Module1 1 introduction-tomatrixms - rajesh sir

Structural Analysis - III
Introduction to
M t i M th dMatrix Methods
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKN
1

Module I
Matrix analysis of structures
Module I
• Definition of flexibility and stiffness influence coefficients –
d l t f fl ibilit t i b h i l h &
development of flexibility matrices by physical approach &
energy principle.
Flexibility method
• Flexibility matrices for truss beam and frame elements –• Flexibility matrices for truss, beam and frame elements –
load transformation matrix-development of total flexibility
matrix of the structure –analysis of simple structures –
l t ti b d l f d l l dplane truss, continuous beam and plane frame- nodal loads
and element loads – lack of fit and temperature effects.
2

Force method and Displacement methodp
•These methods are applicable to discretized structures•These methods are applicable to discretized structures
of all types
• Force method (Flexibility method)
• Actions are the primary unknowns
• Static indeterminacy: excess of unknown
actions than the available number of equationsq
of static equilibrium

• Displacement method (Stiffness method)
• Displacements of the joints are the primary
unknowns
• Kinematic indeterminacy: number of
d d l d ( hindependent translations and rotations (the
unknown joint displacements)
• More suitable for computer programming

Types of Framed Structures
• a. Beams: may support bending moment, shear force and
Types of Framed Structures
axial force
• b. Plane trusses: hinge joints; In addition to axial forces, a
member CAN have bending moments and shear forces if it
5
g
has loads directly acting on them, in addition to joint loads

• c. Space trusses: hinge joints; any couple acting on ac. Space trusses: hinge joints; any couple acting on a
member should have moment vector perpendicular to the axis
of the member, since a truss member is incapable of
ti t i ti tsupporting a twisting moment
• d. Plane frames: Joints are rigid; all forces in the plane of
the frame all couples normal to the plane of the frame
6
the frame, all couples normal to the plane of the frame

G id ll f l t th l f th id ll• e. Grids: all forces normal to the plane of the grid, all
couples in the plane of the grid (includes bending and
torsion))
• f. Space frames: most general framed structure; may
support bending moment shear force axial force and
7
support bending moment, shear force, axial force and
torsion

Deformations in Framed Structures
, ,x y zT M MThree couples:, ,x y zN V VThree forces:
•Significant deformations in framed structures:•Significant deformations in framed structures:
Structure Significant deformationsStructure Significant deformations
Beams flexural
Plane trusses axialPlane trusses axial
Space trusses axial
Plane frames flexural and axialPlane frames flexural and axial
Grids flexural and torsional
Space frames axial flexural and torsional
8
Space frames axial, flexural and torsional

Types of deformations in framed structures
9
b) axial c) shearing d) flexural e) torsional

Static indeterminacy
B
Static indeterminacy
• Beam:
• Static indeterminacy = Reaction components - number ofy p
eqns available
3E R= −
• Examples:
• Single span beam with both ends hinged with
inclined loads
C ti b• Continuous beam
• Propped cantilever
• Fixed beam

11

• Rigid frame (Plane):g d a e ( a e):
• External indeterminacy = Reaction components - number of
il bleqns available 3E R= −
• Internal indeterminacy = 3 × closed frames 3I a=
• Total indeterminacy
= External indeterminacy + Internal indeterminacy
( )3 3T E I R a= + = − +
• Note: An internal hinge will provide an additional eqn
12

Example 1 Example 2Example 1 Example 2
( )3 3T E I R a= + = − +( )3 3T E I R a= + = − + ( )
( )3 3 3 3 2 12= × − + × =( )2 2 3 3 0 1= × − + × =
Example 3
Example 4
( )
( )
3 3T E I R a= + = − + ( )
( )
3 3T E I R a= + = − +
( )3 2 3 3 3 12= × − + × = ( )4 3 3 3 4 21= × − + × =

14

• Rigid frame (Space):g d a e (Space):
• External indeterminacy = Reaction components - number of
il bleqns available
6E R= −
• Internal indeterminacy = 6 × closed frames
Example 1
( )6 6T E I R a= + = − +
Example 1
( )4 6 6 6 1 24= × − + × =
If axial deformations are neglected, static indeterminacy is
not affected since the same number of actions still exist in
not affected since the same number of actions still exist in
the structure

Plane truss (general):
External indeterminacy
= Reaction components - number of eqns available
3E R= −
Minimum 3 members and 3 joints.
A dditi l j i t i 2 dditi l b
Hence, number of members for stability,
Any additional joint requires 2 additional members.
( )3 2 3 2 3m j j= + − = −

