development of truss eqns

7/27/2019 Development of Truss Eqns

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Having set forth the foundation on which the direct stiffness

method is based, we will now derive the stiffness matrix fora linear-elastic bar (or truss) element using the generalsteps outlined in Chapter 1.

We will include the introduction of both a local coordinatesystem, chosen with the element in mind, and a global orreference coordinate system, chosen to be convenient (fornumerical purposes) with respect to the overall structure.

We will also discuss the transformation of a vector from the

local coordinate system to the global coordinate system,using the concept of transformation matrices to expressthe stiffness matrix of an arbitrarily oriented bar element interms of the global system.

Development of Truss Equations


Next we will describe how to handle inclined, or skewed,supports.

We will then extend the stiffness method to include spacetrusses.

We will develop the transformation matrix in three-dimensionalspace and analyze a space truss.

We will then use the principle of minimum potential energyand apply it to the bar element equations.

Finally, we will apply Galerkins residual method to derive thebar element equations.

CIVL 8/7117 Chapter 3 - Development of Truss Equations 1/77


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Stiffness Matrix for a Bar Element

Consider the derivation of the stiffness matrix for the linear-

elastic, constant cross-sectional area (prismatic) barelement show below.

This application is directly applicable to the solution of pin-connected truss problems.


Consider the derivation of the stiffness matrix for the linear-elastic, constant cross-sectional area (prismatic) barelement show below.

where T is the tensile force directed along the axis at

nodes 1 and 2, x is the local coord inate system directedalong the length of the bar.



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Consider the derivation of the stiffness matrix for the linear-

elastic, constant cross-sectional area (prismatic) barelement show below.

The bar element has a constant cross-sectionA, an initiallength L, and modulus of elasticity E.

The nodal degrees of freedom are the local axialdisplacements u1 and u2 at the ends of the bar.


The strain-displacement relationship is:du

Edx

From equilibrium of forces, assuming no distributed loadsacting on the bar, we get:

constantxA T

Combining the above equations gives:

constantdu

AE Tdx

Taking the derivative of the above equation with respect to thelocal coordinate x gives:

0d du

AEdx dx



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The following assumptions are considered in deriving the bar

element stiffness matrix:

1. The bar cannot sustain shear force: 1 2 0y yf f

2. Any effect of transverse displacement is ignored.

3. Hookes law applies; stress is related to strain:x xE

Step 1 - Select Element Type

We will consider the linear bar element shown below.




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Step 2 - Select a Displacement Function


1 2u a a x A linear displacement function u is assumed:

The number of coefficients in the displacement function, ai, isequal to the total number of degrees of freedom associatedwith the element.

Applying the boundary conditions and solving for the unknowncoefficients gives:

2 11

u uu x u

L

1

2

1ux x

u

uL L

Step 2 - Select a Displacement Function


Or in another form:

where N1 and N2 are the interpolation functions gives as:

11 22

uu N N

u

1 21x x

N NL L

The linear displacementfunction plotted over thelength of the bar element isshown below.

u



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Step 3 - Define the Strain/Displacement

and Stress/Strain Relationships


The stress-displacement relationship is:2 1

x

u udu

dx L

Step 4 - Derive the Element Stiffness Matrix and Equations

We can now derive the element stiffness matrix as follows:

xT A

Substituting the stress-displacement relationship into the

above equation gives:

2 1u uT AEL


The nodal force sign convention, defined in element figure, is:

Step 4 - Derive the Element Stiffness Matrix and Equations

therefore,

Writing the above equations in matrix form gives:

1 2x xf T f T

1 2 2 11 2x x

u u u uf AE f AE

L L

1 1

2 2

1 1

1 1x

x

f uAE

f uL

Notice thatAE/L for a bar element is analogous to the springconstant k for a spring element.



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The global stiffness matrix and the global force vectorareassembled using the nodal force equilibrium equations, andforce/deformation and compatibility equations.

Step 5 - Assemble the Element Equations

and Introduce Boundary Condit ions

( ) ( )1 1

n ne e

e e

K F

K k F f

Where k and fare the element stiffness and force matricesexpressed in global coordinates.


Solve the displacements by imposing the boundary conditionsand solving the following set of equations:

Step 6 - Solve for the Nodal Displacements

F Ku

Step 7 - Solve for the Element Forces

Once the displacements are found, the stress and strain in

each element may be calculated from:

2 1x x x

u uduE

dx L



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Consider the following three-bar system shown below. Assumefor elements 1 and 2:A = 1 in2 and E = 30 (106) psi and forelement 3:A = 2 in2 and E = 15 (106) psi.

Example 1 - Bar Problem

Determine: (a) the global stiffness matrix, (b) the displacementof nodes 2 and 3, and (c) the reactions at nodes 1 and 4.


For elements 1 and 2:


For element 3:

6(1) (2) 61 30 10 1 1 1 11030 1 1 1 1

lb lbin in

k k

1 2 node numbers for element 1


6(3) 62 15 10 1 1 1 11030 1 1 1 1

lb lbin in

k


As before, the numbers above the matrices indicate thedisplacements associated with the matrix.



