Introduction: The Direct Stiffness Method - 1 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 1


Instructional Objectives: 
After reading this chapter the student will be able to 
1. Differentiate between the direct stiffness method and the displacement 
method. 
2. Formulate flexibility matrix of member. 
3. Define stiffness matrix. 
4. Construct stiffness matrix of a member. 
5. Analyse simple structures by the direct stiffness matrix. 
 
 
23.1 Introduction 
All known methods of structural analysis are classified into two distinct groups:- 
 
(i) force method of analysis and 
(ii) displacement method of analysis. 
 
In module 2, the force method of analysis or the method of consistent 
deformation is discussed. An introduction to the displacement method of analysis 
is given in module 3, where in slope-deflection method and moment- distribution 
method are discussed. In this module the direct stiffness method is discussed. In 
the displacement method of analysis the equilibrium equations are written by 
expressing the unknown joint displacements in terms of loads by using load-
displacement relations. The unknown joint displacements (the degrees of 
freedom of the structure) are calculated by solving equilibrium equations. The 
slope-deflection and moment-distribution methods were extensively used before 
the high speed computing era. After the revolution in computer industry, only 
direct stiffness method is used. 
 
The displacement method follows essentially the same steps for both statically 
determinate and indeterminate structures. In displacement /stiffness method of 
analysis, once the structural model is defined, the unknowns (joint rotations and 
translations) are automatically chosen unlike the force method of analysis. 
Hence, displacement method of analysis is preferred to computer 
implementation. The method follows a rather a set procedure. The direct stiffness 
method is closely related to slope-deflection equations. 
 
The general method of analyzing indeterminate structures by displacement 
method may be traced to Navier (1785-1836). For example consider a four 
member truss as shown in Fig.23.1.The given truss is statically indeterminate to 
second degree as there are four bar forces but we have only two equations of 
equilibrium. Denote each member by a number, for example (1), (2), (3) and (4). 
Let 
i
a
   be the angle, the i-th member makes with the horizontal. Under the 
                                                         
Page 2


Instructional Objectives: 
After reading this chapter the student will be able to 
1. Differentiate between the direct stiffness method and the displacement 
method. 
2. Formulate flexibility matrix of member. 
3. Define stiffness matrix. 
4. Construct stiffness matrix of a member. 
5. Analyse simple structures by the direct stiffness matrix. 
 
 
23.1 Introduction 
All known methods of structural analysis are classified into two distinct groups:- 
 
(i) force method of analysis and 
(ii) displacement method of analysis. 
 
In module 2, the force method of analysis or the method of consistent 
deformation is discussed. An introduction to the displacement method of analysis 
is given in module 3, where in slope-deflection method and moment- distribution 
method are discussed. In this module the direct stiffness method is discussed. In 
the displacement method of analysis the equilibrium equations are written by 
expressing the unknown joint displacements in terms of loads by using load-
displacement relations. The unknown joint displacements (the degrees of 
freedom of the structure) are calculated by solving equilibrium equations. The 
slope-deflection and moment-distribution methods were extensively used before 
the high speed computing era. After the revolution in computer industry, only 
direct stiffness method is used. 
 
The displacement method follows essentially the same steps for both statically 
determinate and indeterminate structures. In displacement /stiffness method of 
analysis, once the structural model is defined, the unknowns (joint rotations and 
translations) are automatically chosen unlike the force method of analysis. 
Hence, displacement method of analysis is preferred to computer 
implementation. The method follows a rather a set procedure. The direct stiffness 
method is closely related to slope-deflection equations. 
 
