CE220 Reader for 2009

REPORT

UCB/SEMM

STRUCTURAL ENGINEERING MECHANICS AND MATERIALS

CE 220 - STRUCTURAL ANALYSIS,

THEORY AND APPLICATIONS

COURSE NOTES

by

FILIP C. FILIPPOU

FALL SEMESTER 2009 DEPARTMENT OF CIVIL ENGINEERING UNIVERSITY OF CALIFORNIA BERKELEY, CALIFORNIA

Notation

General:

Upper case letters for matrices and vectors refer to variables of the entire structure, e.g. collection of forces, displacements, etc.

Lower case letters for matrices and vectors refer to element variables. A bold face font is used for matrices, a bold face italic font is used for vectors. An

italic font is used for scalars. All variables refer to a global (structural) reference system under otherwise

indicated

General notation:

Static variables

P applied forces at global degrees of freedom (dofs) p element end forces q basic element forces (independent element end forces)

Q collection of all element basic forces for structural model

s section resultant force vector

stress tensor

Kinematic variables

U global dof displacement vector

u element end node displacements

v Element deformations

V collection of all element deformations for structural model

e section deformations strain tensor

Matrices relating static and kinematic variables

B static (equilibrium) matrix for structure

A kinematic (compatibility) matrix for structure

F structure flexibility matrix

f element flexibility matrix

K structure stiffness matrix

k element basic stiffness matrix

Page 1

Subscripts

Subscripts are used to specialize the variables on the preceding page often denoting a subset or subarray of the original variable. They have the following meaning:

Single subscript

Interpretation

b Boolean

c refers to the incipient collapse state for scalars and vectors denotes constraint for arrays

d refers to restrained dofs

e elastic refers to strain based deformations

f refers to free dofs

g global

h hinge

i refers to statically determinate primary system

p refers to the particular solution of a system of equations; denotes plastic in arrays or when used as second subscript

r refers to rotation for arrays denotes resisting when used with force vectors denotes residual when used with deformations in CE221

s collection of element matrices for structural model

u denotes unbalance when used with force vectors

x refers to force redundants

Double subscript

Interpretation

pl plastic capacity

ce elastic at incipient collapse

cp plastic at incipient collapse

Page 2

Variables with diacritical marks

B force influence matrix of structure

p element end forces in local (element) reference system

u element end displacements in local reference system

fA kinematic matrix of structure for constrained free dofs

fU independent free dofs of structural model under constraints

Scalar variables

N axial force

M bending moment

V shear force coefficient of thermal expansion

a normal strain at reference axis of frame element

rotation of element chord

curvature

load factor

Other variables or symbols

variation of (also denotes virtual)

increment of

X, Y, Z global (structure) coordinates

x, y, z local (element) coordinates

Page 3

Variables by section

Statics or Equilibrium

P applied forces at global degrees of freedom (dofs)

fP applied forces at free global dofs

=dP R applied forces at restrained global dofs = support reactions

fwP equivalent nodal forces at free dofsdue to element loading

p element end forces in global (structure) reference system p element end forces in local (element) reference system q basic element forces (independent element end forces)

Q collection of all element basic forces for structural model

s section resultant force vector

stress tensor

N axial force

M bending moment

V shear force

b force transformation matrix from basic to local system

rb force transformation matrix from local to global system

gb force transformation matrix from basic to global system

B static (equilibrium) matrix for structure

fB static (equilibrium) matrix for free dofs of structure

dB static (equilibrium) matrix for restrained dofs of structure

iB submatrix of fB for primary basic forces

xB submatrix of fB for redundant basic forces

B force influence matrix of structure

iB force influence matrix of primary structure under applied nodal forces

xB force influence matrix of primary structure under redundant basic forces

pQ basic forces for particular solution of equilibrium equations

Page 4

iQ basic forces in primary structure

xQ redundant basic forces

plQ plastic capacities for basic forces; superscript + for positive, - for negative

refP reference applied force vector

load factor

cQ basic forces at incipient collapse

ceQ basic forces at incipient collapse at elastic locations

cpQ basic forces at incipient collapse at plastic hinge locations

eB submatrix of fB for elastic basic forces

pB submatrix of fB for plastic basic forces

peQ particular solution for elastic basic forces at incipient partial collapse

xeB force influence matrix of primary structure under redundant basic forces at incipient partial collapse

Kinematics or Compatibility

U displacements at global degrees of freedom (dofs)

fU displacements at free global dofs

dU displacements at restrained global dofs = imposed support displacements

u element end displacements in global (structure) reference system

u element end displacements in local (element) reference system

v element deformations

V collection of all element deformations for structural model

e section deformations strain tensor

a kinematic transformation matrix for extraction of rigid body modes from element end displacements in local reference system

ra kinematic transformation matrix from global to local system

ga kinematic transformation matrix from global to basic system

Page 5

A kinematic (compatibility) matrix relating the element deformations with the displacements at all dofs of the structure

fA submatrix of A relating the element deformations with the displacements at the free dofs of structure

dA submatrix of A relating the element deformations with the displacements at the restrained dofs of structure

dV element deformations due to imposed displacements at restrained dofs

coefficient of thermal expansion

a normal strain at reference axis of frame element

rotation of element chord

curvature

v strain dependent element deformations

hv element deformations due to hinge (release) deformation

V collection of all strain dependent element deformations for structure

hV collection of all hinge dependent element deformations for structure

cA kinematic matrix relating unconstrained with constrained dofs

fU independent free dofs of structural model under constraints

fA kinematic matrix of structure for constrained free dofs

iV subset of element deformations corresponding to basic forces of primary structure

xV subset of element deformations corresponding to redundant basic forces

eV subset of element deformations at elastic locations in the structure

iA submatrix of fA corresponding to element deformations iV

xA submatrix of fA corresponding to element deformations xV

eA submatrix of fA corresponding to element deformations eV

cpA kinematic matrix relating unconstrained dofs with single independent dof of the collapse mechanism

fpA kinematic (compatibility) matrix relating the element deformations with the single free independent dof of the collapse mechanism

pW plastic work increment

Page 6

Force method

F structure flexibility matrix

f element flexibility matrix

E modulus of elasticity

A cross section area

I moment of inertia

sF block diagonal matrix of all element flexibility matrices

0v initial element deformations due to element loading

0V collection of initial element deformations due to element loading

w distributed element load

L element length

ixF xiF xxF intermediate flexibility matrices of force method

VB force influence matrix of structure under initial deformations

Displacement method

K structure stiffness matrix

k element basic stiffness matrix

sK block diagonal matrix of all element stiffness matrices

0q initial or fixed element forces

0Q collection of initial or fixed end basic element basic forces

fP applied forces at free global dofs

0P initial nodal force vector

Page 7

INTRODUCTION - STRUCTURAL MODELING

Objective: systematic description of structural model; ingredients; sign convention; structural variables; loading description

CE220-Theory of Structures Equilibrium Prof. Filip C. Filippou, 2000

Page 8

Introduction; structural model

The structural model is an idealization of the real structure. It consists of nodes in space connected byelements. While the structure is a continuum, the structural model is discrete. The analyst has makesure to represent the actual structure as accurately as warranted by the demands of the analysis, and thentranslate the results of the discrete structural model back to the actual structure for its design and detailing.

In this course we limit ourselves to one-dimensional elements, whose properties depend on a singlevariable x. Moreover, we limit ourselves to illustrating the key concepts with straight, prismatic 2d trussand frame elements, but state these in general form, so as to allow the extension to any type of one-dimensional element (e.g. a circular frame element).

Limiting ourselves to straight, prismatic 2d elements means that the geometry of curvilinear structuresneeds to be approximated with as many elements as the accuracy of the results demands. We showtwo approximation examples below: one is a non-prismatic girder and the other a circular arch.

structure structural model

We distinguish one-dimensional, two-dimensional and three-dimensional elements in a structural model. The dimension refers to the number of variables for describing the element force-deformation properties.Thus, a truss element belongs in the one-dimensional element category, a plate element in the twodimensional element category, and a solid or brick element in the three-dimensional element category.Truss elements used in a 3d (three-dimensional) structural model are called 3d truss elements.

Another example of the approximation involved in the idealization of the actual structure are the structuraljoints: members of frame structures intersect at joints of finite size. In a structural model the nodes arepoints in space. It is not entirely clear how to represent the joint region with one-dimensional elements.We will mention this point in the latter part of the course.