( )2 3I m j= − −Hence, internal indeterminacy,
l ( l d l) dTotal (Internal and external) indeterminacy
( )3 2 3T E I R m j= + = − + − −
2m R j= + −
b f b• m : number of members
• R : number of reaction components
• j : number of jointsj j
• Note: Internal hinge will provide additional eqn
17

Example 1p
2 9 3 2 6 0T m R j= + − = + − × =
3 3 3 0E R= − = − =
0I T E= − =
Example 2
2 15 4 2 8 3T m R j+ + ×2 15 4 2 8 3T m R j= + − = + − × =
3 4 3 1E R= − = − =
2I T E= − =
Example 3
2 6 4 2 5 0T m R j= + − = + − × =
( )3 1 4 4 Hinge at0 AE R= − + = − =
0I T E= − =

Example 4Example 4
2 7 3 2 5 0T m R j= + − = + − × =
3 3 3 0E R= − = − =3 3 3 0E R= = =
0I T E= − =
2 6 4 2 4 2T m R j= + − = + − × =
Example 5
2 6 4 2 4 2T m R j+ +
3 4 3 1E R= − = − =
1I T E 1I T E= − =
Example 6
2 11 3 2 6 2T m R j= + − = + − × =
3 3 3 0E R= − = − =
p
2I T E= − =

• Wall or roof attached pin jointed plane truss (Exception• Wall or roof attached pin jointed plane truss (Exception
to the above general case):
• Internal indeterminacy 2I m j= −
• External indeterminacy = 0 (Since, once the member
forces are determined, reactions are determinable)forces are determined, reactions are determinable)
Example 1 Example 3Example 2Example 1 Example 3Example 2
2
6 2 3 0
T I m j= = −
= − × =
2
5 2 1 3
T I m j= = −
= − × =
2
7 2 3 1
T I m j= = −
20
6 3 0
7 2 3 1= − × =

• Space Truss:
• External indeterminacy = Reaction components -
number of equations available 6E R= −
E l 1
3T m R j= + −Total (Internal and external) indeterminacy
Example 1
Total (Internal and external) indeterminacy
3T m R j= + −
( ) y
12 9 3 6 3T∴ = + − × =
6 9 6 3E R= − = − =

Actions and displacements
•Actions:
p
• External actions (Force or couple or combinations) and
• Internal actions (Internal stress resultants – BM, SF, axial
forces, twisting moments)
22

•Displacements: A translation or rotation at some pointsp ace e ts: t a s at o o otat o at so e po t
•Displacement corresponding to an action: Need not be caused
b th t tiby that action

•Notations for actions and displacements:•Notations for actions and displacements:
D32
D33
24

Equilibrium
•Resultant of all actions (a force, a couple or both) must vanish
for static equilibrium
q
q
•Resultant force vector must be zero; resultant moment vector
must be zero
0xF =∑ 0yF =∑ 0zF =∑
∑ ∑ ∑
must be zero
0xM =∑ 0yM =∑ 0zM =∑
F 2 di i l bl (f i l•For 2-dimensional problems (forces are in one plane
and couples have vectors normal to the plane),
0xF =∑ 0yF =∑ 0zM =∑
•In stiffness method, the basic equations to be solved
25
In stiffness method, the basic equations to be solved
are the equilibrium conditions at the joints

Compatibility
•Compatibility conditions: Conditions of continuity of
displacements throughout the structure
•Eg: at a rigid connection between two members, the
displacements (translations and rotations) of both membersdisplacements (translations and rotations) of both members
must be the same
I fl ibili h d h b i i b l d•In flexibility method, the basic equations to be solved are
the compatibility conditions
26

Action and displacement equations
A SD=D FA=•Spring:
1
S F−
=•Stiffness
1
F S−
=•Flexibility:
•The above equations apply to any linearly elastic
structure
27

•Example 1:
Flexibility and stiffness of a beam subjected to a single load
3
48
L
F
EI
=Flexibility 3
48EI
S
L
=Stiffness
28
48EI L

•Example 2:
Flexibility coefficients of a beam subjected to several loads
A ti th bActions on the beam
Deformations corresponding to actions
29

30
Unit load applied corresponding to each action, separately

1 11 12 13D D D D= + + 2 21 22 23D D D D= + + 3 31 32 33D D D D= + +
1 11 1 12 2 13 3D F A F A F A= + +
2 21 1 22 2 23 3D F A F A F A= + +
3 31 1 32 2 33 3D F A F A F A= + +
11 12 13, ,F F F Flexibility coefficientsetc.
• Flexibility coefficient F12: Displacement corresponding to A1
caused by a unit value of A2.caused by a unit value of A2.
• In general, flexibility coefficient Fij is the displacement
31
In general, flexibility coefficient Fij is the displacement
corresponding to Ai caused by a unit value of Aj.

