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INTERNATIONAL INSTITUTE OF TECHNOLOGY & MANAGEMENT, MURTHAL SONEPATE-NOTES , Subject :Structural analysis—III, Course code –CE-308B Course:-B.Tech ,Branch: Civil Engineering ,

Sem-6th , ( Prepared By: Mrs. Swati kuhar , Assistant Professor , CE)

UNIT- 1

Matrix MethodINTRODUCTION TO MATRIX METHODS

MATRIX METHODS OF ANALYSIS :Broadly the methods of analysis are categorised in two ways

1. Force Methods : Methods in which forces are made unknowns i.e Method of consistent deformation and strain energy method. In both these methods solution of number of simultaneous equations is involved.

2. Displacement Methods in which displacements are made unknowns i.e slope deflection method, Moment distribution method and Kani’s Method (In disguise). In slope deflection method also, the solution of number of simultaneous equations is involved.In both of the above methods, for the solution of simultaneous equations matrix approach can be employed & such Method is called Matrix method of analysis.

FORCE METHOD : Method of consistent deformation is the base and forces are made unknown

∆b = Upward Deflection of point B on primary structure due to all causes∆bo = Upward Deflection of point B on primary structure due to applied

load(Redundant removed i.e condition Xb = 0)∆bb = Upward Deflection of point B on primary structure due to Xb

( i.e Redundant )δbb = Upward Deflection of point B on primary structure due to Xb = 1∆bb = δbb . Xb



∆b = ∆bo + ∆bb Substituting for ∆bb –∆b = ∆bo + δbb . Xb Called Super position equation

Using the compatibility condition that the net displacement at B = 0 i.e

∆b = 0 we get Xb = – ∆bo / δbbTo Conclude we can say, [ ∆ ] = [∆L ] +[ ∆ R]

DISPLACEMENT METHOD:This method is based on slope deflection method and displacements are made unkownswhich are computed by matrix approch instead of solving simultaneous equations and finaly unknown forces are calculated using slpoe deflection equations.Mab = Mab + 2EI / L ( 2 a + b + 3 / L)Mab = Final Moment and may be considered as net force P at the joint Mab = Fixed end moment i.e Force required for the condition of zero displacements & is called locking force. ( i.e. P’)The second term may be considred as the force required to produce the required displacements at the joints. (i.e Pd )Therefore the above equation may be written as [P] = [P’] + [Pd]Thus, there are Two Methods in matrix methods

MATRIX METHODS

FORCE METHOD DISPLACEMENT METHOD

The force method is also called by the names 1) Flexibility Method 2)Static Method 3)Compatibility.Similarly the displacement method is also called by the names 1)Stiffness Method2) Kinematic method 3) Equilibrium Method.In both force method & displacement method there are two different approaches 1) System Approach 2) Element Approach.

To study matrix methods there are some pre-requisites :i) Matrix Algebra - Addition, subtraction ,Multiplication & inversion of

matrices (Adjoint Method )ii) Methods of finding out Displacements i.e. slope & deflection at any

point in a structure, such as a)Unit load method or Strain energy method b) Moment area method etc.

According to unit load method the displacement at any point ‘j’ is given by∆j = ∫0 Mmjds / EIWhere



M – B M due to applied loads & mj – B M due to unit load at j



δij = ∫0 mi.mj.ds / EI



When unit load is applied at i and is called flexibility coefficient.The values of ∆j and δij can be directly read from the table depending upon the combinations of B M diagrams & these tables are called Diagram Multipliers.iii) Study of Indeterminacies – Static indeterminacy & kinematic indeterminacy



INDETERMINATE STRUCTURES

1. Statically Indeterminate Structure2. Kinematically Indeterminate Structure INDETERMINATE STRUCTURESStatically Indeterminate Structure : Any structure whose reaction components or internal stresses cannot be determined by using equations of static equilibrium alone, (i.e.Fx = 0, Fy = 0, Mz = 0) is a statically Indeterminate Structure.The additional equations to solve statically indeterminate structure come from the conditions of compatibility or consistent displacements.

Roller Support : No. of reactions,r = 1

Hinged Support : No. of reactions, r = 2

Fixed Support : No. of reactions, r = 3

1. Pin Jointed Structures i.e. TrussesInternal static indeterminacy : (Dsi) No. of members required for stability is given by – 3 joints – 3 members – every additional joint requires two additional members. m’ = 2 (j – 3)+3 j = No. of joints m’ = 2j – 3 Stable and statically determinate Dsi = m – m’ Where m = No. of members in a structure Dsi = m – (2j – 3)External static indeterminacy (Dse)r = No. of reaction componentsEquations of static equilibrium = 3 (i.e.Fx = 0, Fy = 0, Mz = 0) Dse = r – 3Total static indeterminacy Ds = Dsi + Dse Ds = m –(2j – 3)+(r – 3) Ds = (m+r) – 2jRigid Jointed Structures : No. of reaction components over and above the no. of equations of static equilibrium is called a degree of static indeterminacy.



Ds = r-3Equations of static equilibrium = 3(i.e.Fx = 0, Fy = 0, Mz = 0)

Example 1

No. of reaction components r = 5 (as shown) Ds = r – 3 = 5 – 3 = 2 Ds = 2

Introduce cut in the member BC as shown. At the cut the internal stresses are introduced i.e. shear force and bending moment as shown.Left part : No. of unknowns = 5 Equations of equilibrium = 3 Ds = 5 – 3 = 2Right Part : No. of unknowns = 4 Equations of equilibrium = 2 Ds = 4 – 2 = 2 Ds = Static Indeterminacy = 2Example 2Fig. (A) Fig. (B) Fig. (C)



Another Approach : For Every member in a rigid jointed structure there will be 3 unknowns i.e. shear force, bending moment, axial force.

Let r be the no. of reaction components and m be the no. of membersTotal no. of unknowns = 3m + rAt every joint three equations of static equilibrium are available no. of static equations of equilibrium = 3j (where j is no. of joints)

Ds = (3m + r) – 3j In the example r = 6, m = 6, j = 6Ds = (3 x 6 + 6) – (3 x 6 ) = 6



Kinematic Indeterminacy :

A structure is said to be kinematically indeterminate if the displacement components of its joints cannot be determined by compatibility conditions alone. In order to evaluate displacement components at the joints of these structures, it is necessary to consider the equations of static equilibrium. i.e. no. of unknown joint displacements over and above the compatibility conditions will give the degree of kinematic indeterminacy.

Fixed beam : Kinematically determinate :

Simply supported beam Kinematically indeterminate

Any joint – Moves in three directions in a plane structure Two displacements x in x direction, y in y direction, rotation about z axis as shown.

Roller Support :r = 1, y = 0, & x exist – DOF = 2 e = 1

Hinged Support :r = 2, x = 0, y = 0, exists – DOF = 1 e = 2

Fixed Support :r = 3, x = 0, y = 0, = 0 DOF = 0 e = 3

i.e. reaction components prevent the displacements no. of restraints = no. of reaction components.Degree of kinematic indeterminacy :Pin jointed structure :Every joint – two displacements components and no rotation



Dk = 2j – ewhere,e = no. of equations of compatibility

j = no. of reaction components

Rigid Jointed Structure : Every joint will have three displacement components, two displacements and one rotation.Since, axial force is neglected in case of rigid jointed structures, it is assumed that the members are inextensible & the conditions due to inextensibility of members will add to the numbers of restraints. i.e to the ‘e’ value.

Dk = 3j – ewhere,e = no. of equations of compatibility

j= no. of reaction components + constraints due to in extensibility

Example 1 : Find the static and kinematic indeterminacies

r = 4, m = 2, j = 3

Ds = (3m + r) –3j

= (3 x 2 + 4) – 3 x 3 = 1Dk = 3j – e

= 3 x 3 – 6 = 3i.e. rotations at A,B, & C i.e. a, b & c are

the displacements.(e = reaction components + inextensibility conditions = 4 + 2 = 6)

Example 2 :Ds = (3m+r) – 3j m = 3, r = 6, j = 4 Ds = (3 x 3 + 6) – 3 x 4 = 3

Dk = 3j – e e = no. of reactioncomponents + conditions ofinextensibility

= 6+3 = 9



Dk = 3 x 4 – 9 = 3 i.e. rotation b, c & sway.