Page 9

Structural model: Geometric description, degrees of freedom

X

Yglobal coordinate system

The geometry of the structural model is described by the location of the nodes in an orthogonal right-handed coordinate system. We call this the global or structural coordinate systemand identify its axes by upper case letters. If the global X and Y axis are in the plane of the paper or board, then the Z-axis points toward the viewer.

In a discrete structural model the response of the model is completely described by structural variablesassociated with the nodes. These are of two kinds: generalized forces and generalized displacements.We use the word "generalized" to indicate forces, in the conventional sense, and moments. "Generalized"displacements refers to translations and rotations of a node.

Since a node is a point in space, we can decompose its translation into 3 components along the global X,Y and Z-axes; we can also decompose the rotation into 3 components, i.e. a rotation about the X-axis, arotation about the Y-axis, and a rotation about the Z-axis. Note, however, that the decomposition of therotation into 3 components is only possible for very small (actually infinitesimal) rotations. In this coursewe limit ourselves to such very small rotations, but, moreover, plan to study mostly planar structures forwhich only the rotation about the Z-axis is relevant. With the right-handed coordinate system we usethe right hand grip rule to define positive rotations: with the Z-axis pointing toward the viewer in the abovefigure a counter clockwise rotation (CCW) is positive, and a clockwise (CW) rotation is negative.This is also the consistent convention for moments: a moment acting CCW is positive, and a momentacting CW is negative. Each node of the structural model thus experiences, in general, a translation and a rotation: theseare described by 3 components each. Each component is called a degree of freedom (dof) of the node.By decomposing the translation and rotation into components relative to the global coordinate system, weensure that the degrees of freedom (dofs) are independent.Thus, each node of a 3d structural model has 6 degrees of freedom. In a 2d or planar structural model,there are only two translation components and one rotation in the plane for each node. Thus, each nodehas only 3 degrees of freedom in a planar structural model.

We represent each node with a small black square in this course. This allows us to illustrate its rotation, which a point does not allow us to do. We depict translation dofs with a single arrowhead and rotation dofs with a double arrowhead. The same is true for forces and moments, respectively. In the plane the rotation dof is depicted the same way as the moment, as shown in the figure on the right.

Node dofs may be restrained by devices. We depict such devices with special symbols. Nodes with at least one dof restrained are known as the supports of the structural model. In the following figure of typical node restraint (support) symbols the restrained dofs are shown in gray, and the free dofs are shown in black; to each restrained dof there corresponds a restraining force, which is commonly known as support reaction.

A connectivity array is also required for the elements of themodel indicating which nodes each element connects.


Page 10

Structural model: Partition of global degrees of freedom

We collect the displacement values of all dofs of the structural model in a vector U. Vectors areunderstood in this course in the general sense as mathematical objects endowed with the operationsof addition and multiplication by a scalar and not as geometric objects. Vectors are shown in bold, italictypeface in this course.

We partition the degrees of freedom (dofs) of the structural model into two groups: the free dofs(subscript f), and the restrained dofs (subscript d). We reorder the entries of the displacement vector U, so that the free dofs appear in sequence as the first partition of the vector followed by the restrained dofs.In the structural model we achieve this by numbering the free dofs in sequence first followed by therestrained dofs. We show this in the following example.

Structural model with global coordinate systemand node numbering in arbitrary order

X


1

2

3

4

5

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

Number degrees of freedom in sequence (automatic)1. Start with free dofs and proceed in node order

with translation in X, then Y, then Z (3d), thenrotation about X (3d), Y (3d) and Z

2. Number restrained dofs in the same fashion

is the free dof displacement vector

f

d

=

UU

U

dof numbering

we use the abbreviated notation nf for the number of free dofs and nr for the number of restrained dofs;the total number of dofs then, nt, is equal to the sum of nf and nr, i.e. nt = nf + nr

with the above numbering the displacement vector is partitioned according to

fU

dU is the restrained dof displacement vector (note that restrained displacements are not necessarily zero;instead restrained displacements are of specific value at the start of the analysis with zero a special case; given support settlements thus fall into this category


Page 11

Structural models are subjected to generalized forces at the nodes. We collect the values of the appliedgeneralized forces in a vector P. It is logical to use the same numbering for the generalized forces as forthe generalized displacements, so that we can easily determine the external work by the inner productof the two vectors. The applied force vector is thus partitioned into the vector of the applied forces at thefree dofs of the structural model, and the vector of the applied forces at the restrained dofs. The latter arecommonly known as support reactions. The following example illustrates the applied force vector for aparticular load case. The applied force numbering is the same as the dof numbering on the preceding page.

Structural model: Generalized forces

10

20

15

200 1

2

3

4

5

6

7

8

9

10

11

12

1314

15

f

100002000

152000

=

P

is the vector of the applied forces at the free dofs

f

d

=

PP

Pthe applied force vector is partitioned according to

fP

is the vector of the applied forces at the restrained dofs, commonly known as support reactions

f =

PP

Ror

R


Page 12

Structural model: Loading

Loading1. applied forces or moments at nodes grouped in generalized force vector Pf2. imposed translations or rotations at nodes grouped in support displacement

vector Ud3. element loading

a. applied forces or moments along element axisb. initial deformations (thermal, shrinkage, creep)

element loading

this model has fewer nodes and elementsand thus a smaller number of dofs and equations;it is thus suitable for hand solution, but requirestreatment of concentrated forces as element loading

nodal loading

Distinction between nodal and element forces is a question of discretization

suitable for computer solution if element librarydoes not include concentrated element forces(more equations and dofs than solution to the right)

In this course we will consider loading type 1 and 2. For loading type 3 we limit ourselves to uniformlydistributed transverse forces and uniform initial deformations over the entire element span

We can avoid element loading by approximation

Actual model

requires element with distributed element loading

Approximation

no element loading required,but more computational effort


Page 13

EQUILIBRIUM (STATICS)

Objective: static variables for structure, element and section and their interrelationship through free body equilibrium equations; static matrix of structural model and its properties; degree of redundancy; basic force influence matrix; solution of equilibrium equations


Page 14

Because we deal exclusively with two node elements in this course, i.e. elements that connect to onlytwo nodes, the fictitious cuts separate each element from the structural model at its two ends.The cut reveals element force variables, i.e. a force and a moment at each cut. We decompose theseinto components relative to the global coordinate system and collect them in a generalized element forcevector p. We use the convention in this course that lower case letters refer to element variablesand upper case letters to structural variables. We identify the end points of each element with the lettersi and j, with i referring to the end point at the node with lower number and j to that with higher number.The end points of the element undergo generalized displacements, i.e. a translation and rotation at eachend. We decompose these into components relative to the global coordinate system and collect themin a generalized element displacement vector u. We number the components of the element force anddisplacement vector in sequence starting with the translation (or force) components in X, Y and Z at node i,continuing with the rotation (or moment) components about X, Y and Z at node i and then proceeding inthe same fashion with the translation (or force), and then rotation (or moment) components at node j. Fora two node element we end up with 6 generalized displacement (or force) components in a 2d model, and 12 components in a 3d model.

Structural model: Free bodies and element variablesGeneralized displacements and generalized forces are related by Newton's first law which states thatan object at rest remains at rest if it is acted upon by a set of balanced forces (the case of uniformvelocity is not of interest in this course). To formulate this law we isolate first the nodes of the structuralmodel by fictitious cuts into free bodies. These free bodies also isolate the elements of the model andreveal another important set of force and displacement variables, those belonging to each element.

node and element free bodies; node numbering as before; element numbering in arbitrary order denoted with lower case letters

1

2

3

4

5

a

b c

d

P

( )cp

a

b c

d

( )bp

element b

i

j

1

2

3

4

56 1

2

( ) 3

4

5

6

b

=

uuu

uuuu

1

2

( ) 3

4

5

6

b

=

ppp

pppp

generalized displacement andforce vector for 2d element b;the little white square are thereto indicate that the elementfree body extends from nodeto node, i.e. the full element length


Page 15

Structural model: Node equilibrium equationsWe are now in a position to apply Newton's first law to the node free bodies of the structural model.Instead of writing these equations in vectorial form, we write them in component form, i.e. we write oneequation in the direction of each dof. These are the node equilibrium equations.

