Supplement 1 Networks, trusses, beams and framesme.rice.edu/~akin/Elsevier/Sup_13_Networks.pdfSupplement 1, Networks, trusses, beams and frames 451 13.2 Thermalnetworks There are manyapplications

Supplement 1

Networks, trusses, beams and frames

Topics that do not usually relate to error estimates but are often covered in an

introductory course on finite element analysis include networks, trusses, beams, and

frames. The latter three topics are often covered in books on structural mechanics. To

assist the reader in these areas and to illustrate the use of MODEL for these applications

we will cover some basic theory and a few examples.

13.1 Energy networks

In the previous chapters we saw that groverning principles based on common

differential equations can often be cast into an exact equivalent form based on a

governing integral principal that is frequently based on a stational energy relation, or

’energy balance’. Here we will begin a review of such principals where the primary

nodal unknowns are a scalar quantity and then introduce other considerations that come

into play when they are vector quantities. The emphasis here is on the network concept

where an essentially one-dimensional problem can be conceptually (or actually) extended

to two- or three-dimensions because of available connectivity data that describes how

basic one-dimensional components (elements) in the system are interacting in the energy

balance process overall and at a connecting node of the network. Here we refer to

balance equations as some discipline specific governing physical equations that are

converted to a system of linear algebraic equations of the form SD = C. Examples fromcommon engineering disciplines include: Heat Transfer - Fourier’s Law, Electrical -

Ohm’s and Kirchoff’s Laws, Chemical - Flick’s Law, Mechanics -Newton’s Laws, and

Structures - Minimum Total Potential Energy.

Many balance equations involve two quantities, D and C, in the above system equations

whose product is an energy measure. If one of the quantities is given at a point, then the

second must be computed as a reaction. The balance laws place one of the variables in D

and the other in C. For example, in an electrical resistive circuit network if the voltage,

V , at a point is given, then the current, j, necessary to maintain that voltage is a reaction

that can be computed. The reverse situation is also true. Their product is the energy,

E = jV . Similar related pairs used in common applications are: temperature and heat

Supplement 1, Networks, trusses, beams and frames 449

flow in thermal studies; voltage and current in electrical studies; displacement and force

in stress analysis; and velocity and pressure in fluid flow models. Here we will utilize the

finite element concepts to represent in an energy form the basis laws that engineers are

taught to employ on a more localized basis so they can develop equations that are suitable

to hand solution. We will be solving the same governing concepts but in a process

automated by finite element analysis. For example, our results still satisfy the basis laws:

Thermal equilibrium networks: The algebraic sum of the heat flows into a joint,

including external sources, is zero. The temperature distribution in an assembly of

components minimizes the rate of entropy production.

Electrical resistive networks: The algebraic sum of the currents flowing into a joint,

including external sources, is zero. The voltage (and resulting current) distribution

minimizes the energy in a circuit network.

Elastic structures: The algebraic sum of the force components at a joint, including

external sources, is zero. The equilibrium displacements of a structure in

equilibrium minimize the total potential energy.

Recall from Chapter 2 that the principle of minimum total potential energy states

that the unique set of displacements that occur at equilibrium will satisfy the essential

displacement boundary conditions and minimize the total potential energy of the system.

The total potential energy is the energy stored in the material minus the mechanical work

of the external forces. For the linear spring, the total potential energy, Π, isΠ = kd2 / 2 − Fd for the relative displacement, d of a force, F , in stretching a spring ofstiffness, k. Our alternate equilibrium balance statement was the minimization operation:

∂Π / ∂d = 0, which gives 0 = kd − F , or simply F = kd which is the well known force-displacement equilibrium covered in physics. Likewise, we could represent the balance

of an electrical resistance network as an energy minimization procedure. For a single

resistor, having a resistance R, with a voltage drop of V across it due to a current flow, j,

we recall that the electrical energy is W = GV 2 / 2 − jV where G = 1/R is theconductance of the element in the network. The voltage that corresponds to a minimum

energy state is governed by ∂W / ∂V = 0, so GV = j which we re-arrange to its morefamiliar form V = jR which is known as Ohm’s law. Since energy is a scalar quantity, wecan write the energy contribution from each component in a network and then sum (or

assemble) those to form the total system energy (or rate of energy production) which is to

be minimized. This type of approach usually leads to a system of symmetric linear

algebraic equations for the primary unknown at each node or junction in the network. In

the current discussion we will put off considering transient effects occuring over time.

We will often employ the thermal-electrical-mechanical analogy:

Thermal Electrical Mechanical

k = conductivity G = conductance K = stiffness

T = temperature V = voltage U = displacement

q = heat flow j = current F = force

450 Finite Element Analysis with Error Estimators

2

1

3

5

8

4

6

7 11

10

9

1

2

4

6

75

3

Furnace, T=250 C

400

cm

T= 25 C

T= 25 C

200 cm 200 cm 300 cm

Elem Area Nodes

1 4.0 1 2

2 4.0 1 3

3 4.0 1 4

4 6.0 2 4

5 4.0 4 3

6 4.0 3 5

7 4.0 3 6

8 6.0 4 6

9 4.0 5 6

10 4.0 5 7

11 6.0 6 7

Steel

Figure 13.1 Thermal environment for a truss

0100

200300

400500

600700

−400−350

−300−250

−200−150

−100−50

0

50

100

150

200

250

7

X Coordinates

6

FEA Solution Component_1: 11 Elements, 7 Nodes

4

5−−−−−−max

2

3

Y Coordinates

−−−−−−min 1

Com

pone

nt 1

(m

ax =

250

, min

= 2

5)

A conducting truss system

Figure 13.2 Truss temperatures and mesh


13.2 Thermal networks

There are many applications where a thermal equilibrium problem can be accurately

represented as a two- or three-dimensional assembly of one-dimensional thermal

elements. Two such examples would be the space frame of a aircraft or rocket, and the

copper or gold path printed on a circuit board. In both cases the primary purpose for the

existance of the network (carrying loads, and current, respectively) also gives rise to the

ability, or need, to serve as a thermal network. While such thermal systems can be

modeled in a continuum sense it is much more cost effective to employ a network of line

components. For a thermal network we can directly utilize the conduction and

convection element matrices developed earlier for a pure one-dimensional formulation.

Only the calculation of the physical length of the element needs to be generalized. That

is, we simply allow 2-D or 3-D coordinates to be input. The element connectivity data

still govern where contributions get scattered into the system solution.

Consider the two-dimensional thermal network shown in Fig. 13.1. The far end

supports a furnace that keeps the end nodes at a high temperature compared to the wall

supports at room temperature. In this case the two-dimensional coordinates are mainly

for later use in a structural analysis and are used only to compute the effective thermal

lengths in this example. The resulting temperature distributions are shown in Fig. 13.2.

Here we have limited the element to a linear interpolation between two nodes and no

numerical integration is required. The element matrices are so simple they are hand

coded as given in Fig. 13.3. Note that the comments show that we could allow

convection losses as well, and have the ability to give a heat source per unit length, such

as an electric motor generating heat at a member. If we do include convection then the

true solution along a member is not linear, but a hyperbolic function. It is not practical to

consider an error estimator here for such a network and one would have to consider

refining the mesh. Convection is usually not important in trusses.

13.3 Electrical resistance networks

Consider an electric resistance element connecting two nodes in a DC circuit

network. Ohm’s law giv es the relation between the direct current, j, entering the element

at node 1, the voltage drop E2 − E1 over its length, and the resistance, R, of the material.Specifically j = (E2 − E1) / R so noting that the current flowing in one end is the negativeof that flowing out at the other end (node 2) we can express this relation in matrix form:

1

Re

1

−1−1

1

Ee1

Ee2

=

je1

je2

.

Symbolically this element relation is Ge Ee = je and the corresponding system networkbalance relationship is G E = J (or our previous notation S D = C). Here vector J is theresultant of all nodal currents. That is, at each node J equals the sum of the currents, je,

from all the connecting elements at that node, plus the external current which is usually

zero (except for reactions to applied voltages).

As an example, consider the DC current circuit network, illustrated in Fig. 13.4.

Assembling the system shown gives the system balance equations:


! .............................................................. ! 2! *** ELEM_SQ_MATRIX PROBLEM DEPENDENT STATEMENTS FOLLOW *** ! 3! For required REAL (DP) :: S (LT_FREE, LT_FREE) ! 4! and optional REAL (DP) :: C (LT_FREE) ! 5! .............................................................. ! 6! Define any new array or variable types, then give statements ! 7

! 8! This file always need for a user defined application ! 9

!10! A conducting, convection truss member with heat generation !11! Equation: K*A U,xx - H*P (U - U_e) + Q_e = 0, !12! U = temperature, K = conductivity, A = area, h = convection , !13! P = perimeter, Q_e = heat source per unit length !14! 1 *---(K_e, A_e, h_e, P_e, U_e, Q_e)---* 2, Element in xyz !15

REAL(DP) :: L_BAR ! Length !16REAL(DP) :: K_e, A_e, h_e, P_e, U_e, Q_e ! properties !17

!18IF ( debug_el_sq .or. debug_include ) & !19

WRITE (N_BUG, *) ’Entering my_el_sq_inc’ !20!21

L_BAR = SQRT( SUM( (COORD (2, 1:N_SPACE) & !22- COORD (1, 1:N_SPACE)) **2) ) ! Length !23

K_e = GET_REAL_LP (1) ! thermal conductivity !24A_e = GET_REAL_LP (2) ! area of bar !25h_e = GET_REAL_LP (3) ! convection coefficent on perimeter !26P_e = GET_REAL_LP (4) ! perimeter of area A_e !27Q_e = GET_REAL_LP (5) ! source per unit length, BTU/ hr ft !28U_e = GET_REAL_LP (6) ! convecting temperature, F !29

!30S (1, 1) = K_e * A_e / L_BAR + h_e * P_e * L_BAR / 3.d0 !31S (2, 1) = -K_e * A_e / L_BAR + h_e * P_e * L_BAR / 6.d0 !32S (1, 2) = -K_e * A_e / L_BAR + h_e * P_e * L_BAR / 6.d0 !33S (2, 2) = K_e * A_e / L_BAR + h_e * P_e * L_BAR / 3.d0 !34