Example 3 :Ds = (3m + r) – 3j r = 6, m = 10, j = 9

Ds = (3 x 10 + 6) – 3 x 9 = 9

Conditions of inextensibility : Joint : B C E F H I

1 1 2 2 2 2 Total = 10

Reaction components r = 6

e = 10 + 6 = 16

Dk = 3j – e= 3 x 9 – 16 = 11



FORCE METHOD :

This method is also known as flexibility methodor compatibility method. In this method the degree of static indeterminacy of the structure is determined and the redundants are identified. A coordinate is assigned to each redundant. Thus,P1, P2-----------Pn are the redundants at the coordinates1,2,---------n.If all the redundants are removed , the resulting structure known asreleased structure, is statically determinate. This released structure is also known as basic determinate structure. From the principle of super position the net displacement at any point in statically indeterminate structure is some of the displacements in the basic structure due to the applied loads and the redundants. This is known as the compatibility condition and may be expressed by the equation;

∆1 = ∆1L + ∆1R Where ∆1 - - - - ∆n = Displ. At Co-ord.at 1,2 - -n∆2 = ∆2L + ∆2R ∆1L ---- ∆nL = Displ.At Co-ord.at 1,2---------n| | | Due to aplied loads| | | ∆1R ----∆nR = Displ.At Co-ord.at 1,2---------n∆n = ∆nL + ∆nR Due to Redudants

The above equations may be return as [∆] = [∆L] + [∆R]---------(1)∆1 = ∆1L + δ11 P1 + δ12 P2 +----------δ1nPn

∆2 = ∆2L + δ21 P1 + δ22 P2 +---------δ2nPn

| | | | || | | | | - - - - - (2)

∆n = ∆nL + δn1 P1 + δn2 P2 +---------δnnPn



∆ = [∆ L] + [δ] [P]----------------------(3)

[P]= [δ]-1 {[∆] – [∆ L]}-----------------(4)If the net displacements at the redundants are zero then∆1, ∆2-------∆n =0,Then [P] = - [δ] -1 [∆ L]--------------(5)The redundants P1,P2,---------Pn are Thus determinedDISPLACEMENT METHOD :This method is also known as stiffness or equilibrium. In this method the degree of kinematic indeterminacy (D.O.F) of the structure is determined and the coordinate is assigned to each independent displacement component.

In general, The displacement components at the supports and joints are treated as independent displacement components. Let 1,2,------n be the coordinatesassigned to these independent displacement components ∆1, ∆2-------∆n.

In the first instance lock all the supports and the joints to obtain the restrained structure in which no displacement is possible at the coordinates. Let P’1, P’2 , - - - - P’n be the forces required at the coordinates 1,2,----n in therestrained structure in which the displacements ∆1, ∆2-------∆n are zero. Next, Letthe supports and joints be unlocked permitting displacements ∆1, ∆2-------∆n atthe coordinates. Let these displacements require forces in P1d, P2d,-------Pnd atcoordinates 1,2,-------n respectively.If P1, P2- - - - - - Pn are the external forces at the coordinates 1,2,-------n, then theconditions of equilibrium of the structure may be expressed as:

P1 = P’1 +P1dP2 = P’2 +P2d-------------------------------------(1)| | || | |Pn = P’n + Pnd

or [P ] = [P’] + [ Pd]-------------------(2)

P1 = P’1 +K11 ∆1+ K12, ∆2+ K13∆3 +-----------K1n ∆nP2 = P’2 +K21 ∆1+ K22 ∆2+ K23∆3 +------------K2n ∆n| | | | | || | | | | | - - - - - - - - - (3)Pn = P’n +Kn1 ∆1+ Kn2, ∆2+ Kn3∆3 + - - - - - - - - -Knn ∆n

i.e [P ] = [P’ ] + [ K ] [ ∆ ]---------------------(4) [ ∆ ]= [ K ] –1 {[P] – [P’]}---------------------------------------(5)

Where P = External forces



P’ = Locking forcesPd=Forces due to displacements

If the external forces act only at the coordinates the terms P’1, P’2 ,--------P’nvanish. i. e the Locking forces are zero,then [ ∆ ]= [ K ] –1 [ P ]-------------------------(6)On the other hand if there are no external forces at the coordinates then [P]=0

then [ ∆ ]= – [ K ] –1 [ P’ ]----------------------(7)Thus the displacements can be found out. Knowing the displacements the forces are computed using slope deflection equations:Mab= Mab+ 2EI / L (2a+ b+3 / L) Mba=Mba+ 2EI / L (a+ 2b+3 / L)Where Mab& Mba are the fixed end moments for the member AB due to external loading



FLEXIBILITY AND STIFFNESS MATRICES : SINGLE CO-ORD.

D= δ x P D=PL3/3EI = δ x P δ=L3 /3EIδ=Flexibility Coeff.

P=K x DP=K x PL3 / 3EI K=3EI / L3

K=Stiffness Coeff.

M=K x D =K x ML/EI K=EI / L

K=Stiffness Coeff.δ X K= 1

TWO CO-ORDINATE SYSTEMD1= δ11P1 + δ12 X P2 & D2= δ21P1+ δ22P2 D1 δ11 δ12 P1

D2 δ21 δ22 P2

[δ ]=L3 / 3EI

L2 / 2EI

L2 / 2EIL / EI

Unit Force At Co-ord.(1)δ11=L3 / 3EI δ21=L2 / 2EI

Unit Force At Co-ord.(2) δ12=δ21 =L2 / 2EI δ22=L / EI

D= ML / EID= δ x M=ML/EI δ=L / EIδ=Flexibility Coeff.



Develop the Flexibility and stiffness matrices for frame ABCD with reference to Coordinates shown

The Flexibility matrix can be developed by applying unit force successively at coordinates (1),(2) &(3) and evaluating the displacements at all the coordinates

δij = ∫ mi mj / EI x ds δij =displacement at I due to unit load at j



INVERSING THE FLEXIBILITY MATRIX [ δ ]THE STIFENESS MATRIX [ K ] CAN BE OBTAINED



UNIT-2U

77777



INTERNATIONAL INSTITUTE OF TECHNOLOGY & MANAGEMENT, MURTHAL SONEPAT

E-NOTES , Subject :Structural analysis—III, Course code –CE-308B Course:-B.Tech ,Branch: Civil Engineering , Sem-6th ,

( Prepared By: Mrs. Swati kuhar , Assistant Professor , CE)

Introduction 1INTERNATIONAL INSTITUTE OF TECHNOLOGY & MANAGEMENT, MURTHAL

SONEPATE-NOTES , Subject :Structural analysis—III, Course code –CE-308B Course:-

B.Tech ,Branch: Civil Engineering , Sem-6th , ( Prepared By: Mrs. Swati kuhar , Assistant Professor , CE)

UNIT 3 Direct stiffness method and the global stiffness matrix Although there are several finite element methods, we analyse the Direct Stiffness Method here, since it is a good starting point for understanding the finite element formulation. We consider first the simplest possible element – a 1-dimensional elastic spring which can accommodate only tensile and compressive forces. For the spring system shown, we accept the following conditions:

Condition of Compatibility – connected ends (nodes) of adjacent springs have the same displacements

Condition of Static Equilibrium – the resultant force at each node is zero

Constitutive Relation – that describes how the material (spring) responds to the applied loads

Model spring systemThe constitutive relation can be obtained from the governing equation for an elastic bar loaded axially along its length:

ddu(AEΔll0)+k=0(1)Δll0=ε(2)ddu(AEε)+k=0(3)ddu(Aσ)+k=0(4)dFdu+k=0(5)dFdu=−k(6)dF=−kdu(7)The spring stiffness equation relates the nodal displacements to the applied forces via the spring (element) stiffness. The minus sign denotes that the force is a restoring one, but from here on in we use the scalar version of Eqn.7.