X


1

2

3

4

5

Structural model

a

b c

d

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

Node free bodies

Write equilibrium equations at free dofs in terms of the applied forces P and the element forces p(b)(a) 141 (b)(a) 22 5 (b)(a)3 3 (c)6 1(b)4 4 (c)

2(b)55 (c)

6 3(b)6

7

8

9

10

000

0000 00 00 00 0

ppPpP p

P pp pP ppP

pP p

pPPPP

(d)(c) 14 (d)(c) 25 (d)(c) 36 (d)

6

00

00

00

00

00

000000

0

=

pp

pp

pp

p

which we can write in compact form as f fr = 0P P

where Pfr is the resisting force vector at the free dofsof the structural model; we see that the resisting forcevector for all dofs is the sum of the element forcesand we write symbolically

( )r

el

el=P p with the understanding that each

element force term is assigned tothe appropriate dof equilibrium equation

The equilibrium equations for the restrained dofs result insimilar fashion dr = 0R P

We can combine these two sets of equations in the expression of Newton's first law for the structural model

f fr

dr

=

00

P PR P

or, shorter r = 0P P

if the applied and resisting forces are not in equilibrium, then we are dealing with Newton's second law which states that r

t = MP P U

tUwhere is the dof acceleration vector and M is the mass matrix of the structural model (see CE225)


Page 16

Structural model: Element equilibrium equationsAfter writing the equilibrium equations for the node free bodies we investigate the element free bodies, whichare also separated by the fictitious cuts from the structural model. We note that there are again as manyequilibrium equations as dofs for the element free body: 3 for an element in a planar structural model, and 6 foran element in a 3d model. We write these equations for an element with general orientation in a planar modelunder the assumption that there are no loads acting along the element span. The presence of element loading complicates the equations without any benefit to the argument of comparing the number of available equationsto the number of force unknowns of the problem.

1 4

2 5

3 6 5 4

000X Y

+ =

+ =

+ + =

p p

p p

p p p p

i

j

1

2

3

4

56

j iX X X =

j iY Y Y =

the element equilibriumequations are

In a structural problem the applied forces at the free dofs of the structural model are given. Similarly, thedisplacements of the restrained dofs are also given. It is our task to determine the displacements at thefree dofs and the forces at the restrained dofs (support reactions). We write this in compact form

Given: and

Determine: and

fP dU

fU R

Structural Analysis Problem

Structural model: Degree of redundancy or static indeterminacySince the support reactions are unknown at the start of the analysis, we cannot make any use of the equilibrium equations at the restrained dofs of the structural model. We therefore focus our attention on the equilibrium equations at the free dofs only. There are nf available equations in 6 x ne unknowns where ne is the number of elements. However, the element equilibrium provides another 3 equations for each element, i.e. a total of 3 x ne equations without increasing the number of unknowns. Thus, in summary we have nf + 3 x ne equations in 6 x ne unknowns. The difference between number of force (or static) unknowns of the problem and number of available equations is known as the degree of redundancy or degree of static indeterminacy of the structural model. We use the shorthand NOS to denote the degree of redundancy of the model.

NOS = (3 x ne) - nf NOS = (6 x ne) - (nf + 3 x ne)

Note: instead of adding 3 x ne element equilibrium equations to the nf node equlibrium equations it is much more efficient to use the element equilibrium equations to express the element forces in terms of only three independent or basic element forces. This is what the boxed equation reflects: there are only 3 x ne unknown element forces and nf available node equilibrium equations. This assumes that none of the basic element forces are set to zero through special devices (releases). We will address this issue later on. Note: nothing changes to the above discussion by adding the nr equilibrium equations at the restrained dofs since these equations involve an additional nr unknown support reactions. Thus, it is convenient to leave the support reactions and the corresponding equilibrium equations out of consideration.


Page 17

Element variables: global, local and basic reference system

element coordinate system (e.g. for b)

x

y

i

jb

Definition of element specific or local coordinate system1. Local x-axis points from lower to higher numbered

node (we denote the former node i and latter node j)2. In 2d local y-axis lies in the plane and is normal to x;

in 3d local y-axis is specified by user (e.g. alongprincipal axis of cross section)

3. Local z-axis is normal to the plane formed by x-y

Instead of adding the element equations of equilibrium to the nodal equations of equilibrium it is more efficient and insightful to use the number of element equilibrium equations to reduce the number of unknown element forces*. For a 2-node, 2d element we have 6 element forces and 3 equilibrium equations. Thus, we have just 3 independent or basic element forces. We can use the element equilibrium equations to express the other element forces in terms of the basic forces. We denote the basic element forces by q.

*Note that adding the element equilibrium equations to the node equilibrium equations results in a very large system of equations with many 0 terms, since each element equilibrium equation involves only the forces of the particular element.In working out the relation between dependent and independent element forces it is convenient to workwith an element specific right handed coordinate system. This system is shown in the following figure. Itis called local or element coordinate system, and we use lower case letters to denote its axes.

The force vectors at the end points of the element can be expressed relative to the global or to the localcoordinate system. The former decomposition is shown on the left and the latter on the right of the following figure. We introduce a new symbol for denoting the components of the element force vector inthe local coordinate system.

In the local coordinate system the element equilibrium equations in the absence of element loading become

i

j

2 2L X Y= + 1

2

3

456end forces in local system

i

j

1

2

3

4

56

j iX X X =

j iY Y Y =

end forces in global system

element force vector in local coordinates

1 4

2 5

3 6 5

000L

+ =

+ =

+ + =

p p

p p

p p p

1

2

3

4

5

6

=

ppp

pppp


Page 18

There are some obvious choices for the basic element forces (note that we cannot pick any three!)

Another choice is this group

One common choice is this group

In this course we pick the second choice for the basic forces. We use then the element equilibrium equations to express theother three element forces in terms of the basic.

1 4

3 65

3 62 5

L

L

=

+=

+= =

p p

p pp

p pp p

Because of their importance we introduce a new symbol for the basic element forces: q

with b the force transformation matrix from the basic to the local coordinate system. We call q1 the longitudinal basic force, q2 and q3 the flexural basic forces of the element. q2 acts at node i and q3 at node j

Element variables: basic element forces

We restate the element equilibrium equations in the absence of element loading

1 4

2 5

3 6 5

000L

+ =

+ =

+ + =

p p

p p

p p p

1 2 3p p p

4 3 6p p p

4 3 6p p p 1 2 3q q q

with this definition we have 4 1

3 2

6 3

1 1

2 35

2 32

L

L

=

=

=

=

+=

+=

p q

p q

p q

p q

q qp

q qp

or, in compact form

1

21

32

43

5

6

1 0 01 10

0 1 01 0 0

1 10

0 0 1

L L

L L

= = =

b

pp

qp

p q qp

qpp

Summary: the relation between local components of the element force vector and basic element forces is

= bp q with

1 0 01 10

0 1 01 0 0

1 10

0 0 1

L L

L L

=

b


Page 19

Force component transformation from local to global coordinate system

After having established a relation between the local and basic element forces, we need to relate the global element force components to the local. This is rather straightforward, if we note that the two coordinate systems are related by a rotation in the plane of the model. We use the direction cosines of the element orientation in the undeformed (original) position in lieu of trigonometric functions of the orientation angle.

X

Y

xy L

i

j

X

Y

4p

5p4p

5p

4p

5p

5p

4p

The transformation for the forces at end i is identical. Note that the moments do not transform during the rotation. Thus, we can write for the relation between global components and local components of the element end forces

or, compactly

6 element end forces in global reference

6 element end forces in local reference

3 basic element end forces and3 dependent end forces

ip

jp

(a) ip

jp

(b)

1q

2q

3q

(c)

L

2 3

L+q q

2 3

L+q q

1q

global

local

basic

4 4 5 4 5

5 4 5 4 5

cos sin

sin cos

X YL LY XL L

= =

= + = +

p p p p p

p p p p p

4 4

5 5

X YL LY XL L

=

p pp p

1 1

2 2

3 3r

4 4

5 5

6 6

0 0 0 0

0 0 0 0

0 0 1 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0 1

X YL L

pY XpL LppX YpL L

Y X pL L

= = =

b

ppp

p pppp


Page 20

combining with gives

with

Thus, finally the relation between element force components in the global coordinate system and the basic forces is

r= bp p

Element equilibrium: from basic forces to global element force components

= bp q r g= =b b bp q q

2 2

2 2

g r

2 2

2 2

0 0 0 0 1 0 01 10 0 0 0 0

0 0 1 0 0 0 0 1 0 0 1 01 0 00 0 0 0

1 100 0 0 0

0 0 10 0 0 0 0 1 0 0 1

X Y YX YL L LL LY X XY XLL L L L L L

X Y X Y YL L L L LY X Y X XL LL L L L L

= = =

b b b

2 2

12 2

21

32 g

432 2

5

62 2

0 1 0

0 0 1

X Y YL L LY X XL L L

qX Y Y

qL L LY X XL L L

= = =

b

pp

qp

p qppp

with bg the force transformation matrix from the basic to the globall coordinate system; this matrix can also be regarded as the equilibrium or static matrix of the element.


