The Direct Stiffness Method: Truss Analysis - 6 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 1


                                                         
{}[]
1
72.855
1
' 0.707 0.707 1.688 kN
55.97 7.07
AE
p
AE
??
=- - =-
??
-
??
  (16) 
 
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 
 
 {}[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - =
4
3
8
7
1
07 . 7
'
u
u
u
u
m l m l
AE
p 
 
{} [] {}
1
1
' 0.707 53.825 5.38 kN
7.07
AE
p
AE
==   (17) 
 
 
25.2 Inclined supports 
Sometimes the truss is supported on a roller placed on an oblique plane (vide 
Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is 
zero. . .e i displacement along " y is zero in the present case. 
  
 
Page 2


                                                         
{}[]
1
72.855
1
' 0.707 0.707 1.688 kN
55.97 7.07
AE
p
AE
??
=- - =-
??
-
??
  (16) 
 
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 
 
 {}[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - =
4
3
8
7
1
07 . 7
'
u
u
u
u
m l m l
AE
p 
 
{} [] {}
1
1
' 0.707 53.825 5.38 kN
7.07
AE
p
AE
==   (17) 
 
 
25.2 Inclined supports 
Sometimes the truss is supported on a roller placed on an oblique plane (vide 
Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is 
zero. . .e i displacement along " y is zero in the present case. 
  
 
 
                                                         
 
 
If the stiffness matrix of the entire truss is formulated in global co-ordinate system 
then the displacements along y are not zero at the oblique support. So, a special 
procedure has to be adopted for incorporating the inclined support in the analysis 
of truss just described. One way to handle inclined support is to replace the 
inclined support by a member having large cross sectional area as shown in Fig. 
25.3b but having the length comparable with other members meeting at that joint. 
The inclined member is so placed that its centroidal axis is perpendicular to the 
inclined plane. Since the area of cross section of this new member is very high, it 
does not allow any displacement along its centroidal axis of the jointA. Another 
method of incorporating inclined support in the analysis is to suitably modify the 
member stiffness matrix of all the members meeting at the inclined support. 
 
 
Page 3


                                                         
{}[]
1
72.855
1
' 0.707 0.707 1.688 kN
55.97 7.07
AE
p
AE
??
=- - =-
??
-
??
  (16) 
 
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 
 
 {}[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - =
4
3
8
7
1
07 . 7
'
u
u
u
u
m l m l
AE
p 
 
{} [] {}
1
1
' 0.707 53.825 5.38 kN
7.07
AE
p
AE
==   (17) 
 
 
25.2 Inclined supports 
Sometimes the truss is supported on a roller placed on an oblique plane (vide 
Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is 
zero. . .e i displacement along " y is zero in the present case. 
  
 
 
                                                         
 
 
If the stiffness matrix of the entire truss is formulated in global co-ordinate system 
then the displacements along y are not zero at the oblique support. So, a special 
procedure has to be adopted for incorporating the inclined support in the analysis 
of truss just described. One way to handle inclined support is to replace the 
inclined support by a member having large cross sectional area as shown in Fig. 
25.3b but having the length comparable with other members meeting at that joint. 
The inclined member is so placed that its centroidal axis is perpendicular to the 
inclined plane. Since the area of cross section of this new member is very high, it 
does not allow any displacement along its centroidal axis of the jointA. Another 
method of incorporating inclined support in the analysis is to suitably modify the 
member stiffness matrix of all the members meeting at the inclined support. 
 
 
                                                         
Consider a truss member as shown in Fig. 25.4. The nodes are numbered as 1 
and 2. At 2, it is connected to a inclined support. Let ' 'y x be the local co-ordinate 
axes of the member. At node 1, the global co-ordinate system xy is also shown. 
At node 2, consider nodal co-ordinate system as " "y x , where " y is perpendicular 
to oblique support. Let 
1
' u and
2
' u be the displacements of nodes 1 and 2 in the 
local co-ordinate system. Let 
1 1
,v u be the nodal displacements of node 1 in 
global co-ordinate systemxy. Let 
22
", " uv be the nodal displacements along " x -
and " y - are in the local co-ordinate system " "y x at node 2. Then from Fig. 25.4, 
 
x x
v u u ? ? sin cos '
1 1 1
+ =      
 
" 2 " 2 2
sin " cos " '
x x
v u u ? ? + =      (25.1) 
 
