Truss Lecture (1) - Mechanics of Materials

8/10/2019 Truss Lecture (1) - Mechanics of Materials

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Finite Elements Analysis of Trusses


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Trusses Review of Statics

2

DEFINITION OF A TRUSS

A truss is an assembly of bars (in 2D or 3D) with the following properties:

All members are connected by pin joints (no moment transfer) No member is continuous through a node

The external forces are only applied to the nodes (never on the members) The external constraints are only applied at the nodes

No moment is applied to the structure

FUNDAMENTAL CONSEQUENCE

Each member in a truss is a two- force member

(we already know the direction of all internal forces)


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Externally determinate trusses

3

QUESTION 1 Can we calculate all the external forces using equilibrium alone?

3 unknowns - 3 equationsYES

4 unknowns - 3 equationsNO

The answer depends on the nature of the external constraints (exactly the sameas for a continuous rigid body)


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Statically determinate trusses

4

QUESTION 2 Can we calculate all the internal forces using equilibrium alone?

A TRUSS is STATICALLY DETERMINATE if # of equations = # of unknowns

# of equations = # of equilibrium conditions we can impose

= (# of nodes) X (# of equations per node)

= 2n (in 2 dimensions)

# of unknowns = (# of internal forces) + (# of external reactions)

= m+3 (in 2 dimensions)

n = number of nodes

m = number of membersMAXWELL

S RULE

If a truss is statically determinate, then 2n= m+3 (in 2D)

# of equations = # of unknowns 2n = m+3 (in 2D)


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5

Maxwell

s rule for static determinacy

2n = 2 X 4 = 8

m + 3 = 5 + 3 = 8

MAXWELL

S RULE

If a truss is statically determinate, then 2n=m+3 (in 2D)

2n = m+3

2n = 2 X 4 = 8

m + 3 = 6 + 3 = 9

2n < m+3


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Quiz #1

6

Is the following truss statically determinate?

(A) Yes (B) No


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Statically determinate trussesSolution by method of joints

7

Dismember the truss and create a freebody

diagram for each member and pin.

The two forces exerted on each member areequal, have the same line of action, andopposite sense.

Forces exerted by a member on the pins or joints at its ends are directed along the memberand equal and opposite.

Conditions of equilibrium on the pins provide 2 n equations for 2 n unknowns. For a simple truss,2n = m + 3. May solve for m member forcesand 3 reaction forces at the supports.

Conditions for equilibrium for the entire trussprovide 3 additional equations which are not

independent of the pin equations.


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Statically determinate trussesSolutions by the method of sections

When the force in only one member or the

forces in a very few members are desired, themethod of sections works well.

To determine the force in member BD, pass asection through the truss as shown andcreate a free body diagram for the left side.

With only three members cut by the section,the equations for static equilibrium may beapplied to determine the unknown memberforces, including F BD.


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How can we solve statically indeterminate trusses?

9

OUTLINE OF THE SOLUTION PROCEDURE FOR THE STIFFNESS METHOD

1. Analyze the deformation of one element in a local 1D coordinate system. Deriveits stiffness.

2. Express the stiffness of each element in a common (global) coordinate system.

3. Assemble the stiffness of each element to derive the overall stiffness of the truss(i.e., the relationship between applied external loads and nodal displacements).

4. Account for boundary conditions and apply loads.

5. By linear algebra procedures, calculate the displacement of each node.

6. Solve for the reaction forces at the nodes where displacement BCs are specified.

7. Solve for internal forces (and hence stresses) in each element.

This method shares many common principles with Finite Elements Analysis; the extension fromtrusses to continuum solids only require an alternative definition of the element stiffness.