Page 21

The basic forces of a 2d frame element are:the axial force q1 and the end moments q2 and q3

The end element forces in the global coordinate system can be expressed in terms of these

1q

1q

1XL q

1YL q

1XL

q

1YL

q

3q

3

Lq

3

Lq

32

YL

q

32

XL q

32

XL

q

32

YL q

3q

2q

2

Lq

2

Lq

22

YL

q

22

XL q

22

XL

q

22

YL q

2q

Element equilibrium: from basic forces to global element force components

The figures on the left show each basic force in turnalong with the dependent element force componentsin the local reference system satisfying equilibrium.

The figures on the right show the element forcecomponents in the global reference system for each basic force in turn; each group of force components corresponds to a column of the static matrix bg and satisfies element equilibrium


Page 22

end i

1q

1q2q

Special cases of basic element forces for 2d frame elements

We deal at first only with devices that release one or more of the element basic forces. We will returnlater to deal with more general internal force releases.

longitudinal basic force release

flexural basic force release

truss element: both flexural basicforces released; only one basic force

2d frame element; one flexural basicforce released; two basic forces only

2d frame element; one flexural basicforce released; two basic forces only 1q

2q

2d frame element; longitudinall basicforce released; two basic forces only

1q 2q

end j


Page 23

Relation between element and section static variablesIt is very important to distinguish between element forces and internal forces. The latter arise at a fictitious cut through the element at a distance x from end i, as shown in the following figure. The cut separates the element into two parts, each of which needs to satisfy the equations of equilibrium. For the 3 additional equilibrium equations there are 3 additional unknowns, the internal element forces at the cut at a distancex from end i. These internal forces are the resultants of the normal and shear stresses acting at the cut andare, therefore, known as stress resultants. They are also known as section forces, since they are the static variables of an infinitesimal slice of the frame element.

We collect the section forces in a vector s(x), which for the 2d element in the figure includes the normal force N(x), the shear force V(x) and the bending moment M(x). Note that the section or internal forces N, V, and M (there is another shear force, another bending moment, and a torsional moment in 3d) are continuous functions of x and not discrete variables as the basic element forces. As continous functions of x the internal forces can be differentiated with respect to x. Moreover, the sign convention for the internal forces needs to follow the sign convention for continuous functions: the forces at the face of the cut (section) closest to end i follow the axes orientation of the local coordinate system, while those at the opposite section act in the opposite direction. By contrast, the basic element forces are discrete static variables and follow the sign convention adopted for all structural and element variables.

2d frame element with moment release at xi j

1q2q 2

L xxq

x L-x

We can now introduce releases of the internal forces at any section inside the element. These result in one constaint equation between the basic element forces for each release. Each release, therefore, reduces the number of basic element forces by one. Three common cases follow

2d frame element with shear force releasei j

1q2q

i j

1q 2q

2d frame element with normal force release

2q

( )M x

2q

2 3

L+q q

3q

3q

2 3

L+q q

2 3

L+q q 2 3

L+q q

( )V x

frame element

end jend i

x

y

+ 1q1q

( )N x

1qx

( )M x

( )V x

( )N x slice of length x


Page 24

Example 1: Equilibrium equations for gable frame, properties of static matrix After establishing the relation between basic element forces and element force components in the global coordinate system we return to the equilibrium equations of the node free bodies in the direction of the free dofs. Setting up these equilibrium equations in the undeformed configuration establishes a linear relation between applied forces and basic element forces. The coefficients of the basic element force can be collected in the static matrix of the structural model. We use example 1 to illustrate. The followingprovides a brief overview of process. More details are available in the example description.

X


1

2

3

4

5

Structural model

a

b c

d

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

Node free bodies(b)(a)141(b)(a)22 5(b)(a)3 3 (c)6(b) 14 4 (c)(b) 255 (c)(b)6 36

7 4

8

9

10

000

0000 00 00 00 0

= + +

ppPpP p

P pppP ppP p

P pp

P pPPP

(d)(c) 1

(d)(c) 25 (d)(c) 36 (d)

6

000000

0

+

p

ppppp

(a) (a) (a)g.41 g.42 g.43

1(a) (a) (a)

2 g.51 g.52 g.53

(a) (a) (a)3g.61 g.62 g.63

(a)41

52

63

7

8

9

10

0 0 00 0 00 0 00 0 00 0 00 0 00 0 0

=

b b bPP b b bP b b bP

qP

qP

qPPPP

(b) (b) (b)g.11 g.12 g.13

(b) (b) (b)g.21 g.22 g.23

(b) (b) (b)g.31 g.32 g.33

(b) (b) (b)g.41 g.42 g.43

1(b) (b) (b)

2g.51 g.52 g.53

(b) (b) (b)g.61 g.62 g.63

0 0 00 0 00 0 00 0 0

+

b b b

b b b

b b b

b b b qqb b bqb b b

(c) (c) (c)g.11 g.12 g.13

(b) (c) (c) (c)g.21 g.22 g.23

(c) (c) (c)g.31 g.32 g.33

3 (c) (c) (c)g.41 g.42 g.43

(c) (c) (c)g.51 g.52 g.53

(c) (c) (c)g.61 g.62 g.63

0 0 00 0 00 0 0

0 0 0

+

b b b

b b b

b b b

b b b

b b b

b b b

(c) (d)1 1

2 2(d) (d) (d)

g.11 g.12 g.133 3

(d) (d) (d)g.21 g.22 g.23

(d) (d) (d)g.31 g.32 g.33

(d) (d) (d)g.61 g.62 g.63

0 0 00 0 00 0 00 0 00 0 00 0 0

+

q qq q

b b bq q

b b b

b b b

b b b

Now, collect all basic element forces into a single vector Q for compact notation. We have:

(a)1

12

23

3(b)1 4

2 5

3 6(c)

71

82

93

10(d)1 11

2 12

3

= =

Q

qQ

qQ

qQ

q Qq Qq Q

QqQqQqQ

q Qq Qq


Page 25

The equilibrium equations at the free dofs then become

P1Q2 Q3+

12Q4 0.8

Q5 Q6+

100.6

P2 Q1 0.6 Q4Q5 Q6+

100.8+

P3 Q3 Q5+

P4 0.8 Q4Q5 Q6+

100.6+ 0.8 Q7

Q8 Q9+

100.6+

P5 0.6 Q4Q5 Q6+

100.8 0.6 Q7+

Q8 Q9+

100.8+

P6 Q6 Q8+

P7 0.8 Q7Q8 Q9+

100.6

Q11 Q12+

12+

P8 0.6 Q7Q8 Q9+

100.8 Q10+

P9 Q9 Q11+

P10 Q12

Compare these with the equations set up one at a time in Example 1.

Bf

0

1

0

0

0

0

0

0

0

0

112

0

0

0

0

0

0

0

0

0

112

0

1

0

0

0

0

0

0

0

0.8

0.6

0

0.8

0.6

0

0

0

0

0

0.610

0.810

1

0.610

0.810

0

0

0

0

0

0.610

0.810

0

0.610

0.810

1

0

0

0

0

0

0

0

0.8

0.6

0

0.8

0.6

0

0

0

0

0

0.610

0.810

1

0.610

0.810

0

0

0

0

0

0.610

0.810

0

0.610

0.810

1

0

0

0

0

0

0

0

0

1

0

0

0

0

0

0

0

0

112

0

1

0

0

0

0

0

0

0

112

0

0

1

or, in compact form where is the static matrix of the structural model for the free dofsf f= BP Q fB

(a) (a) (a) (b) (b) (b)g.41 g.42 g.43 g.11 g.12 g.13

(a) (a) (a) (b) (b) (b)g.51 g.52 g.53 g.21 g.22 g.23

1(a) (a) (a) (b) (

2 g.61 g.62 g.63 g.31 g.32

3

4

5

6

7

8

9

10

0 0 0 0 0 0

0 0 0 0 0 0 =

b b b b b b

b b b b b bPP b b b b bPPPPPPPP

b) (b)g.33

(b) (b) (b) (c) (c) (c)g.41 g.42 g.43 g.11 g.12 g.13

(b) (b) (b) (c) (c) (c)g.51 g.52 g.53 g.21 g.22 g.23

(b) (b) (b) (c) (c) (c)g.61 g.62 g.63 g.31 g.32 g.33

(c) (c)g.41 g.42

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

b

b b b b b b

b b b b b b

b b b b b b

b b (c) (d) (d) (d)g.43 g.11 g.12 g.13(c) (c) (c) (d) (d) (d)

g.51 g.52 g.53 g.21 g.22 g.23

(c) (c) (c) (d) (d) (d)g.61 g.62 g.63 g.31 g.32 g.33

(d) (d) (d)g.61 g.62 g.63

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

b b b b

b b b b b b

b b b b b b

b b b

1

2

3

4

5

6

7

8

9

10

11

12

QQQQQQQQQQQQ


Page 26

Static matrix- Stability and degree of redundancy (static indeterminacy) of structure