The general method of analyzing indeterminate structures by displacement 
method may be traced to Navier (1785-1836). For example consider a four 
member truss as shown in Fig.23.1.The given truss is statically indeterminate to 
second degree as there are four bar forces but we have only two equations of 
equilibrium. Denote each member by a number, for example (1), (2), (3) and (4). 
Let 
i
a
   be the angle, the i-th member makes with the horizontal. Under the 
                                                         
action of external loads  and , the joint E displaces to E’. Let u and v be its 
vertical and horizontal displacements. Navier solved this problem as follows. 
x
P
y
P
 
In the displacement method of analysis u and v are the only two unknowns for 
this structure. The elongation of individual truss members can be expressed in 
terms of these two unknown joint displacements. Next, calculate bar forces in the 
members by using force–displacement relation. Now at E, two equilibrium 
equations can be written viz., 
0 =
?
x
F
and 
0 =
?
y
F
 by summing all forces 
in x and y directions. The unknown displacements may be calculated by solving 
the equilibrium equations. In displacement method of analysis, there will be 
exactly as many equilibrium equations as there are unknowns. 
 
Let an elastic body is acted by a force F and the corresponding displacement be 
u in the direction of force. In module 1, we have discussed force- displacement 
relationship. The force (F) is related to the displacement (u) for the linear elastic 
material by the relation 
ku F =
     (23.1) 
 
where the constant of proportionality k is defined as the stiffness of the structure 
and it has units of force per unit elongation. The above equation may also be 
written as 
 
aF u =      (23.2)  
                                                         
Page 3


Instructional Objectives: 
After reading this chapter the student will be able to 
1. Differentiate between the direct stiffness method and the displacement 
method. 
2. Formulate flexibility matrix of member. 
3. Define stiffness matrix. 
4. Construct stiffness matrix of a member. 
5. Analyse simple structures by the direct stiffness matrix. 
 
 
23.1 Introduction 
All known methods of structural analysis are classified into two distinct groups:- 
 
(i) force method of analysis and 
(ii) displacement method of analysis. 
 
In module 2, the force method of analysis or the method of consistent 
deformation is discussed. An introduction to the displacement method of analysis 
is given in module 3, where in slope-deflection method and moment- distribution 
method are discussed. In this module the direct stiffness method is discussed. In 
the displacement method of analysis the equilibrium equations are written by 
expressing the unknown joint displacements in terms of loads by using load-
displacement relations. The unknown joint displacements (the degrees of 
freedom of the structure) are calculated by solving equilibrium equations. The 
slope-deflection and moment-distribution methods were extensively used before 
the high speed computing era. After the revolution in computer industry, only 
direct stiffness method is used. 
 
The displacement method follows essentially the same steps for both statically 
determinate and indeterminate structures. In displacement /stiffness method of 
analysis, once the structural model is defined, the unknowns (joint rotations and 
translations) are automatically chosen unlike the force method of analysis. 
Hence, displacement method of analysis is preferred to computer 
implementation. The method follows a rather a set procedure. The direct stiffness 
method is closely related to slope-deflection equations. 
 
The general method of analyzing indeterminate structures by displacement 
method may be traced to Navier (1785-1836). For example consider a four 
member truss as shown in Fig.23.1.The given truss is statically indeterminate to 
second degree as there are four bar forces but we have only two equations of 
equilibrium. Denote each member by a number, for example (1), (2), (3) and (4). 
Let 
i
a
   be the angle, the i-th member makes with the horizontal. Under the 
                                                         
action of external loads  and , the joint E displaces to E’. Let u and v be its 
vertical and horizontal displacements. Navier solved this problem as follows. 
x
P
y
P
 
In the displacement method of analysis u and v are the only two unknowns for 
this structure. The elongation of individual truss members can be expressed in 
terms of these two unknown joint displacements. Next, calculate bar forces in the 
members by using force–displacement relation. Now at E, two equilibrium 
equations can be written viz., 
0 =
?
x
F
and 
0 =
?
y
F
 by summing all forces 
in x and y directions. The unknown displacements may be calculated by solving 
the equilibrium equations. In displacement method of analysis, there will be 
exactly as many equilibrium equations as there are unknowns. 
 
