structure is calculated first and then coordinate number is

assigned to each redundant force direction. Thus, if F1, F2, …, Fn are the redundant
forces in the coordinate direction 1, 2, …, n respectively. If the restraints to the
entire redundant are removed, the resulting structure is called as basic determinate
structure or Released structure. From the principle of superposition, the net
displacement at any point in a statically determinate structure is the sum of the
displacement in basic determinate structure due to the applied loads and redundant
forces.
Δ1 = Δ1L + δ11F1 + δ12F2 + … + δ1nFn
Δ2 = Δ2L + δ21F1 + δ22F2 + … + δ2nFn
… … … ... … …
… … … ... … …
Δn = ΔnL + δn1F1 + δn2F2 + … + δnnFn
where Δ= displacement in ith
coordinate direction
δij = displacement at i due to unit force at j (flexibility matrix element)
ΔiL= displacement at i due to given loading in released structure in
coordinate direction i.
The above equation can be expressed in matrix form as
      L F    or
  Q QLD D F Q       
where,
DQ = Displacement corresponding to action in original structure
DQL = Displacement corresponding to action in released structure
F = Flexibility coefficient factor matrix
Q = Unknown redundant force matrix
Where

DQL =
0
L
i
x
Mm
d
EI
Fij =
0
L
i j
x
m m
d
EI
In the indeterminate structure, the final displacements [Δ] are either zero or known
values. The solution for [F] from above equation gives all the redundant forces.
Then, bending moment, shear forces at any required point can be calculated by
using equations of statics.
Application________________________________________________________
1. Analysis of pin jointed plane trusses
2. Analysis of continuous beams
3. Analysis of rigid jointed rectangular portal frames
Steps for the solution of Indeterminate Beams by Flexibility Method_ _______
1. Determine the degree of static indeterminacy Dsi
2. Choose the redundant
3. Assign the coordinates to the redundant force direction
4. Remove restraints to redundant forces and get basic determinate structure
5. Determine the deflections in coordinate directions due to given loading in
the basic determinate structure
6. Determine the flexibility matrix
7. Apply the compatibility conditions
8. Calculate the redundant forces
9. Calculate member forces, shear forces and bending moment
10.Draw SFD, BMD

Example No. 3.1 Analyse the continuous beam shown in fig. 3.1 (a) by flexibility
matrix method. Flexural rigidity is constant throughout.
Solution
1) Degree of static indeterminacy = R-3 = 4-3 = 1
Let Q1= RB
2) Displacement analysis
Zone Origin Limit M fig. 3.1(d) m1fig. 3.1(e)
BA B 0-3
2
10
2
x
 x
3) Superposition principle
  
    1 11 10
Q QL
QL
D D F Q
D F Q
       
   
3 3 2
1
0 0
10 101.25
2
i
QL x x
Mm x
D d xd
EI EI EI

    
3 3
11
0 0
9i i
x x
m m xx
F d d
EI EI EI
   
1
1
101.25 9
0
11.25
11.25B
Q
EI EI
Q kN
R kN
   
  
  
4) Reaction calculations
0
0
11.25 30 0
18.75
@ 0
10 3 1.5 11.25 3 0
11.25
A
y
A
A
A
A
H
F
R
R kN
M A
M
M kNm

 
  
 
 
     
 

Example No. 3.2 Analyse the continuous beam shown in figure 3.2(a) by
flexibility matrix method.
Solution
1) Degree of static indeterminacy = R-3 = 4-3 = 1
Let Q1= RB

Zone Origin Limit M fig. 3.2(d) m1fig. 3.2(e)
CD C 0-5 20x 0.5x
DB C 5-10  20 20 5x x  0.5x
BA C 10-20    
2
20 20 5 10x x x     0.5 10x x  
  
    1 11 10
Q QL
QL
D D F Q
D F Q
       
   
        25 10 20
1
0 0 5 10
1
20 0.5 1020 20 5 0.520 ( 0.5 )
2 2
3229.166
L
i
QL x x x x
QL
x x x xx x xMm x x
D d d d d
EI EI EI EI
D
EI
      
   
 
   
 
25 10 202 2
11
0 0 5 10
11
0.5 100.25 0.25
2 2
125
L
i i
x x x x
xm m x x
F d d d d
EI EI EI EI
F
EI

   

   
1
1
3229.166 125
0
25.83
25.83B
Q
EI EI
Q kN
R kN
   
  
  