The equilibrium equations at the free dofs of the structural model including the effect of element loading take the following form

The degree of redundancy (static indeterminacy) of the structural model is given by the difference between the number of basic element forces and the number of linearly independent equations of equilibrium at the free dofs. For a stable structural model the degree of redundancy or static indeterminacy NOS is the difference between the number of columns and the number of rows of matrix Bf.

where Pfw are the equivalent nodal forces due to element loading

Degree of redundancy (static indeterminacy) NOS

STRUCTURE A

NOS = 18-12=6

STRUCTURE B

NOS = 15-11=4

STRUCTURE C

NOS = 15-11=4

NOS = 10-7=3

a b

Examples of statically indeterminate structures

NOS = 10 - 8 = 2

f f fw= +BP Q P

The above equation can be rewritten with the known loading terms on the left and the unknown basic element forces on the right. f fw f

= BP P Q

The solution of this linear system of equations depends on the properties of the static matrix Bf. In a stable structural model the number of linearly independent rows of static matrix Bf should be equal to to the number of free dofs. The number of linearly independent rows of a matrix is known as its rank. Thus, for a stable structural model the rank of the static matrix should be equal to the number of free dofs.


Page 27

CE220 - Theory of Structures Example 1 - Equilibrium for Gable Frame Prof. Filip C. Filippou, 2000

Example 1 - Equilibrium Equations for Gable FrameObjectives: (a) set up the equilibrium equations for a frame structure with general orientation members(b) assess stability and degree of static indeterminacy of structural model(c) automate process of setting up equilibrium equations

X


1

2

3

4

5

Structural model

a

b c

d

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

Node free bodies

8 8

12

6

We isolate node free bodies and number the equations of equilibrium (or, degrees of freedom dofs,according to Newton's second law). We number the free dofs first starting from the lowest numbered node,and number the support or restrained dof's after completing the numbering of the free dof's in the same way.

The equilibrium of the element free bodies is satisfied by selecting 3 basic forces as independent or basicforces and then expressing the element end forces in the global reference system in terms of them, so as tosatisfy the 3 equilibrium equations of the element. We obtain the relation

If we write the equilibrium in the undeformed configuration (first order or linear equilibrium)then the array bg depends only on the geometry of the element (length and direction cosines)

p bg q=

the element equilibrium transformation matrix to global coordinates bg has the form

ip

jp

(a) ip

jp

(b)

1q

2q

3q

(c)

L

2 3

L+q q

2 3

L+q q

1q

2 2

1

2 22

13

2 g4

2 2 35

62 2

0 1 0

0 0 1

X Y YL L LY X X

L L L

X Y YL L LY X X

L L L

pp

qp

p q b qp

qpp

= = =

where X/L and Y/L are the direction cosinesof the element and L is the element length

Page 28


When writing the equilibrium equations we proceed equation by equation and limit ourselves to theequations not involving support reactions, i.e. the equilibrium equations for the free dofs of the structuralmodel. We note in the following figure that the basic forces Q are numbered in sequence starting fromelement a through element d. The node i and node j correspondence for the elements of the model isshown next to the figure.

1

2

3

5

6

7

8

9

10

11

12

13 14

15

12Q

11Q

10Q

7Q

8Q9Q

4Q

5Q6Q

1Q

2Q

3Q

4 element node i node j

a 1 2

b 2 3

c 3 4

d 4 5

P1Q2 Q3+

12Q4 0.8

Q5 Q6+

100.6=

P2 Q1 0.6 Q4Q5 Q6+

100.8+=

P3 Q3 Q5+=

P4 0.8 Q4Q5 Q6+

100.6+ 0.8 Q7

Q8 Q9+

100.6+=

we can write this system ofequations in compact form Pf Bf Q=P5 0.6 Q4

Q5 Q6+

100.8 0.6 Q7+

Q8 Q9+

100.8+=

P6 Q6 Q8+=

P7 0.8 Q7Q8 Q9+

100.6

Q11 Q12+

12+=

P8 0.6 Q7Q8 Q9+

100.8 Q10+=

P9 Q9 Q11+=

P10 Q12=

Page 29


Observations:

1. the basic longitudinal force is always assumed at node j2. we can use the appropriate terms of the element equilibrium matrix bg; the signs will be correct as long aswe remember that X is measured by subtracting the X-coordinate of node i from that of node j and thesame is true for Y. Thus, for element c in the gable frame of the example, Y is negative because node j islocated lower than node i.

Thus, the bg matrix for element c is

bgc

810

610

0

810

610

0

6

102

8

102

1

6

102

8

102

0

6

102

8

102

0

6

102

8

102

1

:= bgc

0.8

0.6

0

0.8

0.6

0

0.06

0.08

1

0.06

0.08

0

0.06

0.08

0

0.06

0.08

1

=

The static matrix Bf for the freedofs of this structural model is

Bf

0

1

0

0

0

0

0

0

0

0

112

0

0

0

0

0

0

0

0

0

112

0

1

0

0

0

0

0

0

0

0.8

0.6

0

0.8

0.6

0

0

0

0

0

0.610

0.810

1

0.610

0.810

0

0

0

0

0

0.610

0.810

0

0.610

0.810

1

0

0

0

0

0

0

0

0.8

0.6

0

0.8

0.6

0

0

0

0

0

0.610

0.810

1

0.610

0.810

0

0

0

0

0

0.610

0.810

0

0.610

0.810

1

0

0

0

0

0

0

0

0

1

0

0

0

0

0

0

0

0

112

0

1

0

0

0

0

0

0

0

112

0

0

1

:=

In the structure equilibrium matrix Bf we see the contribution of element c in columns 7-9. The terms of theequilibrium matrix bg for element c appear in rows 4 through 9. Before discussing the automatic assemblyof the equilibrium equations let us write down the equilibrium equations for the support reactions.

Page 30


For the restrained dofs the equilibrium equations become

1

2

3

5

6

7

8

9

10

11

12

13 14

15

12Q

11Q

10Q

7Q

8Q9Q

4Q

5Q6Q

1Q

2Q

3Q

4 P11Q2 Q3+

12=

P12 Q1=

P13 Q2=

P14Q11 Q12+

12=

P15 Q10=

we can write these equations in compact form as R Bd Q=

with R

P11

P12

P13

P14

P15

= and Bd

0

1

0

0

0

112

0

1

0

0

112

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

1

0

0

0

112

0

0

0

0

112

0

:=

we can see that this matrix contains a lot of zeros, since only one element connects to a support. Thus,the determination of the support reaction by the operation Bd Q is not very efficient. Instead, eachsupport reaction should be evaluated from the corresponding equation.

we can combine the equilibrium equations for the free and restrained dofs in a singleset by stacking the corresponding vector and matrices on top of each other. We get

PfR

BfBd

Q=

the matrix BBfBd

= is the static matrix of the structural model for all dofs

Matrix B can be readily assembled automatically by considering the contribution of one element afteranother. The contribution of each element goes into 3 separate columns, so that the contribution of elementa goes into columns 1 to 3, element b into columns 4 to 6, and so on. The rows of each element equilibriummatrix bg go to the rows of matrix B according to the incidence relation of the dofs of this element. Thisincidence relation expresses the relation between element dof and global dof numbers. Since the 2-node, 2dframe element has 6 dofs, the incidence (id) array for this element has 6 rows. Each row contains the globaldof number that corresponds to the element dof number for that row. We illustrate this for the example

Page 31


Incidence matrices for element dofs

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

Node free bodieselement a element b element c element d

11

12

13

1

2

3

1

2

3

4

5

6

4

5

6

7

8

9

7

8

9

14

15

10

this means that the first row of the equilibrium matrix of element c goes to row 4 of the structure matrix B,the fourth row to row 7, etc. In Matlab this reassignment of rows can be easily accomplished with arrayindexing as the FEDEASLab function B_matrix.m illustrates.