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Assembling the global stiffness matrix by the direct stiffnessmethods gives:


Relating global nodal forces related to global nodaldisplacements gives:

6

1 1 0 0

1 2 1 010

0 1 2 1

0 0 1 1

K

1 1

2 26

3 3

4 4

1 1 0 0

1 2 1 010

0 1 2 1

0 0 1 1

x

x

x

x

F u

F u

F u

F u

E1 E 2 E 3


The boundary conditions are:

Example 1 Bar Problem

1 4 0u u

1 1

2 26

3 3

4 4

1 1 0 0

1 2 1 010

0 1 2 1

0 0 1 1

x

x

x

x

F u

F u

F u

F u

Applying the boundary conditions and the known forces

(F2x = 3000 lb.) gives:

26

3

3000 2 110

0 1 2

u

u



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Example 1 Bar Problem

Solving foru2 and u3 gives:

2

3

0.002

0.001

u in

u in

The global nodal forces are calculated as:

1

2 6

3

4

1 1 0 0 0 2000

1 2 1 0 0.002 300010

0 1 2 1 0.001 0

0 0 1 1 0 1000

x

x

x

x

F

Flbs

F

F


Consider the following guidelines, as they relate to the one-dimensional bar element, when selecting a displacementfunction.

Selecting Approximation Functions for Displacements

1. Common approximation functions are usuallypolynomials.

2. The approximation function should be continuous withinthe bar element.



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3. The approximating function should provide interelementcontinuity for all degrees of freedom at each node fordiscrete line elements, and along common boundary linesand surfaces for two- and three-dimensional elements.




For the bar element, we must ensure that nodes common totwo or more elements remain common to these elementsupon deformation and thus prevent overlaps or voidsbetween elements.

The linear function is then called a conforming (orcompatible) function for the bar element because itensures both the satisfaction of continuity between adjacentelements and of continuity within the element.



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4. The approximation function should allow for rigid-bodydisplacement and for a state of constant strain within theelement.

Completeness of a function is necessary forconvergence to the exact answer, for instance, for

displacements and stresses.


The interpolation function must allow for a rigid-bodydisplacement, that means the function must be capable ofyielding a constant value.

Consider the follow situation:


Therefore:

1 1 1 2u a a u u

1 1 2 2 1 2 1u N u N u N N a

Since u = a1 then:

This means that: 1 2 1N N

1 1 2 1u a N N a

The displacement interpolation function must add to unity atevery point within the element so the it will yield a constantvalue when a rigid-body displacement occurs.



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In many problems it is convenient to introduce both local andglobal (or reference) coordinates.

Local coordinates are always chosen to convenientlyrepresent the individual element.

Global coordinates are chosen to be convenient for the wholestructure.

Transformation of Vectors in Two Dimensions


Given the nodal displacement of an element, represented bythe vectord in the figure below, we want to relate thecomponents of this vector in one coordinate system tocomponents in another.




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Lets consider that d does not coincident with either the localor global axes. In this case, we want to relate globaldisplacement components to local ones. In so doing, we willdevelop a transformation matrix that will subsequently beused to develop the global stiffness matrix for a bar element.



We define the angle to be positive when measuredcounterclockwise from x to x. We can express vectordisplacement d in both global and local coordinates by:


1 1 1 1u v u v d i j i j



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Consider the following diagram:


Using vector addition: a b i

Using the law of cosines, we get: | | | | cos | | cos a i a

Similarly: | | | | sin | | sin b i b




The vectora is in the direction and b is in the direction,therefore:

i j

| | cos | | sin a a i i b b j j



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The vectori can be rewritten as: cos sin i i j

The vectorj can be rewritten as: sin cos i j

Therefore, the displacement vector is:

1 1 1 1 1 1cos sin sin cosu v u v u v i j i j i j i j




Combining like coefficients of the local unit vectors gives:

1 1 1cos sinu v u

1 1 1sin cosu v v

1 1

1 1

cos

sin

u uC S C

v vS C S



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The matrix is called the transformation matrix.

The previous equation relates the global displacement d tothe d local displacements

C S

S C

The figure below shows u expressed in terms of the globalcoordinates x and y.

u Cu Sv


Example 2 - Bar Element Problem

Using the following expression we just derived, we get:

The global nodal displacement at node 2 is u2 = 0.1 in andv2 = 0.2 in for the bar element shown below. Determine thelocal displacement.

2 cos60 (0.1) sin60 (0.2) 0.223o ou in

u Cu Sv



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Global Stiffness Matrix

We will now use the transformation relationship developedabove to obtain the global stiffness matrix for a bar element.



We want to relate the global element forces fto the globaldisplacements d for a bar element with an arbitraryorientation.

We known that for a bar element in local coordinates we have:

1 1

2 2

1 1

1 1

x

x

f uAE

f uL

f k d

1 1

1 1

2 2

2 2

x

y

x

y

f uf v

kf u

f v

f = kd



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Combining both expressions for the two local degrees-of-freedom, in matrix form, we get:

Using the relationship between local and global components,we can develop the global stiffness matrix.

We already know the transformation relationships:

1 1 1 2 2 2cos sin cos sinu u v u u v

1

1 1

2 2

2

0 00 0

u

u vC Su uC S

v

*d = T d

* 0 0

0 0

C S

C S

T



Substituting the global force expression into element force

equation gives:

A similar expression for the force transformation can bedeveloped.

1

11 *

22

2

0 0

0 0

x

yx

xx

y

f

ff C S

ff C S

f

f T f

f = k d

* *T f k T d

*T f k d

*d = T d

Substituting the transformation between local and globaldisplacements gives:



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The matrix T* is not a square matrix so we cannot invert it.

