Chapter 3a – Development of Truss · PDF fileDevelopment of Truss Equations Next we will...

Chapter 3a – Development of Truss Equations

Learning Objectives• To derive the stiffness matrix for a bar element.

• To illustrate how to solve a bar assemblage by the direct stiffness method.

• To introduce guidelines for selecting displacement functions.

• To describe the concept of transformation of vectors in two different coordinate systems in the plane.

• To derive the stiffness matrix for a bar arbitrarily oriented in the plane.

• To demonstrate how to compute stress for a bar in the plane.

• To show how to solve a plane truss problem.

• To develop the transformation matrix in three-dimensional space and show how to use it to derive the stiffness matrix for a bar arbitrarily oriented in space.

• To demonstrate the solution of space trusses.

Having set forth the foundation on which the direct stiffness method is based, we will now derive the stiffness matrix for a linear-elastic bar (or truss) element using the general steps outlined in Chapter 2.

We will include the introduction of both a local coordinate system, chosen with the element in mind, and a global or reference coordinate system, chosen to be convenient (for numerical 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 express the stiffness matrix of an arbitrarily oriented bar element in terms of the global system.

Development of Truss Equations

CIVL 7/8117 Chapter 3a - Development of Truss Equations 1/80

Development of Truss EquationsNext we will describe how to handle inclined, or skewed,

supports.

We will then extend the stiffness method to include space trusses.

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

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

Finally, we will apply Galerkin’s residual method to derive the bar element equations.










Stiffness Matrix for a Bar ElementConsider the derivation of the stiffness matrix for the linear-

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

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




where T is the tensile force directed along the axis at nodes 1 and 2, x is the local coordinate system directed along the length of the bar.



The bar element has a constant cross-section A, an initial length L, and modulus of elasticity E.

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


Stiffness Matrix for a Bar Element

The strain-displacement relationship is:du

Edx

From equilibrium of forces, assuming no distributed loads acting 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 the local coordinate x gives:

0d du

AEdx dx


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. Hooke’s law applies; stress is related to strain: x xE


Step 1 - Select Element Type

We will consider the linear bar element shown below.


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, is equal to the total number of degrees of freedom associated with the element.

Applying the boundary conditions and solving for the unknown coefficients gives:

2 11

u uu x u

L

1

2

1ux x

uuL L


Step 2 - Select a Displacement Function


Or in another form:

where N1 and N2 are the interpolation functions gives as:

11 2

2

uu N N

u

1 21x x

N NL L

The linear displacement function plotted over the length of the bar element is shown here.

u

Step 3 - Define the Strain/Displacement and Stress/Strain Relationships


The stress-displacement relationship is: 2 1x

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 AE

L



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 uAEf uL

Notice that AE/L for a bar element is analogous to the spring constant k for a spring element.


The global stiffness matrix and the global force vector are assembled using the nodal force equilibrium equations, and force/deformation and compatibility equations.

Step 5 - Assemble the Element Equations and Introduce Boundary Conditions

( ) ( )

1 1

n ne e

e e

K F

K k F f

Where k and f are the element stiffness and force matrices expressed in global coordinates.



Solve the displacements by imposing the boundary conditions and 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


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

Example 1 - Bar Problem

Determine: (a) the global stiffness matrix, (b) the displacement of 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 1

1030 1 1 1 1

lb lbin in

k k

1 2 node numbers for element 1

6

(3) 62 15 10 1 1 1 1

1030 1 1 1 1

lb lbin in

k


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



Assembling the global stiffness matrix by the direct stiffness methods gives:


Relating global nodal forces related to global nodal displacements 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


1

2 26

3 3

4

1 1 0 0 0

1 2 1 010

0 1 2 1

0 0 1 1 0

x

x

x

x

F

F u

F u

F


The boundary conditions are:

Example 1 – Bar Problem

1 4 0u u

Applying the boundary conditions and the known forces (F2x = 3,000 lb) gives:

26

3

3,000 2 110

0 1 2

u

u


Example 1 – Bar Problem

Solving for u2 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 2,000

1 2 1 0 0.002 3,00010

0 1 2 1 0.001 0

0 0 1 1 0 1,000

x

x

x

x

F

Flb

F

F



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

Selecting Approximation Functions for Displacements

1. Common approximation functions are usually polynomials.

2. The approximation function should be continuous within the bar element.




3. The approximating function should provide interelement continuity for all degrees of freedom at each node for discrete line elements, and along common boundary lines and surfaces for two- and three-dimensional elements.





