This method is a powerful tool for analyzing indeterminate structures. One of its advantages over the flexibility method is that it is conducive to computer programming. Stiffness method the unknowns are the joint displacements in the structure, which are automatically specified.

1. SUBJECT :- STRUCTURAL ANALYSIS - 2
MATRIX ANALYSIS
(STIFFNESS METHOD)

2. STIFFNESS METHOD
• This method is a powerful tool for analyzing indeterminate structures. One of its
advantages over the flexibility method is that it is conducive to computer
programming.
• Once the analytical model of structure has been defined, no further decisions are
required in the stiffness method in order to carry out the analysis.
• Stiffness method the unknowns are the joint displacements in the structure, which
are automatically specified.
• In the stiffness method the number of unknowns to be calculated is the same as
the degree of kinematic indeterminacy of the structure.
• Stiffness =
Action
Displacement

3. The essential features of stiffness method :-
• This method is also known as displacement method or equilibrium method.
• This method is a matrix version of classical generalized slope-deflection method.
• Kinematically indeterminate structures are solved using this method.
• Joint displacements are treated as primary unknowns in this method.
• Numbers of unknowns id equal to the degree of kinematic indeterminacy of the
structure.
• The unknown joint displacements for a particular structure are uniquely defined.
• Conditions of joint equilibrium are used to form equations in unknown
displacement.

4. ACTIONS AND DISPLACEMENT
• The term action and displacement are the fundamental concepts in structural
analysis. An action(force) is most commonly a single force or couple. However,
an, action may be also a combination of force and couple, a distributed loading, or
a combination of these actions.
• In addition to actions that are external to a structure, it is necessary to deal also
with internal actions. These actions are the bending moment, shear force, axial
force and twisting moment.

5. • The cantilever beam is subjected at end B to
loads in the form of action 𝑃1 and 𝑀1. At the
fixed end A the reactive force and reactive
couple are denoted 𝑅 𝐴 and 𝑀𝐴, respectively.
In calculating the axial force N, bending
moment M, and shear force V at any section
of the beam such as midpoint, it is necessary
to consider the static equilibrium of the
beam. One possibility to construct a free
body diagram of the right-hand half of the
beam, as show in fig-b. in so doing, it is
evident that each of the internal actions
appears in the diagram as a single force or
couple.

6. • There situations, however, in which the
internal actions appear as two forces or
couples. This case occurs most commonly
in structure analysis when a “release” is
made at some point in a structure as shown
in a fig for a continuous beam. If the
bending moment is released at joint B of the
beam, the result is the same as if a hinge
were placed in the beam at the joint. In the
order to take account of B.M. in the beam,
it must be considered as consisting of two
equal and opposite couples 𝑀 𝐵 that act on
the left and right hand positions of the beam
with the hinge at B.

7. • A displacement, which is most commonly a deflection or a rotation at some
point in a structure. A deflection refer to the distance moved by a point in the
structure, and a rotation means the angle of rotation of the tangent to the
elastic curve at a point.

8. • Action is noted by A and displacement is noted by D.
• Portrays a cantilever beam subjected to action 𝐴1, 𝐴2 and 𝐴3. The displacement
corresponding to 𝐴1 and due to all loads acting simultaneously is denoted by 𝐷1
in fig-a, similarly, the displacements corresponding to 𝐴2and 𝐴3 are denoted by
𝐷2 and 𝐷3.

9. • Now consider the cantilever beam subjected to action 𝐴1 only the displacement
corresponding to 𝐴1 in this beam is denoted by 𝐷11. The significance of the two
subscripts is as follows.
• the first subscript indicates that the displacement correspond to action 𝐴1 and
the second indicates that the cause of the displacement is action 𝐴1. In a similar
manner, the displacement corresponding to 𝐴2 in this beam is demoted by
𝐷21, where the first subscript shows that the displacement correspond to 𝐴2 and
the second shows that it is caused by 𝐴1.also show in fig-b is the displacement
𝐷31 corresponding to the couple 𝐴3 and caused by 𝐴1.
• 𝐷11 =
𝐴1 𝐿3
24𝐸𝐼
𝐷21 =
5𝐴1 𝐿3
48𝐸𝐼
𝐷31 =
𝐴1 𝐿2
8𝐸𝐼

10. SUPERPOSITION
• In using the principle of
superposition it is assumed that
certain action and displacements
cause other action and
displacements to be developed
in the structure.
• In general terms principle states
that the effect produced by
several causes can be obtained
by combining the effects due to
the individual causes.

11. • The beam is subjected to load 𝐴1 and 𝐴2 which produce various action and
displacement through out the structure.
• for reaction 𝑅 𝐴, 𝑅 𝐵 and 𝑀 𝐵 are developed at the support, and displacement D
is produced at the midpoint of the beam. The effect of the action 𝐴1 and 𝐴2
acting separately are shows in fig-b and fig-c.

12. • The beam has constant flexural rigidity EI and is subjected to the loads 𝑃1 , M, 𝑃2,
and 𝑃3 . since rotation can occur at joints B and C ,the structure is kinematically
indeterminate to the second degree when axial deformation are neglected. Let the
unknown rotation at these joints be 𝐷1 𝑎𝑛𝑑 𝐷2, respectively, and assume that
counterclockwise rotations are positive . These unknown displacement may be
determined by solving equations of superposition for the action at joint B and C,
described in the following discussion.
• The restrained structure which is obtained by this means is shown in fig-b and
consist of two fixed end beams. The restrained structure is assumed to be acted
upon by all of the loads except those that correspond to the unknown displacement ,
thus, only the loads 𝑃1 , 𝑃2, and 𝑃3 are shows in fig-b. all loads that correspond to
the unknown joints displacement, such as the couple Min this example, are taken
into account later. The moments 𝐴 𝐷𝐿1 and 𝐴 𝐷𝐿2 are the action of the restrained
corresponding to 𝐷1 and 𝐷2, respectively, and caused by loads acting on the
structure .

13. • For example, the
restrained action 𝐴 𝐷𝐿1 Is
the sum of reactive
moments at B due to the
load 𝑝1 acting on member
AB and the reactive
moment at B due to the
𝑃2 Acting on member BC.

15. EXAMPLE
• K.I. = 2
Let, θ 𝐵 = 𝐷1
θ 𝑐 = 𝐷2
AD = actions in actual structure corresponding
to redundant
AD1 = 0
AD2 = 0
ADL = actions in restrained structure due to
loads corresponding to redundant.
ADL1 =
𝑤𝑙
8
-
𝑤𝑙
8
=
24 ∗10
8
-
12 ∗10
8
= 15KN.m
ADL1 =
𝑤𝑙
8
=
12 ∗10
8
= 15KN.m

16. • 𝑠11 =
4 𝐸𝐼
10
+
4 𝐸𝐼
10
= 0.8EI
𝑠21 =
2 𝐸𝐼
10
= 0.2EI
𝑠12 =
2 𝐸𝐼
10
= 0.2EI
𝑠22 =
4 𝐸𝐼
10
= 0.4EI
S = EI
0.8 0.2
0.2 0.4
S = 0.8*0.4 – 0.2*0.2 = 0.28EI
• 𝑠−1 =
1
S
adjS
=
1
0.28EI
0.4 −0.2
−0.2 0.8
D = -𝑠−1 * ADL
=
1
0.28EI
0.4 −0.2
−0.2 0.8
*
15
15
θ 𝐵 = -10.71/EI
θ 𝑐 = -32.14/EI