!35C (1) = (h_e * P_e * U_e + Q_e) * L_BAR / 2.d0 !36C (2) = (h_e * P_e * U_e + Q_e) * L_BAR / 2.d0 !37

Figure 13.3 General matrices for a two-node bar

j1 j2

G = 1 / R21

V1

V2

e

1

4

2

1

3

2

V4 = 0

Elem Nodes R

1 1 2 20

2 2 3 5

3 2 4 6

3

V1 = 140 v

V3 = 90 v

Network Current

4 a

6 a

10 a

Figure 13.4 A simple current driven network


1

20

−1

20

0

0

−1

20

1

20+

1

5+

1

6

−1

5

−1

6

0

−1

51

5

0

0

−1

6

0

1

6

E1 = 140E2

E3 = 90E4 = 0

=

J1

0

J3

J4

Solving the network system equilibrium yields the single unknown voltage at node 2:

(1/20 + 1/5 + 1/6) E2 = 0 + 140/20 + 90/5 or simply E2 = 60 volts. Substitutingall the voltages to determine the ‘reaction’ currents gives J1 = 4, J3 = 6, and J4 = − 10amps, respectively. That is, external current entered the system at nodes 1 and 3 and was

removed at node 4. Post-processing the results gives the current in each element. The

results, for the first node in the topology, are

j(1) = (E2 − E1) / R1 = (60 − 140)/20 = − 4, from 1 to 2,j(2) = (E3 − E2) / R2 = (90 − 60)/5 = + 6, from 3 to 2,j(3) = (E4 − E2) / R3 = (0 − 60)/6 = − 10, from 2 to 4

In a similar manner the electrical power P = EJ = J2 R can be computed at the systemlevel as a matrix product or by summing similar products at the element level. At the

system level P = ET J or P = [140 60 90 0]T [4 0 6 − 10] = 1100 watts, and thisnetwork internal power should match the sum of the power in the elements. That is,

1

1

3

4

2

V2 = 0

Elem Nodes R

1 1 2 36.

2 1 3 1.

3 2 5 2.

4 3 4 10.

5 4 5 20.

6 3 6 3.

7 4 7 5.

8 5 8 4.

9 6 7 7.

10 7 8 6.

10

9

5

8

7

6

V1 = 120 v 3 6

74

852

Figure 13.5 A voltage driven network


! .......................................................... ! 1! ** ELEM_SQ_MATRIX PROBLEM DEPENDENT STATEMENTS FOLLOW *** ! 2! .......................................................... ! 3! Define new array or variable types, then give statements ! 4! NOTE: ELEMENT REACTIONS ARE THE IN & OUT CURRENT FLOWS ! 5

REAL(DP) :: resistance, conductance ! R, 1/R ! 6! 7

! Get resistance ! 8resistance = GET_REAL_LP (1) ! real element property ! 9IF ( resistance > 0.d0 ) THEN !10

conductance = 1.d0 / resistance !11ELSE !12

PRINT *, ’WARNING: Invalid resistance, element ’, IE !13N_WARN = N_WARN + 1 ; conductance = 1.d0 !14

END IF ! valid data !15!16

! Conductance matrix !17S (1,1) = conductance ; S (2,1) = -conductance !18S (1,2) = -conductance ; S (2,2) = conductance !19

! *** END ELEM_SQ_MATRIX PROBLEM DEPENDENT STATEMENTS *** !20

Figure 13.6 A typical DC electric network element

P =eΣ Eet Se Ee

P1 = [140 60]

1

20

−1

20

−1

201

20

140

60

= 320

P2 = [60 90]

1

5

−1

5

−1

51

5

60

90

= 180

P3 = [60 0]

1

6

−1

6

−1

61

6

60

0

= 600

and the total power is P =eΣ Pe = (320 + 180 + 600) = 1100 watts, as expected.

These procedures can be used on DC systems in general, but it can become difficult to

clarify the topology and the boundary conditions. A typical implementation of a static

DC circuit is given in Fig. 13.6. When it is applied to the network in Fig. 13.5 the

computed voltages are shown in Fig. 13.7. They are listed with the element current

(element reactions) in Fig. 13.8. There we see that the external reaction currents are

11.46 amps entering at node 1 and an equal amount leaving at ground node 2. The

element reactions show how the current splits among the elements. For example at node 1


0

0.5

1

1.5

2

0

0.2

0.4

0.6

0.8

10

20

40

60

80

100

120

8

7

5

X Coordinates

6

4

FEA Solution Component_1: 10 Elements, 8 Nodes

−−−−−−min 2

3

Y Coordinates

1−−−−−−max

Com

pone

nt 1

(m

ax =

120

, min

= 0

)

Voltage driven network

Figure 13.7 Computed nodal voltage levels

*** REACTION RECOVERY *** ! 1NODE, PARAMETER, REACTION, EQUATION ! 2

1, DOF_1, 1.1461E+01 1 ! 32, DOF_1, -1.1461E+01 2 ! 4

! 5*** OUTPUT OF RESULTS IN NODAL ORDER *** ! 6

NODE, X-Coord, Y-Coord, DOF_1, ! 71 0.0000E+00 1.0000E+00 1.2000E+02 ! 82 0.0000E+00 0.0000E+00 0.0000E+00 ! 93 1.0000E+00 1.0000E+00 1.1187E+02 !104 1.0000E+00 5.0000E-01 7.3625E+01 !115 1.0000E+00 0.0000E+00 1.6255E+01 !126 2.0000E+00 1.0000E+00 9.8964E+01 !137 2.0000E+00 5.0000E-01 6.8845E+01 !148 2.0000E+00 0.0000E+00 3.7291E+01 !15

!16** ELEMENT REACTION, AND INTERNAL SOURCES ** !17ELEMENT NODE DOF REACTION ELEM_SOURCE !18

1 1 1 3.33333E+00 0.00000E+00 !192 1 1 8.12749E+00 0.00000E+00 !203 2 1 -8.12749E+00 0.00000E+00 !214 3 1 3.82470E+00 0.00000E+00 !225 4 1 2.86853E+00 0.00000E+00 !236 3 1 4.30279E+00 0.00000E+00 !247 4 1 9.56175E-01 0.00000E+00 !258 5 1 -5.25896E+00 0.00000E+00 !269 6 1 4.30279E+00 0.00000E+00 !27

10 7 1 5.25896E+00 0.00000E+00 !28

Figure 13.8 Network voltage and current results


we see that elements 1 and 2 take 3.33 and 8.13 amps (29 % and 71% ), respectively of

the incoming current.

13.4 Trusses as a network

The truss element is a very common structural member. A truss element is a "two

force member". That is, it is loaded by two equal and opposite collinear forces. These

two forces act along the line through the two connection points of the member. In

elementary statics we compute the forces in truss elements as if they were rigid bodies.

However, there was a class of problems, called statically indeterminant, that could not be

solved by treating the members as rigid bodies. With the finite element approach we will

be able to solve both classes of problems. In Sec. 7.4 the equilibrium equation for an

elastic bar was developed. Clearly, the elastic bar is a special form of a truss member. To

extend the previous work to include trusses in two- or three-dimensions basically requires

some review of analytic geometry at this point.

13.4.1 Direction cosines

Consider a directed line segment in global space going from point 1 at (x1 , y1 , z1)

to point 2 at (x2 , y2 , z2 ). Then the length of the line between the two points has

components parallel to the axes of L x = x2 − x1 , L y = y2 − y1 , Lz = z2 − z1 and thetotal length is L2 = (L2x + L2y + L2z). Specifying the end points of a line is a common way

Lx

Ly

X

Y

xL

yL

(X1, Y

1)

(X2, Y

2)

0

0

Y

X0

X

UX

UY

UL

a) Typical truss structure

b) Coordinates and displacement components

Figure 13.9 A truss structure and components


of locating its direction in space. Another common way to describe the direction is to

give the direction angles or the corresponding direction cosines. Let the direction angles

from the x- , y- , and z-axes to the line segment be denoted by θ x , θ y, and θ z , respectively.

Recall the relation between the total magnitude of a vector and its components, i.e.,

L x = L Cos θ x , etc. We generally will find the inverse geometric relation more useful.Specifically the direction cosines become: Cos θ x = L x / L , Cos θ y = L y / L , andCos θ z = Lz / L .

For two-dimensional problems we will assume that the structure lies in the global

x - y plane so that Lz = 0, Cos θ z = 0, and θ z = 90. In that special case only one angle isrequired to describe the direction rather than the usual three. It is common then to select

θ x as the required angle and to omit reference to θ y = 90 − θ x and to replace the seconddirection cosine with the relation Cos θ y = Sin θ x (for θ z = 90 ). This is illustrated inFig. 13.9. For two-dimensional problems one can utilize the simplicity of referring to a

single angle. However, if one wants to automate the analysis for two- and three-

dimensional problems then it is best in the long run to refer to the direction cosines.

To extend the bar element to a general truss element we need to consider the

relations between a local coordinate system that is parallel and perpendicular to the

element and the fixed global coordinate directions. Let the local x-axis lie along the

member, that is, it passes through the two end points of the member. This means that the

direction cosines of the local x-axis are the same as those for the line segment. The bar

element had a single displacement, u, at any point. That local displacement vector will

have components in the global space. Let the global displacements of a point be denoted

by ux , uy, and uz . To be consistent with this, one could also define three local

components of the displacement. For a bar element the local y- and z-components are

identically zero. Later we will consider members that have no zero local components.

Thus, we will consider the general case of transformation of local displacement

components. Referring to Fig. 13.9 again, one finds from geometry that the local x

displacement is related to the two-dimensional global displacements by

uxL = uxg Cos θ x + uyg Cos θ y .

Similarly, if there was a local y-component of displacement it would be related to the

global components by uyL = − uxg Cos θ y + uyg Cos θ x . Writing these identities in amatrix form in terms of θ x = θ

(13.1)

ux

uy

L

=

Cos θ

− Sin θSin θ

Cos θ

ux

uy

g

or symbolically this transformation is uL = t(θ ) ug where t is a nodal transformationmatrix and ug and uL denote the global and local displacement components, respectively,

at a point. If this relation is written at each node of the element it defines the element dof

transformation matrix, T . Specifically,


u1x

u1y

− − −u2x

u2y

e

L

=

Cos θ

− Sin θ− − − −

0

0

Sin θ

Cos θ

− − − −0

0

|

|

|

|

|

0

0

− − − −Cos θ

− Sin θ

0

0

− − − −Sin θ

Cos θ

u1x

u1y

− − − − −u2x

u2y

e

g

or(13.2)ueL = T(θ ) ueg .