For the bar element, we must ensure that nodes common to two or more elements remain common to these elements upon deformation and thus prevent overlaps or voids between elements.

The linear function is then called a conforming (or compatible) function for the bar element because it ensures both the satisfaction of continuity between adjacent elements and of continuity within the element.




4. The approximation function should allow for rigid-body displacement and for a state of constant strain within the element.

Completeness of a function is necessary for convergence to the exact answer, for instance, for displacements and stresses.



The interpolation function must allow for a rigid-body displacement, that means the function must be capable of yielding 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 at every point within the element so the it will yield a constant value when a rigid-body displacement occurs.


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

Local coordinates are always chosen to conveniently represent the individual element.

Global coordinates are chosen to be convenient for the whole structure.

Transformation of Vectors in Two Dimensions



Given the nodal displacement of an element, represented by the vector d in the figure below, we want to relate the components of this vector in one coordinate system to components in another.



Let’s consider that d does not coincide with either the local or global axes. In this case, we want to relate global displacement components to local ones. In so doing, we will develop a transformation matrix that will subsequently be used to develop the global stiffness matrix for a bar element.




We define the angle to be positive when measured counterclockwise from x to x’. We can express vector displacement d in both global and local coordinates by:


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


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 vector a is in the direction and b is in the direction, therefore:

i j

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




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

The vector j can be rewritten as: sin cos j 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



The matrix is called the transformation matrix.

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

C S

S C

The figure below shows u expressed in terms of the global coordinates 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 and v2 = 0.2 in for the bar element shown below. Determine the local displacement.

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

u Cu Sv


Global Stiffness Matrix

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




We want to relate the global element forces f to the global displacements d for a bar element with an arbitrary orientation.

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

1 1

2 2

1 1

1 1x

x

f uAEf uL

f k d

1 1

1 1

2 2

2 2

x

y

x

y

f u

f vk

f u

f v

f = kd



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 0

0 0

u

u vC S

u 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 be developed.

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 global displacements gives:



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

Let’s expand the relationship between local and global displacement.

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 between local and global forces.

Therefore our original local coordinate force-displacement expression

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 uAEf uL

f = k d



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 vAEf 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 of T. It can be shown that: 1 TT 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 above element 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



Local forces can be computed as:

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 vAEf uL

f v

1 1 1 2 2

1

2 1 1 2 2

2

0

0

x

y

x

y

f Cu Sv Cu Sv

f AEf L Cu Sv Cu Sv

f

1

1

2

2

1 0 1 0 0 0

0 0 0 0 0 0

1 0 1 0 0 0

0 0 0 0 0 0

uC S

vS CAE

uC SL

vS C




Therefore:

For the bar element shown below, evaluate the global stiffness matrix. Assume the cross-sectional area is 2 in2, the length is 60 in, and the E is 30 x 106 psi.

3 1

cos30 sin302 2

o oC S

2 2

2 2

2 2

2 2

CS CSC C

S SCS CSAE

L CS CSC C

S SCS CS

k



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 4 460 3 33 3

4 4 4 4

3 31 14 4 4 4

in psi

in

k

6

0.750 0.433 0.750 0.433

0.433 0.250 0.433 0.25010

0.750 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 local displacements by:

1 1

2 2

1 1

1 1x

x

f uAEf uL

The force-displacement equation for is:2xf

12

2

1 1x

uAEf

uL

The stress in terms of global displacement is:

1

1

2

2

0 01 1

0 0

u

vC SEuL C S

v

1 1 2 2

ECu Sv Cu Sv

L



For the bar element shown below, determine the axial stress. Assume the cross-sectional area is 4 x 10-4 m2, the length is 2 m, and the E is 210 GPa.

The global displacements are known as u1 = 0.25 mm, v1 = 0, u2 = 0.5 mm, and v2 = 0.75 mm.

1 1 2 2

ECu Sv Cu Sv

L

6210 10 1 3 1 3(0.25) (0) (0.5) (0.75)

2 2 4 2 4KN

m

2381.32 10 81.32kN

mMPa



Solution of a Plane Truss

We will now illustrate the use of equations developed above along with the direct stiffness method to solve the following plane truss example problems.