Let an elastic body is acted by a force F and the corresponding displacement be 
u in the direction of force. In module 1, we have discussed force- displacement 
relationship. The force (F) is related to the displacement (u) for the linear elastic 
material by the relation 
ku F =
     (23.1) 
 
where the constant of proportionality k is defined as the stiffness of the structure 
and it has units of force per unit elongation. The above equation may also be 
written as 
 
aF u =      (23.2)  
                                                         
 
 
The constant  is known as flexibility of the structure and it has a unit of 
displacement per unit force. In general the structures are subjected to  forces at 
 different locations on the structure. In such a case, to relate displacement at i 
to load at 
a
n
n
j , it is required to use flexibility coefficients with subscripts. Thus the 
flexibility coefficient   is the deflection at  due to unit value of force applied at 
ij
a
i
j. Similarly the stiffness coefficient  is defined as the force generated at i  
ij
k
                                                         
Page 4


Instructional Objectives: 
After reading this chapter the student will be able to 
1. Differentiate between the direct stiffness method and the displacement 
method. 
2. Formulate flexibility matrix of member. 
3. Define stiffness matrix. 
4. Construct stiffness matrix of a member. 
5. Analyse simple structures by the direct stiffness matrix. 
 
 
23.1 Introduction 
All known methods of structural analysis are classified into two distinct groups:- 
 
(i) force method of analysis and 
(ii) displacement method of analysis. 
 
In module 2, the force method of analysis or the method of consistent 
deformation is discussed. An introduction to the displacement method of analysis 
is given in module 3, where in slope-deflection method and moment- distribution 
method are discussed. In this module the direct stiffness method is discussed. In 
the displacement method of analysis the equilibrium equations are written by 
expressing the unknown joint displacements in terms of loads by using load-
displacement relations. The unknown joint displacements (the degrees of 
freedom of the structure) are calculated by solving equilibrium equations. The 
slope-deflection and moment-distribution methods were extensively used before 
the high speed computing era. After the revolution in computer industry, only 
direct stiffness method is used. 
 
The displacement method follows essentially the same steps for both statically 
determinate and indeterminate structures. In displacement /stiffness method of 
analysis, once the structural model is defined, the unknowns (joint rotations and 
translations) are automatically chosen unlike the force method of analysis. 
Hence, displacement method of analysis is preferred to computer 
implementation. The method follows a rather a set procedure. The direct stiffness 
method is closely related to slope-deflection equations. 
 
The general method of analyzing indeterminate structures by displacement 
method may be traced to Navier (1785-1836). For example consider a four 
member truss as shown in Fig.23.1.The given truss is statically indeterminate to 
second degree as there are four bar forces but we have only two equations of 
equilibrium. Denote each member by a number, for example (1), (2), (3) and (4). 
Let 
i
a
   be the angle, the i-th member makes with the horizontal. Under the 
                                                         
action of external loads  and , the joint E displaces to E’. Let u and v be its 
vertical and horizontal displacements. Navier solved this problem as follows. 
x
P
y
P
 
In the displacement method of analysis u and v are the only two unknowns for 
this structure. The elongation of individual truss members can be expressed in 
terms of these two unknown joint displacements. Next, calculate bar forces in the 
members by using force–displacement relation. Now at E, two equilibrium 
equations can be written viz., 
0 =
?
x
F
and 
0 =
?
y
F
 by summing all forces 
in x and y directions. The unknown displacements may be calculated by solving 
the equilibrium equations. In displacement method of analysis, there will be 
exactly as many equilibrium equations as there are unknowns. 
 