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Stiffness of a single truss element in local CS

10

= PL EA

F 2 ' = EA L

q2 ' q1 '( )

F 1 ' = F 2 ' = EA L

q1 ' q2 '( )

x = local element-based coordinate system

We can combine these two equations in:

F 1 '

F 2 '

=

EA L

1 1 1 1

q1 '

q2 '

F '{ }= k ' q '{ }

or alternatively:

k ' = EA L

1 1 1 1

Stiffness matrix for a truss member in local coordinates


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Stiffness of a single truss element in global CS

11

Nodal Displacements

q1' = q

1cos + q

2sin

q2 ' = q3 cos + q4 sin

q1 '

q2 '

=

cos sin 0 00 0 cos sin

q1q2q3

q4

Transformation matrix [L]

x

= local element-based coordinate system(X,Y) = global truss-based coordinate system

Nodal Forces

F 1 = F 1 'cos

F 2 = F 1 'sin

F 3 = F 2 'cos

F 4 = F 2 'sin

F 1

F 2 F 3 F 4

=

cos 0sin 0

0 cos 0 sin

F 1 ' F 2 '

[L]T = Transpose of [L]

F 1 '

F 2 '

= EA L

1 1

1 1

q1 '

q2 '

X

Y


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12

F 1 F 2 F 3 F 4

=

cos 0sin 0

0 cos 0 sin

EA L

1 1 1 1

cos

sin 0 00 0 cos sin

q1q2q3q4

I am now ready to express the stiffness of the truss element in global coordinates

q1 '

q2 '

=

cos sin 0 00 0 cos sin

q1q2q3q4

F 1 F 2 F 3 F 4

=

cos 0sin 0

0 cos 0 sin

F 1 '

F 2 '

F 1 '

F 2 '

=

EA L

1 1 1 1

q1 '

q2 '

Stiffness matrix of truss element in global CS, [k]

k = L T

k ' L

Force-displacement relationship for a truss element in global coordinate system in 2D: F { }= k q{ }


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13

By performing the matrix multiplication, I finally find an explicit expression for [k]

k = EA L

cos 2 sin cos cos 2 sin cos

sin cos sin 2 sin cos sin 2

cos 2 sin cos cos 2 sin cos

sin cos sin2

sin cos sin2

Geometric formulas for the calculation of sines and cosines as a function of the nodal coordinates


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Example Analysis of a 1-element truss

14

This single element truss has four globaldegrees of freedom: Q 1, Q 2, Q 3, Q 4.

Because there is only one element, theglobal degrees of freedom correspond to thedegrees of freedom of element e = 1, that is:

Q1 = q 1, Q 2 = q 2, Q 3 = q 3, Q 4 = q 4

I want to calculate:(i) The nodal displacements Q

i

(ii) The reaction forces at the constraints.

K =

k 1

= EA L

1/ 2 1/ 2 1/ 2 1/ 21/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 1/ 2

= EA2 L

1 1 1 11 1 1 1 1 1 1 1

1 1 1 1

The stiffness matrix for the truss is: The displacement boundary conditions are:

Q1 = Q2 = Q4 = 0

The load boundary conditions are:

F 3 = F


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EA2 L

1 1 1 11 1 1 1 1 1 1 1 1 1 1 1

00

Q30

=

R1 R2 F R4

The governing equation {F}=[K]{Q} then becomes:

where R i are the unknown reaction forces and Q 3 is the unknown displacement.

This easy problem can be conveniently solved in two steps:

(1) Eliminate the rows and columns of [K] corresponding to the degrees of freedom for which thedisplacement is prescribed (1,2,4) and solve for the remaining displacements.

EA2 L

1 1 1 11 1 1 1 1 1 1 1 1 1 1 1

00

Q30

=

R1 R2 F R4

EA2 L

Q3 = F Q3 =2 LF EA


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EA2 L

1 1 1 11 1 1 1 1 1 1 1 1 1 1 1

00

Q30

=

R1 R2 F R4

(2) Now that {Q} is fully known, solve for theunknown reaction forces using the rows that wehad previously eliminated.