Lets expand the relationship between local and globaldisplacement.

where T is:

1 1

1 1

2 2

2 2

0 0

0 0

0 0

0 0

u uC S

v vS C

u uC S

v vS C

d = Td

0 0

0 0

0 0

0 0

C S

S C

C S

S C

T



We can write a similar expression for the relationship betweenlocal and global forces.

Therefore our original local coordinate force-displacementexpression

1 1

1 1

2 2

2 2

0 0

0 0

0 0

0 0

x x

y y

x x

y y

f fC S

f fS C

f fC S

f fS C

f = Tf

1 1

2 2

1 1

1 1x

x

f uAE

f uL

f = k d



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May be expanded:

The global force-displacement equations are:

1 1

1 1

2 2

2 2

1 0 1 0

0 0 0 0

1 0 1 0

0 0 0 0

x

y

x

y

f u

f vAE

f uL

f v

f k d Tf k Td

Multiply both side by T -1 we get: -1f T k Td

where T-1

is the inverse ofT. It can be shown that:

1 T

T T



The global force-displacement equations become:

Where the global stiffness matrix k is:

Expanding the above transformation gives:

We can assemble the total stiffness matrix by using the aboveelement stiffness matrix and the direct stiffness method.

= Tf T k Td

Tk T k T

2 2

2 2

2 2

2 2

CS CSC C

S SCS CSAE

L CS CSC C

S SCS CS

k

( ) ( )1 1

n ne e

e e

K F

K k F f F Kd



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Example 3 - Bar Element Problem

Simplifying the global elemental stiffness matrix is:

The global elemental stiffness matrix is:

2 6

3 33 34 4 4 4

3 31 1(2 ) 30 10 4 4 4 4

60 3 33 34 4 4 4

3 31 14 4 4 4

lbin

in psi

in

k

6

0.750 0.433 0.750 0.4330.433 0.250 0.433 0.250

100.750 0.433 0.750 0.433

0.433 0.250 0.433 0.250

lbin

k


Computation of Stress for a Bar in the x-y Plane

For a bar element the local forces are related to the localdisplacements by:

1 1

2 2

1 1

1 1x

x

f uAE

f uL

The force-displacement equation for is:2xf

122

1 1xuAE

fuL

The stress in terms of global displacement is:

1

1

1 1 2 2

2

2

0 01 1

0 0

u

vC SE ECu Sv Cu Sv

uL LC S

v



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Example 5 - Plane Truss Problem

The plane truss shown below is composed of three barssubjected to a downward force of 10 kips at node 1. Assumethe cross-sectional areaA = 2 in2 and E is 30 x 106 psi for allelements.

Determine the x and y displacement at node 1 and stresses ineach element.



Element Node 1 Node 2 C S

1 1 2 90o 0 1

2 1 3 45o 0.707 0.707

3 1 4 0o 1 0



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The global elemental stiffness matrix are:1 1 2 2

2 6(1)

0 0 0 0

0 1 0 1(2 )(30 10 )0 1

0 0 0 0120

0 1 0 1

lbin

u v u v

in psiC S

in

kelement 1:

1 1 4 4

2 6(3)

1 0 1 0

0 0 0 0(2 )(30 10 )1 0

1 0 1 0120

0 0 0 0

lbin

u v u v

in psiC S

in

k

element 2:

element 3:

1 1 3 3

2 6(2)2 2

2 2

1 1 1 1

1 1 1 1(2 )(30 10 )

1 1 1 1240 2

1 1 1 1

lbin

u v u v

in psiC S

in

k

2 2

2 2

2 2

2 2

CS CSC C

S SCS CSAE

L CS CSC C

S SCS CS

k



The total global stiffness matrix is:

5

1.354 0.354 0 0 0.354 0.354 1 0

0.354 1.354 0 1 0.354 0.354 0 0

0 0 0 0 0 0 0 0

0 1 0 1 0 0 0 0

0.354 0.354 0 0 0.354 0.354 0 0

0.354 0.354 0 0 0.354 0.354 0 0

1 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0

5 10 lb in

K

The total global force-displacement equations are:

5

0 1.354 0.354 0 0 0.354 0.354 1 010,000

0.354 1.354 0 1 0.354 0.354 0 0

2 0 0 0 0 0 0 0 0

2 0 1 0 1 0 0 0 0

0.354 0.354 0 0 0.354 0.354 0 030.354 0.354 0 0 0.354 0.354 0 03

1 0 0 0 0 0 1 04

0 0 0 0 0 0 0 04

5 10

F xF y

F xF y

F xF y

1

1

0

0

0

0

0

0

u

v

element 1

element 2

element 3



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5

01.354 0.354 0 0 0.354 0.354 1 0

10,0000.354 1.354 0 1 0.354 0.354 0 0

2 0 0 0 0 0 0 0 0

2 0 1 0 1 0 0 0 0

0.354 0.354 0 0 0.354 0.354 0 030.354 0.354 0 0 0.354 0.354 0 03

1 0 0 0 0 0 1 04

0 0 0 0 0 0 0 04

5 10

F xF y

F xF y

F xF y

1

1

0

0

0

0

0

0

u

v



Applying the boundary conditions for the truss, the aboveequations reduce to:



Applying the boundary conditions for the truss, the aboveequations reduce to:

15

1

0 1.354 0.3545 10

10,000 0.354 1.354

u

v

Solving the equations gives: 21

21

0.414 10

1.59 10

u in

v in

The stress in an element is: 1 1 2 2E

Cu Sv Cu SvL



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Element Node 1 Node 2 C S

1 1 2 90o 0 1

2 1 3 45o 0.707 0.707

3 1 4 0o 1 0

element 1

element 2

element 3

6

(1)1

30 103,965

120v psi

6(2 )

1 1

30 10

(0.707) (0.707) 1,471120 u v psi

6

(3 )1

30 101,035

120u psi



Lets check equilibrium at node 1:

2 2(1,471 )(2 )(0.707) (1,035 )(2 ) 0xF psi in psi in

2 2(3,965 )(2 ) (1,471 )(2 )(0.707) 10,000 0yF psi in psi in



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Develop the element stiffness matrices and system equationsfor the plane truss below.

Assume the stiffness of each element is constant. Use thenumbering scheme indicated. Solve the equations for thedisplacements and compute the member forces. All elementshave a constant value ofAE/L



Develop the element stiffness matrices and system equationsfor the plane truss below.

Member Node 1 Node 2 ElementalStiffness

1 1 2 k 0

2 2 3 k 3/4

3 1 3 k /2



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Compute the elemental stiffness matrix for each element. Thegeneral form of the matrix is:


1 1 2 k 0

2 2 3 k 3/4

3 1 3 k /2

2 2

2 2

2 2

2 2

CS CSC C

S SCS CSAEk

L CS CSC C

S SCS CS



For element 1:


1 1 2 k 0

2 2 3 k 3/4

3 1 3 k /2

1 1 2 2

1

1(1)

2

2

1 0 1 0

0 0 0 0

1 0 1 0

0 0 0 0

u v u v

u

vk

u

v

k



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2 2 3 3

2

2(2 )

3

3

1 1 1 1

1 1 1 1

2 1 1 1 1

1 1 1 1

u v u v

u

vk

u

v

k



For element 2:


1 1 2 k 0

2 2 3 k 3/4

3 1 3 k /2

1 1 3 3

1

1(3 )

3

3

0 0 0 0

0 1 0 1

0 0 0 0

0 1 0 1

u v u v

u

vk

u

v

k



For element 3:


1 1 2 k 0

2 2 3 k 3/4

3 1 3 k /2



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1 1 2 2 3 3

1

1

2

2

3

3

2 0 2 0 0 0

0 2 0 0 0 2

2 0 3 1 1 1

2 0 0 1 1 1 1

0 0 1 1 1 1

0 2 1 1 1 3

u v u v u v

u

v

uk

v

u

v

K



Assemble the global stiffness matrix by superimposing theelemental global matrices.

element 1

element 2

element 3



The unconstrained (no boundary conditions satisfied)equations are:

11

11

12

22

33

33

2 0 2 0 0 0

0 2 0 0 0 2

2 0 3 1 1 1

2 0 0 1 1 1 1

0 0 1 1 1 1

0 2 1 1 1 3

x

y

x

y

Fu

Fv

Puk

Pv

Fu

Fv

The displacement at nodes 1 and 3 are zero in both directions.

Applying these conditions to the system equations gives:11

11

12

22

33

33

2 0 2 0 0 0

0 2 0 0 0 2

2 0 3 1 1 1

2 0 0 1 1 1 1

0 0 1 1 1 1

0 2 1 1 1 3

x

y

x

y

Fu

Fv

Puk

Pv

Fu

Fv



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Applying the boundary conditions to the system equationsgives:

Solving this set of equations is fairly easy. The solution is:

2 1

2 2

3 1

2 1 1

u Pk

v P

1 2 1 22 2

3P P P Pu v

k k



Member (element) 1:

Using the force-displacement relationship the force in eachmember may be computed.

1 21 2 1 1 0x y

P Pf k P P f

k

1 22 1 2 2 0x y

P Pf k P P f

k



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Member (element) 2:


1 2 1 22 2

31 12

2 2x

P P P Pf k P

k k

1 2 1 23 2

31 12

2 2x

P P P Pf k P

k k

2 30 0y yf f



Member (element) 3:


1 1

3 3

0 0

0 0

x y

x y

f f

f f

The solution to this simple problem can be readily checked byusing simple static equilibrium equations.



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Consider the two bar truss shown below.

Determine the displacement in the y direction of node 1 andthe axial force in each element.

Assume E = 210 GPa andA = 6 x 10-4 m2



The global elemental stiffness matrix forelement 1 is:

8.05

4sin6.0

5

3cos )1()1(

6 4(1)

0.36 0.48 0.36 0.48

0.48 0.64 0.48 0.64210 10 (6 10 )

5 0.36 0.48 0.36 0.48

0.48 0.64 0.48 0.64

k

Simplifying the above expression gives:

1 1 2 2

(1)

0.36 0.48 0.36 0.48

0.48 0.64 0.48 0.6425,200

0.36 0.48 0.36 0.48

0.48 0.64 0.48 0.64

u v u v

k



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1sin0cos )2()2(

6 4(2)

0 0 0 0

0 1 0 1(210 10 )(6 10 )

4 0 0 0 0

0 1 0 1

k

1 1 3 3

(2)