B stack Bf Bd,( ):=

B

0

1

0

0

0

0

0

0

0

0

0

1

0

0

0

0.083

0

0

0

0

0

0

0

0

0

0.083

0

1

0

0

0.083

0

1

0

0

0

0

0

0

0

0.083

0

0

0

0

0.8

0.6

0

0.8

0.6

0

0

0

0

0

0

0

0

0

0

0.06

0.08

1

0.06

0.08

0

0

0

0

0

0

0

0

0

0

0.06

0.08

0

0.06

0.08

1

0

0

0

0

0

0

0

0

0

0

0

0

0.8

0.6

0

0.8

0.6

0

0

0

0

0

0

0

0

0

0

0.06

0.08

1

0.06

0.08

0

0

0

0

0

0

0

0

0

0

0.06

0.08

0

0.06

0.08

1

0

0

0

0

0

0

0

0

0

0

0

0

0

1

0

0

0

0

0

0

1

0

0

0

0

0

0

0.083

0

1

0

0

0

0

0.083

0

0

0

0

0

0

0

0.083

0

0

1

0

0

0

0.083

0

=

Even though the function B_matrix.m assembles the static matrix B for all dofs of the structural model weuse only the static matrix Bf for the free dofs of the model for discussions of stability and staticindeterminacy. This is so because the addition of the support dofs does not change stability or staticindeterminacy considerations. In the case of static indeterminacy this is easy to see: there are as manyequilibrium equations at the restrained dofs as unknown support reactions.

Page 32


Moreover, the total set of equilibrium equations in the form PfR

BfBd

Q= is not very useful

since it contains the unknown support reactions on the left hand side of the equations. We always workonly with the top portion of these equations, and determine the support reactions only after havingdetermined the basic element forces Q, as subsequent examples will illustrate.

We return now to deal with the equilibrium equations at the free dofs of the structure Pf Bf Q=

It is the properties of the static matrix Bf for the free dofs of the structural model that determine whether asolution exists, and whether it is unique, or whether there are multiple solutions. Note that the specificationof a loading is not necessary for the task, except for the case of an unstable structure being stable for aspecific load, which is of no interest to us. Let us look at one important property of static matrix Bf :the rank of the matrix, i.e. the number of linearly independent rows

A necessary and sufficient condition for a structural model to be stable is that the rank of Bf be equal tothe number of rows, i.e. to the number of free dofs; such a matrix is known to have full rank

Let us check the present example: rank Bf( ) 10= which is equal to the number of rows rows Bf( ) 10=A structure is called statically determinate, if it is stable and its static matrix Bf has the same number ofrows as columns (i.e. it is a square matrix). In such case the inverse of the structure equilibrium matrixexists and is called the force influence matrix. We give it the symbol Bbar.

The equilibrium equations have a unique solution for a stable, statically determinate structure. This is animportant class of structures, since the internal stress state does not depend on material properties.

A structure is called statically indeterminate, if it is stable and its structure equilibrium matrix has morecolumns than rows (i.e. more unknowns than available equations). The difference between number ofcolumns or unknowns and number of rows or available equations is called the degree of static indeterminacyand we use the shorthand NOS for this number.

In a stable, statically indeterminate structure the equilibrium equations have multiple solutions. Thesesolutions can be expressed as the sum of a primary basic force field (called primary or particular solution),and NOS self-stress basic force fields with NOS redundant basic forces as parameters (consult Strang, 3rdedition, pp 71-77). The selection of these redundant basic forces can be done automatically with theGauss-Jordan form of the structure equilibrium matrix. We will illustrate this solution process with examples.

The structure in this example has 12 unknown basic forces and 10 equations of equilibrium, i.e. NOS = 2.

In the above system of equilibrium equations, there is one that involves only one unknown basic force, whichcan be therefore determined immediately and independently from the rest, i.e.

P10 Q12= given the value for P10 at the beginning of the analysis we can immediately calculate Q12.Quite often the value for P10 is zero, resulting in a basic force with zero value and the finalresult 0 = 0 for the corresponding equilibrium equation. If the value of P10 is not zero, wecan still calculate Q12 and account for its effect on the remaining equilibrium equations (thiswould be equation 7 for this structure). To economize on the number of equations andbasic forces, it is advisable to identify such equations and basic forces at the beginning ofthe analysis and avoid numbering the corresponding dof and basic force. We should makesure to include the effect of a non-zero value for the corresponding basic force on theremaining equilibrium equations.

Page 33


An interesting case results with the insertion of a moment release at node j of element b, i.e. by releasingthe basic force Q6. Since Q6 = 0, equation 6 now involves a single unknown

P6 Q8= Thus, for this structure we should only set up 8 equations (i.e. eliminate #6 and #10 above)and then use only 9 unknown basic forces, because Q8 and Q12 can be determinedindependently at the start of the analysis from the corresponding equilibrium equation.Note that NOS = 1 in this case, since we have 8 equilibrium equations for 9 basic forces.

8 8

12

6

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

12Q

11Q

10Q

7Q

8Q9Q

4Q

5Q

1Q

2Q

3Q

Page 34

Mat

lab

scri

pt fo

r E

xam

ple

1 in

CE

220

clas

s not

es

% Automatic assembly of equilibrium equations for gable frame with FEDEASLab function B_matrix

% Clear workspace memory and initialize global variables

CleanStart

Cre

ate

mod

el

% specify node coordinates (could only specify non-zero terms)

XYZ(1,:) = [ 0 0]; % first node

XYZ(2,:) = [ 0 12]; % second node, etc

XYZ(3,:) = [ 8 18]; %

XYZ(4,:) = [ 16 12]; %

XYZ(5,:) = [ 16 0]; %

% connectivity array

CON {1} = [ 1 2];

CON {2} = [ 2 3];

CON {3} = [ 3 4];

CON {4} = [ 4 5];

% boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's)

BOUN(1,:) = [1 1 1];

BOUN(5,:) = [1 1 0];

% specify element type

[ElemName{1:4}] = deal('2dFrm'); % 2d frame element

Model = Create_SimpleModel(XYZ,CON,BOUN,ElemName);

Page 35

Equ

ilibr

ium

equ

atio

ns

% set up equilibrium equations for all dof's

B = B_matrix(Model);

% extract rows corresponding to free dof's to form matrix Bf

Bf = B(1:Model.nf,:);

% display equilibrium matrix

disp(Bf)

% check rank of matrix Bf

disp(['the rank of the structure equilibrium matrix is ' num2str(rank(Bf))])

% determine NOS by difference of no of columns and rows (use size command)

ncol = size(Bf,2);

disp(['the no of columns of the structure equilibrium matrix is ' num2str(ncol)])

nrow = size(Bf,1);

disp(['the no of rows of the structure equilibrium matrix is ' num2str(nrow)])

NOS = ncol-nrow;

disp(['the degree of static indeterminacy of the model is ' num2str(NOS)])

0 0.0833 0.0833 -0.8000 -0.0600 -0.0600 0 0 0 0 0 0

1.0000 0 0 -0.6000 0.0800 0.0800 0 0 0 0 0 0

0 0 1.0000 0 1.0000 0 0 0 0 0 0 0

0 0 0 0.8000 0.0600 0.0600 -0.8000 0.0600 0.0600 0 0 0

0 0 0 0.6000 -0.0800 -0.0800 0.6000 0.0800 0.0800 0 0 0

0 0 0 0 0 1.0000 0 1.0000 0 0 0 0

0 0 0 0 0 0 0.8000 -0.0600 -0.0600 0 0.0833 0.0833

0 0 0 0 0 0 -0.6000 -0.0800 -0.0800 1.0000 0 0

0 0 0 0 0 0 0 0 1.0000 0 1.0000 0

0 0 0 0 0 0 0 0 0 0 0 1.0000

the rank of the structure equilibrium matrix is 10

the no of columns of the structure equilibrium matrix is 12

the no of rows of the structure equilibrium matrix is 10

the degree of static indeterminacy of the model is 2

Page 36

When the structural model is stable and its static matrix Bf is square, the structure is statically determinate and a unique solution for the basic forces exists for any applied loading. This solution does not depend on the element properties. We can write this solution in compact form

In a statically determinate model the basic force influence matrix is the inverse of the static matrix for the free dofs. The coefficient (i,j) of the basic force influence matrix expresses the effect of a unit force at dof j on the basic force i of the structural model.