Derivation of the Stiffness Matrix for a Single Spring Element




From inspection, we can see that there are two degrees of freedom in this model, ui and uj. We can write the force equilibrium equations:k(e)ui−k(e)uj=F(e)i(8)−k(e)ui+k(e)uj=F(e)j(9)In matrix form

[ke−ke−keke]{uiuj}={F(e)iF(e)j}(10)The order of the matrix is [2×2] because there are 2 degrees of freedom. Note also that the matrix is symmetrical. The ‘element’ stiffness relation is:[K(e)]{u(e)}={F(e)}(11)Where Κ(e) is the element stiffness matrix, u(e) the nodal displacement vector and F(e) the nodal force vector. (The element stiffness relation is important because it can be used as a building block for more complex systems. An example of this is provided later.)Derivation of a Global Stiffness Matrix

For a more complex spring system, a ‘global’ stiffness matrix is required – i.e. one that describes the behaviour of the complete system, and not just the individual springs.

From inspection, we can see that there are two springs (elements) and three degrees of freedom in this model, u1, u2 and u3. As with the single spring model above, we can write the force equilibrium equations:

k1u1−k1u2=F1(12)−k1u1+(k1+k2)u2−k2u3=F2(13)k2u3−k2u2=F3(14)In matrix form

k1−k10−k1k1+k2−k20−k2k2 u1u2u3 =⎡⎣⎢ ⎤⎦⎥⎧⎩⎨⎪⎪ ⎫⎭⎬⎪⎪ ⎧⎩⎨F1F2F3 (15)⎪⎪ ⎫⎭⎬⎪⎪




The ‘global’ stiffness relation is written in Eqn.16, which we distinguish from the ‘element’ stiffness relation in Eqn.11.[K]{u}={F}(16)Note the shared k1 and k2 at k22 because of the compatibility condition at u2. We return to this important feature later on.Assembling the Global Stiffness Matrix from the Element Stiffness Matrices

Although it isn’t apparent for the simple two-spring model above, generating the global stiffness matrix (directly) for a complex system of springs is impractical. A more efficient method involves the assembly of the individual element stiffness matrices. For instance, if you take the 2-element spring system shown,

split it into its component parts in the following way

and derive the force equilibrium equationsk1u1−k1u2=F1(17)k1u2−k1u1=k2u2−k2u3=F2(18)k2u3−k2u2=F3(19)then the individual element stiffness matrices are:[k1−k1−k1k1]{u1u2}={F1F2}and[k2−k2−k2k2]{u2u3}={F2F3}(20)such that the global stiffness matrix is the same as that derived directly in Eqn.15:




(Note that, to create the global stiffness matrix by assembling the element stiffness matrices, k22 is given by the sum of the direct stiffnesses acting on node 2 – which is the compatibility criterion. Note also that the indirect cells kij are either zero (no load transfer between nodes i and j), or negative to indicate a reaction force.)For this simple case the benefits of assembling the element stiffness matrices (as opposed to deriving the global stiffness matrix directly) aren’t immediately obvious. We consider therefore the following (more complex) system which contains 5 springs (elements) and 5 degrees of freedom (problems of practical interest can have tens or hundreds of thousands of degrees of freedom (and more!)). Since there are 5 degrees of freedom we know the matrix order is 5×5. We also know that it’s symmetrical, so it takes the form shown below:

We want to populate the cells to generate the global stiffness matrix. From our observation of simpler systems, e.g. the two spring system above, the following rules emerge:

The term in location ii consists of the sum of the direct stiffnesses of all the elements meeting at node i

The term in location ij consists of the sum of the indirect stiffnesses relating to nodes i and j of all the elements joining node i to j




Add a negative for reaction terms (–kij) Add a zero for node combinations that don’t interact

By following these rules, we can generate the global stiffness matrix:

This type of assembly process is handled automatically by commercial FEM codes

Drag the springs into position and click 'Build matrix', then apply a force to node 5. You will then see the force equilibrium equations, the equivalent spring stiffness and the displacement at node 5.

Solving for (u)The unknowns (degrees of freedom) in the spring systems presented are the displacements uij. Our global system of equations takes the following form:

To find {u} solveu=F[K]−1(22)Recall that [k][k]−1=I=IdentitiyMatrix=[1001].Recall also that, in order for a matrix to have an inverse, its determinant must be non-zero. If the determinant is zero, the matrix is said to be singular and no unique solution for Eqn.22 exists. For instance, consider once more the following spring system:




We know that the global stiffness matrix takes the following formk1−k10−k1k1+k2−k20−k2k2 u1u2u3 =⎡⎣⎢ ⎤⎦⎥⎧⎩⎨⎪⎪ ⎫⎭⎬⎪⎪ ⎧⎩⎨F1F2F3 (23)⎪⎪ ⎫⎭⎬⎪⎪

The determinant of [K] can be found from:det adgbehcfi =(aei+bfg+cdh)−(ceg+bdi+afh)(24)⎡⎣⎢ ⎤⎦⎥Such that:

(k1(k1+k2)k2+0+0)−(0+(−k1−k1k2)+(k1−k2−k2))(25)det[K]=(k12k2+k1k22)−(k12k2+k1k22)=0(26)Since the determinant of [K] is zero it is not invertible, but singular. There are no unique solutions and {u} cannot be found. If this is the case in your own model, then you are likely to receive an error message!

1Instructional Objectives

After reading this chapter the student will be able to1. Derive member stiffness matrix of a beam element.2. Assemble member stiffness matrices to obtain the global stiffness matrix for a beam.3. Write down global load vector for the beam problem.4. Write the global load-displacement relation for the beam.

Introduction.

In chapter 23, a few problems were solved using stiffness method from fundamentals. The procedure adopted therein is not suitable for computer implementation. In fact the load displacement relation for the entire structure was derived from fundamentals. This procedure runs into trouble when the structure is large and complex. However this can be much simplified provided we follow the procedure adopted for trusses. In the case of truss, the stiffness matrix of the entire truss was obtained by assembling the member stiffness matrices of individual members.In a similar way, one could obtain the global stiffness matrix of a continuous beam from assembling member stiffness matrix of individual beam elements. Towards




this end, we break the given beam into a number of beam elements. The stiffness matrix of each individual beam element can be written very easily.For example, consider a continuous beam ABCD as shown in Fig. 27.1a. Thegiven continuous beam is divided into three beam elements as shown in Fig. 27.1b. It is noticed that, in this case, nodes are located at the supports. Thus each span is treated as an individual beam. However sometimes it is required to consider a node between support points. This is done whenever the cross sectional area changes suddenly or if it is required to calculate vertical or rotational displacements at an intermediate point. Such a division is shown in Fig. 27.1c. If the axial deformations are neglected then each node of the beam will have two degrees of freedom: a vertical displacement (corresponding to shear) and a rotation (corresponding to bending moment). In Fig. 27.1b, numbersenclosed in a circle represents beam numbers. The beam ABCD is divided intothree beam members. Hence, there are four nodes and eight degrees of freedom. The possible displacement degrees of freedom of the beam are also shown in the figure. Let us use lower numbers to denote unknown degrees of freedom (unconstrained degrees of freedom) and higher numbers to denote known (constrained) degrees of freedom. Such a method of identification is adopted in this course for the ease of imposing boundary conditions directly on the structure stiffness matrix. However, one could number sequentially as shown in Fig. 27.1d. This is preferred while solving the problem on a computer.




In the above figures, single headed arrows are used to indicate translational and double headed arrows are used to indicate rotational degrees of freedom.

Beam Stiffness Matrix.

Fig. 27.2 shows a prismatic beam of a constant cross section that is fully restrained at ends in local orthogonal co-ordinate system x' y' z' . The beam endsare denoted by nodes j and k . The x' axis coincides with the centroidal axis ofthe member with the positive sense being defined from j to k . Let L be the lengthof the member, A area of cross section of the member and I zz is the moment ofinertia about z' axis.

