Overview
The stiffness and flexibility methods are the most common techniques for analysing statically indeterminate structures. In the flexibility method the requirement that the deflected shape of the structure being analysed be compatible with the boundary conditions of the structure is used to generate additional equations. In the flexibility approach the number of compatibility equations required (i.e. the size of the flexibility matrix) is equal to the degree of static indeterminacy of the structure. Remember, you introduce releases until the structure is statically determinate
In the stiffness method the equations being solved are equilibrium equations .
However, the unknown variables in the equations are not the member forces (as would be used in a traditional static analysis) but are the nodal displacements . Once the displaced shape of the structure is known the forces in the individual members can be calculated. For example, in a pin-jointed truss change in member length will dictate the member force. Similarly, in a frame structure the relative displacements, lateral and rotational, of the members ends will, in combination with the loads applied to the member, define the bending moments in the member.
In the stiffness method the number of equations equals the number of displacements required to describe the displaced shape of the structure. The number of displacements required to describe the displaced shape of a structure is called the
Degree of Kinematic Indeterminacy, this is also called the number of degrees of freedom .
For example the statically determinate structure shown above has a degree of kinematic indeterminacy of 11. To describe the displacement of each internal requires two coordinates (i.e. a

X and a

Y ) plus the roller support can displace. Thus making a total of 11 degrees of freedom.
Note: the structure shown below, which is statically indeterminate, has the same degree of kinematic indeterminacy, and its solution will involve solving the same number of equilibrium equations as the statically determinate structure.
Stiffness and flexibility methods are related.
1

Characteristics of the stiffness method
Generally requires more equations to be solved than the flexibility method.
It is easy to program because the method can be expressed in algorithm form easily.
(With the flexibility method one must choose appropriate releases). For this reason most simple structural analysis programs are based on the stiffness method.
The stiffness method is a simple form of finite element method F.E.M.
The fundamental equation describing the stiffness method is,
    

F
R

Where,
 
 
is the stiffness matrix, which is know
 
R
is the vector of unknown displacements
is the vector of fictitious restraining forces (For a simple pin- jointed truss

F
R
is the same at the applied loads F )
2

Consider a simple structure that is statically determinate, so it’s easy to calculate the forces in the members using statics alone. To formulate this problem using the stiffness method involves developing equilibrium equations as functions of the unknown displacements

X and

Y .
Force P Force P
Al pha
L1, A1, E1
Delta X
L1, A1, E1
L2, A2, E2 L2, A2, E2
If the node at which load P is applied is displaced laterally by

X compressive strain in member 1, the horizontal member, is,
then the

1


X
L
1
Therefore the compressive stress in the member is,

1

E
1

1

E
1

X
L
1
Therefore the force applied to the node by the horizontal member is
F

A
1
E
1

X
L
1
The vertical member does not apply a horizontal force to the node therefore given that the node is in equilibrium
P cos
 


A
1
E
1
L
1



X

0
This equation can be solved for

X because it is the only unknown.
3

A similar expression can be developed for vertical equilibrium of the node,

P sin
 


A
2
L
E
2
2



Y

0 and this equation can be solved to calculate

Y . Once

X and

Y have been calculated the member forces, stresses and strains can be calculated by back substituting into the earlier equation.
4

For the simple statically determinate problem presented the procedure seems unnecessarily complex. However, consider applying the method to the structure shown below. This time the structure is indeterminate and simple static cannot be used to calculate the forces in members.
Force P
Al pha
However, this structure has the same number of degrees of freedom as the earlier structure and hence solving this problem using the stiffness method is only marginally more difficult that the earlier problem.
Again the objective is to formulate two equilibrium equation for the node that are functions of the unknown horizontal,

X , and vertical,

Y , displacements of the node.
Considering horizontal displacement first. If the node is displaced horizontally by a 1 unit of displacement then the change of length of the lateral unit is simply 1.
However, the change of length of the inclined member is equal to 1 cos

.
1
1
Cos
Beta
Beta
5

This is clearly seen by considering the previous diagram in conjunction with the diagram below.
1
R ed uc tio n i n L ength
However, this relationship is only true if the magnitude of the lateral displacement is very small, so small that the geometry of the structure is not affected. In this case the angle of inclination of the inclined member must be

post deflection. This is an important restriction on the use of the stiffness approach. If the deflections are large the relationship between the force in the inclined member and the displacement will not be linear.
In a similar manner it is easy to calculate the change in length of the inclined member due to a vertical displacement of the loaded node.
1 S in B eta
Be ta
Namely sin

times the nodes vertical displacement. If the displacements are small then the overall change in length of the inclined member is equal to the sum of the change in length due to the lateral nodal displacement and the change in length due to the vertical nodal displacement.

L inclined
 
X cos
  
Y sin

A change in the length of the inclined member gives rise to strains, stresses and as a result the member will apply both a horizontal and vertical force to the node.
6

Thus, the equilibrium equations for the loaded node are,
P cos
 


A horiz .
E horiz .
L horiz .



X
 cos



A inclined
L
E inclined inclined




X cos
 for horizontal equilibrium, and
P sin
 


A vert .
E vert .
L vert .



Y
 sin



A inclined
E inclined




X
L inclined cos

 
Y sin
 

0
 
Y sin
 

0 for vertical equilibrium.
The equations can be rearranged as follows.


P
P cos
 sin













A horiz
L horiz

.


E
A horiz .
.

 inclined
L

E

 inclined inclined
A inclined
L


E sin inclined inclined
 cos


 cos
2




A
 vert .
L


A inclined
E inclined
E vert .
L inclined





A inclined vert .

 cos
E
 sin inclined
L inclined


 sin
2











X
Y


 

0
0


7

1.
Identify the degrees of freedom required to describe the deflected shape of the structure. Assume that all the n degrees of freedom are restrained in position.
2.
Apply the loads to the structure and calculate the forces
 
R
that need to be applied at the restraints to prevent movement. In addition calculate
 
R m

1
, the m required structural actions, for the restrained structure.
3.
Generate the stiffness matrix
 
. The stiffness matrix is generated by, displacing each degree of freedom by one unit while preventing movement at all other degrees of freedom. This is done for each degree of freedom in turn.
Thus, for each degree of freedom i we calculate s , the force applied at i ii associated with a unit displacement at i (this is the force required to cause the displacement) and zero displacement at all other nodes. And we calculate a series of values s ji
, the force applied at j associated with a unit displacement at i and zero displacement at all other nodes (including j ).
In addition, calculate
 
U m
 n
the m structural actions associated with unit displacements at each of the n degrees of freedom.
4.
Find the unknown displacements
 
by solving the equation,
    

F
R

5.
Calculate the member forces and moments from the nodal displacements using the equation,
  m

1
  
R m

1

 
U m
 n
   n

1
8

Example – 1
Note:
For pin-jointed members s
11
 m  i

1
A i
E i cos
2
 i
L i s
21 s
12
 m  i

1
A i
E i cos
 i sin
 i
L i
 m  i

1
A i
E i
L i sin
 i cos
 i s
22
 m  i

1
A i
E i
L i sin
2
 i
  u




 








A
A
A m
L
1
2
L
L
E
1
E
2
E m
1
2
 m cos cos cos



1
2 m



A
1
E
1
L
1 sin

1
A
2
E
2
L
2
 sin
A m
E m sin
L m


2 m









Th eta
9