0
25.83 20 20 0
14.17
@ 0
20 25.83 10 20 15 20 5 0
7.085
7.085
y
A C
A C
A
A
C
F
R R
R R
M C
R
R kN
R kN
 
    
 
 
        
  
 

End moments
0
29.15
29.15
AB CB
BA
BC
M M
M kNm
M kNm
 
 

flexibility matrix method if support B sinks by 25mm. Take EI= 3800kNm2
Solution
1) Degree of static indeterminacy = R-3 = 5-3 =2
Let Q1= RB and Q2= RC
Zone Origin Limit M fig. 3.3(b) m1fig.
3.3(c)
m2fig.
3.3(d)
CD C 0-4 0 0 x
DB C 4-6  30 4x  0 x
BA C 6-12
 
 
2
10 6
30 4
2
x
x

  
 6x  x
  
1 11 12 1
2 21 22 2
0.025
0
Q QL
QL
QL
D D F Q
D F F Q
D F F Q
       
       
       
      
 
 
 
2
12
1
1
0 6
10 6
30 4 6
2 4860
L
QL x x
x
x x
Mm
D d d
EI EI EI
 
    
      
 
 
 
2
6 12
2
2
0 4 6
10 6
30 4
230 4 12740
L
QL x x x
x
x x
x xMm
D d d d
EI EI EI EI
 
   
          

 
 
212
1 1
11
0 6
12
1 2
12 21
0 6
4 6 122 2 2
2 2
22
0 0 4 6
6 72
6 180
576
L
x x
L
x x
L
x x x x
xm m
F d d
EI EI EI
x xm m
F F d d
EI EI EI
m m x x x
F d d d d
EI EI EI EI EI

  

   
    
 
 
   
1
2
1
2
4860 72 180
0.025
0 12740 180 576
50.319
6.395
33.286
B
C
A
QEI EI EI
Q
EI EI EI
Q R kN
Q R kN
R kN
   
      
        
      
      
   
  
  
@ 0
30 2 6.395 6 0
21.63
@ 0
21.63 10 6 3 33.286 6 0
41.346
BC
BC
AB
BC
M B
M
M kNm
M B
M
M kNm
 
    
 
 
      
 
flexibility matrix. Take EI= constant. The stiffness coefficients of spring B and C
are kB = EI kN/m and kC = EI/2 kN/m
Solution
Let Q1= RB and Q2= RC

Zone Origin Limit M fig. 3.4(b)
m1fig.
3.4(c)
m2fig.
3.4(d)
CD C 0-4 0 0 x
DB C 4-6  30 4x  0 x
BA C 6-12  
 
2
10 6
30 4
2
x
x

    6x  x

  
1 11 12 1
2 21 22 2
1
2
Q QL
QLB
QLC
D D F Q
DR F F Q
DR F F QEI
       
       
              
 
 
 
2
12
1
1
0 6
10 6
30 4 6
2 4860
L
QL x x
x
x x
Mm
D d d
EI EI EI
 
    
      
 
 
 
2
6 12
2
2
0 4 6
10 6
30 4
230 4 12740
L
QL x x x
x
x x
x xMm
D d d d
EI EI EI EI
 
   
          
 
 
212
1 1
11
0 6
12
1 2
12 21
0 6
4 6 122 2 2
2 2
22
0 0 4 6
6 72
6 180
576
L
x x
L
x x
L
x x x x
xm m
F d d
EI EI EI
x xm m
F F d d
EI EI EI
m m x x x
F d d d d
EI EI EI EI EI

  

   
    
 
 
   
1
2
1
2
4860 72 180
1
2 12740 180 576
52.67
5.64
31.69
B
C
B
C
A
R QEI EI EI
R QEI
EI EI EI
Q R kN
Q R kN
R kN
   
      
               
      
   
  
  
@ 0
30 2 5.64 6 0
26.16
@ 0
26.16 10 6 3 31.69 6 0
36.3
BC
BC
AB
BC
M B
M
M kNm
M B
M
M kNm
 
    
 
 
      
 

Example No. 3.5 Analyse the beam AB shown in figure 3.5(a) by flexibility
matrix.
Solution
Let Q1= RB and Q2= MBC

Zone Origin Limit
M fig.
3.5(b)
m1fig.
3.5(c)
m2fig.
3.5(d)
BC B 0-4 0 x 1
CA B 4-8  24 4x  x 1
  
1 11 12 1
2 21 22 2
0
0
Q QL
QL
QL
D D F Q
D F F Q
D F F Q
       
      
       