Let an elastic body is acted by a force F and the corresponding displacement be 
u in the direction of force. In module 1, we have discussed force- displacement 
relationship. The force (F) is related to the displacement (u) for the linear elastic 
material by the relation 
ku F =
     (23.1) 
 
where the constant of proportionality k is defined as the stiffness of the structure 
and it has units of force per unit elongation. The above equation may also be 
written as 
 
aF u =      (23.2)  
                                                         
 
 
The constant  is known as flexibility of the structure and it has a unit of 
displacement per unit force. In general the structures are subjected to  forces at 
 different locations on the structure. In such a case, to relate displacement at i 
to load at 
a
n
n
j , it is required to use flexibility coefficients with subscripts. Thus the 
flexibility coefficient   is the deflection at  due to unit value of force applied at 
ij
a
i
j. Similarly the stiffness coefficient  is defined as the force generated at i  
ij
k
                                                         
due to unit displacement at j with all other displacements kept at zero. To 
illustrate this definition, consider a cantilever beam which is loaded as shown in 
Fig.23.2. The two degrees of freedom for this problem are vertical displacement 
at B and rotation at B. Let them be denoted by  and  (=
1
u
2
u
1
?
). Denote the 
vertical force P by  and the tip moment M by . Now apply a unit vertical 
force along  and calculate deflection  and .The vertical deflection is 
denoted by flexibility coefficient  and rotation is denoted by flexibility 
coefficient . Similarly, by applying a unit force along , one could calculate 
flexibility coefficient  and . Thus is the deflection at 1 corresponding to 
due to unit force applied at 2 in the direction of . By using the principle of 
superposition, the displacements and  are expressed as the sum of 
displacements due to loads  and  acting separately on the beam. Thus, 
1
P
2
P
1
P
1
u
2
u
11
a
21
a
1
P
12
a
22
a
12
a
1
P
2
P
1
u
2
u
1
P
2
P
 
2 12 1 11 1
P a P a u + =
 
2 22 1 21 2
P a P a u + =
         (23.3a) 
 
The above equation may be written in matrix notation as 
 
{} [ ] { } P a u =
 
 
where, {} {} ; and {}
1
2
;
u
u
u
??
=
??
??
11 12
21 22
aa
a
aa
??
=
??
??
1
2
P
P
P
? ?
=
? ?
??
 
 
                                                         
Page 5


Instructional Objectives: 
After reading this chapter the student will be able to 
1. Differentiate between the direct stiffness method and the displacement 
method. 
2. Formulate flexibility matrix of member. 
3. Define stiffness matrix. 
4. Construct stiffness matrix of a member. 
5. Analyse simple structures by the direct stiffness matrix. 
 
 
23.1 Introduction 
All known methods of structural analysis are classified into two distinct groups:- 
 
(i) force method of analysis and 
(ii) displacement method of analysis. 
 
In module 2, the force method of analysis or the method of consistent 
deformation is discussed. An introduction to the displacement method of analysis 
is given in module 3, where in slope-deflection method and moment- distribution 
method are discussed. In this module the direct stiffness method is discussed. In 
the displacement method of analysis the equilibrium equations are written by 
expressing the unknown joint displacements in terms of loads by using load-
displacement relations. The unknown joint displacements (the degrees of 
freedom of the structure) are calculated by solving equilibrium equations. The 
slope-deflection and moment-distribution methods were extensively used before 
the high speed computing era. After the revolution in computer industry, only 
direct stiffness method is used. 
 
The displacement method follows essentially the same steps for both statically 
determinate and indeterminate structures. In displacement /stiffness method of 
analysis, once the structural model is defined, the unknowns (joint rotations and 
translations) are automatically chosen unlike the force method of analysis. 
Hence, displacement method of analysis is preferred to computer 
implementation. The method follows a rather a set procedure. The direct stiffness 
method is closely related to slope-deflection equations. 
 