Two degrees of freedom (one translation and one rotation) are considered at each end of the member. Hence, there are four possible degrees of freedom for this member and hence the resulting stiffness matrix is of the order 4 4 . In this method counterclockwise moments and counterclockwise rotations are taken as positive. The positive sense of the translation and rotation are also shown in the figure. Displacements are considered as positive in the direction of the coordinate




axis. The elements of the stiffness matrix indicate the forces exerted on




the member by the restraints at the ends of the member when unit displacements are imposed at each end of the member. Let us calculate the forces developed in the above beam member when unit displacement is imposed along each degree of freedom holding all other displacements to zero. Now impose a unitdisplacement along y' axis at j end of the member while holding all otherdisplacements to zero as shown in Fig. 27.3a. This displacement causes both shear and moment in the beam. The restraint actions are also shown in the figure. By definition they are elements of the member stiffness matrix. In particular they form the first column of element stiffness matrix.In Fig. 27.3b, the unit rotation in the positive sense is imposed at j end of the beam while holding all other displacements to zero. The restraint actions are shown in the figure. The restraint actions at ends are calculated referring totables given in lesson …

⎡ 12 EI z⎢⎢⎢

6 EI z

L2

12 EI z 6 EI z⎤ 1

L3

z 6 EI 4 EI⎢

z

L3

6 EI

L2⎥ z ⎥2 EI ⎥ 2

⎢ 12 EI z

L2

⎢⎢

6 EI z

z

L

⎢ 6 EI L2

L3

z z

L2

12 EI z

L3

6 EI

6 EI z ⎥ 3L ⎥

⎣2 EI

L

L2 L2

⎥⎥

z

L2

4 EIL

z ⎥ 4⎦




In Fig. 27.3c, unit displacement along y' axis at end k is imposed andcorresponding restraint actions are calculated. Similarly in Fig. 27.3d, unitrotation about z' axis at end k is imposed and corresponding stiffnesscoefficients are calculated. Hence the member stiffness matrix for the beam member is

1 2 3 4

k

(27.1)

The stiffness matrix is symmetrical. The stiffness matrix is partitioned to separate the actions associated with two ends of the member. For continuous beam




problem, if the supports are unyielding, then only rotational degree of freedom

⎡ 4EIz⎢

2EIz ⎤⎢ 2EIL⎢ z

4EI ⎥L ⎥

⎣L z ⎥

L ⎦




shown in Fig. 27.4 is possible. In such a case the first and the third rows and columns will be deleted. The reduced stiffness matrix will be,

k (27.2)

Instead of imposing unit displacement along y' at j end of the member in Fig.a, apply a displacement u'1 along y' at j end of the member as shown in

Fig. 27.5a, holding all other displacements to zero. Let the restraining forcesdeveloped be denoted by q11 , q21 , q31 and q41 .

1

q

⎢⎢⎢ ⎥⎥

⎥

⎥⎥

⎦

2




The forces are equal to,

q11 k11u'1 ; q21 k21u'1 ; q31 k31u'1 ; q41 k41u'1 (27.3)

Now, give displacements u'1 , u'2 , u'3 and u'4 simultaneously along displacementdegrees of freedom 1,2,3 and 4 respectively. Let the restraining forces developedat member ends be q1 , q2 , q3

andq4 respectively as shown in Fig. 27.5b along

respective degrees of freedom. Then by the principle of superposition, the force displacement relationship can be written as,

⎡q ⎤⎡ 12EIz 6EIz 12EIz 6EIz ⎤ ⎡u' ⎤

⎢ ⎥ ⎢⎢ ⎥ ⎢ L3

6EIL2

4EIL3

6EI

2 ⎥2EI ⎥ ⎥

⎢ 2 ⎥ ⎢z z z z ⎥ ⎢u' ⎥⎢ ⎥ ⎢ L2 L L2 L ⎥ ⎢ ⎥⎢ ⎥

⎢ ⎥ ⎢ ⎥ (27.4)⎢

q ⎥ ⎢ 12EIz 6EIz 12EIz

6EIz ⎥ ⎢u' ⎥

⎢ 3 ⎥⎢ ⎥⎢ ⎥⎢L3

⎢⎢ 6EI

L2

2EIL3

6EI

2 ⎢ 3 ⎥⎢ ⎥4EI ⎥ ⎢ ⎥

⎢⎣q4 ⎥⎦ z

⎣ L2 z z L L2

z ⎢⎣u'4 ⎥⎦L

This may also be written in compact form as, q k u'

1L

L

⎢




(27.5)




Beam (global) Stiffness Matrix.

The formation of structure (beam) stiffness matrix from its member stiffness matrices is explained with help of two span continuous beam shown in Fig. 27.6a. Note that no loading is shown on the beam. The orthogonal co-ordinate system xyz denotes the global co-ordinate system.

For the case of continuous beam, the x - and x' - axes are collinear and otheraxes ( y and y' , z and z' ) are parallel to each other. Hence it is not required totransform member stiffness matrix from local co-ordinate system to global co

k '

⎣

⎢ 2

⎣

⎥

k




ordinate system as done in the case of trusses. For obtaining the global stiffness matrix, first assume that all joints are restrained. The node and member numbering for the possible degrees of freedom are shown in Fig 27.6b. The continuous beam is divided into two beam members. For this member there are six possible degrees of freedom. Also in the figure, each beam member with its displacement degrees of freedom (in local co ordinate system) is also shown. Since the continuous beam has the same moment of inertia and span, the member stiffness matrix of element 1 and 2 are the same. They are,

Global d .o. f Local d .o. f

11 k ⎡

'11

22

k '12

33

k '13

44

k '14 1 1⎤

⎢k ' 21⎢ k

'22k '23

k '24 2 ⎥2 (27.6a)

k '31⎢ k '32

k '33

k '34 3 3⎥

⎢41 k

'42k '43

⎥k '44 4 4⎦

Global d .o. f 3 4 5 6Local d .o. f

1 k ⎡

211k ⎢221

2

k 212k 222

3

k 213k 223

4k 214 1 3⎤k 224 2 4⎥

k 2 ⎢ k 31 k 232k 233

⎥k 234 3 ⎥5

(27.6b)

2⎢ 41 k 242

k 243

⎥k 244 4 6⎥⎦

The local and the global degrees of freedom are also indicated on the top and side of the element stiffness matrix. This will help us to place the elements of the element stiffness matrix at the appropriate locations of the global stiffness matrix. The continuous beam has six degrees of freedom and hence the stiffness matrixis of the order 6 6 . Let K denotes the continuous beam stiffness matrix of

k '




order 6 6 . From Fig. 27.6b, K may be written as,




⎡⎢⎢ ⎢

Member AB (1)⎤⎥

⎥⎥ (27.7)K ⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎣⎢ ⎥⎦

Member BC (2)

The 4 4 upper left hand section receives contribution from member 1 ( AB)

and

4 4 lower right hand section of global stiffness matrix receives contribution from member 2. The element of the global stiffness matrix corresponding to global degrees of freedom 3 and 4 [overlapping portion of equation 27.7] receiveselement from both members 1 and 2.

Formation of load vector.

Consider a continuous beam ABC as shown in Fig. 27.7.

We have two types of load: member loads and joint loads. Joint loads could be handled very easily as done in case of trusses. Note that stiffness matrix of each member was developed for end loading only. Thus it is required to replace the member loads by equivalent joint loads. The equivalent joint loads must be evaluated such that the displacements produced by them in the beam should be the same as the displacements produced by the actual loading on the beam. This is evaluated by invoking the method of superposition.

k 111 k 1

12 k 113 k 1

14 0 0

k 121 k 1

22 k 123 k 1

240 0

k 131 k 1

32k 1 k 2

33 11k 1 k 2

34 12 k 213 k 2

14

k 141 k 1

42k 1 k 2

43 21k 1 k 2

44 22 k 223 k 2

24

0 0 k 231 k 2

32 k 233 k 2

34

0 0 k 2 k 2 k 2 k 2




The loading on the beam shown in Fig. 27.8(a), is equal to the sum of Fig. 27.8(b) and Fig. 27.8(c). In Fig. 27.8(c), the joints are restrained against displacements and fixed end forces are calculated. In Fig. 27.8(c) these fixed end actions are shown in reverse direction on the actual beam without any load. Since the beam in Fig. 27.8(b) is restrained (fixed) against any displacement, the displacements produced by the joint loads in Fig. 27.8(c) must be equal to the displacement produced by the actual beam in Fig. 27.8(a). Thus the loads shown