      
 8
1
1
0 4
24 4 640
2
L
QL x x
x xMm
D d d
EI EI EI
 
    
 8
2
2
0 4
24 4 96
2
L
QL x x
xMm
D d d
EI EI EI
 
    
4 82 2
1 1
11
0 0 4
4 8
1 2
12 21
0 0 4
4 8
2 2
22
0 0 4
96
2
20
2
1 1 6
2
L
x x x
L
x x x
L
x x x
m m x x
F d d d
EI EI EI EI
m m x x
F F d d d
EI EI EI EI
m m
F d d d
EI EI EI EI
   
    
   
  
  
  
1
2
1
2
640 96 20
0
0 96 20 6
10.909
20.36
13.091
B
CB
A
QEI EI EI
Q
EI EI EI
Q R kN
Q M kNm
R kN
   
      
        
      
      
   
 
  

@ 0
20.36 10.909 4 0
23.276
@ 0
13.09 4 23.276 0
29.084
CB
CB
AC
AC
M C
M
M kNm
M C
M
M kNm
 
   
 
 
    
 
Flexibility Matrix Method for Portal Frames____________________________
The force method of analysis can be employed to analyze the indeterminate
frames. The basic steps in the analysis of indeterminate frame by flexibility method
are the same as that discussed in the analysis of indeterminate beams. Under the
action of external loads, the frames undergo axial and bending deformations. Since
the axial rigidity of the members is much higher than the bending rigidity, the axial
deformations are much smaller than the bending deformations hence are normally
not considered in the analysis.
matrix.
Solution
1) Degree of static indeterminacy = R-3 = 6-3 =3(RDH, RDV, MD)
Let Q1= RDH , Q2= RDV and Q3= MD
m1fig.
3.6(c)
m2fig.
3.6(d)
m3fig.
3.6(e)
DE D 0-2 0 x 0 1
EC D 2-5  5 2x  x 0 1

CB C 0-4 2
15 3 2x  5 x 1
BF B 0-4  24 5 3x    5 x 4 1
FA B 4-10
   24 5 3 10 4x x    
 5 x 4 1
  
1 11 12 13 1
2 21 22 23 2
3 31 32 33 3
0
0
0
Q QL
QL
QL
QL
D D F Q
D F F F Q
D F F F Q
D F F F Q
       
      
             
            

       
     
25 4 4
1
1
0 2 0 0
10
4
1
15 3 2 5 24 5 3 55 2
4 4
24 5 3 10 4 5
4
174.167
L
QL x x x x
x
QL
x x xx xMm
D d d d d
EI EI EI EI
x x x
d
EI
D
EI
             
       

 
   

     
       
34 4 10
2
2
0 0 0 4
2
25 4 4 10
3
3
0 2 0 0 4
3
15 1.5 156 20 4 20
4 4 4
374
15 1.55 2 24 5 15 24 5 15 10 40
4 4 4
125.5
L
QL x x x x
QL
L
QL x x x x x
QL
x x x xMm
D d d d d
EI EI EI EI
D
EI
xx x x xMm
D d d d d d
EI EI EI EI EI
D
EI
    
   

 
          
    

 
   
    
   
2 22 5 4 4 102 2
1 1
11
0 0 2 0 0 4
5 525 87.5
4 4 4
L
x x x x x
x xm m x x
F d d d d d
EI EI EI EI EI EI EI
 
           
   4 4 10
1 2
12 21
0 0 0 4
5 4 5 45 10
4 4 4
L
x x x x
x xm m x
F F d d d d
EI EI EI EI EI
 
        
   2 5 4 4 10
1 3
13 31
0 0 2 0 0 4
5 55 17.5
4 4 4
L
x x x x x x
x xm m x x
F F d d d d d d
 
            
4 4 102
2 2
22
0 0 0 4
16 16 45.333
4 4 4
L
x x x x
m m x
F d d d d
EI EI EI EI EI
       
4 4 10
2 3
23 32
0 0 0 4
4 4 12
4 4 4
L
x x x x
m m x
F F d d d d
EI EI EI EI EI
        
2 5 4 4 10
3 3
33
0 0 2 0 0 4
8.5
4 4 4
L
x x x x x
x x
m m d d d d d
F d d
           
Equation of Compatibility
  Q QLD D F Q       

0 174.167 87.5 10 17.5
1
0 374 10 45.33 12
0 125.5 17.5 12 8.5
DH
DV
D
R
R
EI
M
       
                
              