Row 1:

Row 2:

Row3:

EA2 L

Q3 = R1 R1 = F

EA2 L

Q3 = R2 R2 = F

EA2 L

Q3 = R4 R4 = F

This completes the solution of the problem!


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Assembly procedure and analysis of arbitrary trusses

17

a = 1 m

E = 200 GPa (steel)

A = 2 cm X 2 cm = 0.0004 m 2 for all 5 members

F = 10 kN

How do extend the previous analysis to multi-member trusses?

Let

s see this through an example.

Sequence of steps

1. Carefully number nodes, elements (members) andglobal degrees of freedom

2. Derive the stiffness matrix for each member3. Assemble the global stiffness matrix4. Prescribe boundary conditions (loads and

displacements)5. Write down the complete system of equations6. Remove rows and columns corresponding to DOFs for

which the displacement is specified7. Solve the matrix problem for the unknown noda l

d i sp lacements8. By using the rows that were previously eliminated,

solve for the unknown r eac t i on f o r ce s9. Calculate the in te rna l force (and s t ress ) in each

m em ber

Note: the procedure is the same whether the truss is staticallydeterminate or indeterminate (either internally ox externally)


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STEP 1 Number nodes, elements and global degrees of freedom

Q i = global degrees of freedom

Q{ } =

Q1Q2Q3Q4Q5Q6Q7Q8

x and y displacement at node 1




Connectivity matrix

Element, e Node 1 Node 2

1 1 2

2 2 3

3 3 4

4 4 1

5 1 3


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STEP 2 Derive the stiffness matrix for each member

k 1

= EA

a

1 0 1 00 0 0 0 1 0 1 00 0 0 0

k 2

= EA

a

0 0 0 00 1 0 10 0 0 00 1 0 1

k 3

= EA

a

1 0 1 00 0 0 0 1 0 1 00 0 0 0

k 4

= EA

a

0 0 0 00 1 0 10 0 0 00 1 0 1

k 3

= EA

2 2 a

1 1 1 11 1 1 1 1 1 1 1 1 1 1 1


1 1 2

2 2 3

3 3 4

4 4 1

5 1 3

= 0o

= 90o

= 180o

= 270o

= 45o


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20

STEP 3 Assemble the global stiffness matrix

k 1

= EA

a

1 0 1 00 0 0 0 1 0 1 00 0 0 0


1 1 2

1 2 3 4

1234 Global DOFs

EA/a 0 -EA/a 0

0 0 0 0

-EA/a 0 EA/a 0

0 0 0 0

1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8


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21



2 2 3

3 4 5 6

3456 Global DOFs

0 0 0 0

0 EA/a 0 -EA/a

0 0 0 0

0 -EA/a 0 EA/a

1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

k 2

= EA

a

0 0 0 00 1 0 10 0 0 00 1 0 1


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22



3 3 4

5 6 7 8

5678 Global DOFs

EA/a 0 -EA/a 0

0 0 0 0

-EA/a 0 EA/a 0

0 0 0 0

1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

k 3

= EA

a

1 0 1 00 0 0 0 1 0 1 00 0 0 0


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23



4 4 1

7 8 1 2

7812 Global DOFs

0 0 0 0

0 EA/a 0 -EA/a

0 0 0 0

0 -EA/a 0 EA/a

1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

k 4

= EA

a

0 0 0 00 1 0 10 0 0 00 1 0 1


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24



5 1 3

1 2 5 6

1256

Global DOFs

EA/2V2a

EA/2V2a

-EA/2V2a

-EA/2V2a

EA/2V2a

EA/2V2a

-EA/2V2a

-EA/2V2a

-

EA/2V2a

-

EA/2V2a

EA/2V2

a

EA/2V2

a

-EA/2V2a

-EA/2V2a

EA/2V2a

EA/2V2a

1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

k 3

= EA

2 2 a

1 1 1 11 1 1 1 1 1 1 1 1 1 1 1


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1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