Once the basic forces are determined in a statically determinate structure, the support reactions can be obtained one by one from the equation of equilibrium of the corresponding restrained dof. Finally, the analyst should check global equilibrium between applied forces and support reactions for the free body of the entire structural model.

Consult examples 2, 3 and 4 for applications of the equilibrium equations to statically determinate structures.

Special case for equilibrium equations - Statically determinate structures

( )1f f fw= BQ P P ( )f fw= BQ P P

is the basic force influence matrix of the structural modelB


Page 37

CE 220 - Theory of Structures Ex. 2 - Truss equilibrium Prof. Filip C. Filippou, 2000

Example 2(r) - Equilibrium equations for statically determinate 2d truss

Objectives:

(a) set up and solution of the equilibrium equations for a statically determinate truss(b) determination of force influence matrix; physical interpretation(c) determination of support reactions and check of global equilibrium

In this example we will set up the equilibrium equations at the nodes of a statically determinate truss.The truss is given in Figure 1. The global coordinate system X-Y is also shown.

88

6

12

3

4

a b

c d e

X

Y

Fig. 1: Statically determinate truss

We number the nodes in arbitrary numerical sequence and identify the elements with lower case romanletters (could have also used numbers). The element numbering is also in arbitrary sequence.

Before embarking on the equilibrium equations we note the following: a truss is a structure made upexclusively of truss elements. These are straight elements (see Remark below) with a momentrelease at each end. Thus, each element has only one basic force q1. The following figure displaysall moment releases for the truss model.

1q

1q

truss element

It is quite clear from the above figure that no moments can be applied at the nodes of the truss. With thisassumption the moment equilibrium at the nodes is satisfied automatically, i.e. 0 = 0. Strictly speaking,however, this results in an unstable node. Thus, it is better to represent the model as shown in thefollowing figure and then postulate that the applied moments at the nodes are zero (this is what makessense in computer analysis).

Page 38


in this case there is one element with a rigid (moment) connection at each node so that the stability of thenodes against rotation is ensured. If no moments act at the nodes, there are no end moments in theelements. For convenience we simplify the display of these construction details with a single circle at thenode, as shown in Fig. 1. This indicates a node without any applied or resisting moments.

Remark: The longitudinal basic force q1 in each element should not be confused with the axial force N(x),which is the internal force acting parallel to the element axis at a section located a distance xfrom end i of the element. While N(x) is equal to q1 in the absence of longitudinal element loads,this is not the case in their presence. q1 and N(x) are also not equal if the "truss element" iscurved. It should then be considered as a frame element with end moment releases.

Based on the preceding discussion, there are no moment equilibrium equations at the nodes of a trussstructure. Starting with the first node we number the equilibrium equations at the node free bodies in thefollowing order: equilibrium equations without support reactions are numbered first, in the X-direction andthen Y-direction following the node sequence (we call these equilibrium equations for the free degrees offreedom or dofs). After we finish with the equations without support reactions, we number in the samefashion the equations with support reactions (we call these equilibrium equations for restrained dofs). Thenumbering in Figure 2 results.

12

3

4

67 8

Fig. 2: Dof or equilibrium equation numbering

Page 39


Length of elements La 8:= La 8=

Lb 8:= Lb 8=

Lc 62 82+:= Lc 10=

Ld 6:= Ld 6=

Le 62 82+:= Le 10=

We now set up the equilibrium equations at the free bodies of the nodes after separating these withmental cuts from the elements. Each truss element has a single basic force. Note that we keep theloading general, i.e. we assume that forces may be present at any degree of freedom. We have:

Equilibrium equations for free dofs

12

3

4

5

67 8

1Q 2Q

4Q3Q 5Q

P1 Q1 Q2+ 0=

P2 Q4+ 0=

P3 Q2 0.8 Q5 0=

P4 0.8 Q3 0.8 Q5+ 0=

P5 0.6 Q3 Q4 0.6 Q5 0=

moving the unknown basic forces to the right hand side of the equations gives the following equations

12

3

45

67 8

1Q 2Q

4Q3Q 5Q

P1 Q1 Q2=

P2 Q4=

P3 Q2 0.8 Q5+=

P4 0.8 Q3 0.8 Q5=

P5 0.6 Q3 Q4+ 0.6 Q5+=

Conclusion: To determine the signs of the element forces Q directly on the right hand side of theabove equilibrium equations we can use the element force orientation at the element side of the cut ofthe corresponding node free body.

Page 40


The coefficients of the element forces Q (note that they range in absolute value from 0 to 1, since theyrepresent direction cosines) can be collected in static or equilibrium matrix of the structure for the free dofs.

We can then write the above equations in compact form: after denoting the applied forces at the free dofswith Pf the system of equilibrium equations for the free dofs of the structure becomes Pf = Bf * Q

Pf Bf Q= with Pf

P1

P2

P3

P4

P5

= Bf

1

0

0

0

0

1

0

1

0

0

0

0

0

0.8

0.6

0

1

0

0

1

0

0

0.8

0.8

0.6

:= and Q

Q1

Q2

Q3

Q4

Q5

=

Static (equilibrium) matrix and force influence matrix Equation for dof 1

Equation for dof 2

The static matrix for the free dofs is Bf

1

0

0

0

0

1

0

1

0

0

0

0

0

0.8

0.6

0

1

0

0

1

0

0

0.8

0.8

0.6

= Equation for dof 3

Equation for dof 4

Equation for dof 5

In the static matrix for the free dofs of a truss structure rows correspond to equilibriumequations and columns to the basic force of one element

Note that the static matrix for the free dofs Bf only depends on the geometry of the structural model; it isindependent of the loading. Thus, it can be set up for a given model geometry, without specification of theapplied forces, exactly as we have done in this example.If Bf is a square and nonsingular matrix, we talk of a statically determinate, stable structure.In such case we can invert the matrix to obtain the (basic) force influence matrix Bbar.

Bbar Bf1:= Bbar

1

0

0

0

0

0.667

0.667

0.833

1

0.833

1

1

0

0

0

0.5

0.5

0.625

0

0.625

0.667

0.667

0.833

0

0.833

=

The rows of the basic force influence matrix correspond to the longitudinal basic force in a particularelement (e.g. the coefficients in row 3 correspond to element c), and the columns correspond to aparticular dof (e.g. the second column corresponds to the effect of a force at dof 2). Thus the term (i,j) ofthe basic force influence matrix shows the effect of a unit force at dof j on the basic force i of the structure.This basic force influence matrix coefficient is very useful in design optimization problems, since it showswhich member is most affected by the given loading.

Remark: the static matrix at the free dofs Bf and the basic force influence matrix Bbar depend only on themodel geometry for linear statics, in which case the equilibrium equations are formulated in the undeformed(original) configuration of the structural model. When the deformed shape deviates appreciably from theundeformed configuration, the equilibrium equations should be established in the deformed configuration.This is covered in Nonlinear Structural Analysis.

Page 41


Determination of basic forces for specific load case

For a specific load case the applied forces at the free dofs are given. It is then possible to solve the systemof equilibrium equations for the unknown element forces Q. In a stable, statically determinate structure forwhich Bf is square and nonsingular there is a unique set of element forces Q that equilibrate the givenloading (unique solution of equilibrium equations). Note that the solution does not depend on elementproperties, which means that the member sizes do not need to be known beforehand.

88

6

10

5

Pf

0

5

0

10

0

:=

solve for unknown element forces by Gauss elimination (function lsolve in Mathcad, \ in Matlab)

basic forces in truss elements(+ : tension, -: compression)Q lsolve Bf Pf,( ):= Q

8.33

8.33

2.08

5

10.42

=

Note that if we have spent the effort to determine the basic force influence matrix (getting the inverse of amatrix is more work than performing Gauss elimination!), then we can obtain the element forces also bymultiplying the second column of Bbar by (-5) and the fourth column by 10 and adding up the results.

Q Bbar 2 5( ) Bbar 4 10+:= Q

8.33

8.33

2.08

5

10.42

=

Thus, the end forces result from the linear combination of the 2nd and 5th column of the basic forceinfluence matrix after these are factored by the values of the force acting at the corresponding dof.

Page 42


Support reactions and global equilibriumWe can now determine the support reactions from the equilibrium equations at the restrained dofs

12

3

45

67 8

1Q 2Q

4Q3Q 5Q

Equilibrium equations at restrained dofs

P6 Q1 0.8 Q3=

P7 0.6 Q3=

P8 0.6 Q5=

We do this directly for each support reaction (for convenience we can use the symbol R to denote thesupport reactions; in such case we number these independently).