⎪⎪ ⎝

⎪

L ⎠

⎪⎪




in Fig. 27.8(c) are the equivalent joint loads .Let, p1 , p2 , p3 , p4 , p5 and p6 be theequivalent joint loads acting on the continuous beam along displacement degrees of freedom 1,2,3,4,5 and 6 respectively as shown in Fig. 27.8(b). Thus the global load vector is,

⎧⎪⎪⎧ p1 ⎫ ⎪

Pb ⎫⎪⎪

Pab2 ⎪⎪ ⎪ ⎪ L2 ⎪⎪ p2 ⎪ ⎪ ⎪⎪ ⎪ ⎪

⎛ Pa wL ⎞ ⎪

⎪ p ⎪ ⎪ ⎜ ⎟ ⎪⎪ 3 ⎪ ⎪ ⎝ L 2 ⎠ ⎪ (27.8)⎨ ⎬ ⎨p ⎛ wL2 Pba 2 ⎞⎬⎪ 4 ⎪ ⎪ ⎜ ⎟⎪⎪⎪⎪ p5 ⎪⎪⎪

⎜ 12 2 ⎟⎪ ⎛ wL ⎞ ⎪

⎪⎩ p6

⎪⎭

⎜⎪ ⎝ 2⎪ 2P ⎟⎠ ⎪

⎪ wL2 ⎪⎩⎪ 12 ⎪⎭

Solution of equilibrium equations

After establishing the global stiffness matrix and load vector of the beam, the load displacement relationship for the beam can be written as,

P K u(27.9)

where Pis the global load vector, u is displacement vector and K is the

global stiffness matrix. This equation is solved exactly in the similar manner as discussed in the lesson 24. In the above equation some joint displacements are known from support conditions. The above equation may be written as

⎧⎪pk ⎫⎪ ⎡k11 k12 ⎤⎧⎪uu ⎫⎪⎨ ⎬ ⎢ ⎥⎨⎬

(27.10)

⎪⎩pu ⎪⎭ ⎢⎣k21 k22 ⎥⎦⎪⎩uk ⎪⎭

L




where pk and uk denote respectively vector of known forces and knowndisplacements. And pu , uu denote respectively vector of unknown forces and unknown displacements respectively. Now expanding equation (27.10),




{ pk } k11 {uu } k12 {uk

}

{ pu } k 21 {uu } k 22

{u k }

(27.11a)(27.11b)

Since uk is known, from equation 27.11(a), the unknown joint displacementscan be evaluated. And support reactions are evaluated from equation (27.11b), after evaluating unknown displacement vector.

Let

R1 , R3 and R5 be the reactions along the constrained degrees of freedom as

shown in Fig. 27.9a. Since equivalent joint loads are directly applied at the supports, they also need to be considered while calculating the actual reactions. Thus,

⎧R1 ⎫ ⎧ p1 ⎫⎪ ⎪ ⎪ ⎪R p K u (27.12)⎨ 3 ⎬⎪ ⎪ ⎨ 3

⎬⎪⎪

21 u

⎪⎩R5 ⎪⎭ ⎪⎩ p5 ⎪⎭

The reactions may be calculated as follows. The reactions of the beam shown in Fig. 27.9a are equal to the sum of reactions shown in Fig. 27.9b, Fig. 27.9c and Fig. 27.9d.




From the method of superposition,

R Pb

K u1 L 14 4 K16 u6 (27.13a)

R Pa

K3 L

34

u4

K 36 u6 (27.13b)

R wL

2P K5 2

or

54

u4

K56 u6 (27.13c)

⎥

⎪

⎪⎪




⎧R ⎫1 ⎧ Pb ⎫ ⎡K K ⎤⎪ ⎪⎪ L ⎪ ⎢

14 16 ⎧u ⎫⎪R ⎪ ⎪ Pa

⎪ ⎢K K ⎥⎪ 4 ⎪ (27.14a)

⎨ 3 ⎬ ⎨ L ⎬ ⎢ 34 36 ⎥⎨ ⎬

⎪ ⎪ ⎪ wl ⎪ ⎢ ⎥⎪⎩u6 ⎪⎭⎪⎩R5 ⎪⎭ ⎪

2P⎪⎩ 2

⎭

⎣K54 K56 ⎦

Equation (27.14a) may be written as,

⎧R1 ⎫ ⎧⎫⎪ Pb L ⎪

⎡K14K16 ⎤⎧u ⎫

⎪R ⎪ ⎪Pa L

⎪ ⎢K K ⎥⎨

4 ⎬ (27.14b)⎨ 3 ⎬⎨ ⎬ ⎢ 34 36 ⎥ u⎪R ⎪ ⎪ wl ⎪ ⎢K K ⎥⎩ 6 ⎭⎩ 5 ⎭ ⎪⎩

2 2P⎪⎭

⎣ 54 56 ⎦

Member end actions the first element 1.

q1 , q2 , q3 , q4 are calculated as follows. For example consider

⎧q1 ⎫ ⎧ Pb⎫

⎪ L⎪⎧ 0 ⎫

⎪ ⎪ ⎪ Pab2 ⎪ ⎪ ⎪⎪q2 ⎪ ⎪

2 L ⎪⎪ K ⎪u2

⎪ ⎬(27.16)

⎨ ⎬⎪q3 ⎪⎨ Pa

⎬element1 ⎨⎪ 0 ⎪

⎪ ⎪ ⎪ L ⎪ ⎪ ⎪⎪⎩q4

⎪⎭

⎪ Pa 2b ⎪⎪⎪

⎪⎩u

4 ⎪⎭




⎩In the next lesson few problems are solved to illustrate the method so far discussed.

INTERNATIONAL INSTITUTE OF TECHNOLOGY & MANAGEMENT, MURTHAL SONEPATE-NOTES , Subject :Structural analysis—III, Course code –CE-308B Course:-B.Tech ,Branch: Civil Engineering , Sem-6th ,


UNIT - 4

Introduction

In engineering problems there are some basic unknowns. If they are found, the behaviour of the entire structure can be predicted. The basic unknowns or the Field variables which are encountered in the engineering problems are displacements in solid mechanics, velocities in fluid mechanics, electric and magnetic potentials in electrical engineering and temperatures in heat flow problems.

In a continuum, these unknowns are infinite. The finite element procedure reduces such unknowns to a finite number by dividing the solution region into small parts called elements and by expressing the unknown field variables in terms of assumed approximating functions (Interpolating functions/Shape functions) within each element. The approximating functions are defined in terms of field variables of specified points called nodes or nodal points. Thus in the finite element analysis the unknowns are the field variables of the nodal points. Once these are found the field variables at any point can be found by using interpolation functions.

After selecting elements and nodal unknowns next step in finite element analysis is to assemble element properties for each element. For example, in solid mechanics, we have to find the force-displacement i.e. stiffness characteristics of each individual element. Mathematically this relationship is of the form



[k]e {}e {F}e

where [k]e is element stiffness matrix, { }e is nodal displacement vector of the element and {F}e is nodal force vector. The element of stiffness matrix kij represent the force in coordinate direction ‘i’ due to a unit displacement in coordinate direction ‘j’. Four methods are available for formulating these element propertiesviz. direct approach, variational approach, weighted residual approach and energy balance approach. Any one of these methods can be used for assembling element properties. In solid mechanics variational approach is commonly employed to assemble stiffness matrix and nodal force vector (consistant loads).

Element properties are used to assemble global properties/structure properties to get system equations[k] { } {F}. Then the boundary conditions are imposed. The solution of these simultaneous equations give the nodal unknowns. Using these nodal values additional calculations are made to get the required values e.g. stresses, strains, moments, etc. in solid mechanics problems.Thus the various steps involved in the finite element analysis are:

(i) Select suitable field variables and the elements.(ii) Discritise the continua.

(iii) Select interpolation functions.(iv) Find the element properties.(v) Assemble element properties to get global properties.

(vi) Impose the boundary conditions.(vii) Solve the system equations to get the nodal unknowns.