0.404
7.142
3.850
DH
DV
D
R kN
R kN
M kNm

 

4) Reaction Calculation
0
12
4.858
0
10 5 0
5.404
y
AV DV
AV
x
AH DH
AH
F
R R
R kN
F
R R
R kN
 
 
  
 
   
  
For member AB For member BC For member CD
5.404 10
10 4 4.502 0
18.802
B AB
AB
M M
M kNm
   
   

3 4 2
7.142 4 0
4.562
B BC CB
BC
M M M
M kNm
     
  

3.850 5 3
0.404 5 0
9.130
C CD
BC
M M
M kNm
    
  


matrix.
Solution
1) Degree of static indeterminacy = R-3 = 5-3 =2(RCH, RCV)
Let Q1= RCV , Q2= RCH

m1fig.
3.7(c)
m2fig.
3.7(d)
CB C 0-5 2
20 2x x 0
BD B 0-2 250 5 x
DA B 2-4  250 40 2x   5 x
  
1 11 12 1
2 21 22 2
0
0
Q QL
QL
QL
D D F Q
D F F Q
D F F Q
       
      
       
      
 
 
5 2 43
1
1
0 0 0 2
1
22 4
2
2
0 0 2
2
850 20010 1250
6962.5
170 40250
2266.67
L
QL x x x x
QL
L
QL x x x
QL
xMm x
D d d d d
EI EI EI EI
D
EI
x xMm x
D d d d
EI EI EI
D
EI
  
   

 

  
 
   
  
5 2 42
1 1
11
0 0 0 2
25 25 141.67
L
x x x x
m m x
F d d d d
EI EI EI EI EI
       
2 4
1 2
12 21
0 0 2
5 5 40
4
L
x x x
m m x x
F F d d d
EI EI EI EI
 
      
2 42 2
2 2
22
0 0 2
21.33
L
x x x
m m x x
F d d d
EI EI EI EI
     
Equation of Compatibility
  Q QLD D F Q       

0 6962.5 141.67 401 1
0 2266.67 40 21.33
CV
CH
R
REI EI
        
                
40.68
29.67
CV
CH
R kN
R kN


 
4) Reaction Calculation
0
100
59.32
0
29.97 40 0
10.025
y
AV CV
AV
x
AH
AH
F
R R
R kN
F
R
R kN
 
 
  
 
  
  
For member AB For member BC
10.025 4 40 2 46.6 0
6.7
B AB
AB
M M
M kNm
        

40.68 5 20 5 2.5 0
46.6
46.6
B BC
BC
AB
M M
M kNm
M kNm
       

  

Flexibility Matrix Method for Plane Truss______________________________
The truss is said to be statically indeterminate when the total number of reactions
and member axial forces exceed the total number of static equilibrium equations.
In the simple plane truss, the degree of indeterminacy can be determined from
inspection. Also, following formula is used to evaluate the static indeterminacy of
plane truss  2i m j r  
where m, j and r are number of members, joints and unknown reaction components
respectively. The indeterminacy in the truss may be external, internal or both. A
plane truss is said to be externally indeterminate if the number of reactions exceeds
the number of static equilibrium equations available and has exactly  2 3j 
members. A truss is said to be internally indeterminate if it has exactly three
reaction components and more than  2 3j  members. Finally a truss is both
internally and externally indeterminate if it has more than three reaction
components and also has more than 2 3j  members. The basic method for the
analysis of indeterminate truss by force method is similar to the indeterminate
beam analysis.
Steps in the Flexibility Analysis of Plane Truss___________________________
1. Determine the degree of static indeterminacy of the structure.
2. Identify the number of redundant reactions equal to the degree of
indeterminacy.
3. The redundant must be so selected that when the restraint corresponding to
the redundant are removed, the resulting truss is statically determinate and
stable.

