Introduction: The Direct Stiffness Method - 4 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 3


 
 
 
                                                         
 
 
                                                         
 
 
667 . 1
6
4
4
4
11
= + =
EI EI
k 
 
EI
EI
k 5 . 0
4
2
21
= = 
 
EI
EI
k 166 . 0
6 6
6
31
- =
×
- =      (4) 
 
Similarly, apply a unit rotation along and calculate reactions corresponding to 
three degrees of freedom (see Fig.23.5e) 
2
u
 
EI k 5 . 0
12
= 
 
EI k =
22
 
 
0
32
= k        (5) 
 
Apply a unit displacement along and calculate joint reactions corresponding to 
unknown displacements in the kinematically determinate structure. 
3
u
                                                         
Page 4


 
 
 
                                                         
 
 
                                                         
 
 
667 . 1
6
4
4
4
11
= + =
EI EI
k 
 
EI
EI
k 5 . 0
4
2
21
= = 
 
EI
EI
k 166 . 0
6 6
6
31
- =
×
- =      (4) 
 
Similarly, apply a unit rotation along and calculate reactions corresponding to 
three degrees of freedom (see Fig.23.5e) 
2
u
 
EI k 5 . 0
12
= 
 
EI k =
22
 
 
0
32
= k        (5) 
 
Apply a unit displacement along and calculate joint reactions corresponding to 
unknown displacements in the kinematically determinate structure. 
3
u
                                                         
E
L
EI
k 166 . 0
6
2
13
- = - = 
 
0
23
= k 
 
EI
EI
k 056 . 0
6
12
3
33
= =     (6) 
 
Finally applying the principle of superposition of joint forces, yields 
 
11
22
33
6 1.667 0.5 0.166
24 0.5 1 0
12 0.166 0 0.056
F u
F EI u
F u
- ?? ? ? ?? ? ?
?? ? ? ? ?? ?
=- +
?? ? ? ? ?? ?
?? ? ? ? ?? ?
-
?? ? ? ?? ? ?
 
 
Now, 
 
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
0
0
0
3
2
1
F
F
F
 as there are no loads applied along  and .Thus the  
2 1
,u u
3
u
 
unknown displacements are, 
 
1
1
2
3
10.5 0.166 6
1
0.5 1 0 24
0.166 0 0.056 24
u
u
EI
u
-
- ????
?? ? ?
??
=- -
?? ? ?
?
?? ? ?
?? --
?? ??
??
?
??
  (7) 
 
Solving 
 
1
18.996
u
EI
=
 
 
2
14.502
u
EI
=
 
 
3
270.587
u
EI
=-
      (8) 
 
 
                                                         
Page 5


 
 
 
                                                         
 
 
                                                         
 
 
667 . 1
6
4
4
4
11
= + =
EI EI
k 
 
EI
EI
k 5 . 0
4
2
21
= = 
 
EI
EI
k 166 . 0
6 6
6
31
- =
×
- =      (4) 
 
Similarly, apply a unit rotation along and calculate reactions corresponding to 
three degrees of freedom (see Fig.23.5e) 
2
u
 
EI k 5 . 0
12
= 
 
EI k =
22
 
 
0
32
= k        (5) 
 
Apply a unit displacement along and calculate joint reactions corresponding to 
unknown displacements in the kinematically determinate structure. 
3
u
                                                         
E
L
EI
k 166 . 0
6
2
13
- = - = 
 
0
23
= k 
 
EI
EI
k 056 . 0
6
12
3
33
= =     (6) 
 
Finally applying the principle of superposition of joint forces, yields 
 
11
22
33
6 1.667 0.5 0.166
24 0.5 1 0
12 0.166 0 0.056
F u
F EI u
F u
- ?? ? ? ?? ? ?
?? ? ? ? ?? ?
=- +
?? ? ? ? ?? ?
?? ? ? ? ?? ?
-
?? ? ? ?? ? ?
 
 
Now, 
 
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
0
0
0
3
2
1
F
F
F
 as there are no loads applied along  and .Thus the  
2 1
,u u
3
u
 
unknown displacements are, 
 
1
1
2
3
10.5 0.166 6
1
0.5 1 0 24
0.166 0 0.056 24
u
u
EI
u
-
- ????
?? ? ?
??
=- -
?? ? ?
?
?? ? ?
?? --
?? ??
??
?
??
  (7) 
 
Solving 
 
1
18.996
u
EI
=
 
 
2
14.502
u
EI
=
 
 
3
270.587
u
EI
=-
      (8) 
 
 
                                                         
Summary 
The flexibility coefficient and stiffness coefficients are defined in this section. 
Construction of stiffness matrix for a simple member is explained. A few simple 
problems are solved by the direct stiffness method. The difference between the 
slope-deflection method and the direct stiffness method is clearly brought out.