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Displacement Methods:Stiffness Matrix Method
Classification of Structures..• The primary function of structure is to safely transfer the
external loads acting on it to the foundation. • Structures can be classified in many different ways
• Dimension : 1 – D –>one dimension say length is very large when compared with other two. Ex: Beams, Trusses…
2- D or Surface Structure-> Here length & Breadth are large compared to thickness.
Ex: slabs, Deep Beams, Shells, walls, plates…
3-D or Solid Structure-> here all the three Dimensions L,b,d are prominent Ex: Dams, foundation , retaining walls etc..
• Structures can be classified based on the way they carry loads:
• Beam - carry their loads by developing B.M & S.F• Truss – Carry loads developing axial forces like tension and
compression.• Frames - carry loads by axial force, B.M & S.F.• Arches – carry load by compression & bending• Cables – carry loads by developing axial tension• Two way Grids – are subjected to both bending and twisting.• Thin Plates – are subjected to bending and twisting.• Thin Shells – Transfer stress as membrane stresses.
• Analysis of any structure involves finding out the displacements & internal forces
• Any method is SA should satisfy – • 1. Equilibrium • 2. Compatibility - --- these condition refer to the continuity of
displacement throughout the structure and are some times referred as geometrical conditions.Displacements should be compactable with support conditions for example there can be no translation and rotaion at a fixed support.Compatibility condition should be satisfied at all point in a structure.
• 3. Stress – Strain Law --- Stress ∞(Strain )n
• 4. Boundary condition
Kinematic Indeterminacy (Degree of Freedom)
Minimum independent quantities (displacement quantities like deflection and rotation ) - required to define the displaced geometry of the structure.
No points can go anywhere so kinematic indeterminacy = 0 Or Kinematically Determinate
3 Joints or Nodes = 2 D.O.F per Node
Total D.O.F = 2x3 = 6
Restrained D.O.F = 3
Kinematic Indeterminacy = 6-3 = 3
4 Joints or Nodes = 2 D.O.F per Node
Total D.O.F = 2x4 = 8
Restrained D.O.F = 3
Kinematic Indeterminacy = 8-3 = 5
3 D.O.F per Node
2 Nodes
Total D.O.F = 2x3 = 6
Restrained D.O.F = 2+1 = 3
Actual D.O.F = 6 – 3 = 3
Beams
V
MP
3 D.O.F per Node
2 Nodes
Total D.O.F = 2x3 = 6
Restrained D.O.F = 3
Actual D.O.F = 6 – 3 = 3
FRAMES
3 D.O.F per Node
4 Nodes
Total D.O.F = 4x3 = 12
Restrained D.O.F = 3+3 =6
Actual D.O.F = 12– 6 = 6
Axial RigidityDoesn’t have any effect on static indeterminacy.Beams are considered by default axially rigid.
3 D.O.F per Node - 2 Nodes
Total D.O.F = 2x3 = 6
Restrained D.O.F = 2 + 1 = 3
Constraint = 1 (axial rigidity)
Actual D.O.F = 6-3-1x1 (no of members) = 2
Frames3 D.O.F per Node - 2 Nodes
Total D.O.F = 4x3 = 12
Restrained D.O.F = 6
Constraint = 1 (axial rigidity)
Actual D.O.F = 12-6-1x3 (no of members) = 3
The supports are rigid vertical deflection = 0
Horizontal Deflection != 0 it’s a constant
So D.O.F = 2 Rotation + 1 Hor. Deflection
= 3
=0 =0
Matrix method of SA• In an elastic structure, there are 2 set of interrelated
quantities – forces (incl. moments, stresses, reaction etc..) & displacements (incl. rotations, strains, twist etc..).
• The behavior of a structure can largely be defined by defining the force – displacement relationship in the form of matrix.
Two Methods :
Flexibility Method Stiffness Method
1. Force Method Displacement Method
2. Basic Unknowns are the redundant forces
Basic unknowns are displacement of joints
3. [Δ] = [F][P][Disp Matrix] = [Flexibility Matrix] x [Load Matrix]
[K] [Δ] = [P][Stiffness Matrix] x[Disp. Matrix] = [Load Matrix]
Choice of Method• Usually Stiffness Method is preferred for the complex
structures in this method selection of unknowns is easy.• In force method skill and experience are required in the
selection of redundant.• When static indeterminacy is less than kinematic
indeterminacy the force method is preferred, otherwise displacement method is preferred.
K.I = 5, S.I = 1 (m-2j+3), K.I > S.I (Flexibility Method)
K.I = 2, S.I = 7 (m-2j+3), K.I < S.I (Stiffness Method)
Advantages of Matrix Method…• The flexibility and stiffness method provides a systematic
method for analysis of large structures with high degree of static and kinematic indeterminacy.
• The matrix approach can be easily converted to a computer programs which can be used for obtaining results.
• Matrix method is a generalized method takes into account of all he parameters that may influence a structural system.