0 0 0 00 1.25 0 1.25

25,2000 0 0 0

0 1.25 0 1.25

u v u v

k



The total global equations are:

The displacement boundary conditions are:

1 1

1 1

2 2

2 2

3 3

3 3

0.36 0.48 0.36 0.48 0 0

0.48 1.89 0.48 0.64 0 1.25

0.36 0.48 0.36 0.48 0 025,200

0.48 0.64 0.48 0.64 0 0

0 0 0 0 0 0

0 1.25 0 0 0 1.25

x

y

x

y

x

y

F u

F v

F u

F v

F u

F v

1 2 2 3 3 0u u v u v



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1 1

1 1

2 2

2 2

3 3

3 3

0.36 0.48 0.36 0.48 0 0

0.48 1.89 0.48 0.64 0 1.25

0.36 0.48 0.36 0.48 0 025,200

0.48 0.64 0.48 0.64 0 0

0 0 0 0 0 0

0 1.25 0 0 0 1.25

x

y

x

y

x

y

F u

F v

F u

F v

F u

F v



The total global equations are:

By applying the boundary conditions the force-displacementequations reduce to:

125,200(0.48 1.89 )P v

1

2

2

3

3

x

x

y

x

y

F

P

F

F

F

F



Solving the equation gives:

By substituting P = 1,000 kN and = -0.05 m in the aboveequation gives:

The local element forces for element 1 are:

51 (2.1 10 ) 0.25v P

1 0.0337v m

1

1 1

2 2

2

0.05

0.03371 1 0.6 0.8 0 025,200 1 1 0 0 0.6 0.8

x

x

u

f v

f u

v

The element forces are: 1 276.6 76.7x xf kN f kN



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The local element forces for element 2 are:

The element forces are:

1

1 1

3 3

3

0.05

0.03371 1 0 1 0 031,500

1 1 0 0 0 1x

x

u

f v

f u

v

1 31,061 1,061x xf kN f kN


Transformation Matrix and Stiffness Matrix for

a Bar in Three-Dimensional Space

Lets derive the transformation matrix for the stiffness matrix fora bar element in three-dimensional space as shown below:



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The coordinates at node 1 are x1, y1, and z1, and the coor-dinates of node 2 are x2, y2, and z2. Also, let x, y, and z bethe angles measured from the global x, y, and z axes,respectively, to the local axis.




The three-dimensional vector representing the bar element isgives as:

u v w u v w d i j k i j k



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Taking the dot product of the above equation with gives:i

( ) ( ) ( )u v w u i i j i k i

By the definition of the dot product we get:

2 1 2 1 2 1x y z

x x y y z zC C C

L L L

i i j i k i

where 2 2 22 1 2 1 2 1( ) ( ) ( )L x x y y z z

cos cos cosx x y y z zC C C

where Cx, Cy, and Cz are projections of on to i,j, and k,respectively.

i




Therefore: x y zu C u C v C w

The transformation between local and global displacements is:

*d T d1

1

1 1

2 2

2

2

00 0

0 0 0x y z

yx z

u

v

u wC C C

Cu C C u

v

w

00 0

0 0 0x y z

yx z

C C C

CC C

*T



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The transformation from the local to the global stiffness matrixis:

Tk T k T

0

0

00 0 01 1

0 1 1 0 0 0

0

0

x

y

z x y z

yx x z

y

z

C

C

C C C CAE

CC C CL

C

C

k

2 2

2 2

2 2

2 2

2 2

2 2

x y x yx z x zx x

y yy z y zx y x y

y z y zz zx z x z

x y x yx z x zx x

y yy z y zx y x y

y z y zz zx z x z

C C C CC C C CC C

C CC C C CC C C C

C C C CC CC C C CAE

C C C CC C C CL C C

C CC C C CC C C C

C C C CC CC C C C

k




The global stiffness matrix can be written in a more convenientform as:

AE

L

k

2

2

2

x x y x z

x y y y z

x z y z z

C C C C C

C C C C C

C C C C C



45/77


Example 8 Space Truss Problem

Consider the space truss shown below. The modulus ofelasticity, E = 1.2 x 106 psi for all elements. Node 1 isconstrained from movement in the y direction.

To simplify the stiffness matricesfor the three elements, we willexpress each element in thefollowing form:

AEL

k



Consider element 1:

(1) 2 2( 72) (36) 80.5L in

720.89

80.5

360.45

80.5

0

x

y

z

C

C

C

0.79 0.40 0

0.40 0.20 0

0 0 0



46/77



Consider element 1:

1 1 1 2 2 2

2 6(0.302 )(1.2 10 )

80.5

u v w u v w

lbin

in psi

in

k



Consider element 2:

(2) 2 2 2( 72) (36) (72) 108L in

720.667

108

360.33

108

720.667

108

x

y

z

C

C

C

0.45 0.22 0.45

0.22 0.11 0.45

0.45 0.45 0.45



47/77



Consider element 2:

1 1 1 3 3 3

2 6(0.729 )(1.2 10 )

108

u v w u v w

lbin

in psi

in

k

The boundary conditions are:

2 2 2

3 3 3

4 4 4

0

0

0

u v w

u v w

u v w

1 0v



Consider element 3: (3) 2 2 24 1 4 1 4 1( ) ( ) ( )L x x y y z z

(3) 2 2( 72) ( 48) 86.5L in

720.833

86.5

0

480.550

86.5

x

y

z

C

C

C

0.69 0 0.46

0 0 0

0.46 0 0.30



48/77



Consider element 3:

1 1 1 4 4 4

2 6(0.187 )(1.2 10 )

86.5

u v w u v w

lbin

in psi

in

k



Canceling the rows and the columns associated with theboundary conditions reduces the global stiffness matrix to:

The global force-displacement equations are:

1 1

9,000 2,450

2,450 4,450

u w

K

1

1

9,000 2,450 0

2,450 4,450 1,000

u

w


1 10.072 0.264u in w in



49/77



It can be shown, that the local forces in an element are:

The stress in an element is:

i

i

ix x y z x y z i

jx x y z x y z j

j

j

u

v

f C C C C C C wAE

f C C C C C C uL

v

w

i

i

i

x y z x y z

j

j

j

u

v

wEC C C C C C

uL

v

w



The stress in element 1 is:

0

0

0

264.0

0

072.0

045.089.0045.089.05.80

102.1 6)1(


(1) 955 psi

6

(2)

0.072

00.2641.2 10

0.667 0.33 0.667 0.667 0.33 0.6670108

0

0

(2 ) 1,423 psi



50/77

6

(3)

0.072

0

0.2641.2 100.83 0 0.55 0.83 0 0.55

086.5

0

0




(3 ) 2,843 psi


Inclined, or Skewed, Supports

If a support is inclined, or skewed, at some angle for theglobal x axis, as shown below, the boundary conditions onthe displacements are not in the global x-y directions but inthe x-y directions.



51/77



We must transform the local boundary condition ofv 3 = 0(in local coordinates) into the global x-y system.



Therefore, the relationship between of the components of thedisplacement in the local and the global coordinate systemsat node 3 is:

3 3

3 3

' cos sin

' sin cos

u u

v v

We can rewrite the above expression as:

3 3 3 3cos sin

' [ ] sin cosd t d t

We can apply this sort of transformation to the entiredisplacement vector as:

1 1' [ ] [ ] 'Td T d or d T d



52/77



Where the matrix [T1]T is:

Both the identity matrix [I] and the matrix [t3] are 2 x 2 matrices.

The force vector can be transformed by using the sametransformation.

T1

3

[ ] [0] [0]

[ ] [0] [ ] [0]

[0] [0] [ ]

I

T I

t

1' [ ]f T f

In global coordinates, the force-displacement equations are:

[ ]f K d



Applying the skewed support transformation to both sides ofthe equation gives:

By using the relationship between the local and the globaldisplacements, the force-displacement equations become:

Therefore the global equations become:

1 1[ ] [ ][ ]T f T K d

1 1' [ ][ ][ ] 'Tf T K T d

1 1

1 1

2 2

1 12 2

3 3

3 3

[ ][ ][ ]

' '

' '

x

y

x T

y

x

y

F u

F v

F uT K T

F v

F u

F v



53/77



Determine the stiffness matrixfor each element.

Consider the plane truss shown below. Assume E = 210 GPa,A = 6 x 10-4 m2 for element 1 and 2, andA = (6 x 10-4)m2

for element 3.2




(1) (1)cos 0 sin 1

1 1 2 2

6 2 4 2

(1)

0 0 0 0

0 1 0 1(210 10 / )(6 10 )0 0 0 01

0 1 0 1

u v u v

kN m mm

k



54/77


55/77



Using the direct stiffness method, the global stiffness matrix is:

5.0

5.0

0

0

5.0

5.0

5.0

5.1

0

1

5.0

5.0

0

0

1

0

1

0

0

1

0

1

0

0

5.0

5.0

1

0

5.1

5.0

5.0

5.0

0

0

5.0

5.0

10260,1 5m

NK

We must transform the global displacements into localcoordinates. Therefore the transformation [T1] is:

2 22 2

2 22 2

1 0 0 0 0 0

0 1 0 0 0 00 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0

0 0 0 0

1

3

[ ] [0] [0]

[ ] [0] [ ] [0]

[0] [0] [ ]

I

T I

t



The first step in the matrix transformation to find the product of[T1][K].

The next step in the matrix transformation to find the product of[T

1

][K][T1

]T.

5

0.5 0.5 0 0 0.5 0.5

0.5 1.5 0 1 0.5 0.5

0 0 1 0 1 0

0 1 0 1 0 0

0.707 0.707 0.707 0 1.414 0.707

0 0 0.707 0 0.70 0

1,260 101NT K

m

T

0.5 0.5 0 0 0.707 0

0.5 1.5 0 1 0.707 0

0 0 1 0 0.707 0.707

0 1 0 1 0 0

0.707 0.707 0.707 0 1.5 0.5

0 0 0.707 0 0.5 0.5

51,260 101 1T K NT m



56/77

5

1

1

2

2

3

3

1

1

2

2

3

3

1,260 10

0.5 0.5 0 0 0.707 0

0.5 1.5 0 1 0.707 0

0 0 1 0 0.707 0.707

0 1 0 1 0 0

0.707 0.707 0.707 0 1.5 0.5

0 0 0.707 0 0.5 0.5

'

'

'

'

x

y

x

y

x

y

Nm

F

F

F

F

F

F

u

v

u

v

u

v



The displacement boundary conditions are:1 1 2 3' 0u v v v



By applying the boundary conditions the global force-displacement equations are:

2 25

3 3

1,0001 0.7071,260 10

' ' 00.707 1.5x

x

u F kNN

m u F

Solving the equation gives: 2 311.91 ' 5.61u mm u mm



57/77


58/77


Potential Energy Approach to Derive Bar Element Equations

Summing the differential energy over the whole bar gives:

If we let the volume of the element approach zero, then:

x xdU d dV

0

x

x x

V

U d dV

For a linear-elastic material (Hookes law) as shown below:

x xE



The potential energy of the external forces is:

The internal strain energy statement becomes

1

2x x

V

U dV

s

V S

X u dV T u dS f u

M

b x ix i

i 1

where Xb is the body force (force per unit volume), Tx is thetraction (force per unit area), and fix is the nodal concentratedforce. All of these forces are considered to act in the local xdirection.



59/77



1. Formulate an expression for the total potential energy.

2. Assume a displacement pattern.

3. Obtain a set of simultaneous equations minimizing thetotal potential energy with respect to the displacement

parameters.

Apply the following steps when using the principle of minimumpotential energy to derive the finite element equations.



We can approximate the axial displacement as:

Consider the following bar element, as shown below:

2 2

0

L

p x x 1x 1 x

s

V S

Adx f u f u

2

X u dV T u dS

b x

11 22

uu N N

u

1 21

x xN N

L L



60/77



where N1 and N2 are the interpolation functions gives as:

Using the stress-strain relationships, the axial strain is:

11 2

2

x

udN dNdu

udx dx dx

1

2

1 1x

u

uL L

1 1

B L L

The axial stress-strain relationship is: [ ]x xD

[ ]{ }x B d



The total potential energy expressed in matrix form is:

Where [D] = [E] for the one-dimensional stress-strainrelationship and E is the modulus of elasticity.

Therefore, stress can be related to nodal displacements as:

where {P} represented the concentrated nodal loads.

[ ][ ]x D B d

0

L T T T T

p x x

V S

A dx d P u X dV u T dS2

b x



61/77



If we substitute the relationship between and into theenergy equations we get:

u d

0

LT TT T

p

T TT T

s

V S

Ad B D B d dx d P

2

d N X dV d N T dS

b x

In the above expression for potential energy p is a function ofthe d, that is: p = p( ).1,u 2u

However, [B] and [D] and the nodal displacements u are not afunction ofx.



where

Integration the energy expression with respect to x gives:

[ ] [ ] [ ]2

T TT T

p

ALd B D B d d f

[ ] [ ]V S

f P N X dV N X dS T T

b b

We can define the surface tractions and body-force matricesas:

[ ] xS

f N T dS T

s [ ]V

f N X dV T

b b



62/77



Minimization ofp with respect to each nodal displacementrequires that:

For convenience, lets define the following

1 2

0 0p p

u u

* [ ] [ ] [ ]T T TU d B D B d

1* 1 22

1

1 1[ ]

1

uLU u u E

uL L

L




The loading on a bar element is given as:

* 2 21 1 2 22 2E

U u u u uL

1 1 2 2T

x xd f u f u f

Therefore, the minimum potential energy is:

1 2 11

2 2 02

p

x

AEu u f

u L

1 2 22

2 2 02

p

x

AEu u f

u L



63/77



The above equations can be written in matrix form as:

The stiffness matrix for a bar element is:

This form of the stiffness matrix obtained from the principle ofminimum potential energy is identical to the stiffness matrix

derived from the equilibrium equations.

1 1

2 2

1 10

1 1

p x

x

u fAE

u fd L

1 1

1 1

AEk

L



Consider the bar shown below:

The energy equivalent nodal forces due to the distributed loadare:

0 [ ] xS

f N T dS T 0

0

Lx1

f Lf Cx dx

f x

L

1x

2x



64/77



The total load is the area under the distributed load curve, or:

The equivalent nodal forces for a linearly varying load are:

0

0

0

L2 3

2

L

L 23

Cx Cx CLx1 2 3Lf 6L

Cx dxf x CLCx

L 33L

1x

2x

21( )( )

2 2

CLF L CL

1

1of the total load

3 3x

Ff 2

2 2of the totalload

3 3x

Ff



Consider the axially loaded bar shown below. Determine theaxial displacement and axial stress. Let E = 30 x 106 psi,

A = 2 in2, and L = 60 in. Use (a) one and (b) two elements inthe finite element solutions.



65/77



The one-element solution:

The distributed load can be converted into equivalent nodalforces using:

0 [ ] xS

F N T dS

T 00

L1x

2x

x1

F LF 10x dx

F xL



The one-element solution:

0

2 2 2

L1x

2 22x

10L 10L 10Lx1

F 2 3 6L10x dx

F x 10L 10LL 3 3

6,000

12,0001x

2x

F lb

F lb



66/77


67/77



The two-element solution:

The distributed load can be converted into equivalent nodalforces.