The same type of coordinate transformation will apply to components of the element

force vector, Pe , namely :(13.3)PeL = T(θ ) P

eg .

Notice that the transformation matrix is square. Thus, the inverse transformation can be

found by inverting the matrix T . Therefore,

(13.4)ug = T−1 uL , Pg = T−1 PL .

If one carries out the inversion process, an interesting result is obtained. Specifically, we

find that the inverse of the transformation is the same as its transpose. This is always

true, and it makes our calculations much easier since we can write

(13.5)T−1 = TT .

A matrix with this property is called an orthogonal matrix. Therefore, the simple way to

write the inverse transformation is(13.6)ug = TT uL .

13.4.2 Transformation of element matrices

Our ultimate goal is to solve the global equilibrium equations. This requires that all

elements be referred to a single global coordinate system, and that the assembly of

element contributions be relative to that system. Therefore, before we can assemble the

element stiffness and load matrices they must be written relative to the global axes. This

means that we need to define global versions of the element matrices, say Seg and Ceg.

Clearly, they are somehow related to the corresponding local element matrices, SeL and

CeL . To gain some insight into the relation between the two systems recall that the

element behavior was defined in terms of the total potential energy, Πe , of the element.Since that quantity is a scalar, its value must be the same regardless of whether it is

computed in element coordinates or global coordinates. If we compute the total potential

energy Πe using Eq. 7.7 in local coordinates the result is

(13.7)Πe = 12

ueT

L SeL u

eL − ue

T

L CeL .

By way of comparison, if it is calculated in global coordinates

(13.8)Πe = 12

ueT

g Seg u

eg − ue

T

g Ceg .

The two forms can be more easily compared if Eq. 13.7 is also written in terms of the

global components of the displacements of the element. Before doing that, let us recall


! ........................................................... ! 1! *** ELEM_SQ_MATRIX PROBLEM DEPENDENT STATEMENTS FOLLOW *** ! 2! ........................................................... ! 3! Define any new local array or variable types, then statements ! 4

! 5! A TWO-DIMENSIONAL TRUSS BY DIRECT ENERGY APPROACH ! 6! ELEMENT REAL PROPERTIES: (1) = AREA, (2) = ELASTIC MODULUS ! 7! (3) = TEMP RISE, (4) = COEFF EXPANSION, (5) = WEIGHT DENSITY ! 8

! 9REAL(DP) :: X_I, X_J, Y_I, Y_J ! coordinates !10REAL(DP) :: D_X, D_Y, BAR_L ! lengths !11REAL(DP) :: DELTA_T, ALPHA ! temp rise, expansion !12REAL(DP) :: AREA, GAMMA ! area, wt. density !13REAL(DP) :: M_E, THERMAL ! modulus, thermal strain !14REAL(DP) :: C_X, C_Y ! direction cosines !15

!16! Get geometry !17

X_I = COORD (1, 1) ; X_J = COORD (2, 1) !18Y_I = COORD (1, 2) ; Y_J = COORD (2, 2) !19

!20! Get properties for this element !21

AREA = GET_REAL_LP (1); M_E = GET_REAL_LP (2) !22DELTA_T = GET_REAL_LP (3); ALPHA = GET_REAL_LP (4) !23GAMMA = GET_REAL_LP (5) !24

!25! Find bar length and direction cosines !26

D_X = X_J - X_I ; D_Y = Y_J - Y_I ! lengths !27BAR_L = SQRT (D_X * D_X + D_Y * D_Y) ! total length !28C_X = D_X / BAR_L ; C_Y = D_Y / BAR_L ! cosines !29

!30! Form global strain-displacement matrix !31

B (1, :) = (/ - C_X, - C_Y, C_X, C_Y /) / BAR_L !32!33

! Form global stiffness, S = B’ EAL B !34S = M_E * AREA * BAR_L * MATMUL ( TRANSPOSE (B), B ) !35

!36! Initial (thermal) strain loading !37

THERMAL = ALPHA * DELTA_T ! strain !38C = B (1, :) * M_E * THERMAL * AREA * BAR_L ! force !39

!40! Weight load, in negative Y-direction (wt density * volume) !41

C = C + (/ 0.d0, -0.5d0, 0.d0, -0.5d0 /) & ! components !42* GAMMA * AREA * BAR_L ! total weight !43

!44! Save for stress post-processing (set post_el in keywords) !45

IF ( N_TAPE1 > 0 ) WRITE (N_TAPE1) M_E, B, THERMAL !46! End of application dependent code !47

Figure 13.10 A truss element stiffness and loads

the form of the element stiffness and load matrices for a bar parallel to the x-axis :

SeL =Ee Ae

Le

1

−1−1

1

, CeL =

C1x

C2x

where C1 and C2 represent the resultant loads along the local x-axis. Since the global

structure will have two displacements per node it will be useful to rewrite the element

equations in terms of two local displacements per node. Specifically, the expanded


element equations for the equilibrium of a single element are

Ee Ae

L

1

0

−10

0

0

0

0

−10

1

0

0

0

0

0

u1x

u1y

u2x

u2y

e

L

=

Ce1x

0

Ce2x

0

L

.

Note that the stiffness matrix has been expanded by adding rows and columns of zeros to

correspond to the local y displacement. That was done because the element cannot resist

loads in the local y direction. The above expanded element matrices would be substituted

into Eq. 13.7. Next, substituting the transformation identity of Eq. 13.2 into Eq. 13.7

yields Πe = 12

ueT

g (TeT SeL T

e) ueg − ueg(TeT Ceg) . Comparing this scalar with the same

quantity in Eq. 13.8 gives the desired identities

(13.9)Seg = TeT SeL T

e , and Ceg = TeT CeL .

Of major importance here is that Eq. 13.9 is not restricted to truss elements. For certain

types of elements it would be simpler to form the global element matrices numerically by

matrix multiplication. For the truss element in two dimensions the products in these

transformations are easily written out. The results are

(13.10)Se =Ee Ae

Le

λ λ

λ µ

− λ µ− λ µ

λ µ

µ µ

− λ µ− µ µ

− λ µ− λ µ

λ λ

λ µ

− λ µ− µ µ

λ µ

µ µ

e

and(13.11)Ce

T = λC1x − µC1x λC2x − µC2x e

where λ = Cos θ x = L x / L, µ = Cos θ y = L y / L = Sin θ x . A similar set of transformedglobal stiffness and force vectors can be obtained for a truss element located in three-

dimensional space.

13.4.3 Direct energy approach

The above transformation process is valid in many structural applications and is the

usually way to see truss, frame, plate, and shell elements developed. The truss element

can be expressed directly from the energy approach and this leads to a simpler program.

In Chapter 3 on variational methods we saw that we could define the work and energy

terms in a general integral form. For an axially loaded bar the stiffness and load matrices

are given in Eqs. 3.20-22 and the initial thermal strain effects are in Eq. 3.32. Those

equations need the strain-displacement matrix Be expressed in the global coordinate

system. Here we need to rotate the bar and express its behavior in terms in terms of four

displacement components, instead of two as before. The two axial displacements are

related to the four truss member displacements by a sub-set of Eq 13.2, namely

uaxial = ττ eue


u1x

u2x

e

axial

=

Cos θ

0

Sin θ

0

0

Cos θ

0

Sin θ

u1x

u1y

u2x

u2y

e

g

Thus we can get the global Be matrix as εε = Baxialuaxial = (Baxialττ e)ue = Beue whichsimplifies to

Be = [ − Cos θ − Sin θ Cos θ Sin θ ] / Le

Assuming constant properties we can write the stiffness matrix by inspection as the

matrix product

Se = BeT

Ee Be Ae Le

and the truss member load vector due to any temperature rise (from stress free) is

Cet = BeT Ee α e ∆t Ae Le .

Another common loading condition is the member weight. Here we just need to put half

the weight at each end node and assume that Y is vertical so there are no X components

of this load. We get the weight from the weight density, γ , times the member volume.

This form is much simpler to program than the previous one. A typical implementation is

shown in Fig. 13.10.

The calculation of the stiffness, thermal and weight effects should be clear. In this

case Be has only one row but it usually has one per spatial dimension. The last action, in

line 46, is to save data necessary to recover the strains and stresses in each element. It

gets activated if the data keyword "post_el" is present. If it is present then each element

gets post-processed as shown in Fig. 13.11. There the same data (modulus of elasticity,

Be matrix, and initial strains) are recovered from sequential storage. Multiplying Be by

the gathered displacements yields the mechanical strains, in line 25. We inv oke Hooke’s

law, generalized to include initial strains, to get the one stress component, in line 29, and

then present the three items for output.

The text by Logan [9] gives a detailed hand calculation of a two bar truss with one

vertical heated member, one inclined member with a slope of 5:-4, and no external loads

applied. The two bottom nodes are pinned, while the top one is prevented from

horizontal motion. The input data are shown in Fig. 13.12 and the computed result was

exact, as summarized in the output of Fig. 13.13, and illustrated in Fig. 13.14. Of course,

the heated vertical member was found to be in compression and the unheated inclined

member was in tension.


