Chapter 3 - Global Analysis Equations

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Formation of the Global Analysis Equations 1 Prepared by : Oscar Victor M. Antonio, Jr., D. Eng.

description

lecture on matrix structural analysis using global coordinates equations

Transcript of Chapter 3 - Global Analysis Equations

Page 1: Chapter 3 - Global Analysis Equations

Formation of the Global

Analysis Equations

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Prepared by : Oscar Victor M. Antonio, Jr., D. Eng.

Page 2: Chapter 3 - Global Analysis Equations

Force-displacement relationship

{ } [ ]{ }ΔkF =

Introduction

Element forces

Element stiffness matrix

Element displacement

Eternally applied forces

Global stiffness matrix

Global degrees of freedom

Chapter 2

Chapter 3

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Direct Stiffness Method

Complete set of force-displacement relationships for a framework element with n degrees of freedom (global axes)

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Direct Stiffness Method - The Basic Equations O.V.M.Antonio,Jr.

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- number 1, i, n identify the DOF at the joints of the element, and correspond to a global numbering system

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- when the global analysis equations are formed using the direct stiffness method, all the DOF appear in each of the rows

- once the element force-displacement relationships have been numerically evaluated for all members of the structure, application of the direct stiffness method consists of their combination in an algebraic form that satisfies the requirements of static equilibrium & joint compatibility at all the junction points

Direct Stiffness Method - The Basic Equations O.V.M.Antonio,Jr.

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Consider the formation of the force-displacement equations of the point q in the global x direction

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Direct Stiffness Method - The Basic Equations O.V.M.Antonio,Jr.

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For junction point equilibrium, the applied load must be equal to the sum of the internal forces acting on the bars meeting at the point

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x-direction internal force components

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F à internal force

P à external force

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The force-displacement equations for the elements yield expressions for Fi

A … FiD in terms of the corresponding

element DOF ΔiA … Δ9

D

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Direct Stiffness Method - The Basic Equations

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Satisfying compatibility condition: ΔiA = Δi

B = ΔiC = Δi

D = Δi

Kii , Ki1 , Ki2 , Ki3 ,…, Ki9 are global stiffness coefficients

If the subscripts of the coefficients of two or more different elements are identical (common DOF), such coefficients are added to form one coefficient of the stiffness

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When an additional member (H) is added into the structure, a force Fi

H is added and is supplemented by stiffness terms Kii

H , Ki2H , Ki10

H , Ki11H

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•  joint equilibrium •  joint displacement compatibility

important to satisfy

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Example 3.1 For the system shown :

1. Write the member force-displacement relationships in global coordinates

2. Assemble the global stiffness equations

E = 200,000 MPa

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Example 3.1

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Define the coordinates, degrees of freedom, and external forces:

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Example 3.1

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Recall: Stiffness Equations

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

−−

−−

−−

−−

=

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

2

2

1

1

22

22

22

22

4

3

2

1

vuvu

sincossinsincossincossincoscossincos

sincossinsincossincossincoscossincos

LEA

FFFF

φφφφφφ

φφφφφφ

φφφφφφ

φφφφφφ

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Example 3.1

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Consider member ab

Member force-displacement relationship

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Example 3.1

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Consider member bc

Member force-displacement relationship

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Example 3.1

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Consider member ac

Member force-displacement relationship

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Example 3.1

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Global stiffness equations in matrix form

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Example 3.2 The truss of Example 3.1 is supported and loaded as shown.

1. Calculate the displacements at a and b

2. Calculate the reactions

3. Calculate the bar forces

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Example 3.2

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Define the coordinates, degrees of freedom, and external forces:

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Example 3.2

Direct Stiffness Method - The Basic Equations

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Example 3.2

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Example 3.2

Direct Stiffness Method - The Basic Equations

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Example 3.2

Direct Stiffness Method - The Basic Equations

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Example 3.3 The truss shown is the same as in Example 3.2 except for the addition of horizontal tie ad.

1. Calculate the displacements at a and b

2. Calculate the reactions

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Example 3.3

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Example 3.3

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Example 3.3

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Direct Stiffness Method - The General Procedure

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Automatic approach in calculating the applied load versus displacement equations for the complete structure

1. Each element stiffness coefficient is assigned a double subscript (kij). i à force & j à DOF

2. Provision is made for a square matrix whose size is equal to the number of DOF in the complete system, with a possibility that each force will be related to every displacement in the system.

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3. The procedure of step 2 is continued until all elements have been reached. Each time a coefficient is placed in a location where a value has already been placed, it is added to the latter until the final value is achieved.

4. The process of steps 2 and 3 is repeated for all other rows in order (assembly process). The result is a complete set of coefficients of the stiffness equations for the entire structure (global stiffness equation).

K1i = Σk1i

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5. The support conditions are accounted for by noting which displacements are zero and then removing from the equations the columns of stiffness coefficients multiplying these DOFs.

0 0 0

0 321.4 383.0

Ryb

Ryc

Rxc

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Solve for the unknown DOFs after removing columns 4, 5, and 6

Solve for the unknown support reactions

6. The internal forces acting on the ends of the elements are determined by back substitution of the solved DOFs in the element force-displacement equations.

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Modified approach for doing steps 5 and 6

{ } [ ]{ }ΔKP =

Recall: global stiffness equation

Partition the equation by grouping quantities relating to support conditions and those relating to the remaining DOFs

conformable

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Expanding the partitioned equation and noting that {Δs} = 0

The general solution is obtained symbolically by

[D] : global flexibility matrix

[K] : global stiffness matrix

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After solving the vector of displacements at all unsupported nodes {Δf}, the support reactions {Ps} are then computed

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To obtain the internal force distribution in the ith element [Fi], the calculated DOFs {Δi} for that element is multiplied by the element stiffness matrix [ki]

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Example 3.4 For the system shown:

1. Write the force-displacement relationships in global coordinates

2. Assemble the global stiffness equations

3. Show that the stiffness equations contain rigid-body motion

E = 200,000 MPa

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Example 3.4

Define the coordinates, DOFs, and external forces

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Example 3.4

Member force-displacement relationships

Member ab

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Example 3.4

Member force-displacement relationships

Member cd

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Example 3.4

Member force-displacement relationships

Member ac

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Example 3.4

Member force-displacement relationships

Member bd

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Example 3.4

Member force-displacement relationships

Member bc

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Example 3.4

Global stiffness equations in matrix form

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Rigidity motion: If it can be shown that the stiffness matrix is singular by performing Matrix Row Operations, thus the displacements are indefinite à there may be rigid body motion.

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Example 3.4

Adding rows 1, 5, and 7 of the global stiffness matrix gives a vector

which is the negative of the original row 3, therefore the determinant is zero

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Example 3.5 The truss of Example 3.4 is supported and loaded as shown:

1. Calculate the displacements at b, c, and d

2. Calculate the reactions

3. Calculate the bar forces

E = 200,000 MPa

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Example 3.5

Boundary conditions: Δ1 = Δ2 = Δ4 = 0

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Example 3.5

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Example 3.5

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Example 3.5

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Example 3.5

recall

Substitute the last two equations

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Example 3.5

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Example 3.5

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Example 3.6 The truss of shown is the same as in Example 3.5 except for the addition of horizontal constraints at b and c. Calculate the displacements at c and d.

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Example 3.6

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Example 3.6

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Reference: Matrix Structural Analysis Second Edition by William McGuire, Richard H. Gallagher, and Ronald D. Ziemian

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