(viii) Make the additional calculations to get the required values.(ix) A BRIEF EVPLANATION OF FEA FOR A STRESS ANALYSIS PROBLEM

The steps involved in finite element analysis are clarified by taking the stress analysis of a tension strip with fillets (refer Fig.1.1). In this problem stress concentration is to be studies in the fillet zone. Since the problem is having symmetry about both x and y axes, only one quarter of the tension strip may be considered as shown in Fig.1.2. About the symmetric axes, transverse displacements of all nodes are to be made zero. The various steps involved in the finite element analysis of this problem are discussed below:

Step 1: Four noded isoparametric element (refer Fig 1.3) is selected for the analysis (However note that 8 noded isoparametric element is ideal for this analysis). The four noded isoparametric element can take quadrilateral shape also

Fillet t

b2 C A b1

B D



as required for elements 12, 15, 18, etc. As there is no bending of strip, only displacement continuity is to be ensured but not the slope continuity. Hence displacements of nodes in x and y directions are taken as basic unknowns in the problem.

P

Fig. 1.1 †ypical tension flat

x P

3128252219161310741



Introduction 3

A 15 9 13 17 21 24 29 33 37 41 45

C2

2 5 8 11 14 17 20 23 26 29

43 6 9 12 15 18 21

32 P

B 8 12 16 20 2424 27 30 33 D

32 28 36 40 44 48

Fig. 1.2 Discretisation of quarter of tension flat

n

6 10 4 3

5

7(a) Element no.

5 11 1 2(b) Typical element

Fig. 1.3

Step 2: The portion to be analysed is to be discretised. Fig. 1.2 shows discretised portion. For this 33 elements have been used. There are 48 nodes. At each node unknowns are x and y components of displacements. Hence in this problem total unknowns (displacements) to be determined are 48 × 2 = 96.

Step 3: The displacement of any point inside the element is approximated by suitable functions in terms of the nodal displacements of the element. For the typical element (Fig. 1.3 b), displacements at P are

u Ni ui N1u1 N2u2 N3u3 N4u4

and v Nivi N1v1 N2v2 N3v3 N4v4 …(1.2)

The approximating functions Ni are called shape functions or interpolation functions. Usually they are derived using polynomials. The methods of deriving these functions for various elements are discussed in this text in latter chapters.

Step 4: Now the stiffness characters and consistant loads are to be found for each element. There are four nodes and at each node degree of freedom is 2. Hence degree of freedom in each element is 4 × 2 = 8. The relationship between the nodal displacements and nodal forces is called element stiffness characteristics. It is of the form

[k]e {}e {F}e , as explained earlier.

3



For the element under consideration, ke is 8 × 8 matrix and e and Fe are vectors of 8 values. In solid mechanics element stiffness matrix is assembled using variational approach i.e. by minimizing potential energy. If the load is acting in the body of element or on the surface of element, its equivalent at nodal points are to be found using variational approach, so that right hand side of the above expression is assembled. This process is called finding consistant loads.



4 Finite Element Analysis

Step 5: The structure is having 48 × 2 = 96 displacement and load vector components to be determined. Hence global stiffness equation is of the form

[k] {} = {F}

96 × 96 96 × 1 96 × 1Each element stiffness matrix is to be placed in the global stiffness matrix

appropriately. This process is called assembling global stiffness matrix. In this problem force vector F is zero at all nodes except at nodes 45, 46, 47 and 48 in x direction. For the given loading nodal equivalent forces are found and the force vector F is assembled.

Step 6: In this problem, due to symmetry transverse displacements along AB and BC are zero. The system equation [k] {} {F} is modified to see that the solution for {} comes out with the above values. This modification of system equation is called imposing the boundary conditions.

Step 7: The above 96 simultaneous equations are solved using the standard numerical procedures like Gauss- elimination or Choleski’s decomposition techniques to get the 96 nodal displacements.

Step 8: Now the interest of the analyst is to study the stresses at various points. In solid mechanics the relationship between the displacements and stresses are well established. The stresses at various points of interest may be found by using shape functions and the nodal displacements and then stresses calculated. The stress concentrations may be studies by comparing the values obtained at various points in the fillet zone with the values at uniform zone, far away from the fillet (which is equal to P/b2t).

FINITE ELEMENT METHOD VS CLASSICAL METHODS

1. In classical methods exact equations are formed and exact solutions are obtained where as in finite element analysis exact equations are formed but approximate solutions are obtained.

2. Solutions have been obtained for few standard cases by classical methods, where as solutions can be obtained for all problems by finite element analysis.

3. Whenever the following complexities are faced, classical method makes the drastic assumptions’ and looks for the solutions:

(a) Shape



(b) Boundary conditions(c) Loading

Fig. 1.4 shows such cases in the analysis of slabs (plates).To get the solution in the above cases, rectangular shapes, same boundary condition along a side and regular equivalent loads are to be assumed. In FEM no such assumptions are made. The problem is treated as it is.

4. When material property is not isotropic, solutions for the problems become very difficult in classical method. Only few simple cases have been tried successfully by researchers. FEM can handle structures with anisotropic properties also without any difficulty.

5. If structure consists of more than one material, it is difficult to use classical method, but finite element can be used without any difficulty.

6. Problems with material and geometric non-linearities can not be handled by classical methods. There is no difficulty in FEM.



Hence FEM is superior to the classical methods only for the problems involving a number of complexities which cannot be handled by classical methods without making drastic assumptions. For all regular problems, the solutions by classical methods are the best solutions. Infact, to check the validity of the FEM programs developed, the FEM solutions are compared with the solutions by classical methods for standard problems.

(a) Irregular shaper (b) Irregular boundary condition

(c) Irregular loading

Fig. 1.4

FEM VS FINITE DIFFERENCE METHOD (FDM)

1. FDM makes pointwise approximation to the governing equations i.e. it ensures continuity only at the node points. Continuity along the sides of grid lines are not ensured.

FEM make piecewise approximation i.e. it ensures the continuity at node points as well as along the sides of the element.

2. FDM do not give the values at any point except at node points. It do not give any approximating function to evaluate the basic values (deflections, in case of solid mechanics) using the nodal values.

FEM can give the values at any point. However the values obtained at points other than nodes are by using suitable interpolation formulae.

3. FDM makes stair type approximation to sloping and curved boundaries as shown in Fig.



1.5.FEM can consider the sloping boundaries exactly. If curved elements

are used, even the curved boundaries can be handled exactly.4. FDM needs larger number of nodes to get good results while FEM needs fewer nodes.5. With FDM fairly complicated problems can be handled where as FEM can

handle all complicated problems.



6 Finite Element Analysis

Fig. 1.5 VDW approximation of shape

A BRIEF HISTORY OF FEM

Engineers, physicists and mathematicians have developed finite element method independently. In 1943 Courant[1] made an effort to use piecewise continuous functions defined over triangular domain.

After that it took nearly a decade to use this distribution idea. In fifties renewed interest in this field was shown by Polya [2], Hersh [3] and Weinberger [4]. Argyris and Kelsey [5] introduced the concept of applying energy principles to the formation of structural analysis problems in 1960. In the same year Clough [6] introduced the word ‘Finite Element Method’.

In sixties convergence aspect of the finite element method was pursued more rigorously. One such study by Melesh [7] led to the formulation of the finite element method based on the principles of minimum potential energy. Soon after that de Veubeke [8] introduced equilibrium elements based on the principles of minimum potential energy, Pion [9] introduced the concept of hybrid element using the duel principle of minimum potential energy and minimum complementary energy.

In Late 1960’s and 1970’s, considerable progress was made in the field of finite element analysis. The improvements in the speed and memory capacity of computers largely contributed to the progress and success of this method. In the field of solid mechanics from the initial attention focused on the elastic analysis of plane stress and plane strain problems, the method has been successfully extended



to the cases of the analysis of three dimensional problems, stability and vibration problems, non-linear analysis. A number of books [10 – 20] have appeared and made this field interesting.

NEED FOR STUDYING FEM

Now, a number of users friendly packages are available in the market. Hence one may ask the question ‘What is the need to study FEA?’.

The above argument is not sound. The finite element knowledge makes a good engineer better while just user without the knowledge of FEA may produce more dangerous results. To use the FEA packages properly, the user must know the following points clearly:

1. Which elements are to be used for solving the problem in hand.2. How to discritise to get good results.3. How to introduce boundary conditions properly.



4. How the element properties are developed and what are their limitations.5. How the displays are developed in pre and post processor to understand their limitations.6. To understand the difficulties involved in the development of FEA

programs and hence the need for checking the commercially available packages with the results of standard cases.