! .............................................................. ! 1! *** POST_PROCESS_ELEM PROBLEM DEPENDENT STATEMENTS FOLLOW *** ! 2! .............................................................. ! 3! Define any new array or variable types, then give statements ! 4

! 5! A TWO-DIMENSIONAL TRUSS BY DIRECT ENERGY APPROACH ! 6! ELEMENT REAL PROPERTIES: (1) = AREA, (2) = ELASTIC MODULUS ! 7! (3) = TEMP RISE, (4) = COEFF EXPANSION, (5) = WEIGHT DENSITY ! 8

! 9! STRESS = M_E * (MECHANICAL STRAIN - INITIAL STRAIN) !10

!11REAL(DP) :: THERMAL ! initial strain !12REAL(DP) :: M_E ! modulus of elasticity !13LOGICAL, SAVE :: FIRST = .TRUE. ! printing !14

!15IF ( FIRST ) THEN ! first call !16

FIRST = .FALSE. ; WRITE (6, 5) ! print headings !175 FORMAT (’ E L E M E N T S T R E S S E S’, /, & !18& ’ ELEMENT STRESS MECH. STRAIN THERMAL STRAIN’) !19

END IF ! first call !20!21

!--> Read stress strain data from N_TAPE1 (set by post_el) !22READ (N_TAPE1) M_E, B, STRAIN_0 (1) ! THERMAL = STRAIN_0 !23

!24!--> Calculate mechanical strain, STRAIN = B * D !25

STRAIN (1) = DOT_PRODUCT ( B(1, :), D ) !26!27

!--> Generalized Hooke’s Law !28STRESS (1) = M_E * (STRAIN (1) - STRAIN_0 (1)) !29

!30WRITE (6, 1) IE, STRESS (1), STRAIN (1), STRAIN_0 (1) !311 FORMAT (I5, 3ES15.5) !32

! *** END POST_PROCESS_ELEM PROBLEM DEPENDENT STATEMENTS *** !33

Figure 13.11 Post-processing the truss


title "Logan 3rd Ed. thermal loaded truss" ! 1nodes 3 ! Number of nodes in the mesh ! 2elems 2 ! Number of elements in the system ! 3el_real 5 ! Number of real properties per element ! 4dof 2 ! Number of unknowns per node ! 5el_nodes 2 ! Maximum number of nodes per element ! 6space 2 ! Solution space dimension ! 7b_rows 1 ! Number of rows in the B (operator) matrix ! 8shape 1 ! Element shape, 1=line, 2=tri, 3=quad, 4=hex ! 9post_el ! Require post-processing, create n_tape1 !10remarks 12 ! Number of user remarks !11quit ! keyword input, begin remarks !12Logan Example 15.3, 2-D Truss with temperature rise !13occuring in bar (1) only !14

1 o Y_Roller E = 30,000 ksi, A = 2 inˆ2 !15| \ 8 ft high, 6 ft wide (96 by 72) !16

(1) (2) alpha = 7e-6, rise = 75 F !17| \ Y_1 = 0.0333 inch !18

2 * * 3 Stress: -5,333, + 6,666 psi !19Pin Pin !20

ELEMENT REAL PROPERTIES: !21(1) = AREA, (2) = MODULUS OF ELASTICITY, !22(3) = TEMP RISE, (4) = COEF THERMAL EXPANSION, !23(5) = WEIGHT DENSITY !241 10 0.0 96.0 ! node, bc flag, x, y !252 11 0.0 0.0 !263 11 72.0 0.0 !27

1 1 2 ! element, two nodes !282 1 3 !29

1 1 0.0 ! node, direction, given displacement !302 1 0.0 !312 2 0.0 !323 1 0.0 !333 2 0.0 !34

1 2. 30.e6 75. 7.e-6 0. ! elem, properties !352 2. 30.e6 0.0 7.e-6 0. ! elem, properties !36

Figure 13.12 A simple thermally loaded truss


1, DOF_1, -8.0000E+03 1 ! 32, DOF_1, 0.0000E+00 3 ! 42, DOF_2, 1.0667E+04 4 ! 53, DOF_1, 8.0000E+03 5 ! 63, DOF_2, -1.0667E+04 6 ! 7

! 8*** OUTPUT OF RESULTS IN NODAL ORDER *** ! 9NODE, X-Coord, Y-Coord, DOF_1, DOF_2, !10

1 0.0000E+00 9.6000E+01 0.0000E+00 3.3333E-02 !112 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 !123 7.2000E+01 0.0000E+00 0.0000E+00 0.0000E+00 !13

!14E L E M E N T S T R E S S E S !15

ELEMENT STRESS MECH. STRAIN THERMAL STRAIN !161 -5.33333E+03 3.47222E-04 5.25000E-04 !172 6.66667E+03 2.22222E-04 0.00000E+00 !18

Figure 13.13 Output summary for the two bar truss


−20 0 20 40 60 80 100

0

10

20

30

40

50

60

70

80

90

100

X

YFE Deformed Mesh Geometry (solid): Scale = 200

1

2 3

1 2

Pin Pin

Y Roller

Temperature rise

Figure 13.14 Deformation of the two bar truss

PY

PY

PX

2 @ 4 = 8 4

1

1

22

3

1

2

33'

2

1

b) Symmetric load & shapea) Unsymmetric load, symmetric shape

Figure 13.15 A three-bar truss structure


13.4.4 Example structure calculations

Consider the example three bar truss shown in Fig. 13.15. Assume that all three

members have the same area and modulus of elasticity. The structure is described by

Element L x L y L Topology EA

1 4 3 5 1 2 1000

2 8 0 8 1 3 1000

3 4 −3 5 2 3 1000

The structure is pinned at node 1 and on a horizontal roller at node 3. No distributed

loads or thermal loads are considered on the bars. Thus, for each element Ce = 00. Onlynodal loads are externally applied. Their values, at node 2 are P x = 10, and P y = − 20.From Eq. 13.11 the element stiffness matrices, when transformed to the global axes, have

the values ofe = 1 : Global

Se =1000

125

16

12

−16−12

12

9

−12−9

−16−12

16

12

−12−912

9

1

2

3

4

e = 2 : Global

Se =1000

512

64

0

−640

0

0

0

0

−640

64

0

0

0

0

0

1

2

5

6

e = 3 : Global

Se =1000

125

16

−12−16

12

−129

12

−9

−1612

16

−12

12

−9−12

9

3

4

5

6

.

The assembled system equilibrium equations are

(128 + 125)

symmetric

96

72

−128−96

(128 + 128)

−96−72

(96 − 96)(72 + 72)

−1250

−12896

(125 + 128)

0

0

96

−72−96

72

u1

v1

u2

v2

u3

v3

=

0

0

P x

P y

0

0

+

R1

R2

0

0

0

R3

.

However, three displacements (u1, v1 , and v3) are prescribed to be zero. Modifying the

above equations to include the boundary conditions reduces them to


256

Sym.

0

144

−12896

253

u2

v2

u3

=

P x

P y

0

.

These equations can be inverted by hand yielding:

u2

v2

u3

=1

4. 608 × 106

27216

−1228818432

48384

−24576

Sym.

36864

P x

P y

0

.

Substituting the given load values yields [u2 v2 u3] = [0. 1124 − 0. 2367 0. 1467] .The reader should verify that substituting these three displacements into the original

equilibrium assembly yields reaction values of R1 = − 10. 00, R2 = 6. 25 andR3 = 13. 75. Thus, the resultant values are equal and opposite to the applied loads asexpected. Of course, they also satisfy moment equilibrium at all points in the plane.

Note that if P x had been zero, then u2 = u3 / 2 = 0. 0533, and v2 = − 0. 21. That is, thedeformations would have been symmetric with respect to the center of the truss.

The concepts of symmetry and anti-symmetry are often useful in finite element

analysis. It is common to find half, quarter, or one-eighth order symmetry conditions that

can reduce the analysis cost to the square of the corresponding fractional part of a total

analysis cost. For a truss we have no rotational degrees of freedom so we only have to

consider the displacement components tangent or normal to the symmetry plane. Here

we will apply symmetry to the above truss. First, we view the loads, members, and

supports as viewed relative to a mirror placed at the symmetry section. The resulting

partial model is shown in Fig. 13.15. The applied loads and the stiffness of members

lying in the symmetry plane are reduced by half. The nodes or member midpoints that lie

in the symmetry plane are allowed to move only in that plane. Any supports that are not

on the symmetry plane can be modified to support the structure in a consistent manner

when viewed from the symmetry plane. This means that our simplified structure can be

described as

Element L x L y L Topology EA

1 4 3 5 1 2 1000

2 4 0 4 1 3 1000

The stiffness for the third element is no longer needed. The first member is unchanged.

The length of the second member is cut in half so its stiffness doubles. The assembled

elements give the algebraic equations of equilibrium as:


! *** ELEM_SQ_MATRIX PROBLEM DEPENDENT STATEMENTS FOLLOW *** ! 1! ........................................................... ! 2

! 3! APPLICATION: TWO-DIMENSIONAL TRUSS. ELEMENT REAL PROPERTIES: ! 4! (1) = AREA, (2) = MODULUS OF ELASTICITY, ! 5! (3) = TEMP RISE, (4) = COEF THERMAL EXPANSION ! 6

! 7REAL(DP) :: X_I, X_J, Y_I, Y_J ! coordinates ! 8REAL(DP) :: D_X, D_Y, BAR_L ! lengths ! 9REAL(DP) :: DELTA_T, ALPHA ! temp rise, coeff !10REAL(DP) :: AREA ! area !11REAL(DP) :: M_E ! modulus of elasticity !12REAL(DP) :: C_X, C_Y, C_XX, C_XY, C_YY ! cosines & products !13REAL(DP) :: F, STIF ! forces, stiffness !14REAL(DP) :: THERMAL ! expansion !15

!16! Get geometry !17

X_I = COORD (1, 1) ; X_J = COORD (2, 1) !18Y_I = COORD (1, 2) ; Y_J = COORD (2, 2) !19

!20! Get properties for this element !21

AREA = GET_REAL_LP (1) ! area !22M_E = GET_REAL_LP (2) ! elastic modulus !23DELTA_T = GET_REAL_LP (3) ! temperature rise !24ALPHA = GET_REAL_LP (4) ! coeff thermal expansion !25

!26!--> FIND BAR LENGTH AND DIRECTION COSINES !27

D_X = X_J - X_I ; D_Y = Y_J - Y_I ! lengths !28BAR_L = SQRT (D_X * D_X + D_Y * D_Y) ! total length !29C_X = D_X / BAR_L ; C_Y = D_Y / BAR_L ! cosines !30

!31STIF = M_E * AREA / BAR_L ! AXIAL STIFFNESS, K=E*A/L !32

!33!--> TRANSFORM TO 2-D STIFFNESS (closed form) !34

C_XX = C_X * C_X ; C_XY = C_X * C_Y ; C_YY = C_Y * C_Y !35!36

S (1, 1) = STIF * C_XX ; S (2, 1) = STIF * C_XY !37S (3, 1) = - STIF * C_XX ; S (4, 1) = - STIF * C_XY !38S (1, 2) = STIF * C_XY ; S (2, 2) = STIF * C_YY !39S (3, 2) = - STIF * C_XY ; S (4, 2) = - STIF * C_YY !40S (1, 3) = - STIF * C_XX ; S (2, 3) = - STIF * C_XY !41S (3, 3) = STIF * C_XX ; S (4, 3) = STIF * C_XY !42S (1, 4) = - STIF * C_XY ; S (2, 4) = - STIF * C_YY !43S (3, 4) = STIF * C_XY ; S (4, 4) = STIF * C_YY !44

!45! Form any local loads !46

C = 0.d0 ; THERMAL = 0.d0 ! initialize !47!48

IF ( DELTA_T /= 0.d0 ) THEN ! THERMAL STRAIN EFFECTS !49THERMAL = ALPHA * DELTA_T ! thermal strain !50F = M_E * THERMAL * AREA ! thermal force !51C (1) = - C_X * F ; C (2) = - C_Y * F ! components !52C (3) = C_X * F ; C (4) = C_Y * F ! components !53

END IF ! thermal !54!55

! End of application dependent code !56

Figure 13.16 Hard coding the 2-D truss element


(128 + 250)

symmetric

96

72

−128−96128

−96−72

96

72

−2500

0

0

250

0

0

0

0

0

0

u1

v1

u2

v2

u3′

v3′

=

0

R1

R2

P y / 2R3′

0

.