This may be written as  
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
2
2
1
1
" " 2
1
"
" sin cos
0 0
0 0
sin cos
'
'
v
u
v
u
u
u
x x
x x
? ?
? ?
  (25.2) 
Denoting 
" "
sin " ; cos " ; sin ; cos
x x x x
m l m l ? ? ? ? = = = = 
 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
2
2
1
1
2
1
"
" " "
0 0
0 0 '
'
v
u
v
u
m l
m l
u
u
              (25.3a) 
 
 or  {} []{} u T u ' ' =       
 
where [] ' T is the displacement transformation matrix. 
 
 
Page 4


                                                         
{}[]
1
72.855
1
' 0.707 0.707 1.688 kN
55.97 7.07
AE
p
AE
??
=- - =-
??
-
??
  (16) 
 
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 
 
 {}[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - =
4
3
8
7
1
07 . 7
'
u
u
u
u
m l m l
AE
p 
 
{} [] {}
1
1
' 0.707 53.825 5.38 kN
7.07
AE
p
AE
==   (17) 
 
 
25.2 Inclined supports 
Sometimes the truss is supported on a roller placed on an oblique plane (vide 
Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is 
zero. . .e i displacement along " y is zero in the present case. 
  
 
 
                                                         
 
 
If the stiffness matrix of the entire truss is formulated in global co-ordinate system 
then the displacements along y are not zero at the oblique support. So, a special 
procedure has to be adopted for incorporating the inclined support in the analysis 
of truss just described. One way to handle inclined support is to replace the 
inclined support by a member having large cross sectional area as shown in Fig. 
25.3b but having the length comparable with other members meeting at that joint. 
The inclined member is so placed that its centroidal axis is perpendicular to the 
inclined plane. Since the area of cross section of this new member is very high, it 
does not allow any displacement along its centroidal axis of the jointA. Another 
method of incorporating inclined support in the analysis is to suitably modify the 
member stiffness matrix of all the members meeting at the inclined support. 
 
 
                                                         
Consider a truss member as shown in Fig. 25.4. The nodes are numbered as 1 
and 2. At 2, it is connected to a inclined support. Let ' 'y x be the local co-ordinate 
axes of the member. At node 1, the global co-ordinate system xy is also shown. 
At node 2, consider nodal co-ordinate system as " "y x , where " y is perpendicular 
to oblique support. Let 
1
' u and
2
' u be the displacements of nodes 1 and 2 in the 
local co-ordinate system. Let 
1 1
,v u be the nodal displacements of node 1 in 
global co-ordinate systemxy. Let 
22
", " uv be the nodal displacements along " x -
and " y - are in the local co-ordinate system " "y x at node 2. Then from Fig. 25.4, 
 
x x
v u u ? ? sin cos '
1 1 1
+ =      
 
" 2 " 2 2
sin " cos " '
x x
v u u ? ? + =      (25.1) 
 
This may be written as  
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
2
2
1
1
" " 2
1
"
" sin cos
0 0
0 0
sin cos
'
'
v
u
v
u
u
u
x x
x x
? ?
? ?
  (25.2) 
Denoting 
" "
sin " ; cos " ; sin ; cos
x x x x
m l m l ? ? ? ? = = = = 
 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
2
2
1
1
2
1
"
" " "
0 0
0 0 '
'
v
u
v
u
m l
m l
u
u
              (25.3a) 
 
 or  {} []{} u T u ' ' =       
 
where [] ' T is the displacement transformation matrix. 
 
 
                                                         
 
 
Similarly referring to Fig. 25.5, the force 
1
' p has components along x and 
y axes. Hence  
 
x
p p ? cos '
1 1
=     (25.4a) 
 
x
p p ? sin '
1 2
=     (25.4b) 
 
Page 5


                                                         
{}[]
1
72.855
1
' 0.707 0.707 1.688 kN
55.97 7.07
AE
p
AE
??
=- - =-
??
-
??
  (16) 
 
Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 
 
 {}[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- - =
4
3
8
7
1
07 . 7
'
u
u
u
u
m l m l
AE
p 
 
{} [] {}
1
1
' 0.707 53.825 5.38 kN
7.07
AE
p
AE
==   (17) 
 
 
25.2 Inclined supports 
Sometimes the truss is supported on a roller placed on an oblique plane (vide 
Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is 
zero. . .e i displacement along " y is zero in the present case. 
  