•Example 3:
Stiffness coefficients of a beam subjected to several loads
Actions on the beam
32

Unit displacement applied corresponding to each DOF separately
33
Unit displacement applied corresponding to each DOF, separately,
keeping all other displacements zero

A A A A= + +1 11 12 13A A A A= + +
A S D S D S D+ +1 11 1 12 2 13 3A S D S D S D= + +
2 21 1 22 2 23 3A S D S D S D= + +
3 31 1 32 2 33 3A S D S D S D= + +
11 12 13, ,S S S Stiffness coefficients:etc.
• Stiffness coefficient S12: Action corresponding to D1 caused by
a unit value of D2.a unit value of D2.
• In general, stiffness coefficient Sij is the action corresponding
34
In general, stiffness coefficient Sij is the action corresponding
to Di caused by a unit value of Dj.

•Example 4: Flexibility and stiffness coefficients of a beamExample 4: Flexibility and stiffness coefficients of a beam
Flexibility coefficients:y
35

Stiffness coefficients:
36

•Example 5: Flexibility and stiffness coefficients of a trussExample 5: Flexibility and stiffness coefficients of a truss
37

Flexibility and stiffness matrices
1. Flexibility matrix
Th tibilit ti
1 11 1 12 2 13 3 1... n nD F A F A F A F A= + + + +
oThe compatibility equations are:
2 21 1 22 2 23 3 2...
............................................................
n nD F A F A F A F A= + + + +
1 1 2 2 3 3 ...n n n n nn nD F A F A F A F A= + + + +
oIn matrix form,
1 11 12 1 1
2 21 22 2 2
...
...
n
n
D F F F A
D F F F A
⎧ ⎫ ⎡ ⎤ ⎧ ⎫
⎪ ⎪ ⎢ ⎥ ⎪ ⎪
⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬
1 2
... ... ... ... ......
...n n nn nn F F F AD
⎢ ⎥=⎨ ⎬ ⎨ ⎬
⎢ ⎥⎪ ⎪ ⎪ ⎪
⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭⎩ ⎭
38
1 2
11
n n nn nn
n n nn × ××
⎣ ⎦ ⎩ ⎭⎩ ⎭

D = FA {D} Displacement matrix (vector),{D} Displacement matrix (vector),
[F] Flexibility matrix,
{A} A ti t i ( t ){ } [ ]{ }D F A= {A} Action matrix (vector){ } [ ]{ }D F A
F are the flexibility coefficientsijF are the flexibility coefficients
39

2. Stiffness matrix2. Stiffness matrix
oThe equilibrium equations are:
1 11 1 12 2 13 3 1... n nA S D S D S D S D
A S D S D S D S D
= + + + +
= + + + +2 21 1 22 2 23 3 2...
............................................................
n nA S D S D S D S D
A S D S D S D S D
= + + + +
+ + + +1 1 2 2 3 3 ...n n n n nn nA S D S D S D S D= + + + +
oIn matrix form,
1 11 12 1 1
2 21 22 2 2
...
...
n
n
A S S S D
A S S S D
⎧ ⎫ ⎡ ⎤ ⎧ ⎫
⎪ ⎪ ⎢ ⎥ ⎪ ⎪
⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬
1 2
... ... ... ... ......
...n n nn nn S S S DA
⎢ ⎥=⎨ ⎬ ⎨ ⎬
⎢ ⎥⎪ ⎪ ⎪ ⎪
⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭⎩ ⎭
40
1 2
11
n n nn nn
n n nn × ××
⎣ ⎦ ⎩ ⎭⎩ ⎭

A = SD {A} Action matrix (vector) ,
{ } [ ]{ }A S D=
[S] Stiffness matrix,
{D} Displacement matrix (vector)p ( )
ijS are the stiffness coefficientsijS are the stiffness coefficients
• Relationship between flexibility and stiffness matrices
{ } [ ]{ } [ ][ ]{ }A S D S F A
Relationship between flexibility and stiffness matrices
[ ] [ ]
1
F S
−
∴
{ } [ ]{ } [ ][ ]{ }A S D S F A= =
41
[ ] [ ]F S∴ =

•Example: Cantilever element
3
L 2
L L
11 ;
3
L
F
EI
= 21 12 ;
2
L
F F
EI
= = 22
L
F
EI
=
3 2
L L3 2
1 1 2
3 2
L L
D A A
EI EI
= +
2
2 1 2
2
L L
D A A
EI EI
= +
3 2
L L⎡ ⎤3 2
1 1
2
3 2
L L
D AEI EI
D AL L
⎡ ⎤
⎢ ⎥⎧ ⎫ ⎧ ⎫
⎢ ⎥=⎨ ⎬ ⎨ ⎬
⎢ ⎥⎩ ⎭ ⎩ ⎭2 2
2
D AL L
EI EI
⎢ ⎥⎩ ⎭ ⎩ ⎭
⎢ ⎥⎣ ⎦
42[ ]F