Points in the plane of symmetry must always move in that plane. Thus, u3′ = 0.Conversely, node 1 must be able to move normal to the plane of symmetry. Thus, u1 ≠ 0.Node 2 has an external load, P, applied tangent to the plane of symmetry. Thus, it must

be allowed to move tangent to the plane (v2 ≠ 0) so that the force can do work on thestructure. That is, in a given direction one can specify either the force or the

displacement at a point, but not both. Clearly, node 1 has an unknown reaction that is

parallel to the symmetry plane. Thus, it must be restrained in that direction, v1 = 0. Therestrained structural stiffness is

378

sym.

−9672

0

0

0

u1

v2

v3′

=

0

−20/20

.

However, these equations are still singular after the application of the usually symmetric

conditions. Note that the third row and column are zero. This means that there is no

stiffness associated with the displacement v3′ . From the original structure in Fig. 13.15,

we note that the center of member 2 must have a zero vertical deflection. Employing this

additional physical insight, we can now also state that v3′ = 0. Therefore, for a symmetricstructure with symmetric loads the equilibrium equations, relative to the plane of

symmetry are

378

−96−96

72

u1

v2

=

0

−10

,

u1

v2

=

−0. 05333−0. 21

as before, except for the sign change on u1. This solution shows that for the above

example there are only two degrees of freedom required when symmetry is available

versus the three that were used before. For this simple example there was not much

difference in the computational effort required in the symmetric and non-symmetric

solutions. However, if there are hundreds or thousands of symmetric elements then the

cost saving is very significant when a symmetric analysis can be utilized.

Since the two-dimensional truss is relatively simple it is easy to avoid the matrix

operations used in Fig. 13.10 to form the stiffness matrix and load vectors. Several

authods have giv en the explicit algebraic form of those matrices. For completeness we

also illustrate that approach by implementing Eqs. 13.10-11 in Fig. 13.16.


11

6

7 8

9

2 1

54

3

1

3

4

2

10

L H

LL1 3

2 4 5

6

P5

P3

1 2

34

P3

L

P 13.1 P 13.2

P2

Figure 13.17 Typical planar truss geometries

0 2 4 6 8 10 12 14 16 18 20

−2

0

2

4

6

8

10

12

X for 11 Elements (with 2 nodes)

Y fo

r 6

Nod

es

Deformed Mesh (solid): Components [0.0016667, −0.0044013], Scale = 200

1

2

3

4 5

6 1 2

3 4 5 6 7 8 9

10

11

P

Figure 13.18 Meek’s deformed planar truss example


13.7 Example truss solutions

The truss P13.1 in Fig. 13.17 was used as an example problem by Meek [11]. It has

two L = 10 inch bays with members made of steel, E = 30, 000 ksi, and constant cross-sections of A = 1 in2. The bottom center load is P3 = 10, 000 lb, P5 = 0, and the left andright corners are supported with a pin and roller, respectively. Its deformed shape is

shown in Fig. 13.18. To compute the deflections and the reactions one needs data such as

those listed in Fig. 13.19 to produce the selected outputs in Fig. 13.20 that correspond to

the deformed plot. The latter output figure also includes the member stresses at their

center (positive is tension). No member line loads are present so there is no local bending

and they are pure axial loads of a typical truss. In other words, both sides of the element

see the same constant stress in this example.

Of course, you can use a truss calculation to also solve simple axial beam problems

simply yo setting the y-coordinates and y-displace,ents to zero. Logan [9] gives an axial

bar thermal loading example that can be used to validate the truss code. Employ two

equal length co-linear truss elements to model a bar that is fixed at both ends and

undergoes a uniform temperature increase. Assume numerical values of

E = 30, 000 ksi, A = 4 in2, L = 4 ft = 48 inches, α = 7e − 6 1/F , and a temperature rise of50 F . We want to obtain the displacements (null), reactions (42, 000 lb), and stress. The

input (lines 1-52) and output (lines 53-107) for this problem are given in Fig. 13.22. Note

that all displacements are zero (lines 77-81), but the system reactions are not (lines

67-75) are not, nor are the individual element reactions (lines 83-100). The latter show

the member to be in compression as do the member stresses (lines 102-106).

13.8 Introduction to beams

A common structural system considered in engineering is that of the elastic beam.

Such a beam is shown in Fig. 13.23. In mechanics of materials a number of common

assumptions are made in order to reduce the analysis to a one-dimensional formulation.

The most common assumption is that planes in the beam, normal to a fiber along the

x-axis, remain normal to that fiber in its deformed state. This assumption makes that

axial displacement, u, and the axial strain, ε , vary linearly with the transverse coordinate,

y. Let v denote the transverse displacement, and θ = v′ the slope of the beam. Then theaxial displacement relation, for small slopes, and the axial strain are

(13.12)u(x, y) = − yv′ = − ydv

dx,

(13.13)ε (x, y) =du

dx= − y

d2v

dx2= − yv′′ ,

respectively. For an elastic material the stress, σ , is defined by Hooke’s law as

(13.14)σ (x, y) = E(x) ε (x, y)

where E is the elastic modulus of the material. These quantities could vary with both of

the spatial coordinates. We desire to formulate a one-dimensional model. We will define


title "Sample Meek 2-D truss" ! 1nodes 6 ! Number of nodes in the mesh ! 2elems 11 ! Number of elements in the system ! 3el_homo ! Element properties are homogeneous ! 4el_real 7 ! Number of real properties per element ! 5dof 2 ! Number of unknowns per node ! 6el_nodes 2 ! Maximum number of nodes per element ! 7space 2 ! Solution space dimension ! 8b_rows 1 ! Number of rows in B (operator) matrix ! 9shape 1 ! Element shape, 1=line, 2=tri, 3=quad !10loads ! An initial source vector is input !11el_react ! Compute & list element reactions !12post_el ! Require post-processing !13example 206 ! Source library example number !14data_set 1 ! Data set for example (this file) !15remarks 15 ! Number of user remarks !16quit ! keyword input, begin remarks !17

2 4 5 Meek’s Example 7.2 truss !18*--(10)-*--(11)-* E = 30,000 ksi, A = 1 inˆ2 !19|\(4) /|\(7) /| Two 10 inch bays !20| \ / | \ / | Vertical deflection at 3 !21

(3) X (6) X (9) is -4.4013E-03 inches !22| / \ | / \ | !23|/(5) \|/(8) \| Reactions 5K each at 1, 6 !24

1 #--(1)--*--(2)--o 6 !25Pin 3| Roller !26

v P=10K !27ELEMENT REAL PROPERTIES: !28(1) = AREA, (2) = MODULUS OF ELASTICITY, !29(3) = TEMP RISE, (4) = COEF THERMAL EXPANSION, !30(5) = LINE LOAD, (6) = MOMENT OF INERTIA, !31(7) = HALF DEPTH OF BAR !321 11 0.0 0.0 ! node, bc flag, x, y !332 00 0.0 10.0 !343 00 10.0 0.0 !354 00 10.0 10.0 !365 00 20.0 10.0 !376 01 20.0 0.0 !38

1 1 3 ! element, two nodes !392 3 6 !403 1 2 !414 2 3 !425 1 4 !436 3 4 !447 4 6 !458 3 5 !469 5 6 !47

10 2 4 !4811 4 5 !49

1 1 0.0 ! node, direction, displacement !501 2 0.0 !516 2 0.0 !52

1 1. 30000. 0 0 0 0 0 ! elem, A, E, null properties !533 2 -10. ! node, direction, load !546 2 0.0 ! terminate with last !55

Figure 13.19 Typical data for Meek’s truss example


title "Sample Meek 2-D truss" ! 1! 2


1, DOF_1, 4.4409E-16 1 ! 51, DOF_2, 5.0000E+00 2 ! 66, DOF_2, 5.0000E+00 12 ! 7

REACTION RESULTANTS ! 8PARAMETER, SUM POSITIVE NEGATIVE ! 9DOF_1, 4.4409E-16 4.4409E-16 0.0000E+00 !10DOF_2, 1.0000E+01 1.0000E+01 0.0000E+00 !11

!12*** EXTREME VALUES OF THE NODAL PARAMETERS *** !13PARAMETER MAXIMUM, NODE MINIMUM, NODE !14DOF_1, 1.6667E-03, 2 -1.5639E-04, 5 !15DOF_2, 0.0000E+00, 1 -4.4013E-03, 3 !16

!17*** OUTPUT OF RESULTS IN NODAL ORDER *** !18NODE, X-Coord, Y-Coord, DOF_1, DOF_2, !19

1 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 !202 0.0000E+00 1.0000E+01 1.6667E-03 -9.1153E-04 !213 1.0000E+01 0.0000E+00 7.5514E-04 -4.4013E-03 !224 1.0000E+01 1.0000E+01 7.5514E-04 -2.8910E-03 !235 2.0000E+01 1.0000E+01 -1.5639E-04 -9.1153E-04 !246 2.0000E+01 0.0000E+00 1.5103E-03 0.0000E+00 !25