The global stiffness matrix for the entire truss is then:

K = EA

a

1 + 1/ 2 2 1/ 2 2 1 1/ 2 2 1/ 2 2

1/ 2 2 1 + 1/ 2 2 1/ 2 2 1/ 2 2 1

1 1

1 1

1/ 2 2 1/ 2 2 1 + 1/ 2 2 1/ 2 2 1

1/ 2 2 1/ 2 2 1 1/ 2 2 1 + 1/ 2 2

1 1

1 1


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108 28 -80 0 -28 -28 0 0

28 108 0 0 -28 -28 0 -80

-80 0 80 0 0 0 0 0

0 0 0 80 0 -80 0 0

-28 -28 0 0 108 28 -80 0

-28 -28 0 -80 28 108 0 0

0 0 0 0 -80 0 80 0

0 -80 0 0 0 0 0 80

By plugging in the numerical values, we get:

K = 106 N/m


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STEP 4 Prescribe boundary conditions

Displacement boundary conditions

Q1 = Q2 = Q4 = 0

Load boundary conditions

F 6 = F

F 7 = F


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28

STEP 5 Write down the system of equations

10 6

108 28 80 0 28 28 0 028 108 0 0 28 28 0 80

80 0 80 0 0 0 0 00 0 0 80 0 80 0 0

28 28 0 0 108 28 80 0

28 28 0 80 28 108 0 00 0 0 0 80 0 80 00 80 0 0 0 0 0 80

00

Q30

Q5

Q6Q7Q8

=

R1 R20

R40

F F 0

where R 1, R 2, R 4 are the unknown reaction forces.


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29

STEP 6 Remove rows and columns corresponding to the degrees of freedomfor which the displacement is specified

10 6

108 28 80 0 28 28 0 028 108 0 0 28 28 0 80

80 0 80 0 0 0 0 00 0 0 80 0 80 0 0

28 28 0 0 108 28 80 0

28 28 0 80 28 108 0 00 0 0 0 80 0 80 00 80 0 0 0 0 0 80

00

Q30

Q5

Q6Q7Q8

=

R1 R20

R40

F F 0

10 6

80 0 0 0 00 108 28 80 00 28 108 0 00 80 0 80 00 0 0 0 80

Q3Q5Q6Q7

Q8

=

00

10,00010,000

0


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30

STEP 7 Solve the matrix problem for {Q}

10 6

80 0 0 0 00 108 28 80 00 28 108 0 00 80 0 80 00 0 0 0 80

Q3Q5Q6Q7

Q8

=

00

10,00010,000

0

Q3 = 0 mm

Q5 = 0.6 mm

Q6 = 0.25 mm

Q7 = 0.73 mm

Q8

= 0 mm

At this point, we know the displacement of every node in the truss.


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STEP 8 By using the rows I had eliminated at step 6, solve for the reaction forces

10 6

108 28 80 0 28 28 0 028 108 0 0 28 28 0 80

80 0 80 0 0 0 0 00 0 0 80 0 80 0 0

28 28 0 0 108 28 80 0

28 28 0 80 28 108 0 00 0 0 0 80 0 80 00 80 0 0 0 0 0 80

0000

0.0006

0.000250.00073

0

=

R1 R20

R40

10,00010,000

0

from which I finally get:

R1

= 10 kN R2 = 10 kN R4 = 20 kN


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33

STEP 9 Calculate the internal force (and stress) in each member

This procedure requires careful bookkeeping, as we need totransition from local to global reference systems.

Example: Calculate the internal force in element #5.