For element 1, the total force of the triangular-shapeddistributed load is:

1(30 .)(300 ) 4,500

2

lbinin lb




Based on equations developed for the equivalent nodal force ofa triangular distributed load, develop in the one-elementproblem, the nodal forces are:

(1)

1(1)

2

1(4,500)

1,50032 3,000

(4,500)3

x

x

lbflbf



68/77




For element 2, the applied force is in two parts: a triangular-shaped distributed load and a uniform load. The uniform loadis:

(30 )(300 / ) 9,000in lb in lb

(2 )2

(2 )3

1 1(9,000) (4,500)6,0002 3

7,5001 2(9,000) (4,500)

2 3

x

x

lbf

lbf

The nodal forces for element 2 are:




The final nodal force vector is:

The element stiffness matrices are:

1

2

3 3

1,500

9,000

7,500

x

x

x x

F

F

F R

(1) (2) 1 121 1

AE

L

k k

1 2

3 4



69/77




The assembled global stiffness matrix is:

The assembled global force-displacement equations are:

6

1 1 0

2 10 1 2 1

0 1 1

K

1

62

3

1 1 0 1,500

2 10 1 2 1 9,000

0 1 1 0 7,500x

u

u

R




After the eliminating the row and column associated with d3x,we get:


16

2

1 1 1,5002 10

1 2 9,000

u

u

1

2

0.006

0.00525

u in

u in



70/77




The axial stress-strain relationship is:

1(1)

2

1 1 xx

x

dE

dL L

(2 ) 0.005251 1 5,250 ( )30 30 0

x E psi T

0.0061 1750 ( )

30 30 0.00525

E psi T


Comparison of Finite Element Solution to Exact Solution

In order to be able to judge the accuracy of our finite elementmodels, we will develop an exact solution for the bar elementproblem.

The exact solution for the displacement may be obtained by:

where the force P is shown on the following free-body diagram.

0

1( )

L

u P x dxAE

21

( ) (10 ) 52

P x x x x



71/77



Therefore:

Applying the boundary conditions:

0

1( )

L

u P x dxAE

3

21

1 55

3

x

o

xu x dx C

AE AE

3 3

1 1

5 5( ) 0

3 3

x Lu L C C

AE AE

The exact solution for axial displacement is:

3 35

( )3

u L x LAE

2( ) 5

( )P x x

xA A



A plot of the exact solution for displacement as compared toseveral different finite element solutions is shown below.



72/77



A plot of the exact solution for axial stress as compared toseveral different finite element solutions is shown below.



A plot of the exact solution for axial stress at the fixed end(x = L) as compared to several different finite elementsolutions is shown below.



73/77

Stiffness Matrix for a Bar ElementGalerkins Residual Method and Its Application

to a One-Dimensional BarThere are a number of weighted residual methods.

However, the Galerkins method is more well-known and will bethe only weighted residual method discussed in this course.

In weighted residual methods, a trial or approximate function ischosen to approximate the independent variable (in ourcase, displacement) in a problem defined by a differentialequation.

The trial function will not, in general, satisfy the governingdifferential equation.

Therefore, the substitution of the trial function in the differentialequation will create a residual over the entire domain of theproblem.


to a One-Dimensional Bar

Therefore, the substitution of the trial function in the differentialequation will create a residual over the entire domain of theproblem.

minimumV

RdV

In the residual methods, we require that a weighted value ofthe residual be a minimum over the entire domain of the

problem.The weighting function allows the weighted integral of the

residuals to go to zero.

0V

RW dV



74/77


to a One-Dimensional Bar

Using Galerkins weighted residual method, we require theweighting functions to be the interpolation, Ni. Therefore:

0 1, 2, ,iV

RN dV i n


Example 12 - Bar Element Formulation

Lets derive the bar element formulation using Galerkinsmethod. The governing differential equation is:

Applying Galerkins method we get:

0d du

AEdx dx

0

0 1, 2, ,L

i

d duAE N dx i n

dx dx

We now apply integration by parts using the following generalformula:

udv uv vdu



75/77



If we assume the following:

then integration by parts gives:

ii

dNu N du dx

dx

d du dudv AE dx v AE

dx dx dx

0 0

0L L

ii

dNdu duN AE AE dx

dx dx dx



Recall that:

1 21 2

dN dNduu u

dx dx dx

1

2

1 1 udu

udx L L

Our original weighted residual expression, with theapproximation foru becomes:

1

20

1 1LL

i

io

udN du

AE dx N AEudx L L dx



76/77



Substituting N1 for the weighting function Ni gives:

111

2 00

1 1LL udN du


1

2 00

1 1 1L

x

u duAE dx AE

uL L L dx

1 2 1xAE

u u fL

1

0

x

x

duf AE

dx



Substituting N2 for the weighting function Ni gives:

122

2 00

1 1LL udN du


1

20

1 1 1L

x L

u duAE dx AE

uL L L dx

2 1 2xAE

u u fL

2x

x L

duf AEdx

1 1

2 2

1 1

1 1

x

x

u fAE

u fL



77/77


Problems:

6. Verify the global stiffness matrix for a three-dimensionalbar. Hint: First, expand T* to a 6 x 6 square matrix, thenexpand k to 6 x 6 square matrix by adding the appropriaterows and columns of zeros, and finally, perform the matrixtriple product k = TTkT.

7. Do problems 3.4, 3.10, 3.12, 3.15a,b, 3.18, 3.23, 3.37,3.43, 3.48, 3.50, and 3.55 on pages 146 - 165 in yourtextbook A First Course in the Finite Element Method byD. Logan.

8. Use SAP2000 and solve problems 3.63 and 3.64.

End of Chapter 3


development of truss eqns

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