Disadvantage: Large number of simultaneous equation makes the analysis
tedious or manual computations. So we require computing power.Further the analyst doesn't get an adequate feel for the flow of forces & structural behavior as in classical methods
Stiffness Method of Analysis For Beam • In order to apply stiffness method to beams, we must first
determine how to sub - divide the beam into its component .• In general each element must be free from load and have
prismatic cross-section.• For this reason the nodes of each element are located at a
support or at points where the c/s area suddenly changes, or where the the vertical or rotational displacement at a point is to determined.
• Beam Element Stiffness Matrix• Consider a beam element of uniform c/s area. The
longitudinal axis of the element lies along the x-axis , the element has
constant I , modulus of elasticity E and Length L.
M1 , Θ1M2, Θ2
21
F1 , δ1F2 , δ2
L, EI
42
1 3D.O.F
[K] [Δ] = [P]
Stiffness matrix K is a 4x4 matrix with stiffness coefficients.
Stiffness coefficient Kij means the force developed at ith D.O.F due to unit displacement at jth D.O.F such that other D.O.Fs are arrested or fixed.
42
1 D.O.F
6EI/L2 16EI/L2
12EI/L3
12EI/L3
K11 = force in 1 due to 1 = + 12EI/L3 ( )K21 = - 6EI/L2 K31 = -12EI/L3
K41 = - 6EI/L2
2EI/L
6EI/L26EI/L2
4EI/L
1
MAB = MFAB + 2EI/L(2θA+ θB ) = 4EI/LMBA = MFBA + 2EI/L(θA+ 2θB ) = 2EI/LθA = 1
K12 = -6EI/L2
K22 = 4EI/LK32 = +6EI/L2
K42 = 2EI/L
3
K13 = force in 1 due to 3 = -12EI/L3 K23 = + 6EI/L2 K33 = +12EI/L3
K43 = + 6EI/L2
6EI/L2
6EI/L2
4EI/L
12EI/L
K14 = -6EI/L2
K24 = 2EI/LK34 = +6EI/L2
K44 = 4EI/L
1 2 3 4K11 K12 K13 K14 1K21 K22 K23 K24 2K31 K32 K33 K34 3K41 K42 K43 K44 4[ Stiffness Matrix [K]
For a simple beam element =
F1 = K11. δ1 + K12 . Θ1 + K13. δ2 + K14. Θ2
M1 = K21. δ1 + K22 . Θ1 + K23. δ2 + K24. Θ2
F2 = K31. δ1 + K32 . Θ1 + K33. δ2 + K34. Θ2
M2 = K41. δ1 + K42 . Θ1 + K43. δ2 + K44. Θ2
F1M1F2 =M2
[ K11 K12 K13 K14 K21 K22 K23 K24 K31 K32 K33 K34 K41 K42 K43 K44 [
δ1
Θ1 δ2
Θ2
[Load Matrix
Displacement Matrix
12EI/L3 -6EI/L2 -12EI/L3 -6EI/L2
-6EI/L2 4EI/L 6EI/L2 2EI/L-12EI/L3 6EI/L2 12EI/L3 6EI/L2
-6EI/L3 2EI/L 6EI/L2 4EI/L[=
Properties of stiffness matrix• Symmetric Square Matrix of order n , n is number of
coordinates chosen for solution of problem• The diagonal elements are +ve• The element stiffness matrix is singular i.e determinant = 0 ,
hence inverse cannot be obtained.• The third row is of same magnitudes of first row but opposite
in sign i.e F1 = -F2Load matrix [P]The loads applied are transformed to equivalent joint loads [EJL]
or nodal loads.It can be obtained = Joint load matrix – Support Reaction Matrix
[P] = [Pj] – [PL]
C
P P
A
B
+PL/8
MBC
MCB
PL/2
P -PL/8
-PL/8 + PL/8
PL/2 + PL/2
PL/2
2
31 5
6
0 1 0 2[Pj] = -P 3 0 4 0 5 0 6
+PL/2 -PL/8[PL] = PL/2 +PL/2 PL/8 + -PL/8 PL/2 +PL/8
[P] = [Pj] – [PL]
LOAD MATRIX[ [4
Stiffness of a beam subjected to 2 Clockwise moments at the two ends
A
M A
φA
B
1D.O.F
2
2EI/L4EI/L
1
MAB = MFAB + 2EI/L(2θA+ θB ) = 4EI/LMBA = MFBA + 2EI/L(θA+ 2θB ) = 2EI/LθA = 1
K11 = 4EI/LK12 = 2EI/L
M B
2 – D.O.F K Matrix = 2x2
4EI/L
12EI/L K21 = 2EI/LK22 = 4EI/L
Stiffness Matrix [K] = K11 K12 = 4EI/L 2EI/L K21 K22 2EI/L 4EI/L[ [
RELATION BETWEEN STIFFNESS AND FLEXIBILITY MATRIX
[Δ] = [F][P] [Disp Matrix] = [Flexibility Matrix] x [Load Matrix] ----- > 1[K] [Δ] = [P] [Stiffness Matrix] x[Disp. Matrix] = [Load` Matrix] ---------- >2
Eq 1 x [F] -1 [F] -1 [Δ] = [F] -1 [F][P] [F] -1 [Δ] = [P]
So , [F] -1 = [K]