Unless user has the background of FEA, he may produce worst results and may go with overconfidence.Hence it is necessary that the users of FEA package should have sound knowledge of FEA.

WARNING TO FEA PACKAGE USERS

When hand calculations are made, the designer always gets the feel of the structure and get rough idea about the expected results. This aspect cannot be ignored by any designer, whatever be the reliability of the program, a complex problem may be simplified with drastic assumptions and FEA results obtained. Check whether expected trend of the result is obtained. Then avoid drastic assumptions and get more refined results with FEA package. User must remember that structural behaviour is not dictated by the computer programs. Hence the designer should develop feel of the structure and make use of the programs to get numerical results which are close to structural behaviour.

QUESTIONS

1. Explain the concept of FEM briefly and outline the procedure.2. Discuss the advantages and disadvantages of FEM over

(i) Classical method(ii) Finite difference method.

3. Clearly point out the situations in which FEM is preferred over other methods.4. When there are several FEM packages are available is there need to study this method?

Discuss.

References1. R. Courant, “Variational Methods for the Solutions of Problems of Equilibrium and Vibrations”,

Bulletin of American Mathematical Society, Vol. 49, 1943.2. G. Polya, Estimates for Eigen Values, Studies presented to Richard Von Mises, Academic Press,

New York, 1954.3. J. Hersch, “Equations Differentielles et Functions de cellules”, C.R. Acad. Science, Vol. 240, 1955.4. H.F. Weinberger, “Upper and Lower Bounds for Eigen Values by Finite Difference Method”,

Pure Applied Mathematics, Vol. 9, 1956.



5. J.H. Argyris and S. Kelsey, “Energy Theorems and Structural Analysis”, Aircraft Engineering, Vol. 27, 1955.

6. R.W. Clough, “The Finite Element Method in Plane Stress Analysis”, Proceeding of 2nd ASCE Conference on Electronic Computation, Pittsburg, PA, September, 1960.

7. R.J. Melosh, “Basis for the Derivation for the Direct Stiffness Method”, AIAA Journal, Vol. 1, 1963.

8. B. Fraeijs de Veubeke, “Upper and Lower Bounds in Matrix Structural Analysis”, AGARD ograph72, B.F. de Veubeke (ed). Pergaman Press, New York, 1964.

9. T.H.H. Pian, “Derivation of Element Stiffness Matrices”, AIAA Journal, Vol. 2, 1964. pp. 556–57.



10. O.C. Zienkiewicz, The Finite Element Method in Engineering Science, McGraw-Hill, London 1971.11. K.H. Huebner, The Finite Element Methods for Engineers, John Wiley and Sons, 1971.12. Desai and Abel, Introduction to the Finite Element Method, CBS Publishers & Distributors, 1972.13. H.C. Martin and G. F. Carey, Introduction to Finite Element Analysis- Theory and Applications,

Tata McGraw-Hill Publishing Company Ltd., New Delhi, 1975.14. K.L. Bathe and E.L. Wilson, Finite Element Methods, Prentice Hall, 1976.15. Y.K. Cheuny and M.F. Yeo, A Practical Introduction to Finite Element Analysis, Pitman

Publishers, 1979.16. R.D. Cook, D.S. Makus and M.F. Plesha, Concept and Applications of Finite Element Analysis,

John Wiley and Sons, 1981.17. J.N Reddy, An introduction to the Finite Element Method, McGraw-Hill International Edition,

1984.18. C.S. Krishnamoorthy, Finite Element Analysis, Theory and Programming, Tata McGraw-Hill

Publishing Company Ltd., New Delhi, 1987.19. T.R. Chandrapatla and A.D. Belegundu, Introduction to Finite Elements in Engineering, Prentice

Hall, 1991.20. S. Rajasekharan, Finite Element Analysis in Engineering Design, Wheeler Publisher, 1993.



INTRODUCTION

This chapter summarizes the results from theory of elasticity which are useful in solving the problems in structural and continuum mechanics by the finite element method.

STRESSES IN A TYPICAL ELEMENT

In theory of elasticity, usually right hand rule is used for selecting the coordinate system. Fig. 2.1 shows various orientations of right hand rule of the coordinate systems. Equations derived for any one such orientation hold good for all other orientations of

z

z y

y x x(b)

x

(a)

y

z(c)

Fig. 2.1

coordinate system with right hand rule. In this Chapter orientation shown in Fig. 2.1(a) is used for the explanation. Fig. 2.2 shows a typical three dimensional element of size dx × dy × dz. Face abcd may be called as negative face of x and the face efgh as the positive face of x since the x value for face abcd is less than that for the face efgh. Similarly the face aehd is negative face of y and bfgc is positive face of y. Negative and positive faces of z are dhgc and aefb.

The direct stresses and shearing stresses acting on the negative faces are shown in the Fig. 2.3 with suitable subscript. It may be noted that the first subscript of shearing stress is the plane and the second subscript is the direction. Thus the xy means shearing stress on the plane where x value is constant and y is the direction.



z z

dy

a e

dxx

dz xyyx

b f

d hy

y xzy

yz zx

zy

zc

gx x

Fig. 2.2 Fig. 2.3

In a stressed body, the values of stresses change from face to face of an element. Hence on positive face the various stresses acting are shown in Fig. 2.4 with superscript ‘+’.

All these forces are listed in table 2.1

Note the sign convention: A stress is positive when it is on positive face in positive direction or on negative face in negative direction. In other words the stress is + ve when it is as shown in Figs 2.3 and 2.4.

z

z

zy

zx yz

yxz y

zx

xy

x

xFig. 2.4

Yy

dz

X



Table 2.1 Stresses on a typical element

Face Stress on –ve Face Stresses on +ve Face

x x

xy

xz

x dx

x x x

xy

dxxy xy x

xz dx

xz xz x

y y

yx

yz

y

dyy y y

yx dy

yx yx y

yz

dyyz yz y

z z

zx

zy

z dz

z z z

zx dz

zx zx z

zy

dzzy zy z

x

Fig. 2.5



Let the intensity of body forces acting on the element in x, y, z directions be X, Y and Z respectively as shown in Fig 2.5. The intensity of body forces are uniform over entire body. Hence the total body force in x, y, z direction on the element shown are given by

(i) X dx dy dz in x – direction(ii) Y dx dy dz in y – direction and

(iii) Z dx dy dz in z – direction

Matrix Displacement Formulation

INTRODUCTION

Though mathematicians, physicists and stress analysts worked independently in the field of FEM, it is the matrix displacement formulation of the stress analysts which lead to fast development of FEM. Infact till the word FEM became popular, stress analyst worked in this field in the name of matrix displacement method. In matrix displacement method stiffness matrix of an element is assembled by direct approach while in FEM though direct stiffness matrix may be treated as an approach for assembling element properties (stiffness matrix as far as stress analysis is concerned), it is the energy approached which has revolutionized entire FEM.

Hence in this chapter, a brief explanation of matrix displacement method is presented and solution techniques for simultaneous equations are discussed briefly.

MATRIV DISPLACEMENT EQUATIONS

The standard form of matrix displacement equation is,[k] {} {F}

where [k] is stiffness matrix{} is displacement vector and{F} is force vector in the coordinate directions

The element kij of stiffness matrix maybe defined as the force at coordinate i due to unit displacement in coordinate direction j.

The direct method of assembling stiffness matrix for few standard cases is briefly given in this article.

1. Bar Element



Common problems in this category are the bars and columns with varying cross section subjected to axial forces as shown in Fig. 3.1.