4. Select redundant as the reaction component in excess of three and the rest
from the member forces. However, one could choose redundant actions
completely from member forces.
5. Perform P analysis and K analysis.
6. Apply compatibility equation and calculate unknown redundant forces.
Example No. 3.8 Find the forces in all members of the truss shown in fig. 3.8(a).
Cross sectional area and modulus of elasticity is same for all members.
Solution
Internal    2 6 2 4 3 1siD m j r       
External 3 4 3 1siD R    
Let Q1=RC and Q2=FBD
Analysis of truss
I. P- Analysis

0yF   20..... 1y AV DF R R   
2 10 2 0
10
10
0
10
A D
D
AV
x
AH
M R
R kN
R kN
F
R kN
     
  
  
 
  
Joint B Joint D Joint A
 
 
0
10
10
0
20
20
x
BC
BC
y
BA
BA
F
F kN
F kN C
F
F kN
F kN C
 
  

 
  

 
0
0
0
10
10
x
DA
y
DC
DC
F
F
F
F kN
F kN C
 
 
 
  

 
0
cos45 10 0
14.142
x
AC
AC
F
F
F kN T
 
 
 
II. K1- Analysis

0 1
2 1 2 0
1
x AH
A D
D
F R kN
M R
R kN
    
     
  
0
0
1
y
y AV D
AV
F
F R R
R kN
 
   
  
Joint D Joint C Joint B
 
0
0
0
1
1
x
DA
y
DC
DC
F
F
F
F kN
F kN C
 
 
 
  

 
0
sin 45 0
1.414
0
1 cos45 0
0
y
DC CA
CA
x
BC CA
BC
F
F F
F kN T
F
F F
F
 
 
 
 
   
 
0BA BCF F 
III. K2- Analysis

Joint B Joint C Joint D
 
 
0 1cos45 0
0.707
0.707
0
1sin 45 0
0.707
0.707
x BC
BC
BC
y
BA
BA
BA
F F
F kN
F kN C
F
F
F kN
F kN C
    
 
 
 
  
 
 
 
 
0 0.707 cos45 0
1
0
sin 45 0
0.707 0.707
x AC
AC
y
CD AC
CD
F F
F kN T
F
F F
F kN C
    

 
  
  
 
0 1cos45 0
0.707
0
0.707 1sin 45 0
0
x DA
DA
y
D
D
F F
F kN C
F
R
R
     

 
  

Calculation Table
Member
L
AE
P 1K 2K
1PK L
AE
2PK L
AE
2
1K L
AE
1 2K K L
AE
2
2K L
AE
AB
2
AE
-20 0
-
0.707
0
14.14
AE
0 0
1
AE
BC
2
AE
-10 0
-
0.707
0
14.14
AE
0 0
1
AE
CA
2.828
AE
14.142 1.414 1
56.55
AE
40
AE
5.65
AE
4
AE
2.828
AE
CD
2
AE
-10 -1
-
0.707
20
AE
14.14
AE
2
AE
1.414
AE
1
AE
DA
2
AE
0 0
-
0.707
0 0 0 0
1
AE
BD
2.828
AE
--- --- 1 --- --- --- ---
2.828
AE
1
1
2
2
76.55
76.56
QL
QL
PK L
D
AE AE
PK L
D
AE AE
 
 



2
1
11
1 2
12 21
2
2
22
7.65
5.414
9.656
K L
F
AE AE
K K L
F F
AE AE
K L
F
AE AE
 
  
 



Compatibility Equations
      Q QLD D F Q 
1
2
0 76.55 7.65 5.4141 1
0 96.56 5.414 9.656
Q
QAE AE
      
        
       
1
2
4.856
7.277
C
BD
Q R kN
Q F kN
   
  
Forces in other members can be calculated as follows:
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
AB AB
BC BC
CA CA
CD CD
DA DA
AB AB
BC BC
CA CA
CD CD
DA DA
F P K Q K Q
F P K Q K Q
F P K Q K Q
F P K Q K Q
F P K Q K Q
  
  
  
  
  

Indeterminate Truss with Lack of Fit___________________________________
Lack of fit is a fabrication error in the length of member. Lack of fit is considered
in {DQ} vector.
If member is short in length it is taken as positive.
If member is long in length it is taken as negative.
Indeterminate Truss with Temperature changes_________________________
The change in temperature in the member causes exapansion or contraction of the
member. Change in length is obtained as follows
L L t   
Expansion or contraction is considered in {DQ} vector.
Expansion is taken as positive.
Contraction is taken as negative.
Example No. 3.9 Analyse the plane truss as shown in fig. 3.9(a). It was found that
after fabrication, the member BF is 5mm short. Area for all members is 1000mm2
.
Take E=200kN/mm2
.
Solution
Internal
 
 
2
11 2 6 3
2
si
si
si
D m j r
D
D
  
   


External 3 0siD R  
Let Q1=FBF and Q2=FDF
Analysis of truss
I. P- Analysis
All members are zero force members
II. K1- Analysis
Zero force members
CD=DE=EF=CE=0
Joint B Joint A Joint F