The general method of analyzing indeterminate structures by displacement 
method may be traced to Navier (1785-1836). For example consider a four 
member truss as shown in Fig.23.1.The given truss is statically indeterminate to 
second degree as there are four bar forces but we have only two equations of 
equilibrium. Denote each member by a number, for example (1), (2), (3) and (4). 
Let 
i
a
   be the angle, the i-th member makes with the horizontal. Under the 
                                                         
action of external loads  and , the joint E displaces to E’. Let u and v be its 
vertical and horizontal displacements. Navier solved this problem as follows. 
x
P
y
P
 
In the displacement method of analysis u and v are the only two unknowns for 
this structure. The elongation of individual truss members can be expressed in 
terms of these two unknown joint displacements. Next, calculate bar forces in the 
members by using force–displacement relation. Now at E, two equilibrium 
equations can be written viz., 
0 =
?
x
F
and 
0 =
?
y
F
 by summing all forces 
in x and y directions. The unknown displacements may be calculated by solving 
the equilibrium equations. In displacement method of analysis, there will be 
exactly as many equilibrium equations as there are unknowns. 
 
Let an elastic body is acted by a force F and the corresponding displacement be 
u in the direction of force. In module 1, we have discussed force- displacement 
relationship. The force (F) is related to the displacement (u) for the linear elastic 
material by the relation 
ku F =
     (23.1) 
 
where the constant of proportionality k is defined as the stiffness of the structure 
and it has units of force per unit elongation. The above equation may also be 
written as 
 
aF u =      (23.2)  
                                                         
 
 
The constant  is known as flexibility of the structure and it has a unit of 
displacement per unit force. In general the structures are subjected to  forces at 
 different locations on the structure. In such a case, to relate displacement at i 
to load at 
a
n
n
j , it is required to use flexibility coefficients with subscripts. Thus the 
flexibility coefficient   is the deflection at  due to unit value of force applied at 
ij
a
i
j. Similarly the stiffness coefficient  is defined as the force generated at i  
ij
k
                                                         
due to unit displacement at j with all other displacements kept at zero. To 
illustrate this definition, consider a cantilever beam which is loaded as shown in 
Fig.23.2. The two degrees of freedom for this problem are vertical displacement 
at B and rotation at B. Let them be denoted by  and  (=
1
u
2
u
1
?
). Denote the 
vertical force P by  and the tip moment M by . Now apply a unit vertical 
force along  and calculate deflection  and .The vertical deflection is 
denoted by flexibility coefficient  and rotation is denoted by flexibility 
coefficient . Similarly, by applying a unit force along , one could calculate 
flexibility coefficient  and . Thus is the deflection at 1 corresponding to 
due to unit force applied at 2 in the direction of . By using the principle of 
superposition, the displacements and  are expressed as the sum of 
displacements due to loads  and  acting separately on the beam. Thus, 
1
P
2
P
1
P
1
u
2
u
11
a
21
a
1
P
12
a
22
a
12
a
1
P
2
P
1
u
2
u
1
P
2
P
 
2 12 1 11 1
P a P a u + =
 
2 22 1 21 2
P a P a u + =
         (23.3a) 
 
The above equation may be written in matrix notation as 
 
{} [ ] { } P a u =
 
 
where, {} {} ; and {}
1
2
;
u
u
u
??
=
??
??
11 12
21 22
aa
a
aa
??
=
??
??
1
2
P
P
P
? ?
=
? ?
??
 
 
                                                         
 
 
The forces can also be related to displacements using stiffness coefficients. 
Apply a unit displacement along  (see Fig.23.2d) keeping displacement  as 
zero. Calculate the required forces for this case as  and .Here,  
represents force developed along  when a unit displacement along is 
introduced keeping =0. Apply a unit rotation along (vide Fig.23.2c) ,keeping 
. Calculate the required forces for this configuration and . Invoking 
the principle of superposition, the forces  and are expressed as the sum of 
forces developed due to displacements and acting separately on the beam. 
Thus, 
1
u
2
u
11
k
21
k
21
k
2
P
1
u
2
u
2
u
0
1
= u
12
k
22
k
1
P
2
P
1
u
2
u
 
2 12 1 11 1
u k u k P + =
 
 
2 22 1 21 2
u k u k P + =
     (23.4) 
  
{} [ ] { } u k P =