1 1 2

2 2 3

3 3 4

4 4 1

5 1 3

q1q2q3q4

=

Q1Q2Q5Q6

=

00

0.00061

0.00025

m

from the solution obtained at step 7

Local Global

5 = 45o cos = sin = 2 / 2

s 5 = 200 109

2 1/ 2 1/ 2 1/ 2 1/ 2

00

0.00061 0.00025

= 36 10 6 Pa = 36 MPa

T 5 = A5s 5 = 0.0004 36 106 N = 14.4 kN

Similarly, we obtain:s 1 = 0 MPas 2 = 50 MPas 3 = 25 MPa

s 4 = 0 MPa

T 1 = 0 kN

T 2 = 20 kN

T 3 = 10 kN

T 4

= 0 kN


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34

MATLAB Program for Truss Analysis

Program TRUSS2D

Developed by Chadrupatla and Belegundu

Included in:Chandrupatla and Belegundu, Introduction to Finite Elements in Engineering , Prentice Hall, 2002

Slightly modified by L Valdevit


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Step 1 - Input file and numbering schemeNext line is problem title >TRUSS EXAMPLENN NE NM NDIM NEN NDN4 5 1 2 2 2

ND NL3 2

Node# X Y1 0 02 1 03 1 14 0 1

Elem# N1 N2 Mat# Area1 1 2 1 0.00042 2 3 1 0.0004

3 3 4 1 0.00044 4 1 1 0.00045 1 3 1 0.0004

DOF# Displacement1 02 04 0

DOF# Load6 -100007 10000

MAT# E1 200.E9

SCALING FACTOR FOR PLOTTING100.

-------- LEGEND --------------NN = Number of nodesNE = Number of truss elementsNM = Number of different materialsNDIM = Number of coordinates per node (2 fr 2D, 3 for 3D)NEN = Number of nodes per element (2 for a truss)NDN = Number of degrees of freedom per node (truss: 2 for 2D, 3 for 3D)

ND = # of DOFs along which displacement is specifiedNL = # of DOFs along which load is applied----------------------------------

F i I D


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Function InputData Reads the input file and stores all values

function [] = InputData()global NN NE NM NDIM NEN NDNglobal ND NL NCH NPR NBWglobal X NOC F AREA MAT S

global PM NU U STRESS REACTglobal CNSTglobal TITLE FILE1 FILE2global LINP LOUTglobal NQglobal SF

FILE1 = input(' Input Data File Name ', 's');LINP = fopen(FILE1,' r');FILE2 = input(' Output Data File Name ', 's') ;LOUT = fopen(FILE2,' w') ;

DUMMY = fgets(LINP);TITLE = fgets(LINP);DUMMY = fgets(LINP);TMP = str2num(fgets(LINP));

[NN, NE, NM, NDIM, NEN, NDN] =deal(TMP(1),TMP(2),TMP(3),TMP(4),TMP(5),TMP(6));

NQ = NDN * NN;

DUMMY = fgets(LINP);TMP = str2num(fgets(LINP));[ND, NL]= deal(TMP(1),TMP(2));

NPR=1; % E (Y oung's modulus)

%--- -- Coordinates -----DUMMY = fgets(LINP);for I=1 :NN

TMP = str2num(fgets(LINP));

[N, X(N,:)]=deal(TMP(1),TMP(2:1+NDIM));end

%-- --- Connectivity -----DUMMY = fgets(LINP);for I=1 :NE

TMP = str2num(fgets(LINP));[N,NOC(N,:), MAT(N,:), AREA(N,:)] = ...

deal(TMP(1),TMP(2:1+NEN), TMP(2+NEN), TMP(3+NEN));end

%----- S pecified Displacements -----DUMMY = fgets(LINP);for I=1: ND

TMP = str2num(fgets(LINP));[NU(I,:),U(I,:)] = deal(TMP(1), TMP(2));

end

%--- -- Component Loads -----DUMMY = fgets(LINP);F = zeros(NQ,1);for I=1: NL

TMP = str2num(fgets(LINP));[N,F(N)]=deal(TMP(1),TMP(2));

end

%-- --- Material Properties -----DUMMY = fgets(LINP);for I=1:N M

TMP = str2num(fgets(LINP));[N, PM(N,:)] = deal(TMP(1), TMP(2:NPR +1));

end

%--- -- Scaling factor for plotting -----DUMMY = f gets(LINP);