A plane truss is a structure composed of bar elements all lying in a common plane that are connected together by frictionless pins.

The plane truss also must have loads acting only in the common plane.


Example 5 - Plane Truss Problem

The plane truss shown below is composed of three bars subjected to a downward force of 10 kips at node 1. Assume the cross-sectional area A = 2 in2 and E is 30 x 106 psi for all elements.

Determine the x and y displacement at node 1 and stresses in each 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



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 lbin

K

The total global force-displacement equations are:

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 yF xF yF xF y

1

1

0

0

0

0

0

0

u

v

element 1

element 2

element 3

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 yF xF yF xF y

1

1

0

0

0

0

0

0

u

v



Applying the boundary conditions for the truss, the above equations reduce to:




Applying the boundary conditions for the truss, the above equations 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 2

ECu Sv Cu Sv

L

where is the local node numberi




1 1 2 90o 0 1

2 1 3 45o 0.707 0.707

3 1 4 0o 1 0

element 1 6

(1)1

30 103,965

120v psi

1 1 2 2

ECu Sv Cu Sv

L

element 2 6

(2)1 1

30 10(0.707) (0.707) 1,471

120u v psi


element 3 6

(3)1

30 101,035

120u psi




1 1 2 90o 0 1

2 1 3 45o 0.707 0.707

3 1 4 0o 1 0

1 1 2 2

ECu Sv Cu Sv

L



Let’s check equilibrium at node 1:

(2)1 cos(45 )x xF f

(2)1 sin(45 )y xF f

(3)1xf

(1)1xf 10,000 lb




Let’s 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



Develop the element stiffness matrices and system equations for the plane truss below.

Assume the stiffness of each element is constant. Use the numbering scheme indicated. Solve the equations for the displacements and compute the member forces. All elements have a constant value of AE/L




Develop the element stiffness matrices and system equations for the plane truss below.

Member Node 1 Node 2 Elemental Stiffness

1 1 2 k 0

2 2 3 k 3/4

3 1 3 k /2



Compute the elemental stiffness matrix for each element. The general 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

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

vku

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

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

ukv

u

v

K



Assemble the global stiffness matrix by superimposing the elemental 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

PukPv

Fu

Fv

The displacement at nodes 1 and 3 are zero in both directions. Applying these conditions to the system equations gives:

1

1

12

22

3

3

2 0 2 0 0 0 0

0 2 0 0 0 2 0

2 0 3 1 1 1

2 0 0 1 1 1 1

0 0 1 1 1 1 0

0 2 1 1 1 3 0

x

y

x

y

F

F

PukPv

F

F



Applying the boundary conditions to the system equations gives:

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

2 1

2 2

3 1

2 1 1

u Pkv P

1 2 1 22 2

3P P P Pu v

k k




Member (element) 1:

Using the force-displacement relationship the force in each member may be computed.

1 2xf k Cu

1 1 1 2 2

1

2 1 1 2 2

2

0

0

x

y

x

y

f Cu Sv Cu Sv

fk

f Cu Sv Cu Sv

f

1 0C S

2 0yf

1 0yf 1 2P Pk

k

1 2P P

2 2xf k Cu 1 2P Pk

k

1 2P P



Member (element) 2:


2 2 2xf k Cu Sv

2 2 2 3 3

2

3 2 2 3 3

3

0

0

x

y

x

y

f Cu Sv Cu Sv

fk

f Cu Sv Cu Sv

f

1 1

2 2C S

3 2 2xf k Cu Sv

1 2 1 231 1

2 2

P P P Pk

k k

1 2 1 231 1

2 2

P P P Pk

k k

22P

22P




Member (element) 3:


1 1

3 3

0 0

0 0x y

x y

f f

f f

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



Consider the two bar truss shown below.

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

Assume E = 210 GPa and A = 6 x 10-4 m2




The global elemental stiffness matrix for element 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





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 0

0 1.25 0 1.2525,200

0 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

element 1

element 2

1 1

1 1

2

2

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 0 025,200

0.48 0.64 0.48 0.64 0 0 0

0 0 0 0 0 0 0

0 1.25 0 0 0 1.25 0

x

y

x

y

x

y

F u

F v

F

F

F

F



The total global equations are:

By applying the boundary conditions the force-displacement equations 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 above equation 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 0.8x

x

u

f v

f u

v

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

Tension



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

Compression



Transformation Matrix and Stiffness Matrix for a Bar in Three-Dimensional Space

Let’s derive the transformation matrix for the stiffness matrix for a bar element in three-dimensional space as shown below:



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 be the angles measured from the global x, y, and z axes, respectively, to the local axis.