R1 P6= Q1 0.8 Q3= R1 Q1 0.8 Q3:= R1 10=

R2 P7= 0.6 Q3= R2 0.6 Q3:= R2 1.25=

R3 P8= 0.6 Q5= R3 0.6 Q5:= R3 6.25=

and with these we can check the global equilibrium (a very important check indeed!)

88

6

10

5

10

1.25 6.25

In addition to the trivial sum of the forces in X and sum of the forces in Y (which are satisfied by inspection!)we check the sum of the moments about the left and right support (only one of the two is really necessary)

sum of moments about left support (CCW is +ve) 10 6 5 8 R3 16+ 0=

sum of moments about right support (CCW is +ve) 10 6 5 8+ R2 16 0= Ok!

Page 43


If we are interested in the support reaction influence matrix, then we set up the static matrix for the supportdofs consisting of the coefficients of the basic forces in the equilibrium equations for the restrained dofs ofthe structure. We have from the earlier equations

R1 P6= Q1 0.8 Q3= eq for dof 6thus the complete staticmatrix for the support dofs isR2 P7= 0.6 Q3= Bd

1

0

0

0

0

0

0.8

0.6

0

0

0

0

0

0

0.6

:= eq for dof 7eq for dof 8

R3 P8= 0.6 Q5=

The support reaction influence matrix is the product of Bd and Bbar and contains the coefficients for theeffect of the applied forces at the free dofs on the support reactions.

Bd Bbar

1

0

0

0

0.5

0.5

1

0

0

1

0.375

0.375

0

0.5

0.5

= e.g. a unit force at dof 4 produces R1 = -1, R2 = -0.375and R3 = 0.375

Note: if we had bothered setting up the matrix Bd , we could determine the support reactions from the basicelement forces with the following expression

support reactions R Bd Q:= R

10

1.25

6.25

=

but the direct determination earlier is faster for hand calculations.

CONCLUSIONS:

1. In a stable, statically determinate structure there are as many equilibrium equations for the free dofs ofthe structural model as there are unknown basic element forces. The static matrix for the free dofs Bf is,therefore, square and invertible. Note that the inclusion of the support reactions does not change anythingto this discussion, since there are as many equilibrium equations at the support dofs as unknown supportreactions. Thus, it is better to work with the static matrix for the free dofs only.

2. The static matrix for the free dofs of the structure only depends on the geometry of the structural modelin its original (undeformed) configuration (linear equilibrium).

3. The inverse of the static equilibrium matrix for the free dofs Bf gives the basic force influence matrixBbar. This matrix is useful in many applications, such as design optimization, system identification, damageassessment.

4. For a particular loading, the equilibrium equations give a unique solution for the basic forces of thestructural model. The element properties are not required, thus the element sizes need not be available.

5. The support reactions are calculated after the determination of basic forces.

Page 44

Matlab script for Example 2 in CE220 class notes

% Solution for statically determinate truss

OPTION A: perform each step with Matlab functions

% clear memory clear all % specify static (equilibrium) matrix Bf Bf = [1 -1 0 0 0; 0 0 0 -1 0; 0 1 0 0 0.8; 0 0 0.8 0 -0.8; 0 0 0.6 1 0.6]; % determine and display force influence matrix Bbar Bbar = inv(Bf); disp('the force influence matrix Bbar is'); disp(Bbar); the force influence matrix Bbar is 1.0000 -0.6667 1.0000 0.5000 -0.6667 0 -0.6667 1.0000 0.5000 -0.6667 0 0.8333 0 0.6250 0.8333 0 -1.0000 0 0 0 0 0.8333 0 -0.6250 0.8333

specify applied force vector

Pf = [ 0; -5; 0; 10; 0]; % solve for basic forces Q (Note the use of the backslash or left matrix divide operator which computes the solution for a linear system of equations by Gauss elimination) Q = Bf\Pf; % display result format short disp('the basic forces are'); disp(Q); the basic forces are 8.3333 8.3333 2.0833 5.0000 -10.4167

OPTION B: use FEDEASLab functions

% clear memory; close any open windows CleanStart; % define model geometry XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 8 0]; % second node, etc XYZ(3,:) = [ 16 0]; %

Page 45

XYZ(4,:) = [ 8 6]; % % element connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 1 4]; CON { 4} = [ 2 4]; CON { 5} = [ 3 4]; % boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's) BOUN(1,:) = [ 1 1]; BOUN(3,:) = [ 0 1]; % specify element type ne = length(CON); % number of elements [ElemName{1:ne}] = deal('Truss'); % truss element

create Model

Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName); % plot and label model for checking (optional) Create_Window (0.80,0.80); % open figure window Plot_Model (Model); % plot model Label_Model (Model); % label model

form static (equilibrium) matrix B

B = B_matrix(Model); % extract submatrix for free dofs Bf = B(1:Model.nf,:);

specify loading

Pe(2,2) = -5; % force at node 2 in direction Y Pe(4,1) = 10; % force at node 4 in direction X % generate data object Loading Loading = Create_Loading(Model,Pe); % extract applied force vector Pf at free dofs from Loading field Pref Pf = Loading.Pref;

Page 46

solution for basic forces

solve for basic forces and display the result Q = Bf\Pf; disp('the basic forces are'); disp(Q); the basic forces are 8.3333 8.3333 2.0833 5.0000 -10.4167

determination of support reactions

% the product B*Q delivers all forces at the global dofs % the upper 5 should be equal to the applied forces, the lower 3 are the support reactions disp('B*Q gives'); disp(B*Q); B*Q gives 0 -5.0000 0 10.0000 0.0000 -10.0000 -1.2500 6.2500

post-processing of results

% plot axial force distribution Create_Window(0.80,0.80); Plot_Model(Model); Plot_AxialForces(Model,Q);

Page 47

CE 220 - Theory of Structures Ex. 3 - Beam equilibrium Prof. Filip C. Filippou, 2000

Example 3(r) - Equilibrium equations for statically determinate 2d beam

Objectives:

(a) set up of equilibrium equations for a statically determinate beam(b) systematic reduction of equilibrium equations to exclude longitudinal basic forces for specific loading(c) systematic reduction of equilibrium equations involving a single unknown basic force(d) relation between basic forces and internal forces; bending moment and shear force diagrams(e) influence lines for moving loads

Geometry of structural model

a b c1

2 3 4

10 10 5

Case A: Complete set of equilibrium equations and basic element forces

12

3

4 56

7

8

9

a

b

c

1Q2Q 3Q

4Q5Q 6Q

7Q8Q 9Q

Complete set of equilibrium equations and basic element forces

X-direction Y-direction Moments about Z-axis

P1 Q2=

P2 Q1 Q4= P3Q2 Q3+

10

Q5 Q6+

10+= P4 Q3 Q5+=

P5 Q4 Q7= P6 Q6 Q8+=

P7 Q7= P8Q8 Q9+

5= P9 Q9=

Page 48

CE 220 - Theory of Structures Ex. 3 - Beam equilibrium Prof. Filip C. Filippou, 2000

We distinguish 3 horizontal force equilibrium equations (equations 2, 5 and 7) involving 3 longitudinalbasic forces. We note that the longitudinal basic forces Q1, Q4 and Q7 do not appear in the otherequations. We can therefore uncouple these three equations from the rest and solve themindependently. We could have concluded this independence of equations with longitudinal basic forcesby inspection at the start of the analysis. If we are not interested in the longitudinal basic forces, we canset up only the equilibrium equations involving the flexural basic element forces. This is shown next

Case B: Complete set of equilibrium equations for flexural basic element forces

12

34

5

6

a b c1

2 3 4

10 10 5

a

b

c

1Q 2Q

3Q 4Q

5Q 6Q

Y-direction Moments about Z-axis

P1 Q1=

P2Q1 Q2+

10

Q3 Q4+

10+= P3 Q2 Q3+=

P4 Q4 Q5+=

P5Q5 Q6+

5= P6 Q6=

Looking at these 6 equations and the unknown 6 basic element forces we conclude that equations 1 and 6involve a single unknown basic force each. We can therefore solve for the corresponding unknown basicforce independently from the rest. Assuming that the applied force at this degree of freedom is zero in mostcases, we can proceed to remove the two equations and the corresponding basic forces from the system ofequations. This process can be done by inspection at the start of the analysis once the loading is given. Ifthe applied forces at dofs 1 and 6 are not equal to zero, the corresponding basic forces will not be zero andwe should be careful to include the effect of the non-zero basic force on the remaining equilibriumequations. We will see later in this example how this is done.

Page 49

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