!26** BEGIN ELEMENT APPLICATION POST PROCESSING ** !27

E L E M E N T S T R E S S E S !28ELEMENT MID SECTION (BENDING) STRESS AT: !29NUMBER RIGHT LEFT !30

1 0.2265409E+01 0.2265409E+01 !312 0.2265409E+01 0.2265409E+01 !323 -0.2734591E+01 -0.2734591E+01 !334 0.3867295E+01 0.3867295E+01 !345 -0.3203772E+01 -0.3203772E+01 !356 0.4530818E+01 0.4530818E+01 !367 -0.3203772E+01 -0.3203772E+01 !378 0.3867295E+01 0.3867295E+01 !389 -0.2734591E+01 -0.2734591E+01 !39

10 -0.2734591E+01 -0.2734591E+01 !4011 -0.2734591E+01 -0.2734591E+01 !41

Figure 13.20 Selected output for the Meek truss


TITLE: "Logan 3rd Ed. thermal loaded bar" ! 1! 2

**** PROBLEM CLASS: (DEFAULT) VALUE **** ! 3DIMENSION OF SPACE ............................(1) 2 ! 4NUMBER OF ROWS IN B MATRIX ....................(1) 1 ! 5NUMBER OF ITERATIONS TO BE RUN ................(1) 1 ! 6NUMBER OF NODAL POINTS IN SYSTEM ..............(2) 3 ! 7NUMBER OF ELEMENTS IN SYSTEM ..................(1) 2 ! 8NUMBER OF PARAMETERS PER NODE .................(1) 2 ! 9MAXIMUM NUMBER OF NODES PER ELEMENT ...........(2) 2 ! 10NUMBER OF DIFFERENT ELEMENT TYPES .............(1) 1 ! 11SHAPE 1=LINE 2=TRI 3=QUAD 4=HEX 5=TET 6=WEDG ..(1) 1 ! 12NUMBER OF REAL PROPERTIES PER ELEMENT .........(0) 7 ! 13HOMOGENEOUS ELEM PROPERTIES: 0=FALSE, 1=TRUE ..(0) 1 ! 14LIST ELEMENT REACTIONS: 0=OMIT, 1=LIST ..(0) 1 ! 15

! 16THE NEXT 11 LINES ARE USER REMARKS ! 171 Logan Example 15.1, Fixed bar with temperature rise ! 182 1 2 3 ! 193 *--(1)--*--(2)--* E = 30,000 ksi, A = 4 inˆ2 ! 204 L = 4 ft = 48 inches, alpha = 7e-6, rise = 50 F ! 215 Reactions are 42,000 lb, stress = 10,500 psi ! 226 End points 1 & 3 fixed ! 237 ELEMENT REAL PROPERTIES: ! 248 (1) = AREA, (2) = MODULUS OF ELASTICITY, ! 259 (3) = TEMP RISE, (4) = COEF THERMAL EXPANSION, ! 2610 (5) = LINE LOAD, (6) = MOMENT OF INERTIA, ! 2711 (7) = HALF DEPTH OF BAR ! 28

! 29*** NODAL POINT DATA *** ! 30

NODE, BC_FLAG, X-Coord, Y-Coord, ! 311 11 0.0000 0.0000 ! 322 00 24.0000 0.0000 ! 333 10 48.0000 0.0000 ! 34

! 35ELEMENT TYPE NUMBER = 1 ! 36NUMBER OF NODES PER ELEMENT .......... 2 ! 37NUMBER OF GEOMETRIC CONTROL NODES .... 2 ! 38NUMBER OF PARAMETRIC DIMENSIONS ...... 1 ! 39

! 40*** ELEMENT CONNECTIVITY DATA *** ! 41ELEMENT, 2 NODAL INCIDENCES. ! 42

1 1 2 ! 432 2 3 ! 44

! 45*** CONSTRAINT EQUATION DATA *** ! 46CONSTRAINT TYPE_ONE: (PAR_1 @ NODE_1) = A_1. ! 47EQ. NO. NODE_1 PAR_1 A_1 ! 48

1 1 1 0.00000E+00 ! 492 1 2 0.00000E+00 ! 503 3 1 0.00000E+00 ! 51

! 52

Figure 13.22a Truss model of a thermal stress bar


*** SYSTEM GEOMETRIC PROPERTIES *** ! 53VOLUME = 4.80000E+01 ! 54CENTROID = 2.40000E+01 0.00000E+00 ! 55

! 56*** ELEMENT PROPERTIES *** ! 57ELEMENT, 7 PROPERTY & REAL_VALUE PAIRS ! 581 1 4.00E+00 2 3.00E+04 3 5.00E+01 4 7.00E-06 ! 59

5 0.00E+00 6 0.00E+00 7 0.00E+00 ! 60! 61

*** INPUT SOURCE RESULTANTS *** ! 62ITEM SUM POSITIVE NEGATIVE ! 63

1 0.0000E+00 4.2000E+01 -4.2000E+01 ! 642 0.0000E+00 0.0000E+00 0.0000E+00 ! 65

! 66*** REACTION RECOVERY *** ! 67

NODE, PARAMETER, REACTION, EQUATION ! 681, DOF_1, 4.2000E+01 1 ! 691, DOF_2, 0.0000E+00 2 ! 703, DOF_1, -4.2000E+01 5 ! 71

REACTION RESULTANTS ! 72PARAMETER, SUM POSITIVE NEGATIVE ! 73DOF_1, 0.0000E+00 4.2000E+01 -4.2000E+01 ! 74DOF_2, 0.0000E+00 0.0000E+00 0.0000E+00 ! 75

! 76*** OUTPUT OF RESULTS IN NODAL ORDER *** ! 77NODE, X-Coord, Y-Coord, DOF_1, DOF_2, ! 78

1 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 ! 792 2.4000E+01 0.0000E+00 0.0000E+00 0.0000E+00 ! 803 4.8000E+01 0.0000E+00 0.0000E+00 0.0000E+00 ! 81

! 82** ELEMENT REACTION, INTERNAL SOURCES AND SUMMATIONS ** ! 83

ELEMENT 1 ! 84NODE DOF REACTION ELEM_SOURCE SUMS ! 85

1 1 4.20000E+01 -4.20000E+01 ! 861 2 0.00000E+00 -0.00000E+00 ! 872 1 -4.20000E+01 4.20000E+01 ! 882 2 0.00000E+00 0.00000E+00 ! 89

SUM: 1 0.00000E+00 0.00000E+00 0.00000E+00 ! 90SUM: 2 0.00000E+00 0.00000E+00 0.00000E+00 ! 91

! 92ELEMENT 2 ! 93

NODE DOF REACTION ELEM_SOURCE SUMS ! 942 1 4.20000E+01 -4.20000E+01 ! 952 2 0.00000E+00 -0.00000E+00 ! 963 1 -4.20000E+01 4.20000E+01 ! 973 2 0.00000E+00 0.00000E+00 ! 98

SUM: 1 0.00000E+00 0.00000E+00 0.00000E+00 ! 99SUM: 2 0.00000E+00 0.00000E+00 0.00000E+00 !100

!101E L E M E N T S T R E S S E S !102

ELEMENT MID SECTION STRESS AT: !103NUMBER RIGHT LEFT !104

1 -0.1050000E+02 -0.1050000E+02 !1052 -0.1050000E+02 -0.1050000E+02 !106

ELEMENT POST PROCESSING COMPLETE. !107

Figure 13.22b Truss model of a thermal stress bar


p(x)

x

dv / dxv(x)EI

v(x)

u(x)

x

dF

xdM (x)

stress

dy

y

Figure 13.23 A small deflection elastic beam

a generalized strain and a generalized stress to accomplish this goal.

From calculus we should recall that the quantity v′′ (x) = 1 / ρ is known as thecurvature of the deflected beam and ρ is the radius of curvature. The signs in

Eqs. 13.12-13 have been chosen so that a positive sign denotes tension. From statics one

can show that the resultant axial load from the distributed stress is zero. However, there

is a non-zero resultant moment. Its value is given by

M(x) = ∫ dm = ∫ − y dF = ∫ − y σ da = ∫ − y E ε da =A

∫ E y2 v′′ da = EI v′′ (x)where A is the cross-sectional area of the member and

I (x) =A

∫ y2 dais the moment of inertia of the cross sectional area. We will call this moment our


generalized stress measure since it only depends on x. In mechanics of materials the

deflections of the beam are determined by solving the differential equation of

equilibrium :d2

dx2EI

d2v

dx2

= p(x)

where p(x) is the distributed transverse load per unit length. The four constants of

integration are determined by satisfying the boundary conditions on the deflections, v,

and slopes, θ . Howev er, in our present study we need an integral formulation for our

finite element model, that is equivalent to the solution of the differential equation.

13.9 Variational procedure

A variational formulation for the elastic beam is related to minimizing the total

energy and work in the system. One important term required for the analysis is the strain

energy. That quantity is defined as half the volume integral of the product of the stress

and strain. Here we wish to reduce this quantity to a function of x alone. The scalar

strain energy is

(13.15)U =1

2

V

∫ σ ε dV = 12L

∫A(x)

∫ (E y v′′ ) ( y v′′ ) da dx .

Since only v′′ depends on x this reduces to

(13.16)U =1

2

L

∫ E(x) I (x) (v′′ )2 dx = 12L

∫ M(x) v′′ (x) dx .Comparing Eqs. 13.15-16 suggests that we should select the curvature, v′′ , as ourgeneralized strain measure. Having made this choice we can use Eq. 13.15 to define a

generalized constitutive relation. Define

(13.17)σ = M(x) , ε = v′′ (x) , σ = E ε , E = E(x) I (x)

as the generalized Hooke’s law. The left hand side of the last three equations have been

defined as arrays even though they only contain a single term. This is done to give some

insight into what would happen for a plate or shell where there would be three curvatures

of the surface, three corresponding moments, and E would become a 3 × 3 arrayinvolving the material properties and thickness of the section. If we consider a beam of

width b and thickness h then Eq. 13.17 could be written as E = Eh3b / 12. For a plate thegeneralized stresses and strains would be

σ T = [ M xx M yy M xy ] , ε T = [ w, xx w, yy w, xy ]

where w is the transverse displacement of the plate. The moments are written on a unit

length basis (b = 1), so for the plate

E =Eh3

12(1 − ν 2)

1

0

0

0

1

0

0

0

(1 − ν ) / 2

.