The three-dimensional vector representing the bar element is gives as:

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



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



The transformation from the local to the global stiffness matrix is:

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 CAECC 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 CC CC C C CC C C C

C C C CC CC C C CAEC C C CC C C CL C CC 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 convenient form 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


Example 8 – Space Truss Problem

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

To simplify the stiffness matrices for the three elements, we will express each element in the following form:

AE

L

k




Consider element 1:

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

720.89

80.536

0.4580.50

x

y

z

C

C

C

0.79 0.40 0

0.40 0.20 0

0 0 0

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



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

10836

0.3310872

0.667108

x

y

z

C

C

C

0.45 0.22 0.45

0.22 0.11 0.45

0.45 0.45 0.45

(2) 2 2 23 1 3 1 3 1( ) ( ) ( )L x x y y z z



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




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

86.50

480.550

86.5

x

y

z

C

C

C

0.69 0 0.46

0 0 0

0.46 0 0.30



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

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




Canceling the rows and the columns associated with the boundary 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



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 wAEf C C C C C C uL

v

w

i

i

ix 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

045.089.0045.089.05.80102.1 6

)1(


(1) 955 psi

6

(2)

0.072

0

0.2641.2 100.667 0.33 0.667 0.667 0.33 0.667

0108

0

0

(2) 1,423 psi

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 the global x axis, as shown below, the boundary conditions on the displacements are not in the global x-y directions but in the x’-y’ directions.


Inclined, or Skewed, Supports

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




Therefore, the relationship between of the components of the displacement in the local and the global coordinate systems at 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 3

cos sin' [ ]

sin cosd t d t

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

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



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 same transformation.

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 of the equation gives:

By using the relationship between the local and the global displacements, 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 21 1

2 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



Determine the stiffness matrix for each element.

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

for element 3. 2

2 2

2 2

2 2

2 2

CS CSC C

S SCS CSAEk

L CS CSC C

S SCS CS





(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 m

m

k

2 2

2 2

2 2

2 2

CS CSC C

S SCS CSAEk

L CS CSC C

S SCS CS

(2) (2)cos 1 sin 0




2 2 3 3

6 2 4 2(2)

1 0 1 0

0 0 0 0(210 10 / )(6 10 )

1 0 1 01

0 0 0 0

u v u v

kN m m

m

k

2 2

2 2

2 2

2 2

CS CSC C

S SCS CSAEk

L CS CSC C

S SCS CS


(3) (3)2 2cos sin

2 2




1 1 3 3

6 2 4 2(3)

1 1 1 1

1 1 1 1(210 10 / )(6 2 10 )

1 1 1 12 21 1 1 1

u v u v

kN m m

m

k

2 2

2 2

2 2

2 2

CS CSC C

S SCS CSAEk

L CS CSC C

S SCS CS



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 5

mNK

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

2 22 2

2 22 2

1 0 0 0 0 0

0 1 0 0 0 0

0 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].

1T K

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 0.707 0

1,260 101NT K m

2 22 2

2 22 2

5

1 0 0 0 0 00.5 0.5 0 0 0.5 0.5

0 1 0 0 0 0 0.5 1.5 0 1 0.5 0.50 0 1 0 0 0 0 0 1 0 1 0

1,0 1 0 1 0 00 0 0 1 0 0

0.707 0.707 0.707 0 1.414 0.7070 0 0 00 0 0.707 0 0.707 0

0 0 0 0

260 10 Nm

1T K



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

2 2

2 2

2 22 2

T 5

1 0 0 0 0 00.5 0.5 0 0 0.5 0.5

0 1 0 0 0 00.5 1.5 0 1 0.5 0.50 0 1 0 0 00 0 1 0 1 0

0 1 0 1 0 0 0 0 0 1 0 00.707 0.707 0.707 0 1.414 0.707 0 0 0 0

0 0 0.707 0 0.707 00 0 0 0

1,260 101 1NT K mT

T0.5 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

T1T 1T K


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

FF

FF

FF

uvuv

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 kNNm u F

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




Therefore:

The global nodal forces are calculated as:

1

1

2 2

2

3

3

0.5 0.5 0 0 0.707 0 0

0.5 1.5 0 1 0.707 0 0

0 0 1 0 0.707 0.707 11.91

0 1 0 1 0 0 0

0.707 0.707 0.707 0 1.5 0.5 5.61

0 0 0.707 0 0.5 0.5 0

1,260 10

'

'

x

y

x Nmm

y

x

y

F

F

F

F

F

F

1 1500 500x yF kN F kN

2 30 ' 707y yF F kN

1 1

1 1

2 21 1

2 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


Potential Energy Approach to Derive Bar Element Equations

The differential internal work (strain energy) dU in a one-dimensional bar element is:

Let’s derive the equations for a bar element using the principle of minimum potential energy.

The total potential energy, p, is defined as the sum of the internal strain energy U and the potential energy of the external forces :

p U

( )( )( )x xdU y z x d




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 xV

U d dV

For a linear-elastic material (Hooke’s law) as shown below:

x xE

0

x

x xV

E d dV

21

2 x

V

E dV

1

2 x x

V

U dV



The potential energy of the external forces is:

The internal strain energy statement becomes

1

2 x x

V

U dV

sV S

X u dV T u dS f u

M

b x ix ii 1

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




1. Formulate an expression for the total potential energy.

2. Assume a displacement pattern.

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

Apply the following steps when using the principle of minimum potential 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 2

2

uu N N

u

1 21

x xN N

L L




where N1 and N2 are the interpolation functions gives as:

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

11 2

2x

udN dNduudx dx dx

1

2

1 1x

u

uL L

1 1

BL 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-strain relationship 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

LT T T T

p x x

V S

Adx d P u X dV u T dS

2 b x




If we substitute the relationship between and into the energy 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 of the d, that is: p = p( ). 1,u 2u

However, [B] and [D] and the nodal displacements u are not a function of x.

x x



where

Integration the energy expression with respect to x gives:

[ ] [ ] [ ]2

T TT Tp

ALd B D B d d f

[ ] [ ]V S

f P N X dV N X dS T Tb b

We can define the surface tractions and body-force matrices as:

[ ] x

S

f N T dS Ts [ ]

V

f N X dV Tb b




Minimization of p with respect to each nodal displacement requires that:

For convenience, let’s define the following

1 2

0 0p p

u u

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

1*1 2

2

11 1

[ ]1

uLU u u EuL 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 2

T

x xd f u f u f

Therefore, the minimum potential energy is:

1 2 11

2 2 02

px

AEu u f

u L

1 2 22

2 2 02

px

AEu u f

u L




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 of minimum potential energy is identical to the stiffness matrix derived from the equilibrium equations.

1 1

2 2

1 10

1 1p x

x

u fAEu fd L

1 1

1 1

AEk

L



Consider the bar shown below:

The energy equivalent nodal forces due to the distributed load are:

0 [ ] x

S

f N T dS T 0

0

Lx

1f Lf Cx dxf x

L

1x

2x




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 32

L

L 23

Cx Cx CLx1 2 3Lf 6L Cx dx

f x CLCxL 33L

1x

2x

21( )( )

2 2

CLF L CL

1

1of the total load

3 3x

Ff 2

2 2of the total load

3 3x

Ff



Consider the axially loaded bar shown below. Determine the axial displacement and axial stress. Let E = 30 x 106 psi, A = 2 in2, and L = 60 in. Use (a) one and (b) two elements in the finite element solutions.




The one-element solution:

The distributed load can be converted into equivalent nodal forces using:

0 [ ] x

S

F N T dS T 0

0

10L

1x

2x

x1F LF x dx

F x

L




2 2 2

2 20

10 10 10- + -

2 3 61010 10

- -3 3

L1x

2x

L L Lx1F L x dx

F x L LL

6,000

12,0001x

2x

F lb

F lb





The element equations are:

(1) 6 1 110

1 1

k

16

2

6,0001 110

12,0001 1 0 x

u

R

1 0.006u in

The second equation gives:

61 210 ( ) 12,000xu R 2 18,000xR lb




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

{ } [ ]x E B d

1 2 1

2

1 1 u u uE E

uL L L

6 0 0.00630 10 3,000 ( )

60psi T




The two-element solution:

The distributed load can be converted into equivalent nodal forces.