For such bar with cross section A, Young’s Modulus E and length L (Fig. 3.2 (a)) extension/shortening

is given by PL

EA

P3 A2 P2 A1

P

L3 L2 L1

A3

L1

xL2

1

Fig. 3.2

P3

P2



P1

P EA L

If 1, P EAL

By giving unit displacement in coordinate direction 1, the forces development in the coordinate direction 1 and 2 can be found (Fig. 3.2 (b)). Hence from the definition of stiffness matrix,

k11

EA L

and k21

EAL

Similarly giving unit displacement in coordinate direction 2 (refer Fig. 3.2 (c)), we get

L



k12 EA

Land k22

EA L

EA µ 1 1yThus, k L ¡µ1

1

¡j

…(3.5)

2. Truss Element

Members of the trusses are subjected to axial forces only, but their orientation in the plane may be at any angle to the coordinate directions selected. Figure 3.3 shows a typical case in a plane truss. Figure 3.4 (a) shows a typical member of the truss with Young’s Modulus E, cross sectional area A, length L and at angle to x-axis



k21

k

P sin EA cos sinL

P cos EA cos2

31

k41

L

P sin EA cos sinL

2

1 (a)

1

P P1

L



4P P

3

P 1(b) P

(c)

k42

P sin EA sin2 L

(iii) Unit displacement in coordinate direction 3,Extension along the axis is 1 sin and hence the forces developed are as shown in the Fig. 3.4 (d)

P EA cosL



k13 P cos EA cos2 L



k23

k

P sin EA cos sinL

P cos EA cos2

33

k43

L

P sin EA cos sinL

(vi) Due to unit displacement in coordinate direction 4,Extension of the bar is equal to 1 sin,Fig. 3.4 (e).

and hence the forces developed are as shown in

P EA sin L

P cos EA sin cosL

P sin EA sin2 L

P cos EA sin cosL

P sin EA sin2 L

¡µ cos2 cos sin

cos2

cos sin y¡

EA ¡ cos sin sin2 cos sin –sin2 ¡k L ¡ –cos2 cos sin cos2 cos sin ¡

¡µcos sin –sin2

cos sin

sin2 ¡j



µ¡ l2lml2

lm y¡

EA ¡ lmm2 lm m2 ¡L ¡l2 lm l2 lm ¡ …(3.6)

¡µlm m2 lm m2 ¡j

Where l and m are the direction cosines of the member i.e. l = cos and m = cos (90 – ) = sin .

(v) Beam ElementIn the analysis of continuous beams normally axial deformation is negligible (small deflection theory) and hence only two unknowns may be taken at each end of a element (Fig. 3.5). Typical element and the coordinates of displacements selected are shown in Fig. 3.5 (b). The end forces

xL 8 L

E1, I1 E2, I2 E3, I3 E4, I4

E,I, L



developed due to unit displacement in all the four coordinate directions are shown in Fig. 3.6 (a, b, c, d).

y 1 3 5 7 9

2 L1 4z L2 6 3 4 10

(a)

1 3

From the definition of stiffness matrix and looking at positive senses indicated, we can write(a) Due to unit displacement in coordinate direction 1,

k11 12EI

L3k21

6EI

L2k31

12EI

L3k41

6EIL2

(b) Due to unit displacement in coordinate direction 2,

k12 6EI

L2k22

4EI

Lk32

6EI

L2k42

2EIL



(c) Due to unit displacement in coordinate direction 3,

k13 12EI

L3k23

6EI

L2k33

12EI

L3k43

6EIL2

L3

¡12 6L¡

¡

¡

¡¡ 3

0



(d) Due to unit displacement in coordinate direction 4,

k14 6EI

L2k24

2EI

Lk34

6EI

L2k44

4EIL

µ¡ 12 6L 12 6L y¡ k FI ¡ 6L

µ 6L4L26L2L2

6L126L

2L2 ¡4L2 j

…(3.7)

If axial deformations in the beam elements are to be considered as in case of columns of frames, etc. (Fig. 3.7), it may be observed that axial force do not affect values of bending moment and shear force and vice versa is also true. Hence stiffness matrix for the element shown in Fig. 3.8 is obtained by combining the stiffness matrices of bar element and beam element and arranging in proper locations. For this case

µ¡ EA 0 0 EA

y0 0 ¡

¡

¡k ¡

L0

12EI L3

0 6EIL2

6EI

L2

4EIL

L0

12EI L3

0 6EI L2

6EI L22EIL ¡

¡ EA 00

EA

0 0 ¡

L0 12EI

L

L6EI

0L2

12EI L3

6EI ¡

L2 ¡

…(3.8)

¡ 6EIµ L2

2EIL

0 6EI

L24EI ¡L j¡

¡

¡



(a) (b)

Fig. 3.7

2 51 4

3 6

Fig. 3.8

B



The following special features of matrix displacement equations are worth noting:(i) The matrix is having diagonal dominance and is positive definite. Hence

in the solution process there is no need to rearrange the equations to get diagonal dominance.

(ii) The matrix is symmetric. It is obvious from Maxwell’s reciprocal theorem. Hence only upper or lower triangular elements may be formed and others obtained using symmetry.

(iii) The matrix is having banded nature i.e. the nonzero elements of stiffness matrix are concentrated near the diagonal of the matrix. The elements away from the diagonal are zero. Considerable saving is effected in storage requirement of stiffness matrix in the memory of computers by avoiding storage of zero values of stiffness matrices. The banded nature of matrix is shown in Fig. 3.9.

Fig. 3.9

In this case instead of storing N ×N size matrix only N × B size matrix can be stored.

III

IIIIV

VVI

VII VII

I

IX



no force in the z-direction and no variation of any forces in z-direction. Hence z xz yz 0

The conditions xz yz 0 give xz yz 0 and the condition z 0 gives,

z x y 1 z 0



i.e.

Basic Equations in Elasticity 17

z 1 x y

If this is substituted in equation 2.13 the constitutive law reduces to

j¡ x y¡ µ¡1

E

0 y¡ j¡ x y¡

¡ y ¡ 2 ¡ 1 0 ¡ ¡ y ¡ …(2.14)¡ ¡ 1 ¡0 0 1 ¡ ¡ ¡t xy 1 µ¡

2

¡j t xy 1

(c) (d)

no force in the z-direction and no variation of any forces in z-direction. Hence z xz yz 0

The conditions xz yz 0 give xz yz 0 and the condition z 0 gives,

z x y 1 z 0



i.e. z 1 x y

If this is substituted in equation 2.13 the constitutive law reduces to

j¡ x y¡ µ¡1

E

0 y¡ j¡ x y¡

¡ y ¡ 2 ¡ 1 0 ¡ ¡ y ¡ …(2.14)¡ ¡ 1 ¡0 0 1 ¡ ¡ ¡t xy 1 µ¡

2

¡j t xy 1

y

y

xx

z z



(c) (d)

Fig. 2.7

Axi-Symmetric Problems



z(e)

Axi-symmetric structures are those which can be generated by rotating a line or curve about an axis. Cylinders (refer Fig. 2.8) are the common examples of axisymmetric structures. If such structures are subjected to axisymmetric loadings like uniform internal or external pressures, uniform self weight or live load uniform over the surface,there exist symmetry about any axis. The advantage of symmetry may be made use to simplify the analysis. In these problems cylindrical coordinates can be used advantageously. Because of symmetry, the stress components are independent of the angular ( ) coordinate. Hence all derivatives with respect to vanish i.e. in these cases.

v r z r z 0

z, w

z, w



In these cases stress-strain relation is

j r y

j1

0 y j r y

¡ z ¡ E ¡ 1

0 ¡ ¡ z ¡

(1 )(1 2) ¡

1 0

¡

…(2.17)

¡ ¡¡ 1 2 ¡ ¡ ¡

t rz 1 y 2 j t rz 1Fig. 2.7

Axi-Symmetric Problems



z(e)

Axi-symmetric structures are those which can be generated by rotating a line or curve about an axis. Cylinders (refer Fig. 2.8) are the common examples of axisymmetric structures. If such structures are subjected to axisymmetric loadings like uniform internal or external pressures, uniform self weight or live load uniform over the surface,there exist symmetry about any axis. The advantage of symmetry may be made use to simplify the analysis. In these problems cylindrical coordinates can be used advantageously. Because of symmetry, the stress components are independent of the angular ( ) coordinate. Hence all derivatives with respect to vanish i.e. in these cases.

v r z r z 0

z, w

z, w



In these cases stress-strain relation is

J r yJ1

0 y J r y¡ z ¡ E ¡ 1

0 ¡ ¡ z ¡ (1 )(1 2) ¡ 1 0 ¡

…(2.17)

¡ ¡ ¡ 1 2 ¡ ¡ ¡



t rz 1 y 2 J t rz 1

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