 
 
                                                         
 
 
If the stiffness matrix of the entire truss is formulated in global co-ordinate system 
then the displacements along y are not zero at the oblique support. So, a special 
procedure has to be adopted for incorporating the inclined support in the analysis 
of truss just described. One way to handle inclined support is to replace the 
inclined support by a member having large cross sectional area as shown in Fig. 
25.3b but having the length comparable with other members meeting at that joint. 
The inclined member is so placed that its centroidal axis is perpendicular to the 
inclined plane. Since the area of cross section of this new member is very high, it 
does not allow any displacement along its centroidal axis of the jointA. Another 
method of incorporating inclined support in the analysis is to suitably modify the 
member stiffness matrix of all the members meeting at the inclined support. 
 
 
                                                         
Consider a truss member as shown in Fig. 25.4. The nodes are numbered as 1 
and 2. At 2, it is connected to a inclined support. Let ' 'y x be the local co-ordinate 
axes of the member. At node 1, the global co-ordinate system xy is also shown. 
At node 2, consider nodal co-ordinate system as " "y x , where " y is perpendicular 
to oblique support. Let 
1
' u and
2
' u be the displacements of nodes 1 and 2 in the 
local co-ordinate system. Let 
1 1
,v u be the nodal displacements of node 1 in 
global co-ordinate systemxy. Let 
22
", " uv be the nodal displacements along " x -
and " y - are in the local co-ordinate system " "y x at node 2. Then from Fig. 25.4, 
 
x x
v u u ? ? sin cos '
1 1 1
+ =      
 
" 2 " 2 2
sin " cos " '
x x
v u u ? ? + =      (25.1) 
 
This may be written as  
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
2
2
1
1
" " 2
1
"
" sin cos
0 0
0 0
sin cos
'
'
v
u
v
u
u
u
x x
x x
? ?
? ?
  (25.2) 
Denoting 
" "
sin " ; cos " ; sin ; cos
x x x x
m l m l ? ? ? ? = = = = 
 
 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
2
2
1
1
2
1
"
" " "
0 0
0 0 '
'
v
u
v
u
m l
m l
u
u
              (25.3a) 
 
 or  {} []{} u T u ' ' =       
 
where [] ' T is the displacement transformation matrix. 
 
 
                                                         
 
 
Similarly referring to Fig. 25.5, the force 
1
' p has components along x and 
y axes. Hence  
 
x
p p ? cos '
1 1
=     (25.4a) 
 
x
p p ? sin '
1 2
=     (25.4b) 
 
                                                         
Similarly, at node 2, the force
2
' p has components along " x and " y axes. 
32
"'cos
x
pp ? ' ' =
     (25.5a)  
 
42
"'sin
x
pp ? ' ' =
     (25.5b) 
 
The relation between forces in the global and local co-ordinate system may be 
written as, 
 
1
21
32
4
0 cos
0 ' sin
cos "' 0
sin " 0
x
x
x
x
p
p p
p p
p
?
?
?
?
?? ??
?? ??
? ?
??
??
=
?? ? ?
'' ??
??
??
??
??
''
??
?? ? ?
   (25.6) 
 
{} [] {} ' ' p T p
T
=      (25.7) 
 
Using displacement and force transformation matrices, the stiffness matrix for 
member having inclined support is obtained. 
 
[] [ ] [ ][ ] ' ' ' T k T k
T
= 
 
[]
?
?
?
?
?
?
?
?
?
?
?
?
-
-
?
?
?
?
?
?
?
?
?
?
?
?
=
" "
0 0
0 0 1 1
1 1
" 0
" 0
0
0
m l
m l
L
AE
m
l
m
l
k
 (25.8) 
 
Simplifying, 
 
[]
?
?
?
?
?
?
?
?
?
?
?
?
?
?
- -
- -
- -
- -
=
2
2
2
2
" " " " "
" " " " "
" "
" "
m m l mm lm
m l l ml ll
mm ml m lm
lm ll lm l
L
EA
k
  (25.9) 
 
If we use this stiffness matrix, then it is easy to incorporate the condition of zero 
displacement perpendicular to the inclined support in the stiffness matrix. This is 
shown by a simple example.