12EI 6EI− 4EI
11 3
12
;
EI
S
L
= 21 12 2
6
;
EI
S S
L
= = 22
4EI
S
L
=
1 1 23 2
12 6EI EI
A D D
L L
= −
2 1 22
6 4EI EI
A D D
L L
−
= +
3 2
12 6EI EI
A DL L
−⎡ ⎤
⎢ ⎥⎧ ⎫ ⎧ ⎫3 2
1 1
2 2
2
6 4
A DL L
A DEI EI
L L
⎢ ⎥⎧ ⎫ ⎧ ⎫
= ⎢ ⎥⎨ ⎬ ⎨ ⎬
−⎩ ⎭ ⎩ ⎭⎢ ⎥
⎢ ⎥⎣ ⎦L L⎢ ⎥⎣ ⎦
[ ][ ] [ ][ ] [ ]
1 0⎡ ⎤ [ ]S
43
[ ][ ] [ ][ ] [ ]
1 0
0 1
F S S F I
⎡ ⎤
= = =⎢ ⎥
⎣ ⎦
[ ]S

•Flexibility matrix and stiffness matrix are relating actions and
corresponding displacements
The flexibility matrix obtained for a structure analysed by[ ]Fy y y
flexibility method may not be the inverse of the stiffness matrix
obtained for the same structure analysed by stiffness method
[ ]S
obtained for the same structure analysed by stiffness method
because different sets of actions and corresponding displacements
may be utilized in the two methods.
44

Equivalent joint loads
•Analysis by flexibility and stiffness methods requires that loads
must act only at joints
must act only at joints.
•Thus, loads acting on the members (i.e., loads that are not acting
at the joints) must be replaced by equivalent loads acting at the
joints.
•The loads that are determined from loads on the members are
called equivalent joint loads.
45

•Equivalent joint loads are added to the actual joint loads to get
combined joint loads.
•Analysis carried out for combined joint loads
•Combined joint loads can be evaluated in such a manner that
the resulting displacements of the structure are same as the
di l t d d b th t l l ddisplacements produced by the actual loads
•This is achieved thru the use of fixed end actions to getg
equivalent joint loads
46

•Example 1•Example 1
Beam with actual applied loads
Applied joint loads
47
Applied joint loads

Applied loads other than joint loads
(To be converted to equivalent joint loads)( o be co e ted to equ a e t jo t oads)
48

Member fixed end actions
(Due to applied loads other than joint loads )
49
Fixed end actions for the entire beam

Equivalent joint loads (Negative of fixed end actions)Equivalent joint loads (Negative of fixed end actions)
Combined joint loads
(A li d j i t l d + E i l t j i t l d )
50
(Applied joint loads + Equivalent joint loads)

Fixed end actions
M
M
( )2
2
Mb
a b
l
−
a b ( )2
2
Ma
b a
l
−
3
6Mab
l 3
6Mab
l
M
M
4
M
2
l 4
M
2
l
3
2
M
l
3
2
M
l
2l2l

120 kN40 kN/m 20 kN/m
•Example 2
A
B C
D
/ 20 kN/m
4 m
12 m 12 m 12 m
2
l2
480
12
wl
= 480 240 240
wl 240 120 120
240
2
wl
= 240 120 120
213 33 106 67213.33 106.67
Fi d d ti
5288.89 31.11
Fixed end actions

480 266 67 133.33 240
A
B C
D
480 266.67
B C
240
240 88.89
328 89
+
=
120 31.11
151 11
+
120
328.89= 151.11=
(Opposite of fixed end actions)
Combined joint loads are same as equivalent joint loads here,
i th l d li d t j i t di tlsince there are no loads applied to joints directly
53

•Superposition of combined joint loads and restraint actions gives
the actual loads.the actual loads.
•Superposition of joint displacements due to the combined
j i l d d i i i h di ljoint loads and restraint actions gives the displacements
produced by the actual loads.
•But joint displacements due to restraint actions are zero.
Thus, joint displacements due to the combined joint loads give
th di l t d d b th t l l dthe displacements produced by the actual loads
•But member end actions due to actual loads are obtained byy
superimposing member end actions due to restraint actions and
combined joint loads
54

SummarySummary
• Definition of flexibility and stiffness influence coefficients –
d l t f fl ibilit t i b h i l h &
development of flexibility matrices by physical approach &
energy principle.
55

Module1 1 introduction-tomatrixms - rajesh sir

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Module1 1 introduction-tomatrixms - rajesh sir