While we do not plan to consider plates here we will use the generalized symbolism to

give insight to such problems.


Equation 13.17 shows that our selections for generalized stress-strain measures will

correctly define the strain energy in the system. Next, we need to define the work done

by the applied loads, Pi, or couples, Ci. The work done by a transverse force is the

product of the force and the transverse displacement. Likewise, the work done by a

couple is the product of the couple and rotation (slope) at its point of application. These

contributions define a work term, W , giv en by

W =L

∫ v(x) p(x) dx +iΣ v(xi ) Pi +

jΣ v′(x j) C j .

The last two terms represent work done by concentrated point loads or couples. Thus, the

total potential energy, Π = U − W is

(13.18)Π =1

2

L

∫ EI ( v′′ (x) )2 dx −L

∫ v(x) p(x) dx −iΣ vi Pi −

jΣ v′j C j .

To determine the displacement field, v(x), that corresponds to the equilibrium state we

must minimize Π and satisfy the boundary conditions on v and v′ = θ .

13.10 Hermite element matrices

To introduce our finite elements we select a series of line segments to make up the

region L. There are numerous elements that could be selected. First we will select an

element with two nodes. Next, it is necessary to assume a displacement approximation so

we can evaluate the potential energy in Eq. 13.16. That equation contains second

derivatives and thus we need to assume a solution for v that will at least have both the

deflection, v, and the slope, v′, continuous between elements. The most commonassumption is to select the cubic Hermite polynomial presented in Fig. 3.6. The

unknowns at each of the two element nodes are v and v′ = θ . These quantities will becalled our generalized displacements or the generalized degrees of freedom. Thus, our

element interpolation functions are the Hermite form in Fig. 3.6:

v(x) =

H e1(x) He2(x) H

e3(x) H

e4(x)

v1

v′1v2

v′2

e

or v(x) = He(x) δδ e, where δδ e denotes the generalized displacements of the element. Thederivatives of the displacements are

(13.19)v′(x) = θ (x) = He′(x) δδ e , v′′ (x) = He′′ (x) δδ e .

Since v′′ and δδ e have been selected as our generalized strains and generalizeddisplacements we will use the notation of Eq. 13.19 and write Eq. 13.17 as

ε e = Be δδ e

where ε = v′′ in our present study. In the study of plates and shells additional curvatureterms would be present in ε . Employing our generalized notation the stiffness matrix and

distributed load vector can be written by inspection as


Ke =Le∫ Be(x)T De(x) Be(x) dx , Fep =

Le∫ He(x)T pe(x) dx .

Here we will again use unit coordinates on the element and set r = x / Le so thatd( ) / dx = d( ) / dr × 1 / Le. Thus,

Be = He′′ =1

L2

d2H

dr2

so for the cubic Hermite in Fig. 3.6 this becomes (with L = Le)

Be =1

L2

(12r − 6) L(6r − 4) (6 − 12r) L(6r − 2)

.

Recalling that

L

∫ rm dx =L

(m + 1)

and assuming that Ee is a constant then the stiffness (with L = Le) is

(13.20)Ke =EI

L3

12

6L

−126L

4L2

− 6L2L2

12

−6L

sym.

4L2

.

If the lateral load, pe, is constant then

(13.21)Fep = peLe∫ HeT dx = pe Le

1

0

∫ HeT (r) dr = pe Le

1 / 2Le

T

/12

1 / 2−Le

T

/12

.

Note that the distributed load puts half the resultant load at each end. It also causes equal

and opposite nodal couples at each of the two nodes.

When we wrote Eq. 13.18 we assumed that point loads would only be applied at the

node points. This may not always be true and we should consider such a load condition

as a special case of a distributed load. In that case the length of the distributed load

approaches zero and the magnitude of the force per unit length approaches infinity, but

the resultant load P is constant. That is, we define the load to be p(x) = P δ (x − x0)where δ is the Dirac Delta distribution. Then the generalized load vector is

Fep =Le∫ HeT (x)P δ (x − x0) dx

which is integrated by inspection by using the integral property of the Dirac Delta to yield

Fep = P He(r0) where r0 = x0 / L is the point of application of the load. To check this

concept, assume that the load is at Node 1. Then, r0 = 0 so thatFe

T

p = P [ 1 0 0 0 ]

as expected. That is, all the force goes into Node 1 and no element nodal moments are

generated. Other common loading conditions can be treated in the same way and a

number have been tabulated by Akin [1] and others. As another example, if p(x) varies

linearly from pe1 to pe2 at the nodes of element e then


p(x) = (1 − x / Le) pe1 + x / Le pe2 = [ (1 − r) r ]

p1

p2

e

and

(13.22)Fep =Le∫

1 − 3r2 + 2r3

Le(r − 2r2 + r3)3r2 − 2r3

Le(r3 − r2)

p(x) dx =Le

20

7

Le

3

−2Le/3

3

2Le/3

7

−Le

p1

p2

e

.

If the load is constant so that pe1 = pe2 = pe, then this reduces to Eq. 13.21, as expected.Likewise, if pe1 = 0 and pe2 = p, this defines a triangular load and

(13.23)FeT

p =PL

20[ 3 2L/3 7 − L ] .

It is common to tabulate such results in terms of an applied unit resultant load. That

resultant isRe =

Le∫ pe(x) dx .

For common load variations, such as constant, linear, parabolic, and cubic forms where p

varies in proportion to rn, the resultant loads are Re = pL / (n + 1). The location, x, ofthe resultant applied load is found from

x Re =Le∫ px dx

and the corresponding results are x = L(n + 1) / (n + 2). Thus, if we normalize Eq. 13.23by dividing the resultant load, pL / 2, the result is

feT

p = [ 3/10 L/15 7/10 − L/10 ] .

We can also check the unit load results by applying statics to the data in that figure. To

check the load summary, we first take the sum of the moments about Node 1. This gives

+ 1 = 0 + (7/10)L + L/15 − L/10 = L(21 + 2 − 3) / 30 , OK .

Similarly, the sum of the moments about Node 2 is verified.

13.11 Cantilever with triangular load

To present an analytic example of this element consider a single element solution of

the cantilever beam shown in Fig. 13.24 to determine the deflection and slope at the free

end. Usually the deflected shape of a beam is defined by a fourth or fifth order

polynomial in x. Thus, our cubic element solution will usually give only an approximate

solution. For a single element the system equations are

EI

L3

12

6L

− − − −−126L

6L

4L2

− − − −−6L2L2

|

|

|

|

|

−12−6L

− − − −12

−6L

6L

2L2

− − − −−6L4L2

v1

θ 1

− − −v2

θ 2

=WL

2

3/10

L/15

− − − −7/10

−L/10

+

0

0

F2

M2

.


L

EI

w

v = w L4 / 30 E I

slope = - w L3 / 24 E I

3 R / 10

7 R / 10

R L / 15 R L / 10

R = w L / 2

Figure 13.24 A single element approximate solution

The right side support requires that v2 = 0 = θ 2. The reduced equations become

EI

L3

12

6L

6L

L2

v1

θ 1

=WL

2

3/10

L/15

+

0

0

so that

v1

θ 1

=L3

12 EIL2

4L2

−6L−6L

12

3/10

L/15

WL

2=

WL3

EI

L/30

−1/24

.

The exact solution is 120 EI v = wL4 [ 4 − 5x / L + (x/L)5 ] so that the exact values ofthe maximum deflection and slope are v = WL / (30 EI ) and θ = −WL3 / (24 EI ),respectively. Thus, our single element solution gives the exact values of both v and θ at

the nodes, but is only approximate in the interior of the element. The last two equations

give the exact reactions.

In practical analysis one can often utilize a partial model to reduce the data

preparation and more importantly the analysis cost. One must be alert for planes where

the geometry, material property, supports and loads are symmetric mirror images. Often

the loading conditions occur in an anti-symmetric, or negative mirror image, fashion so

that one can still use a half portion model and simply recognize that the deflections and


stresses on the omitted half have signs opposite from those in the partial model. These

concepts are easy to illustrate for a beam, but are usually employed in much larger and

more complicated structural systems. Even with today’s large memory computers one

ev entually finds a structure that is too big to execute. Then one searches for alternate

procedures such as solving two hav e size problems and combining their results (with the

appropriate sign changes). The corresponding results for the other half can be found by

superposition (with careful attention to their signs) and output. These concepts extend to

two- and three-dimensional problems. By carefully using partial models and carefully

combining their results one can obtain the full analysis results in much less memory and

with substantially reduced computations. This also means reduced wall clock turnaround

time on the analysis.

13.12 Beam matrices via Galerkin’s method

Here we will illustrate the development of the element matrices by applying

Galerkin’s Method to the governing differential equation. Recall that for a beam subject

to a load p(x), the differential equation describing the elastic curve is giv en by Eq. 13.14.

If we substitute an approximate solution, u(x), this gives

(13.24)d2

dx2

EId2u

dx2

− p(x) = R(x)

where R(x) is the residual error term. Interpolating the approximate solution with

u(x) = He(x) δδ e and applying Galerkin’s criterion to the error term gives

0 = ∫ u(x) R(x) dx =eΣ

Le∫ u(x) R(x) dx =

eΣ δδ eT

Le∫ HeT (x) R(x) dx .

But the array δδ e is a vector of arbitrary constants. This implies that we require

0 =Le∫ HeT (x) R(x) dx =

Le∫ HeT

d2

dx2EI

d2u

dx2− p(x)

dx .

Let a prime denote a derivative, then twice integrating the first term by parts gives

(13.25)

L

∫ EI H′′ T u′′ dx + EI HT u′′′

L

0− EI H′T u′′

L

0−

L

∫ HT p dx = 00 .Substituting the interpolation for u the first integral gives

(13.26)Le∫ EI H′′ T (x) H′′ (x) dx δδ e = Se δδ e , where Se =

Le∫ EI H′′ T H′′ dx

is the element stiffness matrix given earlier. The last integral is the consistent force

vector given earlier in Eq. 13.22. The remaining two terms define the natural boundary

conditions on the beam.