SF = str2num(fgets(LINP));

fclose(LINP);


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InputData

Nodal coordinate matrix

Connectivity matrix

X =

0 0

1 01 1

0 1

Line = Node #

x y

Line = Element # NOC =

1 2

2 3

3 4

4

1

1

3

Local

Node 1

Local

Node 2

MAT =

1

1

1

1

1

AREA =

0.0004

0.0004

0.0004

0.0004

0.0004

Material

PM (1)=

200 E 9 Line = Element #

Cross-section


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InputData

Displacement boundary conditions

Applied Loads

Value of the displacementof dof specified in NU

NU =

1

24

Degree of freedom with displacement b.c.

U =0

0

0

F =

0

0

0

0

0 1000010000

0

Line = Degree of F reedom


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F ti Stiff


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40

Function Stiffness Assemble the global stiffness matrix

Example for element #3 (N=3)

I1 = 3 I2 = 4

Node 1 Node 2

SE =

80 E 6 0 80 E 6 00 0 0 0

80 E 6 0 80 E 6 00 0 0 0

NOC =

1 2

2 3

3 44

1

1

3

globX1=NOC(N,1)*2-1;globY1=NOC(N,1)*2;globX2=NOC(N,2)*2-1;globY2=NOC(N,2)*2;

globX1 = 5globY1 = 6

globX2 = 7globY2 = 8S =

80 E 6 0 80 E 6 0 0 0 0 0 80 E 6 0 80 E 6 0 0 0 0 0

S(globX1,globX1) = S(globX1,globX1) + SE(1,1);

S(globX1,globY1) = S(globX1,globY1) + SE(1,2);

S(globX1,globX2) = S(globX1,globX2) + SE(1,3);S(globX1,globY2) = S(globX1,globY2) + SE(1,4);S(globY1,globY1) = S(globY1,globY1) + SE(2,2);S(globY1,globX2) = S(globY1,globX2) + SE(2,3);S(globY1,globY2) = S(globY1,globY2) + SE(2,4);S(globX2,globX2) = S(globX2,globX2) + SE(3,3);S(globX2,globY2) = S(globX2,globY2) + SE(3,4);S(globY2,globY2) = S(globY2,globY2) + SE(4,4);

and the rest follows from the symmetry of SE;

Function ModifyForBC


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41

Function ModifyForBCImpose displacement BCs with the penalty method

Removing lines and columns is easy to do by hand, but rather cumbersome computationally.Changing a matrix (or a vector) dimension within a program is never an advisable operation.The Penalty Method is a very convenient alternative that provides an approximate solution

Suppose we want to impose Q 4 = a 4 (e.g. 0).If you add a large number C to S 44 and Ca 4 toF4, the governing equation becomes:

S 11 S 18

S 41 S 42 S 43 S 44 + C S 45 S 46 S 47 S 48

S 81 S 88

Q1

Q4

Q8

=

F 1

F 4 + Ca 4

F 8

The fourth row yields the equation:

S 41Q1 + S 42Q2 + S 43Q3 + (S 44 + C )Q4 + S 45Q5 + S 46Q6 + S 47Q7 + S 48Q8 = F 4 + Ca 4

Divide by C:

S 41C Q1 +

S 42C Q2 +

S 43C Q3 + (

S 44C + 1)Q4 +

S 45C Q5 +

S 46C Q6 +

S 47C Q7 +

S 48C Q8 =

F 4C + a 4

If C is much larger than any S ij and F j, this is approximately equivalent to: Q4 = a 4which is the BC we want to impose.