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

1(30 .)(300 ) 4,500

2lb

inin lb




Based on equations developed for the equivalent nodal force of a triangular distributed load, develop in the one-element problem, the nodal forces are:

(1)1

(1)2

1(4,500)

1,50032 3,000

(4,500)3

x

x

lbf

lbf





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

(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)1 1

(1) (2)2 2 2

(2)3 3

x x

x x x

x x

F f

F f f

F f

(1) (2) 1 121 1

AEL

k k

1 2

2 3

3

1,500

9,000

7,500xR





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

16

2

3

1 1 0 1,500

2 10 1 2 1 9,000

0 1 1 0 7,500x

u

u

R

element 1

element 2




After the eliminating the row and column associated with u3x, 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

62 32 10 7,500xu R 3 18,000xR

Solving the last equation gives:





The axial stress-strain relationship is:

1(1)

2

1 1 xx

x

dE

dL L

0.0061 1750 ( )

30 30 0.00525E psi T




The axial stress-strain relationship is:

2(2)

3

1 1 xx

x

dE

dL L

0.005251 15,250 ( )

30 30 0E psi T



Comparison of Finite Element Solution to Exact Solution

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

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 ) 5

2P x x x x



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( )

3u L x L

AE

2( ) 5( )

P x xx

A A




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



A plot of the exact solution for axial stress as compared to several 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 element solutions is shown below.

Stiffness Matrix for a Bar ElementGalerkin’s Residual Method and Its Application

to a One-Dimensional Bar

There are a number of weighted residual methods.

However, the Galerkin’s method is more well-known and will be the only weighted residual method discussed in this course.

In weighted residual methods, a trial or approximate function is chosen to approximate the independent variable (in our case, displacement) in a problem defined by a differential equation.

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

Therefore, the substitution of the trial function in the differential equation will create a residual over the entire domain of the problem.




Therefore, the substitution of the trial function in the differential equation will create a residual over the entire domain of the problem.

minimumV

RdV In the residual methods, we require that a weighted value of

the residual be a minimum over the entire domain of the problem.

The weighting function W allows the weighted integral of the residuals to go to zero.

0V

RW dV



Using Galerkin’s weighted residual method, we require the weighting functions to be the interpolation functions Ni. Therefore:

0 1, 2, ,i

V

RN dV i n



Example 12 - Bar Element Formulation

Let’s derive the bar element formulation using Galerkin’s method. The governing differential equation is:

Applying Galerkin’s 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 general formula:

udv uv vdu



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

0

L

i

d duAE N dx

dx dx

udv uv vdu




Recall that:

1 21 2

dN dNduu u

dx dx dx 1

2

1 1 uduudx L L

Our original weighted residual expression, with the approximation for u becomes:

1

2 00

1 1LL

ii

udN duAE dx N AE

udx L L dx



Substituting N1 for the weighting function Ni gives:

111

2 00

1 1LL udN du

AE dx N AEudx L L dx

1

20

1 1 1L uAE dx

uL L L

1 2 1x

AEu u f

L

10

Ldu

N AEdx 0x

duAE

dx

0x x

AE

1xf0x x

A

1 22

AELu u

L




Substituting N2 for the weighting function Ni gives:

122

2 00

1 1LL udN du

AE dx N AEudx L L dx

1

20

1 1 1L uAE dx

uL L L

1 2 2x

AEu u f

L

20

Ldu

N AEdx x L

duAE

dx

x x LAE

2xfx x L

A

1 22

AELu u

L



Writing the last two equations in matrix form gives:

1 1

2 2

1 1

1 1x

x

u fAEu fL

This element formulation is identical to that developed from equilibrium and the minimum potential energy approach.



Problems:

2. Verify the global stiffness matrix for a three-dimensional bar. Hint: First, expand T* to a 6 x 6 square matrix, then expand k to 6 x 6 square matrix by adding the appropriate rows and columns of zeros, and finally, perform the matrix triple product k = TTk’T.

3. 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 your textbook “A First Course in the Finite Element Method” by D. Logan.

4. Use SAP2000 and solve problems 3.63 and 3.64.

End of Chapter 3a


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