13.13 Beam on an elastic foundation *

The stresses in a statically indeterminate structure on an elastic foundation are

influenced by the deformation of the foundation while the pressure distribution on the


foundation is affected by the relative stiffness of the structure and the foundation

medium. To allow for this structure-foundation interaction the finite element method is

ideally suited. The problem has been studied by many authors. Most of those works use

Winkler hypothesis, and assume that the soil adheres to the beam, i.e., the separation

between beam and soil is not allowed. This is not true for many physical cases. For

instance, when a beam or a beam-column rests on the soil foundation with some type of

load on it, some parts of the beam might be lifted up. Because of soils lacking both

adhesive and cohesive properties, gaps occur in those regions. The method presented in

the study represents the foundation by a one-dimensional line finite element. The

foundation is assumed to be of the two-parameter type. Also, the separation between

structure and soil foundation is allowed when tension develops. The location of those

regions is solved by iterating the solutions. The differential equation of equilibrium of a

beam-column on an elastic foundation is

EI d4w / dx4 + N d2w / dx2 − ks d2w / dx2 + kw w = q(x)

in which kw and ks are Winkler modulus and second-parameter foundation coefficients,

respectively, N is the axial tension force, q is the transverse load, and EI is the bending

stiffness. This is the differential equation of an ordinary beam on a Winkler elastic

foundation which is well known.

The analysis of beams, beam-columns and plates on elastic foundations is wide-

spread in engineering. Hetenyi [8] extensively developed the classical differential

equation solutions for this problem. Finite element approaches have been used

extensively in the analysis of beams on an elastic foundation. Most of these works use

the Winkler hypothesis. Thus, the foundation acts as if it consisted of infinitely many

closely spaced linear springs. Bowles [3] formulates a stiffness matrix by combining a

conventional beam element with discrete soil springs at the end of the beam. The degree

of accuracy using this element is highly dependent on the number of elements modeled.

Ting [19] derives the stiffness and flexibility matrices from the exact solution of the

differential equation.

13.13.1 Foundation models

As a result of a line load, q(x), on the upper surface the beam deflects, causing the

foundation to resist with a line load p(x), whose units may be taken as N /mm. Various

foundation models define p(x) in various ways. The Winkler Foundation model has been

used for a century. It assumes that the foundation applies only a reaction p(x) normal to

the beam, and the p(x) is directly proportional to the beam deflection w(x):

p(x) = k w(x). The Winkler foundation modulus k has units N /mm /mm. Effectively,this foundation is a row of closely-packed linear springs. Note that interactions between

the springs are not considered, so it does not accurately represent the characteristics of

many practical foundations. To improve the Winkler model some authors assumed

interactions between the springs and added a second parameter. The Filonenko-Borodich

Foundation assumes that the top ends of the springs are connected to an elastic membrane

that is stretched by a constant tension. The Pasternak Foundation introduces shear

interactions between the springs. He assumed that the top ends of the springs are

connected to an incompressible layer that resists only transverse shear deformation. The


Generalized Foundation model assumes that at each point of contact there is not only a

pressure but also a moment applied to the beam by the foundation. The moment is

assumed to be proportional to the angle of rotation (or slope) of the beam.

Mathematically, all these models are equivalent. The only difference is the definition of

the parameters, so we rewrite these equations in the form

p(x) = k w(x) − ks d2w(x) / dx2

in which k is the first parameter (Winkler’s modulus), and ks is the second parameter.

The integral form mathematically equivalent to this problem is to find the extremum

of the functional

Π =1

2L

∫ EI (w′′ )2 + N (w′ )2 − ks(w′ )2 + k2w2 − 2 q(x) w dxfor all smooth functions w satisfying the boundary conditions. In a finite element

formulation the assumed interpolation functions should be such that w and w′ arecontinuous between elements. This assures that Π is defined and that we can write Π asthe sum of contributions from the elements into which the region is divided. Substituting

the usual interpolation matrices :

Π =1

2 eΣ u

T K1u + uT K2u + uT K3u + uT K4u − 2 uT C

e

allows us to define the total element stiffness matrix as K = K1 + K2 + K3 + K4 whereK1 is the classical beam stiffness matrix, given earlier, which involves the beam flexural

stiffness EI , K2 is the stiffness matrix which involves the axial load N , called geometric

stiffness matrix, K3 is the second-parameter foundation stiffness matrix, and K4 is the

Winkler foundation stiffness matrix. They are defined by the relations

(13.27)uT K1u = ∫ EI (w′′ )2 dx , uT K2u = ∫ − N (w′)2 dx ,uT K3u = ∫ − ks(w′)2 dx , uT K4u = ∫ kw w2 dx , uT C = ∫ q(x) w dx

where the integrations are carried out over the element and where u denotes the element

degrees of freedom. Assembling the elements stiffness matrices according to the

connectivity of elements, the system equations can be obtained as K U = F.

13.13.2 Cubic beam-foundation matrices

The practical application of the finite element method depends on the use of various

interpolation functions and their derivatives. Most of the interpolation functions for C0

and C1 continuity elements are well known. The C0 functions are continuous across an

inter-element boundary while the C1 functions also have their first derivatives continuous

across the boundary. Here we begin with a C1 continuity element for the beam. This

element is obtained by using Hermite interpolation polynomial and nodal variables that

include derivatives as well. The geometrical nodes, function H and the Jacobian matrix J

remain identical to those of the linear element ( C0 bar element). The resulting matrices

are (with L = Le):


(13.28)K1 = EI

L3

12

6L

−126L

4L2

−6L2L2

12

−6L

Sym.

4L2

K2 =−N30L

36

3L

−363L

4L2

−3L−L2

36

−3L

Sym.

4L2

, K3 =ks

NK2

K4 =kw

420

156

22L

54

−13L

4L2

13L

−3L2156

−22L

Sym.

4L2

.

13.13.3 Fifth order beam element

The Hermite family includes members with additional derivatives at the two ends

(or with interior nodes with function and derivative values). To seek a more accurate

solution we will next consider a two noded fifth order Hermite polynomial. This has

three variables per node : wi, w′i, w′′i. The corresponding generalized parameters are u1,u2, and u3, respectively. For a two nodes line element, the equation of deflection is:

w(x) = H1u1 + H2u2 + H3u3 + H4u4 + H5u5 + H6u6 .

The shape function in local coordinates are given in Fig. 3.6. By using numerical

integration and cubic interpolation one obtains the numerical form equivalent to the

element stiffness matrices given in Eq. 13.28. Likewise, if we employ the fifth order

interpolation functions, the corresponding numerically integrated (6 × 6) matrices areobtained. Both forms can be used in later examples.

For fourth or sixth order differential equations it is common to use these same

higher order continuity elements when the "data" are everywhere smooth. If the data,

p(x), in Eq. 13.24 is smooth then we expect the exact solution (with R(x) = 0) to havesmooth fourth derivatives. Then we are justified in using the higher continuity elements.

In the section on eigen-solutions to the Schrodinger equation we will illustrate the

increased accuracy obtained for using higher continuity elements.


Table 13.1 Homogeneous solution interpolation for

semi-infinite beam on foundation

a) PDE :d2

dx2

EId2v

dx2

− Nd2v

dx2+ k v = q, |N | ≠ 2√ k EI

b) Homogeneous Interpolation : H =1

a − b

( a ebx − b eax ) ( eax − ebx )

c) Stiffness Matrix : c = a2 + 3 ab + b2

d) Force Vector : F =q

a b

−a − b1

e) Mass Matrix : M =ρ

2 ab(a + b)

−csym.

(a + b)−c

d = √ √ k4 EI + N4 EI , e = √ √ k4 EI − N4 EIa = − d + e, b = − d + 2e.

K =1

2 ab(a + b)

(−EI a3 b3 − N a2 b2 − c k)symmetric

(EI a2 b2 − k)(a + b)(−c EI a3 b3 − N a2 b2 − k)

13.14 Nodally exact beam elements *

Recall that the analytic solution to a differential equation is generally viewed as the

sum of a homogeneous solution and a particular solution. It has been proved by

Tong [20] that if the finite element interpolation functions are the exact solution to the

homogeneous differential equation, then the finite element solution of a non-

homogeneous (no zero) source term will always be exact at the nodes. Clearly, this also

means that if the source is zero, then this type of solution would be exact everywhere. It

is well known that the exact solution of the non-homogeneous equations for the beam on

an elastic foundation will generally involve the products and sums of trigeometric and

hyperbolic functions. Therefore, if we replaced the previous polynomial interpolations

with a more complicated set, we can assure ourselves of results that are at least exact at

the nodes. In practice, using hyperbolic functions with large arguments can break down

due to the way their values are computed in the operating system mathematics library.

For a semi-infinite beam-column on an elastic foundation the element matrices obtained

from interpolating with the homogeneous solutions are given in Table 13.1.


13.15 Plane frames

If we combine the bar member, which carrys only loads parallel to its axis, and a

beam which carries only loads transferse to its axix we get the so-called beam-column

element. The one dimensional version of that has three nodal degrees of freedom: an

axial displacement vector, a transverse displacement vector, and a rotation vector

(perpendicular to the other two). Analogous to a planar truss, which is the extension of a

simple two force axial member to a planar assembly that resists loads by axial loads only,

a frame is an assembly of beam-column members that adds the ability to also transmit

moments (couples) through its joints. In other words, a frame is a combination of

individual beam-column elements that resists loadings by a combination of bending, axial

member forces, and transverse (shear) forces. Therefore it is a more efficient structure

than a truss.

Weav er [22] has considered the detailed programming aspects of several one-

dimensional structural elements (nn = 2). Among them is the plane frame element whichhas two displacements and one rotation per node (nd = 3). A typical element is similar tothat shown in Fig. 13.9 where xL and yL are the member axes and x, y are the global axes

(ns = 2). The difference is that at each node there is a moment vector perpendicular tothe paper (in the direction of the z axis). A frame element with an area of A, having a

cross-sectional moment of inertia I , a length L, and made of a material having a modulus

of E has an element square matrix (stiffness matrix) that is a combination of those from

the axial bar and

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Supplement 1 Networks, trusses, beams and framesme.rice.edu/~akin/Elsevier/Sup_13_Networks.pdfSupplement 1, Networks, trusses, beams and frames 451 13.2 Thermalnetworks There are manyapplications