0
cos36.86 0
0.8
0
sin36.86 0
0.6
x
BC
BC
y
BA
BA
F
F
F kN
F
F
F kN
 
 
  
 
  
  
0
sin36.86 0
1
0
cos36.86 0
0.8
y
AC BA
AC
x
AF AC
AF
F
F F
F kN
F
F F
F kN
 
 
 
 
 
  
0
1sin36.86 0
0.6
y
CF
CF
F
F
F kN
 
 
  
III. K2- Analysis
Zero force members
AB=BC=AF=AC=0
Joint D Joint E Joint F
0
1cos36.86 0
0.8
0
1sin36.86 0
0.6
x
DC
DC
y
DE
DE
F
F
F kN
F
F
F kN
 
  
  
 
  
  
0
sin36.86 0
1
0
cos36.86 0
0.8
y
DE CE
CE
x
EF CE
EF
F
F F
F kN
F
F F
F kN
 
 

 
  
 
0
1sin36.86 0
0.6
y
CF
CF
F
F
F kN
 
 
  

Calculation Table
Member
L
AE 1K 2K
2
1K L
AE
1 2K K L
AE
2
2K L
AE
CF 0.015 -0.6 -0.6 0.0054 0.0054 0.0054
AB 0.015 -0.6 0 0.0054 0 0
BC 0.01 -0.8 0 0.0064 0 0
CD 0.01 0 -0.8 0 0 0.0064
DE 0.015 0 -0.6 0 0 0.0054
EF 0.01 0 -0.8 0 0 0.0064
FA 0.01 -0.8 0 0.0064 0 0
AC 0.025 1 0 0.025 0 0
CE 0.025 0 1 0 0 0.025
BF 0.025 1 0 0.025 0 0
FD 0.025 0 1 0 0 0.025
Ʃ=0.0736 Ʃ=0.0054 Ʃ=0.0736
      Q QLD D F Q 
1
2
5 0 0.0736 0.0054
0 0 0.0054 0.0736
Q
Q
      
        
       
 
 
1
2
68.30
5.011
BF
DF
Q F kN T
Q F kN C
  
 

Forces in other members can be calculated as follows:
 
 
 
 
 
 
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
1 1 2 2
37.97
40.98
54.64
4.0
3.0
4.0
54.6
CF CF
AB AB
BC BC
CD CD
DE DE
EF EF
AC AC
CF CF
AB AB
BC BC
CD CD
DE DE
EF EF
AC AC
F P K Q K Q kN C
F P K Q K Q kN C
F P K Q K Q kN C
F P K Q K Q kN C
F P K Q K Q kN C
F P K Q K Q kN C
F P K Q K Q
   
   
   
   
   
   
     
 1 1 2 2
4
68.30CE CECE CE
kN C
F P K Q K Q kN C   
Example No. 3.10 Analyse the plane truss as shown in fig. 3.10(a). If member AC
is subjected to temperature rise of 20o
C. Take E=200GPa and coefficient of
thermal expansion α= 1.2 x 10-5
/0
C. Cross sectional area of each member is
indicated in bracket.
Solution
Internal  2 1siD m j r   
External 3 0siD R  
Let Q1=FAC

Analysis of truss
I. P- Analysis
All members are zero force
members
II. K- Analysis
Joint C Joint A Joint B
0
cos53.13 0
0.6
0
sin53.13 0
0.8
x
DC
DC
y
BC
BA
F
F
F kN
F
F
F kN
 
  
  
 
  
  
0
cos53.13 0
0.6
0
sin53.13 0
0.8
x
AB
AB
y
AD
AD
F
F
F kN
F
F
F kN
 
 
  
 
 
  
0
cos53.13 0
1
x
AB BD
BD
F
F F
F kN
 
  
 

Calculation Table
Member
L
AE K
2
K L
AE
Final Forces
F=KQ
AB 0.01 -0.6 0.0036 18kN(T)
BC 0.01 -0.8 0.0064 24kN(T)
CD 0.01 -0.6 0.0036 18kN(T)
DA 0.01 -0.8 0.0064 24kN(T)
DB 0.01 1 0.01 30kN(C)
AC 0.01 1 0.01 30kN(C)
Ʃ=0.04
Expansion prevented due to rise in temperature=
5
5000 1.2 10 20 1.2L t mm 
      
      Q QLD D F Q 
      1.2 0 0.04 ACF   
 1 30ACQ F kN C  

Unit 3 flexibility-anujajape

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