Hence all I need to do to enforce a displacement BC to dof N is add to the diagonal stiffness term S NN alarge number C; if the prescribed displacement is different from 0, I will also need to add the term C.a N

to the force term F N. Typically C is chosen as 10,000 times the largest value of S ij

Function ModifyForBC


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Function ModifyForBCImpose displacement BCs with the penalty method

This is trivially implemented in the MATLAB code

%----- Decide Penalty Parameter CNST -----

CNST = max(max(S))*10000;

%-- --- Modify for Boundary Conditions -----% --- Displacement BC ---

for I = 1:NDN = NU(I);

S(N,N) = S(N,N) + CNST;

F(N) = F(N) + CNST * U(I);end


NU =1

24


U =0

0

0


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Function StressCalc


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Function StressCalcCalculate the stress in each member

for I = 1:NEI1 = NOC(I, 1);I2 = NOC(I, 2);I3 = MAT(I);X21 = X(I2, 1) - X(I1, 1);Y21 = X(I2, 2) - X(I1, 2);EL = sqrt(X21 * X21 + Y21 * Y21);CS = X21 / EL;SN = Y21 / EL;J2 = 2 * I1;J1 = J2 - 1;K2 = 2 * I2;K1 = K2 - 1;DLT = (F(K1) - F(J1)) * CS +

+ (F(K2) - F(J2)) * SN;STRESS(I) = PM(I3, 1) * (DLT / EL);

end

Example for element #3 (I=3)

I1 = 3 I2 = 4 I3 = 1

Node 1 Node 2 Material

X21 = x(4) - x(3) = 0-1 = -1Y21 = y(4) - y(3) = 1-1 = 0

EL = Sqrt((-1)2

+(0)2

) = 1

EAL = 200E9 * 0.0004 / 1 = 80E6

Sine and Cosine of the angle CS = X21 / EL = -1/1 = -1SN = Y21 / EL = 0/1 = 0

X =

0 0

1 0

1 10 1

NOC =

1 2

2 3

3 4

4

1

1

3

Element length

EA/L

J2 = 6 J1 = 5 K2 = 8 K1 = 7Global DOFs corresponding to element #3

Elongation of element #3DLT = (Q 7-Q 5)*CS+ (Q 8-Q 6)*SNStress in element #3

STRESS(3) = E*DTL/EL = -25 MPa (compression)

Function ReactionCalc


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Function ReactionCalcCalculate the external reaction forces

%----- Reaction Calculation ----- for I = 1:ND

N = NU(I);REACT(I) = CNST * (U(I) - F(N));end


NU =1

2

4


U =0

0

0

The reaction forces follow simply from the Penalty Method.

Let

s look for R 4 = F 4 S 11 S 18

S 41 S 42 S 43 S 44 + C S 45 S 46 S 47 S 48

S 81 S 88

Q1

Q4

Q8

=

F 1

F 4 + Ca 4

F 8

S 41Q1 + S 42Q2 + S 43Q3 + (S 44 + C )Q4 + S 45Q5 + S 46Q6 + S 47Q7 + S 48Q8 = F 4 + Ca 4

F 4 = Ca 4 S 41Q1 + S 42Q2 + S 43Q3 + (S 44 + C )Q4 + S 45Q5 + S 46Q6 + S 47Q7 + S 48Q8( )=

Ca 4

CQ 4 =

C a 4

Q4( )

The fourth equation is:

which can be solved for F 4:


46/46

Functions Output and Draw

Output for Input Data from file TRUSS2D_lv.INP

TRUSS EXAMPLE

Node# X-Displ Y-Displ1 9.2350E-09 9.2350E-092 9.2350E-09 -1.8470E-083 6.0359E-04 -2.5002E-044 7.2859E-04 9.2350E-09

Elem# Stress1 1.3235E-12

2 -5.0000E+073 -2.5000E+074 -0.0000E+005 3.5355E+07

DOF# Reaction1 -1.0000E+042 -1.0000E+044 2.0000E+04

Truss Lecture (1) - Mechanics of Materials

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